Einstein's gravity theory passes toughest test yet: Bizarre binary star system pushes study of relativity to new limits

Apr 25, 2013 by Dave Finley
This artist's impression shows the exotic double object that consists of a tiny, but very heavy neutron star that spins 25 times each second, orbited every two and a half hours by a white dwarf star. The neutron star is a pulsar named PSR J0348+0432 that is giving off radio waves that can be picked up on Earth by radio telescopes. Although this unusual pair is very interesting in its own right, it is also a unique laboratory for testing the limits of physical theories. Credit: ESO

(Phys.org) —A strange stellar pair nearly 7,000 light-years from Earth has provided physicists with a unique cosmic laboratory for studying the nature of gravity. The extremely strong gravity of a massive neutron star in orbit with a companion white dwarf star puts competing theories of gravity to a test more stringent than any available before.

Once again, Albert Einstein's , published in 1915, comes out on top.

At some point, however, scientists expect Einstein's model to be invalid under extreme conditions. General Relativity, for example, is incompatible with . Physicists hope to find an alternate description of gravity that would eliminate that incompatibility.

A newly-discovered pulsar—a spinning neutron star with twice the mass of the Sun—and its white-, orbiting each other once every two and a half hours, has put gravitational theories to the most extreme test yet. Observations of the system, dubbed PSR J0348+0432, produced results consistent with the predictions of General Relativity.

The tightly-orbiting pair was discovered with the National Science Foundation's Green Bank Telescope (GBT), and subsequently studied in visible light with the Apache Point telescope in New Mexico, the Very Large Telescope in Chile, and the William Herschel Telescope in the Canary Islands. Extensive radio observations with the Arecibo telescope in Puerto Rico and the Effelsberg telescope in Germany yielded vital data on subtle changes in the pair's orbit.

In such a system, the orbits decay and are emitted, carrying energy from the system. By very precisely measuring the time of arrival of the pulsar's radio pulses over a long period of time, astronomers can determine the rate of decay and the amount of emitted. The large mass of the neutron star in PSR J0348+0432, the closeness of its orbit with its companion, and the fact that the companion white dwarf is compact but not another neutron star, all make the system an unprecedented opportunity for testing alternative theories of gravity.

Under the extreme conditions of this system, some scientists thought that the equations of General Relativity might not accurately predict the amount of gravitational radiation emitted, and thus change the rate of orbital decay. Competing gravitational theories, they thought, might prove more accurate in this system.

Superdense neutron star, emitting beams of radio waves as a pulsar, center, is closely paired with a compact white-dwarf star. Together, the two provide physicists with an unprecedented natural, cosmic "laboratory" for studying the nature of gravity. The grid background illustrates the distortions of spacetime caused by the gravitational effect of the two objects. Credit: Antoniadis, et al.

"We thought this system might be extreme enough to show a breakdown in , but instead, Einstein's predictions held up quite well," said Paulo Freire, of the Max Planck Institute for Radioastronomy in Germany.

That's good news, the scientists say, for researchers hoping to make the first direct detection of gravitational waves with advanced instruments. Researchers using such instruments hope to detect the gravitational waves emitted as such dense pairs as and black holes spiral inward toward violent collisions.

Gravitational waves are extremely difficult to detect and even with the best instruments, physicists expect they will need to know the characteristics of the waves they seek, which will be buried in "noise" from their detectors. Knowing the characteristics of the waves they seek will allow them to extract the signal they seek from that noise.

"Our results indicate that the filtering techniques planned for these advanced instruments remain valid," said Ryan Lynch, of McGill University.

Freire and Lynch worked with a large international team of researchers. They reported their results in the journal Science.

Explore further: Acoustic tweezers manipulate cell-to-cell contact

More information: "A Massive Pulsar in a Compact Relativistic Binary," by J. Antoniadis et al. Science, 2013

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ValeriaT
1 / 5 (22) Apr 25, 2013
the orbits decay and gravitational waves are emitted
It's somewhat bizarre, just the phenomena which confirms general relativity most firmly violates it most pronouncedly - the gravitational waves aren't waves from perspective of AWT at all, but the CMBR noise.
Torbjorn_Larsson_OM
5 / 5 (14) Apr 25, 2013
"General Relativity, for example, is incompatible with quantum theory."

No, or you couldn't porpose gravitons. GR is actually perfectly compatible as long as it is analogous to any vacuum field at low energies. It is when you run up against its non-linearities it turns out to be a mere effective description.

In fact, it remains for an upgraded LISA to test GR low-energy compatibility (existence of gravitons).

@Crackpot: Since aether theory isn't science at all of the last century, it is *that* which is bizarre. As gravitational waves are tested (caused) by many pulsar systems as per the article, we know it isn't caused by CMB "noise".

Really, don't comment on science as long as you can't make stick to the subject. What you are claiming is equivalent to that since we see airplanes fly, therefore unicorns. Even a typical 5 year old would squirm in your deluded presence, they are both smarter and more experienced than you.
vacuum-mechanics
1 / 5 (22) Apr 25, 2013
That's good news, the scientists say, for researchers hoping to make the first direct detection of gravitational waves with advanced instruments. Researchers using such instruments hope to detect the gravitational waves emitted as such dense pairs as neutron stars and black holes spiral inward toward violent collisions…

The bad news is that even scientists could detect the gravitational waves; they still cannot explain (by General relativity) how the wave could propagate though the empty space! Maybe this physical view could help…
http://www.vacuum...18〈=en
mike_smoth
1.3 / 5 (13) Apr 25, 2013
It seems to me that more tests are being made against Einstein's theory than against quantum theory. Thus far Einstein's theory has held against amazing experiments. It seems to me that Quantum theory needs to be reevaluated and not Einstein's. After all, Einstein's theory can predict where a star will appear after the gravitation effects on its light. It predicted weird phenomena like black holes long before they where found. It even predicted things that not Eisenstein believed and he called it his greatest blunder yet the theory was right. At this point quantum theory can't even predict with certainty where and atom could be at any moment.
Tektrix
4.6 / 5 (10) Apr 25, 2013
"At this point quantum theory can't even predict with certainty where and atom could be at any moment."

LOL, nice one. Subtle jokes are the best kind :)
Q-Star
3.9 / 5 (26) Apr 25, 2013
The bad news is that even scientists could detect the gravitational waves; they still cannot explain (by General relativity) how the wave could propagate though the empty space!


No matter how many times ya post that, it remains just as untrue today as any other. Just because ya can't understand their explanations, doesn't mean they can't explain it.

All it means is they can't explain it to ya (and that is your shortcoming, not theirs.)
antialias_physorg
4.7 / 5 (18) Apr 25, 2013
It seems to me that Quantum theory needs to be reevaluated and not Einstein's.

It's not really a question of "either this OR that is incomplete". In the end both have to be revised.

More precisely: a unified theory of quantum gravitation will be different from either. However, it will render either as a result of some simplification which applies at large sacles (which will give you GR) and another simplification at small scales (which will give you QM).

The thing we have to figure out is: which are these implicit assumptions we currently make.
Whydening Gyre
1.9 / 5 (13) Apr 25, 2013
Anti is correct. Most opposing positions find agreement somewhere around a center. However, GR has been tested the most with no significant margin of error, ergo...
ekim
5 / 5 (7) Apr 25, 2013
The bad news is that even scientists could detect the gravitational waves; they still cannot explain (by General relativity) how the wave could propagate though the empty space!

Empty space is not empty. It is filled with a sea of virtual particles.
Tektrix
4.2 / 5 (10) Apr 25, 2013
Certainly one key measure of a theory's validity is it's predictive power. Another is the ability to design new physical systems based on the theory, and make them real. In both regards, Relativity and QM are spectacular successes. Both are behind technology nearly everyone uses daily. For instance, components that exploit quantum effects receive and process GPS signals that are accurately corrected for relativistic effects; a very practical blend of both theories. These two theories are definitely not competing for superiority, rather, they are seeking harmony.
Whydening Gyre
1.3 / 5 (13) Apr 25, 2013
The bad news is that even scientists could detect the gravitational waves; they still cannot explain (by General relativity) how the wave could propagate though the empty space!

Empty space is not empty. It is filled with a sea of virtual particles.

And an ALMOST equal number of empty spaces...
Osiris1
1 / 5 (10) Apr 26, 2013
I wonder if a better gravitational field and wave detector will await the discovery and use of 'dark energy' fields. In truth, since we do not know about it might it not be a number of fundamental energy fields in an expanded 'new standard model' now in three dimensions that may include temporal fields as well. Left or right hand rule manipulations of these forces may generate gravitational positive and negative fields, and useful long distance communication inasmuch as these can propagate independent of spacial constraint......brane theory of gravitational fields being felt from hidden other universes at specific places in our universe.......Just ruminating... but then a lot of ideas come from such.
Sanescience
1 / 5 (7) Apr 26, 2013
Question: I haven't found any information on if gravity waves would alter starlight passing near the source, or if there would be an effect on light emitted from a nebula that could be detected. Anyone seen such an article or discussion?
antialias_physorg
5 / 5 (7) Apr 26, 2013
However, GR has been tested the most with no significant margin of error, ergo...

..only in its sphere of applicability (i.e. anything that is 'large' for a given value of 'large')

QM has also been tested with no significant error observed within its sphere of applicability (i.e. for anything 'small' for a given value of 'small')

The point is that GR requires smooth space with definite spatial relations while QM requires uncertainty. These two demands are mutually exclusive. So something (and probably both) has to give.
ValeriaT
1 / 5 (13) Apr 26, 2013
As gravitational waves are tested (caused) by many pulsar systems as per the article, we know it isn't caused by CMB "noise"
The trick is, the energy is radiated in superluminal form, but in the intensity which general relativity predicts. So that you cannot prove the existence of gravitational waves with pulsar experiments, only radiation of gravity field energy into outside.
Since aether theory isn't science at all of the last century, it is *that* which is bizarre
Dense aether model just enables to understand these seeming paradoxes. You can model it like the radiation of energy with splash at the water surface, but the energy will be released in form of underwater waves, so it will propagate like the noise at the water surface.
ant_oacute_nio354
1 / 5 (12) Apr 26, 2013
Relativity theory is wrong and the spacetime doesn't exist.
Gravitational waves are not waves of spacetime, they are
waves of acceleration or force and must be detected with
accelerometers. Gravitational waves can be generated and
detected in a lab. c^2t^2 - x^2 = 1.9121x10^-34m2
ValeriaT
1 / 5 (9) Apr 26, 2013
The pulsar pair PSR J0348+0432 discussed is an interesting system because it is composed of compact objects in a tight, highly relativistic binary (v/c~=0.002) and the mass ratio is large. That last part is important for testing dark matter theories like scalar-tensor theories of gravity. Unfortunately, the popular article headline totally overstates the importance of this system for general relativity confirmation. The measurement of orbital change perirod represents only a 10% observation, which is uninteresting. Secondly the binary has a very low eccentricity (10E-6 ). The lowest order post-Keplerian effect would precession of the pericenter of the binary, which you can't measure when the eccentricity is so low. That measurement is way better for say the Hulse-Taylor pulsar binary than the change of orbital period measurement. The precession measurement in Hulse-Taylor pulsar is good to 6 decimal places, whereas the change of orbital period in H-T pulsar is only a few percent.
ValeriaT
1 / 5 (9) Apr 26, 2013
The conclusion is, the PSR J0348+0432 is a cool system, but isn't interesting enough for tests of GR right now.

Relativity theory is wrong and the spacetime doesn't exist
Why to throw out the baby with the bath water? The fact, some aspects of greneral relativity were misunderstood with dumb Einstein's followers (even Einstein opposed the gravitational waves originally) doesn't mean, everything about relativity is wrong. I do agree, that the gravitational waves can be detected with accelerometers in the lab (Podkletnov, Taimar and others), but they do represent the scalar waves (Tesla, Dollard, Meyl etc.) as well.
theon
1.4 / 5 (11) Apr 26, 2013
Sigh. The article contains many words but does not explain clearly what is really measured and how much the accuracy is better than before. Treating readership for good-news-worshippers is an insult.
ValeriaT
1 / 5 (13) Apr 26, 2013
@Theon: It's not an insult, but an intention. You laymen are supposed to pay the experiments and safe life for physicists. But not to understand it, which would enable the critique.
brt
3.6 / 5 (13) Apr 26, 2013
@Theon: It's not an insult, but an intention. You laymen are supposed to pay the experiments and safe life for physicists. But not to understand it, which would enable the critique.


It's sad that you are such a combination of arrogance, stupidity, and delusion. There's nothing I can tell you that you haven't already been told, so I'll leave it at that.
brt
1.8 / 5 (5) Apr 26, 2013
Anti is correct. Most opposing positions find agreement somewhere around a center. However, GR has been tested the most with no significant margin of error, ergo...


except for Dark Matter... until we find particle candidates anyway.
Dileep_Sathe
1 / 5 (5) Apr 26, 2013
Troubling Solar System: Dennis Sciama wrote a small but very useful book: The Physical Foundations of General Relativity in 1969. In the preface he says: Isaac Newton's laws of motion are logically incomplete and that situation leads us, step by step, to full complexity of Albert Einstein's General Relativity. However, it must be noted that Prof. Frank Wilczek (the present 62-year old American leader of theoretical physics and a Nobel Laureate) had lot of trouble in learning "classical" mechanics about 45 years ago – when he was in school / college, in Physics Today, October 2004. My research in physics education shows that the "logical" incompleteness of Newton's laws has not been fully understood as yet and so i) still there are some global and chronic problems in learning a very basic topic – that is circular motion and ii) that is why physics is not adequately popular in society – as indicated by the events to popularize physics in the Einstein Year: 2005. So, I hope, although "ex
Koen
1 / 5 (12) Apr 26, 2013
GR is by David Hilbert, not by Albert Einstein. Not many know that also Newton's gravity theory predicts correctly the deflection of sun light during a solar eclipse. Einstein was mistaken about Soldner's Newtonean calculation of the deflection angle (misused a factor 2, in his favour). Einstein was the biggest plagiarist and academic fraud (by misinterpretation of the scientific results of other scientists) in the history of science. So many good experiments were performed that show results in total disagreement with Hilbert's GR, it is a scam.
Q-Star
3.6 / 5 (20) Apr 26, 2013
Anti is correct. Most opposing positions find agreement somewhere around a center. However, GR has been tested the most with no significant margin of error, ergo...


except for Dark Matter... until we find particle candidates anyway.


That's a true thing ya say, as long as ya remember that Maxwell's equations for electromagnetism worked just as 40 years before the discovery of the electron and proton as they do today some 100 plus years after their discovery.

The good theories always predict the particles to be found. If a theory or model is working ya keep it. If the particle it predicts is found and fits, that's good. If the particle doesn't fit the theory ya modify the theory. So far GR works better than any other. It has worked so well, the best bet is trust in the "dark particle" which is predicted at least until a better general theory is found

Quantum theory lead to the discovery of the fundamental particles, it wasn't the particles leading to the theory.
brt
2 / 5 (4) Apr 26, 2013
yeah, I guess that's true. I may have been intentionally provocative in my statement because I lean towards a modification of gravity for dark matter.
Q-Star
3.4 / 5 (17) Apr 26, 2013
yeah, I guess that's true. I may have been intentionally provocative in my statement because I lean towards a modification of gravity for dark matter.


MOND and TeVeS are very constructive avenues of inquiring. The science and method of them is generally very good. The only thing in their disfavor is that they have yet to home-in on the consistent applicability across different phenomena. The answer may in deed be found somewhere there. Personally I don't think so, but that is more opinion than a "certain" pronouncement.

Any of the modified gravity theorists can take heart in the fact that the most successful of all models in physics, QM/QT was in the exact same position 80 years ago. (Having to use different formulas/models in applications to particular phenomena)
tadchem
1 / 5 (3) Apr 26, 2013
QM requires equations of motion to be solved by separation of variables with time as an orthogonal coordinate. GR works at scales where the velocities are a significant fraction of c by utilizing rotations in Minkowski 4-space, so time is not an orthogonal coordinate.
johanfprins
1 / 5 (5) Apr 26, 2013
Any of the modified gravity theorists can take heart in the fact that the most successful of all models in physics, QM/QT was in the exact same position 80 years ago. (Having to use different formulas/models in applications to particular phenomena)


It is STILL in that position! Wake up my boy!

Q-Star
3 / 5 (14) Apr 26, 2013
Any of the modified gravity theorists can take heart in the fact that the most successful of all models in physics, QM/QT was in the exact same position 80 years ago. (Having to use different formulas/models in applications to particular phenomena)


It is STILL in that position! Wake up my boy!



Ya are correct. It is a humbling awaking indeed.
Fleetfoot
3.7 / 5 (3) Apr 26, 2013
yeah, I guess that's true. I may have been intentionally provocative in my statement because I lean towards a modification of gravity for dark matter.


It's interesting and may lead to new finds in maths but it's unlikely to replace dark matter, in fact I think it's been shown by observations of some clusters that MOND would still need a dark matter component as well.

More importantly, the large scale density measurements need it and the growth of structure was bottom up (stars first, then small proto-galaxies merging into larger galaxies, clusters and finally super-clusters) whereas it would be the other way round without dark matter and I believe the details of nucleogenesis also require dark matter.
IronhorseA
1 / 5 (4) Apr 27, 2013
Anti is correct. Most opposing positions find agreement somewhere around a center. However, GR has been tested the most with no significant margin of error, ergo...


except for Dark Matter... until we find particle candidates anyway.


Dark matter is a based on assumed orbital velocities calculated using Newtonian gravity. Simulations using GR put an upper limit on dark matter, or in some simulations eliminate the need for it at all. Now for the next generation of supercomputers to increase the simulation size.
rkilburn81
1 / 5 (4) Apr 27, 2013
Gravity and the other three fundamental forces are all part of the same spectrum. One unified force, waves and eddies in the fabric of space/time. We have different names for each wave and eddie, just as we have names for each cloud, but in the end aren't all clouds just water vapor?
Fleetfoot
4 / 5 (4) Apr 28, 2013
Gravity and the other three fundamental forces are all part of the same spectrum. One unified force, waves and eddies in the fabric of space/time. We have different names for each wave and eddie, just as we have names for each cloud, but in the end aren't all clouds just water vapor?


A receiver that can detect EM at 100Hz will not detect gravitational waves of the same frequency. Each phenomenon has its own spectrum.
Fleetfoot
4 / 5 (4) Apr 28, 2013
Dark matter is a based on assumed orbital velocities calculated using Newtonian gravity.


They are based on use of the virial theorem and since the gravitational effects are "weak field", GR and the Newtonian approximation give equivalent results.

Simulations using GR put an upper limit on dark matter, ..


It also sets a lower limit.

or in some simulations eliminate the need for it at all.


Not with GR, only MOND based theories like TeVeS get close but even they still need DM when tested accurately enough.

http://arxiv.org/abs/0901.3932
ant_oacute_nio354
1 / 5 (8) Apr 29, 2013
Spacetime doesn't exist.
Relativity theory is all wrong.

Antonio Jose Saraiva
Fleetfoot
4 / 5 (4) Apr 29, 2013
Relativity theory is all wrong.


It fits the experimental results, that makes it right.
johanfprins
1 / 5 (5) Apr 29, 2013
Relativity theory is all wrong.


No, it is not ALL wrong! Only some deductions from it are wrong: Like time-dilation and length contraction.

Fleetfoot
3.9 / 5 (7) Apr 29, 2013
Relativity theory is all wrong.


No, it is not ALL wrong! Only some deductions from it are wrong: Like time-dilation and length contraction.


Both of those are observed phenomena, in the Michelson-Morley and Ives-Stilwell experiments respectively for example.
Q-Star
3.4 / 5 (17) Apr 29, 2013
Relativity theory is all wrong.


No, it is not ALL wrong! Only some deductions from it are wrong: Like time-dilation and length contraction.


Both of those are observed phenomena, in the Michelson-Morley and Ives-Stilwell experiments respectively for example.


Exactly so, and those two experiments point directly to the HOW "deductions from it are wrong" confuse most people. They mix up the place where SR leaves off and GR picks up.

And most people don't realize the magnitude of problems that have been caused when basic science education teaches mechanics in term of "mass" and "forces" rather than the approach both Newton and Einstein used,,, i.e. "inertia", "momentum" and "change in momentum". Real physics, the deep stuff, requires that most people go back and relearn the basics all over again in terms of momentum.

At least for relativistic classical physics. QT is not my area so I am not sure if that is as true there.
ValeriaT
1.8 / 5 (5) Apr 29, 2013
Only some deductions from it are wrong: Like time-dilation and length contraction.
Time dilation was proven for muons in colliders, for example: when they're moving with respect to observer, they're more stable. The time dilation in gravity field is routinely measured with few meters altitude elevation.
johanfprins
1 / 5 (6) Apr 30, 2013
Relativity theory is all wrong.


No, it is not ALL wrong! Only some deductions from it are wrong: Like time-dilation and length contraction.


Both of those are observed phenomena, in the Michelson-Morley and Ives-Stilwell experiments respectively for example.


Incorrectly interpreted:

The time difference given by the Lorentz transformation is NOT simultaneously on the two clocks since they keep exactly the same time. This is easy to prove by deriving this time difference correctly by means of Einstein's clock. This does model the Michelson Morley experiment correctly.

It is also easy to prove from the Lorentz transformation that a rod passing at a speed v MUST become longer owing to its de Broglie wavelength. Length contraction is only valid when the speed of light is NOT the same within all inertial reference frames. This was already proved by Lorentz LOOONG before Einstein postulated his Special Theory of Relativity.
johanfprins
1 / 5 (5) Apr 30, 2013
Only some deductions from it are wrong: Like time-dilation and length contraction.
Time dilation was proven for muons in colliders, for example: when they're moving with respect to observer, they're more stable. The time dilation in gravity field is routinely measured with few meters altitude elevation.


It is NOT time dilation, but the DIFFERENCE IN TIME (ON TWO CLOCKS KEEPING TIME AT EXACTLY THE SAME RATE) at which the same event is observed within two inertial reference-frames moving with a speed v relative to one another. I refer you AGAIN to: http://www.cathod...tion.pdf

The "time-dilation" of a muon is correctly derived in this document. Not that a BIGOT like you will read it. This is proved by the fact that I have posted this reference at least 4 times before and you have not yet read it so that you can come back with valid criticisms. All you are able to do is to parrot mainstream dogma!
johanfprins
1 / 5 (4) Apr 30, 2013
It is NOT time dilation, but the DIFFERENCE IN TIME
This is just a semantical game. Of course it's possible to derive the time dilatation with various means (which are mutually equivalent on background) but at the moment, when these results aren't distinguishable experimentally, -


Wrong again! The results ARE distinguishable experimentally and logically! As usual Natello, ValeriaT AKAK does not want facts to confuse his/her prejudices. AGAIN I post that YOU should read, if you can and can follow logic, the following:
http://www.cathod...tion.pdf

then the more consistent and comprehensive theory (Einstein's one) should be preferred.


Einstein's postulates for STR are indeed consistent and comprehensive: It is just a pity that his derivation of "time-dilation", "length-contraction" and his thought experiment to explain non-simultaneity of simultaneous events, all violate his second postulate.

johanfprins
1 / 5 (5) Apr 30, 2013
results ARE distinguishable experimentally and logically!


After then your theory is wrong, because the life-time of muon was measured in colliders repeatedly and it fits the Einstein's theory well.


It also fits my derivation which does not violate Einstein's second postulate as Einstein's derivation does. How can Einstein's concept of "time-dilation" be correct if it violates Einstein's own postulate on which he based his Special Theory of Relativity?

BTW if fast muon decays faster, it just means, that the time runs faster for it from its beginning to end - it's not just "shifted".


WRONG! Why do you refuse to read my correct derivation of the increase in the muon's lifetime? I AGAIN give you the reference: http://www.cathod...tion.pdf

And the AWT explains, why is it so.


Oh my God! Again garbage from Cloud Cuckoo Land. There are no electromagnetic aether waves: Therefore light and matter waves do not require an aether!!
johanfprins
1 / 5 (5) Apr 30, 2013
This is inherent property of all theories. It can be demonstrated, every testable theory must be based on mutually inconsistent postulate set.


I believe that in your case this is so since you dwell in Cloud Cuckoo Land. Einstein's two postulates are NOT mutually inconsistent at all. The first postulate gives the reason why the speed of light must be the same in all inertial reference frames. And the second postulate states that this must be so since the first postulate requires that it must be so! Where are they mutually inconsistent?

If the postulates would be consistent, they could be substituted each other and whole the theory would change into tautology. I'm illustrating it with gravitational lensing here in trivial way.


You are using words that you are incapable of understanding and will NEVER understand with your limited brain capacity!

johanfprins
1 / 5 (5) Apr 30, 2013
Do you think, that the gravitational lens are possible in theory, in which the light spreads with constant speed, so we cannot observe any refractive phenomena?


If you knew any physics you would have known that the speed of light slows down within a gravitational field, that is why you get lensing. It is only in STR that the speed of light is constant.

Why do you refuse to read my correct derivation of the increase in the muon's lifetime
How it differs from Einstein's prediction? If it doesn't differ, why I should read about it?


It differs in the same way from Einstein's prediction as Kepler's laws differed from Ptolemy's epicycles: Both gave the same paths for the planets as viewed from earth but one of them is wrong physics.

If it differs, then it violates the experimental confirmations of it - so I shouldn't read about it


You would have made a perfect inquisitor in the time of Galileo. Another demonstration of your narrow-minded bigotry!

Fleetfoot
3.7 / 5 (3) Apr 30, 2013
Only some deductions from it are wrong: Like time-dilation and length contraction.


Both of those are observed phenomena, in the Michelson-Morley and Ives-Stilwell experiments respectively for example.


Incorrectly interpreted:

The time difference given by the Lorentz transformation is NOT simultaneously on the two clocks since they keep exactly the same time.


For time dilation, in the Ives-Stilwell experiment, the moving ions demonstrate a frequency shift greater than predicted by classical Doppler. The difference is what we call the time dilation factor. There is no "interpretation" involved, just a simple empirical observation.

For length contraction, see any text book on the MMX. While there is degeneracy with time dilation in the basic version, the Kennedy-Thorndike experiment resolved that.
johanfprins
1 / 5 (6) Apr 30, 2013
For time dilation, in the Ives-Stilwell experiment, the moving ions demonstrate a frequency shift greater than predicted by classical Doppler. The difference is what we call the time dilation factor. There is no "interpretation" involved, just a simple empirical observation.


Of course there is an interpretation involved: Your interpretation is that a "moving clock keeps slower time than a stationary clock": Interpretation A. The correct interpretation is that an event occurring within the "moving" IRF is observed within the moving IRF BEFORE it is observed within the "stationary" IRF. The equations for the relativistic Doppler shift remains the same without violating Einstein's first postulate as interpretation A does!

For length contraction, see any text book on the MMX. While there is degeneracy with time dilation in the basic version, the Kennedy-Thorndike experiment resolved that.


I know the text books far better than you with your feeble mind can ever know them.
johanfprins
1 / 5 (7) Apr 30, 2013
In order to derive length-contraction the nose and tail coordinates are transformed from the IRF relative to which the rod is moving AS IF the rod is stationary within this reference frame: WHICH it is NOT.

To find out what the rod will be within the IRF relative to which it is moving you MUST transform its nose and tail coordinates from the IRF within which it is stationary into the IRF relative to which it is moving. You then find that the rod is LONGER, and that there is a time difference across it caused by the phase relating to its coherent de Broglie wavelength.

See http://www.cathod...tion.pdf

Or do you also do not want facts to confuse your irrational dogmatic beliefs?
johanfprins
1 / 5 (6) Apr 30, 2013
: Your interpretation is that a "moving clock keeps slower time than a stationary clock": Interpretation A


And it is correct interpretation, because if you rotate the clock hanging on the rope, it will run more slowly than the stationary clock during it and it will remain delayed even after such an experiment - for ever.


Why do you have to rotate the clock to get this result? This can only mean that the clocks you are using are not perfect since their capability of keeping time is controlled by their orientations relative to one another. Are you using grandfather clocks? This can hardly be a relativistic effect!

You're ignoring the deBroglie wave effect here in the same way, like in your naive "all is the wave" interpretation of QM.


Where am I ignoring the de Broglie wave effect? In fact the derivation of length contraction is ignoring the de Broglie wave effect: My length dilation does NOT!

johanfprins
1 / 5 (4) Apr 30, 2013
My length dilation does NOT!
LOL, your "length dilation" even doesn't contain the "Broglie" word...:-)


Why do you refuse to read and check my calculations derived with impeccable mathematics from the Lorentz transformation. I refer you AGAIN to: http://www.cathod...tion.pdf

Why are you such a closed-minded bigot who is not willing to even look at any evidence which do not suit your religious beliefs?
Fleetfoot
4 / 5 (4) Apr 30, 2013
For time dilation, in the Ives-Stilwell experiment, the moving ions demonstrate a frequency shift greater than predicted by classical Doppler. The difference is what we call the time dilation factor. There is no "interpretation" involved, just a simple empirical observation.


Of course there is an interpretation involved: Your interpretation is that a "moving clock keeps slower time than a stationary clock": Interpretation A.


The observation is that the moving ions' spectral line is shifted. You could interpret that with aether theory or SR, but I didn't do either.

The correct interpretation is that an event occurring within the "moving" IRF is observed within the moving IRF BEFORE it is observed within the "stationary" IRF.


There are no measurements of times of events involved in that experiment. There is no comparison of discrete times so 'before' has no meaning in the context. Perhaps you should find out a bit more about the experiment before commenting.
johanfprins
1 / 5 (7) Apr 30, 2013
The observation is that the moving ions' spectral line is shifted. You could interpret that with aether theory


How? Please give the equations.

or SR, but I didn't do either.


Oh my God it is again Natello, ValeriaT, and now "Fleetfood" AKAK! The same shit over and over again.

There are no measurements of times of events involved in that experiment. There is no comparison of discrete times so 'before' has no meaning in the context.


The time relationship that explains this shift is derived from the Lorentz-transformation and the Lorentz transformation for different times is NOT different clocks keeping time at different rates, but different times on different clocks that keep the SAME time-rate. To an idiot it will seem like Einstein's time-dilation but it is NOT.
thefurlong
1.5 / 5 (8) Apr 30, 2013
@johanfprins

I have taken a brief look at the papers you wrote. Something you said in http://www.cathod...tion.pdf confused me. You wrote,
At that instant in time any point-position within IRF=K, which coincides with a point-position within IRF=K', has the same values for its coordinates within both IRF=K and IRF=K'.


To me, this sounds like you are saying that an object that exists at -3 meters away from the origin in K, also at exists at -3 meters away from the origin in K', when t = t' = 0. Is that what you are saying? If not, what are you saying?
Fleetfoot
2.3 / 5 (3) Apr 30, 2013
Please give the equations.


The observed empirical formula is:

f'/f = c/(c+v) * sqrt(1-(v/c)^2)

The first term is the classical effect, the second is the time dilation factor.

The time relationship that explains this shift is derived from the Lorentz-transformation


In the Ives-Stilwell experiment, all measurements are made in the lab frame so no transforms are used at all. It is simply an observed result without any theoretical derivation.

and the Lorentz transformation for different times is NOT different clocks keeping time at different rates


Clocks running at different rates in the ratio of the time dilation factor would be an aether-based interpretation.

but different times on different clocks that keep the SAME time-rate.


Clocks running at the same rate with the time dilation factor caused by geometric projection would be the SR interpretation.

To an idiot it will seem like Einstein's time-dilation but it is NOT.


Both match the result.
johanfprins
1 / 5 (5) May 01, 2013
@johanfprins

I have taken a brief look at the papers you wrote.


Thank you: This is the only way in which one must discuss physics!

To me, this sounds like you are saying that an object that exists at -3 meters away from the origin in K, also at exists at -3 meters away from the origin in K', when t = t' = 0. Is that what you are saying? If not, what are you saying?


Yes that is what I am saying: of the instantaneous position of any object. These coordinates are determined by the Galilean transformation.

For example, after a time t has elapsed in K the distance between the origins is D=v*t. If the time that elapsed on the clock in K/ is t/, the distance between the origins is D=v*t/, and since this distance is a SINGLE distance one MUST have that v*t/=v*t: i.e. that t/=t. Thus the clocks MUST keep the same time.

If, however, at this instant t/=t an event occurs at (say) the origin of K/, this event is observed at a a LATER time t(e)=(gamma)*t within K.
johanfprins
1 / 5 (4) May 01, 2013
If, however, at this instant t/=t an event occurs at (say) the origin of K/, this event is observed at a a LATER time t(e)=(gamma)*t within K.


This is so since the information that an event has occurred at the origin of K/ cannot reach the origin of K at a speed faster than light-speed.

In fact, if the origin of K/ moves towards the origin of K (not away from it) and an event occurs at the origin of K/ at the same synchronous time on both clocks, the time at which the event is recorded within K is BEFORE the clocks reach the time t. This sounds like a breach of causality, but it is not since the event will not be recorded within K unless it occurs within K/.

I have gone great pains in my manuscripts to prove this by direct derivations from the Lorentz transformation.

johanfprins
1 / 5 (6) May 01, 2013
Please give the equations.


The observed empirical formula is:

f'/f = c/(c+v) * sqrt(1-(v/c)^2)

The first term is the classical effect, the second is the time dilation factor.


The second term is NOT caused by two clocks keeping time at different rates but by a difference in time on both clocks which keep the same time-rate. If the two clocks did keep time at different rates, it will violate both postulates on which Einstein based his Special Theory of Relativity.

Clocks running at the same rate with the time dilation factor caused by geometric projection would be the SR interpretation.


What do you mean by geometric projection? Stop posting nonsense and first read my manuscripts if you can understand mathematics. An PLEASE if you want to use aether, derive and post the mathematical formulas to prove that this non-theory works!

ValeriaT
1 / 5 (6) May 01, 2013
If the two clocks did keep time at different rates, it will violate both postulates on which Einstein based his Special Theory of Relativity
It's easy to say, but why?
johanfprins
1 / 5 (6) May 01, 2013
If the two clocks did keep time at different rates, it will violate both postulates on which Einstein based his Special Theory of Relativity
It's easy to say, but why?


Einstein's first postulate: The laws of physics are the SAME within ALL inertial reference frames.

If two clocks within two different inertial reference frames do not keep the same time, the laws of physics cannot be the same within the two inertial reference frames. The mechanisms of the clocks are determined by the laws of physics. So if the two clocks are identical, how can they keep different times if the laws of physics are not different within the two inertial reference frames?

It is so simple that any idiot, except YOU of course, should be able to understand the logic!
ValeriaT
1 / 5 (5) May 01, 2013
Einstein's first postulate: The laws of physics are the SAME within ALL inertial reference frames.
Well, this is the actual source of your problem with contemporary physics: you even don't know the special relativity - yet you're determined to correct it. The first postulate of special relativity actually sounds:
The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other
BTW You've been reported for usage of rude language.
johanfprins
1 / 5 (6) May 01, 2013
Einstein's first postulate: The laws of physics are the SAME within ALL inertial reference frames.


Well, this is the actual source of your problem with contemporary physics: you even don't know the special relativity - yet you're determined to correct it. The first postulate of special relativity actually sounds: The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other

BTW You've been reported for usage of rude language.


The consequences of the two statements are exactly the same as can be verified by reading different text books on physics; and not just depending on WIKI like you are doing. May I again suggest that you at least take a simple course on elementary physics and read more than WIKI?. Quite clearly you do not even understand the physics which is nowadays being taught in secondary schools.
Fleetfoot
1 / 5 (1) May 01, 2013
Please give the equations.


The observed empirical formula is:

f'/f = c/(c+v) * sqrt(1-(v/c)^2)

The first term is the classical effect, the second is the time dilation factor.


The second term is NOT caused by ...


I emphasized that the above relationship is empirical, it presumes no specific cause. You asked for the equation and that's what I gave you.

Clocks running at the same rate with the time dilation factor caused by geometric projection would be the SR interpretation.


What do you mean by geometric projection? Stop posting nonsense ..


That is the standard SR explanation for time dilation. If you don't even recognise it, perhaps you should read a textbook on the subject.

An PLEASE if you want to use aether, ...


I have only a historical interest in obsolete aether theories but you contrasted two explanations, one was that based on an aether and the other was that from SR, hence the second and third parts in my reply.
thefurlong
1.7 / 5 (11) May 01, 2013

Einstein's first postulate: The laws of physics are the SAME within ALL inertial reference frames.

If two clocks within two different inertial reference frames do not keep the same time, the laws of physics cannot be the same within the two inertial reference frames.


I have to read the papers you submitted before commenting further on them, but I would like to comment on this. You are misinterpreting the first postulate and its consequences in a variety of ways.

This postulate arose from Einstein's thought experiments regarding Maxwell's equations, not as is commonly thought from the Michaelson-Morley experiments. Specifically, Einstein realized that Maxwell's equations yield a set of differential equations whose form would change depending on the inertial reference frame. To elucidate, these equations predict that any oscillating electromagnetic field must travel at the speed of light in a vacuum. [This will be continued in my next comment.]
thefurlong
1.7 / 5 (11) May 01, 2013
[part 2]
This expression is a function of the constants, magnetic permeability, and electric permittivity. Therefore, the speed that light is measured travelling at directly determines the values that these constants should take, which, in turn, affects what Maxwell's equations should be. Since these are constants, then, they can't depend on anything, including a change of coordinate system. Hence, Einstein postulated that they, and indeed, the differential equations describing the laws of physics must not depend on inertial reference frame.
So, when you read that postulate, you should take it to mean that the differential equations governing physics don't vary. That doesn't, however, mean that we can't measure the same thing differently depending on our IRF.
This brings me to the second place where I think you are confused. You are correct in saying that the same physical situation cannot have two different outcomes (at least in the continuum limit). [To be continued]
johanfprins
1 / 5 (7) May 01, 2013
I emphasized that the above relationship is empirical, it presumes no specific cause.


Nonsense, this result is a direct consequence of the postulates of the Special Theory of Relativity: And the experiment was done to test this. The result was just incorrectly interpreted as a proof of "time-dilation" which does NOT occur! To claim the result has been empirically discovered is another one of your blatant lies!

That is the standard SR explanation for time dilation. If you don't even recognise it, perhaps you should read a textbook on the subject.


No it is NOT! YOU should read some elementary physics!

I have only a historical interest in obsolete aether theories


Another lie since AWT is YOUR mantra.

but you contrasted two explanations, one was that based on an aether and the other was that from SR, hence the second and third parts in my reply.


I most certainly DID NOT! Why would I contrast ANYTHING with an absurd aether which DOES NOT exist?
thefurlong
2.2 / 5 (13) May 01, 2013
[part 3]
However, two different people can measure same thing and get different results. We see this all the time when cars pass each other on the highway. If I am on a sidewalk, and I see two cars approach and pass each other at 60 mph, each car will measure the other's speed at 120 mph. Therefore, some values depend on the coordinate system. There are, however, values that are invariant of coordinate system. In regular old galilean relativity, this is just Euclidean distance. In that case, no matter how fast Alice and Bob are moving relative to eachother, they will always measure a mile to be the same size. In relativity, because of the first postulate, Euclidean distance is no longer invariant. However, a new measure, x^2 + y^2 + z^2 -(ct)^2 is invariant, hence that is what guarantees that physics remains consistent. I have more to say but I have to get back to work now...
johanfprins
1 / 5 (6) May 01, 2013
This postulate arose from Einstein's thought experiments regarding Maxwell's equations, not as is commonly thought from the Michaelson-Morley experiments.


I have NOT stated this since ANY FOOL who have read Einstein paper will know that he did NOT quote the MM experiment, although he referred to it in an oblique manner; which unfortunately reflects badly on Einstein's scientific integrity.

Specifically, Einstein realized that Maxwell's equations yield a set of differential equations whose form would change depending on the inertial reference frame. To elucidate, these equations predict that any oscillating electromagnetic field must travel at the speed of light in a vacuum.


Not correct: Einstein argued that according to Galileo's explanation of relativity, the speed of light must be constant or else you will be able to do an experiment within an IRF to determine whether the IRF is moving or not. He extrapolated Galileo's logic to include light speed!
johanfprins
1 / 5 (6) May 01, 2013
Thus to keep laws of physics independent of the motion of the IRF the speed of light must be the same within ALL IRF's.

So, when you read that postulate, you should take it to mean that the differential equations governing physics don't vary. That doesn't, however, mean that we can't measure the same thing differently depending on our IRF.


Where have I stated this? I am NOT a moron! Obviously if you look at the physics happening in an IRF from a passing IRF you will not see the same physics. Why do you think that I do NOT know this?

However, two different people can measure same thing and get different results. We see this all the time when cars pass each other on the highway. If I am on a sidewalk, and I see two cars approach and pass each other at 60 mph, each car will measure the other's speed at 120 mph. Therefore, some values depend on the coordinate system.


WE ALL know this! It does not change one iota what I have claimed above.
johanfprins
1 / 5 (5) May 01, 2013
In relativity, because of the first postulate, Euclidean distance is no longer invariant.


WRONG: Euclidean distance is still instantaneously invariant. If you suddenly stop all motion of the IRF's, the coordinate relationship of distance will be exactly the same as in the Euclidean-Galilean case.

However, a new measure, x^2 + y^2 + z^2 -(ct)^2 is invariant, hence that is what guarantees that physics remains consistent.


Wrong! It cannot be so since the coordinates x,y,z and ict are NOT linearly independent and NEVER will be linearly independent. Thus, unique "time-space" intervals cannot exist within such a space. There is NO Minkowski-space and NO Lorentz-group of transformations. It is not mathematically allowed!

I have more to say but I have to get back to work now..


I suggest that you read an elementary book on linear algebra and especially concentrate on the linear independence of coordinates.

ValeriaT
1 / 5 (5) May 01, 2013
@johanprins: my experience with you from many discussions about particle-wave duality here is, you're refusing some concept (particles), whereas you're introducing another ones (boundary conditions, etc.), which are impossible without re-introduction of particle concept on background - or which are even directly equivalent to it (the boundary condition for flat wave is just the density gradient of vacuum, which represents the particle in AWT). So that - despite I still don't understand your point regarding SR clearly - I'm pretty sure, that after many hours of vigorous discussion we will all again realize, you're describing the same stuff, like Einstein did - just with your own words. The inability of yours to present some experimental falsification of your interpretation of special relativity speaks for itself in this direction. For me it's the simply waste of time, because I do understand your naive flat-world philosophy already - so I can see its limits in advance.
johanfprins
1 / 5 (6) May 01, 2013
@johanprins: my experience with you from many discussions about particle-wave duality here is, you're refusing some concept (particles), whereas you're introducing another ones (boundary conditions, etc.), which are impossible without re-introduction of particle concept on background -


Prove to me mathematically why it is "impossible without introduction of particle concept on background".

(the boundary condition for flat wave is just the density gradient of vacuum, which represents the particle in AWT).


Again prove to me mathematically that this absurd assertion of yours is logically valid!

I'm pretty sure, that after many hours of vigorous discussion we will all again realize, you're describing the same stuff, like Einstein did - just with your own words.


How can you be sure if you refuse to even read my mathematically quantitative arguments that I am NOT deriving the "same stuff" like "time dilation" and "length contraction" as Einstein did!

johanfprins
1 / 5 (6) May 01, 2013
The inability of yours to present some experimental falsification of your interpretation of special relativity speaks for itself in this direction.


How must I falsify something that is correct?

For me it's the simply waste of time, because I do understand your naive flat-world philosophy already - so I can see its limits in advance.


Your "time is valueless"! I wish you would keep it to yourself: Nobody with brains is interested in your ideas. You cannot even see that you are a certifiable moron: How can you judge anybody else's insights. Please go and play with your rubber ducks in your foam bath while your mommy washes you stinking asshole while telling you what a "genius" you are.
ValeriaT
1 / 5 (6) May 01, 2013
How must I falsify something that is correct?
You have to present some feasible easy to follow experiment, which will prove your interpretation of Lorentz/Einstein transforms valid and the classical interpretation invalid. The juggling with equations will not impress anybody here, as the people have their own experience with forty years standing development of string theory equations already.
go and play with your rubber ducks in your foam bath while your mommy washes you stinking asshole while telling you what a "genius" you
I'm not a genius. Just because I'm inherently silly, I'm experienced with handling of theories like the black box while looking for their results and experimental predictions. It's surprisingly powerful approach: you can recognize an aged egg without being hen, or even without being able to lay eggs. The details of theories don't bother me, I don't even need to know, how they're built in - I only need to become familiar with their logics and arguments.
Fleetfoot
3.7 / 5 (3) May 01, 2013
I emphasized that the above relationship is empirical, it presumes no specific cause.


Nonsense, this result is a direct consequence of the postulates of the Special Theory of Relativity:


The experiment was designed as a test as you say but the fact remains that the result can be obtained from the result empirically and the confirmed formula is independent of your presumptions about the cause.

To claim the result has been empirically discovered is another one of your blatant lies!


I didn't claim it was, but you can run the same experiment today and check that the formula applies REGARDLESS of what explanation you espouse.

That is the standard SR explanation for time dilation. If you don't even recognise it, perhaps you should read a textbook on the subject.


No it is NOT!


Clocks in SR produce ticks at the same interval of proper time irrespective of its velocity and those then project onto the observer's coordinate time axis.
Fleetfoot
3 / 5 (2) May 01, 2013
I have only a historical interest in obsolete aether theories


Another lie since AWT is YOUR mantra.


Not me, AWT is an acronym used by "Valeria", there is actually no such theory and his comments on it are nonsensical as I regularly point out to him.

but you contrasted two explanations, one was that based on an aether and the other was that from SR, hence the second and third parts in my reply.


I most certainly DID NOT!


Well maybe I misunderstood, here is what you said:

the Lorentz transformation for different times is NOT different clocks keeping time at different rates, but different times on different clocks that keep the SAME time-rate.


Differently moving clocks running at different rates is the aether explanation for the Lorentz transforms while differently moving clocks running at the same rate but with a scaling factor from the projection onto coordinate axes is SR. If not that, what were you contrasting?
thefurlong
1.4 / 5 (9) May 01, 2013

WRONG: Euclidean distance is still instantaneously invariant. If you suddenly stop all motion of the IRF's, the coordinate relationship of distance will be exactly the same as in the Euclidean-Galilean case.

Yes, if everything suddenly moves at rest with everything else, everything will agree on length, but I don't see how that applies. I have the nagging feeling that you think that because you can look at the evolution of a differential equation as a continuum of distinct "snapshots" that each of those snapshots can be treated as static--that is to say lacking any velocity component.
Fleetfoot
3 / 5 (2) May 01, 2013
However, a new measure, x^2 + y^2 + z^2 -(ct)^2 is invariant, ..


Wrong! It cannot be so since the coordinates x,y,z and ict are NOT linearly independent and NEVER will be linearly independent....

I suggest that you read an elementary book on linear algebra and especially concentrate on the linear independence of coordinates.


You are the one who needs to open a textbook, 'thefurlong' is completely correct. The value is called the "invariant interval" for that reason:

http://en.wikiped...ntervals
Q-Star
2.8 / 5 (18) May 01, 2013
However, a new measure, x^2 + y^2 + z^2 -(ct)^2 is invariant, ..


Wrong! It cannot be so since the coordinates x,y,z and ict are NOT linearly independent and NEVER will be linearly independent....

I suggest that you read an elementary book on linear algebra and especially concentrate on the linear independence of coordinates.


You are the one who needs to open a textbook, 'thefurlong' is completely correct. The value is called the "invariant interval" for that reason:

http://en.wikiped...ntervals


I'm just chiming in to agree. That why Einstein's SR & GR have been so successful, and the greatest gift to modern physics, Space is relative to the observer. Time is relative to the observer, spacetime is invariant for any and all observers. He leveled the cosmic playing field.
Whydening Gyre
1 / 5 (9) May 01, 2013
I have the nagging feeling that you think that because you can look at the evolution of a differential equation as a continuum of distinct "snapshots" that each of those snapshots can be treated as static--that is to say lacking any velocity component.


Yeah, but... I am getting the nagging feeling you think a series of "snapshots" don't actually represent that velocity component... Call them a series of "experiments", if you will.
johanfprins
1 / 5 (6) May 02, 2013
You have to present some feasible easy to follow experiment, which will prove your interpretation of Lorentz/Einstein transforms valid and the classical interpretation invalid.


I agree: And the only feasible experiment would be to compare two clocks which has moved linearly relative to one another after they have moved a distance apart. The only problem is to bring the two clocks back together again: This involves deceleration and acceleration. That time might change during acceleration and deceleration is possible if Einstein's principle of equivalence is really valid. The flying clocks experiment can thus not test time dilation for Special Relativity without making extra assumptions.

