Once again, Albert Einstein's General Theory of Relativity, 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 quantum theory. 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-dwarf companion, 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 gravitational waves 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 gravitational radiation 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.

"We thought this system might be extreme enough to show a breakdown in General Relativity, 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 neutron stars 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:**
Pulsars: The Universe's gift to physics

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

## ValeriaT

## Torbjorn_Larsson_OM

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

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

## Tektrix

LOL, nice one. Subtle jokes are the best kind :)

## Q-Star

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

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

## ekim

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

## Tektrix

## Whydening Gyre

And an ALMOST equal number of empty spaces...

## Osiris1

## Sanescience

## antialias_physorg

..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

## ant_oacute_nio354

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

## ValeriaT

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

## ValeriaT

## brt

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

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

## Dileep_Sathe

## Koen

## Q-Star

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

## Q-Star

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

## johanfprins

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

## Q-Star

Ya are correct. It is a humbling awaking indeed.

## Fleetfoot

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

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

## Fleetfoot

A receiver that can detect EM at 100Hz will not detect gravitational waves of the same frequency. Each phenomenon has its own spectrum.

## Fleetfoot

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.

It also sets a lower limit.

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

Relativity theory is all wrong.

Antonio Jose Saraiva

## Fleetfoot

It fits the experimental results, that makes it right.

## johanfprins

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

## Fleetfoot

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

## Q-Star

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

## johanfprins

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

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

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

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

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?

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

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

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?

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

## johanfprins

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.

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.

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

## Fleetfoot

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

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!

I know the text books far better than you with your feeble mind can ever know them.

## johanfprins

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

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!

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

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

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.

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

How? Please give the equations.

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

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

I have taken a brief look at the papers you wrote. Something you said in http://www.cathod...tion.pdf confused me. You wrote,

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

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.

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.

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

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

Both match the result.

## johanfprins

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

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

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

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.

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

## johanfprins

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

## johanfprins

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

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.

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.

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

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

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

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!

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

Another lie since AWT is YOUR mantra.

I most certainly DID NOT! Why would I contrast ANYTHING with an absurd aether which DOES NOT exist?

## thefurlong

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

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.

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

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?

WE ALL know this! It does not change one iota what I have claimed above.

## johanfprins

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.

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

## ValeriaT

## johanfprins

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

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

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

How must I falsify something that is correct?

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

## Fleetfoot

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.

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.

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

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.

Well maybe I misunderstood, here is what you said:

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

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

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

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

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

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

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!

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

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 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!

"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

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

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

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

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.

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

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

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

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

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

http://en.wikiped...tionship

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

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

That maybe the source of the previous confusion, there is no absolute time in SR but it is the fundamental to aether theory.

## johanfprins

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."

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

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

Thus you can formulate Maxwell's equations as if time is not absolute, but not space-time itself!

## Fleetfoot

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

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

Apparently you don't.

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

## johanfprins

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

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

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

"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

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

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.

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

## Fleetfoot

No problem.

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

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

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

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

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

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

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

Light clocks,,, ya are correct, because "c" is "c" for everyone.

## thefurlong

Only when the laws of physics depend on the IRF.

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.

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

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

Nothing, just teaching you some basics.

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

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.

I AM USING A LIGHT CLOCK!!! PLEASE first do your homework before proving time and again that you are a certifiable moron!

## Fleetfoot

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.

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

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

## johanfprins

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.

Clearly it is not overly difficult for you to post claptrap!

## Fleetfoot

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

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

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

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

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

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

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

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!

Basics? YOU?!!! LOL!

And YOU want to teach ME basics?

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

## thefurlong

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.

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

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

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

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

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

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

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

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.

Why are you assuming they should?

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

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".

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

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.

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

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

## Fleetfoot

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.

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

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

Thank you

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

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

## johanfprins

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

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

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.

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.

Understanding special relativity IS difficult. That's why so many people get it wrong.

## thefurlong

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.

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

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.

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

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

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

Which brings us back to - how do you change change? by stopping it, of course.

