A question about relativity of simultaneity

[DmitryS]

I'm not sure what's to prove.

To prove your own words, "Those are intermediate variables. He ends up with the LT further on in the paper". To prove that Einstein's model of 1905 leads to LT, and that there's an actual derivation of the LT in 1905.

Instead, you provide a link to another source with the LT derivation - I can derive the LT myself on different grounds. That's not the point I'm making.

There is an issue if you take Einstein's 1905 paper as some sort of Gospel. It was the first paper on SR, but the theory has been developed and refined for over 100 years since then. I love the 1905 paper, but it's not an ideal source from which to learn SR as a student. Your confusions bear this out to some extent.

I just said that in this paper there is no derivation of LT. You said I was wrong. Now you say I was right? I take it for a scientific paper, not for a Gospel.

I don't want to get into a game of having to justify everything Einstein wrote in 1905. We all see flaws in that paper, when it comes to it.

Good.

But, an incompatibility between the LT and RoS? That's just nonsensical.
PS both @vanhees and I have proved the compatibility of the LT and RoS in various posts in this thread.

You're missing my point again. I never said LT and RoS are incompatible. What's more, it's easy to see how RoS works through the LT. I only said that the kind of model of RoS presentedd through the light rays experiments brings us only to the kind of transforms obtained by Einstein in his 1905 paper. And those are incompatible with the LT.

Once again. I take no issue with LT, RoS, or their incompatibility. I take issue with the way RoS is introduced through those experiments.

Are we going to go through all this again? I might retort: "You aren't trolling me, if you don't mind me asking?" But I won't. I will just tell you that it seems to me that instead of the discussion of the relevant points I'm making I sometimes have to struggle with the paternal attitude as if to a would-be renegade.

Maybe points I'm making are not relevant. If so, you can demonstrate that, I hope? So far, I haven't seen in this thread a proof of the light-ray experiments leading to LT which did not contain the assumption of LT in the first place. Einstein so far was the only one who tried to do it - and seems to have failed.

 

[PeroK]

Maybe points I'm making are not relevant. If so, you can demonstrate that, I hope? So far, I haven't seen in this thread a proof of the light-ray experiments leading to LT which did not contain the assumption of LT in the first place. Einstein so far was the only one who tried to do it - and seems to have failed.

I have to say, in all honesty, I doubt I am able to help you understand SR.

 

[DmitryS]

I have to say, in all honesty, I doubt I am able to help you understand SR.

But I do understand it! I'm not after that!

 

[PeroK]

But I do understand it! I'm not after that!

Anyone who posts something like this:

So far, I haven't seen in this thread a proof of the light-ray experiments leading to LT which did not contain the assumption of LT in the first place. Einstein so far was the only one who tried to do it - and seems to have failed.

Is just wasting everyone's time. Every competent student of SR in the last 100 years has derived the LT themselves from only the invariance of the speed of light.

 

[TSny]

To me, that looks like he assumed the LT to be true, whatever his own reasoning might have led to. If you can derive the LT from the relations above, show me how - and I will confess I was wrong.

You appear to be asking how Einstein gets from these equations:

to these equations:

Take, for example, the equation for $\xi$. Recall that Einstein uses the symbol $x'$ to denote $x-vt$. (See near the top of page 6 in your link.) Thus,

$$\xi = a \frac{c^2}{c^2-v^2}x' = a \frac{c^2}{c^2-v^2}\left( x-ct \right) = \frac{a}{1-v^2/c^2}\left( x-ct \right) = \frac{a}{\sqrt{1-v^2/c^2}}\frac{1}{\sqrt{1-v^2/c^2}}\left( x-vt \right)$$

The symbol $a$ represents a yet-to-be-determined function of $v$. Einstein originally used the notation $\phi(v)$ to denote this unknown function. (See near the bottom of page 6 in your link.) However, later when Einstein writes his results at the bottom of page 7, he redefines the notation $\phi(v)$ to denote $\frac{a}{\sqrt{1-v^2/c^2}}$. This is certainly OK, but it could be a bit confusing. $\phi(v)$ is still a yet-to-be-determined function of $v$.

Einstein uses $\beta$ to denote $\frac{1}{\sqrt{1-v^2/c^2}}$. So, using these definitions of $\phi(v)$ and $\beta$, we arrive at Einstein's result $$\xi = \phi(v) \beta \left( x-ct \right).$$The results for $\tau, \eta$, and $\zeta$ can be obtained in a similar manner.

Einstein later argues that the unknown function $\phi(v)$ is actually independent of $v$ and equals 1. At this point, he has finally finished his derivation of the LT equations given at the bottom of page 9.

 

[Ibix]

That's exactly the point that in my opinion needs a proof on your part. What I see from the text - and I provided the link to the online PDF - is that Einstein simply writes down the LT after these transforms.