But even without being able to do such an experiment, logic tells you that Einstein's first postulate can ONLY be valid if the two clocks keep exactly the same time.This has also been repeatedly pointed out by other scientists. Nobody with brains can get past this argument EVER!

johanfprins
1 / 5 (6) May 02, 2013
I'm experienced with handling of theories like the black box while looking for their results and experimental predictions.


You are unduly flattering yourself since it is clear that you do not and NEVER has understood the black box approach. It is useful but should be handled with care by somebody who has brains: This excludes YOU!

The details of theories don't bother me, I don't even need to know, how they're built in - I only need to become familiar with their logics and arguments.


A good summary of why you should not attempt to even argue physics. If your model cannot explain the details it is not a model, but like your AWT, nothing else than pie in the sky! Nonsense, claptrap and the hallucinations of a crackpot!

And please accept that you have proved time and again on this forum that you are not able to understand what logic is!

johanfprins
1 / 5 (6) May 02, 2013
The experiment was designed as a test as you say but the fact remains that the result can be obtained from the result empirically and the confirmed formula is independent of your presumptions about the cause.


The formula will not be "empirically" measured if does not have a cause. Do you want to argue that it does not have a cause?

I didn't claim it was, but you can run the same experiment today and check that the formula applies REGARDLESS of what explanation you espouse.


This is the way physics is: Why are you raising this stupid argument as if this is not the case when doing other measurements? You are really confused!

Clocks in SR produce ticks at the same interval of proper time irrespective of its velocity and those then project onto the observer's coordinate time axis.


"Proper" time in SR is the exact same time that is simultaneously kept by ALL clocks in a gravity-free universe, no matter with what speed they are moving relative to one another!
johanfprins
1 / 5 (6) May 02, 2013
I have only a historical interest in obsolete aether theories


Another lie since AWT is YOUR mantra.


Not me, AWT is an acronym used by "Valeria", there is actually no such theory and his comments on it are nonsensical as I regularly point out to him.


Thank God that you are NOT ValeriaT under another name: You nearly fooled me!!

Differently moving clocks running at different rates is the aether explanation for the Lorentz transforms while differently moving clocks running at the same rate but with a scaling factor from the projection onto coordinate axes is SR. If not that, what were you contrasting?


You are a bit more coherent here. I could not follow what you meant by a projection. Obviously it comes from your religious belief in a non-existing Minkowski space. There is NO "proper time" as defined within the Miinkowski paradigm. So there is no projection from this "proper time" onto the coordinate axes of SR. There is only absolute time!

johanfprins
1 / 5 (5) May 02, 2013
The difference in time demanded by the Lorentz transformation comes purely from the fact that the information that an event has occurred at a coordinate position within another inertial "moving" reference frame cannot reach the origin of the "stationary" reference frame faster than the speed of light.

There are no unique space-time distances as Minkowski has claimed, since the coordinates x,y,z,and ict are NOT linearly independent: This means that there are more than one coordinate point which have coordinates 0,0,0,0: To have unique four-dimensional distances ONLY the origin of a four-dimensional space can be 0,0,0,0. This IS NOT SO IN MINKOWSKI SPACE!
johanfprins
1 / 5 (5) May 02, 2013
Yes, if everything suddenly moves at rest with everything else, everything will agree on length, but I don't see how that applies.


It does apply: At any instant in time an event occurs at the same coincident space and time coordinates as determined by the Galilean transformation. An observer at the origin of the "stationary" IRF observes this event at a different position and time since the information that the event occurred cannot reach the observer faster than the speed of light. If the information could have reached him instantaneously, he would would have seen the event as if the Galilean transformation applies.

I have the nagging feeling that you think that because you can look at the evolution of a differential equation as a continuum of distinct "snapshots" that each of those snapshots can be treated as static.


This is what calculus does: You take instantaneous "static snapshots" at times (delta)t apart and let (delta)t go to zero to get the velocity!
johanfprins
1 / 5 (8) May 02, 2013
You are the one who needs to open a textbook, 'thefurlong' is completely correct. The value is called the "invariant interval" for that reason:


An interval within a four-dimensional space can only be invariant when the four coordinates are linearly independent so that just one "position" in that space has the coordinates 0,0,0,0. This is simple first year mathematics.

This means that when x^2+y^2+z^2+(ict)^2=0, one MUST have that if x=0, y=0, z=0, and t=0. This expression cannot be zero for any other set of coordinates if invariant distances exist.

This condition for invariant ditances is NOT VALID in Minkowski space since for any point on a spherical wave-front one must have that x^2+y^2+z^2+(ict)^2=0, even though in this case x need not be zero, y need not be zero, z need not be zero and t need not be zero; as is demanded that they must be for invariant space-time distances to exist.
johanfprins
1 / 5 (4) May 02, 2013
I'm just chiming in to agree. That why Einstein's SR & GR have been so successful, and the greatest gift to modern physics, Space is relative to the observer. Time is relative to the observer, spacetime is invariant for any and all observers. He leveled the cosmic playing field.


I know that this is the official dogma; but my analyses indicates that it STR and GTR are successful in spite of being wrongly interpreted. Einstein used time-dilation and length contraction to motivate his GTR. But it is easy to show that time dilation and length contraction cannot be derived from the Lorentz transformation. I refer you again to: http://www.cathod...tion.pdf and
http://www.cathod...tion.pdf

So why does Einstein's GTR work if the arguments that Einstein used to justify non-Euclidean coordinates are wrong?
johanfprins
1 / 5 (6) May 02, 2013
I believe that this was fortuitous since the non-Euclidean coordinates are NOT required owing to any relativity effects, but are required since matter consists of waves which distort space-time around them. This distortion is wrongly interpreted as "tunneling tails".

I am at present busy investigating this possibility.

Nonetheless, it does not remove the fact that Einstein used "length contraction" and "time dilation" in STR, to justify his theory of GTR, even though the Lorentz transformation cannot be used to prove that "time-dilation" and "length contraction" are possible; unless Einstein's second postulate is wrong: i.e. unless the speed of light IS NOT the same within all IRF's.

Fleetfoot
2.3 / 5 (3) May 02, 2013
The experiment was designed as a test as you say but the fact remains that the result can be obtained from the result empirically and the confirmed formula is independent of your presumptions about the cause.


The formula will not be "empirically" measured if does not have a cause. Do you want to argue that it does not have a cause?


This is the sense in which I am using the term:

http://en.wikiped...tionship

Clocks in SR produce ticks at the same interval of proper time irrespective of its velocity and those then project onto the observer's coordinate time axis.


"Proper" time in SR is the exact same time that is simultaneously kept by ALL clocks in a gravity-free universe, no matter with what speed they are moving relative to one another!


No, proper time is the line integral of tau accumulated along a worldline and applies equally well to accelerated motion and in gravity:

http://en.wikiped...per_time
Fleetfoot
3 / 5 (4) May 02, 2013
Differently moving clocks running at different rates is the aether explanation for the Lorentz transforms while differently moving clocks running at the same rate but with a scaling factor from the projection onto coordinate axes is SR. If not that, what were you contrasting?


You are a bit more coherent here. I could not follow what you meant by a projection. Obviously it comes from your religious belief in a non-existing Minkowski space. There is NO "proper time" as defined within the Miinkowski paradigm.


As I said in another reply, proper time is defined as the line integral along a path in any metric solution.

So there is no projection from this "proper time" onto the coordinate axes of SR. There is only absolute time!


That maybe the source of the previous confusion, there is no absolute time in SR but it is the fundamental to aether theory.
johanfprins
1 / 5 (6) May 02, 2013
The formula will not be "empirically" measured if does not have a cause. Do you want to argue that it does not have a cause?


This is the sense in which I am using the term:

http://en.wikiped...tionship


Fom your own reference: "Sometimes theoretical explanations for what were initially empirical relationships are found, in which case the relationships are no longer considered empirical."

No, proper time is the line integral of tau accumulated along a worldline and applies equally well to accelerated motion and in gravity:


I know the official dogma better than you can EVER know it. This definition is not physically nor mathematically possible since one cannot define a "world-line" unless the time and position coordinates are linearly independent. This is not the case in Galilean space and also not within Minkowski space. Time is absolute and the same at all positions and within all inertial reference frames at the same instant in time.
Fleetfoot
3 / 5 (4) May 02, 2013
There are no unique space-time distances as Minkowski has claimed, since the coordinates x,y,z,and ict are NOT linearly independent: This means that there are more than one coordinate point which have coordinates 0,0,0,0: To have unique four-dimensional distances ONLY the origin of a four-dimensional space can be 0,0,0,0. This IS NOT SO IN MINKOWSKI SPACE!


Here's an analog: Place a cocktail stick on a table and place a transparent sheet of graph paper over it. Read off the coordinates of each end as (x1, y1) and (x2, y2), then calculate

s^2 = (x2 - x1)^2 + (y2 - y1)^2

Move and rotate the sheet and read off the new values. The new value of s will be the same as the first, obviously it is just the length of the stick calculated using Pythagoras. The origin of the graph may well have moved, it doesn't matter. In SR, the invariant interval is the same thing but in 4 dimensions, it has a single value between events regardless of the origin or rotation of the axes.
johanfprins
1 / 5 (6) May 02, 2013
It can only argued that time is only "not absolute" within the body of a coherent wave since the wave has a phase angle that gives different phase-times along the wave. But time does not change from position to position within space itself!

Thus you can formulate Maxwell's equations as if time is not absolute, but not space-time itself!
Fleetfoot
2.3 / 5 (3) May 02, 2013
The formula will not be "empirically" measured if does not have a cause. Do you want to argue that it does not have a cause?


This is the sense in which I am using the term:

http://en.wikiped...tionship


Fom your own reference: "Sometimes theoretical explanations for what were initially empirical relationships are found, in which case the relationships are no longer considered empirical."


First line: "In science, an empirical relationship is one based solely on observation rather than theory."

No, proper time is the line integral of tau accumulated along a worldline and applies equally well to accelerated motion and in gravity:


I know the official dogma better than you can EVER know it.

Apparently you don't.

This definition is not physically nor mathematically possible ..


You opinion is irrelevant, the fact remains that that IS the definition of proper time.
johanfprins
1 / 5 (6) May 02, 2013
Here's an analog: Place a cocktail stick on a table and place a transparent sheet of graph paper over it. Read off the coordinates of each end as (x1, y1) and (x2, y2), then calculate

s^2 = (x2 - x1)^2 + (y2 - y1)^2

Move and rotate the sheet and read off the new values. The new value of s will be the same as the first, obviously it is just the length of the stick calculated using Pythagoras. The origin of the graph may well have moved, it doesn't matter.


Although the position of the origin does not matter, one must have an origin from which you measure the coordinates x1, x2, y1 and y2. This requires that the coordinates must be linearly independent: i.e. that s^2=x^2+y^2 can only give s=0 when x=0 and y=0. Only then your expression for s is unique.

This is not the case for the coordinates x, y, z and ict, which supposedly define unique distances within M-space: Since within M-space you can have that s=0 while x,y,z,ict need not be zero. It is just simple mathematics!

johanfprins
1 / 5 (5) May 02, 2013
First line: "In science, an empirical relationship is one based solely on observation rather than theory."


Only a fool will only read the first line and think that this contains the whole definition!

No, proper time is the line integral of tau accumulated along a worldline and applies equally well to accelerated motion and in gravity:


I know the official dogma better than you can EVER know it.


Apparently you don't.


Oh I do! I just do not agree with it since it violates the most basic rules of mathematics.

This definition is not physically nor mathematically possible .


You opinion is irrelevant, the fact remains that that IS the definition of proper time.


"Opinion"? I have proved above that a space-time distance s within M-space can be zero without having that x, y, z, and t must all also be zero. Thus, these coordinates are NOT linearly-independent and can thus not define unique four-dimensional distances and a "proper time".
johanfprins
1 / 5 (7) May 02, 2013
Consider a four-dimensional linear space with coordinates x,y,z,u. The distance of a coordinate point s from the origin at (0,0,0,0) is thus given by:

s^2=x^2+y^2+z^2+u^2. If s=0 without x=0, y=0, z=0 and u=0, one will have a coordinate point x,y,z,u which is not the origin, but which is situated at the origin since s=0. This is obviously nonsense.

In the case of Minkowski space any point on a spherical wave-front around the spatial origin x=0, y=0, z=0, is given by the equation:

x^2+y^2+z^2=(ct)^2 which demands that

x^2+y^2+z^2+(ict)^2=0 without also demanding that x must be 0, y must be 0, z must be zero AND t must be zero. Thus for such a point on the wave-front you have that the corresponding coordinates within M-space need not be zero for the distance of these coordinates to be zero as measured from the origin (0,0,0,0). Thus, clearly one does not have a unique space-time distance from the origin for all points within M-space. It is thus nonsensical to define a "world-line"
Fleetfoot
3.7 / 5 (3) May 02, 2013
First line: "In science, an empirical relationship is one based solely on observation rather than theory."


Only a fool will only read the first line and think that this contains the whole definition!


Only a fool will cherry pick a single part and disregard the main body. If you compared the article as a whole with what I wrote, it should be obvious that my point is that you can perform the experiment, plot the frequency shift versus the velocity and confirm the equation empirically, without recourse to ANY theory. It then becomes a separate question as to whether or not any particular theory is compatible with that equation.

No, proper time is the line integral of tau accumulated along a worldline and applies equally well to accelerated motion and in gravity:


I know the official dogma better than you can EVER know it.


Apparently you don't.


Oh I do!


Then why make yourself look ignorant by deliberately getting it wrong?
Fleetfoot
3.7 / 5 (3) May 02, 2013
Place a cocktail stick on a table and place a transparent sheet of graph paper over it. Read off the coordinates of each end as (x1, y1) and (x2, y2), then calculate

s^2 = (x2 - x1)^2 + (y2 - y1)^2

Move and rotate the sheet and read off the new values. The new value of s will be the same as the first, obviously it is just the length of the stick calculated using Pythagoras. The origin of the graph may well have moved, it doesn't matter.


Although the position of the origin does not matter, one must have an origin from which you measure the coordinates x1, x2, y1 and y2.


No problem.

This requires that the coordinates must be linearly independent: i.e. that s^2=x^2+y^2 can only give s=0 when x=0 and y=0.


You have lost track, s is not zero. Try an example:

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?

Move the origin by (-2,-1):

If (x1,y1) = (4,2) and (x2,y2) = (8,5), what is s?

Rotate about (4,2):

If (x1,y1) = (4,2) and (x2,y2) = (7,6), what is s?
johanfprins
1.4 / 5 (9) May 02, 2013
Only a fool will cherry pick a single part and disregard the main body. If you compared the article as a whole with what I wrote, it should be obvious that my point is that you can perform the experiment, plot the frequency shift versus the velocity and confirm the equation empirically, without recourse to ANY theory.


Where have I disputed this. Are YOU really SO stupid? YOU must be ValeriaT.

It then becomes a separate question as to whether or not any particular theory is compatible with that equation.


This is so in ALL physics! So what NEW are you trying to state? Please get brain-transplant!

So why make yourself look ignorant by deliberately getting it wrong?


Where did I get it wrong: Except within your demented mind. Do you not agree that a distance from the origin within a four-dimensional "space" is s=SQRT(x^2+y^2+z^2+u^2) and that s can only be zero when it is the origin. And in M-space you have that s=0 when the coordinates are NOT the origin?
johanfprins
1.5 / 5 (8) May 02, 2013
This requires that the coordinates must be linearly independent: i.e. that s^2=x^2+y^2 can only give s=0 when x=0 and y=0.


You have lost track, s is not zero. Try an example:

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?

Obviously not in two-dimensional Euclidean space, but that is NOT what we are discussing We are discussing four-dimensional Minkowski space and in this case one does have FOR ANY "SPACE-TIME" POINT ON THE surface of a spherical wave-front surrounding the THREE-DIMENSIONAL SPATIAL ORIGIN that s=SQRT(x^2+y^2+z^2+(ict)^2) MUST BE =0, even though the coordinates need NOT be individually ZERO!!

PLEASE buy a brain somewhere: You obviously have NOTHING between your ears!
thefurlong
1 / 5 (8) May 02, 2013
Yeah, but... I am getting the nagging feeling you think a series of "snapshots" don't actually represent that velocity component... Call them a series of "experiments", if you will.


So what? It all just depends on what you use to measure time. You need to be careful that the time told by the device you use doesn't have unambiguous meaning, which is why we use light clocks. I mean, yes, you can look at things as a series of snapshots, but the only way you can do that is by designating a moment for each of those snapshots. That's important if you are going to use that snapshot to calculate a correct velocity value, but this is where we come full circle. You are ultimately using a light clock to measure relative velocity, but the behavior of the light clock depends on the relative velocity at every instant. Therefore, it seems that relative velocity must have a physical component at every instant.
Q-Star
3 / 5 (14) May 02, 2013
You have lost track, s is not zero. Try an example:

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?


Obviously not in two-dimensional Euclidean space, but that is NOT what we are discussing We are discussing four-dimensional Minkowski space and in this case one does have FOR ANY "SPACE-TIME" POINT ON THE surface of a spherical wave-front surrounding the THREE-DIMENSIONAL SPATIAL ORIGIN that s=SQRT(x^2+y^2+z^2+(ict)^2) MUST BE =0, even though the coordinates need NOT be individually ZERO!!

PLEASE buy a brain somewhere: You obviously have NOTHING between your ears!


He has a very good brain. This exactly why GR and QT can't be made to blend. Ya are using a very important variable incorrectly, the ict is subtracted in calculating Minkowsky spacetime.

s^2 = x^2 plus y^2 plus z^2 minus ict^2. Pythagoras won't work ya must subtract the time element. It's only way ALL observers will agree with "s". It's not overly difficult, give it a try.
Q-Star
3 / 5 (14) May 02, 2013
Yeah, but... I am getting the nagging feeling you think a series of "snapshots" don't actually represent that velocity component... Call them a series of "experiments", if you will.


So what? It all just depends on what you use to measure time. You need to be careful that the time told by the device you use doesn't have unambiguous meaning, which is why we use light clocks. I mean, yes, you can look at things as a series of snapshots, but the only way you can do that is by designating a moment for each of those snapshots. That's important if you are going to use that snapshot to calculate a correct velocity value, but this is where we come full circle. You are ultimately using a light clock to measure relative velocity, but the behavior of the light clock depends on the relative velocity at every instant. Therefore, it seems that relative velocity must have a physical component at every instant.


Light clocks,,, ya are correct, because "c" is "c" for everyone.
thefurlong
1 / 5 (8) May 02, 2013

At any instant in time an event occurs at the same coincident space and time coordinates as determined by the Galilean transformation.

Only when the laws of physics depend on the IRF.

An observer at the origin of the "stationary" IRF observes this event at a different position and time since the information that the event occurred cannot reach the observer faster than the speed of light.

You need to be careful not to confuse witnessing an event with measuring when it happened. If I fire a gun, a person standing a half a mile a way will hear the shot a split second after it has gone off. That doesn't mean the person won't be able to measure that it happened exactly when I fired it.

You take instantaneous "static snapshots" at times (delta)t apart and let (delta)t go to zero to get the velocity

Well, duh! But you still need to be able to measure time and distance correctly to take the right limit. Those values depend on relative velocity.
Fleetfoot
1 / 5 (1) May 03, 2013
.. If you compared the article as a whole with what I wrote, it should be obvious that my point is that you can perform the experiment, plot the frequency shift versus the velocity and confirm the equation empirically, without recourse to ANY theory.


Where have I disputed this.


In your previous post where you insisted the formula I gave was derived from theory.

So what NEW are you trying to state?


Nothing, just teaching you some basics.

Do you not agree that a distance from the origin within a four-dimensional "space" is s=SQRT(x^2+y^2+z^2+u^2) and that s can only be zero when it is the origin.


Your error is that the point that is the origin in the first coordinate system is no longer (0,0) after the translation or rotation. The invariant interval is defined between two physically identified points (such as the ends of the cocktail stick) in 2D or between two events in 4D.
johanfprins
1 / 5 (6) May 03, 2013
So what? It all just depends on what you use to measure time. You need to be careful that the time told by the device you use doesn't have unambiguous meaning, which is why we use light clocks.


ValeriaT: You refuse to read my manuscripts and are therefore wasting everybody's time on this forum. I am using a light clock to derive the difference in time correctly. AGAIN I refer you to http://www.cathod...ion.pdf. The correct derivation using a light clock proves that the two clocks keep the same time.

You are ultimately using a light clock to measure relative velocity,
I AM USING A LIGHT CLOCK!!! PLEASE first do your homework before proving time and again that you are a certifiable moron!

Fleetfoot
1 / 5 (1) May 03, 2013
This requires that the coordinates must be linearly independent: i.e. that s^2=x^2+y^2 can only give s=0 when x=0 and y=0.


You have lost track, s is not zero. Try an example:

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?


Obviously not in two-dimensional Euclidean space, but that is NOT what we are discussing ..


I'm splitting the problem into two parts, it's easy to see your error in 2D and once we agree that part (which should be trivial) then I'll give you an equivalent example in 4D. Go along with the game for a moment and answer the three questions.

We are discussing four-dimensional Minkowski space


Specifically we are discussing the invariant interval in the Minkowski Metric.

FOR ANY "SPACE-TIME" POINT ON THE surface of a spherical wave-front surrounding the THREE-DIMENSIONAL SPATIAL ORIGIN that s=SQRT(x^2+y^2+z^2+(ict)^2) MUST BE =0


Only if the wavefront was emitted from the origin and that need not be true when you change frames.
johanfprins
1 / 5 (5) May 03, 2013
He has a very good brain. This exactly why GR and QT can't be made to blend. Ya are using a very important variable incorrectly, the ict is subtracted in calculating Minkowsky spacetime.

s^2 = x^2 plus y^2 plus z^2 minus ict^2


No it is NOT: According to Minkowski space-time s^2=x^2+y^2+z^2-(ct)^2: NOT -(ict)^2

Thus you can have many points, other than the origin, for which s=0, even though the coordinates x, y,z, and t need not be zero. In such a four-dimensional space one CANNOT have unique space-time distances as is claimed that one has within MST.

Pythagoras won't work ya must subtract the time element. It's only way ALL observers will agree with "s". It's not overly difficult, give it a try.


Clearly it is not overly difficult for you to post claptrap!
Fleetfoot
1 / 5 (1) May 03, 2013
We are discussing four-dimensional Minkowski space and in this case one does have FOR ANY "SPACE-TIME" POINT ON THE surface of a spherical wave-front surrounding the THREE-DIMENSIONAL SPATIAL ORIGIN that s=SQRT(x^2+y^2+z^2+(ict)^2) MUST BE =0, even though the coordinates need NOT be individually ZERO!!


Let's correct your error (and I'm setting i^2 = -1 and c=1 just to clarify the typing):

s^2 = (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 - (t2 - t1)^2 = 0

where (x2, y2, z2) is a point on the surface at time t2 and (x1, y1, z1) was the location from which the wavefront was emitted at time t1.

If you changes coordinate systems, all eight values must be transformed, not just the four you list. If (x1, y1, z1) happens to be the origin in your first frame, it may not be after the transform is applied.
johanfprins
1 / 5 (5) May 03, 2013
Light clocks,,, ya are correct, because "c" is "c" for everyone


Correct and it is exactly for this reason why two clocks moving relative to one another must keep exactly the same time BUT why the same event occurs at different absolute times within two inertial reference frames which move relative to one another. PLEASE FOR GOD's sake first study my derivation using a light clock before flouting your massive ignorance further!
Fleetfoot
1 / 5 (1) May 03, 2013
Thus you can have many points, other than the origin, for which s=0, even though the coordinates x, y,z, and t need not be zero. In such a four-dimensional space one CANNOT have unique space-time distances as is claimed that one has within MST.


There seems to be some confusion here. The value of s is invariant (which is what we were discussing) but it doesn't identify a unique event. In fact s=0 defines the whole set of events which lie on the surface of the future and past light cones.
johanfprins
1 / 5 (3) May 03, 2013
At any instant in time an event occurs at the same coincident space and time coordinates as determined by the Galilean transformation.

Only when the laws of physics depend on the IRF.


Nope! When the event occurs at the instant that the two origins coincide, it occurs instantaneous-simultaneously within BOTH IRF's. Since the origins can be chosen at will it means that if you chose the origins to coincide at any event, this event will be instantaneous-simultaneously occurring within both IRF's.

It is only in the case where the origin is chosen not coincide with the event, which still occurs instantaneous-simultaneously within the two IRF's, that the event is referenced relative to this origin to occur at a different time and a different position than it is still actually and instantaneously occurring within both IRF's

I will be back later today!.

johanfprins
1 / 5 (7) May 03, 2013
You need to be careful not to confuse witnessing an event with measuring when it happened.If I fire a gun, a person standing a half a mile a way will hear the shot a split second after it has gone off. That doesn't mean the person won't be able to measure that it happened exactly when I fired it.


A very good analogy: Let us apply it to STR.

Consider an event where a light switches on within an IRFK/ moving relative to an IRFK with speed v. If one chooses the two origins at the position of the light when it switches on, then it switches on simultaneously within both IRF's. (You are at the position of the gun being fired off within your example). Since one can choose the origins at will, one can in principle choose it at any event so that this event occurs simultaneously at the coincident coordinates of the light within both IRF's; which must thus also happen even when you are not at the position of the light (or gun in your example). Continued below
johanfprins
1 / 5 (4) May 03, 2013
If your origin is not coincident with the moving light when it switches on, you have to determine from the speed of light and the time it reaches you when and where it has been switched on. If the gun is moving away from you the sound will approach you at a speed c-v.

But in STR, the light is approaching you with a speed c so that when you determine the position and time, where the light was switched on, you do NOT get the coincident time and position at which it actually switched on. Your calculation will give you a later time and a further position. This is the relativistic reality that you then experience. And it is astounding that this experience is real within IRFK.

If you could simultaneously have marked the coincident coordinate within IRFK at the time that the light switched on in IRFK/, you can afterwards take a tape measure and measure this position. You will then obviously find that this position IS NOT the longer relativistic distance given by the Lorentz transformation.
johanfprins
1 / 5 (6) May 03, 2013
You take instantaneous "static snapshots" at times (delta)t apart and let (delta)t go to zero to get the velocity


Well, duh! But you still need to be able to measure time and distance correctly to take the right limit. Those values depend on relative velocity.


Correct! That is why you need an origin at which you also choose time to be zero, even in the Galilean case.

I assume that the "duh" occurred when the dummy fell out of your mouth.

johanfprins
1 / 5 (7) May 03, 2013
Where have I disputed this.

In your previous post where you insisted the formula I gave was derived from theory.


You claimed that the formula "empirically" proved time dilation. You cannot claim this unless you can prove by theoretical derivation that this is so; and then the formula is NOT empirical anymore!

So what NEW are you trying to state?


Nothing, just teaching you some basics.


Basics? YOU?!!! LOL!

Your error is that the point that is the origin in the first coordinate system is no longer (0,0) after the translation or rotation.


And YOU want to teach ME basics?

The invariant interval is defined between two physically identified points (such as the ends of the cocktail stick) in 2D or between two events in 4D.


You cannot "identify two different points unless each of these points is referenced to an origin. In MST, points with different coordinates can be situated at a a distance s=0 from the origin.
johanfprins
1 / 5 (6) May 03, 2013
This means that there are different four-dimensional coordinate points which are not separated by invariant intervals!!
thefurlong
1.4 / 5 (10) May 03, 2013

Nope! When the event occurs at the instant that the two origins coincide, it occurs instantaneous-simultaneously within BOTH IRF's.


If the event occurs at both origins, then yes, but you can't assume that it occurs at the same time for points at other places. We can see this directly from the Lorentz Transformation. t' = gamma*(t- vx/c^2), where gamma's the lorentz factor.

It is only in the case where the origin is chosen not coincide with the event [...] that the event is referenced relative to this origin to occur at a different time and a different position than it is still actually and instantaneously occurring within both IRF's

Physics doesn't care which origins you use. You and I are only guaranteed to agree when and where the event happens when you, I, and it all occupy the same position simultaneously. We only use the origin because it allows for simpler calculations. We can use non-zero positions if we want to.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
So

Yeah, but... I am getting the nagging feeling you think a series of "snapshots" don't actually represent that velocity component... Call them a series of "experiments", if you will.


So what? It all just depends on what you use to measure time. You need to be careful that the time told by the device you use doesn't have unambiguous meaning, which is why we use light clocks.

What do you mean by "unambiguous meaning"?
If you set 2 clocks (light clocks or Seiko's - I don't care) for the same time, send them in two different directions at different velocities keeping one on your desk for reference, won't they all at any given simultaneous moment still be set to the same time? We just can't measure that because of the boundary of c (limiting our ability to make that observation), right? It's not the clocks that have changed, it's just our measurements of them. So, assuming that time has no variance, time dilation is just our way of describing what it LOOKS like...
johanfprins
1 / 5 (5) May 03, 2013
I'm splitting the problem into two parts, it's easy to see your error in 2D

There is no error in 2D since you only use space coordinates.

and once we agree that part (which should be trivial) then I'll give you an equivalent example in 4D.

The example can ONLY be equivalent if the fourth coordinate is also a space-coordinate NOT when it is a complex time-coordinate.

Go along with the game for a moment and answer the three questions.
Specifically we are discussing the invariant interval in the Minkowski Metric.

You cannot have an invariant interval between two points without first referencing the two points relative to an origin and then take the difference.

Only if the wavefront was emitted from the origin and that need not be true when you change frames.


A change in the origin CANNOT change non-linear coordinates into linear ones. If the coordinates allow unique distances it should also allow them when the wave-front is emitted from the origin.
johanfprins
1 / 5 (5) May 03, 2013
Let's correct your error (and I'm setting i^2 = -1 and c=1 just to clarify the typing):

s^2 = (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 - (t2 - t1)^2 = 0

where (x2, y2, z2) is a point on the surface at time t2

Which point on the wave-front? The wave front is a surface and there are an infinite number of points on it at the same time.

and (x1, y1, z1) was the location from which the wavefront was emitted at time t1.


If you changes coordinate systems, all eight values must be transformed,


And this should be the case for any values of x1,y1, and z1: Also when they are all ZERO. You then get that the distance from the origin in Minkowski space-time to the spherical wave-front must be zero for all the points on the wave-front. Thus you cannot have invariant space-time distances between the points on a wave-front.

PLEASE! Take a course in the simple algebra of linear spaces!

I am taking a break. It tires me out to try and explain simple mathematics and physics to you
thefurlong
1 / 5 (8) May 03, 2013

What do you mean by "unambiguous meaning"?

Well, you and I can both agree that if we see a stationary clock, it keeps correct track of the time we experience. Of course, if we both see it as stationary, then we are also at rest with each other.

If you set 2 clocks (light clocks or Seiko's - I don't care) for the same time, send them in two different directions at different velocities keeping one on your desk for reference, won't they all at any given simultaneous moment still be set to the same time?

Why are you assuming they should?

It's not the clocks that have changed, it's just our measurements of them. So, assuming that time has no variance, time dilation is just our way of describing what it LOOKS like...

Well, if the clock measures that amount of time passing, so will your physical body, as long as it is at rest with the clock, so it's more than just how it "looks."
thefurlong
1.4 / 5 (9) May 03, 2013

Correct! That is why you need an origin at which you also choose time to be zero, even in the Galilean case.

As I said in my last comment, you don't have to use the origin if you don't want to. This seems to be one major source of your incorrect thinking. I can see myself as standing 3 meters from the origin. All that matters is that when you and I occupy the same position, and a balloon pops there too, we both agree that the balloon popped at the same time. This has nothing to do with which position I designate as "0".

I assume that the "duh" occurred when the dummy fell out of your mouth.

For a person who wants his non-mainstream theories to be taken seriously, you certainly make very little effort to encourage it. I mean, how do you think someone like Planck got his theories accepted? Do you think he flipped the bird at his peers and told them to go read basic linear algebra textbooks?
Fleetfoot
1 / 5 (1) May 03, 2013
You claimed that the formula "empirically" proved time dilation.


Not true, you asked me for the formula and I gave you only what you asked for. I explained identified the terms to give us a common basis for discussion. I made no claim that it was proof of anything.

The invariant interval is defined between two physically identified points (such as the ends of the cocktail stick) in 2D or between two events in 4D.


You cannot "identify two different points unless each of these points is referenced to an origin.


Obviously, they both have values from the same coordinate system, but you still need two to define an interval. Humour me, here is the example again:

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?

Move the origin by (-2,-1):

If (x1,y1) = (4,2) and (x2,y2) = (8,5), what is s?

Rotate about (4,2):

If (x1,y1) = (4,2) and (x2,y2) = (7,6), what is s?

This is trivial mental arithmetic for both of us but will illustrate a key point so please play along.
Fleetfoot
1 / 5 (1) May 03, 2013
Let's correct your error (and I'm setting i^2 = -1 and c=1 just to clarify the typing):

s^2 = (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 - (t2 - t1)^2 = 0

where (x2, y2, z2) is a point on the surface at time t2


Which point on the wave-front?


All of them, it is a parametric equation defining a surface, not a point.

and (x1, y1, z1) was the location from which the wavefront was emitted at time t1.

.. the distance from the origin in Minkowski space-time to the spherical wave-front must be zero for all the points on the wave-front.


Correct, that's why any path taken by light is called a "null geodesic".

Thus you cannot have invariant space-time distances between the points on a wave-front.


That is wrong but it will be easier to explain why if you answer my three questions on the cocktail stick example.
Fleetfoot
1 / 5 (1) May 03, 2013
A change in the origin CANNOT change non-linear coordinates into linear ones.


The question doesn't arise, we are using a simple orthogonal cartesian coordinate scheme in each frame and the translation and rotation I applied in the 2D example are both linear operations.

If the coordinates allow unique distances it should also allow them when the wave-front is emitted from the origin.


The parametric equation (x2-x1)^2 + (y2-y1)^2 = R^2 defines the set of points (x2,y2) as a circle of radius R about centre (x1,y1), not a unique point. Setting s=0 is the parametric equation defining the light cones in SR. I would be surprised if you weren't aware of that method of defining a surface.
Fleetfoot
1 / 5 (1) May 03, 2013
It tires me out to try and explain simple mathematics and physics to you.


Part of the problem is that there is a disparity in our understanding of some terms. If you do the simple example, it will highlight that discrepancy and we can resolve it. It will make the conversation easier.
thefurlong
1 / 5 (9) May 03, 2013

A very good analogy

Thank you

If one chooses the two origins at the position of the light when it switches on, then it switches on simultaneously within both IRF's. (You are at the position of the gun being fired off within your example).
Since one can choose the origins at will, one can in principle choose it at any event so that this event occurs simultaneously at the coincident coordinates of the light within both IRF's; which must thus also happen even when you are not at the position of the light (or gun in your example).

I cannot stress enough that this is a major source of your error. We only necessarily agree on the time and position of an event when it happens at 0 distance away from both of us. It has nothing to do with where I have designated the origin any more than if I use feet instead of meters.
thefurlong
1 / 5 (8) May 03, 2013
If your origin is not coincident with the moving light when it switches on, you have to determine from the speed of light and the time it reaches you when and where it has been
switched on.

Only if the origin is not in the same place as me. If you and I designate the origin to be -3 feet away (hence, I am at position x = 3 feet), and the light goes off at x = 3 while I and I occupy the same position, you can, without conviction, say that I experienced it happen at the same time. Otherwise, you have to do measurements, as you said.
I think the error you are making is as follows. C sees A and B are at rest with each other and measures them to be a distance d away from each other. C and A share the same position at what they agree to be time t = t'. C then sees B jump at this time. If you were C, you would believe that A also thinks that B jumps at that time, but that just isn't an assumption you can make. That would only be necessarily true if d = 0 feet.
thefurlong
1 / 5 (8) May 03, 2013
Well, also A, B, and C would agree that B jumped at the same time if they were all moving at the same velocity.
johanfprins
1 / 5 (8) May 03, 2013
There seems to be some confusion here. The value of s is invariant (which is what we were discussing) but it doesn't identify a unique event. In fact s=0 defines the whole set of events which lie on the surface of the future and past light cones.


The value of s cannot be invariant since s can also be zero for coordinates that are NOT the origin.

The diagram for a light cone is done within a single inertial reference frame where the time is exactly the same at all points within this reference frame. It says NOTHING about different times within other inertial reference frames. Since the light cone construction can be done within any of all the possible inertial reference frames, you must have that time must be instantaneously the same within all inertial reference frames. Thus, clocks cannot run at different rates within different inertial reference frames. Thus the light cone violates your arguments.

I am now signing off for the rest of my day here in SA!
Whydening Gyre
1.4 / 5 (10) May 03, 2013
Well, also A, B, and C would agree that B jumped at the same time if they were all moving at the same velocity.


so, if a, b and c were all moving at different velocities when b jumped, a and c would see it at different times. what's so difficult about that? With time being the invariant, B still jumped. A and C must triangulate WHEN according to their positions which are a result of their velocities.
This is simple Euclidian triangulation that I don't understand why you guys are trying to make so difficult.
thefurlong
1.7 / 5 (12) May 03, 2013

so, if a, b and c were all moving at different velocities when b jumped, a and c would see it at different times. what's so difficult about that?

Ok, again, we must distinguish between when A, B, and C, witness an event, and when they measure it to occur. The process of measuring is were the fundamental disagreement in physical values is.

With time being the invariant, B still jumped. A and C must triangulate WHEN according to their positions which are a result of their velocities.

Yes, and how are they going to measure the times? The canonical way to do it is to use light clocks, which leads to time dilation and length contraction.

This is simple Euclidian triangulation that I don't understand why you guys are trying to make so difficult.

Understanding special relativity IS difficult. That's why so many people get it wrong.
thefurlong
1.6 / 5 (12) May 03, 2013
The value of s cannot be invariant since s can also be zero for coordinates that are NOT the origin.

The value of s IS invariant; it doesn't change. It's value is always 0 as long as the spacetime point lies in the light cone. Everyone can always agree that such points are always at s away from the given space-time coordinate, regardless of change of coordinate systems and IRF. What you mean to say is that the point that has space-time "distance" 0 to the origin is no longer unique. This distance is a spatial-temporal one. There is no law of physics or math saying that multiple points can't be 0 light-meters away from the origin.


The diagram for a light cone is done within a single inertial reference frame where the time is exactly the same at all points within this reference frame.

Sure, but if you do any lorentz transformation to switch IRF's, the space-time "distance" between two points will still be the same. Try it.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
Ok, again, we must distinguish between when A, B, and C, witness an event, and when they measure it to occur. The process of measuring is were the fundamental disagreement in physical values is.

That is BS doubletalk. You measure to WITNESS. Measuring/witnessing an exact(timewise) event allows prediction/forecasting of future events that include only what we are looking at - 3 clocks. PICK something to be your reference point and stick to it.
With time being the invariant, B still jumped. A and C can triangulate WHEN according to their positions which are a result of their velocities.

Yes, and how are they going to measure the times? The canonical way to do it is to use light clocks, which leads to time dilation and length contraction.

Still BS. The light clock is just a triangulation point on another triangulation.
From all our experimentation the one thing that EVERYTHING else visible is in sync with - is time.
Fleetfoot
3 / 5 (2) May 03, 2013
There seems to be some confusion here. The value of s is invariant (which is what we were discussing) but it doesn't identify a unique event. In fact s=0 defines the whole set of events which lie on the surface of the future and past light cones.


The value of s cannot be invariant since s can also be zero for coordinates that are NOT the origin.


You have a misunderstanding of the meaning of "invariant". If you answer my simple three question example, we can clear that up.

The diagram for a light cone is done within a single inertial reference frame where the time is exactly the same at all points within this reference frame. It says NOTHING about different times within other inertial reference frames.


The cones are defined in a single frame but cover a surface which includes different times but that is just wording, basically we agree. That's not where our disagreement lies.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
All three (or 4, if you prefer) triangulation points are in motion, otherwise we wouldn't be able to quantify the others. The only way possible to make the next conjecture is to "step outside of time". And once again, we are back to triangulation (guess we could call it quadrilation) in order to observe. Our brains are subject to this "triangulation methodology" until we (and our surrounding universe) evolve - and change...
Which brings us back to - how do you change change? by stopping it, of course.
Q-Star
2.6 / 5 (15) May 03, 2013
This is simple Euclidian triangulation that I don't understand why you guys are trying to make so difficult.


Because they aren't talking about Euclidean space. That would wrongly assume that space is absolute, and time is absolute.

Euclidean geometry doesn't work for spacetime. In Minkowskian spacetime, space is NOT absolute. time is NOT absolute, spacetime IS absolute. That is relativity reduced to a single sentence.
thefurlong
1.7 / 5 (12) May 03, 2013

That is BS doubletalk. You measure to WITNESS.

No. I can witness thunder seconds after the lightning, but I would be wrong to think that the thunder occurred after the lightning. In the context of special relativity, even when we correct for this time lag, we still disagree on when the lightning struck if our relative velocity is > 0.
With time being the invariant, B still jumped. A and C can triangulate WHEN according to their positions which are a result of their velocities.


A and C will experience time differently, not just their clocks. Why should they pick one time over another?

Still BS. The light clock is just a triangulation point on another triangulation.
From all our experimentation the one thing that EVERYTHING else visible is in sync with - is time.

No. The whole point of special relativity is to drop the assumption that everyone experiences time and space the same way.
thefurlong
1.4 / 5 (11) May 03, 2013
The only way possible to make the next conjecture is to "step outside of time".

You are assuming we can "step outside of time." Give me a physical mechanism for how to do that, and I will believe you.
And once again, we are back to triangulation (guess we could call it quadrilation) in order to observe. Our brains are subject to this "triangulation methodology" until we (and our surrounding universe) evolve - and change...
Which brings us back to - how do you change change? by stopping it, of course.

Well yes, we can treat the time we experience as a sequence of snap shots, but each of us will have a different measurement of the relationship between two points and two moments as the time interval goes to 0. If I see the points moving, but you see them stationary, I will always see their distance shrunken by the lorentz factor. Of course, I won't think they've shrunken.
Fleetfoot
1 / 5 (1) May 03, 2013
Euclidean geometry doesn't work for spacetime.


In SR, it works for any spatial foliation, i.e. a 3D slice perpendicular to any given time axis. Remember in the absence of gravity, there is no curvature.
Q-Star
2.2 / 5 (13) May 03, 2013
Euclidean geometry doesn't work for spacetime.


In SR, it works for any spatial foliation, i.e. a 3D slice perpendicular to any given time axis. Remember in the absence of gravity, there is no curvature.


Correct, as usual,,, but only within an inertial reference frame. If two observers are in separate reference frames, then Euclid fails, whether or not they are accelerating (or in the presense of a gravitational mass) or not.

We know that Minkowski spacetime was one of the "AH HAH" moments while Einstein was working on Special Relativity. Along with the other big "AH HAH" moment,, "c" is the velocity of all things through spacetime regardless of their relative velocities through space only.
Fleetfoot
1 / 5 (1) May 03, 2013
Euclidean geometry doesn't work for spacetime.


In SR, it works for any spatial foliation, i.e. a 3D slice perpendicular to any given time axis. Remember in the absence of gravity, there is no curvature.


Correct, as usual,,, but only within an inertial reference frame.


Well I did say "in SR" ;-)

If two observers are in separate reference frames, then Euclid fails, ..


It works for each but of course you cannot mix values from different frames.

We know that Minkowski spacetime was one of the "AH HAH" moments while Einstein was working on Special Relativity.


Wasn't that an Ah-hah moment for Minkowski ;-)

Along with the other big "AH HAH" moment,, "c" is the velocity of all things through spacetime regardless of their relative velocities through space only.


Be VERY careful with that one or you end up turning the "block universe" into a "moving spotlight" interpretation which leads to all sorts of philosophical confusion.
Q-Star
2.2 / 5 (13) May 03, 2013
We know that Minkowski spacetime was one of the "AH HAH" moments while Einstein was working on Special Relativity.


Wasn't that an Ah-hah moment for Minkowski ;-)


Not for Minkowski, to him Riemann was a god, he wasn't sure it was actually a real thing rather than a fun math,,, but he also had a very poor opinion of Einstein's math abilities.

Along with the other big "AH HAH" moment,, "c" is the velocity of all things through spacetime regardless of their relative velocities through space only.


Be VERY careful with that one or you end up turning the "block universe" into a "moving spotlight" interpretation which leads to all sorts of philosophical confusion.


Well as soon as that happens, I'm leaving the conversation!!!!! (Before I get philosophically confused.) Seriously, it was the "all things move at "c" in spacetime" that finally made it come together for me. And I've kept that approach ever since.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
The only way possible to make the next conjecture is to "step outside of time".

You are assuming we can "step outside of time." Give me a physical mechanism for how to do that, and I will believe you.

I made no assumption. I made only a statement of possibility (one of many, as you are aware) based on the realities science has accumulated. Ergo, you are word dancing.
As for showing you the mechanism, either;
A. I can't because I don't know it.
B. I won't because you wouldn't understand it and would use it to see what makes it tick, regardless of the unintended consequences.
C. It's not possible.
D. Hey - that's what scientists are for...:-)
Q-Star
2.2 / 5 (13) May 03, 2013
As for showing you the mechanism, either;


Which to choose?

A. I can't because I don't know it.


Then why suggest it?

B. I won't because you wouldn't understand it and would use it to see what makes it tick, regardless of the unintended consequences.


What?

C. It's not possible.


This is a science site, not a science fantasy site, so insert it to the discussion?

D. Hey - that's what scientists are for...:-)


Then why ya arguing with him then? Ya are telling him that his explanations are wrong, when in fact they are right on point and correct. If ya can't understand what he is saying, that doesn't mean it's wrong, it only means ya don't understand him.
Whydening Gyre
1.3 / 5 (12) May 03, 2013
Well as soon as that happens, I'm leaving the conversation!!!!! (Before I get philosophically confused.) Seriously, it was the "all things move at "c" in spacetime" that finally made it come together for me. And I've kept that approach ever since.

Well, then. it seems you've picked a single IRF for yourself...
Q-Star
2.3 / 5 (14) May 03, 2013
Well as soon as that happens, I'm leaving the conversation!!!!! (Before I get philosophically confused.) Seriously, it was the "all things move at "c" in spacetime" that finally made it come together for me. And I've kept that approach ever since.

Well, then. it seems you've picked a single IRF for yourself...


No,,, I've pick the only reference frame that contains any and all observers. Regardless of their separation in space or separation in time. In spacetime there is only a single reference frame. The reference frame where spacetime is absolute. It's not a thought experiment, it's the reality & universe we occupy.
Whydening Gyre
1.4 / 5 (10) May 03, 2013

Then why ya arguing with him then? Ya are telling him that his explanations are wrong, when in fact they are right on point and correct. If ya can't understand what he is saying, that doesn't mean it's wrong, it only means ya don't understand him.

Who's arguing?
Damn... now I'm pulled into the dance with TWO of you.
Q-Star
2.6 / 5 (15) May 03, 2013
Who's arguing?


Well, let's see,

Yeah, but... I am getting the nagging feeling you think a series of "snapshots" don't actually represent that velocity component...


He's correct, they only represent that velocity component for one observer. But space and time aren't absolute. That is the beauty of Einstein's relativity. It is absolute.

What do you mean by "unambiguous meaning"?
If you set 2 clocks (light clocks or Seiko's - I don't care) for the same time, send them in two different directions at different velocities keeping one on your desk for reference, won't they all at any given simultaneous moment still be set to the same time?


No they won't. It's been experimentally shown.

With time being the invariant


Time is not invariant, that is the point.

This is simple Euclidian triangulation that I don't understand why you guys are trying to make so difficult.


It is not Euclidean triangulation. It is Riemann - Minkowski geometry.
Whydening Gyre
1 / 5 (9) May 03, 2013
No,,, I've pick the only reference frame that contains any and all observers. Regardless of their separation in space or separation in time. In spacetime there is only a single reference frame. The reference frame where spacetime is absolute. It's not a thought experiment, it's the reality & universe we occupy.

Wait a minute - aren't scientists doing all these observations for the purpose of establishing postulates on further possibilities? What happens IF science discovers space or time or spacetime is not absolute?
Anyway, I don't think we live in an absolute universe. Almost, maybe...
And I think it was you, Q, who said the only possible open system was our universe. (Which I believe absolutely). But maybe that was Anti...
Q-Star
2.4 / 5 (14) May 03, 2013
Who's arguing?


Who?

That is BS doubletalk. You measure to WITNESS. Measuring/witnessing an exact(timewise) event allows prediction/forecasting of future events that include only what we are looking at - 3 clocks. PICK something to be your reference point and stick to it.


He is sticking with ONE reference point, it's just that ya don't understand it.

Yes, and how are they going to measure the times? The canonical way to do it is to use light clocks, which leads to time dilation and length contraction.

Still BS. The light clock is just a triangulation point on another triangulation.


Not BS, it's very well tested physics.

