The synchronization of clocks and the relativity of motion

[AlMetis]

This is a mistake. They are indeed synchronized in the stationary system.

I should have said, they are not according to observers at A and B.

 

[AlMetis]

What are you asking here?

The principle of relativity defines the reciprocity of all kinematics between inertial frames. The diagrams show otherwise.

 

[Sagittarius A-Star]

I took your advice and read the section 1.3.1 of David Morin’s book.
...
In order to demonstrate the one-way and two-way time of flight I have replaced David's detectors with mirrors at C and D.

In your diagram 1.0, I added the observer C, who is at rest with reference to the ground, in the middle between the locations of the two reflection events. Because he receives the signal from the left event at R_C first, he concludes, that, with reference to his rest-frame, the reflection event at location R_C happend before the reflection event at location R_D.

 

[Sagittarius A-Star]

No it does not, but to get to the relativity of simultaneity, Einstein has to point out the synchronicity is not an “absolute”.

Yes, but that is what he did:

So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

Source:
https://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Dale]

I should have said, they are not according to observers at A and B.

Or better, you should have said “according to the moving frame”

The principle of relativity defines the reciprocity of all kinematics between inertial frames. The diagrams show otherwise.

How do they show that? Which law of physics is shown to be different in the two diagrams? Please write down the specific law of physics in each frame.

Did you not construct your diagram by using the same law of physics in each frame? Specifically:

##ds^2=-c^2dt^2+dx^2=0## for light

 

[AlMetis]

I added the observer C, who is at rest with reference to the ground, in the middle between the locations of the two reflection events.

In my diagram 1.0, B is at rest with the platform at all times beginning with the flash at time = 0, as it was in David Morin’s example.
You have moved B (your C) to the right after the light flash.

 

[AlMetis]

Because he receives the signal from the left event at RC first, he concludes, that, with reference to his rest-frame, the reflection event at location RC happend before the reflection event at location RD.

I agree that the reflection from the left mirror (C) arrives at B first, that’s why I noted ACB≠ADB

If we deduce in David’s thought experiment, the one-way times of flight as Einstein did in his thought experiment, we can also say the one-way time of flight A to B is less( reflection happens first) than the one-way time of flight A to D in both observers frames A (on train) and B (on platform), thus as I noted: AC≠AD

Yet in diagram 2.0 the identical kinematics with B moving, A finds AC=AD.

 

[AlMetis]

Yes, but that is what he did:

Yes, I know. But he did it by showing that while the clocks at A and B are synchronous in the stationary frame, they are not in the rest frame of the rod. Which as Dale pointed out is “the moving frame”.
To get to this conclusion requires a test of the one-way time of light in both directions AB and BA in the rest frame of AB the rod.

If you use his two-way convention you will find AB=BA. It is a mathematical forgone conclusion by definition.
As you cannot measure the one way time of flight directly, you must deduce it from kinematic law as Einstein did in the equations above #29.
When you deduce the one-way times of flight you find as did Einstein AB≠BA and in David Morin’s experiment, AB≠AC.

 

[AlMetis]

How do they show that?

As I mentioned above in response to Sagittarius A-Star, the one-way times of flight are not reciprocated.

 

[AlMetis]

Which law of physics is shown to be different in the two diagrams?

The laws are not shown to differ, the observations are shown to differ, i.e. they are not reciprocal kinematics as the principle of relativity claims.

 

[AlMetis]

Did you not construct your diagram by using the same law of physics in each frame?

There is no change in the laws between diagrams, as far as I know. If you you see such a change, please point it out.

 

[Sagittarius A-Star]

In my diagram 1.0, B is at rest with the platform at all times beginning with the flash at time = 0, as it was in David Morin’s example.
You have moved B (your C) to the right after the light flash.

No, I did not move B. I added a 3^rd observer.

 

[Sagittarius A-Star]

I agree that the reflection from the left mirror (C) arrives at B first, that’s why I noted ACB≠ADB

That is irrelevant, because B is not located in the middle between the locations of the reflection-events, with reference to the rest-frame of the ground.