It's just algebra to derive the Lorentz transforms from the setup Einstein gives (as @TSny has just laid out). I find Einstein's derivation a bit messy because he has these intermediate values with Greek letters, but it's easy enough to do something cleaner.

Assert that everything happens in the $z=0$ plane. Consider two light pulses from the origin, one parallel to $x$ and one parallel to $y$, let them reflect off two mirrors equidistant from the origin and return simultaneously to the origin. Write down the coordinates, $(x,y,t)$ and $(x',y',t')$, of the two reflection events and the return event in two frames, one where the experimental gear is stationary and one where it is moving in the $+x$ direction at constant speed. Equate the primed coordinates of each event to the unprimed coordinates after applying the general linear transform. The general linear transform (a 3$\times$3 matrix) has nine parameters and you have nine constraints. Solve for the parameters. You then have a bit more work to do because you can't assume that the distances to the mirrors in the primed frame are the same as they are in the unprimed one (and, indeed, one turns out not to be the same), but noting that if the forward transform is $\Lambda(v)$ the inverse transform must be $\Lambda(-v)$ and hence $\Lambda(v)\Lambda(-v)=\mathbf{I}$ gets you out of that. Finally you observe from rotational symmetry that the $z$ transform must be the same as the $y$ transform.

It's worth grinding through it once.

 

[DmitryS]

You appear to be asking how Einstein gets from these equations:
View attachment 315075

to these equations:

View attachment 315076

Take, for example, the equation for $\xi$. Recall that Einstein uses the symbol $x'$ to denote $x-vt$. (See near the top of page 6 in your link.) Thus,

$$\xi = a \frac{c^2}{c^2-v^2}x' = a \frac{c^2}{c^2-v^2}\left( x-ct \right) = \frac{a}{1-v^2/c^2}\left( x-ct \right) = \frac{a}{\sqrt{1-v^2/c^2}}\frac{1}{\sqrt{1-v^2/c^2}}\left( x-vt \right)$$

The symbol $a$ represents a yet-to-be-determined function of $v$. Einstein originally used the notation $\phi(v)$ to denote this unknown function. (See near the bottom of page 6 in your link.) However, later when Einstein writes his results at the bottom of page 7, he redefines the notation $\phi(v)$ to denote $\frac{a}{\sqrt{1-v^2/c^2}}$. This is certainly OK, but it could be a bit confusing. $\phi(v)$ is still a yet-to-be-determined function of $v$.

Einstein uses $\beta$ to denote $\frac{1}{\sqrt{1-v^2/c^2}}$. So, using these definitions of $\phi(v)$ and $\beta$, we arrive at Einstein's result $$\xi = \phi(v) \beta \left( x-ct \right).$$The results for $\tau, \eta$, and $\zeta$ can be obtained in a similar manner.

Einstein later argues that the unknown function $\phi(v)$ is actually independent of $v$ and equals 1. At this point, he has finally finished his derivation of the LT equations given at the bottom of page 9.

Thank you very much for talking to the point, but there's a problem.
I cannot see where $\phi(v)$ denotes $\frac{a}{\sqrt{1-v^2/c^2}}$ Maybe you refer to a different edition, or a later issue of the article? As far as I can see, in the article at my link it is universally $\phi(v) = a$
Certainly, $\phi(v)=\frac{a}{\sqrt{1-v^2/c^2}}$ would solve the problem. But how would one justify such definition of $\phi(v)$? I don't see it can be substantiated - maybe you can help me here.

 

[DmitryS]

It's just algebra to derive the Lorentz transforms from the setup Einstein gives (as @TSny has just laid out). I find Einstein's derivation a bit messy because he has these intermediate values with Greek letters, but it's easy enough to do something cleaner.

Assert that everything happens in the $z=0$ plane. Consider two light pulses from the origin, one parallel to $x$ and one parallel to $y$, let them reflect off two mirrors equidistant from the origin and return simultaneously to the origin. Write down the coordinates, $(x,y,t)$ and $(x',y',t')$, of the two reflection events and the return event in two frames, one where the experimental gear is stationary and one where it is moving in the $+x$ direction at constant speed. Equate the primed coordinates of each event to the unprimed coordinates after applying the general linear transform. The general linear transform (a 3$\times$3 matrix) has nine parameters and you have nine constraints. Solve for the parameters. You then have a bit more work to do because you can't assume that the distances to the mirrors in the primed frame are the same as they are in the unprimed one (and, indeed, one turns out not to be the same), but noting that if the forward transform is $\Lambda(v)$ the inverse transform must be $\Lambda(-v)$ and hence $\Lambda(v)\Lambda(-v)=\mathbf{I}$ gets you out of that. Finally you observe from rotational symmetry that the $z$ transform must be the same as the $y$ transform.

It's worth grinding through it once.

That's something, thanks.