From all our experimentation the one thing that EVERYTHING else visible is in sync with - is time.


Then YOUR experimentation hasn't been enough. Time has been experimentally been shown to be relative, not absolute.
Q-Star
2.2 / 5 (13) May 03, 2013
No,,, I've pick the only reference frame that contains any and all observers. Regardless of their separation in space or separation in time. In spacetime there is only a single reference frame. The reference frame where spacetime is absolute. It's not a thought experiment, it's the reality & universe we occupy.

Wait a minute - aren't scientists doing all these observations for the purpose of establishing postulates on further possibilities? What happens IF science discovers space or time or spacetime is not absolute?
Anyway, I don't think we live in an absolute universe. Almost, maybe...
And I think it was you, Q, who said the only possible open system was our universe. (Which I believe absolutely). But maybe that was Anti...


No I said quite the opposite, ya need to pay more attention. I've said many times: "The only possible truly CLOSED system is the Universe entire."
Q-Star
2.2 / 5 (13) May 03, 2013
No,,, I've pick the only reference frame that contains any and all observers. Regardless of their separation in space or separation in time. In spacetime there is only a single reference frame. The reference frame where spacetime is absolute. It's not a thought experiment, it's the reality & universe we occupy.

Wait a minute - aren't scientists doing all these observations for the purpose of establishing postulates on further possibilities? What happens IF science discovers space or time or spacetime is not absolute?


If it passes the test(s) then I'll except the "new and improved" model of reality. So far Einstein's spacetime, SR & GR are the best tested, most all inclusive of observation, and most applicable to most uses as anything that ever was.

These things have withstood the most rigorous and conclusive testing that has ever been done in the entire history of science,,,, well except maybe QT & the Standard Model of particles
Fleetfoot
1 / 5 (1) May 03, 2013
Well as soon as that happens, I'm leaving the conversation!!!!! (Before I get philosophically confused.) Seriously, it was the "all things move at "c" in spacetime" that finally made it come together for me. And I've kept that approach ever since.

Well, then. it seems you've picked a single IRF for yourself...


No, he's talking about 4-velocity:

http://en.wikiped...velocity

It's magnitude is independent of the choice of frame.
Whydening Gyre
1.4 / 5 (10) May 03, 2013

Then YOUR experimentation hasn't been enough. Time has been experimentally been shown to be relative, not absolute.

to what degree of certainty? and relative to what? Space? Thusly making spacetime absolute?
Was spacetime here before our matter universe appeared? If not, what was the absolute BEFORE that? Membranes? Energy strings? What were they made of? so on and so on...
I don't deny science - it gives me answers to questions.
Don't deride those that question - they give you guys something to do...
Einstein said - imagination is more important than knowledge. You saying he was wrong?
Whydening Gyre
1.4 / 5 (11) May 03, 2013
If it passes the test(s) then I'll except the "new and improved" model of reality. So far Einstein's spacetime, SR & GR are the best tested, most all inclusive of observation, and most applicable to most uses as anything that ever was.

These things have withstood the most rigorous and conclusive testing that has ever been done in the entire history of science,,,, well except maybe QT & the Standard Model of particles

I accept that...:-)
thefurlong
1 / 5 (10) May 03, 2013

I made no assumption. I made only a statement of possibility (one of many, as you are aware) based on the realities science has accumulated. Ergo, you are word dancing.

You said
The only way possible to make the next conjecture is to "step outside of time".

It is reasonable to interpret that as you assuming. Well, if you aren't assuming it, you aren't actually arguing anything.
Q-Star
2.2 / 5 (13) May 03, 2013
to what degree of certainty? and relative to what? Space? Thusly making spacetime absolute?


To the degree of certainty our technology allows. As the technology has improved the level of certainty has only gone up.

Was spacetime here before our matter universe appeared?


Don't know. But THIS the universe that IS.

If not, what was the absolute BEFORE that? Membranes? Energy strings? What were they made of? so on and so on...


Fun to ponder, but it is still not subject to falsification. It's not science until ya can test it. Why waste time formulating unanswerable questions while there are plenty of questions that can be answered still begging our attention.

Einstein said - imagination is more important than knowledge. You saying he was wrong?


Yeah, I suppose I'm saying he was wrong, if he said that, not knowing the context. Like all great scientists, bar none, Einstein could be wrong, & was wrong about somethings.

thefurlong
1 / 5 (10) May 03, 2013

As for showing you the mechanism, either;
A. I can't because I don't know it.
B. I won't because you wouldn't understand it and would use it to see what makes it tick, regardless of the unintended consequences.
C. It's not possible.
D. Hey - that's what scientists are for...:-)

A. I won't argue with that.
B. In other words, "I don't know what I am talking about so I am going to save face by coyly suggesting the possibility I am a masterful genius secretly constructing his scientific opus, all while being mercilessly shunned by my peers in the scientific community."
C. If modern physics has anything to say about it, yes.
D. Well, I am a scientist, how about you? You don't actually need a science degree to do science (though acquiring one usually helps).
Whydening Gyre
1.4 / 5 (11) May 03, 2013

I made no assumption. I made only a statement of possibility (one of many, as you are aware) based on the realities science has accumulated. Ergo, you are word dancing.

You said
The only way possible to make the next conjecture is to "step outside of time".

It is reasonable to interpret that as you assuming. Well, if you aren't assuming it, you aren't actually arguing anything.

correct - I was postulating. Neither were you, were ya? Meaning - we have nothing to argue about - time for a cocktail :-)
thefurlong
1 / 5 (10) May 03, 2013

Then why ya arguing with him then? Ya are telling him that his explanations are wrong, when in fact they are right on point and correct. If ya can't understand what he is saying, that doesn't mean it's wrong, it only means ya don't understand him.

Yeah, what a total cop-out! It's like he/she said "My opinions are so strong that I must tell you how wrong you are, but please don't get me to defend them because I am not qualified to form those opinions."
thefurlong
1 / 5 (10) May 03, 2013

correct - I was postulating. Neither were you, were ya? Meaning - we have nothing to argue about - time for a cocktail :-)

Neither was I what? Making the claim everyone experiences time and space differently? I most certainly was, and I'll keep making it regardless of the fact that I don't yet have a PhD. That's the beauty of science. It doesn't care who you are.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
A. I won't argue with that.
B. In other words, "I don't know what I am talking about so I am going to save face by coyly suggesting the possibility I am a masterful genius secretly constructing his scientific opus, all while being mercilessly shunned by my peers in the scientific community."
C. If modern physics has anything to say about it, yes.
D. Well, I am a scientist, how about you? You don't actually need a science degree to do science (though acquiring one usually helps).

Not a scientist. Artist.
Let's find some common ground. Beauty in science, beauty in art. Both complement eachother, like stairs.
You guys have trained yourselves to see the next step, I'm trained to wonder about the step after that.
And, after all, don't we all come here as diversion from what we do to pay the rent?
Just as an aside, 25 years in computer field before I took on Art.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
Neither was I what?

Arguing.
Making the claim everyone experiences time and space differently? I most certainly was, and I'll keep making it regardless of the fact that I don't yet have a PhD.

I agree with that claim, we do.
I don't care if you have 3 PhD's or a GED. I come to this site to let my head roam. Why do you come here? Same reason, I suspect.

So... lighten up, Francis...:-)
thefurlong
1 / 5 (10) May 03, 2013

Not a scientist. Artist.
Let's find some common ground. Beauty in science, beauty in art. Both complement eachother, like stairs.
You guys have trained yourselves to see the next step, I'm trained to wonder about the step after that.
And, after all, don't we all come here as diversion from what we do to pay the rent?
Just as an aside, 25 years in computer field before I took on Art.

Telling someone that his explanation is BS is arguing.
I consider myself both an artist (I'd to think I'm pretty good at drawing) and a scientist, and I currently work as a software engineer. I guess I just don't see the endeavors I am qualified for as having any boundaries.
thefurlong
1 / 5 (10) May 03, 2013
Neither was I what?

Arguing.
Making the claim everyone experiences time and space differently? I most certainly was, and I'll keep making it regardless of the fact that I don't yet have a PhD.

I agree with that claim, we do.
I don't care if you have 3 PhD's or a GED. I come to this site to let my head roam. Why do you come here? Same reason, I suspect.

It would be easier to do that if you didn't spend the last few comments telling me that my arguments were BS, then acting as if you weren't contesting what I was saying once I called you out on it. That may not have been your intention, but it certainly seems like it was.
Whydening Gyre
1.4 / 5 (10) May 03, 2013
Neither was I what?

Arguing.
Making the claim everyone experiences time and space differently? I most certainly was, and I'll keep making it regardless of the fact that I don't yet have a PhD.

I agree with that claim, we do.
I don't care if you have 3 PhD's or a GED. I come to this site to let my head roam. Why do you come here? Same reason, I suspect.

It would be easier to do that if you didn't spend the last few comments telling me that my arguments were BS, then acting as if you weren't contesting what I was saying once I called you out on it. That may not have been your intention, but it certainly seems like it was.

Which all kinda wraps back around to the concept of IRF...:-)
Fleetfoot
1 / 5 (1) May 04, 2013
Time has been experimentally been shown to be relative, not absolute.


to what degree of certainty? and relative to what? Space? Thusly making spacetime absolute?


Think of spacetime as a flat salt lake. Your future is everything generally north of your present location, to be exact everything between north-west and north-east.

Your past is everything generally south of your present location, everything between south-west and south-east.

The regions to the east and west of you are "elsewhere" and can be "now" depending on your motion.

In this analogy, inertial motion means the track you leave on the salt will be a straight line while acceleration makes it curved. A clock you carry with you works like an odometer, your personal ("proper") time is a measure of the length of the track.

Einstein said - imagination is more important than knowledge. You saying he was wrong?


It is important when solving new problems. Once the solution is found, we move on.
Fleetfoot
1 / 5 (1) May 04, 2013
(Contd.)

Time has been experimentally been shown to be relative, not absolute.


... and relative to what? Space?


A clock you carry with you works like an odometer, your personal ("proper") time is a measure of the length of the track.


You need a method to measure the time between events that do not lie on your track which in general may be curved. First at the point you are considering "now", draw the tangent to your track, that is the "time axis" of your coordinate system. Then draw a line tpassing through the remote point in question and perpendicular to that time axis. Where the perpendicular crosses the axis defines the "time at which the event happened".

You ask "relative to what". You should now see that if you accelerate (a curve in your track) then the tangent to the track will be turned to a new direction. Projecting the remote event onto that new line via a perpendicular will cross it at a different point, hence give a new time coordinate.
Fleetfoot
1 / 5 (1) May 04, 2013
We know that Minkowski spacetime was one of the "AH HAH" moments while Einstein was working on Special Relativity.


Wasn't that an Ah-hah moment for Minkowski ;-)


Not for Minkowski, to him Riemann was a god, he wasn't sure it was actually a real thing rather than a fun math,,, but he also had a very poor opinion of Einstein's math abilities.


I just had a look at the biography of Minkowski on Wikipedia:

"Minkowski is perhaps best known for his work in relativity, in which he showed in 1907 that his former student Albert Einstein's special theory of relativity (1905), presented algebraically by Einstein, could also be understood geometrically as a theory of four-dimensional space-time. Einstein himself at first viewed Minkowski's treatment as a mere mathematical trick, before eventually realizing that a geometrical view of space-time would be necessary in order to complete his own later work in general relativity (1915)."
johanfprins
1 / 5 (8) May 04, 2013
This is how the AWT models the clock slowing stuff (http://www.aether...ns.htm). The clocks (laser resonators) are modeled with laser resonators, where the light is bouncing back and forth. It's evident, the speed of clock is not just delayed -


Light within a laser resonator is a STATIONARY wave as can be easily obtained from Maxwell's equations. That is why the laser cavity has to have a certain size and geometry. As usual your insight into physics is infantile; just like your AWT hallucinations of ducks paddling in the aether!
johanfprins
1 / 5 (4) May 04, 2013
There seems to be some confusion here. The value of s is invariant (which is what we were discussing) but it doesn't identify a unique event. In fact s=0 defines the whole set of events which lie on the surface of the future and past light cones.


Whoa!! If your coordinates are (x,y,z,t), what do they signify unless one can describe them as an instantaneous "event" occurring at time t at position x,y,z? Two sets of coordinates (x1,y1,z1,t1) and (X2,y2,z2,t2) can only be interpreted as two "events" occurring at two different positions (x1,y1.z1) and (x2,y2,z2) at two different times t1 and t2. Or do you have a different term that does not need the term "event"? I would like to hear if you have this, since this will give a different direction to this discussion!

You people have posted so much nonsense since I last visited this thread that it will take a lot of time to rectify it. I also have other issues to attend to, I will return when I can to rectify your misconceptions.
ValeriaT
1 / 5 (7) May 04, 2013
@JohanPrins: You still didn't answered the question for me, what will happen with local time of photon, if the time of body in motion is delayed instead of slowed in your numerology. How much it will remain delayed?
Light within a laser resonator is a STATIONARY wave
How it can be stationary, when the laser is in motion?
johanfprins
1 / 5 (6) May 04, 2013
@JohanPrins: You still didn't answered the question for me, what will happen with local time of photon, if the time of body in motion is delayed instead of slowed in your numerology. How much it will remain delayed?


There is no local time for a photon since a photon cannot be stationary within an inertial reference frame. And no clock can move with the speed c.

Light within a laser resonator is a STATIONARY wave


How it can be stationary, when the laser is in motion?


Relative to the moving cavity the light wave is stationary: In addition the cavity is stationary within the IRF moving along with it! All motion IS RELATIVE! In fact ALL bodies which we observe moving with any velocity with magnitude less than light speed are not really moving since they are ALL stationary within their own inertial reference frames.

There are more important posts on this thread that I need to respond to: So please stop posting your usual nonsense.
ValeriaT
1 / 5 (5) May 04, 2013
You can replace the photon with electron moving with 0.9999c, it doesn't change the principle. You're claiming, the time at this electron will be delayed with respect to stationary observer instead slowed/dilated according to standard interpretation of special relativity. Well, let say it's 10:00 A.M. and I will shot the electron with speed 0.999999c. How much such delay of local time of electron will be at the moment of start?
So please stop posting your usual nonsense.
I see, famous crackpot whining and writing ignored books about arrogance of mainstream science doesn't behave any better... ;-) You're just trying to cover the fact, that your theory is logical nonsense - exactly in the way, which the mainstream physicists do behave with respect to you. I'd say, you deserve their hostility in full extent.
johanfprins
1 / 5 (5) May 04, 2013
If the event occurs at both origins, then yes, but you can't assume that it occurs at the same time for points at other places. We can see this directly from the Lorentz Transformation. t' = gamma*(t- vx/c^2), where gamma's the lorentz factor.


I so wish that you will read section 3 of http://www.cathod...tion.pdf before shooting your mouth off. In that section I consider three synchronized clocks spaced a distance L/ apart passing by an observer with his own clock. When the middle clock coincides with the observer's clock, simultaneous events occur at all three passing clocks. Thus the event at the middle clock occurs at the coincident position and coincident time t=t/=0. According to the observer, the simultaneous event on the clock which has already passed him occurs at a distance L=(gam)*L/ from him at the time t=(v/c^2)*(gam)*L/. Which IS NOT your time dilation formula.
johanfprins
1 / 5 (5) May 04, 2013
According to the event at the clock approaching him also occurs a distance L=(gam)*L/ away from him but at a time t=MINUS(v/c^2)*(gam)*L. Thus according to the observer this event has already occurred before the event occurs which is coincident with the observer.

According to the interpretation of time dilation, the clock approaching him must keep time faster within the moving IRF, while the clock that moves away from him must keep time slower within the moving IRF. But if this were to be the case, the three events cannot EVER be simultaneous within the IRF within which the three clocks are stationary: Which we know is absurd since the derivation was started by assuming that the three events do occur simultaneously within the reference frame within which the three clocks are stationary.

Thus the only logical interpretation is that according to the observer and his clock, both the distances and times are different when the events do not coincide with him.
ValeriaT
1 / 5 (6) May 04, 2013
According to the interpretation of time dilation, the clock approaching him must keep time faster within the moving IRF, while the clock that moves away from him must keep time slower within the moving IRF
This is complete and utter nonsense. The time dilatation in special relativity is proportional only to the relative speed of object with respect to its observer (and the time, which resides in its motion). The direction of motion has no role in it. For example the time dilatation of twin traveling into distant cosmic space will not reverse, when this twin will return back to Earth. How the hell did you come into it? You should visit the school again.
johanfprins
1 / 5 (5) May 04, 2013
Now consider three observers with three synchonized clocks so that each observer coincides with one of the passing clocks when the events occur simultaneously at then positions of the passing clocks at t=t/=0.

The event at each clock occurs instantaneously simultaneously relative to the observer with which it coincides. But each observer will observe the events at the clocks which do not coincide with him at different distances and times on his clock. However after the events have occurred, the three observers can compare their clocks and all three will have to agree that there readings demand that all three events also occurred simultaneously at the coincident positions of the clocks within the IRF of the observers.

According to the readings on the clocks, all three events must have occurred instantaneously-simultaneously at the coincident positions of the three clocks within both inertial reference frames.
johanfprins
1 / 5 (5) May 04, 2013
They will have to conclude that the only reason why each of them could not see that these events did actually occur simultaneous at coincident positions within their inertial reference frame, must be since a single person cannot be simultaneously present at three different positions: This has thus NOTHING to do with clocks keeping different time-rates.

But if three different persons with synchronized clocks are present at these coincident positions, they can afterwards verify that the events also occurred simultaneously within their own IRF by comparing the readings on their clocks!
johanfprins
1 / 5 (6) May 04, 2013
Physics doesn't care which origins you use. You and I are only guaranteed to agree when and where the event happens when you, I, and it all occupy the same position simultaneously. We only use the origin because it allows for simpler calculations. We can use non-zero positions if we want to.


As I have just proved above that, in the case of Special Relativity, the choice of origin does matter very much indeed. It is exactly for this reason why Minkowki's space time is not physically tenable: In fact MST is nothing else than absurd nonsense. As a mathematics professor MInkowski should have known better. No wonder Einstein bunked his lectures. It is just a pity that Einstein later relented and incorporated MST into his theory of gravity.

Please read

http://www.cathod...tion.pdf

before keeping on making an idiot of yourself!
ValeriaT
1 / 5 (5) May 04, 2013
Look, you can always feel like the winner, if you will correct your nonsensical ideas about special relativity. Because you don't understand, what the special relativity actually predicts, you cannot never recognize, how your extrapolations differs from standard special relativity (if at all). Your babbling about abstract clocks therefore has no meaning to analyze here. Just at the moment, when you explicitly say, what the standard relativity predicts I can correct you, because I do know well, what this theory predicts and what it doesn't not. The rest of your posts is confused twaddling of senile oldie, who just decided to reinvent the wheel in his old age.
ValeriaT
1 / 5 (5) May 04, 2013
In fact MST is nothing else than absurd nonsense
It leads to the predictions, which are routinely verified with different speed of clocks inside of gravity wells (Pound-Repka experiments and later). The sensitivity of contemporary atomic clock enables to validate the predictions of space-time concept at few decimeter scale already. The concept of space-time has a very good meaning even in dense aether model, as it represents the density gradient of space-time, where we are living in. If you get toward massive body, the vacuum is more dense, the energy is spreading more slowly there and the clock exhibit the time dilation. Therefore I've absolutely no problem with the Minkovski interpretation of space-time. It's one of the best achievements of theoretical physics of the last century, despite it has its limits at wider space-time scale, where the higher dimensions of space-time manifest itself. But this is well outside of your level of "flat-land" comprehension.
johanfprins
1 / 5 (4) May 04, 2013
@ ValeriaT: As usual you are too stupid to understand physics and mathematics and to follow logic. Furthermore you refuse to read anything that could prove how demented you really are. I just wish that you will realize that your mental development is only that of a baby who plays with plastic ducks while sitting in a foam bath.
ValeriaT
1 / 5 (5) May 04, 2013
Furthermore you refuse to read anything that could prove how demented you really are.
At the moment, when you for example say that special relativity predicts the time dilatation dependent on the direction of motion, then I'm not required to read you anymore, because it's simply not true. I'm not required to become a broody hen laying eggs for being able to recognize an aged egg. When this egg smells roted at one apparent place of it, then it cannot remain healthy in every piece of (sh)it - and I'm not required to bother with technology of laying eggs anymore.
are too stupid to understand physics and mathematics and to follow logic
When one assumption in the chain of logical deductions is wrong, the the whole rest of deductions is probably wrong (until you indeed make another mistake, which would compensate it - which I don't really expect in your case). Can you understand it? Probably not - but this is again not my problem.
johanfprins
1 / 5 (5) May 04, 2013
When one assumption in the chain of logical deductions is wrong, the the whole rest of deductions is probably wrong (until you indeed make another mistake, which would compensate it - which I don't really expect in your case).


You have proved over-and-over that you are far too stupid to judge any assumption and also far to stupid to reason step-by-step logic.

Can you understand it? Probably not - but this is again not my problem.


Then why do you not stay out of it? Your biggest problem is to get a brain implant since there is no grey matter whatsoever between your dirty ears!
johanfprins
1 / 5 (4) May 04, 2013
I only now noted this gem of stupidity:
At the moment, when you for example say that special relativity predicts the time dilatation dependent on the direction of motion, then I'm not required to read you anymore, because it's simply not true.


Firstly I did NOT claim any time dilation since there is NONE. Secondly the times I quoted I have derived correctly from the Lorentz-transformation. So if they are wrong, so must be the Lorentz transformation. So if what can be impeccably derived from the Lorentz trasformation " is not true" what transformation equations must I use? Oh, sorry I forgot you do not believe in mathematics at all! Probably because you are too stupid to understand it!

ValeriaT
1 / 5 (5) May 04, 2013
I did NOT claim any time dilation since there is NONE
I said, you wrote it about "interpretation of time dilation (by official relativity)". Sorry, what the special relativity officially predicts is not defined with you, but with textbooks and you simply cannot change it. You can propose alternative theory instead.
I have derived correctly from the Lorentz-transformation
In this animation the accelerated object not only runs time slower, but it even runs it slowly in just the way, which provides that the speed of light within of object's reference frame remains constant (actually invariant) - as special relativity requires. If the time wouldn't slow down for muon in its own reference frame, then this requirement couldn't be maintained anymore. As you can see, the time dilatation for moving objects is therefore the direct consequence of special relativity. You therefore cannot derive something different.
ValeriaT
1 / 5 (5) May 04, 2013
I forgot you do not believe in mathematics at all
I do believe in mathematics of physical theories only when it doesn't violate its own postulates and axioms. In this particular case it's just your derivation, which violates such an logics - not Einstein's one - because from the logics of the above animation clearly follows, that the time dilatation for object in motion must occur, or the speed of light couldn't remain invariant anymore (which is the postulate of Einstein's theory). If your theory says otherwise, then sorry - it MUST be wrong, no matter how exactly did you actually derive it.

BTW You openly have contempt for my illustrative analogies and drawings - but if you would draw such an animation for yourself, you would clearly realize, that the result of Lorentz transform cannot be any different - no matter how large pile of math you can write about it.
johanfprins
1 / 5 (4) May 04, 2013
I did NOT claim any time dilation since there is NONE
I said, you wrote it about "interpretation of time dilation (by


I have NOT claimed that I am defining what it means. Its meaning is determined by the Lorentz transformation and my results are mathematically-correctly deduced from these equations.

but with textbooks and you simply cannot change it.


You thus believe that a textbook is "God's own Word" and cannot be wrong? Another proof of what a bigot you are!

And please stop posting your AWT animations: They are just plain nonsensical claptrap!
johanfprins
1 / 5 (6) May 04, 2013
I do believe in mathematics of physical theories only when it doesn't violate its own postulates and axioms.
This is exactly what the mainstream interpretation of "time-dilation" does: It violates Einstein's first postulate, and you believe it even though it violates Einstein's postulate: So you ARE lying again.

BTW You openly have contempt for my illustrative analogies and drawings - but if you would draw such an animation for yourself, you would clearly realize, that the result of Lorentz transform cannot be any different - no matter how large pile of math you can write about it.


I do look at your "illustrative analogies and drawings" and find that they do not do what you claim that they do. I am not a bigot like you who refuses to look at other people's ideas and derivations only because you do not want these ideas and derivations to exist. If your cartoons were what you claim they are, I would have acknowledged it. You on the other hand has a closed-mind!
ValeriaT
1 / 5 (4) May 04, 2013
I did NOT claim any time dilation since there is NONE
LOL, you're claiming the "time delay" instead. But how this time delay could be formed, if not with time dilatation? How we could distinguish it experimentally from time dilatation after then, you troll? You're playing the same semantic game here, like when you're saying, that the particle concept in quantum mechanics doesn't exist - only "boundary condition" for stationary waves. You're just replacing one well established world connection with another one without any experimentally testable ways, how to distinguish it - and this is whole the "science" of yours of the last twenty years. Do you really believe, I didn't realize it?
ValeriaT
1 / 5 (6) May 04, 2013
I do look at your "illustrative analogies and drawings" and find that they do not do what you claim that they do
This cartoon illustrates the accelerated object, for which its local time runs more slowly with increasing speed for to maintain the speed of light invariant for any observer of it. What do you think this cartoon illustrates for you?
They are just a plain nonsensical claptrap!
It would be too easy for you...:-) You should prove every claim of yours or I'm simply not obliged to believe you at all. Do you think, that the Pythagoras theorem is BS, just because my dog cannot understand it? Now you're just barking and playing such a dull dog here.
johanfprins
1 / 5 (5) May 04, 2013
I did NOT claim any time dilation since there is NONE
LOL, you're claiming the "time delay" instead. But how this time delay could be formed, if not with time dilatation?


It is derived from the Lorentz-transformation in my manuscript http://www.cathod...tion.pdf that this is so. That the time differences are NOT clocks running at different rates but differences in time on ALL the clocks that run at the same rate. If you could have done mathematics you would have been able to read it for yourself. Except that you are too much of a bigot and a troll to be interested in work which does not fit your infantile preconceived ideas.

Please go and ply with your plastic ducks in your bubble bath: I have better things to do than to argue with an infantile idiot like you!

thefurlong
1.8 / 5 (10) May 04, 2013
As I have just proved above that, in the case of Special Relativity, the choice of origin does matter very much indeed.

Either you are crazy, or your definition of origin does not agree with the mathematical and physical one. The origin is a point that is selected to be 0 in all coordinates, that's all. A person selects it. If we wanted to, we could frame physics in a completely origin-less way. Indeed, physics can be done without actually choosing a specific coordinate system. It's called Lagrangian mechanics.

Please read

http://www.cathod...tion.pdf

before keeping on making an idiot of yourself!

I did, and it makes no sense. Don't you remember me asking you about it? You make all kinds of unfounded assumptions, for example, your assumption that when origins coincide, then an event that happens ANYWHERE in either space, must happen at the SAME TIME for both IRFs. I completely disagree with that assumption, and I challenge you to defend it.
thefurlong
1.4 / 5 (9) May 04, 2013

I so wish that you will read section 3 of http://www.cathod...tion.pdf before shooting your mouth off.

I have no interest in reading the rest of your paper, because the part I read makes no sense to me. Your assumptions, on their face, are wrong, and you refuse to address them. Why should I waste my time reading the rest of a paper seemingly founded on nonsense, which nobody else but you considers to be correct?
thefurlong
1.4 / 5 (9) May 04, 2013

In that section I consider three synchronized clocks spaced a distance L/ apart passing by an observer with his own clock. When the middle clock coincides with the observer's clock, simultaneous events occur at all three passing clocks.

Not correct, because you can't assume that all three clocks experience the event at the same time. Sure, if you are stationary, and you see three moving clocks, and lightning strikes the first one, then while it might look to you as if the other two clocks measure the lightning as striking the middle clock at the same time, it's a whole other story for the clocks themselves. There is absolutely no reason to assume that the other two clocks will agree on when the lightning struck. They won't even hear the resulting thunder at the same time! Surely you will at least agree with that!
Now, I'm sorry, but I refuse to read the rest of your paper because your assumptions are just wrong. Convince me they aren't and I will read the rest of it.
ValeriaT
1 / 5 (6) May 04, 2013
If you could have done mathematics you would have been able to read it for yourself.
It's like to say the Galileo: "Look, you dumbass - your animation of Venus phases is nonsensical - and I've math of epicycles, which you could verify, if you would be able to understand it". It just makes me wonder, why the people like you can never learn from history and why they repeat the very same mistakes again and again. The predicate logics is superior to math, because whole the math is based on predicate logics - not vice-versa.
But Your problem rather is, that your derivations are incomprehensible for laymen and formally thinking physicists already have robust model, which they trust and they will not bother with crackpots like you. The fact, whole your model is fringe from perspective of Einsteinian physics (and dense aether model as well) is your least problem from this perspective. Nobody will actually argue with you, because nobody will be actually interested about it - that's it.
thefurlong
1.4 / 5 (9) May 04, 2013
@johanfprins

To further drive home my point, you aren't even correct in assuming that just because you, the stationary observer sees the three clocks ticking simultaneously, that they must see each other ticking simultaneously. For one thing, how are you going to synchronize the clocks? Even if they start out synchronized and at rest, accelerating them might put them out of synch. You can't synchronize them while you and they are moving because there's no way to guarantee that the stationary observer agrees that you've synchronized them. All either of you can agree on is that if you pass each other while both your clocks show the same time, then you will both agree that you passed each other at that time. This simple assumption is the seed from which the rest of SR grows. It's difficult to wrap your head around, but that's what makes special relativity prone to seeming paradoxes, and why, to this day, people like you claim that it must be incorrect.
johanfprins
1 / 5 (5) May 04, 2013
Either you are crazy, or your definition of origin does not agree with the mathematical and physical one.


I chose different origins in exactly the same way as it is ALWAYS chosen when doing the Special Theory of Relativity; and then I find that three simultaneous events within the moving IRF give results that are DIFFERENT for each choice of origin.

The origin is a point that is selected to be 0 in all coordinates, that's all. A person selects it.


Provided you can ignore the fact that the speed of light is the same within all IRF's. When it is not, the choice of origin changes the position and time at which an event occurs. That is what the Lorentz transformation gives and that is what I accept is the case unless you can prove to me that the Lorentz transformation is wrong.

If we wanted to, we could frame physics in a completely origin-less way.
Bullshitt!! Are you sure you are not ValeriaT AKAK with yet another name? Yes I
johanfprins
1 / 5 (5) May 04, 2013
Indeed, physics can be done without actually choosing a specific coordinate system. It's called Lagrangian mechanics.


You must be ValeriaT since you obviously do not have a clue what you are talking about. When you solve the Lagrange differential equations you get integration constants and these demand that you must choose an origin.

I did, and it makes no sense.


Cannot follow logic and mathematics ValeriaT?

You make all kinds of unfounded assumptions, for example, your assumption that when origins coincide, then an event that happens ANYWHERE in either space, must happen at the SAME TIME for both IRFs. I completely disagree with that assumption, and I challenge you to defend it.


I have just done so above and it is done in more detail in my manuscript. I cannot help it that you are too stupid to follow simple logic and mathematics. It is clear that you are just an incompetent buffoon!

johanfprins
1 / 5 (6) May 04, 2013
I have no interest in reading the rest of your paper, because the part I read makes no sense to me.


Obviously it will not make sense to a buffoon who cannot understand simple physics and mathematics.

Your assumptions, on their face, are wrong, and you refuse to address them.
I did not make ANY assumptions: All I did was to use the Lorentz transformation to derive results. Thus the ONLY assumption that I made is to assume that the Lorentz transformation is correct and that I can use a light clock to derive the different times.

Why should I waste my time reading the rest of a paper seemingly founded on nonsense, which nobody else but you considers to be correct?


I did not force you to read the manuscript, but only pointed out that only a closed-minded bigot will attack results without first studying the manuscript. I have NEVER acted like this because I am a world-renowned physicist with integrity! There are many of my colleagues who agree that I am correct.
johanfprins
1 / 5 (5) May 04, 2013
say the Galileo: "Look, you dumbass - your animation of Venus phases is nonsensical - and I've math of epicycles,


Please do not be so arrogant to compare your pathetic animations with Galoleo's models! Galileo was a genius with brains. You are a moron without ANY brains!

The predicate logics is superior to math, because whole the math is based on predicate logics - not vice-versa


You should not even mention logic since you do not even understand what the word means.

Your problem rather is, that your derivations are incomprehensible for laymen


Only for laymen who did not have mathematics in high school. Here in South Africa we have laymen who can follow my derivations perfectly. So you are again making false assumptions to suit your irrationality!

Nobody will actually argue with you, because nobody will be actually interested about it - that't


So why do you keep on coming back under various false disguises?
thefurlong
1.4 / 5 (9) May 04, 2013
I chose different origins in exactly the same way as it is ALWAYS chosen when doing the Special Theory of Relativity; and then I find that three simultaneous events within the moving IRF give results that are DIFFERENT for each choice of origin.

Yes, I know you find them, but you are wrong because your assumption that everybody agrees that they are simultaneous is wrong. Can you read English? I disagree with your fundamental assumption that when two origins coincide, every single event at every single point witnessed by both parties coincide.
The origin is a point that is selected to be 0 in all coordinates, that's all. A person selects it.

Provided you can ignore the fact that the speed of light is the same within all IRF's.

Uh, no... I can do relativity just fine if I assume that A is at x = 3 and sees B moving at x = 3.
johanfprins
1 / 5 (6) May 04, 2013
@johanfprins

To further drive home my point, you aren't even correct in assuming that just because you, the stationary observer sees the three clocks ticking simultaneously, that they must see each other ticking simultaneously.


That argument you have to take up with Einstein since Einstein accepted that events can occur simultaneously at different positions within an IRF. This will not be possible if the clocks at these positions do not keep the same time.

For one thing, how are you going to synchronize the clocks?


That is easy! They are all stationary within the SAME IRF and you can measure the distances between them with a long tape. You can then sen out a light pulse from one of them to the others: Since the observers at these clocks know haw far they are from your clock, and since they know the speed of ligh, it is easy tp synchronise the clocks!

accelerating them might put them out of synch.


Where in my derivation did I accelerate the clocks?
thefurlong
1.4 / 5 (9) May 04, 2013
That is what the Lorentz transformation gives and that is what I accept is the case unless you can prove to me that the Lorentz transformation is wrong.

It doesn't matter if you fundamentally misunderstand how to use it. The Lorenz transformation is derived assuming that the origins coincide at the moment A and B pass. It's just a standard. If you choose A to be at x = 3, when B passes her, then your Lorentz transformation changes. It doesn't mean that physics fundamentally changes.
The Lorentz transformation is a consequence of assuming the first postulate of relativity, not the other way around, and it is derived by first deriving time dilation and length contraction, not the other way around. I can easily show you how to derive it from those principles, my boy. Don't make the mistake of assuming that just because most people disagree with you that everyone else is an idiot. Perhaps you should consider the alternative.
johanfprins
1 / 5 (6) May 04, 2013
no way to guarantee that the stationary observer agrees that you've synchronized them.


Why must he a free with that? Are you REALLY so incredibly stupid?

All either of you can agree on is that if you pass each other while both your clocks show the same time, then you will both agree that you passed each other at that time


I have not used anything else in my derivation except that the observer within the moving IRF has synchronized his clocks since he is stationary relative to these clocks.

It's difficult to wrap your head around,


No it is not difficult when you derive your results from the Lorentz transformation as I have done.

but that's what makes special relativity prone to seeming paradoxes,
The only "parodoxes" are incorrect derivations from the Lorentz transformation. When you do the derivations correctly thre are no paradoxes.

to this day, people like you claim that it must be incorrect
I have not claimed that STR is incorrect!
thefurlong
1.8 / 5 (10) May 04, 2013
You must be ValeriaT since you obviously do not have a clue what you are talking about. When you solve the Lagrange differential equations you get integration constants and these demand that you must choose an origin.

LOL! Go find my exchange with ValeriaT in the comments of http://phys.org/n...um.html. According to you, I like making unsubstantiated arguments so that I can prove myself wrong with 10 minutes of internet research. For heaven's sake! We are still on the internet. You can easily verify that two people are likely the same person by visiting their profiles and reading their comments!
johanfprins
1 / 5 (6) May 04, 2013
That is what the Lorentz transformation gives and that is what I accept is the case unless you can prove to me that the Lorentz transformation is wrong.

It doesn't matter if you fundamentally misunderstand how to use it.


That is impossible to misunderstand "how to use it". Al;l you do is plug in the position and time coordinates of an event within the moving IRF and calculating the corresponding position and time coordinates within the IRF relative to which it is moving. My goldfish can do this, even though YOU are too stupid to follow it when I do this in my manuscript.

The Lorenz transformation is derived assuming that the origins coincide at the moment A and B pass.
My manuscript is NOT about the derivation of the Lorentz transformation but based on the assumption that the Lorentz transformation is correct and then deriving the results by plugging in the correct coordinates and calculating the transformed ones: That is all!
thefurlong
1.8 / 5 (10) May 04, 2013

That argument you have to take up with Einstein since Einstein accepted that events can occur simultaneously at different positions within an IRF. This will not be possible if the clocks at these positions do not keep the same time.

Well, sure, in a single IRF, events can occur simultaneously! The context is when two events can occur simultaneously in two different IRF's.
johanfprins
1 / 5 (8) May 04, 2013
The Lorentz transformation is a consequence of assuming the first postulate of relativity, not the other way around,


Where have I stated otherwise. For God's sake ValeriaT AKAK stop flaunting your lack of brains!

and it is derived by first deriving time dilation and length contraction,


No this is not what Einstein did! Stop lying!

I can easily show you how to derive it from those principles, my boy.


Obviously since this is the way in which Lorentz derived them "my boy" before it was realized that you must derive them from Einstein's postulates. The need for time-dilation and length contraction thus fell away. It was stupid of Einstein to then again derive these wrong, and unneeded, concepts from the Lorentz transformation. They do not occur when light speed is the same within all IRF's.

Don't make the mistake of assuming that just because most people disagree with you that everyone else is an idiot.


Not everyone else: ONLY YOU!
Fleetfoot
3.7 / 5 (3) May 04, 2013
The value of s is invariant (which is what we were discussing) but it doesn't identify a unique event. In fact s=0 defines the whole set of events which lie on the surface of the future and past light cones.


Whoa!! If your coordinates are (x,y,z,t), what do they signify unless one can describe them as an instantaneous "event" occurring at time t at position x,y,z? Two sets of coordinates (x1,y1,z1,t1) and (X2,y2,z2,t2) can only be interpreted as two "events" occurring at two different positions (x1,y1.z1) and (x2,y2,z2) at two different times t1 and t2. Or do you have a different term that does not need the term "event"? I would like to hear if you have this, since this will give a different direction to this discussion!


That's exactly what I mean too, there's no problem there.

However, while a coordinate pair (x,y) defines a single point on a flat sheet of paper, the equation:

x^2 + y^2 = R^2

defines a set of points which form a circle.
thefurlong
1.4 / 5 (9) May 04, 2013
@johanfprins
All right, think about this. Assume Fred spaces 2 clocks, clock 1 and 2, out by a light year and synchronizes them. Imagine that clock 1 can send a signal to clock 2, the moment it is set, so that clock 2 can synchronize itself. So Fred sets clock 1 to 0 s. This sends a signal traveling at the speed of light, c, to clock 2. Eventually, clock 2 gets the signal and sets its time to 0 s + ((1 light year)/c). You'll agree that now, Fred will always see that both clocks are synchronized. According to Fred, every time clock 1 ticks, clock 2 ticks, and they always read the same thing.
Now, during this process, Barney, who is moving at high speed with respect to Fred, observes his buddy doing this. Barney has a watch that reads 0s when he passes Fred the moment he sets clock 1. Our task, starting from as few assumptions as possible, but at least including the postulates of relativity, is to describe what Barney sees. So, let's try to figure it out...
ValeriaT
1.5 / 5 (8) May 04, 2013
t is derived by first deriving time dilation and length contraction, No this is not what Einstein did! Stop lying!
This is how Einstein derived the time dilation and length contraction. Who is the desperate liar here? Anyway, your comments here are just semantic babbling, because the time dilation manifest itself just with delay of clock at the reference frame of moving objects.
ValeriaT
1 / 5 (6) May 04, 2013
BTW The Lorentz transform were derived long before Einstein on background of Maxwell's theory, based on transverse wave model of aether. Early approximations of the transformation were published by Voigt (1887) and Lorentz (1895). They were completed by Larmor (1897, 1900) and Lorentz (1899, 1904) and were brought into their modern form by Poincaré (1905), who gave the transformation the name of Lorentz. The constant speed of light follows from Maxwell's equations in rather straightforward way, because the invariant speed is native for all transverse waves in any particle environment, which is observed with its own ripples. As such it can serve as a justification/confirmation of dense aether model in which most of energy is propagating just in the form of transverse waves. The spreading of longitudinal waves through vacuum is constrained to rather exotic situations.
thefurlong
1.4 / 5 (9) May 04, 2013
@johanfprins
[continued]
Now, every sane person will agree that when Barney passes Fred, both of them agree that it happened at 0s. That's given from the initial conditions. However, we can't assume that Barney will measure that clock 1 read (1 lyr/c) when he measures that clock 2 finally received that signal, because that's what we're trying to find out. So, we drop that assumption. It might turn out to be true, but we don't know. So, we use the postulates of relativity in conjunction with the initial conditions I gave. From the postulates of relativity, both Fred and Barney agree that the signal clock 1 sends travels at speed c. From this, if Fred and Barney are using light clocks, then we can derive that Barney always measures clock 1 to read t*gamma, where gamma is the lorentz factor and t is the time on Barney's watch. To be continued...
Fleetfoot
3 / 5 (2) May 04, 2013
Or do you have a different term that does not need the term "event"? I would like to hear if you have this, since this will give a different direction to this discussion!


OK, I have answered your question and we have confirmed that we mean the same thing by "an event". Now how about answering my question, you just need to give me three values:

Here is the 2D example again:

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?

Move the origin by (-2,-1):

If (x1,y1) = (4,2) and (x2,y2) = (8,5), what is s?

Rotate about (4,2):

If (x1,y1) = (4,2) and (x2,y2) = (7,6), what is s?

Get that out of the way and I'll do a 4D example and things will become clearer.
ValeriaT
1 / 5 (6) May 04, 2013
They do not occur when light speed is the same within all IRF's
They do - and this animation is illustrating it clearly - wanna bet? I do understand EXACTLY your (silly) problem. A hint: I explained it here at least five times, especially to you.
thefurlong
1.4 / 5 (9) May 04, 2013
@johanfprins
[continued]
Now, we just need to figure out the distance L' between clocks 1 and 2, as measured by Barney and compare it with the same distance L as measured by Fred. Now, we can assume by symmetry that if Fred sees Barney move at speed V, then Barney sees Fred move at speed V. If symmetry doesn't satisfy you, we can use other arguments, but I assume you agree with this assumption. Therefore, to determine distance between 1 and 2, Barney needs only send a signal out to clock 2 when he passes Fred and clock 1, and then wait for it to be reflected back to him. The same goes for Fred. After some algebra, we get that Barney thinks that the distance between 1 and 2 is gamma*L. Hence, we get length contraction. From these arguments, if you do the actual calculations, you'll see that Barney will measure that clock 2 disagrees with clock 1 the moment that clock 2 receives the signal from clock 1. To be continued...
thefurlong
1.4 / 5 (9) May 04, 2013
@johanfprins
[continued]
So, using this method, Barney will never be able to guarantee that both he and Fred see clock 1 and 2 as synchronized. But, if we can't even synchronize them for everybody by sending a light signal, how can we hope to synchronize them using other means? If Fred walks over to clock 2, then he, and his watch will be subject to relativistic laws, so Fred won't be guaranteed that his watch remains synchronized to clock 1. If Fred synchronizes clock 2, then moves it a light year away, because clock 2 now moves with respect to clock 1, it might become unsynchronized. No matter what you try to do, it just isn't possible for both Fred and Barney to gaurantee that they all agree that the clocks are synchronized.
thefurlong
1.8 / 5 (10) May 04, 2013

That is impossible to misunderstand "how to use it". Al;l you do is plug in the position and time coordinates of an event within the moving IRF and calculating the corresponding position and time coordinates within the IRF relative to which it is moving.

LOL! Well, therein lies your problem, son. There are plenty of equations in physics that will give you bogus answers if you just "plug it in" without understanding what you are plugging in and what result you are getting. GIGO and all that...
Anyway, it's right there in the Lorentz transformation that t' = gamma*(t-xv/c^2). When t = 0, surely, you agree that the two origins coincide. If an event E0 happens at (x,0), then the other guy will think E occured at t' = gamma*(-xv/c^2). On the other hand, if E1 happens at (0,0), the other guy thinks E1 happened at 0s. They didn't happen simultaneously for the moving guy. To the stationary guy, these even happened at both origins, so even Lorentz proves you wrong!
thefurlong
1.8 / 5 (10) May 04, 2013
and it is derived by first deriving time dilation and length contraction,


No this is not what Einstein did! Stop lying!

Who cares what Einstein did? All that matters is correct physical reasoning, whether done by Einstein or Alfred E. Neumann. Einstein was correct not because he was a genius but because he made the right argument. There are plenty of smart people who make wrong arguments. If you don't agree with the common approach of using time dilation and length contraction, prove it wrong from first principles, (not Lorentz Transformations).
johanfprins
1 / 5 (6) May 04, 2013
That's exactly what I mean too, there's no problem there.

However, while a coordinate pair (x,y) defines a single point on a flat sheet of paper, the equation:

x^2 + y^2 = R^2

defines a set of points which form a circle.


They are still different points at a distance R from the origin! They ARE NOT different points at a distance R=0 from the origin. Please ValeriaT just stop it and accept that you are an idiot.
johanfprins
1 / 5 (5) May 04, 2013
@johanfprins
All right, think about this. Assume Fred spaces 2 clocks, clock 1 and 2, out by a light year and synchronizes them. Imagine that clock 1 can send a signal to clock 2, the moment it is set, so that clock 2 can synchronize itself. So Fred sets clock 1 to 0 s. This sends a signal traveling at the speed of light, c, to clock 2.


Fred does not have to do this since all time has been synchronized at t=0 when space was created. They still keep the same time within gravity free space. So why go to all this trouble? It is insane!

Barney has a watch that reads 0s when he passes Fred the moment he sets clock 1. Our task, starting from as few assumptions as possible, but at least including the postulates of relativity, is to describe what Barney sees. So, let's try to figure it out...


Although this is irrelevant I will play along: Maybe it will help to clear your head!
johanfprins
1 / 5 (7) May 04, 2013
@johanfprins
[continued]
Now, every sane person will agree that when Barney passes Fred, both of them agree that it happened at 0s. That's given from the initial conditions. However, we can't assume that Barney will measure that clock 1 read (1 lyr/c) when he measures that clock 2 finally received that signal, because that's what we're trying to find out.


So you make the assumption that the two clocks cannot keep the same rate of time until the signal has been sent and received. But this is irrelevant since it is not what the clock reads, but whether the clocks are keeping and has been keeping time at the same rate. Here in South Africa our clocks are 7 hours ahead of the clocks in New York but the actual time in SA and NY is exactly the same. So why does one have to synchronize the positions of the arms on the clocks in SA and NY to conclude that at any instant in time the time in NY and SA is exactly the same? Are you REALLY all there?

johanfprins
1 / 5 (5) May 04, 2013
From this, if Fred and Barney are using light clocks, then we can derive that Barney always measures clock 1 to read t*gamma, where gamma is the lorentz factor and t is the time on Barney's watch.


He measures that an event which occurs at time t on BOTH his own clock and clock 1 occurs within HIS own (Barney's) reference frame at a later time t*gamma at which it actually occurs simultaneously within BOTH IRF's

Now, we just need to figure out the distance L' between clocks 1 and 2, as measured by Barney and compare it with the same distance L as measured by Fred. Now, we can assume by symmetry that if Fred sees Barney move at speed V, then Barney sees Fred move at speed V. If symmetry doesn't satisfy you, we can use other arguments, but I assume you agree with this assumption


Very Good! The first statement by you with which I agree.
johanfprins
1 / 5 (6) May 04, 2013
Therefore, to determine distance between 1 and 2, Barney needs only send a signal out to clock 2 when he passes Fred and clock 1, and then wait for it to be reflected back to him. The same goes for Fred.


Still suprisingly logical. I cannot believe it!

After some algebra, we get that Barney thinks that the distance between 1 and 2 is gamma*L. Hence, we get length contraction.


WRONG!!!!! If you are able to do the algebra correctly you will find that Barney will see a longer distance between the the clocks: i.e. a distance of L/(gamma). This is what the Lorentz transformation gives: No matter what you want!

From these arguments, if you do the actual calculations, you'll see that Barney will measure that clock 2 disagrees with clock 1 the moment that clock 2 receives the signal from clock 1.


