Does this thought experiment disprove special relativity?

That is not what SR predicts. The situation you have described is symmetrical. Neither observer accelerated in the past. You would use Lorentz transformations to transform the coordinates from one frame to the other, and find that the situation is symmetrical. It will be indistinguishable, which observer is moving and which is at rest.

This is different from the Twin Paradox, where one twin undergoes acceleration where the other one does not. This is an asymmetrical problem, where one twin will age less than the other. It is not the same as the symmetrical problem.

I agree with Stam. You must do the math correctly. Words are ambiguous, and confusing. You're word problem did not describe how the clocks were synchronized in the first place, so it is not worth the effort to recreate a SR chapter on clock synchronization.

Jerry Decker

It is possible to create a paradox where none is needed. Always there is one assumption that is not true hidden in the logic sequence. My favorite paradox is the triplet story.

One triplet stayed at home and was not greatly accelerated. The other two took identical long journeys but in opposite directions and by prearrangement returned at the same time. All three agreed that the one who stayed home should be much older than the other two, although SR doesn't specifically say so, Einstein recommended the time dilation to apply only for an accelerated traveler.

Now we have two travelers who should be the same age by symmetry. They don't agree with each other about predicted ages because of the SR observations each has made of the other during the journey. An extension of Einstein advice is that equal acceleration should yield equal ages. Of course the travelers should discover this on comparing notes back home and decide the SR calculations they made of each other were not appropriate for the situation. Improper use of SR was the false assumption.

Xinhang Shen your example is of this type. Even when the logic is perfect, the assumption is wrong. SR has been proven with great accuracy in many experiments, but in situations where it is appropriate.

In fact a great many paradox can be created of this type. They don't disprove SR. They define the methods and cases where SR should be used.

So I'm in agreement with Todd, and with a compelling example, a famous story not of my invention.

I think you want to prove SR isn't a valid theory. In order to get this result, from my viewpoint, isn't necessary a thought experiment because it would mean to repeat the same errors on which SR is based. I would propose instead a true and real experiment: measurement of the speed of a particle, for instance electron, into a particle accelerator. Nobody has measured those speeds but they are only the result of a calculation based on SR. That calculation isn't able to prove the postulate of the constancy of light and the fact that the speed of light is a maximum value. It needs instead a real measurement that would prove without doubt the real behaviour of particles into an accelerator. The scientific method is based on real experiments.

OK, everybody! Please use special relativity to answer the following question:

There are two observers (A and B) in two inertial reference frames respectively. Each observer has an identical clock with him. The two clocks are synchronized to 0:00 when they meet. At 3:00 of clock A, observer A sends the image of clock A to observer B. At 3:10 of clock A, observer B receives the image of clock A and immediately sends the image of clock B back to observer A. At 3:20 of clock A, observer A receives the image of clock B. Now is the question: what time should be shown on the received image of clock B?

You are not specifying the relative velocity between the two frames, but you are specifying that it takes 10 minutes, at the speed of light for the signal from Observer A to reach Observer B. If the velocity between the two inertial frames is zero, then the received image will say 3:10. Whatever the initial speed was, in the frame of Observer A, Observer B is 10 light-minutes distance from Observer A when B receives the message. I calculated the speed between the two frames to be only 5.2% of c, or a gamma of 1.001. Therefore, relativistic effects are negligible. The time on Observers B's clock will be 3:10.

Todd, if the image of clock B shows a time larger than 3:10 no matter how small the difference is, then observer A thinks that clock B runs faster than clock A. On the other hand, observer B also thinks that clock B runs faster than clock A. That is, there is no such a thing that each observer sees the other observer's clock runs slower as predicted by special relativity.

At first, I found the biggest puzzle to be why you asked this question on ResearchGate.

However, the answer is quite pretty, when considered in general. Let te be the time of emission (measured from 00:00), and to the time when A observes the reflected signal from B. Both measured by A. Then the time of reflection, measured in the inertial frame of A, is the arithmetic mean of te and to, while the time shown on the reflected signal is the geometric mean of te and to. With the given numbers, this translates to a time 03:09:44.2 on the received image.

Moreover, this question is a good starting point for further education. What happens if A sends out signals at regular intervals dt ? What happens if B reverses her velocity after receiving signal number N ? How does these results compare with the formulas of relativistic Doppler effect?

Todd: If someone lured 0.14% of a years income from you, you wouldn't consider the amount negligible?

Kåre, the reason I asked this question is to show you that special relativity will fail to answer the question. As I presented on my papers (available on my profile here), the clock time is absolute while relativistic time is simply an artificial time irrelevant to clock time. If you use clock time as relativistic time, you will immediately get a contradiction.