 

[DmitryS]

Anyone who posts something like this:Is just wasting everyone's time. Every competent student of SR in the last 100 years has derived the LT themselves from only the invariance of the speed of light.

You really don't see the difference between "deriving the LT from the invariance of the speed of light" and "deriving the LT from Einstein's setup"?

And please, don't bring up that wasting somebody's time thing again. I hope participation in this forum is not mandatory. I wouldn't take offence at all if nobody wrote a reply to my post. That's life. I feel grateful to you for all your effort and to anyone else who participated. Sometimes it seemed to me that you either ignored my questions or took them for something which they weren't - well, maybe it was partly my fault. Maybe I needed to word them in a different way. That's life again. Thank you for your time and effort.

 

[DmitryS]

It's worth grinding through it once.

I never thought of that algebra from this perspective, that you could actually set it as an experiment with light.

 

[Ibix]

I never thought of that algebra from this perspective, that you could actually set it as an experiment with light.

That's why he says "[f]rom the origin of system k let a ray be emitted at the time $\tau_0$ along the
X-axis".

 

[DrGreg]

Thank you very much for talking to the point, but there's a problem.
I cannot see where $\phi(v)$ denotes $\frac{a}{\sqrt{1-v^2/c^2}}$ Maybe you refer to a different edition, or a later issue of the article? As far as I can see, in the article at my link it is universally $\phi(v) = a$
Certainly, $\phi(v)=\frac{a}{\sqrt{1-v^2/c^2}}$ would solve the problem. But how would one justify such definition of $\phi(v)$? I don't see it can be substantiated - maybe you can help me here.

As commented in post #75, $a$ is initially described as "$\text{a function }\phi(v)$", but he later redefines $\phi$ implicitly by the four equations for $\tau$, $\xi$, $\eta$ and $\zeta$ near the bottom of page 7 which you must compare with the previous equations for those four quantities. E.g. compare $$\zeta=a \frac{c}{\sqrt{c^2-v^2}}z$$ with $$\zeta=\phi(v)z.$$It would have been better if he'd used two different symbols instead of using $\phi$ with two different meanings without explicitly saying so.

 

[Sagittarius A-Star]

I cannot see where $\phi(v)$ denotes $\frac{a}{\sqrt{1-v^2/c^2}}$ Maybe you refer to a different edition, or a later issue of the article? As far as I can see, in the article at my link it is universally $\phi(v) = a$

There exists a German site about the Einstein 1905 paper with comments and explanations. If you are using Chrome and right-click on one of the text parts with comments and explanations, then you can get it translated to English.

2. Einstein introduces a new function that is not identical to the function $\varphi( v ) = a$ mentioned on page 899 . Rather is:

$$ \varphi (v)=\frac {a}{\sqrt {1-(\frac {v}{V})^{2}}}$$

Source:
https://de.wikibooks.org/wiki/A._Ei...namik_bewegter_Körper:_Kinematischer_Teil:_§3

 

[TSny]

I cannot see where $\phi(v)$ denotes $\frac{a}{\sqrt{1-v^2/c^2}}$ Maybe you refer to a different edition, or a later issue of the article? As far as I can see, in the article at my link it is universally $\phi(v) = a$
Certainly, $\phi(v)=\frac{a}{\sqrt{1-v^2/c^2}}$ would solve the problem. But how would one justify such definition of $\phi(v)$? I don't see it can be substantiated - maybe you can help me here.

There's nothing much going on here. Suppose Einstein had used the notation $\bar{\phi}(v)$ instead of $\phi(v)$ at the bottom of page 6. That is, suppose he had written:

Then, we are at liberty to define a new function of $v$ which we may denote $\phi(v)$ as follows:

$\large \phi(v) = \frac{a}{\sqrt{1-v^2/c^2}} = \frac{\bar{\phi}(v)}{\sqrt{1-v^2/c^2}}$.

This is the $\phi(v)$ that Einstein uses at the bottom of page 7. Since $\bar{\phi}(v)$ denoted "a function at present unknown", $\phi(v)$ can also be taken to be "a function at present unknown". Einstein proceeds to show that $\phi(v) = 1$. That means that $a$ as a function of $v$ turns out to be

$a = \bar{\phi}(v) = \sqrt{1-v^2/c^2}$.

 

[PeterDonis]

I wrote the opposite

I know what you wrote; I am just saying that I think what you wrote misdescribes the correct reasoning. But this is really tangential to the main topic of the thread.

 

[PeterDonis]

I don't agree that he assumes what you say, or rather, that he only assumes what you say, but let that be for a moment.