I have NEVER disputed that Barney will conclude that the signal will reach the clocks at different times on Barney's clock.
Fleetfoot
3.7 / 5 (3) May 04, 2013
That's exactly what I mean too, there's no problem there.

However, while a coordinate pair (x,y) defines a single point on a flat sheet of paper, the equation:

x^2 + y^2 = R^2

defines a set of points which form a circle.


They are still different points at a distance R from the origin!


Correct, that's what I said. You are learning.

They ARE NOT different points at a distance R=0 from the origin.


Don't try to jump ahead, you have more to learn first. I'm still waiting to see if you can answer my three questions. I though they would be trivial for you but it seems you are struggling.

Please ValeriaT just stop it and accept that you are an idiot.


Don't be so clueless, go look at our other exchanges, for example:

phys.org/news/2013-05-fermi-swift-shockingly-bright.html
johanfprins
1 / 5 (5) May 04, 2013
What I dispute is that these time differences are caused by the clocks (including Barney's clock) keeping time at different rates. They do not since this will violate Einstein's first postulate. And by doing a bit of algebra it is easy to prove that it also violates his second postulate.

So, using this method, Barney will never be able to guarantee that both he and Fred see clock 1 and 2 as synchronized.


Only two fools will use this method: The fact is that ALL clocks in Fred's IRF MUST keep time at the same rate or else simultaneous events will not be possible within his IRF. Similarly all clocks within Barney's IRF must keep time at the same rate whether you synchronise them with clocks within Fred's IRF or not. And since the laws of physics must be the same within both IRF's the time rate must be the same. The time differences derived from the Lorentz transformation are thus different times which are simultaneously present on ALL the clocks in a gravity-free universe.
johanfprins
1 / 5 (6) May 04, 2013
Well, therein lies your problem, son. There are plenty of equations in physics that will give you bogus answers if you just "plug it in" without understanding what you are plugging in and what result you are getting.


Well then go to my manuscript and show me what I "did not understand" when I plugged in coordinates into the equations of the Lorentz transformation. If you cannot do this it would be better for you to just SHUT UP!

Anyway, it's right there in the Lorentz transformation that t' = gamma*(t-xv/c^2).


It is ONLY 1/2 of the transformation Only a fool will make a deduction by not also doing the transformation of the position coordinates. Unfortunately Einstein sullied his reputation of being a genius by NOT doing the full transformation from which it is easy to prove that the two clocks are actually keeping exactly the SAME time!

Signing off for now!! It tires one out to reason with a bigot who is not willing to really argue logic and mathematics.
Q-Star
2.2 / 5 (13) May 04, 2013
Not for Minkowski, to him Riemann was a god, he wasn't sure it was actually a real thing rather than a fun math,,, but he also had a very poor opinion of Einstein's math abilities.


"Minkowski is perhaps best known for his work in relativity, in which he showed in 1907 that his former student Albert Einstein's special theory of relativity (1905), presented algebraically by Einstein, could also be understood geometrically as a theory of four-dimensional space-time.


I don't have it right at hand, but I'll attempt to paraphrase accurately. In a biography of Einstein (A. Pais), one of the quotes of Minkowski's assessment of Einstein when seeking his Doctorate, was, "his mathematics are only mediocre, and certainly not sufficient for one seeking his Doctor of Physics",,,,, that was when Einstein was applying for Doctorate,,,, which he didn't wasn't awarded until 1906, after his "miracle year" 1905.
Whydening Gyre
1 / 5 (10) May 04, 2013

That argument you have to take up with Einstein since Einstein accepted that events can occur simultaneously at different positions within an IRF. This will not be possible if the clocks at these positions do not keep the same time.

Well, sure, in a single IRF, events can occur simultaneously! The context is when two events can occur simultaneously in two different IRF's.

If the clocks were synchronized at same origin location, making the assumption that they were made to not lose or gain time - then you send them off in two different directions at two different velocities, they will always read the same time at any given simultaneous moment. It's the measurement from a third IRF that makes them appear to be different times, because of the c boundary. If those clocks have been set at same origin, they are "entangled". Even if they have 2 different origins but are synced to allow for c measurement time, they will still read the same.
(cont)
Whydening Gyre
1 / 5 (10) May 04, 2013
Some of you are considering the IRF of the 2 clocks, while others are viewing it from the third observer IRF . 2 different IRF's. If you all agreed on the same IRF vantage point, there would be no quarrels here. So, that gives the appearance that somebody is intentionally skewing the argument - couldn't say who, tho.
Anyway - that's MY IRF...
johanfprins
1 / 5 (8) May 05, 2013
Let us try and cut through the knot. I have developed a question-answer approach, and first tried it out on my grandson in grade 7 who has not yet had a course on the Special Theory of Relativity. He could straight-away answer the mathematics involved. So let us see how thefurlough, fleetfoot and ValeriaT AKAK will compare.

Question 1. Clock2 passes clock1 with speed v and at that instant they synchronize their readings: After a time t/ on clock2 how far is clock2 from clock1, I will even give you a hint which I did not need to do in the case of my grandson: i.e.

distance=(speed)*(time)

ValeriaT just do the mathematics, if you can, without any silly cartoon please!
Fleetfoot
3.7 / 5 (3) May 05, 2013
"Minkowski is perhaps best known for his work in relativity, in which he showed in 1907 that his former student Albert Einstein's special theory of relativity (1905), presented algebraically by Einstein, could also be understood geometrically as a theory of four-dimensional space-time.


I don't have it right at hand, but I'll attempt to paraphrase accurately. In a biography of Einstein (A. Pais), one of the quotes of Minkowski's assessment of Einstein when seeking his Doctorate, was, "his mathematics are only mediocre, and certainly not sufficient for one seeking his Doctor of Physics",,,,, that was when Einstein was applying for Doctorate,,,, which he didn't wasn't awarded until 1906, after his "miracle year" 1905.


Yes, and that goes back to the original comment, I understand it was Minkowski who had the "ah-hah moment" when he realised the maths could be derived from Riemann geometry, not Einstein, and that Einstein was actually slow to accept it.
Fleetfoot
3.7 / 5 (3) May 05, 2013
If the clocks were synchronized at same origin location, making the assumption that they were made to not lose or gain time - then you send them off in two different directions at two different velocities, they will always read the same time at any given simultaneous moment.


That's the presumption in aether theories, it is not true in SR. It would avoid some of the confusion in this thread if you stuck to talking about just one or the other or at least it make clear when bringing aether theory into the conversation.
ValeriaT
1 / 5 (8) May 05, 2013
But this cartoon explains whole the controversy of yours. The trick is, the object in motion is surrounded with deBroglie wave of vacuum, which A) increases its effective mass, as SR requires B) it slows down the energy propagation around it (the time dilatation). The object in motion therefore can see his clock and light running normally inside of resulting blob of dense vacuum around it, but from perspective of other observers he is already delayed.

The rational core of your objections is, in the universe fulfilling the special relativity quite strictly we couldn't have any observers, gravity lenses and local clock differences, the light would always spread along linear path with constant speed. It points to the many worlds concept, i.e. the fact, from strictly low-dimensional perspective the moving observers are causally separated. I'm just providing the hyperdimensional way, in which consistency can be maintained.
johanfprins
1 / 5 (8) May 05, 2013
That's the presumption in aether theories, it is not true in SR.


It does not require an aether for this to be true: As usual you are way off!

Please answer my simple first question above!
Fleetfoot
3 / 5 (2) May 05, 2013
Some of you are considering the IRF of the 2 clocks, while others are viewing it from the third observer IRF . 2 different IRF's. If you all agreed on the same IRF vantage point, there would be no quarrels here. So, that gives the appearance that somebody is intentionally skewing the argument - couldn't say who, tho.
Anyway - that's MY IRF...


Obviously, you can approach this any way you like but generally it is very difficult to use just two frames, the simplest illustration is usually the Twins Paradox using three ships which sync their clocks at the moment of passing (zero separation in the direction of motion at that moment and no accelerations).

I'm tackling a different aspect, Johan seems to be using the term "invariant" in an ambiguous or erroneous way so I want to make sure we both have the same understanding of the term before applying it to any examples. Without that, I suspect it won't be possible to agree what times clocks show.
ValeriaT
1 / 5 (6) May 05, 2013
The trick is, the introduction of time dilatation and mass increasing are actually hyperdimensional quantum effects, as I realized before years already. They have nothing to do in strictly special relativistic universe, which will always remain fully flat and empty. Only such an universe can fully provide the constant speed of light in all directions and situations. At the moment, when the observer moves inside of some dense blob of vacuum, then the complementarity of black holes can be applied to him and he will see somewhat different universe, than the observer outside of that blob. His strictly deterministic low-dimensional math will become inconsistent with the math/reference frames of all other observers inside of universe. In special relativity it will manifest itself with retardation of local clock.
Fleetfoot
3.7 / 5 (3) May 05, 2013
That's the presumption in aether theories, it is not true in SR.


It does not require an aether for this to be true:


Indeed, but aether theories do start from that presumption.

As usual you are way off!


Both statements I made above are true.

Please answer my simple first question above!


You don't run this forum so I'll do that if you first answer the one I have been asking repeatedly, it shouldn't take you more than a minute, the numbers have been chosen to make it trivial:

Place a cocktail stick on a table and place a transparent sheet of graph paper over it. Read off the coordinates of each end as (x1, y1) and (x2, y2), then calculate

s^2 = (x2 - x1)^2 + (y2 - y1)^2

If (x1,y1) = (2,1) and (x2,y2) = (6,4), what is s?

Move the origin by (-2,-1):

If (x1,y1) = (4,2) and (x2,y2) = (8,5), what is s?

Rotate about (4,2):

If (x1,y1) = (4,2) and (x2,y2) = (7,6), what is s?
Fleetfoot
3 / 5 (2) May 05, 2013
Please answer my simple first question above!


.. so I'll do that if you first answer the one I have been asking repeatedly, it shouldn't take you more than a minute, the numbers have been chosen to make it trivial:


p.s. This isn't just about who asks the questions, I intend to use your answer to my question as part of my answer to yours.
johanfprins
1 / 5 (6) May 05, 2013
It does not require an aether for this to be true:


Indeed, but aether theories do start from that presumption. I have not disputed this but you have created (deliberately?) the misconception that this can ONLY be the case when there is an aether. This is disengenious. In fact it is just plain dishonesty on your part!

Please answer my simple first question above!


You don't run this forum so I'll do that if you first answer the one I have been asking repeatedly, it shouldn't take you more than a minute, the numbers have been chosen to make it trivial:


These "questions" of yours do NOT relate to the concept of space-time: I have told you repeatedly that I do not disagree with your calculations within a a 2D space-manifold which does not have an imaginary time-coordinate. So your questions are irrelevant and contributes NOTHING to the discussion!

All you are proving is that you cannot answer my simple question which my grandson in grade 7 could answer!
johanfprins
1 / 5 (7) May 05, 2013
p.s. This isn't just about who asks the questions, I intend to use your answer to my question as part of my answer to yours.


It is clear that you are too stupid to answer a question that a grade 7 pupil can. I will thus answer the question for you:

Question: After a time t/ on clock2 how far is clock2 from clock1,


Answer: the distance D/ between clock 2 and clock 1 must be D/=(speed)*(time)=v*t/

Now question 2: What is the Lorentz-transformed distance D between clock2 and clock1 .

I will give you another hint: The Lorentz formula to use is:

D=(gamma)*(x/+v*t/) where x/ is the coordinate of clock 2 within its own inertial reference frame. Since this will probably also be too difficult for you to figure out, I will give you an additional hint: The value of x/ is zero: i.e. x/=0.

So what is D?

My grandson got it immediately!

Fleetfoot
3 / 5 (2) May 05, 2013
It does not require an aether for this to be true:


Indeed, but aether theories do start from that presumption.
I have not disputed this but you have created (deliberately?) the misconception that this can ONLY be the case when there is an aether. This is disengenious. In fact it is just plain dishonesty on your part!


You read more into the comment than I said.

I'll do that if you first answer the one I have been asking repeatedly, it shouldn't take you more than a minute, the numbers have been chosen to make it trivial:


These "questions" of yours do NOT relate to the concept of space-time: I have told you repeatedly that I do not disagree with your calculations ..


You are wrong, the question DOES relate to your argument but as long as you refuse to face it, you will fail to see why.

It's no concern to me, I'll leave you to continue repeating your error.

johanfprins
1 / 5 (7) May 05, 2013
These "questions" of yours do NOT relate to the concept of space-time: I have told you repeatedly that I do not disagree with your calculations.


You are wrong, the question DOES relate to your argument but as long as you refuse to face it, you will fail to see why.


Since I cannot understand what you are trying to ask me how about giving the answer that you expect me to give? As I am doing with my questions which you refuse to answer.

So let me proceed with my second question:
Now question 2: What is the Lorentz-transformed distance D between clock2 and clock1


The answer my grandson immediately gave was that:

D=(gamma)*(x/+v*t/) =(gamma)*(0+v*t/)=(gamma)*(v*t/)=(gamma)*D/

Astonishing that YOU cannot do this simple algebra.

Lets try the third question.

Since the distance between the clocks is now D, what must the time t be on the clock for this distance to be possible?

Hint: (time)=(distance)/(speed).

Are you also too stupid to do this simple maths?
Fleetfoot
2.3 / 5 (3) May 05, 2013
These "questions" of yours do NOT relate to the concept of space-time: I have told you repeatedly that I do not disagree with your calculations.


You are wrong, the question DOES relate to your argument but as long as you refuse to face it, you will fail to see why.


Since I cannot understand what you are trying to ask me how about giving the answer that you expect me to give? As I am doing with my questions which you refuse to answer.


My questions are clear, I expect you to give just three numbers.

I won't tell you the answers because I know from past experience that that approach results in two people each proceeding with a monologue and talking past each other, the whole exercise becomes pointless. If you answer my questions and I then answer yours, we have a chance of maintaining a dialogue. I will happily answer yours but it has to be a mutual approach. I have answered other questions of yours before, now it's your turn, see it as an act of good faith.
johanfprins
1 / 5 (6) May 05, 2013
OK I will humor you, even though you make no sense. In all three cases s=5: This is what it must be in a two dimensional SPACE where the coordinate axes are REAL numbers: So what?

BTW: I hope you note that in order to do the rotation you first had to translate ALL the points so that the position (4,2) becomes the origin: Then do the rotation. And then again translate all the points so that the point (4,2) ends up at the origin around which you rotated.
Fleetfoot
3.7 / 5 (3) May 05, 2013
OK I will humor you, even though you make no sense. In all three cases s=5: This is what it must be in a two dimensional SPACE where the coordinate axes are REAL numbers: So what?


Thank you.

The value in the first coordinate system is 5. After changing to either of the other coordinate systems, the value is the same, still 5. What that illustrates is the meaning of "invariant", that s (at least for the 2D case) has the same value in the various coordinate systems.

Now apply that definition to the 4D case. Two events (x1,y1,z1,t1) and (x2,y2,z2,t2) in frame K are separated by interval s defined as:

s^2 = (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 - (t2 - t1)^2

The "dogma" as you put it is that this is invariant. To test that, transform those coordinates to frame K' using the Lorentz Transforms and find:

s'^2 = (x'2 - x'1)^2 + (y'2 - y'1)^2 + (z'2 - z'1)^2 - (t'2 - t'1)^2

The interval is invariant if s' = s for all values of the initial coordinates.

[contd.]
Fleetfoot
3.7 / 5 (3) May 05, 2013
[contd.]

The interval is invariant if s' = s for all values of the initial coordinates.


What you said some time ago was:

The value of s cannot be invariant since s can also be zero for coordinates that are NOT the origin.

The diagram for a light cone is done within a single inertial reference frame where the time is exactly the same at all points within this reference frame. It says NOTHING about different times within other inertial reference frames.


Invariant means that the function has the same value in different frames.

Now you check that if you like (it's just linear algebra) or just look up the derivation in a textbook, I'll leave that to you. We'll be able to apply it to your question though.
Fleetfoot
3 / 5 (2) May 05, 2013
You answered my question so I'll answer yours:

... So let us see how thefurlough, fleetfoot and ValeriaT AKAK will compare.

Question 1. Clock2 passes clock1 with speed v and at that instant they synchronize their readings: After a time t/ on clock2 how far is clock2 from clock1


In the coordinate system of clock1, and assuming that the positive x axis is aligned with the direction of motion of clock2 (hence y=z=constant), the distance between the clocks is given by:

x2(t1) = x1 + v * t1

where:

x1 is the location of clock1 (a constant)

t1 is the time measured by clock1

x2(t1) is the location of clock2 at time t1

v is the speed of clock2 as measured in the frame of clock1.
Fleetfoot
3 / 5 (2) May 05, 2013
BTW: I hope you note that in order to do the rotation you first had to translate ALL the points so that the position (4,2) becomes the origin: Then do the rotation. And then again translate all the points so that the point (4,2) ends up at the origin around which you rotated.


Not necessarily but that's one way of doing it. If you stick a pin through the sheet at (4,2) and rotate it, you should be able to find a set of linear equations that relate the coordinates before and after describing that operation without using those intermediate steps.
johanfprins
1 / 5 (7) May 05, 2013
The value of s cannot be invariant since s can also be zero for coordinates that are NOT the origin.

The diagram for a light cone is done within a single inertial reference frame where the time is exactly the same at all points within this reference frame. It says NOTHING about different times within other inertial reference frames.


Invariant means that the function has the same value in different frames.

I know what "invariance" means but it is a nonsensical mathematical AND physics concept when coordinates, which are not (0,0,0,0) are a distance s=0 away from the origin. It does not rectify the mathematics and physics to invent nonsensical names like "null geodesic" and time-like geodesic". And it is nonsensical since it is based on the concept that different clocks keep time at different rates: They do not and cannot do so since the Lorentz-transformation does not allow them top do so.
johanfprins
1 / 5 (9) May 05, 2013
It is quite clear that you are not going to answer my questions honestly. So let me return to question 3 which you have NOT answered: i.e.

Lets try the third question.

Since the distance between the clocks is now D, what must the time t be on the clock for this distance to be possible?

Hint: (time)=(distance)/(speed).


As my grandson calculated t=D/v, and since D=(gamma)*D/ and D/=v*t/ one obtains for the time on clock1 that

t=(gamma)*t/ where t/ is the time on clock2 when the distance between the clocks are D/ while t is the distance between the two clocks when the time is t on clock1.

Note that t=(gamma)*t/ is interpreted by YOU as time dilation: i.e. as the simultaneous times on clock1 and clock2 when the event occurs at clock2. But this, in turn demands, that at that instant in time the distance between clok1 and clock2 must be simultaneously D and D/: Which should be absurd to anybody with a sane mind.

johanfprins
1 / 5 (8) May 05, 2013
The correct interpretation of this so-called "time-dilation" formula is thus that when the event occurs at time t/ at clock1 BOTH CLOCKS must simultaneously read the same time t/, or else the distance between the clocks is not uniquely defined as it must be to model sane physics.

Clock1 only registers that an event has occurred at clock2 at a LATER time t, which must at that instant in time also be simultaneously the same on both clocks; since clock 1 only records this after the two clocks have moved from the distance D/ at which the event actually occurred at clock1 to be at a further distance D from each other when clock1 records the event. This is confirmed by the fact that D/t=v: The relative speed with which the clocks are moving away from one another. Thus t/ and t are two different times, each of which is displayed simultaneously on both clocks!

Is this "time-dilation"?
Whydening Gyre
1 / 5 (9) May 05, 2013
I'm tackling a different aspect, Johan seems to be using the term "invariant" in an ambiguous or erroneous way so I want to make sure we both have the same understanding of the term before applying it to any examples. Without that, I suspect it won't be possible to agree what times clocks show.

So... tell me what IS the one true invariant? My guess is that there are none, but that's what makes it all work TOGETHER. It's all balanced with ever varying degrees of feedback.
ValeriaT
1 / 5 (8) May 05, 2013
Thus t/ and t are two different times, each of which is displayed simultaneously on both clocks! Is this "time-dilation"?
Of course, why not.
Whydening Gyre
1 / 5 (9) May 05, 2013
Fermat, Goddet(?), Einstein and Minkowski (to name but a few) must be rolling in their graves in laughter...
johanfprins
1 / 5 (7) May 05, 2013
So... tell me what IS the one true invariant? My guess is that there are none, but that's what makes it all work TOGETHER. It's all balanced with ever varying degrees of feedback.


An invariant can only be defined if you have two equivalent linear spaces with linearly independent coordinates so that you can have a coordinate transformation from one coordinate space into the other where each coordinate point in the one space only has a single companion point within the other space. This is not the case for a Minkowski four-space since many points with different coordinates are not spatially separate.

Only in such a space, when you transform a physics-equation from one coordinate frame into the other and the equation does not change, the equation is invariant. For example, the equation defining acceleration within Galilean space (d/dt)v=a, gives the same value for a when transforming this equation by means of the Galilean transformation. Acceleration is then an invariant.
johanfprins
1 / 5 (8) May 05, 2013
Thus t/ and t are two different times, each of which is displayed simultaneously on both clocks! Is this "time-dilation"?
Of course, why not.


So you are saying that when two clocks moving relative to one another show simultaneously the exact same time t/ at that same instant in time; and then later show simultaneously the exact later time t at this later instant in time, the clocks keep time at different rates? LOL.

You have proved over and over on this thread that you are brainless, but it is even worse than I thought!
Fleetfoot
3.7 / 5 (3) May 05, 2013
Invariant means that the function has the same value in different frames.


I know what "invariance" means


Good. In that case, you should follow this (economically stated):

If a photon moves at speed c in frame K then for any two events on its path, the interval s has the value 0. If the function s is invariant then the same two events measured in frame K' are also separated by interval s'=0 hence the photon must move at speed c in frame K'. That is entirely consistent with Einstein's postulates.

but it is a nonsensical mathematical AND physics concept when coordinates, which are not (0,0,0,0) are a distance s=0 away from the origin.


s=0 is not a distance. I'll explain tomorrow but I have to leave to run an event now and won't get back until midnight.
Fleetfoot
3 / 5 (2) May 05, 2013
It is quite clear that you are not going to answer my questions honestly.


I have answered your question 1 honestly and accurately. If you think there is an error in my reply, point it out, I'm not infallible and typos sometimes creep in too.

So let me return to question 3 which you have NOT answered: i.e.


I haven't even seen your question 3 yet, it's been a busy day, but I'll look at your question 2 next if you confirm you have no objections to my answer to question 1. If you think that's wrong, there's no point in my building on it.
Whydening Gyre
1 / 5 (9) May 05, 2013
Acceleration is then an invariant.

So, it is the ACT of acceleration that is the constant, not necessarily the thing that is accelerating. The only way you can get acceleration is by applying energy.
Interesting dichotomy...
johanfprins
1 / 5 (8) May 05, 2013
You answered my question so I'll answer yours:

Question 1. Clock2 passes clock1 with speed v and at that instant they synchronize their readings: After a time t/ on clock2 how far is clock2 from clock1


In the coordinate system of clock1, and assuming that the positive x axis is aligned with the direction of motion of clock2 (hence y=z=constant), the distance between the clocks is given by:

x2(t1) = x1 + v * t1

We do not yet know what the time on clock1 is only the time on clock2, so why do you bring in the time on clock1. The distance D/ from clock2 to clock1 is simply (if you want to call it x2)

D/=x2=v*t/: Or if you want to use t2 for t/ you have that the distance must be

x2=v*t2

Now proceed with the other questions if you can do the simple mathematics which I doubt.

johanfprins
1 / 5 (7) May 05, 2013
If a photon moves at speed c in frame K then for any two events on its path, the interval s has the value 0.


It can only have this in a space with coordinates that are NOT linearly independent and "invariance" cannot be defined for ANY transformation into or out of such a space.

If the function s is invariant then the same two events measured in frame K' are also separated by interval s'=0 hence the photon must move at speed c in frame K'. That is entirely consistent with Einstein's postulates.


This is the same as dividing by zero, since the result is undefined!

but it is a nonsensical mathematical AND physics concept when coordinates, which are not (0,0,0,0) are a distance s=0 away from the origin.


s=0 is not a distance.


It must be the "distance of the origin to itself" within a space spanned by four linearly independent coordinates; or else it is not mathematically allowed and is also physics-claptrap!
johanfprins
1 / 5 (7) May 05, 2013
Acceleration is then an invariant.

So, it is the ACT of acceleration that is the constant, not necessarily the thing that is accelerating. The only way you can get acceleration is by applying energy.
Interesting dichotomy...


It is the value of acceleration that remains the same when doing the transformation between two coordinate systems spanned by LINEARLY INDEPENDENT COORDINATES. In contrast, velocity, momentum and kinetic energy are not the same within the different IRF's WITH LINEARLY INDEPENDENT COORDINATES and are thus NOT invariant under a transformation of these coordinates.
thefurlong
1.7 / 5 (12) May 05, 2013
Fred does not have to do this since all time has been synchronized at t=0 when space was created.

So, you've never had to synchronize a clock that was wrong?
They still keep the same time within gravity free space. So why go to all this trouble? It is insane!

If you are at rest with them, they certainly do! We don't know if they keep the same time in another IRF. The point of this thought experiment is to find out if they do.

Although this is irrelevant I will play along: Maybe it will help to clear your head!

Thank you.
johanfprins
1 / 5 (7) May 05, 2013
Fred does not have to do this since all time has been synchronized at t=0 when space was created.

So, you've never had to synchronize a clock that was wrong?


In STR it is assumed that the clocks are perfect and do not gain or lose time. Thus the synchronization has nothing to do with a clock being "wrong".

They still keep the same time within gravity free space. So why go to all this trouble? It is insane!


If you are at rest with them, they certainly do!


Only when"I" am at rest with them? Gee I am important am I not? It is more sane to claim that every clock has at least one human being at rest with the clock, and since, according to Einstein's first postulate, every human being will experience the SAME physics all clocks MUST keep the same time within their respective reference frames.

The point of this thought experiment is to find out if they do.


Einstein's first postulate already demands that it MUST be so!
thefurlong
1.7 / 5 (11) May 05, 2013

So you make the assumption that the two clocks cannot keep the same rate of time until the signal has been sent and received.

I am saying we don't know if they do or they don't. That's why we need to establish standards. Everyone can agree that if Fred and two clocks are at rest with each other, Fred is guaranteed to synchronize the two clocks in the way I described. I am trying to make as few assumptions as possible, so as not to come to the wrong conclusion. Since you'll at least agree that conventional SR deals with spatial movement affecting passage of time, I want to limit Fred's movement in spacing the clocks out and setting them to the same time, so I made it so he doesn't have to move at all. Both of us can now happily agree that Fred has, in his rest frame, synchronized the clocks?
johanfprins
1 / 5 (7) May 05, 2013

So you make the assumption that the two clocks cannot keep the same rate of time until the signal has been sent and received.

I am saying we don't know if they do or they don't.


Maybe we do not know this, but if this is not so Einstein's postulates must be wrong and so must his STR. Is that what you are arguing?

Both of us can now happily agree that Fred has, in his rest frame, synchronized the clocks?


And Barney also did this in his IRF: so what?

Stop being Daft!
thefurlong
1.7 / 5 (12) May 05, 2013

He measures that an event which occurs at time t on BOTH his own clock and clock 1 occurs within HIS own (Barney's) reference frame at a later time t*gamma at which it actually occurs simultaneously within BOTH IRF's

No need to use convoluted language. Just picture two upright light clocks of the same height, one which is at rest, and one which is traveling. The traveling one will tick at a lower rate because the photon has a longer path to travel. In this case, Barney's watch is at rest with him, but Clock 1 isn't, so Barney will measure Clock 1 to tick slower. That's all.

Very Good! The first statement by you with which I agree.

You know, in case you actually end up being proven wrong, you should consider arguing with more humility, lest you end up eating crow--not that I think you'd ever admit you were wrong, or even attempt to question your own conclusions from time to time (which I do all the time, believe me).
thefurlong
1.7 / 5 (12) May 05, 2013
Still suprisingly logical. I cannot believe it!

Perhaps you are of diminutive stature...


After some algebra, we get that Barney thinks that the distance between 1 and 2 is gamma*L. Hence, we get length contraction.


WRONG!!!!! If you are able to do the algebra correctly you will find that Barney will see a longer distance between the the clocks: i.e. a distance of L/(gamma). This is what the Lorentz transformation gives: No matter what you want!


I still don't understand where you think the Lorentz transformation came from--not that that is correct, since you actually derive it from time dilation and length contraction!
ValeriaT
1 / 5 (9) May 05, 2013
..thus t/ and t are two different times, each of which is displayed simultaneously on both clocks! Is this "time-dilation"?

"Of course, why not".

So you are saying that when two clocks moving relative to one another show simultaneously the exact same time t/ at that same instant in time?
Nope, if the t/ and t are two different times, each of which is displayed simultaneously on both clocks. The discussions with you are somewhat bizarre, as our senile mind even cannot remember the last question of yours... But they're still very funny...:-)
Q-Star
2.4 / 5 (14) May 05, 2013
I'm tackling a different aspect, Johan seems to be using the term "invariant" in an ambiguous or erroneous way so I want to make sure we both have the same understanding of the term before applying it to any examples. Without that, I suspect it won't be possible to agree what times clocks show.

So... tell me what IS the one true invariant? My guess is that there are none, but that's what makes it all work TOGETHER. It's all balanced with ever varying degrees of feedback.


"s" is invariant if ya don't mix apples with oranges. "s" is world-line in spacetime. In spacetime space IS variant. And time is variant. Spacetime is the same for ALL observers regardless of frame.

johan is trying force the variance by applying the maths of spacetime in a way they weren't intended. He knows QT, but everyone knows ya can't get relativity and QT to place nice at all scales. People way smarter than ANY of us have been working on that for a long time.
Q-Star
2.3 / 5 (12) May 05, 2013
Acceleration is then an invariant.

So, it is the ACT of acceleration that is the constant, not necessarily the thing that is accelerating. The only way you can get acceleration is by applying energy.
Interesting dichotomy...


Not energy,,,, FORCE. Ya get an acceleration by applying a force.
Noumenon
1.3 / 5 (30) May 05, 2013
LOL, I've been over this with johan many times, here, here, and here, yet he continues to reject time dilation and does not use the Lorentz transformation properly to determine space-TIME distance, which is invariant. He bores me.
thefurlong
1.3 / 5 (12) May 05, 2013
What I dispute is that these time differences are caused by the clocks (including Barney's clock) keeping time at different rates.

Ok, let's clear something up. The clocks not being synchronized, doesn't mean that they don't keep the same rate. I am merely saying that if Fred sees the spatially separated clocks as showing the same time, Barney will see a constant time difference between them.
Noumenon
1.1 / 5 (29) May 05, 2013

johanfprins rejects Minkowski space-time, but in that context time dilation can be derived, as in this post I made last year,... dm/tm = moving IRF, ds/ts = stationary IRF,

dm = y(ds - v*ts) and ds = y(dm + v*tm) ,.... where y = gamma,...

Substitute the 2nd equation into the 1st to find ts , where y = (1 - v^2/c^2)^-1/2 = gamma

dm = y(y(dm + v*tm) - v*ts),...

dm = y^2*dm + y^2*v*tm - y*v*ts,.... divide by y^2 ,...

dm/y^2 = dm + v*tm - v*ts/y,..... now since 1/y^2 = (1-v^2/c^2),...

dm - v^2*dm/c^2 = dm + v*tm - v*ts/y,.... subtract out dm,...

- v^2*dm/c^2 = v*tm - v*ts/y,... divide out by v,...

- v*dm/c^2 = tm - ts/y,....

ts/y = tm + v*dm/c^2,... multiply both sides by y (gamma) ,...

ts = y(tm + v*dm/c^2) ----> the time on stationary clock of the event (the event being the time tm on moving clock, say of a explosion).

ts will show a later time than tm.
thefurlong
1.3 / 5 (12) May 05, 2013
@johanfprins
Again, I don't see how you don't see this from the Lorentz Transformation itself.
t' = gamma*(t-vx/c^2). x' = gamma*(x-vt).
So, let's do the math.
Let's say that for Fred, clock 1 reads t = 1 year, and clock 2 reads t = 1 year, and that they are 1 light year away from each other. Barney is traveling right at 0.1c, and passed Fred when both their clocks read 0s.
Now, from the transformation, Fred will say Clock 1 reads 1 year when his watch says (1 yr - 0.1c*0/c^2)/sqrt(1- 0.01) = 1.00504 yr. Now when does Barney measure Clock 2 to read one year? (1 yr - 0.1c*(1 lyr)/c^2)/sqrt(1-0.01) = 0.90453 yr. So, Barney sees that the clocks aren't synchronized. Now, keep in mind that I am saying he MEASURED these values, which means that even when he accounted for the fact that information took time to get to him, and that he is moving relative to Fred, he STILL got those values.
How on Earth can you now still disagree with me that for Barney the clocks aren't synchronized?
thefurlong
1 / 5 (11) May 05, 2013
LOL, I've been over this with johan many times, http://phys.org/n...ry.html, yet he continues to reject time dilation and does not use the Lorentz transformation properly to determine space-TIME distance, which is invariant. He bores me.

Yeah. At first it was an interesting challenge, but it is starting to get tiresome.
thefurlong
1.3 / 5 (12) May 05, 2013
Maybe we do not know this,

YES!
but if this is not so Einstein's postulates must be wrong and so must his STR. Is that what you are arguing?

NO! You just admitted that maybe we do not know this. The whole point is to see, using as few assumptions as possible, which one of us is correct!

Both of us can now happily agree that Fred has, in his rest frame, synchronized the clocks?


And Barney also did this in his IRF: so what? [/q[]

Stop being Daft!

Did what? Synchronize clock 1 and clock 2? Uhhh....no....
Barney did nothing to Fred's clocks (though he probably took his Fruity Pebbles).
thefurlong
1.3 / 5 (12) May 05, 2013
In STR it is assumed that the clocks are perfect and do not gain or lose time. Thus the synchronization has nothing to do with a clock being "wrong".

Well, if a real lock works properly, then it should reflect the actual passage of time, should it not? You accept that once set real clock correctly, then it should closely approximate an ideal clock, do you not? Once that real clock has been set it should always show what an ideal clock would read, should it not? Then, there's nothing to argue about!
Noumenon
1.4 / 5 (30) May 05, 2013
LOL, I've been over this with johan many times, http://phys.org/n...ry.html, yet he continues to reject time dilation and does not use the Lorentz transformation properly to determine space-TIME distance, which is invariant. He bores me.

Yeah. At first it was an interesting challenge, but it is starting to get tiresome.


It's not that he doesn't understand it, it's that he rejects what he determines as conceptual absurdities, like time dilation and so the failure of simultaneity, the Born interpretation of |Ψ|², etc.

Imo, he mistakes concepts for physical realities and seems to "save" those concepts. For example, time is not a physical entity, it is observer dependent, which is why there can be a disagreement wrt simultaneity. What is physically real is invariants, symmetries, and conservation laws,.... not the conceptual form in which they can be known.
thefurlong
1.9 / 5 (14) May 05, 2013
If you are at rest with them, they certainly do!

Only when"I" am at rest with them? Gee I am important am I not?

Jesus Christ! This is precisely why you aren't important! Other people can now disagree with you on the time, and their point of view will just be as valid.

It is more sane to claim that every clock has at least one human being at rest with the clock


I bet Voyager has a clock on board. Are there any humans at rest with its clock?


and since, according to Einstein's first postulate, every human being will experience the SAME physics all clocks MUST keep the same time within their respective reference frames.


All clocks AT REST in the IRF must keep the same time in their respective IRF. Also, (and this will really blow your mind...or maybe not) but if the situation were symmetric and Barney also had two synchronized clocks, 3 and 4, spaced 1 l yr apart at rest with him, Fred would think that 3 and 4 were unsynchronized!
ValeriaT
1.6 / 5 (7) May 05, 2013
according to Einstein's first postulate, every human being will experience the SAME physics all clocks MUST keep the same time within their respective reference frames
Nope, by Einstein's first postulate "The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translation motion relative to each other." Not a single word about time is here.

If you do want to understand the reality, you shouldn't twist it.
Noumenon
1.1 / 5 (28) May 05, 2013
LOL, I've been over this with johan many times, [...] yet he continues to reject time dilation and does not use the Lorentz transformation properly to determine space-TIME [interval], which is invariant. He bores me.


Corrected post.
Whydening Gyre
1 / 5 (9) May 05, 2013
Acceleration is then an invariant.

So, it is the ACT of acceleration that is the constant, not necessarily the thing that is accelerating. The only way you can get acceleration is by applying energy.
Interesting dichotomy...


Not energy,,,, FORCE. Ya get an acceleration by applying a force.

You're right, I skipped a step. Can I say that we require energy to apply the force?
Fleetfoot
1 / 5 (1) May 06, 2013
Acceleration is then an invariant.

So, it is the ACT of acceleration that is the constant, not necessarily the thing that is accelerating. The only way you can get acceleration is by applying energy.
Interesting dichotomy...


Not energy,,,, FORCE. Ya get an acceleration by applying a force.

You're right, I skipped a step. Can I say that we require energy to apply the force?


Note though that Johan is talking about Galilean Invariance which is aether theory again, coordinate acceleration is not invariant in SR. If you are looking at basic Newtonian mechanics for low speeds:

Power = force * velocity.

and of course energy is the integral of power.

If you are looking for an invariant, you need proper acceleration:

http://en.wikiped...leration
johanfprins
1 / 5 (4) May 06, 2013
In that section I consider three synchronized clocks spaced a distance L/ apart passing by an observer with his own clock. When the middle clock coincides with the observer's clock, simultaneous events occur at all three passing clocks.


I have missed the following dumb-ass remarks by you: But it is NOT surprising since all your remarks are those of a dumb-ass

Not correct, because you can't assume that all three clocks experience the event at the same time.


Where did I say single even?: I am talking about three identical events occurring simultaneously at the positions of the three clocks.

There is absolutely no reason to assume that the other two clocks will agree on when the lightning struck.


Your lack of brain power is really becoming boring. When three events occur simultaneously at the three clocks then the three clocks at that instant in time MUST show the SAME time or else the events will not be simultaneous. Even my goldfish can understand this!
johanfprins
1 / 5 (5) May 06, 2013
[Invariant means that the function has the same value in different frames.


Provided that both frames have the same number of linearly independent dimensions. A coordinate transformation that preserves invariant distance-intervals CANNOT be done between manifolds which do not have linearly independent coordinates.

Minkowski space-time is such a manifold. It is thus totally stupid to even talk about invariant distances. in Minkowski space-time. It only proves that you do not understand elementary linear algebra.

Now you check that if you like (it's just linear algebra) or just look up the derivation in a textbook, I'll leave that to you. We'll be able to apply it to your question though.


If you think that there exists a linear coordinate transformations between manifolds which do are not spanned by linearly independent coordinates, then YOU URGENTLY need a course in linear algebra
johanfprins
1 / 5 (5) May 06, 2013
So you make the assumption that the two clocks cannot keep the same rate of time until the signal has been sent and received.


I am saying we don't know if they do or they don't.


Of course we know that they must be keeping the same TIME-RATE since the laws of physics is the same at any position within an IRF. You do not have to synchronize ANY clocks to show the same time within any IRF to derive the Lorentz equations. The only difference is that when you do not do this the Lorentz equations become far more complex. It is thus easier to do what Einstein did, and accept that since all the clocks within an IRF keep the same time-rate we can assume without changing the physics that they are also showing the exact same time.

Everyone can agree that if Fred and two clocks are at rest with each other, Fred is guaranteed to synchronize the two clocks in the way I described


But it is not necessary to do go to all this trouble.
johanfprins
1 / 5 (4) May 06, 2013
The clocks need not be synchronized for simultaneous events to occur at their positions, they only need to keep the same time-rate, which we know that they must do. For example, perfect clocks in New York do not show the same time as the same perfect clocks in South Africa; but this DOES not mean that two different events cannot occur simultaneously in South Africa and New York.

Both of us can now happily agree that Fred has, in his rest frame, synchronized the clocks?


But he need not have done this to ensure that two events can occur simultaneously at the positions of the two clocks; It can still happen even when the clock arms show different times on the two clocks: All that is required is that the clocks keep the same time rate ; and according to the postulates of relativity they must do so when they are identical perfect clocks within the same IRF.
johanfprins
1 / 5 (4) May 06, 2013
No need to use convoluted language.
Then why do YOU keep on using convoluted language?

Just picture two upright light clocks of the same height, one which is at rest, and one which is traveling.
I am picturing and have done this problem in http://www.cathod...ion.pdf. If you were not such a bigot and dumb-ass you would have look at my derivation and point out where I am, according to you wrong. This is how physics should be discussed; Not in terms of your dumb-ass presumptions.

The traveling one will tick at a lower rate because the photon has a longer path to travel.


WRONG: Within its own "moving" IRF it keeps time at EXACTLY the same time-rate as the "stationary" clock relative to which it is moving with a speed v. They keep time at exactly the same rate: As you will see when you look at my derivation in my manuscript. But of course I cannot expect YOU to have this humility: You are a bigot who have already decided that I must be wrong.
johanfprins
1 / 5 (5) May 06, 2013
You know, in case you actually end up being proven wrong, you should consider arguing with more humility, lest you end up eating crow--not that I think you'd ever admit you were wrong, or even attempt to question your own conclusions from time to time (which I do all the time, believe me).
No you do not! You are not even willing to study a manuscript lest it might force you to question your own conclusions. YOU will die claiming that there are not mountains on the moon.

You must know that all the claptrap that you are posting here I have also believed in at some stage and taught to my students. If I were like you who is TOTALLY unable to question my own conclusions and what I have been indoctrinated with, I would NOT have changed my mind, but would have like bigot (i.e YOU) defend the mainstream dogma come what may!
johanfprins
1 / 5 (4) May 06, 2013
I still don't understand where you think the Lorentz transformation came from--not that that is correct, since you actually derive it from time dilation and length contraction!


The Lorentz transformation was derived from the formula for "length-contraction" LOOOONG before Einstein postulated the Special Theory of Relativity. This derivation of the Lorentz transformation is based on the assumption that there IS a unique aether in which light moves.

When you derive the Lorentz transformation from Einstein's postulates, these equations are NOT derived in terms of "length-contraction" since the need for this concept is replaced by Einstein's two postulates. "Length-contraction" does not feature if there is NOT an aether-drag. When you Lorentz-transform the two ends of a stationary rod into an IRF relative to which the rod is moving, you will find that the rod actually becomes LONGER! See:
http://www.cathod...tion.pdf

johanfprins
1 / 5 (5) May 06, 2013
"s" is invariant if ya don't mix apples with oranges. "s" is world-line in spacetime.


Not in Minkowski's space-time: It is a line within a time-position graph which is just as valid when using Galilean-space and absolute time.

johan is trying force the variance by applying the maths of spacetime in a way they weren't intended.


Correct! Minkowski did not intend that his space-time should violate the most fundamental rules on which linear algebra MUST be based. But unfortunately it does!

He knows QT, but everyone knows ya can't get relativity and QT to place nice at all scales.


Precisely because they believe in claptrap like Minkowski space-time. In fact, Bohr's quantum rules, the de Broglie wavelength and the Schroedinger equation derive directly from the Special Theory of Relativity. The two theories dovetail completely with Schroedinger's equation: Not with Dirac's equation!

johanfprins
1 / 5 (5) May 06, 2013
Ok, let's clear something up. The clocks not being synchronized, doesn't mean that they don't keep the same rate. I am merely saying that if Fred sees the spatially separated clocks as showing the same time, Barney will see a constant time difference between them.


Very Good! I have NEVER argued with this.

But Barney will not see these "time differences" between these clocks at the same instant in time on his clock; as is implied by the concept of "time-dilation". He will see the events which occur simultaneously within Fred's IRF, and record that an event first at the clock coinciding with him and then at the clock that is moving away from him. These are two different times on Barney's clock NOT two different times on the moving clocks at the SAME instant in time on Barney's clock! Get it?

johanfprins
1 / 5 (5) May 06, 2013
johanfprins rejects Minkowski space-time, but in that context time dilation can be derived, as in this post I made last year,... dm/tm = moving IRF, ds/ts = stationary IRF,

dm = y(ds - v*ts) and ds = y(dm + v*tm) ,.... where y = gamma,...

Substitute the 2nd equation into the dm = y(y(dm + v*tm) - v*ts),...etc. etc.


Where did you use Minkowski space-time? As far as I can make out you used the Lorentz transformation. I know you are VERY slow in the uptake: So I repeat again: I have NEVER disagreed with the Lorentz transformation, only that it is NOT a linear transformation over Minkowski space. Neither have I disagreed that one can derive the so-called "time-dilation" formula:

t1=(gamma)*t2

What I am stating is that these two times are NOT simultaneously present on clock1 and clock2. And I have derived and proved above that this cannot be the case: And I have given a reference to a detailed manuscript: http://www.cathod...tion.pdf

Sheesh!
johanfprins
1 / 5 (3) May 06, 2013
@johanfprins
Again, I don't see how you don't see this from the Lorentz Transformation itself.
t' = gamma*(t-vx/c^2). x' = gamma*(x-vt).
So, let's do the math.
Let's say that for Fred, clock 1 reads t = 1 year, and clock 2 reads t = 1 year, and that they are 1 light year away from each other. Barney is traveling right at 0.1c, and passed Fred when both their clocks read 0s. etc., etc., etc.


I wish I could but I cannot follow this garbled nonsense. I have not disagreed that Barney will not experience two simultaneous events at the two clocks occurring at different times. So I just cannot figure out what your point is that you want to make!
johanfprins
1 / 5 (3) May 06, 2013
Maybe we do not know this,

YES!
but if this is not so Einstein's postulates must be wrong and so must his STR. Is that what you are arguing?

NO! You just admitted that maybe we do not know this.


AND I followed it up by stating that if the clocks do not keep the same time-rate Einstein's postulates must be wrong. Do you accept Einstein's postulates? If you do, you will not state that we do not know whether the two clocks keep the same time-rate.

Both of us can now happily agree that Fred has, in his rest frame, synchronized the clocks?


As I have already pointed out this step is not necessary for the argument you are trying to make!

johanfprins
1 / 5 (6) May 06, 2013
In STR it is assumed that the clocks are perfect and do not gain or lose time. Thus the synchronization has nothing to do with a clock being "wrong".

Well, if a real lock works properly, then it should reflect the actual passage of time, should it not? You accept that once set real clock correctly, then it should closely approximate an ideal clock, do you not? Once that real clock has been set it should always show what an ideal clock would read, should it not? Then, there's nothing to argue about!


This is irrelevant to the argument once you have accepted that when discussing STR it is assumed that the clocks NEVER gain or lose time.
johanfprins
1 / 5 (4) May 06, 2013
All clocks AT REST in the IRF must keep the same time in their respective IRF.


At last you have seen the point!


Also, (and this will really blow your mind...or maybe not) but if the situation were symmetric and Barney also had two synchronized clocks, 3 and 4, spaced 1 l yr apart at rest with him, Fred would think that 3 and 4 were unsynchronized!


I have NEVER disagreed with this: If two simultaneous events occur at two separate clocks stationary within the same IRF or even positions without clocks, these events will not be observed simultaneously within a passing IRF. So what are you REALLY trying to say?
johanfprins
1 / 5 (5) May 06, 2013
Nope, by Einstein's first postulate "The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translation motion relative to each other." Not a single word about time is here.


"undergo change" and "changes of state" can according to YOU occur without changes in time? LOL. Please go and play with your ducks in your bubble bath!

If you do want to understand the reality, you shouldn't twist it.


Even when the statement is not "twisted" you are still not able to understand it!
johanfprins
1 / 5 (4) May 06, 2013
Note though that Johan is talking about Galilean Invariance which is aether theory again, coordinate acceleration is not invariant in SR.


I have not stated that it is!

johanfprins
1 / 5 (6) May 06, 2013
Derivation of the "time-difference" formula: Two observers with clock1 and clock2 respectively passes one anther with speed v and synchronize their clocks.

After a time t2 on clock2 an event occurs at the position of this clock: Since this clock has moved with a speed v the distance between the clocks when the event occurs is according to observer 2 D2=v*t2.

Transform event to find what observer1 sees of this event:

Ob1 sees the event at a distance D1=(gamma)*(0+v*t2)=(gamma)*(v*t2).

What must be the time on clock1 when the observers have moved a distance D1 apart?

It must be t1=D1/v= {(gamma)*(v*t2)}/v=(gamma)*t2.

t1=(gamma)*t2 is what is interpreted in the literature to be the "time-dilation" formula: i.e. it is claimed that t1 and t2 are simultaneously the readings on clock1 and clock2 respectively.