It's interesting that you got 03:09:44.2. How did you get it? When the image of clock A arrives at observer B, if clock B is 03:09:44.2, then observer B will think clock B was slower than clock A, contradicting the claim of special relativity that the stationary clock should run faster.

Xinhang> is to show you that special relativity will fail to answer the question.

Well, Todd and I have demonstrated that you failed with that mission.

Xinhang> How did you get it?

By doing Pythagoras in my head, from which it follows that the signal from A is reflected from B when the clock at B shows sqrt[(180+10)2 - 102] minutes after midnight.

Xinhang> When the image of clock A ...

The rest of your post only proves that you are throughly confused. There are no claims in special relativity that stationary clocks run faster. Obviously, since being stationary is a relative concept. In your example, there is perfect symmetry between A and B.

Read at little more, follow Todd's link, solve the additional problems I indicated. Don't rush ahead publishing your misconceptions for (in principle) the whole world. Millions of physics students must have struggled with the same questions over the last century; most of them managed to get it right.

KO Asked: Todd: If someone lured 0.14% of a years income from you, you wouldn't consider the amount negligible?

A year? No! In this case however, we're only talking about 15 seconds, so yes it's negligible. :) My iPhone doesn't display time in seconds so okay, 3:09. Happy?

Rounded to your iPhone 3:10 is the right answer. However, an accurate GPS system works with much better accuracy than 0.1% (and that accuracy should in principle be available in all smartphones with GPS).

In another thread I visit, the relevant difference is of order 1 picosecond after 100 seconds. From that background your approximation seemed inexcusably crude :-)

Note added: I now understand why you said 3:09. Of course, on the image of the clock the minute has not changed yet.

Thanks. My point was more along the lines of; Word problems in special relativity are ambiguous. The question posed is not a well defined problem, therefore I can cheat and get a quick an dirty answer that is "close enough" for all practical purposes.

Kåre>There are no claims in special relativity that stationary clocks run faster. Obviously, since being stationary is a relative concept. In your example, there is perfect symmetry between A and B.

Yes, because A and B is completely in perfect symmetry. Relative to the frame of observer B, according to special relativity, clock B is stationary and should run faster than clock A, but in your calculation, the time of clock B is slower than the time of clock A observed in the reference frame of observer B when the image of clock A arrives at observer B. This directly disproves special relativity.

Kåre>Read at little more, follow Todd's link, solve the additional problems I indicated. Don't rush ahead publishing your misconceptions for (in principle) the whole world. Millions of physics students must have struggled with the same questions over the last century; most of them managed to get it right.

Please be a bit more polite and don't make baseless conclusions!

Xinhang> baseless conclusions!

My polite and sincere advice was based on your writings.

Of course, I should have followed my initial solution to the puzzle of why such a question of elementary relativity was asked on ResearchGate, when a high-school class is more appropriate.

Added:

Todd> Word problems in special relativity are ambiguous.

The problem you and I ended up solving is actually a very good problem -- to give after the first lecture in relativity theory. It requires deductive and mathematical capabilities as expected at final high-school exams. [In Norway, I have recently seen a similar problem in the (easy) first part of such an exam.] It has an explicit answer (which can be used to expose the notion of accuracy, as you did with your amusing solution). The general analytic answer is quite pretty, as I mentioned. The problem is straightforward to extend in several directions, and through this use to build very solid understanding of special relativity.

But it does not belong on ResearchGate.

Daniele> measurement of the speed of a particle, for instance electron, into a particle accelerator. Nobody has measured those speeds but they are only the result of a calculation based on SR.

Why don't you contact the accelerator people at CERN, and propose to give a performance for them at their Xmas party in December? I am sure they plan some entertainment in that connection.

Dear Xinhang Shen,

talking about accelerators: I am right now sitting at one of the institutions devoted at delivering synchrotron radiation and doing measurements with the xrays kindly provided there. Would you be willing to accept/admit that the spectrum of the radiation emitted e.g. by electrons passing a so-called bending magnet (in the electron storage ring) can be accurately computed using SRT, including its frequency dependent angular spread and polarization properties (over many orders of magnitude of the frequency of the radiation emitted, and - of course - by taking into account the highly relativistic electron motion {electron energies of a couple GeV}) ?!

Actually, institutions like the German equivalent of NIST (it is called PTB) maintain their own synchrotron and use that radiation to calibrate xray sensors, because the source characteristics can so accurately be calculated (and experimentally controlled).

The minimum to be done - if you want to criticize SRT - is to come up with a clean and conclusive explanation as to why its application then is so utterly successful! [success is not a proof of being right - but if you claim SRT to be unrealistic, then you have to explain its success.]