## Q-Star

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

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.

## thefurlong

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

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

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

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

Well I did say "in SR" ;-)

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

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

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

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.

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

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

Which to choose?

Then why suggest it?

What?

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

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

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

## Q-Star

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

Who's arguing?

Damn... now I'm pulled into the dance with TWO of you.

## Q-Star

Well, let's see,

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.

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

Time is not invariant, that is the point.

It is not Euclidean triangulation. It is Riemann - Minkowski geometry.

## Whydening Gyre

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

Who?

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

Not BS, it's very well tested physics.

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

## Q-Star

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

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

No, he's talking about 4-velocity:

http://en.wikiped...velocity

It's magnitude is independent of the choice of frame.

## Whydening Gyre

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

I accept that...:-)

## thefurlong

You said

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

## Q-Star

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

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

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.

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

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

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

## thefurlong

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

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

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

Arguing.

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

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

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

Which all kinda wraps back around to the concept of IRF...:-)

## Fleetfoot

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.

It is important when solving new problems. Once the solution is found, we move on.

## Fleetfoot

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

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

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

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

## johanfprins

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.

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

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

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

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

## johanfprins

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

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

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

## ValeriaT

## johanfprins

## ValeriaT

## johanfprins

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.

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

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

## ValeriaT

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

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.

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

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

## ValeriaT

## johanfprins

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

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.

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

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

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

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

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

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.

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.

Bullshitt!! Are you sure you are not ValeriaT AKAK with yet another name? Yes I

## johanfprins

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.

Cannot follow logic and mathematics ValeriaT?

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

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

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.

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

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!

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

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!

So why do you keep on coming back under various false disguises?

## thefurlong

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.

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

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.

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!

Where in my derivation did I accelerate the clocks?

## thefurlong

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

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

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.

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

The only "parodoxes" are incorrect derivations from the Lorentz transformation. When you do the derivations correctly thre are no paradoxes.

I have not claimed that STR is incorrect!

## thefurlong

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

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.

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

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

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

No this is not what Einstein did! Stop lying!

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.

Not everyone else: ONLY YOU!

## Fleetfoot

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

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

## ValeriaT

## thefurlong

[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

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

## thefurlong

[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

[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

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

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

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

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!

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

## johanfprins

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

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

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

## johanfprins

Still suprisingly logical. I cannot believe it!

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 have NEVER disputed that Barney will conclude that the signal will reach the clocks at different times on Barney's clock.

## Fleetfoot

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

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.

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

phys.org/news/2013-05-fermi-swift-shockingly-bright.html

## johanfprins

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

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!

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

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

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

Anyway - that's MY IRF...

## johanfprins

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

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

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

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

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

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

## Fleetfoot

Indeed, but aether theories do start from that presumption.

Both statements I made above are true.

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

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

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!

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

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:

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

You read more into the comment than I said.

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

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:

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

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

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

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

What you said some time ago was:

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

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

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

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

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

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

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

## Whydening Gyre

## johanfprins

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

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

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.

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

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.

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

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

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

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.

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

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

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

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

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.

Thank you.

## johanfprins

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".

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.

Einstein's first postulate already demands that it MUST be so!

## thefurlong

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

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?

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

Stop being Daft!

## thefurlong

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.

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

Perhaps you are of diminutive stature...

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

## Q-Star

"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

Not energy,,,, FORCE. Ya get an acceleration by applying a force.

## Noumenon

## thefurlong

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

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

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

Yeah. At first it was an interesting challenge, but it is starting to get tiresome.

## thefurlong

YES!

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!

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

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

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

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.

I bet Voyager has a clock on board. Are there any humans at rest with its clock?

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

If you do want to understand the reality, you shouldn't twist it.

## Noumenon

Corrected post.

## Whydening Gyre

You're right, I skipped a step. Can I say that we require energy to apply the force?

## Fleetfoot

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

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

Where did I say single even?: I am talking about three identical events occurring simultaneously at the positions of the three clocks.

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

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.