However, as can be seen above t2 is the time on clock2 when the distance between the clocks is D2=v*t2 while t1 is the time on clock1 when the distance is D1=v*t1>D2.
johanfprins
1 / 5 (7) May 06, 2013
Thus, the times t1 and t2 cannot be simultaneously on the two clocks since t2 is only on clock2 when the distance is D1 and t2 is only on clock1 at a lter time after the distance between the two clocks became D1 which is larger than D2.

Does clock 2 keep slower time? Obviously not since for any distance D both the time t and t/ on both clocks must give the same distance D, so that

D=v*t=v*t/:

i.e. t=t/:

The clocks MUST this keep the EXACT same time after synchronization. This, in turn means that ALL clocks, no matter within which IRF they are or what speeds they are moving relative to one anther, MUST keep exactly the SAME time-rate!

QED.!
johanfprins
1 / 5 (8) May 06, 2013
@johan: It's evident, you're mental case.


YOU are the mental case; I have just now derived the "time-difference" formula correctly. It must be correct since you have not pointed out where I have made a mistake.But you rather choose to attack me personally. You are beneath contempt.

The purpose of this thread isn't flooding of discussion with nonsenses.
]

Then why have YOU been doing it under different aliases for years. And not just "nonsense" but absurd demented hallucinations based on a single mantra AWT, AWT, AWT, AWT!!!

You can never understand it with analysis of equations, until you have no physical insight into situation (which considers, you have drawn the picture of the whole situation).
.

If you cannot quantify your theory it is a useless theory; if one can call your infantile cartoons a theory: Which they are not and NEVER will be. Your cartoons are not "physical insights" into ANYTHING and also not a "picture of the whole situation".
Whydening Gyre
1 / 5 (9) May 06, 2013
The only "invariant" I am seeing here is - time, space, and acceleration are variant relative to each other.
Fleetfoot
1 / 5 (1) May 06, 2013
Question 1. Clock2 passes clock1 with speed v and at that instant they synchronize their readings: After a time t/ on clock2 how far is clock2 from clock1


In the coordinate system of clock1, and assuming that the positive x axis is aligned with the direction of motion of clock2 (hence y=z=constant), the distance between the clocks is given by:

x2(t1) = x1 + v * t1


We do not yet know what the time on clock1 is only the time on clock2, so why do you bring in the time on clock1.


I was rushing out and misread the question - my error. The location of x1 in the rest frame of clock2 is:

x1(t2) = x2 - v * t2

where:

x2 is the location of clock2 (a constant)

t2 is the time measured by clock2

x1(t2) is the location of clock1 at time t2

v is the speed of clock2 as measured in the frame of clock1 hence -v is the speed of clock1 as measured by clock2.

Speed is a measure of the angle between the worldlines hence has the same magnitude but opposite sign.
johanfprins
1.5 / 5 (8) May 06, 2013
I was rushing out and misread the question - my error. The location of x1 in the rest frame of clock2 is:

x1(t2) = x2 - v * t2

where:

x2 is the location of clock2 (a constant)


Owing to synchronisation this "constant" is zero since clock2 is at the origin of its rest frame.

t2 is the time measured by clock2


x1(t2) is the location of clock1 at time t2

v is the speed of clock2 as measured in the frame of clock1 hence -v is the speed of clock1 as measured by clock2.

Speed is a measure of the angle between the worldlines hence has the same magnitude but opposite sign.

So far so good!

Have you read the rest of the derivation above? You just now get the Lorentz transformed distance of clock2 from clock 1 etc.

In either case I have given the derivation above: Can you see that the the expression t1=(gamma)*t2 gives the time t1 (on both clocks) when the event occurs at clock2, and t2 is a later time(on both clocks) when the event is recorded by clock1?
ValeriaT
1 / 5 (8) May 06, 2013
YOU are the mental case; I have just now derived the "time-difference" formula correctly. ... ALL clocks, no matter within which IRF they are or what speeds they are moving relative to one anther, MUST keep exactly the SAME time-rate!
This is apparently not correct time-difference, as the time dilatation is continuously verified in every GPS clock many thousand-times per day. In addition, the velocity time dilatation has been verified with Hafele-Keaton, Ives and Stilwell, Rossi and Hall, Hasselkamp, Mondry, and Scharmann and others at speeds less than 10 meters per second with using optical atomic clocks connected by 75 meters of optical fiber. You cannot beat such a pile of experimental evidence: it serves as an evidence of your senility instead.
Richard Feynman: "It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong".
Q-Star
2.4 / 5 (14) May 06, 2013
Acceleration is then an invariant.

So, it is the ACT of acceleration that is the constant, not necessarily the thing that is accelerating. The only way you can get acceleration is by applying energy.
Interesting dichotomy...


Not energy,,,, FORCE. Ya get an acceleration by applying a force.

You're right, I skipped a step. Can I say that we require energy to apply the force?


It would be more correct to say the force transfers energy.
Fleetfoot
3.7 / 5 (3) May 06, 2013
s=0 is not a distance.


It must be the "distance of the origin to itself" ...


For s>=0 It is not a distance, it is a time. For example, a rocket coasts at 0.8c between space stations near Earth and Alpha Centauri. They are exactly 4 light years apart so the journey in Earth coordinates takes 5 years. The interval is timelike so:

s^2 = 5^2 - 4^2

s=3

That means the captain looking at the ship's clock will see a journey time of 3 years, s is ship's time. That value is invariant, it doesn't matter who looks at the ship's clock, they will see it register the same time for the trip.

If the ship's speed were higher, the duration would be less and for a photon traveling at the speed of light, the trip time is zero. What it means is that a photon travels in "suspended animation". Since the duration is zero at all points along its path, simply stating s=0 doesn't identify a single location, it is a parametric equation describing the whole path.
Noumenon
1.1 / 5 (28) May 06, 2013
If you think that there exists a linear coordinate transformations between manifolds which do are not spanned by linearly independent coordinates, then YOU URGENTLY need a course in linear algebra


All four axis are validly orthogonal, and linearly independent in Minkowski space-time. What is wrong with the notion of hyperbolic orthogonality?
Noumenon
1.1 / 5 (28) May 06, 2013
johanfprins rejects Minkowski space-time, but in that context time dilation can be derived, as in this post I made last year,... dm/tm = moving IRF, ds/ts = stationary IRF,

dm = y(ds - v*ts) and ds = y(dm + v*tm) ,.... where y = gamma,...

Substitute the 2nd equation into the dm = y(y(dm + v*tm) - v*ts),...etc. etc.


Where did you use Minkowski space-time? As far as I can make out you used the Lorentz transformation. [...] I have NEVER disagreed with the Lorentz transformation, only that it is NOT a linear transformation over Minkowski space. Neither have I disagreed that one can derive the so-called "time-dilation" formula:


But I just showed you that the two times will be different implying the clocks ran at different rates, though they were once synchronized.

All four axis are validly orthogonal, and linearly independent in Minkowski space-time. What is wrong with the notion of hyperbolic orthogonality? I may read you write-up if i get time.
thefurlong
1 / 5 (11) May 06, 2013

I wish I could but I cannot follow this garbled nonsense.

What couldn't you follow?

I have not disagreed that Barney will not experience two simultaneous events at the two clocks occurring at different times. So I just cannot figure out what your point is that you want to make!

The feeling's mutual. First, you say that clocks 1 and 2 won't be synchronized according to Barney. Then you agree that, yes Barney sees them unsynchronized. So, which is it?

Anyway, since you now admit that Barney doesn't measure them to be synchronized, let's return to the example you gave involving 3 synchronized clocks. I will address this in my next comment.
thefurlong
1.3 / 5 (12) May 06, 2013
You said,

I consider three synchronized clocks spaced a distance L/ apart passing by an observer with his own clock. When the middle clock coincides with the observer's clock, simultaneous events occur at all three passing clocks. Thus the event at the middle clock occurs at the coincident position and coincident time t=t/=0. According to the observer, the simultaneous event on the clock which has already passed him occurs at a distance L=(gam)*L/ from him at the time t=(v/c^2)*(gam)*L/. Which IS NOT your time dilation formula.

Now, for whom are they simultaneous? The observer at rest with the clocks or the one to whom they are moving? I just showed that they can't be simultaneous for both if they occupy different positions.

I am still not reading your 35 page paper. You can't seriously expect me to want to read it when:
1) I currently have no reason to believe that it is is grounded in correct assumptions
2) I have no vested interest in it
3) You constantly insult me
Fleetfoot
3.7 / 5 (3) May 06, 2013
I just showed you that the two times will be different implying the clocks ran at different rates, though they were once synchronized.


The clocks produce the same number of ticks per unit of proper time, hence run at the "same rate" even though their accumulated times differ.

Try the "other" twins paradox: Two twin brothers walk towards a wood. They have identical stride lengths. One goes through the wood, the other goes round, each counting how many steps he takes. When they meet at the other side, the brother one went around the wood has taken more steps.

In spacetime, the Riemann geometry means the curved path is shorter rather than longer, but otherwise the explanation is the same, the clocks in the relativistic Twins Paradox show different total times because the path lengths differ even though the time between ticks is the same for both clocks, i.e. they run at the same rate.
Noumenon
1.1 / 5 (28) May 06, 2013
I just showed you that the two times will be different implying the clocks ran at different rates, though they were once synchronized.


The clocks produce the same number of ticks per unit of proper time, hence run at the "same rate" even though their accumulated times differ.


I was speaking with reference to an Observer and so coordinate time, but good post anyway. ACCORDING to one Observer and his clock, it may be that another relatively moving observer's clock is measured to tick at a different rate.
Whydening Gyre
1 / 5 (9) May 06, 2013
But I just showed you that the two times will be different implying the clocks ran at different rates, though they were once synchronized.


I see it all so clearly now... Dave Bowman - 2001: A space Odyssey
Whydening Gyre
1 / 5 (9) May 06, 2013
I'm done here, now. I humbly thank ALL of you for your contributions to my new Understanding. I truly wish all of you the very best in all your future endeavors. I WILL be watching...:-)
Plus 1.
johanfprins
1 / 5 (6) May 06, 2013
All four axis are validly orthogonal, and linearly independent in Minkowski space-time. What is wrong with the notion of hyperbolic orthogonality?


They are not! If you knew any linear algebra , which you are obviously too stupid to know, you will know that four axes that are "validly orthogonal" cannot have any point with s=0 UNLESS x=0, y=0, z=0 and u=0. This means that a coordinate point for which not all the coordinates are zero, but which has that s^2=x^2+y^2+z^2+u^2=0, cannot be a point in such an orthogonal space EVER!!!!!!

There is no thing like "hyperbolic orthogonality" except within a demented mind.

Furthermore the transformation matrix for the Lorentz transformation IS NOT a 4x4 matrix, as it would be if all four coordinates were linearly independent, but ONLY a 2x2 matrix.
It ONLY changes the time and the coordinates along the direction of motion NOT the coordinates perpendicular to the direction of motion.

Signing off for the day!
Q-Star
2.4 / 5 (14) May 06, 2013
There is no thing like "hyperbolic orthogonality" except within a demented mind.


If ya really believe that, then maybe ya shouldn't be talking about relativity. Or plane geometry. Or Minkowski spacetime. Ya realize that ya called 99.99999999 % of all physicists demented? Not to mention 100 % of all mathematicians ya just called demented.

Ya took a giant leap out of your comfort zone as an "expert" on physics with that simple statement.
Noumenon
1.4 / 5 (30) May 06, 2013
four axes that are "validly orthogonal" cannot have any point with s=0 UNLESS x=0, y=0, z=0 and u=0. This means that a coordinate point for which not all the coordinates are zero, but which has that s^2=x^2+y^2+z^2+u^2=0, cannot be a point in such an orthogonal space

There is no thing like "hyperbolic orthogonality" except within a demented mind.


"Since Hermann Minkowski's foundation for spacetime, the concept of points in a spacetime plane being hyperbolic-orthogonal to a timeline (tangent to a World line) has been used to define simultaneity of events relative to the timeline."

Two vectors, x, y, z, t and x', y', z', t', are normal (hyperbolic orthogonal), when
-c²tt' + xx' + yy' + zz' = 0. Notice here the Lorentz signature {-1, 1, 1, 1} which is implicit in the Minkowski metric. So the interval s² = Δx² + Δy² + Δz² - c²Δt² .
johanfprins
1 / 5 (5) May 07, 2013
But I just showed you that the two times will be different implying the clocks ran at different rates, though they were once synchronized.


And I just showed you TWICE on this thread that the two times are registered on the clocks when they are at TWO different distances from one another . Obviously they cannot be simultaneously at two different distances from one another, so the two times on the clocks are also not simultaneously on the two clocks, but are the exact times you expect for these distances when two clocks KEEPING THE SAME time are moving apart at a speed v.

Even my grandson in grade 7 could derive this: Why can YOU not? The only reason, except being an absolute moron, is that you do not WANT to know that this is so. Just as the Cardinal did not WANT to see mountains on the moon when he looked through Galileo's telescope.

you refuse to have an open mind and to try and point out a mistake in my derivation, which I am sure you would have if you could have!
johanfprins
1 / 5 (5) May 07, 2013
The feeling's mutual. First, you say that clocks 1 and 2 won't be synchronized according to Barney. Then you agree that, yes Barney sees them unsynchronized. So, which is it?


The fact that Barney sees the simultaneous events at the two clocks at different times does not mean that the clocks are not synchronized. Even when the Lorentz transformation is not valid, two events occurring simultaneously at two clocks will not be observed by Barney when the distances between Barney and the two clocks are not exactly the same. If Barney is stupid, he will conclude that the two clocks are keeping different times. The Lorentz transformation transforms ALSO distances to the clocks, and THEREFORE Barney can only see what happens at the two clocks at different times: IT DOES NOT MEAN THAT THE TWO CLOCKS ARE NOT KEEPING THE EXACT SAME TIME! WHICH IS WHAT THEY ACTUALLY DO.

johanfprins
1 / 5 (5) May 07, 2013
Now, for whom are they simultaneous? The observer at rest with the clocks or the one to whom they are moving? I just showed that they can't be simultaneous for both if they occupy different positions.


It is an a priori assumption that different events can occur simultaneously at different positions within the same IRF. Einstein used this assumption to model non-simultaneity of simultaneous events

When events at different positions occur simultaneously, the time at each position must be the same: So one can assume that there could be three clocks at these positions which must show the same time.

You are thus deluded when you claim that three such clocks cannot be simultaneous for the observer relative to which they are stationary. If they cannot be simultaneous, there cannot be simultaneous events: Simultaneity can then not exist, also not for different times on two clocks moving relative to one another so that there can be "time-dilation"; as you argue that there must be.
johanfprins
1 / 5 (5) May 07, 2013
I am still not reading your 35 page paper. You can't seriously expect me to want to read it when:
1) I currently have no reason to believe that it is is grounded in correct assumptions


An honest man with integrity will not come to such a conclusion before studying the manuscript. Physics IS NOT based on what one wants to believe, as you obviously reason that it must be.

2) I have no vested interest in it


Then why are you posting on this thread?

3) You constantly insult me


Your comments make it impossible to respond without insulting you. You should not cry when you constantly come out of your corner leading with your chin.

johanfprins
1 / 5 (5) May 07, 2013
The clocks produce the same number of ticks per unit of proper time, hence run at the "same rate" even though their accumulated times differ.


Why do you bring in MST and the concept of "proper time" if I derived the result directly from the Lorentz transformation? Or are you claiming that the Lorentz equations are wrong when they give a result that differs from MST and "proper time"?

In spacetime, the Riemann geometry means the curved path is shorter rather than longer, but otherwise the explanation is the same, the clocks in the relativistic Twins Paradox show different total times because the path lengths differ even though the time between ticks is the same for both clocks, i.e. they run at the same rate.


My derivation does NOT require Riemann geometry: It is derived directly from the Lorentz equations, which prove, as I have derived above, that the clocks MUST keep the SAME time-rate. If I am wrong then point out WHERE I have gone wrong in my derivation.

johanfprins
1 / 5 (3) May 07, 2013
If ya really believe that, then maybe ya ..


The correct spelling is "you".

Ya realize that ya called 99.99999999 % of all physicists demented? Not to mention 100 % of all mathematicians ya just called demented.


Is it impossible that they could all have been wrong? Only a fool will judge physics and mathematics in this way. When Ptolemy's model held sway the physicists and mathematicians were all wrong for more than 1000 years.

Ya took a giant leap out of your comfort zone as an "expert" on physics with that simple statement.


I know: But this is what a real scientist is expected to do. New paradigms can only be created by scientists who are willing to jump out of their comfort zones. This is what physics research is all about. It has become a major problem that modern "expert scientists" doggedly, against all reason, defend mainstream dogma in order to remain within their comfort zones.

And you are all yapping along in this thread!

johanfprins
1 / 5 (3) May 07, 2013
"Since Hermann Minkowski's foundation for spacetime, the concept of points in a spacetime plane being http://en.wikiped...gonality to a timeline (tangent to a World line) has been used to define simultaneity of events relative to the timeline."


And this is clearly in violation of the mathematics of linear spaces; and has also been the reason why relativity and quantum mechanics have not been and cannot be reconciled. All you need to understand Special Relativity and how it relates to quantum mechanics is the Lorentz transformation: Nothing else. You do not need a "time-line" or Minkowski space to "define" simultaneous events.

Two vectors, x, y, z, t and x', y', z', t', are normal (hyperbolic orthogonal), when
-c²tt' + xx' + yy' + zz' = 0.


Only when they are vectors within a space with linearly-independent coordinates. x,y,z and ict ARE NOT linearly independent coordinates as is mathematically DEMANDED!!
johanfprins
1 / 5 (6) May 07, 2013
Event at "stationary" clock:

Consider again clock2 passing clock1 being synchronized: But let now an event occur at the "stationary" clock1 at time t1. According to clock1, clock2 has moved a distance D1=v*t1 from it.

An observer with clock2 will see the event occurring at a distance D2 from clock2 which is given by the Lorentz transformation as:

D2=-(gamma)(0-v*t1)=(gamma)*(v*t1)=(gamma)*D1

The time t2 that the clocks must move apart to be at this distance from one another must be:

t2=(distance)/v=(gamma)*t1.

Thus according to "time-dilation" it is now the stationary clock that keeps slower time. It is clearly absurd. The difference in times is determined by whether the event occurs at clock1 or at clock2.

If it occurs at clock2 one has that t1>t2; and when it occurs at clock1, one has that t2>t1. It has NOTHING to do with which clock is moving and which clock is stationary: And has nothing to do with one clock keeping time at a slower rate than the other clock.
Noumenon
1.1 / 5 (28) May 07, 2013
...Thus according to "time-dilation" it is now the stationary clock that keeps slower time. It is clearly absurd. [...] If it occurs at clock2 one has that t1>t2; and when it occurs at clock1, one has that t2>t1. It has NOTHING to do with which clock is moving and which clock is stationary:


Of course there is no 'stationary' and 'moving' as they are relative, each see's the other as moving, correct. Each observer measures the other observer's clock to be ticking time at a different rate than his own,.... the difference is Symmetrical between the two observers.

Since johanfprins has no Special IRF to judge this as absurd, that conclusion is invalid.

The physical effects of time dilation can be observed once the observers and their clocks are compared, but only when the conditions are asymmetrical, which requires that one of the observers changes his IRF (by which I mean accelerates), on return to the other to make the comparison.
Noumenon
1.1 / 5 (28) May 07, 2013
Of course, during observation of remote clock their time is delayed, but this is not what the velocity driven time dilatation is about [..]You're apparently confusing the delay of clock induced with their distance from observer with delay of clock induced with their relative motion with respect to observer.


Exactly.
Fleetfoot
3.7 / 5 (3) May 07, 2013
The clocks produce the same number of ticks per unit of proper time, hence run at the "same rate" even though their accumulated times differ.


Why do you bring in MST and the concept of "proper time" if I derived the result directly from the Lorentz transformation?


I mentioned it because there are two meanings being used for "time" in the various replies and "Noumenon" had not identified which applied. He and I knew fine which he meant but there are some others following the conversation for whom it may not have been clear. You are not the only person reading this.

Or are you claiming that the Lorentz equations are wrong when they give a result that differs from MST and "proper time"?


The Lorentz transforms and the Minkowski Metric are both correct, your idea that having an invariant interval of zero defines a point at the origin is what is wrong as I have pointed out a couple of times already. There may be other errors which I'll point out as I find them.
johanfprins
1 / 5 (7) May 07, 2013

The Lorentz transforms and the Minkowski Metric are both correct,


They cannot be since, as I have proved impeccably-correctly above (by sing the Lorentz-transformation) that all clocks within a gravity free universe MUST keep time at exactly the same rate. According to MST, there is a so-called "proper" time which is not required when all the clocks keep the SAME time. So either the Lorentz-transformation must be wrong (which I doubt) or the MST must be wrong, which is obvious that it must be according to the fundamental rules of linear spaces with linearly-independent coordinates.

your idea that having an invariant interval of zero defines a point at the origin is what is wrong as I have pointed out a couple of times already.


I do not know why you harp on this aspect since my derivation of the "time-difference" formula (what you claim is "time-dilation") does not make use of this concept.
johanfprins
1 / 5 (6) May 07, 2013
All you have to do is thus to show me where I have made a mistake in my derivations of the "time-distance" formula above; which you avoid doing since you know that you cannot, but are too much of a fundamentalist bigot to acknowledge this.

There may be other errors which I'll point out as I find them.


There are no errors in my derivation of the "time-interval" formula. If there are you should be able to see such an error right away!

Is it so difficult to see the mountains on the moon? Or have you published further nonsense by using the MST and are not willing to admit that you can be wrong? Tsk! tsk.
thefurlong
1.4 / 5 (10) May 07, 2013
Even when the Lorentz transformation is not valid, two events occurring simultaneously at two clocks will not be observed by Barney when the distances between Barney and the two clocks are not exactly the same. If Barney is stupid, he will conclude that the two clocks are keeping different times.

That's why I keep distinguishing between MEASURING and WITNESSING. It's simple math that Barney will always witness the two events at different times in galilean relativity. However, as I said, even when he accounts for his relative velocity and the speed of light, he will still MEASURE them to be at different times in SR. That doesn't stop him from doing some math and concluding that Fred measures them to be synchronized, if he knows anything about relativity.
But then, what do we have? Two different experiences of time. Why should we assume that one experience of time is more valid than the other?


Q-Star
2.5 / 5 (13) May 07, 2013
If ya really believe that, then maybe ya ..


The correct spelling is "you".


Ya realize that ya called 99.99999999 % of all physicists demented? Not to mention 100 % of all mathematicians ya just called demented.


So even though ya are proud of being in the 0.0000000001 % of physicists, and in the 0.0 % of all mathematicians, ya are branching out and now giving spelling tuition? A true polymath indeed Sir.

Is it impossible that they could all have been wrong? Only a fool will judge physics and mathematics in this way. When Ptolemy's model held sway the physicists and mathematicians were all wrong for more than 1000 years.


For someone who casts around words like "hallucinations", "demented", "delusional", "idiot", "moron", and who would have us believe he is also an expert with word meanings and proper spelling, maybe ya could help me with something?

With your 0.00000001 % wisdom, what does "delusions of grandeur" mean?

thefurlong
1 / 5 (9) May 07, 2013

An honest man with integrity will not come to such a conclusion before studying the manuscript. Physics IS NOT based on what one wants to believe, as you obviously reason that it must be.

I DID start studying your manuscipt and found that it made bad assumptions.

2) I have no vested interest in it


Then why are you posting on this thread?


I meant that I have no vested interest in reading your paper, you child! Jesus Christ!


Your comments make it impossible to respond without insulting you. You should not cry when you constantly come out of your corner leading with your chin.


LOL!
johanfprins
1 / 5 (8) May 07, 2013
I DID start studying your manuscipt and found that it made bad assumptions.


Name them! You are a liar an a fraud!
thefurlong
1 / 5 (9) May 07, 2013

You are thus deluded when you claim that three such clocks cannot be simultaneous for the observer relative to which they are stationary. If they cannot be simultaneous, there cannot be simultaneous events: Simultaneity can then not exist, also not for different times on two clocks moving relative to one another so that there can be "time-dilation"; as you argue that there must be.

You didn't answer my question. Are they simultaneous for the person who is moving relative to the clocks, or not?
johanfprins
1 / 5 (6) May 07, 2013
You didn't answer my question. Are they simultaneous for the person who is moving relative to the clocks, or not?


I have answered that in detail above: Two simultaneous events on the clocks, which keep identical time relative to the observer travelling with the clocks, will be perceived, by the observer moving relative to the clocks, to occur at two different times on his own clock.

This obviously does not mean that the time on the moving observer's clock and the other clocks show simultaneously different times at the instant in time when the simultaneous events occur at the position of the two clocks; which are at that instant showing identically the same time.
Noumenon
1.4 / 5 (30) May 07, 2013
Two simultaneous events on the clocks, which keep identical time relative to the observer travelling with the clocks, will be perceived, by the observer moving relative to the clocks, to occur at two different times on his own clock.


Correct.

This obviously does not mean that the time on the moving observer's clock and the other clocks show simultaneously different times "AT THE INSTANT IN TIME"
when the simultaneous events occur at the position of the two clocks; which are "AT THAT INSTANT" showing identically the same time.


I have added caps in your quote above for reference below,...

At what "at the instant in time" are you referring to here?! Gods special Universal omnipotent Clock?! The speed of the signal used in observation by the moving observer, is already taken into account in STR.

Operationally, there are ONLY the two observers clocks. That's it. There is NO universal reference for Time, period.
Noumenon
1.3 / 5 (29) May 07, 2013
Two vectors, x, y, z, t and x', y', z', t', are normal (hyperbolic orthogonal), when
-c²tt' + xx' + yy' + zz' = 0.


Only when they are vectors within a space with linearly-independent coordinates. x,y,z and ict ARE NOT linearly independent coordinates as is mathematically DEMANDED!!


That is what hyperbolic orthogonal implies here. If the above two vectors are "perpendicular" to each other, their inner product will be zero. Associating √-1 (i) with the time basis axis is not necessary in STR.

You're simply factually wrong. Please review this history on the matter.
Fleetfoot
1 / 5 (1) May 07, 2013
The Lorentz transforms and the Minkowski Metric are both correct,


They cannot be since, as I have proved impeccably-correctly above (by sing the Lorentz-transformation)


I must have missed that but I've missed dozens of posts as I've been out most of the weekend.

that all clocks within a gravity free universe MUST keep time at exactly the same rate. According to MST, there is a so-called "proper" time which is not required when all the clocks keep the SAME time. So either the Lorentz-transformation must be wrong (which I doubt)


Proper time is just the integral of Lorentz-transformed coordinate time increments so either both are right or both are wrong. Personally I agree that the Lorentz Transforms are valid so your proof must be flawed. I'll look back over the thread and see if I can find it.
Fleetfoot
3.7 / 5 (3) May 07, 2013
You didn't answer my question. Are they simultaneous for the person who is moving relative to the clocks, or not?


I have answered that in detail above: Two simultaneous events on the clocks, which keep identical time relative to the observer travelling with the clocks, will be perceived, by the observer moving relative to the clocks, to occur at two different times on his own clock.


Right, and if those clocks were previously synchronised when co-located, then they must have run at different (coordinate) rates as judged by that observer.

This obviously does not mean that the time on the moving observer's clock and the other clocks show simultaneously different times at the instant in time when the simultaneous events occur at the position of the two clocks; which are at that instant showing identically the same time.


It obviously does, you have to ask "at the instant in time" as measured by which clock?
Noumenon
1.4 / 5 (30) May 07, 2013
Euclidean geometry doesn't work for spacetime.


In SR, it works for any spatial foliation, i.e. a 3D slice perpendicular to any given time axis. Remember in the absence of gravity, there is no curvature.


Correct, as usual,,, but only within an inertial reference frame. If two observers are in separate reference frames, then Euclid fails, whether or not they are accelerating (or in the presense of a gravitational mass) or not.

We know that Minkowski spacetime was one of the "AH HAH" moments while Einstein was working on Special Relativity. Along with the other big "AH HAH" moment,, "c" is the velocity of all things through spacetime regardless of their relative velocities through space only.


Minkowski space-time is a pseudo-Euclidean space, which means it has a Lorentz metric signature {-1, 1, 1, 1} or {1, -1, -1, -1}. This seems to be what johanfprins ignores.
thefurlong
1 / 5 (9) May 07, 2013
I have answered that in detail above: Two simultaneous events on the clocks, which keep identical time relative to the observer travelling with the clocks, will be perceived, by the observer moving relative to the clocks, to occur at two different times on his own clock.

You use the word "perceived". Does that mean that you think that the moving observer is incorrect, and the stationary observer is correct?

This obviously does not mean that the time on the moving observer's clock and the other clocks show simultaneously different times at the instant in time when the simultaneous events occur at the position of the two clocks; which are at that instant showing identically the same time.

"at the instant" in whose time? The stationary observer's or the moving observer's?
Fleetfoot
3.7 / 5 (3) May 07, 2013
Minkowski space-time is a pseudo-Euclidean space, which means it has a Lorentz metric signature {-1, 1, 1, 1} or {1, -1, -1, -1}. This seems to be what johanfprins ignores.


I think he is locked in to Galilean Invariance and therefore needs to presume an unobservable absolute time but hasn't realised how strongly that implies a Lorentz-style aether.
johanfprins
1 / 5 (7) May 07, 2013

At what "at the instant in time" are you referring to here?!


The simultaneous instant in time when the two events occur simultaneously: Or are you arguing that two simultaneous events do not occur at the same instant in time? If they do not then PLEASE define simultaneous for me.

Gods special Universal omnipotent Clock?!


This is, as I have proved above, what the Lorentz-transformation demands: Or are you saying that the Lorentz transformation is wrong?

The signal used in observation by the moving observer, is already taken into account in STR.


What you mean by this statement only YOUR demented mind can understand.

Operationally, there are ONLY the two observers clocks. That's it. There is NO universal reference for Time, period.


WRONG: The Lorentz-transformation DEMANDS a universal reference for time as my derivations above impeccably prove. Show me where my derivations are wrong before making such a stupid statement!
Fleetfoot
2.3 / 5 (3) May 07, 2013
So far so good!

Have you read the rest of the derivation above? You just now get the Lorentz transformed distance of clock2 from clock 1 etc.

In either case I have given the derivation above: Can you see that the the expression t1=(gamma)*t2 gives the time t1 (on both clocks) when the event occurs at clock2, and t2 is a later time(on both clocks) when the event is recorded by clock1?


I've skimmed over it but while the maths is trivial, I don't see the point you're trying to make. Let A be an event on the worldline of clock2. The coordinates in that frame are (0, tA2). To find the distance, you need the coordinates of clock1 at the same time in the rest frame of clock2. If that is event B, tB2=tA2 so they are (-v.tA2, tA2). You then ask at what time the distance has that value which is obviously tA2 so to say they are the same is a tautology hence I don't see your point.

What you next need to consider, as mentioned in later posts, are the event times measured by clock1.
johanfprins
1 / 5 (6) May 07, 2013
That is what hyperbolic orthogonal implies here.


The term "hyperbolic orthogonal" is not just an oxymoron but also a contradiction in terms. Like "military intelligence"?

If the above two vectors are "perpendicular" to each other, their inner product will be zero. Associating ��š-1 (i) with the time basis axis is not necessary in STR.


You cannot have an "inner product" unless the coordinates are linearly independent, and the coordinates x,y,z and ict ARE NOT linearly independent!

You're simply factually wrong. Please review http://en.wikiped...#History on the matter.


Why do I have to review the claptrap that I taught my students when I did not know better?
Fleetfoot
3 / 5 (2) May 07, 2013
At what "at the instant in time" are you referring to here?!


The simultaneous instant in time when the two events occur simultaneously: Or are you arguing that two simultaneous events do not occur at the same instant in time? If they do not then PLEASE define simultaneous for me.


Two events are simultaneous if the occur at the same time as measured by two synchronised clocks which are mutually at rest. In your example, events which are simultaneous as determined by a set of clocks synchronised to clock1 will not be simultaneous as determined by a set of clocks synchronised to clock2 and vice versa.

You can derive that directly from the Lorentz transforms as shown here:

http://en.wikiped...rmations

The geometric equivalent is shown as an animation at the top of the same page.
johanfprins
1 / 5 (6) May 07, 2013
Proper time is just the integral of Lorentz-transformed coordinate time increments so either both are right or both are wrong. Personally I agree that the Lorentz Transforms are valid so your proof must be flawed. I'll look back over the thread and see if I can find it.


Good: Find the flaw and post it! In contrast to your bigotted attitude I will be happy to be proved wrong. But if you cannot find the flaw, will you have the integrity to accept that you are wrong? Or to make it easier for you: That you "might be" wrong?

I do not think that you have the honesty, integrity and courage to ever admit that you might be wrong or are wrong! You are a fundamentalist bigot, and NOTHING will EVER change that!
ValeriaT
2.2 / 5 (10) May 07, 2013
The Lorentz-transformation DEMANDS a universal reference for time as my derivations above impeccably prove
You mean a classical aether of preEinsteinian era?
johanfprins
1 / 5 (5) May 07, 2013
Right, and if those clocks were previously synchronised when co-located, then they must have run at different (coordinate) rates as judged by that observer.


Why? Perfect clocks in SA and New York are not synchronized but they still run at the SAME rate.

It obviously does, you have to ask "at the instant in time" as measured by which clock?


By the two clocks at which the two events occur simultaneously: Or do you believe that the two clocks cannot simultaneously show the same time when the two events at their positions are simultaneous? And since one of the two clocks coincide with the clock within the IRF relative to which the two clocks are moving, the clock relative to which they are moving, MUST also exist at the same time.
Q-Star
2.3 / 5 (12) May 07, 2013
Right, and if those clocks were previously synchronised when co-located, then they must have run at different (coordinate) rates as judged by that observer.


Why? Perfect clocks in SA and New York are not synchronized but they still run at the SAME rate.


Ya are starting to sound like the AWT guy and the Plasma, blah, blah, blah Plasma guy.
johanfprins
1 / 5 (5) May 07, 2013
Minkowski space-time is a pseudo-Euclidean space, which means it has a Lorentz metric signature {-1, 1, 1, 1} or {1, -1, -1, -1}. This seems to be what johanfprins ignores.


I am not ignoring it: I am just simply stating the incontrovertible mathematical fact that a "space" with such a metric does not have linearly independent coordinates and therefore one cannot get linear transformations of coordinates from one "space" with this metric into another space within such a metric. Simply stated, One cannot use the mathematics of linear spaces on such a "space" as Minkowski has done. It is complete nonsense!
Q-Star
2.5 / 5 (13) May 07, 2013
I will be happy to be proved wrong.


Somehow that beggars my credulity.

Or to make it easier for you: That you "might be" wrong?


I "might be" wrong, but I suspect the only time ya have ever been wrong is when ya disagreed with yourself. Out of your many, very many, thousands of posts, I've yet to see one where ya were wrong, except the ones where ya contradict your own self sure.
ValeriaT
1.9 / 5 (9) May 07, 2013
Ya are starting to sound like the AWT guy
I'd hardly say, that desynchronized clock run at the same rate...
johanfprins
1 / 5 (6) May 07, 2013
You use the word "perceived". Does that mean that you think that the moving observer is incorrect, and the stationary observer is correct?


If you know anything about relativity you will know that a person on the ground will "perceive" a bomb to follow a parabolic path while a person on the plane will "perceive" the bomb to follow a linear path. Both paths are real within their respective IRF's.

"at the instant" in whose time? The stationary observer's or the moving observer's?

Both. The events occur simultaneously on all clocks. The moving observer only record these events at different times on his clock. This has NOTHING to do with the clocks keeping different times: At any instantaneous instant in time all the clocks within a gravity-free universe show exactly the same time. The fact that the observer, which are moving relative to the simultaneous events, sees these events at different times does no change the fact that all the clocks keep the exact time!
Q-Star
2.3 / 5 (12) May 07, 2013
Ya are starting to sound like the AWT guy
I'd hardly say, that desynchronized clock run at the same rate...


No I don't think ya Zeph would, today and yesterday, ya have made a lot more sense than this fellow. (And ya are always more civil than he is.)
johanfprins
1 / 5 (6) May 07, 2013
I am signing off for the night!
thefurlong
1.4 / 5 (10) May 07, 2013
If you know anything about relativity you will know that a person on the ground will "perceive" a bomb to follow a parabolic path while a person on the plane will "perceive" the bomb to follow a linear path. Both paths are real within their respective IRF's.

Agreed.

Both. The events occur simultaneously on all clocks. The moving observer only record these events at different times on his clock. This has NOTHING to do with the clocks keeping different times: At any instantaneous instant in time all the clocks within a gravity-free universe show exactly the same time.

I'm going to go out on a limb here and say that you think that the universe has some absolute time, t. Is that correct?
Fleetfoot
1 / 5 (1) May 07, 2013
Right, and if those clocks were previously synchronised when co-located, then they must have run at different (coordinate) rates as judged by that observer.


Why? Perfect clocks in SA and New York are not synchronized but they still run at the SAME rate.


They can do, and if they show "15:00" and "16:00" at the same time (whatever that means) then they could still be running at the same rate, one is just an hour ahead. However, if they moved to those locations at the same speed from a location halfway between, and when they were co-located, they were set to show the same time (say "09:00"), then surely you would agree the one now showing "16:00" must be running slower.

In your example, you set t=0 on both clocks as they momentarily passed. You need to ask question 4: In your example, according to the Lorentz Transforms, what time does clock1 show when the time is t/ on clock2?

Then add question 5: Repeat questions 1 to 4 but exchange the roles of the clocks.
Fleetfoot
1 / 5 (1) May 07, 2013
It obviously does, you have to ask "at the instant in time" as measured by which clock?


By the two clocks at which the two events occur simultaneously: Or do you believe that the two clocks cannot simultaneously show the same time when the two events at their positions are simultaneous? And since one of the two clocks coincide with the clock within the IRF relative to which the two clocks are moving, the clock relative to which they are moving, MUST also exist at the same time.


That seems completely muddled, it sounds as though you are describing three clocks when there were only two in your questions, and if you use the Lorentz transforms, you will find they didn't show the same time simultaneously. Your proof only showed clock2 showed the same value as itself, you never calculated the value of clock1.
Fleetfoot
1 / 5 (1) May 07, 2013
Proper time is just the integral of Lorentz-transformed coordinate time increments so either both are right or both are wrong. Personally I agree that the Lorentz Transforms are valid so your proof must be flawed. I'll look back over the thread and see if I can find it.


Good: Find the flaw and post it!


Well if I found the "proof" in the form of your three questions, I already posted the error, you didn't calculate the time according to clock1 so your answers were a tautology, clock2=clock2. However, I may have missed some other post, if so just point me at it.

if you cannot find the flaw, will you have the integrity to accept that you are wrong? Or to make it easier for you: That you "might be" wrong?


The flaw appears to be that your proof is an identity, it is not wrong but proves nothing. I will admit I am wrong if you can show that you can uniquely identify simultaneous events rather than just those that are simultaneous according to an arbitrary clock.
johanfprins
1 / 5 (6) May 08, 2013
I at first wanted to answer all the objections one by one, but it will be a waste of time since not a single objection has any relevance.

The only message that is coming through is that, even though you cannot prove that my derivation is wrong, you are going to refuse to accept my derivation since, if I am correct, it must mean that the interpretation of the Special Theory of Relativity in terms of Minkowski space-time must be wrong: And this is just too horrible for you to even contemplate since it will mean that most of the theoretical physics that has been done during the past century will be wrong.

It is similar to the reaction when Galileo argued that the earth is a planet that circles the sun like all the other planets do. This meant that the planets do not move on epicycles and this was too horrible for Galileo's peers to even contemplate; since it meant that most of the theoretical that had been done during the 1000 years that preceded Galileo, must be wrong.

johanfprins
1 / 5 (4) May 08, 2013
The guiding principle for a scientist with integrity is to accept that it is possible that the physics we revere as being correct might be proved to be wrong in future, and that it is incumbent on a scientist to continuously check whether this is not so; and to accept that it is wrong when a flaw in the logic is found. As Einstein stated: "Only one small fact is required". If such a flaw is found it is demanded that the physics community must accept it and rectify the physics involved, or else physics will not be self-correcting as physicists claim that it is. If this is not the attitude of physicists they are committing High Treason against the interests of humankind.

What I am thus going to do is to first derive and interpret "time-dilation" as anybody can find in any textbook, and then do my derivation and compare the two:

There are two IRF's K and K/ where K/ moves with a speed v relative to K. At the origins of K and K/ there are two clocks clock1 and clock2 respectively:
johanfprins
1 / 5 (6) May 08, 2013
The clocks are synchronized when they move "through" one another so that t=t/=0. After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1. The coordinates of clock2 within K/ are:

x/=0 and t/=t2

The Lorentz-transformation from K/ into K is given by:

x=(gamma)*(x/-v*t/)
y=y/
z=z/
t=(gamma)*(t/-(v/c^2)*t/)

In text books that equation for the transformation of time is used to transform the time coordinate of clock2 into K, by setting t/=t2 and x/=0

One then finds that:

t=t1=(gamma)*t2

It is then argued in the text books that the time t1 is simultaneously on clock1 when the time on clock2 is t2, and therefore clock2 must be keeping time at a slower rate than clock1. It is just assumed that these times are SIMULTANEOUSLY present on the two clocks without ANY proof that it is so.

Furthermore, one should use ALL the Lorentz equations to REALLY find out what the physics is: Obviously the y and z coordinates are not affected.
johanfprins
1 / 5 (6) May 08, 2013
But the x-coordinates are: So that the position of clock1 as measured from clock2 must be -D1 given by:

x=-D1=(gamma)*(0-v*t2)=(gamma)*(-v*t2)=(gamma)*(-D1), so that

D1=(gamma)*D2

where D2 is the distance between the clocks when the time on clock2 is t2, while D1 is the distance between the clocks when the time on clock1 is t1. Two different distances which proves that the times on clock2 and clock1 are times after the clocks have moved two different distances from one another. Thus these two times are NOT simultaneously on the two clocks as is assumed, WITHOUT ANY PROOF, within textbooks.

This means that when an event occurs at the origin of K/ at time t2, the event only occurs within K at a later time t1>t2 after the origins have moved further apart to be at a distance D1 which is larger than the distance D2 when the event occurred at clock2 at the time t2.

This must be so since D2/t2=v and D1/t1=v.
johanfprins
1 / 5 (4) May 08, 2013
Now let us look at this statement by Fleetwood:

The flaw appears to be that your proof is an identity, it is not wrong but proves nothing.


At least you used the word "appears". Why would my proof be an identity when I use the FULL Lorentz transformation instead of just the time transformation as is done in textbooks?

I will admit I am wrong if you can show that you can uniquely identify simultaneous events rather than just those that are simultaneous according to an arbitrary clock.


I have NOT used simultaneity in my derivation: This contrasts with the interpretation of the time formula in the text books as two times which are SIMULTANEOUS on the two clocks. It is really YOU who should prove that the two different times are uniquely simultaneously on the the clocks. But in order prove this you will also have to prove that the two clocks can simultaneously be at two different distances from one another.

So good luck, my boy!
Noumenon
1.3 / 5 (27) May 08, 2013
This comment refers to posts by johan older than an hour as I have not read the above..

Your derivation is based on faulty reasoning. You effectively admitted above that the Lorentz transformation demands a Omniscient God's Universal Clock,.... even though I was careful to ask you "operationally", in other words, experimentally, which clock do you use to determine simultaneity of the two clocks in question. You then blurted out, 'either one' as they run at the same rate, as if that was already operationally demonstrated in your analysis,.. but was NOT.
johanfprins
1 / 5 (6) May 08, 2013
I apologize: The correct equations from K/ to K are:

x=(gamma)*(x/ plus v*t/)
y=y/
z=z/
t=(gamma)*(t/ plus (v/c^2)*x/)

So that the coordinates of clock 2 x/=0 and t/=t2 Lorentz transform to become:

x=D1=(gamma)*(v*t2)=(gamma)*D2

and

t=t1=(gamma)*t2

The conclusions stay the same.
Fleetfoot
1 / 5 (1) May 08, 2013
There are some confusing points in the way you have written this but if you clarify I'll continue:

The clocks are synchronized when they move "through" one another so that t=t/=0.


OK, we have a reference event O where x = x/ = t = t/ = 0 at which both clocks are present.

A minor point, it would be better to use t' rather than t/.

After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.


That is wrong. If you are talking about frame K/, you next statement is correct:

The coordinates of clock2 within K/ are:

x/=0 and t/=t2


That is correct, call it event A. Since x/ = 0, clock2 has not moved. If you were considering frame K then clock2 has moved a distance of D = v*t1 where t1 is the time on clock1 which occurs simultaneous with t2 appearing on clock2. As yet you haven't applied the transforms but in general t1 /= t2 (i.e. they are not equal).

"Clock2 moved D2=v*t2" is wrong but in K/ clock1 has moved that distance.
Fleetfoot
1 / 5 (1) May 08, 2013
I apologize: The correct equations from K/ to K are: ....


Our posts crossed. I think there's an earlier error in the derivation as I identified in the previous post but it may just have been a typo. Until we clear up any confusions like these, it wouldn't be fair of me to judge your proof so all I'm asking for at the moment is clarification.
johanfprins
1 / 5 (4) May 08, 2013
This comment refers to posts by johan older than an hour as I have not read the above..

Your derivation is based on faulty reasoning. You effectively admitted above that the Lorentz transformation demands a Omniscient God's Universal Clock,....


If you read what I have posted you will see that my derivation does not assume simultaneity of the two clocks. The fact that they are simultaneous follows logically from the derivation.

In contrast, the the standard text book interpretation requires without proof that the times t1 and t2 must be simultaneously present on clock1 and clock2. It is thus the standard interpretation that requires an Omniscient God's Universal clock!
johanfprins
1 / 5 (5) May 08, 2013
After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.


That is wrong.


What is wrong? OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2. Clock 1 also sees clock 2 moving away from it at a speed v so that after an elapsed time which I will call t1# on clock1 the distance from clock1 to clock 2 is also D. D is thus invariant as it must be. One must thus have that:

D=v*t2=v*t1#

The v cancels so that you have that t2=t1#

At the same distance D, as measured by the corresponding times on both clocks, these times must be simultaneously the exact same time. The Lorentz transformed distance is DL=(gamma)*(v*t2)=(gamma)*D, which is larger than D and for this reason the Lorentz transformed time t1 is also larger than t2: Not because clock2 keeps slower time.
Fleetfoot
1 / 5 (1) May 08, 2013
After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.


That is wrong.


What is wrong?


I explained in the previous post, you said x/ = 0 which is correct therefore the distance moved by clock2 is zero, not D2 = v*t2.

OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.


Almost, D2=v*t2 in frame K/

Clock 1 also sees clock 2 moving away from it at a speed v so that after an elapsed time which I will call t1# on clock1 the distance from clock1 to clock 2 is also D.


Almost, D1 = v * t1 in frame K

D is thus invariant


No, you have defined two different distances, D1 and D2 that aren't in the same coordinate system or even between the same events.

If you're going to give a formal proof, you need to address all these loopholes.
johanfprins
1 / 5 (5) May 08, 2013
I explained in the previous post, you said x/ = 0 which is correct therefore the distance moved by clock2 is zero, not D2 = v*t2.


There is not absolute motion the when the distance between the clocks is D, you can either say clock2 moved relative to clock1 through this distance, or clock1 moved relative to clock2 through this distance.

OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.


Almost, D2=v*t2 in frame K/


Adding a 2 so that D=D2 does not change the argument.

Clock 1 also sees clock 2 moving away from it at a speed v so that after an elapsed time which I will call t1# on clock1 the distance from clock1 to clock 2 is also D.


Almost, D1 = v * t1 in frame K


johanfprins
1 / 5 (3) May 08, 2013
This is not what I have calculated: I specifically used the symbol t1# to indicate that this time is not the L:orentz-transformed time t1 of the time t2, but is the time on clock1 after clock2 has moved away from it to reach a distance D which you preferred to call D2 above; for no logical reason whatsoever. So what I have calculated is

D=D2=v*t1# from which it follows that t2=t1# and since D=D2 is the same within both reference frames D is invariant as it must be since owing to the symmetry inherent in relative motion, the one clock cannot at the SAME instant in time (t2=t1#) be situated at a different distance from the other clock than the other clock is situated from the first clock.

No, you have defined two different distances, D1 and D2


No I have defined the SAME distance D between the two clocks within both IRF's, as it must be, and then found that the times on the clocks must be also be the same for the same distance D.
ValeriaT
1.4 / 5 (9) May 08, 2013
For what the math actually is, if it needs so lotta words for its explanation/interpretation?
It is thus the standard interpretation that requires an Omniscient God's Universal clock!
The special relativity is just based on absence of such universal clock (as it name implies).
johanfprins
1 / 5 (5) May 08, 2013
For what the math actually is, if it needs so lotta words for its explanation/interpretation?
It is thus the standard interpretation that requires an Omniscient God's Universal clock!
The special relativity is just based on absence of such universal clock (as it name implies).


