A question about relativity of simultaneity

[Peter Strohmayer]

There is no essential difference to Einstein's arrangement: With Einstein, the two emission events at the ends of the train are defined from the point of view of the railroad embankment observer (S).

For the train observer (S'), who passes by S at the time when the emission events take place from the point of view of S (origin coverage at 0/0; 0/0), the emission events can only be determined with the help of the L-T.

Why should I assume with Einstein's arrangement a second pair of light pulses which are simultaneous for S'? That would be other events of the emission than for S.

It is always only about sending light pulses from events to events which are identical for both observers. From the ratio of the lengths of the signal paths (from the view of the two observers) the L-T can be derived.

 

[Ibix]

the emission events can only be determined with the help of the L-T.

This is not correct. You do not need to know the Lorentz transforms to be able to derive them from Einstein's thought experiment, as already laid out in posts 75 and 76.

 

[Peter Strohmayer]

I never denied that Einstein accurately derived the L-T from an introductory "Gedankenexperiment". I only pointed out that it is an abstract-mathematical derivation (cf. #75), where his thought experiment is not very helpful (#99), because it reminds too much of tennis balls (of Newtonian mechanics) (#89).

 

[vanhees71]

Well, if you are very rigorous you have to think in terms of the wave front of electromagnetic waves, which always moves with the speed of light in vacuum.

 

[Peter Strohmayer]

This is the reason why wave fronts (also of spherical waves), which were emitted by observers moving towards each other, propagate together.

 

[vanhees71]

What do you mean bye "propagate together"?

 

[Peter Strohmayer]

A photon cannot overtake another one. If two photons are emitted in the same direction at origin coverage, they propagate together. Even if the respective observer emits his photon in quite certain angle to the axis of motion, they propagate together (compare e.g. the signal paths in the moving light clock).

 

[Mister T]

But that is the problem I'm asking about. The relative speed is greater than the speed of light.

As others have noted, it's the closing speed, not the relative speed, that's greater than $c$.

The person in the center of the train car will still observe the light beam moving at speed $c$. The person on the embankment will also see the speed of the light pulse as moving at speed $c$. But he will see the person in the center of the train car moving at speed $v$. He will observe them moving towards each other at speed $c+v$ but he will not observe either of them moving at a speed greater than $c$.

The point of this thought experiment starts with the sentence highlighted above. Prior to Einstein it was thought that the embankment observer would see the light pulse moving at a speed other than $c$ so that the two light pulses would have been generated simultaneously according to both observers. The point of the thought experiment is to show that if the speed of light is independent of the speed of its source, then simultaneity is relative rather than absolute.

 

[PeterDonis]

With Einstein, the two emission events at the ends of the train are defined from the point of view of the railroad embankment observer (S).

No, they are defined to happen at the same events at which the ends of the train pass the corresponding points on the embankment (the points where the lightning strikes happen and emit the two light pulses). Those events are invariant.

For the train observer (S'), who passes by S at the time when the emission events take place from the point of view of S (origin coverage at 0/0; 0/0), the emission events can only be determined with the help of the L-T.

Wrong. They are defined as above. Since those events take place at the ends of the train, their spatial locations in the train frame are determined by that condition. The train observer can then determine the times of those events in his frame by observing the time on his clock when he receives the light pulses. No LT is required.

Why should I assume with Einstein's arrangement a second pair of light pulses which are simultaneous for S'?

I didn't say you should, I only said you could.

That would be other events of the emission than for S.

Of course, I have already said that.

It is always only about sending light pulses from events to events which are identical for both observers.

This is one way of looking at the thought experiments, yes.

From the ratio of the lengths of the signal paths (from the view of the two observers) the L-T can be derived.

Yes, you can derive the LT from the information provided in the thought experiments, since that information (which includes not only the lengths of the signal paths but the times involved, which can be computed from the fact that the speed of light is the same in all inertial frames) is sufficient to determine the coordinates in both frames of all events.

But you are claiming that you need to assume the LT in order to determine the coordinates in both frames of all events. That is false.

 

[PeterDonis]

I never denied that Einstein accurately derived the L-T from an introductory "Gedankenexperiment".

You have repeatedly claimed that you need to assume the LT in order to determine the coordinates of some of the events in the thought experiment:

I have assumed the L-T for the conversion of the coordinates of only one light pulse (#93). The L-T must be derived mathematically before.