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

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.

But it is not necessary to do go to all this trouble.

## johanfprins

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

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.

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

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

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

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.

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!

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

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

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

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

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.

As I have already pointed out this step is not necessary for the argument you are trying to make!

## johanfprins

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

At last you have seen the point!

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

"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!

Even when the statement is not "twisted" you are still not able to understand it!

## johanfprins

I have not stated that it is!

## johanfprins

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

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

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.

]

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!!!

.

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

## Fleetfoot

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

Owing to synchronisation this "constant" is zero since clock2 is at the origin of its rest frame.

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

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

It would be more correct to say the force transfers energy.

## Fleetfoot

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

All four axis are validly orthogonal, and linearly independent in Minkowski space-time. What is wrong with the notion of hyperbolic orthogonality?

## Noumenon

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

What couldn't you follow?

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

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

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

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

I see it all so clearly now... Dave Bowman - 2001: A space Odyssey

## Whydening Gyre

Plus 1.

## johanfprins

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

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

"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

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

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

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

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.

Then why are you posting on this thread?

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

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"?

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

The correct spelling is "you".

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.

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

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.

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

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

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

Exactly.

## Fleetfoot

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.

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

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.

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

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

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

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.

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

I DID start studying your manuscipt and found that it made bad assumptions.

I meant that I have no vested interest in reading your paper, you child! Jesus Christ!

LOL!

## johanfprins

Name them! You are a liar an a fraud!

## thefurlong

You didn't answer my question. Are they simultaneous for the person who is moving relative to the clocks, or not?

## johanfprins

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

Correct.

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

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

I must have missed that but I've missed dozens of posts as I've been out most of the weekend.

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

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.

It obviously does, you have to ask "at the instant in time" as measured by which clock?

## Noumenon

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

You use the word "perceived". Does that mean that you think that the moving observer is incorrect, and the stationary observer is correct?

"at the instant" in whose time? The stationary observer's or the moving observer's?

## Fleetfoot

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

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.

This is, as I have proved above, what the Lorentz-transformation demands: Or are you saying that the Lorentz transformation is wrong?

What you mean by this statement only YOUR demented mind can understand.

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

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

The term "hyperbolic orthogonal" is not just an oxymoron but also a contradiction in terms. Like "military intelligence"?

You cannot have an "inner product" unless the coordinates are linearly independent, and the coordinates x,y,z and ict ARE NOT linearly independent!

Why do I have to review the claptrap that I taught my students when I did not know better?

## Fleetfoot

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

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

## johanfprins

Why? Perfect clocks in SA and New York are not synchronized but they still run at the SAME rate.

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

Ya are starting to sound like the AWT guy and the Plasma, blah, blah, blah Plasma guy.

## johanfprins

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

Somehow that beggars my credulity.

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

## johanfprins

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

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

## thefurlong

Agreed.

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

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

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

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.

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

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

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

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

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

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 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

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

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

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/.

That is wrong. If you are talking about frame K/, you next statement is correct:

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

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

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

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

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.

Almost, D2=v*t2 in frame K/

Almost, D1 = v * t1 in frame K

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

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.

Adding a 2 so that D=D2 does not change the argument.

## johanfprins

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 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

## johanfprins

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

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.

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

Some people need more than maths, let's see if a picture is worth more than 1000 characters.

## Q-Star

Sorry to interupt, could ya send a link to the place where ya set-up that graphics application?

Thanks. Q

## johanfprins

So why can I not cal it D?

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

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

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

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

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

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.

I think we now have a common understanding of your thought experiment which is what I felt was needed so please carry on.

## johanfprins

Thanks

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

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

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.

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

## johanfprins

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.

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

These were my responses:

It is important that you understand that your numbers measure times and distances between different events.

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

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

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

I think this bears additional emphasis - since it seems the key point/idea he's missing.

## johanfprins

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

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

## Noumenon

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

## johanfprins

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

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

That should have been "as A-B in frame K/." of course.

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

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

Doing it that way allows you to check I am only using the Lorentz Transforms.