Thus, before mankind made clocks there was no time and no change in time in Nature?
As usual ValeriaT you do not have a clue about what you are posting? I can only pity you.
Fleetfoot
1 / 5 (1) May 08, 2013
I specifically used the symbol t1# to indicate that this time is ... the time on clock1 after clock2 has moved away from it to reach a distance D which you preferred to call D2 above; for no logical reason whatsoever.


You called it D2 in your previous post so I was maintaining that notation and it is a distance measured in frame K/ which is the rest frame of clock2.

So what I have calculated is

D=D2=v*t1# from which it follows that t2=t1# and since D=D2 is the same within both reference frames D is invariant ...


No, the value is the same, you simply selected t1# to ensure that, but you don't have an invariant measure. Say in K/ it is the distance between events A and B and that in frame K at time t1# it lies between events C and D, the following diagrams plot the events in both frames:

https://sites.goo...graphics

D and D2 may have the same numerical value but they are not the lengths of the same lines.
Fleetfoot
1 / 5 (1) May 08, 2013
For what the math actually is, if it needs so lotta words for its explanation/interpretation?


Some people need more than maths, let's see if a picture is worth more than 1000 characters.
Q-Star
2.2 / 5 (10) May 08, 2013
the following diagrams plot the events in both frames:

https://sites.goo...graphics
.


Sorry to interupt, could ya send a link to the place where ya set-up that graphics application?

Thanks. Q
johanfprins
1 / 5 (4) May 08, 2013
You called it D2 in your previous post so I was maintaining that notation and it is a distance measured in frame K/ which is the rest frame of clock2.


So why can I not cal it D?

So what I have calculated is

D=D2=v*t1# from which it follows that t2=t1# and since D=D2 is the same within both reference frames D is invariant ...


No, the value is the same, you simply selected t1# to ensure that, but you don't have an invariant measure.


YOU are the one who does not have an invariant measure! So let us change the time notation:

We are simply talking about two clocks being synchronized and passing one another. Do you agree that when time T/ has passed on clock 2 the distance that clock 2 calculates that clock 1 has moved away from it is

D/=v*T/?

Yes or no?

Do you agree that when the time on clock 1 is T, the distance that clock 2 has moved away from clock 1 is

D=v*T.

Yes or no?

Answer these two questions honestly and then I will proceed!
johanfprins
1 / 5 (2) May 08, 2013
No, the value is the same, you simply selected t1# to ensure that, but you don't have an invariant measure. Say in K/ it is the distance between events A and B and that in frame K at time t1# it lies between events C and D, the following diagrams plot the events in both frames:

https://sites.goo...graphics


These diagrams are based on extra assumptions than just the Lorentz equations. So let us first go as far as we can without them by just using the Lorentz equations and then analyze the diagrams. My arguments are based only on the Lorentz equations: Not on any diagrams that might, or might not be commensurate with these equations.
Fleetfoot
1 / 5 (1) May 08, 2013
Sorry to interrupt, could ya send a link to the place where ya set-up that graphics application?

Thanks. Q


The application was written by Mikko Levanto many years ago, I find it very useful for such discussions.

http://www.reagen...est.html

The snapshot is on a Google site I set up in a few minutes:
Fleetfoot
1 / 5 (2) May 08, 2013
Say in K/ it is the distance between events A and B and that in frame K at time t1# it lies between events C and D, the following diagrams plot the events in both frames:

https://sites.goo...graphics


These diagrams are based on extra assumptions than just the Lorentz equations. ...


The diagram is a snapshot of a Java applet which implements the Lorentz Transforms, nothing more. The slider at the top selects the speed of the boost. I added the "Frame" notations with Windows Paint but that's all. The diagram conveyed what would have been hard in words. I don't expect you to trust it but now that you can understand what I have been saying, you can apply the LT's to the coordinates and check what it shows for yourself.

The point is that in your posts, you set the length of C-D in the bottom diagram equal to A-B in the top diagram, but so what, I can't see what you think that proves.
Fleetfoot
1 / 5 (2) May 08, 2013
Do you agree that when time T/ has passed on clock 2 the distance that clock 2 calculates that clock 1 has moved away from it is

D/=v*T/?

Yes or no?


Yes, that is shown as clock1 being at event B and clock2 at event A, the distance D/ being their separation in frame K/, the top panel of my graphic.

https://sites.goo...graphics

Do you agree that when the time on clock 1 is T, the distance that clock 2 has moved away from clock 1 is

D=v*T.

Yes or no?


Yes, that is shown as clock1 being at event C and clock2 at event D, the distance D being their separation in frame K, the lower panel of my graphic.

Answer these two questions honestly and then I will proceed!


I think we now have a common understanding of your thought experiment which is what I felt was needed so please carry on.
johanfprins
1 / 5 (5) May 09, 2013
Do you agree that when time T/ has passed on clock 2 the distance that clock 2 calculates that clock 1 has moved away from it is

D/=v*T/?

Yes or no?

Yes,
Thanks!

Do you agree that when the time on clock 1 is T, the distance that clock 2 has moved away from clock 1 is

D=v*T.

Yes or no?

Yes,
Thanks

I think we now have a common understanding of your thought experiment which is what I felt was needed so please carry on.
Not yet!

Now do you agree that when one chooses a specific distance D, then there must be a time t on clock1 when clock2 is at this distance from clock1 so that:

D=v*t

Yes or no?

Do you also agree that there must be a time t/ on clock2 when clock1 is at the same distance D from clock2, so that

D=v*t/

Yes or no?

Do you agree that since D is the same within the last two expressions that one must have for these two times that:

t=t/

Yes or no?

Please answer, so that we can try to reach an understanding.
Noumenon
1.1 / 5 (28) May 09, 2013
No, the value is the same, you simply selected t1# to ensure that, but you don't have an invariant measure. Say in K/ it is the distance between events A and B and that in frame K at time t1# it lies between events C and D, the following diagrams plot the events in both frames:

https://sites.goo...graphics


These diagrams are based on extra assumptions than just the Lorentz equations. So let us first go as far as we can without them by just using the Lorentz equations and then analyze the diagrams. My arguments are based only on the Lorentz equations: Not on any diagrams that might, or might not be commensurate with these equations.


That diagram shows what a relatively moving observers IRF is measured to be to a stationary observer, and is derived by use of the LT only. Both IRF, the stationary and the moving, are proper 4-vectors, with orthogonal basis. The stationary observer sees that the moving observers IRF is hyperbolic-orthogonal as a consequence of being LT
johanfprins
1 / 5 (5) May 09, 2013
That diagram shows what a relatively moving observers IRF is measured to be to a stationary observer, and is derived by use of the LT only.


No it is not derived by the use of LT "only": It is derived in terms of a specific interpretation of what the LT equations are: Not by just using the LT equations without any interpretation added to obtain the diagram.

Both IRF, the stationary and the moving, are proper 4-vectors, with orthogonal basis.


You see: This is the interpretation that is added by you to the LT equations. I use the LT equations as they are without adding the interpretation that they "are proper 4-vectors" to derive my result. And the result that I obtain is that they ARE NOT "proper 4-vectors" as you claim they are.
Noumenon
1.3 / 5 (29) May 09, 2013
OK, I don't want to interrupt you and fleetfoot line of reasoning, so as an aside,... your last question would be a Yes, that there is an equal numerical time within each respective IRF when a specific distance A-B = C-D separates them,.... however, with the proviso that the numerical time is only meaningful within each IRF and not outside it, thus it is meaningless to compare them. Why?, because the Space distance is NOT the same according to the other observer and since the Space-Time interval is invariant physically (the same), it must be the coordinates that change,... i.e. the time is slowed.
johanfprins
1 / 5 (5) May 09, 2013
OK, I don't want to interrupt you and fleetfoot line of reasoning, so as an aside,... your last question would be a Yes, that there is an equal numerical time within each respective IRF when a specific distance A-B = C-D separates them,....


Thanks for acknowledging that the answer is yes. But the rest of your argument is still based on extra assumptions not "only" on the equations of LT. And this leads to absurd conclusions.

however, with the proviso that the numerical time is only meaningful within each IRF and not outside it, thus it is meaningless to compare them. Why?, because the Space distance is NOT the same according to the other observer ..


You are claiming that at any instant in time (on any of the clocks), clock2 is further away from clock1 than clock1 is from clock2. And that when using the reverse LT equations one must have that at any instant in time, clock1 is further away from clock2 than clock2 is from clock1. Absurd is it not?
Fleetfoot
1 / 5 (2) May 09, 2013
Yes,
Thanks!

Yes,
Thanks

I think we now have a common understanding of your thought experiment which is what I felt was needed so please carry on.
Not yet!


These were my responses:

Yes, that is shown as clock1 being at event B and clock2 at event A, the distance D/ being their separation in frame K/, the top panel of my graphic.

Yes, that is shown as clock1 being at event C and clock2 at event D, the distance D being their separation in frame K, the lower panel of my graphic.


It is important that you understand that your numbers measure times and distances between different events.

Do you agree that since D is the same within the last two expressions that one must have for these two times that:

t=t/


I understand that is given in the question, hence events C-D in frame K are at the same time coordinate as A-B in frame K.

If you want to reach a common understanding, don't remove the qualifications, deleting my words won't change them.
Fleetfoot
1 / 5 (2) May 09, 2013
That diagram shows what a relatively moving observers IRF is measured to be to a stationary observer, and is derived by use of the LT only.


No it is not derived by the use of LT "only": It is derived in terms of a specific interpretation of what the LT equations are: Not by just using the LT equations without any interpretation added to obtain the diagram.


Noumenon is correct, the diagram makes no extra assumptions. The values are obtained solely using the transforms, the diagram is nothing more than a plot of the values.

The 'event' names make it easy to identify which coordinate pairs we are discussing and what the values such as D and D/ etc. represent (i.e. between which pairs they lie) WITHOUT making additional physical assumptions.
Noumenon
1.5 / 5 (30) May 09, 2013
You are claiming that at any instant in time (on any of the clocks), clock2 is further away from clock1 than clock1 is from clock2. And that when using the reverse LT equations one must have that at any instant in time, clock1 is further away from clock2 than clock2 is from clock1. Absurd is it not?


No, they are the SAME space-time distance from each other always as I said that is a physical invariant. They do NOT agree wrt components of that invariant, i.e. spacial distances or time distances separately.
antialias_physorg
4.2 / 5 (5) May 09, 2013
They do NOT agree wrt components of that invariant, i.e. spacial distances or time distances separately.

I think this bears additional emphasis - since it seems the key point/idea he's missing.
johanfprins
1 / 5 (5) May 09, 2013
Do you agree that since D is the same within the last two expressions that one must have for these two times that:

t=t/


I understand that is given in the question, hence events C-D in frame K are at the same time coordinate as A-B in frame K.
AB in K or in K/?

The distance CD in K/ and the distance BA in the K cannot be drawn unless you already assumed that time-dilation occurs. Thus these chords cannot be used as proof that time dilation occurs since this is a circular argument: Thus your diagrams prove nothing! You did not just use the equations of the LT without making this assumption as you claim that you have.

You must prove to me that CD in K/ and AB in K cannot be horizontal and parallel without first postulating that they cannot be horizontal but must be a space-time interval. The latter requires a Minkowski space-time with linearly-independent coordinates: Minkowski space-time is not spanned by such coordinates.
johanfprins
1 / 5 (4) May 09, 2013
Noumenon is correct, the diagram makes no extra assumptions. The values are obtained solely using the transforms, the diagram is nothing more than a plot of the values.
REALLY? I have just now pointed out that if you did not make the additional a priori assumption that time-dilation occurs, the chord CD in K/ and the chord AB in K would not have been drawn to have non-zero slopes!

The 'event' names make it easy to identify which coordinate pairs we are discussing and what the values such as D and D/ etc. represent (i.e. between which pairs they lie) WITHOUT making additional physical assumptions.


But you have made the additional assumption that time-dilation must be occurring in order to draw the sloped chords. So you have assumed what you claim that you are proving. This is nonsensical!
johanfprins
1 / 5 (5) May 09, 2013
No, they are the SAME space-time distance from each other always as I said that is a physical invariant.
A four-dimensional interval within a four-dimensional space can only be PHYSICALLY invariant if the coordinates are linearly-independent which they are NOT in Minkowski space-time. It is even mathematically meaningless!
Noumenon
1.1 / 5 (27) May 09, 2013
No, they are the SAME space-time distance from each other always as I said that is a physical invariant.
A four-dimensional interval within a four-dimensional space can only be PHYSICALLY invariant if the coordinates are linearly-independent which they are NOT in Minkowski space-time. It is even mathematically meaningless!


I provided a link above which shows you're factually incorrect. None of the basis can be expressed via a combination of the other axis. Each IRF, the one WITH the observer, and the one the observer SEES of a moving IRF is (hyperbolic) orthogonal, separately. They are not being mixed.
Noumenon
1.1 / 5 (27) May 09, 2013
,.. which is to say, an observer uses His IRF .....OR..... the Other's LT IRF for each measurement, not both mixed at the same time for any given measurement.
johanfprins
1 / 5 (3) May 09, 2013
No, they are the SAME space-time distance from each other always as I said that is a physical invariant.
A four-dimensional interval within a four-dimensional space can only be PHYSICALLY invariant if the coordinates are linearly-independent which they are NOT in Minkowski space-time. It is even mathematically meaningless!


I provided a link above which shows you're factually incorrect. None of the basis can be expressed via a combination of the other axis. Each IRF, the one WITH the observer, and the one the observer SEES of a moving IRF is (hyperbolic) orthogonal, separately. They are not being mixed.


There is no such a thing as" hyperbolic orthogonality": It is an oxymoron concept. Your link provided no evidence or proof that it is NOT an oxymoron concept!
johanfprins
1 / 5 (7) May 09, 2013
,.. which is to say, an observer uses His IRF .....OR..... the Other's LT IRF for each measurement, not both mixed at the same time for any given measurement.


This means in your own words that time-dilation is not possible since this is exactly what the concept of time-dilation does: It "mixes" measurements.
Fleetfoot
1 / 5 (1) May 09, 2013
I understand that is given in the question, hence events C-D in frame K are at the same time coordinate as A-B in frame K.
AB in K or in K/?


That should have been "as A-B in frame K/." of course.

.. You did not just use the equations of the LT without making this assumption as you claim that you have.

You must prove to me that CD in K/ and AB in K cannot be horizontal and parallel ..


I've also done the same thing as a spreadsheet. I'll add a screenshot of that to the graphic page later tonight and you can then replicate it yourself to check that I haven't cheated.
johanfprins
1 / 5 (5) May 09, 2013
You must prove to me that CD in K/ and AB in K cannot be horizontal and parallel ..


I've also done the same thing as a spreadsheet. I'll add a screenshot of that to the graphic page later tonight and you can then replicate it yourself to check that I haven't cheated.


It is not a question of you cheating but drawing lines by assuming time-dilation and then saying that this proves time dilation. Doing it on a spreadsheet will not change this in any manner. The fact is that on your diagrams both lines AB and CD must be horizontal and parallel. The lower horizontal time-line gives the simultaneous identical times on both clocks when the event occurs at one clock, and the later time-line gives the simultaneous identical times on both clocks when the event is recorded by the other clock.

Only within an EM-wave does the phase-time change with position along the direction of motion of the wave. That is why the wave equation can be written in an invariant form.
Fleetfoot
1 / 5 (1) May 09, 2013
I've also done the same thing as a spreadsheet. I'll add a screenshot of that to the graphic page later tonight and you can then replicate it yourself to check that I haven't cheated.


It is not a question of you cheating but drawing lines by assuming time-dilation and then saying that this proves time dilation. Doing it on a spreadsheet will not change this in any manner.


Doing it that way allows you to check I am only using the Lorentz Transforms.

The fact is that on your diagrams both lines AB and CD must be horizontal and parallel.


No, my diagrams only plot the values what is necessary is that those values must be obtained using the Lorentz transforms and nothing else. Your problem is that time dilation effects are an unavoidable consequence of the transforms.
johanfprins
1 / 5 (5) May 09, 2013
The fact is that on your diagrams both lines AB and CD must be horizontal and parallel.


No, my diagrams only plot the values what is necessary is that those values must be obtained using the Lorentz transforms and nothing else. Your problem is that time dilation effects are an unavoidable consequence of the transforms.


No you are not doing that, since you assume that the time t2 at which an event occurs at clock 2 is simultaneously displayed on clock 2 than the Lorentz-transformed time t1 is displayed on clock 1; and you have no proof of this assumed "simultaneity". As I have impeccably proved above, the distance of clock 2 when the event occurs is D2=v*t2 where t2 is simultaneously displayed on clock 2 and clock 1, while the Lorentz-transformed time t1=(gamma)*t2 is also simultaneously displayed on clock 2 and clock 1 when the distance between them is D1=(gamma)*D2: A larger distance which requires a later time t1 than the time t2; so that D1/t1=v..
Fleetfoot
3 / 5 (2) May 09, 2013
It is not a question of you cheating but drawing lines by assuming time-dilation and then saying that this proves time dilation.


Nobody said anything about time dilation, you outlined some questions and my diagram illustrates what I think you are asking.

Doing it on a spreadsheet will not change this in any manner.


Doing it that way allows you to check I am only using the Lorentz Transforms.

The fact is that on your diagrams both lines AB and CD must be horizontal and parallel.


My diagrams only plot the values, what is necessary is that those values must be obtained using the Lorentz transforms and nothing else and that is the case. Whether they are horizontal or not is determined by the LTs.

The screenshot is here:

https://sites.goo...eadsheet

If you want a copy, PM me.

The transform equations are these (though of course we aren't using y or z):

https://en.wikipe...irection
Fleetfoot
3 / 5 (2) May 09, 2013
It is not a question of you cheating but drawing lines by assuming time-dilation and then saying that this proves time dilation.


Nobody said anything about time dilation, you outlined some questions and my diagram illustrates what I think you are asking.

Doing it on a spreadsheet will not change this in any manner.


Doing it that way allows you to check I am only using the Lorentz Transforms.

The fact is that on your diagrams both lines AB and CD must be horizontal and parallel.


My diagrams only plot the values, what is necessary is that those values must be obtained using the Lorentz transforms and nothing else and that is the case. Whether they are horizontal or not is determined by the LTs.

The screenshot is here:

https://sites.goo...eadsheet

If you want a copy, PM me.

The transform equations are these (though of course we aren't using y or z):

https://en.wikipe...irection
Fleetfoot
3 / 5 (2) May 09, 2013
It is not a question of you cheating but drawing lines by assuming time-dilation and then saying that this proves time dilation.


Nobody said anything about time dilation, you outlined some questions and my diagram illustrates what I think you are asking.

Doing it on a spreadsheet will not change this in any manner.


Doing it that way allows you to check I am only using the Lorentz Transforms.

The fact is that on your diagrams both lines AB and CD must be horizontal and parallel.


My diagrams only plot the values, what is necessary is that those values must be obtained using the Lorentz transforms and nothing else and that is the case. Whether they are horizontal or not is determined by the LTs.

The screenshot is here:

https://sites.goo...eadsheet

If you want a copy, PM me.

The transform equations are these (though of course we aren't using y or z):

https://en.wikipe...irection
Fleetfoot
1 / 5 (1) May 09, 2013
It is not a question of you cheating but drawing lines by assuming time-dilation and then saying that this proves time dilation.


Nobody said anything about time dilation, you outlined some questions and my diagram illustrates what I think you are asking.

Doing it on a spreadsheet will not change this in any manner.


Doing it that way allows you to check I am only using the Lorentz Transforms.

The fact is that on your diagrams both lines AB and CD must be horizontal and parallel.


My diagrams only plot the values, what is necessary is that those values must be obtained using the Lorentz transforms and nothing else and that is the case. Whether they are horizontal or not is determined by the LTs.

The screenshot is here:

https://sites.goo...eadsheet

If you want a copy, PM me.

The transform equations are these (though of course we aren't using y or z):

https://en.wikipe...irection
johanfprins
1 / 5 (6) May 09, 2013
Nobody said anything about time dilation, you outlined some questions and my diagram illustrates what I think you are asking.

Doing it that way allows you to check I am only using the Lorentz Transforms.


Then you should only show the points where and when the event occurred in one IRF and when and where the event occurred in the other IRF WITHOUT connecting these points with a line that has a slope. By connecting them in this manner you are assuming, without any experimental proof, hat the clocks are connected by an actual REAL space-time distance which can only exist when you have time-dilation within a linearly independent four-dimensional space. Which you do not have.

My diagrams only plot the values, what is necessary is that those values must be obtained using the Lorentz transforms and nothing else and that is the case.
No you are not doing this, since you are using lines with slopes to connect points: This is misleading.

johanfprins
1 / 5 (4) May 09, 2013
Whether they are horizontal or not is determined by the LTs.
No it is not: It is determined by the interpretation of the LT equations. The more logical and non-insane manner to interpret these points would be to accept that the clocks keep time at the same rate as I have proved above directly by ONLY using the LT equations. When using this sane interpretation the event-lines must all be horizontal.

Thus your interpretation of the LT equations give you your sloped lines, and my interpretation which I derived directly from the LT equations, without drawing confusing diagrams, gives horizontal lines. So which one is correct? Yours cannot be correct since it demands that the two clocks must simultaneously show two different times while at the SAME instant in time being at two different distances apart from one another: This is obviously absurd physics!
Noumenon
1.1 / 5 (27) May 09, 2013
Is someone going to end up buying a time-share after all of this?

Yours cannot be correct since it demands that the two clocks must simultaneously show two different times while at the SAME instant in time being at two different distances apart from one another: This is obviously absurd physics!


"at the SAME instant in time" of what time are you referring here to have determined that? You can't simply say "at the same instant in time". The reference standard MUST be a physical clock, otherwise you miss the entire point of STR. Einstein was very careful to speak in operational or instrumentalist terms. There is NO universal absolute time operationally determinable. Thus it is invalid to refer to such a thing.
Noumenon
1.1 / 5 (27) May 09, 2013
drawing confusing diagrams


He never drew anything. That is a plot resulting from the LT calculations and nothing more. You are the one presuming an absolute Time reference without any basis in observational fact for having done so, and as a result you miss the entire point of STR.
ValeriaT
1 / 5 (5) May 09, 2013
would be to accept that the clocks keep time at the same rate
It would violate the invariant speed of light: the time gets dilated just for maintaining its invariance.
This is obviously absurd physics!
Nope, such a physics happens at every water surface.
johanfprins
1 / 5 (3) May 09, 2013
Yours cannot be correct since it demands that the two clocks must simultaneously show two different times while at the SAME instant in time being at two different distances apart from one another: This is obviously absurd physics!


"at the SAME instant in time" of what time are you referring here to have determined that?


Exactly! According to time- dilation one has that clock2 shows the time t2 when the event occurs and clock1 shows the time t1=(gamma)*t2 when the event occurs. Thus, the inherent assumption is clear: that this these times must be simultaneouslypresent on th two clocks when the event occurs. Which is obviously absurd; as you just pointed out!

The only correct interpretation is the one I have derived above and that is that the event occurs at clock2 at the time t2, and is observed at clock 1 at a distance and time at which clock1 was not present when the event ocurred. This is exactly what the LT equations give as the correct answer.

johanfprins
1 / 5 (4) May 09, 2013
Einstein was very careful to speak in operational or instrumentalist terms. There is NO universal absolute time operationally determinable. Thus it is invalid to refer to such a thing.


So our Universe is at its oldest within our galaxy since all other galaxies moving with incredible speeds relative to our galaxy must be far younger than our galaxy. So how is it possible that older galaxies exist which have been moving at incredible speeds relative to us since our Universe was created? Please explain!
johanfprins
1 / 5 (6) May 09, 2013
drawing confusing diagrams


He never drew anything. That is a plot resulting from the LT calculations and nothing more.


It IS more since he did not just plot the results as dots, but connected the dots by assuming space-time intervals between them: The worst is that he then claims that his diagrams prove that such intervals exist! LOL!

You are the one presuming an absolute Time reference without any basis in observational fact for having done so, and as a result you miss the entire point of STR.
I am not presuming anything since the equations of the LT demands absolute time in a gravity-free space.

From the symmetry of relativity, I derive that the distance between the clocks is less when the event occurs at clock2 than it is when the event is recorded at clock1. And that this increase in distance occurs at a speed v which DEMANDS that the two clocks MUST keep time at the SAME rate! I do not have to make any assumptions.
johanfprins
1 / 5 (4) May 09, 2013
would be to accept that the clocks keep time at the same rate


It would violate the invariant speed of light: the time gets dilated just for maintaining its invariance.


As usual you do not have a clue. Light speed is maintained since the position and time at which a source emits light and a detector detects light, can be different in different IRF's. It is not maintained by a clock keeping different time-rates within different IRF's.

This is obviously absurd physics!
Nope, such a physics happens at every water surface.

Quack, quack, quack quack!
Noumenon
1.4 / 5 (29) May 09, 2013
Einstein was very careful to speak in operational or instrumentalist terms. There is NO universal absolute time operationally determinable. Thus it is invalid to refer to such a thing.


So our Universe is at its oldest within our galaxy since all other galaxies moving with incredible speeds relative to our galaxy must be far younger than our galaxy. So how is it possible that older galaxies exist which have been moving at incredible speeds relative to us since our Universe was created? Please explain!


I'm not sure I understand the point of this question. Empirically, time is some physical process that cycles, like a light-clock or cesium atom. That's it, nothing more.
Fleetfoot
1 / 5 (1) May 09, 2013
.. my diagrams only plot the values what is necessary is that those values must be obtained using the Lorentz transforms and nothing else. ..


No you are not doing that, since you assume that the time t2 at which an event occurs at clock 2 is simultaneously displayed on clock 2 than the Lorentz-transformed time t1 is displayed on clock 1; and you have no proof of this assumed "simultaneity".


I have assumed no such thing. You stated as a given in your question that the distance to clock1 in the frame of clock2 was t2. That value in my example is 7 units up the vertical axis on the top panel labelled "Frame K/" at location "A". That scale represents clock2 readings.

The x coordinate measured by a ruler is -5 at that clock2 reading so (-5, 7) is where "B" is plotted. There is no assumption there about time at all, (however I understand we have been working on the understanding that we are discussing perfect clocks, not faulty ones, but that's a separate matter).
Fleetfoot
3 / 5 (2) May 09, 2013
No you are not doing that, since you assume that the time t2 at which an event occurs at clock 2 is simultaneously displayed on clock 2 ..


Reading that again, I think you are confused, t2 is the number which appears on the clock2 at event A, we call that value "time".

As I have impeccably proved above, ...


Previously:

D is thus invariant


No, you have defined two different distances, D1 and D2 that aren't in the same coordinate system or even between the same events.


You have not addressed that error in your proof yet.
Fleetfoot
3 / 5 (2) May 10, 2013
Apologies for the multiple posts yesterday, my ISP had problems.

Noumenon is correct, the diagram makes no extra assumptions. The values are obtained solely using the transforms, the diagram is nothing more than a plot of the values.


REALLY? I have just now pointed out that if you did not make the additional a priori assumption that time-dilation occurs, the chord CD in K/ and the chord AB in K would not have been drawn to have non-zero slopes!


A minor correction of terminology, the line A-B for example is horizontal in frame K/ hence the events are "simultaneous" while sloping in K hence "not simultaneous". That illustrates an effect called "relativity of simultaneity", not time dilation (though they are related).

The main point though is that the coordinates of the events A and B are determined by the LTs. Whether I draw a line between them or not is irrelevant, the vertical coordinates are equal in frame K/ but differ in frame K, the slopes inevitably follow.
johanfprins
1 / 5 (4) May 10, 2013
So our Universe is at its oldest within our galaxy since all other galaxies moving with incredible speeds relative to our galaxy must be far younger than our galaxy. So how is it possible that older galaxies exist which have been moving at incredible speeds relative to us since our Universe was created? Please explain!


I'm not sure I understand the point of this question. Empirically, time is some physical process that cycles, like a light-clock or cesium atom. That's it, nothing more.


So are you claiming that time did not exist before the first clock was built by mankind?

Time is a measure of evolution and we know that our universe evolved simultaneously similarly at all positions. Thus, time-rate cannot be different within a galaxy that have been moving for billions of years at an incredible speed away from our galaxy. It is just a simple fact that time within gravity-free space must be absolute; or else we would have had a very absurd Universe.
johanfprins
1 / 5 (2) May 10, 2013
A minor correction of terminology, the line A-B for example is horizontal in frame K/ hence the events are "simultaneous" while sloping in K hence "not simultaneous". That illustrates an effect called "relativity of simultaneity", not time dilation (though they are related).
GOOD! You are slowly getting there!

The main point though is that the coordinates of the events A and B are determined by the LTs. Whether I draw a line between them or not is irrelevant, the vertical coordinates are equal in frame K/ but differ in frame K, the slopes inevitably follow.


First take the chords AB and CD away from your diagram for K/. The two clocks start of at the common origin of the two paths Now without considering any events at any of the two clocks, where will clock1 and clock2 be after a time t has elapsed along the vertical axis which is both the path of clock1 an the time-axis? Please do the diagram without chords AB and CD, and show the positions of the clocks after a time t.
Fleetfoot
3 / 5 (2) May 10, 2013
A minor correction of terminology, ... an effect called "relativity of simultaneity", not time dilation (though they are related).
GOOD! You are slowly getting there!


I got there about 15 years ago. I don't correct all your errors when they arise, it would be too disruptive.

First take the chords AB and CD away from your diagram for K/.


I'll not waste time updating it until we agree a real change but ignore them for the moment.

The two clocks start of at the common origin of the two paths Now without considering any events at any of the two clocks, where will clock1 and clock2 be after a time t has elapsed along the vertical axis which is both the path of clock1 an the time-axis? Please do the diagram without chords AB and CD, and show the positions of the clocks after a time t.


The events A and B will be where they are shown, both must be at the same vertical coordinate, and each must lie on the worldline of the relevant clock.

What about C and D?
johanfprins
1 / 5 (6) May 10, 2013
First take the chords AB and CD away from your diagram for K/.


I'll not waste time updating it until we agree a real change but ignore them for the moment.


How did you construct this diagram? By first plotting the paths of clocks 1 and 2 and then adding the chords, or by first plotting the chords and then adding the paths. I do not think it is the latter. So by asking that you at first remove the chords, I am asking to construct the diagram as it has be constructed. Do you think that this is an unreasonable request?

The events A and B will be where they are shown, both must be at the same vertical coordinate, and each must lie on the worldline of the relevant clock.


I have not asked about ANY events but asked you what the positions of the clocks will be on their respective paths after ANY lapse in time if the clocks start off from the origin at a time t=0.

What about C and D?


D is an event at clock1 and the question does not involve an event
Fleetfoot
3 / 5 (2) May 10, 2013
I'll not waste time updating it until we agree a real change but ignore them for the moment.


How did you construct this diagram? By first plotting the paths of clocks 1 and 2 and then adding the chords, or by first plotting the chords and then adding the paths. I do not think it is the latter. So by asking that you at first remove the chords, I am asking to construct the diagram as it has be constructed. Do you think that this is an unreasonable request?

Asking me to repeat the work I've already done is unreasonable, just ignore all but what is agreed.

I have not asked about ANY events but asked you what the positions of the clocks will be on their respective paths after ANY lapse in time if the clocks start off from the origin at a time t=0.


You asked "where will clock1 and clock2 be after a time t has elapsed". In general that is answered by the lines labelled "clock1" and "clock2". For a specific value of t, events A and B are typical.
Noumenon
1.2 / 5 (26) May 10, 2013
So our Universe is at its oldest within our galaxy since all other galaxies moving with incredible speeds relative to our galaxy must be far younger than our galaxy. So how is it possible that older galaxies exist which have been moving at incredible speeds relative to us since our Universe was created?


I'm not sure I understand the point of this question. Empirically, time is some physical process that cycles, like a light-clock or cesium atom. That's it, nothing more.


So are you claiming that time did not exist before the first clock was built by mankind?

Time is a measure of evolution and we know that our universe evolved simultaneously similarly at all positions.


All I'm saying here, is that in order to perform measurements wrt time, by necessity we have to make use of some physical device. The simplest would be a photon between two mirrors, and call that a clock. Einsten was careful not to make presumptions & took an instrumentist position.
johanfprins
1 / 5 (6) May 10, 2013
All I'm saying here, is that in order to perform measurements wrt time, by necessity we have to make use of some physical device. The simplest would be a photon between two mirrors, and call that a clock. Einsten was careful not to make presumptions & took an instrumentist position.


Fine with me: I have used the light clock and came to exactly the same conclusion I am advocating here, namely that the two clocks do not show a time dilation but simply a difference in time after they have moved further from one another while keeping exactly the same time-rate. I have referred you time and again to this derivation but you refuse to read it. This is the attitude of a closed-minded bigot. So again, I urge you to look at page 22, Figure 8 in the manuscript http://www.cathod...tion.pdf

Please, at least try and act like a real scientist with an open-mind and integrity is supposed to act!
johanfprins
1 / 5 (3) May 10, 2013
I am asking to construct the diagram as it has be constructed. Do you think that this is an unreasonable request?


Asking me to repeat the work I've already done is unreasonable, just ignore all but what is agreed
Why is this unreasonable when we are arguing about this construction?

I have not asked about ANY events but asked you what the positions of the clocks will be on their respective paths after ANY lapse in time if the clocks start off from the origin at a time t=0.


You asked "where will clock1 and clock2 be after a time t has elapsed". In general that is answered by the lines labelled "clock1" and "clock2". For a specific value of t, events A and B are typical.


There need not be "events". The paths are there and the time is there: I assume that what you are trying to say is that after any elapse of time t,, clock 1 and clock 2 will be on a horizontal line that cuts through both paths and through the time axis at this time t.

Correct?
Fleetfoot
3 / 5 (2) May 10, 2013
Why is this unreasonable when we are arguing about this construction?


Because we are both intelligent enough to ignore the parts that aren't agreed.

.. In general that is answered by the lines labelled "clock1" and "clock2". For a specific value of t, events A and B are typical.


There need not be "events". The paths are there and the time is there: I assume that what you are trying to say is that after any elapse of time t,, clock 1 and clock 2 will be on a horizontal line that cuts through both paths and through the time axis at this time t.

Correct?


You are going farther than we have agreed. The generic equation D = -v*t describes the location of clock1 for all t, that is the black line labelled "clock1" through the origin.

For a specific time such as your t2 reading on clock2, by construction, we plot only a single point on that line at a vertical scale of the t2 value, i.e. point B at (-v*t2, t2) for clock1. The point for clock2 is point A at (0, t2).
johanfprins
1 / 5 (4) May 10, 2013
Because we are both intelligent enough to ignore the parts that aren't agreed.


You are not so stupid not to realize that you are completely dishonest and acting disgracefully and unscientifically.

You are going farther than we have agreed.
What have we agreed that it is so "holy" that it is a heresy to transgress it. I do not know of such an agreement.

The generic equation D = -v*t describes the location of clock1 for all t, that is the black line labelled "clock1" through the origin.


Where have I denied this? Furthermore the generic equation D/=t gives the black line labelled clock 2 which gives the corresponding time-"position" at clock2 for ALL t.

For a specific time such as your t2 reading on clock2, by construction, we plot only a single point on that line at a vertical scale of the t2 value, i.e. point B at (-v*t2, t2) for clock1. The point for clock2 is point A at (0,t2).


johanfprins
1 / 5 (5) May 10, 2013
So what makes the time t2 at A any different from any other time point on the vertical time-axis?Why is it for example different from (say) time tB at position B on that axis. tB is ALSO a specific time on clock2; clearly so we must also "plot only a single point (say F) at (-v*tB,tB) for clock1, must we not?

This means that at the instant that an event occurs at clock2 at time tB, clock1 MUST be at the position F (-v*tB,tB). The Lorentz transformation of the event gives it at point C on the path of clock1. So you are arguing that as soon as the event occurs at clock2 at time tB, clock1 "teleports instantaneously" to the position C (-v*t1,t1) without having to move with speed v from F to C during the time interval tB-t1.

Neither does it seem possible that after clock1 has moved from F to C, that clock 2 will still be at position D (0,v*tB): It must be at position (say) G (0,t1). You are thus claiming that clock 1 can teleport instantaneously from position F to position C: Wow!
Fleetfoot
3 / 5 (2) May 10, 2013
You are not so stupid not to realize that you are completely dishonest and acting disgracefully and unscientifically.


On the contrary, I am the one stopping you skipping over key scientific points.

You are going farther than we have agreed.


What have we agreed that it is so "holy" that it is a heresy to transgress it. I do not know of such an agreement.


See below ...

The generic equation D = -v*t describes the location of clock1 for all t, that is the black line labelled "clock1" through the origin.


Where have I denied this?


Sounds like an agreement.

Furthermore the generic equation D/=t gives the black line labelled clock 2 which gives the corresponding time-"position" at clock2 for ALL t.


Yep, we agree that too. So now the question is do you also agree that we plot the locations for a specific time at the points on the chart that I indicated? If not, why not?
johanfprins
1 / 5 (4) May 10, 2013
You are not so stupid not to realize that you are completely dishonest and acting disgracefully and unscientifically.


On the contrary, I am the one stopping you skipping over key scientific points.


A lie. If you do not want me to skip over "key scientific" points you would not mind to revisit those points. The fact that you refuse to do so proves that YOU are the one "skipping over key scientific points" and refuse to admit that you are doing it!

You are going farther than we have agreed.


What have we agreed that it is so "holy" that it is a heresy to transgress it. I do not know of such an agreement.


See below ...etc.

Sounds like an agreement.


So if we agree what did I do so that you accused me that I am going further than to what we agreed?

So now the question is do you also agree that we plot the locations for a specific time at the points on the chart that I indicated?
Only after including the points F and G.
Fleetfoot
3 / 5 (2) May 10, 2013
So what makes the time t2 at A any different from any other time point on the vertical time-axis? Why is it for example different from (say) time tB at position B on that axis.


By construction. We plot the point at the value read from the clock so if tB /= tA they will be different points.

tB is ALSO a specific time on clock2; clearly so we must also "plot only a single point (say F) at (-v*tB,tB) for clock1, must we not?


Correct.

This means that at the instant that an event occurs at clock2 at time tB, clock1 MUST be at the position F (-v*tB,tB).


Obviously, if x = -v*t is generally true, so is that specific case.

The Lorentz transformation of the event gives it at point C on the path of clock1.


No, the Lorentz Equations convert plot coordinates in K/ to the corresponding values in K and vice versa. They only allow you to draw lines or dots on one of the panels of my graphic from the coordinates in the other.
johanfprins
1 / 5 (5) May 10, 2013
BTW: If you want to see the correct derivation and interpretation of the Lorentz-transformation, I have just now uploaded it at http://www.cathod...tzT1.pdf
Fleetfoot
3 / 5 (2) May 10, 2013
On the contrary, I am the one stopping you skipping over key scientific points.


A lie. If you do not want me to skip over "key scientific" points you would not mind to revisit those points.


I'm happy to revisit anything you like, ask away.

You are going farther than we have agreed.


So if we agree what did I do so that you accused me that I am going further than to what we agreed?


You said I wanted to draw a horizontal line between the dots. I had only suggested the lines labeled "Clock1" and "Clock2" and the two dots at A and B on the top panel. I think you agree with the lines but perhaps not the location of the dots.

So now the question is do you also agree that we plot the locations for a specific time at the points on the chart that I indicated?


Only after including the points F and G.


We can add more points next but if you disagree with my locations of A and B, where do you think they should be plotted.
johanfprins
1 / 5 (4) May 10, 2013
So what makes the time t2 at A any different from any other time point on the vertical time-axis? Why is it for example different from (say) time tB at position B on that axis.


[By construction. We plot the point at the value read from the clock so if tB /= tA they will be different points.


Of course they are different points on the same time axis: Are you trying to be funny, or are you just plain stupid. My question refers to the fact that when t2 at A gives the coordinates for clock 1 at (-v*t2,t2), why is tB so different that it does not give the coordinates of clock1 as (-v*tB,tB)? There is no physics reason why point A must be an exception to all other points on the timeline!

The Lorentz transformation of the event gives it at point C on the path of clock1.

No, the Lorentz Equations convert plot coordinates in K/ to the corresponding values in K and vice versa.
johanfprins
1 / 5 (4) May 10, 2013
They only allow you to draw lines or dots on one of the panels of my graphic from the coordinates in the other.


So what? You have shown that the Lorentz transformation gives the point C on the path of clock 1 in panel 1. Are you now claiming that this position is not on the path of clock1?

You are now becoming totally irrational.

Fleetfoot
3 / 5 (2) May 10, 2013
So what makes the time t2 at A any different from any other time point on the vertical time-axis? Why is it for example different from (say) time tB at position B on that axis.


By construction. We plot the point at the value read from the clock so if tB /= tA they will be different points.


Of course they are different points on the same time axis: Are you trying to be funny, or are you just plain stupid.


I'm answering the question you asked. If you don't want stupid answers, don't ask stupid questions.

My question refers to the fact that when t2 at A gives the coordinates for clock 1 at (-v*t2,t2), why is tB so different that it does not give the coordinates of clock1 as (-v*tB,tB)?


It does!

There is no physics reason why point A must be an exception to all other points on the timeline!


Of course. I have no idea why you think I disagree with that, two different times just give two different points, obviously.
Fleetfoot
3 / 5 (2) May 10, 2013
They only allow you to draw lines or dots on one of the panels of my graphic from the coordinates in the other.


So what? You have shown that the Lorentz transformation gives the point C on the path of clock 1 in panel 1.


Yes, but that is Lorentz Transformed from the lower panel which we have not yet discussed. Remember I said we could just ignore everything not yet agreed? If I had redrawn the graphic as you requested, there would be no "C" or "D" points on it yet.

Are you now claiming that this position is not on the path of clock1?

You are now becoming totally irrational.


No, you just forgot what you asked me to do. We'll get to C and D later, and the lower panel which, for the moment, we have to treat as entirely blank.

{Taking a break, back later.}
johanfprins
1 / 5 (4) May 10, 2013
You said I wanted to draw a horizontal line between the dots. I had only suggested the lines labeled "Clock1" and "Clock2" and the two dots at A and B on the top panel. I think you agree with the lines but perhaps not the location of the dots.


What have the dots A and B in common with the time at D? Why bring in an extra later time at A? It is nonsensical. You should draw a horizontal line through D which cuts the path of clock1 at F (-v*t2,t2). This gives the position of clock1 on your top graph when an event occurs at D. The Lorentz transformation gives a later position (-v*t1,t1) at C; not the clock1's position at F when the event occurred.

We can add more points next but if you disagree with my locations of A and B, where do you think they should be plotted.
In order to be relevant with the event at D they should not be plotted at all, since the points D and F on the same horizontal line are the relevant points.
johanfprins
1 / 5 (4) May 10, 2013
Yes, but that is Lorentz Transformed from the lower panel which we have not yet discussed. Remember I said we could just ignore everything not yet agreed? If I had redrawn the graphic as you requested, there would be no "C" or "D" points on it yet.
And when I asked you start from the start you refused in a very unscientific manner. Now that that you are in a corner you suddenly want to retrace your steps. I was hoping that I finally struck an intelligent respondent, but is seems that it is just not in the human nature to see mountains on the moon if the person does not want to see mountains on the moon.

No, you just forgot what you asked me to do. We'll get to C and D later, and the lower panel which, for the moment, we have to treat as entirely blank.
No I did not forget.

Fleetfoot
3 / 5 (2) May 10, 2013
Yes, but that is Lorentz Transformed from the lower panel which we have not yet discussed. Remember I said we could just ignore everything not yet agreed? If I had redrawn the graphic as you requested, there would be no "C" or "D" points on it yet.


And when I asked you start from the start you refused in a very unscientific manner.


There is nothing "scientific" about wasting time, I just overestimated your ability to keep track.

Now that that you are in a corner you suddenly want to retrace your steps.


Nope, I'm happy to move on from where we had reached. We had agreed the the lines for the clocks on the top panel and nothing more.

No, you just forgot what you asked me to do. We'll get to C and D later, and the lower panel which, for the moment, we have to treat as entirely blank.


No I did not forget.


Ah, so are you now claiming your error was deliberate?

If you want to have an intelligent conversation, stop playing word games.

(contd.)
johanfprins
1 / 5 (6) May 10, 2013
Ah, so are you now claiming your error was deliberate?
Which error? I have not made a single error. That is you forte.

If you want to have an intelligent conversation, stop playing word games.
I have NOT been playing word games anywhere: This is also your forte!

Fleetfoot
3 / 5 (2) May 10, 2013
I was hoping that I finally struck an intelligent respondent, ...


Maybe you have, but you must then expect to justify your claims. Here is how you started your attempt at a proof:

OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.


Since you can't keep track, here is the resulting construction step:

https://sites.goo...truction

If you don't agree the top version, say what you think is wrong with it. Otherwise, let's move on. The bottom version includes the representation of your quoted words so if that's not what you meant, you can correct your "impeccable" proof and we'll try again.
Fleetfoot
3 / 5 (2) May 10, 2013
Ah, so are you now claiming your error was deliberate?
Which error? I have not made a single error.


You brought C and D in after saying we should start from the beginning. You then said you hadn't forgotten so it must have been deliberate.

Anyway, I've put up what you wanted, a diagram that "starts from the start" to show the construction so I'll wait for your comments on that, there are no "C" or "D" to confuse you this time, they'll get added later (if we ever get that far).
Fleetfoot
3 / 5 (2) May 10, 2013
Which error? I have not made a single error.


Let me remind you then:

After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.


That is wrong. If you are talking about frame K/, you next statement is correct:

The coordinates of clock2 within K/ are:

x/=0 and t/=t2


That is correct, call it event A. Since x/ = 0, clock2 has not moved.


It is your corrected version that we are now discussing:

OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.


That is what is shown in the "next step" diagram.
johanfprins
1 / 5 (5) May 10, 2013
I was hoping that I finally struck an intelligent respondent, ...


Maybe you have, but you must then expect to justify your claims.
Which I have done impeccably.

Here is how you started your attempt at a proof:


OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.


Since you can't keep track, here is the resulting construction step:

https://sites.goo...truction


If you don't agree the top version, say what you think is wrong with it.
It only contains the two paths: So far so good.

Otherwise, let's move on. The bottom version includes the representation of your quoted words so if that's not what you meant, you can correct your "impeccable" proof and we'll try again.
To avoid the confusion, replace A with D (event) and B with F.
johanfprins
1 / 5 (4) May 10, 2013
Let me remind you then:

After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.

That is wrong. If you are talking about frame K/, you next statement is correct:


It is not wrong, since the distance between the clocks MUST be the same no matter in terms of which clock you calculate it, or else the two clocks must be at different distances from one another, which is so absurd that only an insane mind will accept that it is possible!

It is your corrected version that we are now discussing:


OK let me put it in another way, after synchronization clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.


This is not a "corrected" version: It is the SAME version since the first clock cannot be further from the second clock than the second clock is from the first clock. So it is a blatant lie that I gave two different versions!

johanfprins
1 / 5 (5) May 10, 2013
We have lightning: To save my router I am signing off for now!
Fleetfoot
3 / 5 (2) May 10, 2013
Let me remind you then:

After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.

That is wrong. If you are talking about frame K/, you next statement is correct:


It is not wrong, since the distance between the clocks MUST be the same no matter in terms of which clock you calculate it, or else the two clocks must be at different distances from one another, which is so absurd that only an insane mind will accept that it is possible!


The distance is fine, the error was that you said clock2 had moved in frame K/ but K/ is the rest frame of clock2. In the second version, you corrected that to say that clock1 had moved.

Without that correction you might have meant clock2 moved in frame K.

We now have an agreed version but don't claim you have never made an error, that isn't true.
Fleetfoot
3 / 5 (2) May 10, 2013
.. here is the resulting construction step:

https://sites.goo...truction

If you don't agree the top version, say what you think is wrong with it.


It only contains the two paths: So far so good.


Excellent, that's a beginning.

Otherwise, let's move on. The bottom version includes the representation of your quoted words so if that's not what you meant, you can correct your "impeccable" proof and we'll try again.


To avoid the confusion, replace A with D (event) and B with F.


The applet adds letters in sequence when you right-click and since these are the first two we have defined, I'll stick with A and B. They are just identifiers with no other meaning so which letters are used is irrelevant. If there is any confusion, I'll change the subsequent labels.

What matters now is whether you think having them at the same vertical coordinate is correct or not (I assume you are happy that they must lie on the clock lines).
johanfprins
1 / 5 (5) May 10, 2013
The distance is fine, the error was that you said clock2 had moved in frame K/ but K/ is the rest frame of clock2.
That is a blatant LIE! I have NOT stated that clock2 moved relative to K/. Why do you have to be dishonest to try and score cheap shots?

