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

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.

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

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

*Science*, 2013

**Citation**: Einstein's gravity theory passes toughest test yet: Bizarre binary star system pushes study of relativity to new limits (2013, April 25) retrieved 24 April 2019 from https://phys.org/news/2013-04-einstein-gravity-theory-toughest-bizarre.html

## User comments

ValeriaTTorbjorn_Larsson_OMNo, 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-mechanicsThe 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_smothTektrixLOL, nice one. Subtle jokes are the best kind :)

Q-StarNo 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_physorgIt'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 GyreekimEmpty space is not empty. It is filled with a sea of virtual particles.

TektrixWhydening GyreAnd an ALMOST equal number of empty spaces...

Osiris1Sanescienceantialias_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.

ValeriaTant_oacute_nio354Gravitational 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

ValeriaTValeriaTWhy 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.

theonValeriaTbrtIt'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.

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

Dileep_SatheKoenQ-StarThat'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.

brtQ-StarMOND 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)

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

Q-StarYa are correct. It is a humbling awaking indeed.

FleetfootIt'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.

IronhorseADark 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.

rkilburn81FleetfootA receiver that can detect EM at 100Hz will not detect gravitational waves of the same frequency. Each phenomenon has its own spectrum.

FleetfootThey 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_nio354Relativity theory is all wrong.

Antonio Jose Saraiva

FleetfootIt fits the experimental results, that makes it right.

johanfprinsNo, it is not ALL wrong! Only some deductions from it are wrong: Like time-dilation and length contraction.

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

Q-StarExactly 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.

ValeriaTjohanfprinsIncorrectly 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.

johanfprinsIt 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!

johanfprinsWrong 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.

johanfprinsIt 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!!

johanfprinsI 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!

johanfprinsIf 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!

FleetfootFor 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.

johanfprinsOf 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.

johanfprinsTo 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?

johanfprinsWhy 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!

johanfprinsWhy 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?

FleetfootThe 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.

johanfprinsHow? 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.

thefurlongI 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?

FleetfootThe 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.

johanfprinsThank 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.

johanfprinsThis 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.

johanfprinsThe 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!

ValeriaTjohanfprinsEinstein'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!

ValeriaTjohanfprinsThe 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.

FleetfootI 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.

thefurlongI 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.]

thefurlongThis 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]

johanfprinsNonsense, 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?

thefurlongHowever, 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...

johanfprinsI 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!

johanfprinsWhere 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.

johanfprinsWRONG: 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.

ValeriaTjohanfprinsProve 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!

johanfprinsHow 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.

ValeriaTFleetfootThe 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.

FleetfootNot 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?

thefurlongYes, 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.

FleetfootYou 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-StarI'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 GyreYeah, 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.

johanfprinsI 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!

johanfprinsYou 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!

johanfprinsThe 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!

johanfprinsThank 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!

johanfprinsThere 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!

johanfprinsIt 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!

johanfprinsAn 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.

johanfprinsI 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?

johanfprinsI 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.

FleetfootThis 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

FleetfootAs 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.

johanfprinsFom 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.

FleetfootHere'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.

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

FleetfootFirst 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.

johanfprinsAlthough 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!

johanfprinsOnly 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".

johanfprinss^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"

FleetfootOnly 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?

FleetfootNo 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?

johanfprinsWhere 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?

johanfprinsYou 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!

thefurlongSo 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-StarHe 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-StarLight clocks,,, ya are correct, because "c" is "c" for everyone.

thefurlongOnly 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.

FleetfootIn 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.

johanfprinsValeriaT: 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!

FleetfootI'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.

johanfprinsNo 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!

FleetfootLet'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.

johanfprinsCorrect 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!

FleetfootThere 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.

johanfprinsNope! 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!.

johanfprinsA 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

johanfprinsBut 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.

johanfprinsCorrect! 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.

johanfprinsYou 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.

johanfprinsthefurlongIf 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 GyreWhat 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...

johanfprinsThere 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.

johanfprinsAnd 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

thefurlongWell, 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."

thefurlongAs 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?

FleetfootNot 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.

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

FleetfootThe 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.

FleetfootPart 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.

thefurlongThank 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.

thefurlongOnly 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.

thefurlongjohanfprinsThe 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 Gyreso, 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.

thefurlongOk, 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.

thefurlongThe 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 GyreThat 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.

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

Q-StarBecause 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.

thefurlongNo. 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.

thefurlongYou 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.

FleetfootIn 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-StarCorrect, 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.

FleetfootWell 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-StarNot 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 GyreI 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-StarWhich 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 GyreWell, then. it seems you've picked a single IRF for yourself...

Q-StarNo,,, 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 GyreWho's arguing?

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

Q-StarWell, 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 GyreWait 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-StarWho?

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-StarNo 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-StarIf 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

FleetfootNo, he's talking about 4-velocity:

http://en.wikiped...velocity

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

Whydening Gyreto 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 GyreI accept that...:-)

thefurlongYou said

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

Q-StarTo 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.

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

thefurlongYeah, 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."

thefurlongNeither 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 GyreNot 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 GyreArguing.

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...:-)

thefurlongTelling 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.

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

FleetfootThink 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.

FleetfootYou 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.

FleetfootI 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)."

johanfprinsLight 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!

johanfprinsWhoa!! 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.

ValeriaTjohanfprinsThere 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.

ValeriaTI 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.

johanfprinsI 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.

johanfprinsAccording 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.

ValeriaTjohanfprinsThe 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.

johanfprinsBut 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!

johanfprinsAs 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!

ValeriaTValeriaTjohanfprinsValeriaTjohanfprinsYou 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!

johanfprinsFirstly 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!

ValeriaTValeriaTBTW 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.

johanfprinsI 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!

johanfprinsI 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!

ValeriaTValeriaTjohanfprinsIt 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 an