Daniele,

I obviously cannot know what you would be accepting as a speed measurement of electrons in an accellerator. Maybe the following, which is being used on a daily routine basis at synchrotron radiation laboraties all over the world (and then - maybe you wouldn't accept this. In this case, please let me know why.)

Electrons in a synchrotron storage ring travel there not as a quasi-continuous particle stream but rather in bunches (typically some tens of cm long). As a result, the synchrotron radiation is not continuous in time but does exhibit pulse structure. (Durations of the pulses can be controlled to some extent by controlling the bunch shape). Typically, there are a couple hundred bunches travelling in the storage rings, but one can also choose to have just a single bunch in there. (In daily operation with many bunches, one of them is loaded with more electrons, giving rise to a more intense pulse, which people may use for timing purposes in time resolved experiments. There are other tricks that can be played in bunch shaping, e.g. producing deformations induced by interaction with fs-laser pulses).

Take the single bunch operation for simplicity. You get one light pulse per round travelled by the electron bunch. To the extent of precision with which the path length can be controlled and/or determined (which I actually ignore) you will obtain a pulse-to-pulse distance (in time) compatible with the electrons having (nearly) reached the speed of light - irrespective of whether the electron energy is 0.8 GeV or 5.4 GeV. (These are the upper and lower limits of electron energies at storage rings I have personally done experiments at.)

The other point (for me at least) in favor of SRT - in this context - is the precise computability of the radiation characteristics I have mentioned in my previous post.

Kai,

you write, if I understand well, the maximum value of energy for electron is 5.4 GeV. You are certainly able to calculate the speed of electron at that energy. I say only you would have to measure besides to calculate that speed. I don't understand because postmodern physicists don' t consider this possibility that is normal in physics.

Daniele,

I actually made a proposition of how that speed can be measured! You would have to tell, wether to you this is a measurement of speed. To me it is.

Again in short: in a storage ring, bunches of electrons travel in circles. The circumference depends on the installation and varies between several tens of meters and more than a km (e.g. ESRF: www.esrf.org). Every time the bunch passes a bending magnet, radiation is being emittet tangentially to the (locally circular) orbit. These pulses of light can be detected. the time delay between two consecutive pulses is the taime taken by the electrons to make it around the storage ring once. You have the distance travelled and the time needed to do so. Thousands of experimenters rely on these things working on a daily basis and use them for all sort of characterization, from materials to dynamic processes.

It would be up to you to say, whether the prescription given above qualifies as a measurement of speed for you. If it doesn't, please propose an alternative.

Daniele:

and no - you misunderstood what I was saying about electron energy. I have worked at a couple of different synchrotron radiation centers. Each of them may use a different energy of electrons in their storage ring. That is a design parameter of such machines. The lowest energy source I have used is no longer in operation (has been replaced by a new one 16 yrs ago). It was BESSY, operating at 800 or 850 MeV, if I remember right. On the other end is ESRF with 5.4 GeV. The other locations I have used had energies in between.

So my statement was about electron energies in use at those synchrotron radiation centers I have actually visited myself.

It was, in particular, not a statement about maximum electron energies whatsoever.

Daniele> you would have to measure besides to calculate

To my knowledge, the operation of accelerators is beyond calculations and measurements, it is about controlling the behavior (which incorporates the former two as a prerequisite).

I vaguely recall hearing/reading that in LEP/LHC the property of each particle bunch is measured on one side of the ring, with a correction signal being sent diagonal across (to arrive in time). Regardless the placements of detectors and correctors, one has to know when to apply a correction to a specific bunch, hence one has to know how fast is moves.

Christian should correct me if I am inaccurate (or wrong) here.

Kåre,

at the Swiss Light Source (PSI), dimensions are of course smaller than at CERN. There also, an orbit correction scheme exists which creates feedback from some 250 or so sensors to correct the settings of the power supplies. If I remember right, some ten or so years ago this system operated at 400 Hz, which was the fastest I knew of at that time. But it is still orders of magnitudes slower than the round trip frequency.

Concerning my post: I just wanted to propose one simple and understandable measure of electron speed in these rings - and one which is actually routinely being used (e.g. for purposes of experiment timing and control).

Sergey Shevchenko

“…That is not what SR predicts. The situation you have described is symmetrical. … You would use Lorentz transformations to transform the coordinates from one frame to the other, and find that the situation is symmetrical…”

- including, for example, that every observer in every of both frames according to the SR must believe that in the vis-à-vis frame clocks tick slower then in the her/his frame; what is evident absurd – classical Dingle problem in the SR.

Besides from the SR postulates that there is no absolute Matter’s spacetime and so all/every inertial reference frames are totally equivalent any number of other evidently absurd consequences follow…

Cheers

Kai,

I don' t have criticized your experimental work, I wrote only I wanted know the value of speed that you calculate at 5.4GeV or at different value for electron. Besides I think that thoretical value has to be compared with the measured value. In physics the normal procedure consists in a calculation and after in a measure that confirms the theoretical value. The fact that all postmodern physicists use the same procedure doesn't prove that procedure is correct.