In the second version, you corrected that to say that clock1 had moved.
I also stated this in the first version like anybody with a modicum of rains will attest to.

Without that correction you might have meant clock2 moved in frame K.
I have stated very clearly that clock 1 moves within K/ and clock2 moves within K so that according to the symmetry whene the clocks are a distance D from one another one must have that D=v*t on clock1 and D=v*t/ on clock2 and therefore t must be thae same as t/.

We now have an agreed version but don't claim you have never made an error, that isn't true.
I do claim that, since I have NOT made any serious errors in the physics I have posted on this thread.
Fleetfoot
3 / 5 (2) May 10, 2013
The distance is fine, the error was that you said clock2 had moved in frame K/ but K/ is the rest frame of clock2.


That is a blatant LIE! I have NOT stated that clock2 moved relative to K/.


Here is the quote again, it is quite clear:

The clocks are synchronized when they move "through" one another so that t=t/=0. After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.


Why do you have to be dishonest to try and score cheap shots?


You accused me of dishonesty which is a lie, I have made mistakes but I have been honest throughout. If you are going to act that way, put your own house in order first.

We now have an agreed version but don't claim you have never made an error, that isn't true.


I do claim that, since I have NOT made any serious errors in the physics I have posted on this thread.


I didn't say it was serious or in the physics, just a minor error in your statement of the question.
johanfprins
1 / 5 (4) May 10, 2013
The applet adds letters in sequence when you right-click and since these are the first two we have defined, I'll stick with A and B. They are just identifiers with no other meaning so which letters are used is irrelevant. If there is any confusion, I'll change the subsequent labels.


What matters now is whether you think having them at the same vertical coordinate is correct or not (I assume you are happy that they must lie on the clock line


If you mean by A the event D (the event at t2) in your previous diagram and by B the label F, I have proposed for the simultaneous coordinates (v*t2,t2) for clock 1 then fine, proceed!
johanfprins
1 / 5 (5) May 10, 2013
The clocks are synchronized when they move "through" one another so that t=t/=0. After a lapse of time of t2 on clock2, this clock has moved a distance D2=v*t2 from clock1.
Motion is relative: So do not dishonestly blame me that I have made a "mistake". An observer with clock2 will after a time t2 conclude that he and his clock has moved a distance v*t2 from clock1: Unless the observer is brain-damaged.

You accused me of dishonesty which is a lie, I have made mistakes but I have been honest throughout. If you are going to act that way, put your own house in order first.
If you think that you have been honest, I must accept that you think so; but you are wrong: Your arguments have all along been dishonest.

I didn't say it was serious or in the physics, just a minor error in your statement of the question.
There is no error, not even a minor one, as your own diagrams are proving.

Signing off for tonight!

Fleetfoot
3 / 5 (2) May 10, 2013
What matters now is whether you think having them at the same vertical coordinate is correct or not (I assume you are happy that they must lie on the clock line


If you mean by A the event D (the event at t2) in your previous diagram and by B the label F, I have proposed for the simultaneous coordinates (v*t2,t2) for clock 1 then fine, proceed!


Yes, almost, remember the x coordinate of B is negative as you stated the problem.

Just to confirm some detail:

Label A tags the coordinates (0, t2) which is the location of clock2 when it reads t2.

Label B tags the coordinates (-v*t2, t2) which is the location of clock1 when clock2 reads t2.

Both coordinate pairs are defined in frame K/, the rest frame of clock2.

I believe that reflects your description:

.. clock2 sees clock1 moving away from it at a speed -v so that within K/ after the elapse of time t2, the distance between clock1 and clock 2 must be D=v*t2.
Fleetfoot
3 / 5 (2) May 10, 2013
Motion is relative: So do not dishonestly blame me that I have made a "mistake". An observer with clock2 will after a time t2 conclude that he and his clock has moved a distance v*t2 from clock1: Unless the observer is brain-damaged.


Distances are absolute but the frame depends on which clock1 is deemed at rest and the sign of the coordinate changes too. You got that wrong again as I noted in the last post. It may seem trivial but when you put it into the transforms, the sign matters.

Signing off for tonight!


OK, I'll update the diagram later, move the "proposed" to "agreed" and add one or two more steps. E&OE, on this, I'm watching the football at the same time.
Noumenon
1.1 / 5 (28) May 10, 2013
I could have taught relativity to my dog by now.
Fleetfoot
3 / 5 (2) May 10, 2013
Signing off for tonight!


OK, I'll update the diagram later, move the "proposed" to "agreed" and add one or two more steps. E&OE, on this, I'm watching the football at the same time.


OK, the construction page has been updated with two new diagrams. What has been agreed so far is at the top now. The next diagram just adds the clock lines in the lower panel so should not be a problem.

The third diagram then shows where A and B plot when the Lorentz Transforms are applied and shows clearly that the distance between them has increased, distance is not invariant.

The other two events you defined in that frame will be the subject of the next diagram if it is needed.
Fleetfoot
1 / 5 (2) May 11, 2013
Distances are absolute ...


By which I meant we always treat them as positive while the x coordinate can be negative, not that "space is absolute" in case there is any confusion.
johanfprins
1 / 5 (5) May 11, 2013
I could have taught relativity to my dog by now.


You are not able to teach any physics, even to a genius!
johanfprins
1 / 5 (5) May 11, 2013
OK, the construction page has been updated with two new diagrams. What has been agreed so far is at the top now.
Provided that you agree that point A is an event that occurs within K/ at the time t2 within K/.

Yes or no?

After you have answered I will proceed.

ValeriaT
1 / 5 (5) May 11, 2013
This discussion is deeply annoying and apparently dedicated to PM thread - but it illustrates clearly, what the AWT-like pictures and animations are good for. The formally thinking physicists are getting stuck in description of mutualy dual perspectives often.
johanfprins
1 / 5 (5) May 11, 2013
This discussion is deeply annoying
Annoying to you since you are incapable of arguing logic. What I find refreshing in the case of Fleetfoot is that he is mostly arguing, and trying to argue his point in a manner that a physicist must do.

and apparently dedicated to PM thread -
What PM thread?

but it illustrates clearly, what the AWT-like pictures and animations are good for.
Yes it does: It illustrates clearly that the AWT-like pictures have NOTHING to do with physics and never will have anything to do in physics.

The formally thinking physicists are getting stuck in description of mutualy dual perspectives often.
Not so! I think that Fleetfoot and I are not yet stuck but have been making progress towards a mutual understanding: This is a REAL scientific debate which is not possible when using your AWT hallucinations! No wonder you find it annoying, since you do not have a clue about what physics is really about!
ValeriaT
1.6 / 5 (8) May 11, 2013
What PM thread?
A Personal Messaging functionality, indeed.
I think that Fleetfoot and I are not yet stuck but have been making progress towards a mutual understanding
. What you think is irrelevant, only what you can prove. You're no more to mutual understanding, than before six days.
this is a REAL scientific debate
This is clueless dispute. I'm far closer to understanding of your (biased) perspective, than every mainstream physicist in this thread. If you don't believe, just try to simply ask Fleetfoot for reproduction of your ideas in his own words - and we will see...;-) You're living in illusion about comprehensibility and logical consistency of your ideas.
Noumenon
1.3 / 5 (28) May 11, 2013
I could have taught relativity to my dog by now.


You are not able to teach any physics, even to a genius!


I didn't mean to offend Feetfoot (by using an insult that johanfprins has used on previous occasions towards others), as I don't think he is really trying to 'teach johanfprins physics' per se, as much as he is trying to corner an angry greased duck.
johanfprins
1 / 5 (6) May 11, 2013
@ValeriaT,

You are definitely the most pathetic human being with whom I have ever had contact. Prayers will not help. Maybe you should be shot to put you out of your misery!

If you don't believe, just try to simply ask Fleetfoot for reproduction of your ideas in his own words - and we will see...;-)
Is this not what Fleetfoot is doing? So, when he returns and we conclude our discussion, we will truly see! If he can prove me wrong I will obviously admit it. I am not as stupid as you are to believe like a fanatic in my own ideas. AWT, AWT, AWT ,AWT LOL!

ValeriaT
1.7 / 5 (6) May 11, 2013
If he can prove me wrong I will obviously admit it
The absence of time dilatation is not matter of Fleetfoot (or someone else) agreement - but the falsification of many years standing robust experiments. No theory can beat the experimental facts.
johanfprins
1 / 5 (3) May 11, 2013
@Fleetfoot: It seems to me that you are not coming back. So I will proceed to analyze your diagrams posted on https://sites.goo...ruction.

1. Your first two diagrams: K/ has the paths and the position A of clock1 and the position B of clock2 with the correct distance D between them: K has nothing on it: So far no problem.

2. Your second two diagrams: K/ still the same: The concomitant paths added to K; So far no problem.

3. Your third two diagrams: K/ still the same. In K a Lorentz-transform for an event at clock 1 is shown at time t2. We have NOT discussed a Lorentz transform from clock1 to clock2 at time t2 on clock1, but from clock2 at t2. So the diagram is irrelevant to the discussion.

johanfprins
1 / 5 (3) May 11, 2013
If you represented the LT from clock2 at time t2, as you should have, also on the diagram for K, then even in this diagram, A on the path of clock2 should have been at a time t2, and the Lorentz transformed time at a later time on the path of clock1. Thus B on the vertical line must be replaced by a transformation from A on the line of clock2 into a point D on the line for clock 1 where the time tD will be higher than the time t2.

For A and B to correspond in the diagram for K with the diagram for K/ they should also both have been on the horizontal line going through the same time t2. The LT time-coordinates should be denoted by D on both diagrams at the same higher time-value tD than t2 on the path-lines of clock1 on both diagrams.

Representing the path-line of clock 2 as vertical with clock1 as moving away with speed -v, MUST give the same result when you represent the path-line of clock 1 as vertical and clock 2 moving away at a speed v. The motion IS not absolute.
johanfprins
1 / 5 (3) May 11, 2013
If he can prove me wrong I will obviously admit it
The absence of time dilatation is not matter of Fleetfoot (or someone else) agreement - but the falsification of many years standing robust experiments. No theory can beat the experimental facts.


There has not been a SINGLE experiment ever where two clocks have been compared after having moved all the way LINEARLY relative to one another within gravity-free space. So there is NO experimental proof that the time-dilation predicted by Einstein does actually occur! You are lying again!
ValeriaT
1.8 / 5 (8) May 11, 2013
there has not been a SINGLE experiment ever where two clocks have been compared after having moved all the way LINEARLY relative to one another within gravity-free space
You're willingly lying or you're trolling loudly (or both). It's done routinely with GPS satellites every day. For example Hasselkamp, Mondry, and Scharmann measured the Doppler shift of atoms moving at right angles to the line of sight (so no transverse Doppler effect should exist there) - yet the light was redshifted in an agreement with special relativity theory. You're crackpot, who just cannot face the facts - so he pretends, no such facts exists.
Fleetfoot
3 / 5 (2) May 11, 2013
OK, the construction page has been updated with two new diagrams. What has been agreed so far is at the top now.


Provided that you agree that point A is an event that occurs within K/ at the time t2 within K/.

Yes or no?

After you have answered I will proceed.


I've been out all day so I see you have moved on anyway but just for the record, yes that is what is shown. The vertical arrow marked "t2" was intended to indicate that, and the horizontal arrow marked "D" indicates the distance between the clocks at that time.

https://sites.goo...truction
Fleetfoot
3 / 5 (2) May 11, 2013
I didn't mean to offend Feetfoot (by using an insult that johanfprins has used on previous occasions towards others), ..


None taken, I just thought you were commenting on how slowly the conversation is developing.

as I don't think he is really trying to 'teach johanfprins physics' per se, as much as he is trying to corner an angry greased duck.


I think I understand Johan's argument, I'm just trying to show him where the problem in it lies. I suspect there may be other lurkers who will be interested in the development of the SR explanation too though.
Fleetfoot
3 / 5 (2) May 11, 2013
1. Your first two diagrams: K/ has the paths and the position A of clock1 and the position B of clock2 with the correct distance D between them: K has nothing on it: So far no problem.

2. Your second two diagrams: K/ still the same: The concomitant paths added to K; So far no problem.


Good, I'll take that as the new baseline agreement from now on.

3. Your third two diagrams: K/ still the same. In K a Lorentz-transform for an event at clock 1 is shown at time t2.


That is event B. However, you need to count the scale, t2 is seven units up but B is shown at 5 units and A at nearly 10.

We have NOT discussed a Lorentz transform from clock1 to clock2


To be pedantic, the LTs transform between frames, not clocks, though you have defined the frames based on the clocks in this case.

at time t2 on clock1, but from clock2 at t2.


I mentioned those at the bottom of the page, I'll do another diagram for them later but let's discuss where A and B are in K first.
Fleetfoot
3 / 5 (2) May 11, 2013
If you represented the LT from clock2 at time t2, as you should have, also on the diagram for K, ..


The Lorentz Transforms are:

t' = g (t - vx)

x' = g (x - vt)

where g (gamma) is:

g = 1 / sqrt(1 - v^2)

The coordinates B in K/ are (-5, 7), near enough, just apply the equations and see what coordinates you get for B in K. That's where I and the applet plot it.
ValeriaT
1 / 5 (5) May 11, 2013
I think I understand Johan's argument, I'm just trying to show him where the problem in it lies. I suspect there may be other lurkers who will be interested in the development of the SR explanation
Explanation of blind alleys in it, be more specific? Nope, I don't think I do want to be involved in things, which just cannot work at the first sight. My life is too short and pretty for being wasted in this way at public. JFPrins is old already, he has nothing to lose (which explains, why so many old chaps suffer with sudden flash of creativity and opposition against mainstream ideas before their end).
johanfprins
1 / 5 (6) May 11, 2013
You're willingly lying or you're trolling loudly (or both). It's done routinely with GPS satellites every day. What is done routinely with GPS satellites? I do not know of ANY GPS satellite which is moving linearly through gravity free space. Give me an example of such a GPS satellite please!

For example Hasselkamp, Mondry, and Scharmann http://adsabs.har...89..151H of atoms moving at right angles to the line of sight (so no transverse Doppler effect should exist there) - yet the light was redshifted in an agreement with special relativity theory.
Of course it will be so according to the Lorentz transformations which, however, DO NOT model time-dilation. So this is NO PROOF of time dilation whatsoever.

I can derive the Doppler effect for you correctly but you will no understand it since I cannot model it in terms of a paddling duck farting bubbles into a non-existing aether.
johanfprins
1 / 5 (4) May 11, 2013
If you represented the LT from clock2 at time t2, as you should have, also on the diagram for K, ..


The Lorentz Transforms are:

t' = g (t - vx)

x' = g (x - vt)

where g (gamma) is:

g = 1 / sqrt(1 - v^2)

The coordinates B in K/ are (-5, 7), near enough, just apply the equations and see what coordinates you get for B in K. That's where I and the applet plot it.


The fact still remains that we are talking about a LT from the position of clock2 within K/ into K, while you have now, for no logical reason whatsoever brought in a LT from the position of clock1 within K into K/. What has this to do with our discussion? We are ONLY talking about an event at clock2 within K/ at time t2 being LT into K. This is what must be shown on both diagrams since this is the action that must be compared on both diagrams.
johanfprins
1 / 5 (4) May 11, 2013
JFPrins is old already, he has nothing to lose (which explains, why so many old chaps suffer with sudden flash of creativity and opposition against mainstream ideas before their end).
But fortunately not young and brainless like you! You will NEVER have a "flash of creativity" since you are far too stupid!

ValeriaT
1 / 5 (5) May 11, 2013
I do not know of ANY GPS satellite which is moving linearly through gravity free space.
You're twisting facts again. Are we still talking about special relativity?The special relativity (postulates) do apply to inertial motion, not to the "linear motion through gravity free space".
I can derive the Doppler effect for you correctly
I do care just about result. At the moment, when you say, no time dilatation can be observed, then you're just plain wrong and I'm not required (to be able) to lay the egg, which is already smelling clearly.

BTW Do you believe in relativistic increase of mass? Because this is just the effect, which is responsible for relativistic dilatation of time in AWT (illustratively speaking, the time runs more slowly in increased gravitational potential, as the general relativity requires). And the relativistic increasing of mass can be observed very easily. So if you deny the time dilatation, you must deny the relativistic increase of mass as well.
Fleetfoot
1 / 5 (1) May 11, 2013
If you represented the LT from clock2 at time t2, as you should have, also on the diagram for K, ..


The Lorentz Transforms are:

t' = g (t - vx)

x' = g (x - vt)

where g (gamma) is:

g = 1 / sqrt(1 - v^2)

The coordinates B in K/ are (-5, 7), near enough, just apply the equations and see what coordinates you get for B in K. ..


The fact still remains that we are talking about a LT from the position of clock2 within K/ into K,


The coordinates of A are (0, 7) in K/. We have both events A and B to process, but start with that if you prefer.

clock1 within K into K/. What has this to do with our discussion?


No, I only discussed transforming from K/ to K, not the other way.

We are ONLY talking about an event at clock2 within K/ at time t2 being LT into K.


We need to transform BOTH your events in K/ into K, A and B. I did not intend to restrict the conversation to either one, I just used B as an example.
angelhkrillin
1 / 5 (5) May 11, 2013
Einstein's theories won't be proven wrong...they are just incomplete or should I say people aren't thinking about the bigger picture...
Fleetfoot
1 / 5 (1) May 12, 2013
Einstein's theories won't be proven wrong...they are just incomplete or should I say people aren't thinking about the bigger picture...


They only address one part and for what they do, they are accurate. The question is whether they are intrinsic or emergent from QM and if the latter, how do we construct a theory that merges the two.
Fleetfoot
1 / 5 (2) May 12, 2013
The Lorentz Transforms are:

t' = g (t - vx)

x' = g (x - vt)

where g (gamma) is:

g = 1 / sqrt(1 - v^2)

...

The coordinates of A are (0, 7) in K/. We have both events A and B to process, but start with that if you prefer.


We are ONLY talking about an event at clock2 within K/ at time t2 being LT into K.


OK, I have updated the construction page for the new agreed baseline and then split out just A.

If you can deal with that on its own, we can apply the LTs to B next, that shouldn't be hard as it's also from K/ to K. I've put that on the page as well as the existing merged diagram just to save editing time but just deal with the first diagram for A only first.

Once we have agreement on those two, we can discuss how you intend to deal with the reverse transforms from K to K/ but if you cant handle applying the forward transforms to both A and B, I suspect we'll have even more trouble with the reversing them. One thing at a time then.
johanfprins
1 / 5 (5) May 12, 2013
I'm not required (to be able) to lay the egg, which is already smelling clearly.
You are an expert on laying rotten eggs. Your next statement requires a response.

BTW Do you believe in relativistic increase of mass?
I do not have to believe it since one can derive it directly from Newton's second law.

Because this is just the effect, which is responsible for relativistic dilatation of time in AWT
No it is NOT!

(illustratively speaking, the time runs more slowly in increased gravitational potential, as the general relativity requires).
That is true but it has NOTHING to do with AWT.

And the relativistic increasing of mass can be observed very easily. So if you deny the time dilatation, you must deny the relativistic increase of mass as well
Nonsense as usual. The increase of relativistic mass with linear speed has NOTHING to do with time dilation. When a body moves along a circle under the action of a central force, its mass is its rest mass.
johanfprins
1 / 5 (6) May 12, 2013

No, I only discussed transforming from K/ to K, not the other way.
You have posted the following equations:

t' = g (t - vx) x' = g (x - vt)
They are clearly LT's from K into K/: Not K/ to K as you are claiming.

We need to transform BOTH your events in K/ into K, A and B.
I do not know of two events. We are only discussing a SINGLE event at clock2 at time t2. Where is the other event occurring?

OK, I have updated the construction page for the new agreed baseline and then split out just A.
Would you be kind in enough in future to post the lead to the construction page so that I do not have to scan back to get to it? I have looked at it and it still makes no sense to me. But I think I now can explain more cogently where you are making a mistake. I will do this in my following post below.

But first I want to thank you for your willingness to argue physics in such a detailed manner. It helped to sharpen my thoughts. I am indebted to you.
ValeriaT
1 / 5 (6) May 12, 2013
That is true but it has NOTHING to do with AWT
Of course it has, because the AWT explains the nature of the relativistic increase of mass. In AWT it's not mass of particle itself, but the mass of the vacuum dragged with motion of object - it's de Broglie/wake wave, what gains the mass of object in motion. And because this environment is more dense, it explains, why the energy is propagating more slowly in it, which is interpreted like the dilatation of time.
The increase of relativistic mass with linear speed has NOTHING to do with time dilation. When a body moves along a circle under the action of a central force, its mass is its rest mass.
The relativistic increase of mass is independent to geometry of object motion, it only depends on relative speed of object with respect to observer.

At any case, if the relativistic increase is undeniable fact for you, then the gravitational dilatation of time is required as well for the sake of consistency with general relativity.
johanfprins
1 / 5 (4) May 12, 2013
A relativistic coordinate transformation is more than just a coordinate transformation, since it transforms the physics that is occurring within one IRF to take on the form as it will be experienced within a passing IRF. Without anything happening within an IRF from which the transformation is done, the coordinate transformation is only a mathematical curiosity. Thus to treat it as physics on its own is nonsensical.

Assume we live in a universe where everything moves so slowly that we can use the Galilean transformation. When an IRF=K/ passes an IRF=K in gravity-free space, and within IRF=K/ a ball is launched at 0/ perpendicular to the direction of motion, the ball will recede perpendicularly from the direction of motion. When transforming this motion into IRF=K, the ball will move at an angle to the direction of motion away from 0. This sloped motion is CAUSED by the event of launching the ball within the IRF=K/.
ValeriaT
1 / 5 (5) May 12, 2013
It helped to sharpen my thoughts. I am indebted to you
As I explained already, the moving object surrounded with area of more dense vacuum can be described from two dual perspectives (intrinsic and extrinsic one) - it's subject of black hole complementarity so to say. Unfortunately, you just chosen just the perspective, which is inconsistent with the results of physical experiment, because its unobservable from perspective of stationary observer.
johanfprins
1 / 5 (4) May 12, 2013
It is thus ridiculous physics to transform this sloped motion back into IRF=K/, unless an actual ball is launched within IRF=K at the the same angle relative to the direction of motion so that it is then seen from IRF=K/ to move vertically away from the direction of motion: The event (which I will call the primary event) for this motion has not occurred within IRK=K/. Thus the Galilean transformation transforms primary events within IRF=K/ into IRF=K, and the reverse Galilean transformation transforms primary events within IRF=K into IRF=K/.

Although the transformation equations and the reverse transformation equations are mathematically inverse to one another, they are not physically inverse. They are only physically inverse when two identical events occur simultaneously at the same coincident position within IRF=K/ and IRF=K respectively: For example: If two balls are simultaneously launched perpendicular to the direction of motion from 0/ and 0 at coincidence.
johanfprins
1 / 5 (5) May 12, 2013
Consider thus two balls, one within IRF=K/ and one within IRF=K, which are simultaneously launched from 0 and 0/ when the coincide Within IRF=K/ the ball launched within IRF=K will move at an angle while the ball launched within IRF=K/ will not move at an angle. Similarly within IRF=K the ball launched within IRF=K/ will now move at an angle, while the ball launched within IRF=K will move perpendicularly.

When the Galilean transformation applies, the balls can be simultaneously launched at any time t from any coincident coordinate and observers at 0 and 0/ will agree that the balls were launched at the same instant in time at the same position in space.

This is not the case when the Lorentz-transformation applies: When, in this case, the two balls are launched at exactly the same time t from exactly the same coincident point P, the observer at 0 will see that his ball has been launched from P at time t, but the observer at 0/ will disagree.
johanfprins
1 / 5 (4) May 12, 2013
Similarly the observer at 0/ will see that his ball has been launched from P at exactly the same time t, but the observer at 0 will disagree. The reason for this is that the Lorentz transformation transform simultaneous events not to be simultaneous anymore after the transformation. If two separated primary events within IRF=K/ occurs simultaneously they are not simultaneous within IRF=K, and are not observed within IRF=K at the coincident positions that they actually occurred, and vice versa.

If two simultaneous primary events within IRF=K coincides with the two primary events within IRF=K/ thjese events will not be seen within IRF=K/ at the coincident positions and time at which they occurred: they will be seen at two different positions and at two different time on the clock within IRF=K/, even though this clock keeps time at exactly the same rate that the clock within IRF=K keeps time.

johanfprins
1 / 5 (4) May 12, 2013
If one could have left a mark within IRF=K at the coincident position at which an event occurs within IRF=K/, and after the event which is observed relative to 0 to happen at a different position and time within IRF=K, take a tape measure and measure the mark within IRF=K which the event within IRF=K/ has left when it occurred, this mark will not be at the LOrentz transformed distance at which the event has been experienced within IRF=K, but will be at the coincident position at which the event within IRF=K/ has occurred.

Thus the difference in times obtained from the Lorentz transformation is caused by the fact that simultaneous events are being transformed to be not simultaneous. It has nothing to do with one clock keeping time at a slower rate than the other.

johanfprins
1 / 5 (4) May 12, 2013
Simply stated, what is really occurring is that two events which occur simultaneously at coincident coordinates within IRF=K/ and IRF=K, are not simultaneously observed within the two IRF's.

For example, an event within IRF=K which occurs at the same time t and position in space P as an event within IRF=K/, will be seen within IRF=K at this the time t and position P, but the event that occurred simultaneously at time t and coincident at position P within IRF=K/ will not be seen at this position and time within IRF=K.

Similarly, an event within IRF=K/ which occurs at the same time t and position P in space as an event within IRF=K, will be seen within IRF=K/ at the time t and position P, but the event that occurred simultaneously at time t and coincident at position P within IRF=K will not be seen at this position and time within IRF=K/.

Thanks for your patience.
Fleetfoot
3 / 5 (2) May 12, 2013
You have posted the following equations:
t' = g (t - vx) x' = g (x - vt)
They are clearly LT's from K into K/: Not K/ to K as you are claiming.

That is wrong. Event A is when time t2 appears on clock2 as has been agreed and you defined frame K/ as that in which clock2 is at rest hence the known values are in K/ which the unknowns are in K.

It is conventional to put the knowns in an equation on the right hand side with the unknowns on the left hence x and t are the values in frame K/ while x' and t' are the values in frame K. The equations obviously transform from K/ to K.

We need to transform BOTH your events in K/ into K, A and B.
We are only discussing a SINGLE event at clock2 at time t2. Where is the other event occurring?

Your statement of the problem was that, in frame K/, the distance between clock1 and clock2 was D at time t2 on clock2. The second event B is the location of clock1 in K/ at time t2. You can't have a distance "between" one point!
Fleetfoot
1 / 5 (1) May 12, 2013
OK, I have updated the construction page for the new agreed baseline and then split out just A.


Would you be kind in enough in future to post the lead to the construction page so that I do not have to scan back to get to it?


I've had to split this reply because of the 1000 character limit so including it every time isn't realistic. Here it is again so you can bookmark it, the address won't change, only the contents:

https://sites.goo...truction

I have looked at it and it still makes no sense to me.


If you don't even understand the question you asked, you're not going to be able to work out how to apply the transforms to it.

But I think I now can explain more cogently where you are making a mistake. I will do this in my following post below.


I think that's pointless until you understand the question you have asked, especially since you didn't know which side of an equation the known and unknown values lie on.
Noumenon
1.3 / 5 (28) May 12, 2013
Similarly, an event within IRF=K/ which occurs at the same time t and position P in space as an event within IRF=K


This premise is faulty reasoning. You can't say this. You can't say "at the same time and position" in BOTH IRF's. That is operationally (experimentally) meaningless, unless you brought in yet another IRF, at which point it would be true ONLY from his K// POV, and K and K/ will still disagree wrt simultaneity .

Even though the two IRF's overlap in space, because of their relative velocity, the two times and spaces must be considered and used only as separate coordinates, either K, OR K/ OR K observes LT(K/), OR K/ observes LT(K),.. for any given measurement.

I think Fleetfoot was kind enough to step through your reasoning with you, so please continue.
ValeriaT
1.6 / 5 (7) May 12, 2013
This premise is faulty reasoning. You can't say this. You can't say "at the same time and position" in BOTH IRF's
Yep, exactly. The distant universe is causally separated from nearby one, it's not just a matter of distance, but the time dilatation too. The same mistake is quite widespread inside of mainstream physics, I'm explaining it by the example of black hole paradox for example here. The observer, falling into black hole would see the speed of light unchanged and straightforward even at the case, he would encircled the black hole together with photons, thus effectively staying at rest from more general perspective. But his perspective is not causally exchangeable with perspective of extrinsic observer from outside. Both observers are residing in somewhat different multiverse.
johanfprins
1 / 5 (6) May 12, 2013
That is wrong. Event A is when time t2 appears on clock2 as has been agreed and you defined frame K/ as that in which clock2 is at rest hence the known values are in K/ which the unknowns are in K.


Go to ANY textbook and you will find:

that x=g(x/+vt/) and t=(t/+(v/c^2)x/) are the equations when you transform a primary event KNOWN to occur at x/ and t/ within K/ (NOT within K) from K/ into K

And

that x/=g(x-vt) and t/=g(t-(v/c^2)x) are the equations when you transform a primary event KNOWN to occur at x and t within K (NOT within K/) from K into K/

Thus, your first equation t/=g(t-vx) is totally meaningless since it is NOT an equation of any Lorentz transformation. Your second equation x/=g(x-vt) is the transformation of an event KNOWN to be at position x and at time t within K. Not an event within K/ as you are claiming that it must be.

Can you REALLY not see that you are posting nonsense?
johanfprins
1 / 5 (3) May 12, 2013
Similarly, an event within IRF=K/ which occurs at the same time t and position P in space as an event within IRF=K


This premise is faulty reasoning. You can't say this. You can't say "at the same time and position" in BOTH IRF's. That is operationally (experimentally) meaningless,


Please be kind and answer the following question:

Can a position-coordinate point P/ as referenced within IRF=K/ coincide with a position-coordinate point P within IRF=K? Yes or no?

If the answer is no please explain how two reference frames can have position coordinates of which not a single point-coordinate within the one can coincide with any point coordinate within the other.

If the answer is yes, then define the conditions that must prevail for coordinate point P/ to coincide with coordinate point P. Remember that points P and P/, whether the coincide or not, are moving relative to one another.

I am looking forward to your answer.
johanfprins
1 / 5 (5) May 12, 2013
The relativistic increase of mass is independent to geometry of object motion, it only depends on relative speed of object with respect to observer.


Prove this statement for circular motion: And do it mathematically not by using a paddling duck farting bubbles in a non-existing aether.

ValeriaT
1 / 5 (5) May 12, 2013
Why mathematically? The physics is experimental science and the experiment has always the latest word in it. It was confirmed experimentally in Hafele-Keating experiment and all its later replications. From this animation it's evident, that the time dilatation is independent on the clock motion direction. Every circular path can be approximated with arbitrary number of linear paths - if some difference should appear there, then just during object rotation in space. Why do you think, that the speed of clock should change during rotation of object?
Fleetfoot
3 / 5 (2) May 12, 2013
That is wrong. Event A is when time t2 appears on clock2 as has been agreed and you defined frame K/ as that in which clock2 is at rest hence the known values are in K/ which the unknowns are in K.


Go to ANY textbook and you will find:


I just checked and you're right, however, for convenience I was using the Wikipedia page which describes them the other way round:

https://en.wikipe...irection

One set are at the top, what they call the inverse transforms are a bit lower down.

That's the same error you made some time ago when you subsequently said:

I apologize: The correct equations from K/ to K are:

x=(gamma)*(x/ plus v*t/)
y=y/
z=z/
t=(gamma)*(t/ plus (v/c^2)*x/)


Anyway, the point is that the next step in our discussion is for you to apply whichever equations you want to event A and if you get a different answer to mine, we can investigate the cause. Do you want to continue or give up?
johanfprins
1 / 5 (4) May 12, 2013
Why mathematically? The physics is experimental science and the experiment has always the latest word in it. It was confirmed experimentally in Hafele-Keating experiment and all its later replications.


So according to you Haefele -Keating did not try confirm a mathematical model?

Every circular path can be approximated with arbitrary number of linear paths -
This what calculus tells you: You get the circular path by allowing the linear paths to increase and each in the limit top go to zero. It is then NOT an approximation anymore. That is why mathematics is successful and your farting duck is nonsense.

Why do you think, that the speed of clock should change during rotation of object?
I am not "thinking" but doing mathematics to check the logic of my thinking: Only an idiot like you will think it is not necessary. And I have NOT posted that the speed of a clock will change during rotation. Time only changes within a gravitational field.
johanfprins
1 / 5 (5) May 12, 2013
That is wrong. Event A is when time t2 appears on clock2 as has been agreed and you defined frame K/ as that in which clock2 is at rest hence the known values are in K/ which the unknowns are in K.


Go to ANY textbook and you will find:


I just checked and you're right,
Thank you.

However, for convenience I was using the Wikipedia page which describes them the other way round:

The mistake made by Wikipedia and also by you, and the mainstream physicists over the past century, is to think that you can transform a single event in space into both IRF's without first ascertaining within which IRF the event is occurring. What this means is to determine whether the cause for the event is stationary within IRF=K or stationary within IRF=K/. Only such an event can be transformed from the IRF within which it is caused into another IRF. The transformed event cannot be transformed back onto itself by using the reverse equations.

johanfprins
1 / 5 (5) May 12, 2013
In your analysis you are transforming a transformed event back into the IRF within which it was caused. Although mathematically possible, it is not physically possible since it destroys causality.

I have derived the Lorentz equations by using wave-fronts emitted from laser beams when 0 and 0/ coincide, instead of a spherical wave front. The result is different when the laser source is stationary within K and when the laser source is stationary within K/; proving that you cannot transform a transformed event back into the IRF from which it has been transformed.

I have already posted where this manuscript can be read, and I am doing so again:

See: http://www.cathod...tzT1.pdf

So please read it. You will then see where you are making your mistake and where this same mistake has no been made for more than a century.
johanfprins
1 / 5 (4) May 12, 2013
Anyway, the point is that the next step in our discussion is for you to apply whichever equations you want to event A and if you get a different answer to mine, we can investigate the cause. Do you want to continue or give up?
It is not a question of giving up: The fact is that one is doing nonsensical physics when you use the reverse equations to transform a transformed event back into the IRF from which you have transformed it in the first place.

Although it is mathematically possible, this is not physically possible and it is for this reason that Voodoo conclusions like time-dilation and length contraction have been incorrectly derived. And it is for this reason why it has not been realized to date that Bohr's quantization-rule for atomic orbits and Schroedinger's equation, are both consequences of Einstein's Special Theory of Relativity and Maxwell's equations.
Fleetfoot
1 / 5 (1) May 12, 2013
That is wrong. ...


Go to ANY textbook and you will find:


I just checked and you're right,
Thank you.


As I said, I have no problem admitting when I make mistakes. In this case, the labels are just swapped. I've checked the spreadsheet version and I used the correct equations on that so my numbers are correct.

The mistake made by Wikipedia and also by you, and the mainstream physicists over the past century, is to think that you can transform a single event in space into both IRF's without first ascertaining within which IRF the event is occurring.


It occurs in both frames, they are just different coordinate descriptions. You do have to know however, which frames have the known and unknown values.

Anyway, we agree that the values for A are known in clock2's rest frame so carry on, what values do you get for the other frame?
Fleetfoot
1 / 5 (1) May 12, 2013
In your analysis you are transforming a transformed event back into the IRF within which it was caused. Although mathematically possible, it is not physically possible since it destroys causality.


No, if you carry on you'll find that's not true.

I have derived the Lorentz equations by using wave-fronts emitted from laser beams when 0 and 0/ coincide, instead of a spherical wave front. The result is different when the laser source is stationary within K and when the laser source is stationary within K/; proving that you cannot transform a transformed event back into the IRF from which it has been transformed.


Or proving your derivation is flawed. It doesn't matter though, we only have to apply them, not derive them.

I have already posted where this manuscript can be read, ...


I'll look once we finish this discussion, what are your transformed coordinates for event A in frame K?
Fleetfoot
1 / 5 (1) May 12, 2013
Although it is mathematically possible, this is not physically possible ..


Think back to the cocktail stick example I gave some days ago. We can mathematically convert from one set of grid locations to the other in either direction, the cocktail stick is not affected by either so of course converting either way is equivalent. Changing coordinate schemes has no physical impact whatsoever.

Anyway, the bottom line is that we have spent several days agreeing your thought experiment setup, it's time for you to apply the transforms to event A and see what you get.
Noumenon
1.1 / 5 (27) May 12, 2013
Similarly, an event within IRF=K/ which occurs at the same time t and position P in space as an event within IRF=K

This premise is faulty reasoning. You can't say this. You can't say "at the same time and position" in BOTH IRF's. That is operationally (experimentally) meaningless,


Please be kind and answer the following question:

Can a position-coordinate point P/ as referenced within IRF=K/ coincide with a position-coordinate point P within IRF=K? Yes or no?


Not unless the two coordinate systems are congruent along each axis. We are assuming each observer is moving relative to the other, so this is not the case, so no.

If the answer is no please explain how two reference frames can have position coordinates of which not a single point-coordinate within the one can coincide with any point coordinate within the other.


Because, a coordinate system is arbitrary in the since that it is dependent upon the observers orientation, or even his choice.
johanfprins
1 / 5 (6) May 12, 2013
In your analysis you are transforming a transformed event back into the IRF within which it was caused. Although mathematically possible, it is not physically possible since it destroys causality.


No, if you carry on you'll find that's not true.
It is true! You force me to do wrong physics just because you believe that it is correct. Physics does not allow LT of a LT event back into the IRF from which from which it has been LT'd in the first place.

Or proving your derivation is flawed. It doesn't matter though, we only have to apply them, not derive them.
Only if they are applied correctly: You are applying them incorrectly and are trying to bully me into doing the same.

I have already posted where this manuscript can be read, ...


I'll look once we finish this discussion, what are your transformed coordinates for event A in frame K?


We can only finish this discussion after you understand the correct physics involved.
Noumenon
1.1 / 5 (27) May 12, 2013
[]Continued[]....... Now, if by "position-coordinate point" or "point-coordinate", you mean an event,.. well then, that is a physical thing, and CAN be referenced by each observer, but the components (x,y,z,t values) of that reference will NOT coincide, because each guy chose freely the orientation of their apparatus and coordinates.

If measuring "distances" in space-time between events, they WILL agree even though their particular x,y,z,t values are different. Once they both perform the "Pythagorean theorem" calculation in 4-space with Lorentz signature, they will obtain the exact same result, of that invariant quantity, the "space-time interval", but may disagree wrt individual component dependent quantities, like simultaneity.

(This is why Fleetfoot told you above, you must speak of distances, not points.)
johanfprins
1 / 5 (6) May 12, 2013
Can a position-coordinate point P/ as referenced within IRF=K/ coincide with a position-coordinate point P within IRF=K? Yes or no?


Not unless the two coordinate systems are congruent along each axis. We are assuming each observer is moving relative to the other, so this is not the case, so no.

In the special theory of relativity IT IS ALWAYS assumed that the positional coordinate axes are congruent along each axis and have the same distance scale. So try again without being so blatantly dishonest and scaly. You disgust me!

Signing off for now so that I can first vomit!
Noumenon
1.1 / 5 (27) May 12, 2013
The mistake made by Wikipedia and also by you, and the mainstream physicists over the past century, is to think that you can transform a single event in space into both IRF's without first ascertaining within which IRF the event is occurring.

What this means is to determine whether the cause for the event is stationary within IRF=K or stationary within IRF=K/.

Only such an event can be transformed from the IRF within which it is caused into another IRF. The transformed event cannot be transformed back onto itself by using the reverse equations.


If you consider the event as a light flash, then since c is invariant, it would not matter in which IRF it was caused, and your objection would be nil. In fact in STR it is a light clock that is considered for this reason.
Noumenon
1.1 / 5 (27) May 12, 2013
Can a position-coordinate point P/ as referenced within IRF=K/ coincide with a position-coordinate point P within IRF=K? Yes or no?


Not unless the two coordinate systems are congruent along each axis. We are assuming each observer is moving relative to the other, so this is not the case, so no.


In the special theory of relativity IT IS ALWAYS assumed that the positional coordinate axes are congruent along each axis and have the same distance scale. So try again without being so blatantly dishonest and scaly. You disgust me!


Please carefully reread my comment, I said "we are assuming each observer is moving relative to the other", as in one coordinate system is moving relative to the other,.. this implies that a linear transformation is required and therefore the two systems are NOT congruent. The components of the linear transformation are velocity dependent.
Noumenon
1.1 / 5 (27) May 12, 2013
,... In any case, wrt axis directions, where the transformation matrix components are an identity, your objection is merely for simplification purposes, and is not general, as are arbitrary coordinates.
ValeriaT
1 / 5 (5) May 12, 2013
I am not "thinking" but doing mathematics to check the logic of my thinking: Only an idiot like you will think it is not necessary.
Nope, the doing mathematics is just mechanical rewriting of your thinking in formal language - no less, no more. The math itself cannot prove the logical invalidity of your thinking (after all, as the mathematics of Ptolemy's epicycles or Euler's Hollow Earth theory demonstrated sufficiently). Mathematics will not prohibit you in fringe usage and confusion of variables and reference frames. It just enables to formalize such a nonsense and write it into paper in reproducible way. But the reproducible nonsense is still nonsense - sorry.
Fleetfoot
3 / 5 (2) May 12, 2013
In your analysis you are transforming a transformed event back into the IRF within which it was caused. Although mathematically possible, it is not physically possible since it destroys causality.


No, if you carry on you'll find that's not true.
It is true!


So carry on and prove it.

You force me to do wrong physics just because you believe that it is correct.


Nope, I'm asking you to do ONLY the half you said was valid to transform coordinates from the frame in which the clock is at rest. Clock2 is at rest in frame K/ so coordinates relative to it can be transformed according to your claim. I'm not asking you to do the inverse transform at all.

Physics does not allow LT of a LT event back into the IRF from which from which it has been LT'd in the first place.


I'm not asking you to do that, just apply it in the forward direction. So far you haven't applied a single transform to anything so you have no proof and we have nothing more to discuss.
johanfprins
1 / 5 (5) May 12, 2013
I am still vomiting since I have never in my life found so much lying and dishonesty as I have found on this thread. I will be back after I have recovered since it is in the interest of physics and humankind to stop the treason against everything that physics should stand for!

For example, it is absolutely criminal for Fleetfoot to ask me to "apply the forward direction" by using the equations which he must know are for the for the reverse direction. In the forward direction the equations from K/ for the event at x/=0 and t/=te are, and ONLY are:

x=g(0+v*te) and t=g(te+0) : I have given these equations; so why are you asking me to again give them? That is all there is as far as the LT is concerned: So what else do you want?. If you want anything else then post it yourself and we will discuss it!

See you again when I feel better and not so sick to my stomach anymore!
ValeriaT
1.5 / 5 (8) May 12, 2013
I am still vomiting since I have never in my life found so much lying and dishonesty as I have found on this thread
Do you mean your dishonesty, be more specific? Nobody is asking you to stay here.
.. it is absolutely criminal for Fleetfoot to ask me to "apply the forward direction" by using the equations
Yep, the Fleetfoot is a killing machine in this regard...;-) I'd ask you to "apply the outside direction" instead, because it has no meaning to feed the trolls.
Fleetfoot
3 / 5 (2) May 12, 2013
I am still vomiting since I have never in my life found so much lying and dishonesty as I have found on this thread.


If you can't follow what I said, it is your own problem.

For example, it is absolutely criminal for Fleetfoot to ask me to "apply the forward direction" by using the equations which he must know are for the for the reverse direction.


I am asking you to apply the forward equations, not the ones I posted. I accepted that Wikipedia and the textbooks were at odds which caused my error.

In the forward direction the equations from K/ for the event at x/=0 and t/=te are, and ONLY are:

x=g(0+v*te) and t=g(te+0) : I have given these equations; so why are you asking me to again give them?


I am not asking you to give the equations again, just USE them.

So what else do you want?


Instead of just talking about the transforms, apply them. We have agreed the coordinates of event A in K/ are (0, 7) so what are the transformed coordinates in frame K?
johanfprins
1 / 5 (7) May 12, 2013
I am asking you to apply the forward equations, not the ones I posted. I accepted that Wikipedia and the textbooks were at odds which caused my error.
You could have fooled me! But nevertheless thank you!

I am not asking you to give the equations again, just USE them.

Instead of just talking about the transforms, apply them.

We have agreed the coordinates of event A in K/ are (0, 7)
Where have I agreed to this? I have NOT agreed to any magnitudes of the coordinates whatsoever! On which diagram are the coordinates of event A supposedly (0,7) within K/? And where have I agreed to these values for these coordinates? Stop being dishonest! You are making me feel sicker.

Again, if you want me to apply them in a certain way, which I cannot understand, why do YOU not apply them in the way that YOU want me to apply them? That should shorten this discussion and we will stop wasting everybody's time on this thread.s
Fleetfoot
3.7 / 5 (3) May 12, 2013
I am asking you to apply the forward equations, not the ones I posted. I accepted that Wikipedia and the textbooks were at odds which caused my error.
You could have fooled me! But nevertheless thank you!


Look back 5 hours and you will find this exchange:

Go to ANY textbook and you will find:


I just checked and you're right,


Thank you.


As I said, I have no problem admitting when I make mistakes. In this case, the labels are just swapped. I've checked the spreadsheet version and I used the correct equations on that so my numbers are correct.


Feigning some sort of offense when none has been given so that they can leave the conversation when the questions expose the errors in theirs claims is the behaviour I expect from cranks. I had hoped you were above that so I'll assume you just forgot that and we hopefully we can proceed examining just the science.
Fleetfoot
3 / 5 (2) May 12, 2013
We have agreed the coordinates of event A in K/ are (0, 7)
Where have I agreed to this?

See the next post.

I have NOT agreed to any magnitudes of the coordinates whatsoever! On which diagram are the coordinates of event A supposedly (0,7) within K/?


You have been looking at the construction page, it has been in the top panel of the first diagram every time since the first diagram that had only A and B. You agreed that and each subsequent addition. The actual numerical values are only an arbitrary example but if your maths is valid, it must apply to any such values.

Again, if you want me to apply them in a certain way, which I cannot understand, why do YOU not apply them in the way that YOU want me to apply them?


Because I am checking YOUR "impeccable proof". If you want to admit you don't have one, that's fine, but if you still think it works, you should be prepared to go through it one step at a time.
Fleetfoot
1 / 5 (1) May 12, 2013
Where have I agreed to this?


Here is what was said:

@Fleetfoot: It seems to me that you are not coming back. So I will proceed to analyze your diagrams posted on https://sites.goo...truction

1. Your first two diagrams: K/ has the paths and the position A of clock1 and the position B of clock2 with the correct distance D between them: K has nothing on it: So far no problem.

2. Your second two diagrams: K/ still the same: The concomitant paths added to K; So far no problem.


Good, I'll take that as the new baseline agreement from now on.


Provided that you agree that point A is an event that occurs within K/ at the time t2 within K/.

Yes or no?

After you have answered I will proceed.


.. just for the record, yes that is what is shown. The vertical arrow marked "t2" was intended to indicate that, and the horizontal arrow marked "D" indicates the distance between the clocks at that time.

johanfprins
1 / 5 (6) May 13, 2013
@ Fleetfoot: I am trying to follow you but find the way you are stating the issues confusing: Maybe that is also the case when you try to follow what I am stating.

So what I will assume is that in the top view K/, you are claiming that the coordinates for A (clock2) are (0,7), and that the coordinates for B are (D,7) where D=-v*t2=-7*v.

Now let is LT the coordinates of A into K. The corresponding time on clock 1 is thus t1 where:

t1=g*(0+v*t2)=g*(v*t2)=7*g*v

And since clock 1 is at the origin of K, its coordinates within K are now (0,7*g*v)

The corresponding position of clock2 within K thus follows as:

X=g*(t2+(v/c^2)*0)=g*t2=7*g

And since the time within K is 7*g*v the coordinates of A within K must be (7*v,7*g*v).

We know that g must be larger than unity, say 2 and let's choose v=5, then the coordinates of A within K/ are ( 0,7) and the corresponding LT coordinates are (35,70). So please plot these points on the view from K for A,

johanfprins
1 / 5 (6) May 13, 2013
I apologize: I made typing errors above: Correction:

Now let is LT the coordinates of A into K. The corresponding time on clock 1 is thus t1 where:

t1=g*(t2+(v/c^2)*0)=g*t2=7*g

And since clock 1 is at the origin of K, its coordinates within K are now (0,7*g)

The corresponding position of clock2 within K thus follows as:

X=g*(0 +v*t2)=g*v*t2=7*g*v

And since the time within K is 7*g the coordinates of A within K must be (7*v*g,7*g).