 

[vanhees71]

The entire discussion is just besides the point. The problem with Einstein's original paper of 1905 is that at this time nobody was aware about the mathematical structure which is adequate to formulate spacetime models, and that's why Einstein has to use pretty subtle gedanken experiments to establish the synchronization convention. Thanks to his math professor, Minkowski, to day we know the adequate description: In special relativity spacetime is a affine pseudo-Euclidean manifold with signature (1,3) (west-coast convention), and the symmetry group is the Poincare group, i.e., the group of translations and ##\mathrm{SO}(1,3)^{\uparrow}## transformations of four-vectors (Lorentz transformations).

So, all there is, is that there are preferred pseudo-Cartesian bases (Lorentz bases), for which the line element is given by
$$\mathrm{d} s^2 = \eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
Such Lorentz coordinates define then inertial reference frames and the Poincare transformations, building a group, give the transformation law to change from one inertial frame to another.

Einstein's gedanken experiments are much simpler to formulate mathematically within this adequate mathematical frame work of the (tensor) algebra of Minkowski space, and the most adequate formuation of physics within such a spacetime are field theories, including Maxwell's classical electrodynamics, which is such a relativistic field theory.

 

[Sagittarius A-Star]

You are asking me to define what Einstein did not.

If so, so what? It is always a good idea to explain SR in a better understandable way than Einstein did in 1905.

 

[PeterDonis]

reciprocal kinematics as the principle of relativity claims

Where are you getting this "reciprocal kinematics" as a requirement of the principle of relativity? What reference is this from? Is it from Einstein's paper? If so, please give a specific quote.

 

[Dale]

The laws are not shown to differ, the observations are shown to differ,

There is no change in the laws between diagrams, as far as I know.

Then it does not violate the principle of relativity.

The principle of relativity claims only that the laws of physics are the same in different frames, not that what you are describing as observations are different. Btw, I would not describe those as observations. I would reserve the term "observation" for the actual measured result on some specified experimental measurement device. That is always necessarily an invariant quantity.

they are not reciprocal kinematics as the principle of relativity claims.

This is a big misrepresentation of what relativity claims. You need to rescind this erroneous statement; misinformation is not tolerated here on PF.

 

[Sagittarius A-Star]

As you cannot measure the one way time of flight directly, you must deduce it from kinematic law as Einstein did in the equations above #29.

That's irrelevant. The definition of a standard inertial coordinate-system contains a time-coordinate, that is based on the definition of time as the reading of a standardized clock and on Einstein's definition, that the one-way speed is isotropic.

The relativity of simultaneity between such coordinate systems is a physical effect. It would not exist, if time were absolute.

 

[AlMetis]

This is a misrepresentation of what relativity claims. You need to rescind this claim, misinformation is not tolerated here on PF.

I will rescind this claim.
Can you please explain why?

As I understand it, reciprocal kinematics follow from the principle of relativity of motion between inertial frames. If observations of kinematics between inertial frames are not reciprocal, those that fail reciprocity are distinguishing one frame from another, thus (uniform) motion is no longer a relative measure as the principle claims.

What part of this is incorrect?

 

[AlMetis]

That is irrelevant, because B is not located in the middle between the locations of the reflection-events, with reference to the rest-frame of the ground.

B is at the mid point between the mirrors at the time of the flash, which is the time that the distance to the mirrors is relevant.

 

[Sagittarius A-Star]

B is at the mid point between the mirrors at the time of the flash, which is the time that the distance to the mirrors is relevant.

Why?

 

[AlMetis]

Why?

Because we are measuring from the flash to the mirrors, and the flash is at the midpoint at time =0

 

[Sagittarius A-Star]

Because we are measuring from the flash to the mirrors, and the flash is at the midpoint at time =0

What is the related role of observer B?

 

[PeterDonis]

As I understand it, reciprocal kinematics follow from the principle of relativity of motion between inertial frames.

As I asked before, where are you getting this from? Please give a reference.

 

[Dale]

Can you please explain why?

The only thing that the principle of relativity states is that the laws of physics are the same in any reference frame.

reciprocal kinematics follow from the principle of relativity of motion between inertial frames. If observations of kinematics between inertial frames are not reciprocal, those that fail reciprocity are distinguishing one frame from another, thus (uniform) motion is no longer a relative measure as the principle claims.

What you call reciprocal kinematics does not follow from the principle of relativity. I have already explained why, but I will do so again in more depth.