Kai> I just wanted to propose one simple and understandable measure of electron speed in these rings

I agree that these are much better demonstrations (and tests) of the accurate workings of relativistic dynamics and electrodynamics. But few, if any, of those who find this simple and understandable will question relativity anyway.

At LHC, the rotation frequency is about 11 kHz. But a particle bunch of 4.3 cm length will pass a point in 0.14 ns. So it seems unrealistic to think that one can control individual bunches(?).

Siva Prasad Kodukula

Special theory of relativity emphasises on time interval in reference frame.(flow of time).

A clock is nothing but an instrument to calculate the time intervals with "ticks of clock".

As per relativity B is relative to observer A . Relativity assures that A is relative to B.

All the effects will be same for B relative to A and A relative to B.

Calculation can be done as per the affect of relativistic factor.

There is no confusion in it.

Daniele> I wrote only I wanted know the value of speed that you calculate at 5.4GeV or at different value for electron.

For particles with much higher energies than their rest energy mc2, it is simpler and more instructive to calculate how much the particle speed v differs from the speed of light c (as measured in the stationary frame). For electrons, the expression is

c - v = (0.511/E[GeV])2 x 150 m/s

which gives c-v = 1.5 m/s for a 5.11 GeV electron, and c-v = 1.2 cm/s for a 57 GeV electron (which was about the fastest moving electron at LEP).

Dear Daniele,

for some reason, we seem not to be able to communicate efficiently.

Of course, you have not critisized my work (which btw. is - besides the use of synchrotron radiation and the study of magnetism in materials - unrelated to SRT as such), and there is nothing in what you wrote which could be seen as any kind of critique, so no need to worry.

You wrote, however, hat nobody had ever measured the speed of electrons in accellerators. I proposed you, that some type of measurements which are routinely done can be taken as a measurement of the electron speed. I do want to know from you whether you would regard the proposed method as a valid measurement of electron speed (and hopefully to get a conclusive answer in case you disagree). Unfortunately you are not at all reacting to that question.

When using SRT to calculate that speed (in the lab frame) then you see from Kåre Olaussen's answer that irrespective of actual energy in the GeV range, you obtain a result which is somewhat smaller yet close to the speed of light in vacuum.

If that is the case, then irrespective of the actual electron energy, the time taken for the round trip in an electron strorage ring will be essentially proportional to the length of the trajectory (essentially the circumference of the ring structure). Taking the time interval between light pulses emitted from an electron bunch as the time required for a round trip, this is exactly what is observed. For a ring twice as large the round trip time is doubled, irrespective of whether the electron energy is 1,2,5 or 10 GeV. All these electrons have nearly the same speed.

If Newtonian mechanics were valid, we would have v=sqrt(2E/m) (E = electron kinetic energy, m= electron rest mass, v = velocity), and the conclusions about round trip times in storage rings would be a) qualitatively very different b) v>c should easily be possible. [a) ensures, that the actual path length need not be known with extreme accuracy to follow the agrument. I am not talking about precision measurements here.]

Especially a) shows, that since we have storage rings that are vastly different in size and use sufficiently different electron energies, Newtonian mechanics cannot be applicable, whatever the replacement would be.

SRT seems able to satisfy the finding v

Dear Kai,

I agree with Kare that it is suitable to express speeds of particles at high speeds in function of c. According my calculations, in SR an electron at the energy of 5.4GeV, has a speed v that is smaller and nearest to c. In The Theory of Reference Frames (TR) the same electron at the same energy is unstable and it has a speed v≈140c. You see there is a great difference between the theoretical values in SR and in TR. A simple experiment, performed without prejudices, could extablish which value is correct. It is not important that the accelerator where you work is small with respect to accelerators of CERN, it is important that in your laboratory it is possible to make that experiment.

For a proton at 5.4GeV in SR the speed is v=0.985c while in TR the speed is v=3.69c.

Sensible differences of values allow a certain differentiating measure.

Daniele,

you have still not decided whether you would regard my proposition as a valid measurement of electron speed. That is something we should agree upon before proceding. I may then contact people and find out about the electron round trip travel times. I am pretty sure I can get or find the data for three or for sources. Maybe they are even published.

Be assured, that there is not a single instance with v>c, though, excluding maybe uncertainities which I already mentioned. But the predictions you mention are safely distinct, which is good. I can assure you that from taking the figures you mention at face value you can most probably put TR at rest for good.