We know that g must be larger than unity, say 2 and let's choose v=5, then the coordinates of A within K/ are ( 0,7) and the corresponding LT coordinates within K are (70,14). So please plot these points on the view within K for A.

johanfprins
1 / 5 (4) May 13, 2013
In the case of B, the coordinates within K/ are thus, (35,7) and the corresponding LT coordinates within K are thus (0,14). So please plot these coordinates for B within K/ and K respectively.

I have just plotted these coordinates and find that in both diagrams A and B lie on the same horizontal line: Thus, in both diagrams the clocks are showing the exact same time. According to me this proves that there is no time-dilation but that the event which occurs within K/ at time=7, occurs within K at the time=14: A much later time which occur after the clocks have moved a distance 70 apart.

Thus according to your diagrams the event at A occurs within K/ at a time t2=7 when the two clocks are a distance 35 apart and within K at the time 14 after the clocks have moved further apart so that the are at a distance 70 from one another.

I do not see any time dilation after having plotted these coordinates!
Fleetfoot
3 / 5 (2) May 13, 2013
@ Fleetfoot: I am trying to follow you but find the way you are stating the issues confusing: Maybe that is also the case when you try to follow what I am stating.


Indeed, and thanks for making this attempt, it gives a way to resolve the differences in undrstanding.

So what I will assume is that in the top view K/, you are claiming that the coordinates for A (clock2) are (0,7)


Yes, if you count the grey grid lines, you can check that. In other words, the scale is in units of t2/7.

The speed of light in these diagrams is always at 45 degrees so the horizontal scale is in units of c*t2/7.

and that the coordinates for B are (D,7) where D=-v*t2=-7*v.

Right but also note that for the example I chose D=5 on the grid (again you can check the diagram). That means D=5*(c*t2/7) and since v=D/t2, we have v=(5/7)c.

Gamma = 1/sqrt(1 - (5/7)^2) = 1.429

Your final numbers are way off so I suspect a minor slip up somewhere in your calculations, can you check again please.
Fleetfoot
3 / 5 (2) May 13, 2013
We know that g must be larger than unity, say 2 and let's choose v=5,


See my clarification, v=5/7 (it can't be more than 1) and gamma = 1.429

then the coordinates of A within K/ are ( 0,7) and the corresponding LT coordinates within K are (70,14).


My coordinates for A in K are (7.14, 10.0) but I suspect that's because you used v=5 rather than v=5/7. See what you get now.
johanfprins
1 / 5 (4) May 13, 2013
Yes, if you count the grey grid lines, you can check that. In other words, the scale is in units of t2/7.
My eyes are not good enough to count grey grid lines; and how must I know what the horizontal scale must be? This is my problem with you: You are playing games instead of having a decent open discussion.

The speed of light in these diagrams is always at 45 degrees so the horizontal scale is in units of c*t2/7.
I assume that you claim that the horizontal scale is c*(t2/7). Is this correct? I wish you would be more clear in what you are posting.

Indeed if you are choosing the units as c*((t2/7) and t2=7, your units are those of the speed of light c. I cannot see the reason why you choose the units on the horizontal scale to be light speed.

Right but also note that for the example I chose D=5 on the grid (again you can check the diagram).
Thus the distance D=5*c. Correct? And v=(5/7)*c. Correct? and as you then derive: gamma=1.429. Correct? Proceeding:
johanfprins
1 / 5 (3) May 13, 2013
Now let is LT the coordinates of A into K. The corresponding time on clock 1 is thus t1 where:

t1=g*(t2+(v/c^2)*0)=g*t2=7*g=7*1.429=10

And since clock 1 is at the origin of K, its coordinates within K are now (0,10)

The corresponding position of clock2 within K thus follows as:

X=g*(0 +v*t2)=g*v*t2=1.429*((5/7)*c)*7=(1.429*5)*c=7.146*c

And since the time within K is 10 and the distance in units of c, the coordinates of A within K must be (7.146,10).

Thus the coordinates of A within K/ are ( 0,7) and the corresponding LT coordinates within K are (7.146,10). So please plot these points on the view within K for A.

In the case of B, the coordinates within K/ are thus, (5,7) and the corresponding LT coordinates within K are thus (0,10). So please plot these coordinates for B within K/ and K respectively.

johanfprins
1 / 5 (6) May 13, 2013
Thus according to YOUR diagrams the event at A occurs within K/ at a time t2=7 when the two clocks are a distance 5*c apart and within K at the time 10 after the clocks have moved further apart to be at a distance 7.146*c from one another.

Plotting the coordinates for A and B on your diagrams I find that in both diagrams A and B lie on the same horizontal line. Thus, within both diagrams the clocks are showing the exact same time. According to me this proves that there is no time-dilation but that the event which occurs within K/ at time=7, occurs within K at the time=10: A later time which is present on BOTH clocks after these clocks have moved to be a distance 7.146*c apart.

I do not see ANY time dilation within K/ or within K.

I think our derivations are in agreement and therefore you are obliged to conclude that time-dilation does not occur.

Fleetfoot
3 / 5 (2) May 13, 2013
.. how must I know what the horizontal scale must be?


Any textbook should explain space-time diagrams or see for example:

http://www.astro....iags.htm

The speed of light in these diagrams is always at 45 degrees so the horizontal scale is in units of c*t2/7.
I assume that you claim that the horizontal scale is c*(t2/7). Is this correct? I wish you would be more clear in what you are posting.


Other than redundant brackets, those two sentences are identical!

I cannot see the reason why you choose the units on the horizontal scale to be light speed.


The speed of light is conventionally 45 degrees so if the vertical scale is time in seconds, the horizontal scale is distance in light-seconds, or they could be years and light-years or whatever.

And v=(5/7)*c. Correct? and as you then derive: gamma=1.429. Correct?


Correct. That speed is the only value constrained by the diagram (any other value
Fleetfoot
3 / 5 (2) May 13, 2013
Now let is LT the coordinates of A into K. The corresponding time on clock 1 is thus t1 where:

t1=g*(t2+(v/c^2)*0)=g*t2=7*g=7*1.429=10


Right but that's the coordinate of event A measured on clock1 so for clarity call it t1_A (the underscore means subscript).

And since clock 1 is at the origin of K, its coordinates within K are now (0,10)


No, we are only looking at clock2 so far, be patient ;-).

The corresponding position of clock2 within K thus follows as:

X=g*(0 +v*t2)=g*v*t2=1.429*((5/7)*c)*7=(1.429*5)*c=7.146*c


Correct.

Thus the coordinates of A within K/ are ( 0,7) and the corresponding LT coordinates within K are (7.146,10). So please plot these points on the view within K for A.


Excellent! Now, if you look at the second diagram, that's where I already have it.

The drawing is slightly off because the applet doesn't snap to the grid but the spreadsheet version is exact.

The good news is that we now agree the second diagram, right?
johanfprins
1 / 5 (6) May 13, 2013
Excellent! Now, if you look at the second diagram, that's where I already have it.

The drawing is slightly off because the applet doesn't snap to the grid but the spreadsheet version is exact.
I am not going to search for your spreadsheet.

The good news is that we now agree the second diagram, right?


No we do not! Because the position of the origin of K ,which is the position of clock1 (marked by B), must be plotted at the coordinates (0,10) while you have plotted it at (0,7). First correct this obvious mistake and then we will be in total agreement. It is impossible for clock 1 to show a lower time than the time that exists within K. In fact, the time in K is the time on clock 1 within K.

Fleetfoot
3 / 5 (2) May 13, 2013
Excellent! Now, if you look at the second diagram, that's where I already have it.

The drawing is slightly off because the applet doesn't snap to the grid but the spreadsheet version is exact.
I am not going to search for your spreadsheet.


No need to search, just click "Spreadsheet" on the left hand navigation bar, but don't worry there's no need to look at the moment.

The good news is that we now agree the second diagram, right?


No we do not! Because the position of the origin of K ,which is the position of clock1 (marked by B), must be plotted at the coordinates (0,10) while you have plotted it at (0,7).


I think you jumped ahead, th second graphic doesn't have "B" on it. You asked me to do them separately so the second graphic only has "A" at (7.146,10) which I think we agree. We haven't talked about B yet.
Fleetfoot
3 / 5 (2) May 13, 2013
No we do not! Because the position of the origin of K ,which is the position of clock1 (marked by B), must be plotted at the coordinates (0,10) while you have plotted it at (0,7).


You have successfully applies the LTs to point A, you started with coordinates (0,7) in K/ and converted them to (7.146,10) in K and that's what I get too.

Having established the correct method, what you need to do next is apply that same technique to event B. It's coordinates in K/ are (-5,7). If you apply the same equations and technique that you used for A, you should get 0 for the x coordinate, as you say it must be at the origin but do the calculation just to check. Then we reach the interesting one, you need to calculate the t1 coordinate of B in the same way that you did for A. If you get 10.0, you are right, but I get 4.9 so I am saying the correct place to plot B in frame K is at (0.0, 4.9). That illustrates that the LTs result in the effect called the relativity of simultaneity.
johanfprins
1 / 5 (5) May 13, 2013
You have successfully applies the LTs to point A, you started with coordinates (0,7) in K/ and converted them to (7.146,10) in K and that's what I get too.


Having established the correct method, what you need to do next is apply that same technique to event B. It's coordinates in K/ are (-5,7).
No! If you do it in the same way you do not get the position of the origin of K at the same instant in time when the event occurs within K. The time coordinate at any position in K must be the same as the time on the clock at the origin of K. Both the time and position coordinates have been synchronized when 0/ coincided with 0, so that for any point x within K you MUST have the coordinates (x,t) where t is the same as the time on the clock at 0.

Thus my derivation is correct: If the transformed event occurs within K at time t=10, the time on the clock at 0 MUST be simultaneously also t=10.
Fleetfoot
3 / 5 (2) May 13, 2013
Having established the correct method, what you need to do next is apply that same technique to event B. It's coordinates in K/ are (-5,7).
No! If you do it in the same way you do not get the position of the origin of K at the same instant in time when the event occurs within K.


"At the same time in K" is not the same as "at the same time in K/" which is the whole point, simultaneity is dependent on velocity.

The time coordinate at any position in K must be the same as the time on the clock at the origin of K. Both the time and position coordinates have been synchronized when 0/ coincided with 0, so that for any point x within K you MUST have the coordinates (x,t) where t is the same as the time on the clock at 0.


Correct, and in frame K, the LTs tell you that at event B, 4.9 appears on clock1.

Thus my derivation is correct ...


You haven't derived anything, you assumed the result you want. Deriving it means applying the LTs as you did for event A.
johanfprins
1 / 5 (4) May 13, 2013
"At the same time in K" is not the same as "at the same time in K/" which is the whole point, simultaneity is dependent on velocity.


So you are telling me that the transformed event at A occurs at t=10 within K while SIMULTANEOUSLY the time in K is 4.9? it is NOT possible that the clock at 0 can SIMULTANEOUSLY show a different time (4.9) than a clock at the event (10). The word simultaneously demands the SAME time at both positions WITHIN THE SAME IRF. You just cannot have your cake and eat it!

Correct, and in frame K, the LTs tell you that at event B, 4.9 appears on clock1.
Correct! It means that an event at B (within K/) which is simultaneous with the event A within K/ occurs at a time 4.9 on clock1 at 0 within K, BEFORE the event A occurs at a LATER time 10 on the same clock within K. It does NOT mean that clock1 shows a time 4.9 time SIMULTANEOUSLY with the time 10 when event A occurs within K.
Fleetfoot
1 / 5 (1) May 13, 2013
"At the same time in K" is not the same as "at the same time in K/" which is the whole point, simultaneity is dependent on velocity.


So you are telling me that the transformed event at A occurs at t=10 within K while SIMULTANEOUSLY the time in K is 4.9?


No, in frame K, B happens first and A happens later as you say next. They only happen simultaneously in K/.

Correct, and in frame K, the LTs tell you that at event B, 4.9 appears on clock1.


Correct! It means that an event at B (within K/) which is simultaneous with the event A within K/ occurs at a time 4.9 on clock1 at 0 within K, BEFORE the event A occurs at a LATER time 10 on the same clock within K.


Yes, that's what I was saying too.

It does NOT mean that clock1 shows a time 4.9 time SIMULTANEOUSLY with the time 10 when event A occurs within K.


Exactly. Strangely, we seem to be in complete agreement.
johanfprins
1 / 5 (5) May 13, 2013
Correct! It means that an event at B (within K/) which is simultaneous with the event A within K/ occurs at a time 4.9 on clock1 at 0 within K, BEFORE the event A occurs at a LATER time 10 on the same clock within K.


Yes, that's what I was saying too.
Great!

It does NOT mean that clock1 shows a time 4.9 time SIMULTANEOUSLY with the time 10 when event A occurs within K.


Exactly. Strangely, we seem to be in complete agreement.
We thus agree that there is no time dilation, only differences in time due to non-simultaneity of transformed simultaneous events: i.e. All clocks keep exactly the same time as is demanded by Einstein's first postulate. Thanks. This is obviously the ONLY logical interpretation of Einstein's STR!
Fleetfoot
3 / 5 (2) May 13, 2013
Strangely, we seem to be in complete agreement.


We thus agree that there is no time dilation,


No. We now know that clock1 reads 4.9 at event B but in frame K/, event A is simultaneous with event A, and we know that reads 7 at that time. On its own, that could just be an error of synchronisation but since we also know they both showed time zero when they crossed, there must also be time dilation. This isn't a very good thought experiment to look at time dilation but it does exhibit it.

only differences in time due to non-simultaneity of transformed simultaneous events:


That's correct, as I said some time ago, the two are closely related.

i.e. All clocks keep exactly the same time as is demanded by Einstein's first postulate.


Yes they do, but this setup doesn't really show how time dilation and ticking at the same rate can be reconciled.
johanfprins
1 / 5 (5) May 13, 2013
We now know that clock1 reads 4.9 at event B but in frame K/, event A is simultaneous with event A, and we know that reads 7 at that time. On its own, that could just be an error of synchronisation but since we also know they both showed time zero when they crossed, there must also be time dilation.
Why?

Not a very good thought experiment to look at time dilation but it does exhibit it.

It is an excellent thought experiment that proves impeccably that time-dilation is Voodoo!

only differences in time due to non-simultaneity of transformed simultaneous events:


as I said some time ago, the two are closely related.
They are NOT related at all!

i.e. All clocks keep exactly the same time as is demanded by Einstein's first postulate.


Yes they do, but this setup doesn't really show how time dilation and ticking at the same rate can be reconciled.
They cannot be reconciled since "time-dilation" demands different "ticking rates"!
Fleetfoot
3 / 5 (2) May 13, 2013
We now know that clock1 reads 4.9 at event B but in frame K/, event A is simultaneous with event A, and we know that reads 7 at that time. On its own, that could just be an error of synchronisation but since we also know they both showed time zero when they crossed, there must also be time dilation.
Why?


In frame K/, when t2=0, t1=0 but when t2=7, t1=4.9. You have to compare the time differences and there is a time dilation factor of 4.9/7 = 1/gamma.

The same applies in frame K.

.. this setup doesn't really show how time dilation and ticking at the same rate can be reconciled.
They cannot be reconciled since "time-dilation" demands different "ticking rates"!


If you measure the gaps between ticks ALONG the clock lines in each diagram instead of projecting ticks horizontally onto the vertical scale, you find the ticks have the same period. As I said a week ago, time dilation is a consequence of the geometric projection, not a change in tick rate.
ValeriaT
1 / 5 (4) May 13, 2013
All clocks keep exactly the same time as is demanded by Einstein's first postulate. Thanks. This is obviously the ONLY logical interpretation of Einstein's STR!
It would become inconsistent with relativistic increase of mass and with general relativity in this way. The general relativity predicts, that the time runs slower for heavier objects. You've spent two weeks with nonsense, just face it.
johanfprins
1 / 5 (4) May 14, 2013
In frame K/, when t2=0, t1=0 but when t2=7, t1=4.9. You have to compare the time differences and there is a time dilation factor of 4.9/7 = 1/gamma.
I was waiting and hoping that you will bring this up. I discuss this aspect in detail in my manuscripts.

The only possible correct, but counter-intuitive interpretation is as follows. Although the event at B occurs within K/ at time 7 on both the clocks in K and K/, it occurs within K at the time 4.9 on on both clocks in K and K/. This seems to violate causality since our intuition tells us that an event which causes another event must happen before the other event happens. So how can the event in K/ "cause" an event in K to occur BEFORE the event occurs within K/. The fact is, however, that if there is no event at B within K/ at time 7, there will also be no event within K at time 4.9. Thus although the event occurs at the universal time 7 within K/, this SAME event occurs at time 4.9 within K.

johanfprins
1 / 5 (4) May 14, 2013
I struggled with this quite a bit until I realized that this is exactly what Special Relativity is all about owing to non-simultaneity of transformed simultaneous events. It is not the event within K/ that "causes" the event within K but the same event that occurs at two non-simultaneous times (as measured by all the clocks which keep the exact same time) within K/ and K. So there is no time-dilation on the clocks to explain these two different times.

If you measure the gaps between ticks ALONG the clock lines in each diagram instead of projecting ticks horizontally onto the vertical scale, you find the ticks have the same period.


On any graph of position as a function of time within Euclidean space-time the time coordinate along the time axis must be perpendicular to the space coordinates.The time at all positions on any space-axis perpendicular to the time axis MUST have the SAME time-coordinate: You must thus project all ticks horizontally: NO "time-dilation".
Fleetfoot
3 / 5 (2) May 14, 2013
Although the event at B occurs within K/ at time 7 on both the clocks in K and K/, it occurs within K at the time 4.9 on on both clocks in K and K/.


That doesn't even make grammatical sense.

This seems to violate causality since our intuition tells us that an event which causes another event must happen before the other event happens.


You seem badly confused here. Imagine there is a firecracker attached to clock1 and it detonates when clock1 reads 4.9 creating a flash of light and leaving a cloud of blue smoke at B. There is no question of causality being involved at all because there is only one firecracker and one detonation. That single detonation has different coordinate values in different frames, those coordinate sets are called "events", so there are multiple coordinates pairs but they describe the same, single detonation.
johanfprins
1 / 5 (4) May 14, 2013
All clocks keep exactly the same time as is demanded by Einstein's first postulate. Thanks. This is obviously the ONLY logical interpretation of Einstein's STR!


It would become inconsistent with relativistic increase of mass
A: usual you are clueless: Relativistic mass-increase has NOTHING to do with time-dilation.

and with general relativity in this way. The general relativity predicts, that the time runs slower for heavier objects.


No it predicts that the speed of light decreases within a gravity-field. And for this reason time slows down. But this has nothing to do with the Special Theory of Relativity: STR does not require curved space time and therefore ALL the clocks keep exactly the same universal time.

You've spent two weeks with nonsense, just face it!
And you have spent years posting NOTHING else than claptrap on this forum. Just face it ducky! Quack, quack, quack, quack!
Fleetfoot
3 / 5 (2) May 14, 2013
If you measure the gaps between ticks ALONG the clock lines in each diagram instead of projecting ticks horizontally onto the vertical scale, you find the ticks have the same period.


On any graph of position as a function of time within Euclidean space-time ..


Space is Euclidean, spacetime is Reimann with signature (+++-) or a similar variant.

the time coordinate along the time axis must be perpendicular to the space coordinates.


Correct, and since the time axis is the tangent to the clock's worldline, if the clock line isn't vertical, the perpendicular CANNOT be horizontal.

You must thus project all ticks horizontally


A horizontal CANNOT be perpendicular to a straight line which is not vertical.
johanfprins
1 / 5 (5) May 14, 2013
Although the event at B occurs within K/ at time 7 etc.


That doesn't even make grammatical sense.
I admit that it is counter-intuitive, but it is the ONLY possible interpretation.

This seems to violate causality since our intuition tells us that an event which causes another event must happen before the other event happens.


Imagine there is a firecracker attached to clock1 and it detonates when clock1 reads 4.9 creating a flash of light and leaving a cloud of blue smoke at B. There is no question of causality being involved at all because there is only one firecracker and one detonation.


To do this the firecracker must be detonated within K. The smoke within K is NOT simultaneously observed in K/. The time it is observed in K/ follows from the LT of this event from K into K/. And is given by:

t/=g*(4.9)=7. The smoke is only seen later within K/. No "time-dilation is involved" at all!
johanfprins
1 / 5 (6) May 14, 2013
There is no question of causality being involved at all because there is only one firecracker and one detonation.
Correct! It is the same event which is seen at different times within different IRF's. When fire-cracker explodes at clock1 at time 4.9, it is seen at a later time 7 within K/. If the fire-cracker explodes within K/ at the position of clock1 at the time 7, the explosion has already occurred within K at the time 4.9,

That single detonation has different coordinate values in different frames,
That is why the times are different for the same event occurring.

those coordinate sets are called "events", so there are multiple coordinates pairs but they describe the same, single detonation.
I have not disagreed with this. But what these coordinates are is determined by whether the event occurs within the one IRF or the other IRF. I describe this in detail in http://www.cathod...tzT1.pdf
johanfprins
1 / 5 (5) May 14, 2013
Space is Euclidean, spacetime is Reimann with signature (+++-) or a similar variant.
This is where Einstein out of desperation went wrong: He resisted MST and should have stuck to his intuition. When he found that gravity is curved space-time he incorrectly changed his mind to interpret MST as a Rieman manifold with (+,+,+,-), which it cannot be, since a Rieman manifold, although curved in space and time, has linearly independent space-time coordinates while MST does not.

the time coordinate along the time axis must be perpendicular to the space coordinates.


Correct, and since the time axis is the tangent to the clock's worldline,
There is no such thing as a "proper time" in STR. There is only the SAME time within ALL IRF's.

You must thus project all ticks horizontally


A horizontal CANNOT be perpendicular to a straight line which is not vertical.


The time-line is vertical since "proper time" does not physically manifest.
Noumenon
1.2 / 5 (26) May 14, 2013
those coordinate sets are called "events", so there are multiple coordinates pairs but they describe the same, single detonation.

I have not disagreed with this. But what these coordinates are is determined by whether the event occurs within the one IRF or the other IRF.


As I pointed out above one can consider LIGHT FLASHES as events, and since light is constant c irrespective of IRF orientation or velocity, your above statement is potato salad.
Fleetfoot
3 / 5 (2) May 14, 2013
This seems to violate causality since our intuition tells us that an event which causes another event must happen before the other event happens.


Imagine there is a firecracker attached to clock1 and it detonates when clock1 reads 4.9 creating a flash of light and leaving a cloud of blue smoke at B. There is no question of causality being involved at all because there is only one firecracker and one detonation.


To do this the firecracker must be detonated within K.


"K" is not a box, it is the readings from one ruler and clock. Another ruler and clock give different numbers for the same detonation. I repeat, there is only one firecracker at one place exploding once which we are calling "B" so that single occurrence is describable in both K and K/.

You seem to be making the mistake of thinking there are two explosions, both called "B" which would explain why your first sentence makes no sense.
Noumenon
1.2 / 5 (26) May 14, 2013
When he found that gravity is curved space-time he incorrectly changed his mind to interpret MST as a Rieman manifold with (+,+,+,-), which it cannot be, since a Rieman manifold, although curved in space and time, has linearly independent space-time coordinates while MST does not.


You have this backwards, he interpreted the Riemann manifold as a Minkowski space-time. It operates the same wrt IRF except there are extra terms in the covariant derivative that describe how the coordinates themselves change due to mass-energy.

The tangent space to the Riemann manifold in the limit IS a Minkowski space time. In fact when there is no mass-energy present, the extra (Christoffel) term is zero.

In STR the Lorentz transformed coordinates, are hyperbolic orthogonal, and therefore are linearly independent. You are not entitled to deny this.
Fleetfoot
3.7 / 5 (3) May 14, 2013
There is no question of causality being involved at all because there is only one firecracker and one detonation.


Correct! ... If the fire-cracker explodes within K/ at the position of clock1 at the time 7, the explosion has already occurred within K at the time 4.9,


No, it hasn't "already occurred" because it only happens once. You have two clocks showing two different readings for the SAME occurrence.

those coordinate sets are called "events", so there are multiple coordinates pairs but they describe the same, single detonation.
I have not disagreed with this.

The words you have written disagree with it, they imply there are two separate detonations in each frame.
johanfprins
1 / 5 (5) May 14, 2013
But what these coordinates are is determined by whether the event occurs within the one IRF or the other IRF.


As I pointed out above one can consider LIGHT FLASHES as events, and since light is constant c irrespective of IRF orientation or velocity, your above statement is potato salad.


It is gain YOU who are, like usual, a moron. Although the speed of light is c relative to any object, light is emitted by a source and observed by a detector. These are events, and the coordinates of these events are determined by whether these events occur within the one IRF or the other IRF.

If you have taken my advice and have by now read http://www.cathod...zT1.pdf,
you would have stopped to make such a big crater of yourself.

BTW, what are your qualifications? Have you EVER had ANY course in physics and to what level?
Fleetfoot
3 / 5 (2) May 14, 2013
.. a Rieman manifold, although curved in space and time, has linearly independent space-time coordinates while MST does not.


MST is a Reimann manifold with zero curvature.

the time coordinate along the time axis must be perpendicular to the space coordinates.


Correct, and since the time axis is the tangent to the clock's worldline,


There is no such thing as a "proper time" in STR.


I didn't mention "proper time", however you are wrong, it is simply the measurement along the clock's line.

There is only the SAME time within ALL IRF's.

Nope, times are readings from clocks, there are two clocks so there are two times.

You must thus project all ticks horizontally


A horizontal CANNOT be perpendicular to a straight line which is not vertical.


The time-line is vertical ..


Nope, you can see them on the diagrams you have agreed, they are the lines marked "clock1" and "clock2". In each frame, one is vertical but the other is not.
Noumenon
1.2 / 5 (26) May 14, 2013
But what these coordinates are is determined by whether the event occurs within the one IRF or the other IRF.


As I pointed out above one can consider LIGHT FLASHES as events, and since light is constant c irrespective of IRF orientation or velocity, your above statement is potato salad.


It is gain YOU who are, like usual, a moron. Although the speed of light is c relative to any object, light is emitted by a source and observed by a detector. These are events, and the coordinates of these events are determined by whether these events occur within the one IRF or the other IRF.


Given some of you statements above, one must be careful about how you phrase things. Events don't occur within IRF's, they are described by IRF's, which implies dependency on observers wrt components of that description.

Apparently, even had I received my education from the back of match covers, I've managed better than you. My qualifications are that I am right.
johanfprins
1 / 5 (6) May 14, 2013
"K" is not a box,
I have not said this at all!

it is the readings from one ruler and clock.
Correct!

Another ruler and clock give different numbers for the same detonation.
Exactly! And therefore the explosion is seen at DIFFERENT TIMES WITHIN DIFFERENT REFERENCE FRAMES. Thus, if within the reference frame K, the explosion is seen at clock1 situated at the origin at time 4.9, this SAME explosion is seen within K/ at the position of the clock 5c at the time 7, and vice-versa; even though the clocks are keeping the same time as they MUST according top Einstein's first postulate.

I repeat, there is only one firecracker at one place exploding once which we are calling "B" so that single occurrence is describable in both K and K/.
I have NOT disagreed with this at all. But although it is a SINGLE EVENT it is observed on clock1 and clock2 (which keep the SAME time) at two different times and positions.

johanfprins
1 / 5 (4) May 14, 2013
You seem to be making the mistake of thinking there are two explosions,
If there were two explosions at the same coincident point, one in K and the other in K/ you will see the two explosions at different times within both K and K/. Thus I am NOT "thinking there are two explosions" at all.
Fleetfoot
3 / 5 (2) May 14, 2013
even though the clocks are keeping the same time as they MUST according top Einstein's first postulate.


They produce the same number of ticks per unit of length measured ALONG the clock line in each diagram which is what the first postulate requires. They do not produce the same number per unit when geometrically projected onto vertical.

I repeat, there is only one firecracker at one place exploding once which we are calling "B" so that single occurrence is describable in both K and K/.


I have NOT disagreed with this at all. But although it is a SINGLE EVENT it is observed on clock1 and clock2 (which keep the SAME time) at two different times and positions.


They do not keep the same COORDINATE time as your thought experiment proved.

If you calculate the ticks using the formula for proper time, then you will find they DO tick at the same rate. That is how the first postulate is satisfied.
johanfprins
1 / 5 (6) May 14, 2013
When he found that gravity is curved space-time he incorrectly changed his mind to interpret MST as a Rieman manifold with (+,+,+,-), which it cannot be, since a Rieman manifold, although curved in space and time, has linearly independent space-time coordinates while MST does not.


You have this backwards, he interpreted the Riemann manifold as a Minkowski space-time.
You cannot do that since the 4D Rieman manifold has four linearly-independent coordinates while MST does NOT!

It operates the same wrt IRF except there are extra terms in the covariant derivative that describe how the coordinates themselves change due to mass-energy.
To a moron like you it will sound as simple as all that: It is, however, not. Trust me I have taught Riemann manifolds for may years!

The tangent space to the Riemann manifold in the limit IS a Minkowski space time.
This was a WRONG deduction by Einstein. It cannot be since MST does not have linearly independent coordinates.
johanfprins
1 / 5 (4) May 14, 2013
In STR the Lorentz transformed coordinates, are hyperbolic orthogonal,
As I have posted before this is an oxymoron concept.

and therefore are linearly independent. You are not entitled to deny this.
Why am I not "entitled" to deny this if I can prove from fundamental mathematics that the coordinates of MST ARE NOT LINEARLY INDEPENDENT!
johanfprins
1 / 5 (3) May 14, 2013
Correct! ... If the fire-cracker explodes within K/ at the position of clock1 at the time 7, the explosion has already occurred within K at the time 4.9,

No, it hasn't "already occurred" because it only happens once.
Where did I disagree with this? All I am pointing out is that according to the coordinates within K and K/ this single event is NOT simultaneously observed within K and K/. This has NOTHING to do with "time-dilation"
those coordinate sets are called "events", so there are multiple coordinates pairs but they describe the same, single detonation.
I have not disagreed with this.


The words you have written disagree with it, they imply there are two separate detonations in each frame.
The same event is "observed" to occur at two different times within K and K/, while the clocks within K and K/ show exactly the same time when this event is observed in K and also exactly the same time when the same event is observed within K/.
johanfprins
1 / 5 (4) May 14, 2013
MST is a Reimann manifold with zero curvature.
No it is not, since a 4D Riemann manifold with zero curvature MUST have a signature of (+,+,+,+).
the time coordinate along the time axis must be perpendicular to the space coordinates.

Correct, and since the time axis is the tangent to the clock's worldline,
I didn't mention "proper time", however you are wrong, it is simply the measurement along the clock's line.
No it is not: There is only one time axis in your diagrams and it is perpendicular to the space coordinates as it must be when the coordinates are linearly independent as they must be to span a 4D space-time.

There is only the SAME time within ALL IRF's.
Nope, times are readings from clocks, there are two clocks so there are two times.


You are arguing like the Irish station master who answered the German tourist after the tourist pointed out that the two station-clocks show different times: " That is why we have two".
johanfprins
1 / 5 (6) May 14, 2013
You must thus project all ticks horizontally
A horizontal CANNOT be perpendicular to a straight line which is not vertical.
The time-line is vertical
Nope, you can see them on the diagrams you have agreed, they are the lines marked "clock1" and "clock2". In each frame, one is vertical but the other is not.
Since the other line IS NOT a time line within that reference frame and CANNOT be time-line when the four coordinates are linearrly independent as THEY MUST BE to span 4D Riemann space WITHOUT ANY CURVATURE!

johanfprins
1 / 5 (6) May 14, 2013
Given some of you statements above, one must be careful about how you phrase things. Events don't occur within IRF's, they are described by IRF's, which implies dependency on observers wrt components of that description.
To say they occur within an IRF is short-hand for what you have just now stated. Anybody with some grey matter who have read books on relativity would know this.

Apparently, even had I received my education from the back of match covers, I've managed better than you. My qualifications are that I am right.
Just again proves how demented you are: I will never be so arrogant to judge another person's views without letting him/her know whether I have the qualifications to do so. Only a person with no integrity and self-shame can act in such a criminal manner!
johanfprins
1 / 5 (6) May 14, 2013
They produce the same number of ticks per unit of length measured ALONG the clock line in each diagram which is what the first postulate requires.
Thus, at any instant in time, if you can stop the Universe, they MUST all show the same global time. Thus, when an event occurs all the clocks in all the IRF's show the same time for the event. But this event is only observed at this time within one of all the IRF's which I call the primary IRF, and is observed within all the other IRF's at different global times.

They do not keep the same COORDINATE time as your thought experiment proved.
They do keep the same global time: An event is, however, observed at different global times within the different IRF's.

If you calculate the ticks using the formula for proper time, then you will find they DO tick at the same rate. That is how the first postulate is satisfied.
This does NOT satisfy the first postulate.
johanfprins
1 / 5 (6) May 14, 2013
The first postulate states that the laws of physics must be the same when doing the same experiment within ANY IRF. The rate at which a clock keeps time is determined by the laws of physics. Thus the time cannot be different within different IRF's. This will violate Einstein's first postulate.

I am signing off for now!
johanfprins
1 / 5 (7) May 14, 2013
From a manuscript I am working on:

Thus, although the same experiments within IRF 1 and IRF 2 are identical as observed within each IRF respectively, a transformed experiment need not and most probably will not be identical to an non-transformed experiment. It should be emphasized here that the transformed-coordinates of an experiment do not replicate what is actually happening within the IRF from which they are being transformed, but gives a distorted view within the IRF into which they are being transformed. Nonetheless, this distorted view is not an illusion but is actually real within the IRF into which the coordinates are being transformed. The latter is a very fundamental, and misunderstood concept which has led, and still leads to confusion.

Goodbye for now!
Noumenon
1.2 / 5 (26) May 14, 2013
Given some of you statements above, one must be careful about how you phrase things. Events don't occur within IRF's, they are described by IRF's, which implies dependency on observers wrt components of that description.

To say they occur within an IRF is short-hand for what you have just now stated. [insult]..


Ok, then why did you feel compelled to make a distinction in response to Fleetfoot as follows?,....

those coordinate sets are called "events", so there are multiple coordinates pairs but they describe the same, single detonation. - Fleetfoot


I have not disagreed with this. But what these coordinates are is determined by whether the event occurs within the one IRF or the other IRF. - johanfprins


Again, some of your above statements seem to imply a universal absolute time because you make reference to "at the same instant in time" with no means of interpreting that operationally, so it is valid to call the above into question.
thefurlong
1.4 / 5 (9) May 14, 2013
Jumping back into the fray...
Thus, at any instant in time, if you can stop the Universe, they MUST all show the same global time.

How do you stop the entire universe? How do you get everyone to agree to stop at the "same time" when everybody measures time differently?
Thus, when an event occurs all the clocks in all the IRF's show the same time for the event. But this event is only observed at this time within one of all the IRF's which I call the primary IRF, and is observed within all the other IRF's at different global times.

You need to explain this concept of primary IRF more. Do you mean that for every event that occurs, there is a primary IRF, or that there is a single, absolute IRF?
Also, are you saying that there is one absolute (true) time?
thefurlong
1 / 5 (8) May 14, 2013
@johanfprins
From what I remember with the brief time I spent reading your impeccable manuscripts, I would guess that you identify a primary IRF wrt an event E as the IRF that you consider "at rest" with the event in question. More specifically, suppose that a bomb detonates. You would consider a primary IRF wrt the detonation, the one that always measures the epicenter (from first flash to smoking crater) to be in the same place. Do you agree with this definition?
ValeriaT
1 / 5 (4) May 14, 2013
Thus the time cannot be different within different IRF's. This will violate Einstein's first postulate
Such a claim is very easy to dismiss at empirical basis. Try to imagine the pair of satellites flying around Earth in different altitude. Both they're moving inertially, both they will experience the local difference in speed of time, albeit quite minute one.
Fleetfoot
3 / 5 (2) May 14, 2013
Correct! ... If the fire-cracker explodes within K/ at the position of clock1 at the time 7, the explosion has already occurred within K at the time 4.9,


No, it hasn't "already occurred" because it only happens once.


Where did I disagree with this?


In the line I quoted where you said "the explosion has already occurred within K". In frame K, the firecracker explodes only once. In frame K/, the firecracker explodes only once. There is no coordinate scheme you can choose where it has already happened when it happens.

All I am pointing out is that according to the coordinates within K and K/ this single event is NOT simultaneously observed within K and K/.


"Simultaneously" means two things happen at the same time coordinate in some frame, but whatever frame you choose event B is a single occurrence.

This has NOTHING to do with "time-dilation"


Correct, it has nothing to do with anything other than your inability to express what you are thinking.
Fleetfoot
3 / 5 (2) May 14, 2013
the time coordinate along the time axis must be perpendicular to the space coordinates.


Correct, and since the time axis is the tangent to the clock's worldline,


There is no such thing as a "proper time" in STR.


I didn't mention "proper time", however you are wrong, it is simply the measurement along the clock's line.


No it is not:


You are wrong, that is the DEFINITION of proper time.

There is only one time axis in your diagrams and it is perpendicular to the space coordinates ..


The points (0,0), (0,1), (0,2), (0,3), etc. define the positive time axis in K/. Transform each of those to find that axis in K.

The points (0,0), (1,0), (2,0), (3,0), etc. define the positive space axis in K/. Transform each of those to find that axis in K.

The axes are still perpendicular of course, but they rotate in the opposite directions because our screens are Euclidean and spacetime isn't, that's one effect of the Reimann signature.
Fleetfoot
3 / 5 (2) May 14, 2013
You must thus project all ticks horizontally


A horizontal CANNOT be perpendicular to a straight line which is not vertical.


The time-line is vertical


Nope, you can see them on the diagrams you have agreed, they are the lines marked "clock1" and "clock2". In each frame, one is vertical but the other is not.


Since the other line IS NOT a time line within that reference frame and CANNOT be time-line when the four coordinates are linearrly independent ...


They are independent if you use the correct signature but (++++) is Euclidean and spacetime is not.
Fleetfoot
3 / 5 (2) May 14, 2013
They produce the same number of ticks per unit of length measured ALONG the clock line in each diagram which is what the first postulate requires.


Thus, at any instant in time, ...


"at any instant in time" according to which clock?

if you can stop the Universe, they MUST all show the same global time.


Your application of the LT's showed they will have different time readings.

... within one of all the IRF's which I call the primary IRF,


Don't start bringing silly aether theories into this, there is no such thing in SR.

If you calculate the ticks using the formula for proper time, then you will find they DO tick at the same rate. That is how the first postulate is satisfied.
This does NOT satisfy the first postulate.


Every experiment that tests this has confirmed that it is true.
johanfprins
1 / 5 (6) May 15, 2013
Ok, then why did you feel compelled to make a distinction in response to Fleetfoot as follows?,....

I have not disagreed with this. But what these coordinates are is determined by whether the event occurs within the one IRF or the other IRF. - johanfprins
Again, some of your above statements seem to imply a universal absolute time
They all imply that, since this IS the case within a gravity-free universe. The time is globally universal no matter at which speed a clock is moving relative to another one. In fact it is the concept of "time-dilation" which requires that two clocks show different times at the same instant in time without defining this instant in time.

But I must admit that I muddied the water when I responded to fleetfoot that I am not interpreting each IRF as a separate box. I should have responded that this is the only way in which the principle of relativity can be correctly interpreted. (continued below).

johanfprins
1 / 5 (4) May 15, 2013
Galileo, who DISCOVERED the principle of relativity, to which Einstein later added the speed op light, explained this principle by considering a ship which is moving with a uniform velocity. He then stated that within an enclosed cabin on this ship (a box) it is impossible to do any mechanical experiment from which one can deduce that the ship is moving.

Now consider such a cabin: If there are no events within this cabin, everything will be stationary unless a force appears from WITHIN the cabin to move its contents. Such a force I am calling a primary event within the cabin, since it has NOTHING to do with what is happening outside the cabin. Similarly to have light within the cabin one has to switch on a light. The source of the light is within the cabin. This is thus a primary-event within the cabin.

The cabin itself is a stationary IRF within which the coordinates are stationary. If nothing moves one will not be able to measure time.
johanfprins
1 / 5 (4) May 15, 2013
Assume that there is a light-clock within the cabin, with a distance L between the mirrors, the time for each tick of the clock will be (delta)t=(2*L)/c. It also does not matter at which position you place the light clock, it will keep the same time. Anybody living within the cabin will thus experience a Newton-type space-time.

Within a passing ship there is an identical cabin so that within this cabin there is no experiment possible to determine whether the cabin is moving or not. It also has a light clock and if the speed of light were to be c* which is different from c within the first cabin, the time between its ticks would be (delta)t*=(2L)/c*. BUT according to Einstein the speed of light within this cabin must also be c: Thus by setting c*=c, the clock within this cabin MUST keep exactly the same time as the clock within the first cabin. So Einstein's own second postulate PROHIBITS time dilation. The clocks MUST always keep the same universal time.

johanfprins
1 / 5 (6) May 15, 2013
How do you stop the entire universe?
You can do this in a thought experiment to illustrate what will be the situation if you can do this.
How do you get everyone to agree to stop at the "same time" when everybody measures time differently?
As I have just proved above by using Galileo's own description of the principle of relativity, everybody with a clock in his/her hand measures exactly the same time.

Thus, when an event occurs all the clocks in all the IRF's show the same time for the event. But this event is only observed at this time within one of all the IRF's which I call the primary IRF, and is observed within all the other IRF's at different global times.


You need to explain this concept of primary IRF more. Do you mean that for every event that occurs, there is a primary IRF, or that there is a single, absolute IRF?
Every event that occurs is primary to a single RF which I suggest must be called the proper
RF for this event.
johanfprins
1 / 5 (4) May 15, 2013
From what I remember with the brief time I spent reading your impeccable manuscripts, I would guess that you identify a primary IRF wrt an event E as the IRF that you consider "at rest" with the event in question.
Borrowing from Feynman I will define a primary event as an event that is observed within its proper IRF, when not looking outside of this IRF.
More specifically, suppose that a bomb detonates. You would consider a primary IRF wrt the detonation, the one that always measures the epicenter (from first flash to smoking crater) to be in the same place. Do you agree with this definition?
Yes this is the case when the event is a bomb that detonates within its proper IRF. Looking at the bomb detonate from another IRF, changes the coordinates for the event so that it is observed from outside its proper IRF to detonate at a different global time than the global time at which it detonated within its proper IRF.
johanfprins
1 / 5 (6) May 15, 2013
@fleetfoot,

My recent posts answering Noumenon and the Furlough should be enough for you to answer your own subsequent posts. If you still cannot do this, you are welcome to post more questions.

@ValeriaT, Natello AKAK: You are again as usual clueless when it comes to physics!
Noumenon
1.3 / 5 (28) May 15, 2013
Within a passing ship there is an identical cabin so that within this cabin there is no experiment possible to determine whether the cabin is moving or not.


Yes, I agree. The same goes for the 'stationary' cabin as you mentioned.

It also has a light clock and if the speed of light were to be c* which is different from c within the first cabin, the time between its ticks would be (delta)t*=(2L)/c*. BUT according to Einstein the speed of light within this cabin must also be c:


That's correct, c is in fact invariant, so the effect for the physical reasons you sited in fact don't occur. That is to say each IRF time between ticks is determined based on a constant c. But we have not compared the clocks yet,...
thefurlong
1 / 5 (9) May 15, 2013

Thus by setting c*=c, the clock within this cabin MUST keep exactly the same time as the clock within the first cabin. So Einstein's own second postulate PROHIBITS time dilation. The clocks MUST always keep the same universal time.

Now, hold on. This is the point where things go horribly wrong in the discussion, and you get flustered and start insulting people. Now, do you agree that a person at rest with a light clock will have experienced the amount of time shown on the it?
Assuming you do, then you need to be really careful about identifying who is measuring time. Consider 4 people, A, B, and C, and D.
A is at rest with cabin 1. B is at rest with cabin 2. C measures cabin 1 and 2 to be approaching each other at equal speed. C also measures D to be moving in the direction of cabin 1, but at half the speed. [To be continued]
thefurlong
1.4 / 5 (10) May 15, 2013
@johanfprins
Now, obviously, C will see both A and B's clocks tick at the same rate. If C has a clock at rest with him, then his clock will not agree with A and B's clocks. We know this because C will measure the paths the photons follow for A and B's light clocks between "ticks" as longer than the paths that the photons follow for his own clock. Likewise, C will see D's clock tick at a rate different to those of A, B, and his own. Now, consider this from D's perspective. D will not see A and B's clock ticking at the same rate because he will no longer measure them to be moving at equal (but opposite) velocities. He will also see C's clock tick at different rates.
So, now who is correct about whether A and B's clocks keep the same rate, C or D, and why?
Noumenon
1.3 / 5 (28) May 15, 2013
,... cont. from above,....

Thus by setting c*=c, the clock within this cabin MUST keep exactly the same time as the clock within the first cabin. So Einstein's own second postulate PROHIBITS time dilation. The clocks MUST always keep the same universal time.


You can't draw this conclusion yet! You jumped to this conclusion without any experimental means of checking it! You can't draw such assumptions between IRF's, without there being a possible means of empirical verification.

How to compare the light-clocks? The only way, is for Cabin-S to observe Cabin-M's light clock to compare with his own tick rate. It is perfectly valid for Cabin-S to observe Cabin-M's light clock in this way,... as physically valid as he sees his own!!

Since Cabin-M is moving, Cabin-S will see the path of Cabin-M's light reflecting between Cabin-M's mirrors, as a 'saw-tooth' pattern, giving the following LT(Cabin-M Tick Rate),...

Δt=2L/c(1-v²/c²)^(- ½)

Noumenon
1.3 / 5 (28) May 15, 2013
....As you can see this is purely a consequence of Cabin-M's moving light clock and the constancy of c, therefore,... Δtm=2L/c(1-v²/c²)^(- ½) is a longer tick rate, as Cabin-S observes Cabin-M's light-clock,.. than his own light clock, of Δts=2L/c.

There must be an experimental means of comparing the two clocks. In fact the Lorentz Transform was derived based on experimental facts of the negative Michelson results wrt the earths velocity relative to the aether. Einstein took the next logical step,... absolute space and time are meaningless notions.
Fleetfoot
3.7 / 5 (3) May 15, 2013
Thus by setting c*=c, the clock within this cabin MUST keep exactly the same time as the clock within the first cabin. So Einstein's own second postulate PROHIBITS time dilation. The clocks MUST always keep the same universal time.


That is wrong. There are three effects commonly described for relativity which are a consequence of the Lorentz Transforms, time dilation, length contraction and the relativity of simultaneity. When you apply all three, they cancel out so the speed of light is the same in both cabins even though the clocks tick at different coordinate rates.

You can't win an argument by just claiming your beliefs, the only progress that has been made in this thread is when we applied the LTs to your example to check my diagrams. The maths can be checked and is unarguable, everything else is just opinion until demonstrated.
Fleetfoot
3.7 / 5 (3) May 15, 2013
@fleetfoot,

My recent posts answering Noumenon and the Furlough should be enough for you to answer your own subsequent posts. If you still cannot do this, you are welcome to post more questions.


I only asked one question in the three posts, " 'at any instant in time' according to which clock?" did you want to stop the universe?

If you answer that, even just to yourself, it is obvious that different parts of the universe stopped at different times according to the other clock.

Other than that I just corrected more of your errors.

You seem to be hung up on time dilation for some reason. What we have done so far has proved the relativity of simultaneity so if you want to continue, I suggest we continue with the diagrams and calculations and I'll add something to it so that we can investigate time dialtion in the same manner. I think that approach was being productive.

@ValeriaT, Natello AKAK: You are again as usual clueless when it comes to physics!
Fleetfoot
3.7 / 5 (3) May 15, 2013
@ValeriaT, Natello AKAK .....


Apologies for that Zephyr, the comment was accidentally quoted from Johan's post and I didn't notice until after the edit time expired.
johanfprins
1 / 5 (7) May 15, 2013
To even try and argue against all the illogical nonsense that has been posted above today will take a Sage at least two lifetimes. I suggest that I ask questions that can be found in any textbook and ask Feetfood, the furlong, and Noumenon to answer by voting yes or no. I will appreciate it if you will bear with me.

1. Do you accept Galileo's postulation of the Principle of Relativity according to which one can do no experiment within an enclosed IRF to determine whether the IRF is moving?

Yes or no?

2. Do you accept that this means that the laws of physics MUST be exactly the same when referenced within ANY IRF?

Yes or no?

3. Do you accept that an observer who is stationary within an IRF will observe a gravity-free universe as if it is only he and his IRF which are uniquely sta