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

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

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.

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

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.

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

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.

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

Doing it that way allows you to check I am only using the Lorentz Transforms.

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

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.

No you are not doing this, since you are using lines with slopes to connect points: This is misleading.

## johanfprins

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

"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

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

## johanfprins

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

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

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!

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

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.

Nope, such a physics happens at every water surface.

Quack, quack, quack quack!

## Noumenon

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

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

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".

Previously:

You have not addressed that error in your proof yet.

## Fleetfoot

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

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

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

I got there about 15 years ago. I don't correct all your errors when they arise, it would be too disruptive.

I'll not waste time updating it until we agree a real change but ignore them for the moment.

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

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?

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.

D is an event at clock1 and the question does not involve an event

## Fleetfoot

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.

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

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

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

Why is this unreasonable when we are arguing about this construction?

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

Because we are both intelligent enough to ignore the parts that aren't agreed.

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

You are not so stupid not to realize that you are completely dishonest and acting disgracefully and unscientifically.

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.

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.

## johanfprins

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

On the contrary, I am the one stopping you skipping over key scientific points.

See below ...

Sounds like an agreement.

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

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!

So if we agree what did I do so that you accused me that I am going further than to what we agreed?

Only after including the points F and G.

## Fleetfoot

By construction. We plot the point at the value read from the clock so if tB /= tA they will be different points.

Correct.

Obviously, if x = -v*t is generally true, so is that specific case.

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

## Fleetfoot

I'm happy to revisit anything you like, ask away.

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.

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

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!

## johanfprins

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

I'm answering the question you asked. If you don't want stupid answers, don't ask stupid questions.

It does!

Of course. I have no idea why you think I disagree with that, two different times just give two different points, obviously.

## Fleetfoot

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.

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

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.

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

No I did not forget.

## Fleetfoot

There is nothing "scientific" about wasting time, I just overestimated your ability to keep track.

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.

Ah, so are you now claiming your error was deliberate?

If you want to have an intelligent conversation, stop playing word games.

(contd.)

## johanfprins

I have NOT been playing word games anywhere: This is also your forte!

## Fleetfoot

Maybe you have, but you must then expect to justify your claims. Here is how you started your attempt at a proof:

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

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

Let me remind you then:

It is your corrected version that we are now discussing:

That is what is shown in the "next step" diagram.

## johanfprins

It only contains the two paths: So far so good.

To avoid the confusion, replace A with D (event) and B with F.

## johanfprins

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!

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

## Fleetfoot

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

Excellent, that's a beginning.

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

I also stated this in the first version like anybody with a modicum of rains will attest to.

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/.

I do claim that, since I have NOT made any serious errors in the physics I have posted on this thread.

## Fleetfoot

Here is the quote again, it is quite clear:

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.

I didn't say it was serious or in the physics, just a minor error in your statement of the question.

## johanfprins

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

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.

There is no error, not even a minor one, as your own diagrams are proving.

Signing off for tonight!

## Fleetfoot

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:

## Fleetfoot

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.

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

## Fleetfoot

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

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

You are not able to teach any physics, even to a genius!

## johanfprins

Yes or no?

After you have answered I will proceed.

## ValeriaT

## johanfprins

What PM thread?

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.

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

## Noumenon

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

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!

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

## johanfprins

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

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

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

## Fleetfoot

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

None taken, I just thought you were commenting on how slowly the conversation is developing.

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

Good, I'll take that as the new baseline agreement from now on.

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.

To be pedantic, the LTs transform between frames, not clocks, though you have defined the frames based on the clocks in this case.

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

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

## johanfprins

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

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

## ValeriaT

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

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.

No, I only discussed transforming from K/ to K, not the other way.

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

## Fleetfoot

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

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

I do not have to believe it since one can derive it directly from Newton's second law.

No it is NOT!

That is true but it has NOTHING to do with AWT.

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

You have posted the following equations:

They are clearly LT's from K into K/: Not K/ to K as you are claiming.

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?

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

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