Dear Kai,

reading better the description of your observations, I note you write "For a ring twice as large the round trip time is doubled, irrespective of whether the electron energy is 1,3,5 or 10 GeV. All these electrons have nearly the same speed".

Your conclusion is wrong, because it is manifest that doubling the path also time doubles at any speed. It doesn't mean nevertheless speeds are the same.

Daniele,

on their website (link added below) BESSY II indicates that in single bunch operation a pulse is obtained every 800ns (I don't know the precision of this number)

The electron energy used at BESSY is 1.7 GeV

The circumference of the storage ring is quoted to amount to 240m

From these pieces of information maybe you can discriminate the models ?!

240m / 8E-7s = 3E8 m/s

I don't know about the numbers' precision, but I can try to find out. That may be useful since I collected these numbers on their webpages, which partly address the layman public.

https://www.helmholtz-berlin.de/quellen/bessy/betrieb-beschleuniger/betriebsmodi_en.html

https://www.helmholtz-berlin.de/mediathek/info/beschleunigerphysik-fuer-anfaenger/der-speicherring_en.html

https://www.helmholtz-berlin.de/mediathek/info/beschleunigerphysik-fuer-anfaenger/weshalb-elektronen_en.html

Hi everybody, please discuss my question only! Let me repeat my question here:

There are two observers in two inertial reference frames with a constant relative speed: A and B. Each observer has an identical clock with him. At 3:00 of clock A, observer A sends the image of clock A to observer B. At 3:10 of clock A, observer B receives the image of clock A and immediately sends the image of clock B back to observer A. At 3:20 of clock A, observer A receives the image of clock B. Now is the question: what time should be shown on the received image of clock B?

According to Kåre's calculation, the time shown on the received image of clock B (3:09:44.2) is earlier than the time of clock A (3:10) when the image of clock A reaches observer B. That is, when observer B receives the image of clock A, he finds that clock A (the moving clock) runs faster than clock B (the stationary clock) while special relativity claims that the moving clock should run slower than the stationary clock.

Who can explain the contradiction?

kai,

if your data are sure, the electron speed is about c. But you know in SR length contraction at the speed c is zero and time dilation is infinite.

Anyway because Shen doesn't like this discussion I am available to continue it in my question

https://www.researchgate.net/post/What_is_the_relative_velocity_of_two_particle_beams_that_are_accelerated_into_reverse

Xinhang> At 3:10 of clock A, observer B receives the image of clock A

This statement is by itself incomplete, due to the relativity of simultaneity (which you must have heard of). I did not actually use it to solve the problem, since the information "At 3:20 of clock A, observer A receives the image of clock B", a frame independent statement, is sufficient. This latter information further implies that you meant to say "At 3:10 of clock A, as described in the inertial frame of A, observer B receives the image of clock A" . It is this implicit reference to the inertial frame of A which breaks the symmetry between A and B.

Xinhang> That is, when observer B receives the image of clock A, he finds that

You did not mention anything about A also sending clock images, but if A did, B would receive an image showing 3:00. That is sufficient information to compute their relative velocity and other stuff. But the value of the clock at A when B receives the image is meaningful only when the inertial frame has been specified.

If you mark a point on a blank sheet of paper, and ask for the y-coordinate of that point, nobody can answer before you have specified the coordinate system.

XS: Hi everybody, please discuss my question only!

That would be polite, if everybody else wants to discuss a side-issue, they should start a separate thread.

XS:There are two observers in two inertial reference frames with a constant relative speed: A and B. Each observer has an identical clock with him. At 3:00 of clock A, observer A sends the image of clock A to observer B. At 3:10 of clock A, observer B receives the image of clock A and immediately sends the image of clock B back to observer A. At 3:20 of clock A, observer A receives the image of clock B. Now is the question: what time should be shown on the received image of clock B?

As has been said, there is not sufficient information in your question to give a definitive answer. Kare solved that by assuming that A and B passed each other and synchronised the clocks to 00:00:00 so 3 hours elapsed after that event before A transmitted his image. Accepting that and including that in the question would complete the picture.

XS: According to Kåre's calculation, the time shown on the received image of clock B (3:09:44.2) is earlier than the time of clock A (3:10) when the image of clock A reaches observer B. That is, when observer B receives the image of clock A, he finds that clock A (the moving clock) runs faster than clock B (the stationary clock) ...

No he doesn't, you haven't completed B's calculations.

When clock A says "03:00:00.0" a message is sent containing that value. A considers it takes 10 minutes to reach B (a distance of 600 light-seconds) and it is received at 3:10:00.0 but clock B only shows 3:09:44.2 so A determines that B is running slower by 0.139%.

B receives the message 3:09:44.2 so calculates that the message has travelled a distance of 569.21 light seconds and was actually emitted when the time was 3:00:15.0 on clock B so B determines that A is running slower by 0.139%.