If only S' emits a photon (event 1) which subsequently arrives at the end of the train (event 2), then from the point of view of observer S, the coordinates of events 1 and 2 can only be determined via the L-T.

For the train observer (S'), who passes by S at the time when the emission events take place from the point of view of S (origin coverage at 0/0; 0/0), the emission events can only be determined with the help of the L-T.

If these claims were true, they would be inconsistent with the claim that Einstein correctly derived the LT from his thought experiment. You can't derive something by assuming it. But that's what you are claiming. You can't have it both ways. Either the LT can be validly derived from Einstein's thought experiment, or it has to be assumed. The former is in fact correct.

 

[Ibix]

I never denied that Einstein accurately derived the L-T from an introductory "Gedankenexperiment". I only pointed out that it is an abstract-mathematical derivation (cf. #75), where his thought experiment is not very helpful (#99), because it reminds too much of tennis balls (of Newtonian mechanics) (#89).

This is not true. You've said things like "the emission events can only be determined with the help of the L-T" (in post #106), which is a lot stronger than the claim you now say you were making, and is also incorrect. It is perfectly possible to derive the Lorentz transforms from Einstein's version of the thought experiment. I don't personally find his exposition particularly clear, so here is my version.

A train travelling at speed $v$ passes a stationary embankment of length $2l$. As the nose and tail of the train pass the end and start of the embankment, respectively, they are struck by lightning. An observer standing at the center of the embankment sees the flashes arrive at his location simultaneously and concludes that the strikes were simultaneous. Can we deduce the Lorentz transforms?

Let there also be an observer on the train. She measures the train's length to be $2l'$. Both observers zero their clocks when they pass one another, and both measure distance from their own locations. Since both frames are global inertial frames the transform (whatever it is) between the $x,t$ coordinates of the embankment observer and the $x't'$ coordinates of the train observer must take the form$$\left(\begin{array}{cc}A&B\\C&D\end{array}\right)$$
The embankment observer assigns coordinates $x=l, t=0$ to the front strike. The train observer assigns coordinates $x'=l', t'=T_+'$. Applying our matrix yields:$$\left(\begin{array}{c}T_+'\\l'\end{array}\right)=\left(\begin{array}{cc}A&B\\C&D\end{array}\right)\left(\begin{array}{c}0\\l\end{array}\right)$$from which we deduce that $B=T_+'/l$ and $D=l'/l$. The rear strike is at $x=-l,t=0$ for the embankment observer and $x'=-l', t'=T_-'$ for the train observer, and the equivalent calculation merely adds that $T_-'=-T_+'$.

Now we can turn our attention to the events where the train observer receives light from the flashes. A straightforward intercept calculation tells us that the embankment observer assigns coordinates $x=lv/(c+v), t=l/(c+v)$ to the arrival of the pulse from the front while the train observer assigns $x'=0, t'=l'/c+T_+'$. Again using the matrix on the embankment observer's coordinates and equation this with the train observer's coordinates gives us that $A=\frac{(c+v)l'+c^2T_+'}{cl}$ and $C=-l'v/l$. For the arrival of the flash from the rear, another intercept calculation tells us that the embankment observer assigns coordinates $x=lv/(c-v), t=l/(c-v)$ and the train observer assigns $x'=0,t'=l'/c-T_+'$ (where we have used our result for $T_-'$). Again we apply our matrix to the embankment coordinates, recovering that $T_+'=-l'v/c^2$.

Putting all this together, we have found that the Lorentz transforms expressed as a matrix are$$\left(\begin{array}{cc}l'/l&-l'v/c^2l\\-l'v/l&l'/l\end{array}\right)$$We only need to work out what is the relationship between $l$ and $l'$. We can do this by noting that the principle of relativity says that the inverse transform must be the same except for the sign of $v$. Inverting our matrix gives us$$\gamma^2\left(\begin{array}{cc}l/l'&lv/c^2l'\\lv/l'&l/l'\end{array}\right)$$where $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}$ is the Lorentz gamma factor, and the condition that this is the same except for the sign of $v$ tells us that $l'=\gamma l$ and hence that the Lorentz transform as a matrix is$$\left(\begin{array}{cc}\gamma&-\gamma v\\-\gamma v&\gamma\end{array}\right)$$And multiplying this by the generic coordinates $(x,t)^T$ gives us the usual form.

That is a nothing more than a process of writing down coordinates and deducing a relationship. If that's abstract maths, all of physics is abstract maths.

 

[PeterDonis]

This thread has run its course and is now closed. Thanks to all who participated.