The laws of physics are expressed as differential equations, or (even better) as a Lagrangian or Hamiltonian. When you solve a differential equation, by itself, you do not get a single solution. Instead, you get a family of solutions. In order to get a single solution you must also supply what is called boundary conditions or initial conditions. Additionally, if the differential equation itself has some parameters, you will need to specify those parameters too.

For example, for a projectile the law of physics is:
##m \ddot z = -mg## or (even better) ##\mathcal L = \frac{1}{2}m \dot z^2 - mgz##.

If you solve this (in either form) you get the equation of motion:
##z(t)=\frac{1}{2} g t^2 + v_0 t + z_0## where ##v_0## and ##z_0## are the initial conditions.

Those initial conditions do not come from the law of physics, and any initial conditions satisfy the equation of motion and therefore the law of physics.

What you are calling the "kinematics" is the equation of motion, not the law of physics. So it depends not only on the law of physics but also the boundary conditions. The principle of relativity requires that the law of physics be the same in both frames, but does not require that the boundary conditions be the same. In you example, the boundary conditions are different in the different frames, so the different kinematics is completely compatible with the principle of relativity.

Only a difference in the laws would be a violation of the principle, which as you stated earlier you were deliberately not doing. Since you used the same laws of physics in both frames in your scenario there is simply no logical way that you could possibly obtain a violation of the principle of relativity.

 

[Ibix]

Diagram 1.0 shows that does not happen.

The two-way convention calculates the one-way as a mean of each leg which sets the one-way time of flight the same for both observers. But as is shown in diagram 1.0 the one way time is not the same for both observers.

Once again, you are failing to understand the difference between asymmetries in experimenal results that are due to asymmetries you chose to put in your experimental setup, and the symmetry of the physical laws. (Which is the same point @Dale makes about boundary conditions in the post above this one).

 

[AlMetis]

What you call reciprocal kinematics does not follow from the principle of relativity. I have already explained why, but I will do so again in more depth. ...

Thank you, that was an excellent explanation.
I agree with everything except your claim that the boundary conditions are different in my diagrams.
Time=0 shows identical boundary conditions except the arrow indicating the relative motion of A and B.

Please point out what boundary condition/s differ.

 

[Dale]

except the arrow indicating the relative motion of A and B.

That is a different boundary condition

Please point out what boundary condition/s differ.

The initial velocities of A, B, C, and D differ between the frames. Also, although it is not relevant for this problem, the initial times on the clocks would be different as would the length of the train. All of those are boundary conditions as described above

 

[Nugatory]

The entire discussion is just besides the point. The problem with Einstein's original paper of 1905 is that at this time nobody was aware about the mathematical structure which is adequate to formulate spacetime models, and that's why Einstein has to use pretty subtle gedanken experiments to establish the synchronization convention.

And it is for this reason that reading Einstein's papers is an exercise in the history of science, not the science itself. History of science is a fascinating discipline in its own right (and arguably more valuable to a well-rounded layperson than the science itself) but it is a different discipline with different goals and methods.

 

[AlMetis]

That is a different boundary condition

Relative motion is a single condition, how can it differ between itself?

 

[AlMetis]

The initial velocities of A, B, C, and D differ between the frames.

There is only one velocity, the relative velocity of A and B.
A, C and D are the same frame, how can they differ from themselves.

 

[Ibix]

There is only one velocity, the relative velocity of A and B.

No, there are the velocities of the various things relative to your frames.

A, C and D are the same frame, how can they differ from themselves.

If I am in a car I might consider myself at rest. A pedestrian watching me drive by says I'm doing 30mph. Do you understand that? If so, what's the problem with your observers having different velocities with respect to different frames?

 

[Sagittarius A-Star]

There is only one velocity, the relative velocity of A and B.
A, C and D are the same frame, how can they differ from themselves.

C and D are at rest in the train's rest-frame and move in the ground's rest-frame. This creates different boundary conditions for the (frame-dependent) travel distances of the light and does not violate the 1st postulate.

 

[PeterDonis]

There is only one velocity, the relative velocity of A and B.

No. Suppose that, relative to A, B is moving to the right. Then, relative to B, A is moving to the left. These are two different relative velocities. They have the same magnitude, but opposite directions.

I think you have not really thought through what actually changes when you change frames.

 

[AlMetis]

No, there are the velocities of the various things relative to your frames.

There are 2 frames A and B, one relative velocity.