XS: Who can explain the contradiction?

The time dilation factors are symmetrical, there is no contradiction.

George> Kare solved that by assuming that A and B passed each other and synchronised the clocks to 00:00:00 so 3 hours elapsed after that event before A transmitted his image.

That information was given by Xinhang in an earlier post on page 1: "The two clocks are synchronized to 0:00 when they meet.", so I didn't assume it.

By the way, you shouldn't let it bother you if you are downvoted by fools, for good and thoughtful answers.

KO: That information was given by Xinhang ..

Sorry I missed that post.

KO: By the way, you shouldn't let it bother you if you are downvoted by fools

It never bothers me, though I wondered if someone might be suggesting I shouldn't have told him the answer but let him work it out for himself.

F. Leyvraz

Easy: formulate the paradox with sufficient clarity that you can write it down in formulae. Apply mechanically and as correctly as possible the Lorentz transform to the situation you have. If you still have a contradiction, send me the formulae, I will show you where you went wrong.

But now everything is both complicated and imprecise. A very rough and ready answer is the fact that there is no unambiguous way to synchronise clocks in relative motion. The Einstein synchronisation you seem to apply is only assumed to work for clocks at rest with respect to each other. This is then enough to compute, via LT, what moving clocks do.

Dwight Hoxie

I don't think that the answer is simple. Presumably the relative speed of the two observers and their clocks is relativistic and so sending signals between the two observers is not straightforward unless the signal is sent between two coincident points, one in each reference frame.

Siva Prasad Kodukula

what time should be shown on the received image of clock B?

Now the Answer is--------

3:20 only. Theory of relativity emphasises on time dilation. But 3:20 is inclusive of time dilation. The 20 minutes time will be affected by relativistic factor for time dilation. But you have said that the time shown at A after completion of event as 3:20. So 3:20 is the answer. Relativistic factor depends on relative velocity.(which is not mentioned for calculation). Relative velocity is same for both A and B. As far as theory of relativity concerns it is correct .

I am accepting Special Relativity as it is.... But for me it is not sufficient.... I have a modified theory for me ...Which can be verified experimentally by accelerators.....I am waiting for appropriate forum for it.....

George Dishman>B receives the message 3:09:44.2 so calculates that the message has travelled a distance of 569.21 light seconds and was actually emitted when the time was 3:00:15.0 on clock B so B determines that A is running slower by 0.139%.

XS: No!

My numbers are correct, at least to the accuracy of an Excel spreadsheet ;-)

XS: 10 minutes of clock A (600 seconds) transformed to the frame of observer B is 600 times gamma ( 1.001) = 600.6 seconds as seen by observer B. That is, observer B should find that the image of clock A (3:00) is sent at 2:59:54 of clock B, i.e., clock A runs faster than clock B.

A sends the image at 3:00 and in his frame it arrives at 3:10 exactly. He see his signal catching up with clock B and arriving at 3:10 so calculates their separation at that time.

B on the other hand sees the message arrive at 3:09:44.2 on his clock so he calculates how far away A was when the message was sent. That means the distances are different and the time for the message to travel is different. You need to fully calculate all these effects, skipping the steps is risky and has given you an inaccurate answer.

If you want to do it phenomenologically, you have to include all the factors, assigning coordinates and using the Lorentz Transforms is less prone to error.

SPK: Now the Answer is-------- 3:20 only.

That is not even close.

SPK: I am accepting Special Relativity as it is.

Perhaps but you will need to learn how to apply it. My post above gives you the correct answers from the theory.

DH: Presumably the relative speed of the two observers and their clocks is relativistic

No, it's only ~10 light minutes in 3 hours 10 minutes so little more than 0.5% of the speed of light. The solution is straightforward, see my previous post for the numbers.

Dwight> I don't think that the answer is simple.

I disagree (I would agree, for that particular post, if the last 5 words of the sentence were removed).

George's analysis is perfectly correct. It is disappointing to see that his answer cannot be reproduced even when known and explained. It is even more disappointing to see people applying powerful formulas without first contemplating how (and if) they should be used.

But for this simple problem, which is a straightforward application of plane geometry (involving time- and light-like lines), I find it more instructive to work without use of gamma-factors and numerical calculations (at first). Let Ta be the time when A sent a clock image to B, Tb the time when A receives the image from B, where the clock shows Tc. All times are measured since synchronization. Note that these are all physical events; hence they can be analyzed in any frame, with the same results. A simple application of Pythagoras in the restframe of A, using the fact that light is light-like and A and B are time-like, gives the beautiful result

Tc = sqrt[((Tb + Ta)/2)2 - ((Tb - Ta)/2)2] = sqrt(Tb Ta)

I encourage anyone to re-derive this result in the restframe of B, and any other frame. That will increase your faith in the validity of analytic geometry in Minkowski space.

Now assume that A's transmission was an immediate reaction of receiving an image from B, showing a time Td

Hi everybody, many thanks for contributing your calculations. George Dishman has perfectly solved the problem without any contradiction for special relativity.

Demetris Christopoulos

Dear Xinhang,

Your question is too close to Dingle's arguments,

See link for details.

http://blog.hasslberger.com/Dingle_SCIENCE_at_the_Crossroads.pdf

Dear Xinhang,

sorry for having gone somewhat off-topic on the road.

(I do still think that the accelerator/radiation topic is generally an important and interesting one in the context of SRT discussions.)

But I am curious to know whether Georges solution has consequences for how you think about SRT. Are you willing to share?

Best, KF

Dwight Hoxie

Dear Xinhang,

Your statement "Each person has a clock with him, and thus the position of each observer is clearly known by the other observer in his reference frame." In relativity this statement is not true. The two observers can set their clocks to zero at the instant they pass each other but thereafter they can only send signals to each other via light at its speed c. The two observers do not know the position in space and time of each other except at that one instant in which they are coincident.

Dear friends,

Now I have realized that there is no logical contradiction in the kinematics of special relativity if we strictly use relativistic time and length everywhere, just as use Newton's kinematics with Galilean time and length everywhere.

In fact all inertial reference frames can be described by both coordinate systems: one is based on Lorentz Transformation (called relativistic system) and the other is based on Galilean Transformation (called Galilean system). Relativistic system is mathematically equivalent to Galilean system. But the two coordinate systems result in two different interpretations of time, length and the speed of light: for relativistic system, time and length are relative but the speed of light is absolute, and for Galilean system, time and length are absolute but the speed of light is relative. Both interpretations are logically intact. That is, to describe the motions of objects (i.e. in kinematics), using relativistic system will produce exactly the same results as using Galilean system if we use strictly their respective definitions of time and length.

The major advantage of using special relativity is that it creates a set of electromagnetic equations invariant in inertial reference frames. If these equations are not correct, then special relativity is meaningless because of the difficulty in handling its relative time. According to special relativity, times of clocks with relative velocities such as those on GPS satellites flow at different rates and can never been universally synchronized. Because the time is relative in special relativity, there is no clear sequence of time flow in the universe so that we will never be able to correctly record history in relativistic time.

The advantage of Galilean coordinate system is that length and time are invariant in all inertial reference frames. Thus, time keeping becomes straightforward. For example, all clocks on GPS satellites show Galilean times that are synchronized not only relative to earth but also relative to each other no matter what relative velocities between them though in the synchronization process, the calculation of the speed of light should strictly follows Newton's velocity addition formula. With the absolute Galilean time, we have a clear sequence of time so that all historic events in the universe can be recorded in order. In the Galilean coordinate system, many modern physics theories are not developed yet, though so called relativistic theories of modern physics may completely wrong.

Up to now, all clocks we use in the world are universally synchronized. Thus, we are all using absolute Galilean time. It is a mistake to use our current universally synchronized time as relativistic time with relativistic theories to describe physical phenomena.

XS: Both interpretations are logically intact.

Both are self-consistent, that is true.

XS: Relativistic system is mathematically equivalent to Galilean system

That is not true, they produce different results.

XS: That is, to describe the motions of objects (i.e. in kinematics), using relativistic system will produce exactly the same results as using Galilean system if we use strictly their respective definitions of time and length.

You also need to include relativity of simultaneity and relativistic mass but then you get the equivalence between SR and a Lorentzian aether.

Without the different definitions however, Galilean relativity fails, predicting a fringe shift in the Michelson-Morley experiment for example.

XS: Up to now, all clocks we use in the world are universally synchronized

No, that is not true. Most clocks working to public time are synchronised to UTC with a time-zone addition which (roughly described) synchronises them to a single hypothetical clock located at the centre of the Earth but a gravitational potential equal to the surface. However, while all the clocks are synchronised to that one, they cannot be synchronised to each other.

George Dishman>

XS: Relativistic system is mathematically equivalent to Galilean system

That is not true, they produce different results.

This is true. I have found the mathematical relations between Galilean spacetime and relativistic spacetime (see my papers in my profile here). Every relativistic spacetime point within light reach can be uniquely mapped to a point of Galilean spacetime, and every point of Galilean spacetime within light reach can be uniquely mapped to a relativistic spacetime point. They don't produce any differences in kinematics. That is, the result of any kinematic problem calculated based on relativistic spacetime can be finally converted to the result in Galilean spacetime with the mathematical relationship I presented on the papers, which will be exactly the same as the result directly calculated based on Galilean kinematics, and vice versa.

Since special relativity is simply a mathematical trick, this trick can also be applied to sound speed based relativity which uses sound instead of light to synchronize clocks. In this sound based relativity, everything in kinematics will be logically intact too until the speed of an object exceeds the sound speed.

George Dishman>

You also need to include relativity of simultaneity and relativistic mass but then you get the equivalence between SR and a Lorentzian aether.

Without the different definitions however, Galilean relativity fails, predicting a fringe shift in the Michelson-Morley experiment for example.

Of course, in Galilean spacetime, simultaneity of events is defined by same Galilean time while in relativistic space time, simultaneity of events is defined by same relativistic time. All I discussed is kinematics without touching mass and force. Yes, regarding the speed of light, Galilean spacetime model needs a medium aether which may be fluid like, relative to which the speed of light is constant.

M-M experiment simply tells us that aether on the earth surface is dragged by the earth. H-K experiment shows that clocks run at different rates when they fly in different directions on high altitude airplanes. It means that on these high altitudes, M-M experiment may show fringe shift just like clocks showing different time rates because aether there is only partially dragged.

George Dishman>

XS: Up to now, all clocks we use in the world are universally synchronized

Here, the synchronization mean the time flow rate difference rather than the time zone difference because the main difference between relativistic time and Galilean time is the difference between time flow rates of clocks. Since the earth is rotating, there is a relative velocity between each other of the clocks on the earth. If you think they show relativistic times, then they are not synchronized to each other, but if you think they show Galilean times, then they are perfectly synchronized, because the time differences of relativistic times disappear among Galilean times if we include the effects of the differences of speed of light used in the synchronization. The obvious case is in the clocks on GPS satellites. These clocks have been adjusted faster than their corresponding relativistic times to be synchronized with the ground clocks. Then, what times are showing on these moving clocks? Relativity believers will be stuck. The answer is these moving clocks are synchronized to absolute Galilean time with which everything must be calculated in Galilean spacetime and relative speed of light. Therefore, our clocks including clocks on GPS satellites are showing universally synchronized absolute Galilean time.

We should never say time without mentioning "relativistic" or "Galilean", because they are two different definitions of time.

XS: M-M experiment simply tells us that aether on the earth surface is dragged by the earth.

As far as I recall this is not the concept of the aether pondered by Lorentz at the time. If true it would imho also entail new observable effects none of which I have ever heard of. For example, if the aether were dragged by local celestial objects then Doppler kind of shifts of Fraunhofer lines should result due to the relative motion betwen the aether at the location of absorption (sun, stars) and earth.

Aether drag was tested by many people and found not to exist, any aether theory has to work without drag hence Lorentz's version suitably extended is the only valid model.

Marcoen J.T.F. Cabbolet

Dear Xinhang,

Your original question was whether your thought experiment disproves Special Relativity (SR).

The answer is: no. The point is that you go off on a tangent already at the beginning. I'll try to explain this. I'll use the abbreviation IRF for inertial reference frame.

We have the following four events:

Event O: observers A and B meet and synchonize their clocks

Event E1: observer A sends an image of his clock to observer B

Event E2: observer B receives the image and sends an image of his clock

Event E3: observer A receives back the image sent by B

Obviously, the event O has coordinates (0,0) both in the IRF of A and in the IRF of B: there is no misunderstanding about that.

Event E1 has coordinates t = 3:00:00 and x = 0 in the IRF of A. It is, however, a mistake to think that the time coordinate of event E1 in the IRF of B is measured by the clock that observer B holds in his hand. Event E1 has coordinates (t', x') in the IRF of B, and the time coordinate is measured by a clock that at t' = 0 in the IRF of B was already at that x-coordinate x'. The point is that this clock will indicate a different time t', that is, a time t' not equal to 3:00:00.

So when B receives the image of the clock of A at the event E2, his reaction is: the clocks of A are not running correctly, because the time-coordinate of event E1 is not at all 3:00:00. Observer A, on the other hand, will say that the clocks of B are not running correctly: the event where the clock of B that measured the time-coordinate of event E1 was at t' = 0 in the IRF of B, is not at all at t = 0 in the IRF of A.

The point is that at the event O, the two clocks that A and B held in their hands were synchronized, but that does not mean that all the other clocks at the x-axis in the IRF of A and at the x'-axis of the IRF of B were synchronized too.

Xinhang, if you apply this reasoning to events E2 and E3 also then you'll see that this thought experiment doesn't disprove SR. You can do the math with the Lorentz transformations yourself to get the precise coordinates of all the relevant events in both IRFs. You have to be very carefull in applying SR when it comes to measuring coordinates of events.

Marcoen