No, let's not let that be. If you disagree, then what do you think Einstein's assumptions are? If we can't agree on that initial point, how can you expect to have a useful discussion of the rest of the paper?

as I was saying, the starting point of his derivation is the same setup as that of the simultaneity test

Sort of. But even if it is, that doesn't mean Einstein is assuming anything about simultaneity in order to derive the LT. He just happens to be using a similar scenario.

that setup leads to some transforms, which are not LT

I realize you've been exchanging posts with others who have tried to correct you here, but just to add my input to that discussion, at the end of Section 3, he obtains final transformation equations which most certainly are the LT. I did not spell out every single step of his reasoning because (a) you were asking mainly about his assumptions, and (b) I assumed you would be able to apply your intelligence to see the rest without my having to walk you through every step.

I just said that in this paper there is no derivation of LT. You said I was wrong. Now you say I was right?

No. He is just saying that the derivation of the LT in Einstein's 1905 paper is not the most efficient derivation that we now know of, after more than a century of having other people look at the issue and come up with more efficient derivations. But that in no way means that Einstein's derivation is wrong. It's not.

 

[PeterDonis]

I think I made my point abundantly clear from the start.

I don't. This thread started out with you asking about a common scenario that is used to illustrate relativity of simultaneity; then it morphed into you claiming that Einstein's 1905 paper somehow does not derive the Lorentz transformations. So I think you are mistaken if you believe your point is clear to others who are in this discussion. I think you need to clarify what, exactly, you want to discuss.

 

[vanhees71]

There is an issue if you take Einstein's 1905 paper as some sort of Gospel. It was the first paper on SR, but the theory has been developed and refined for over 100 years since then. I love the 1905 paper, but it's not an ideal source from which to learn SR as a student. Your confusions bear this out to some extent.

I'd say, the kinematic part is a masterpiece. It derives the Lorentz transformation by very plausible physical arguments from the two empirical foundations of RT, i.e., the special principle of relativity and the independence of the phase velocity of em. waves of the motion of their source. Also the electromagnetical part (LT of fields, Maxwell's equations, aberration and Doppler shift for em. waves in vacuo) are ok but hard to read since we are used to the vector notation rather than writing out all equations for components. The only thing I'd say hasn't been fully understood in this paper is point-particle mechanics, which was correctly formulated for the first time (afaik) by Planck in 1906. The unfortunate result is that the idea of "relativistic mass" has been established in the 1905 paper and is still not eliminated from the textbook literature (particularly in high-school textbooks).

I don't want to get into a game of having to justify everything Einstein wrote in 1905. We all see flaws in that paper, when it comes to it.

But, an incompatibility between the LT and RoS? That's just nonsensical. The LT encapsulates the RoS along with time dilation and length contraction. That I can prove, but not by dissecting the 1905 paper.

PS both @vanhees and I have proved the compatibility of the LT and RoS in various posts in this thread.

I think, now we have all three types of train gedanken experiments in this thread:

(a) Signals sent from both ends of the train simultaneously wrt. the embankment rest frame (Einstein's original one in "The meaning of relativity").
(b) Signals sent from both ends of the train simultaneously wrt. the train rest frame.
(c) One signal sent from the center of the train.

I think that (c) is the best one. The most straight-forward way to analyze either of them is the use of the LT, because the LT is the result of the Einstein synchronization convention or simply the realization of the symmetry transformations of Minkowski spacetime (proper orthochronous Poincare transformations) in the most convenient coordinates (Lorentzian ones).

For illustration you can also draw the Minkowski diagrams for all three cases. It's a very good exercise in learning, how to read those diagrams.

 

[Peter Strohmayer]

Einstein's thought experiment is confusing because S is supposed to "observe" two light pulses at the platform, which S' has emitted in the middle of the train. The similarity with tennis balls is too great.
It is better to imagine that S and S' each emit a light pulse in different directions when they meet, so that the two photons at the tips of the light pulses propagate together (no photon can overtake another).
The signal path of the light pulse of S (from the point of view of S) can never be of the same length as the signal path of the light pulse of S' (from the point of view of S'), because although the photons at the tips of the light pulses are always on the same level, the observers S and S' have moved away from each other. For each observer, the length of his signal path is also the length of his signal time (light-like distance).
Thus, if the signals arrive at the ends of the train at the same time from the point of view of S' when the signal paths are of equal length (simultaneous events), then they cannot possibly arrive at the same time from the point of view of S when the signal paths are of unequal length (non-contemporaneous events).
This is sufficient to derive the L-T.

Peter Strohmayer

 

[PeterDonis]

the two photons at the tips of the light pulses propagate together

What do you mean by this? The photons are propagating in different directions so they are never "together" after the point of emission.

the photons at the tips of the light pulses are always on the same level

What do you mean by "on the same level"?

 

[Peter Strohmayer]

At their encounter, both S and S' emit a light pulse to the left along the x-axis. These photons propagate together. The same happens to the right.

 

[PeterDonis]

At their encounter, both S and S' emit a light pulse to the left along the x-axis. These photons propagate together. The same happens to the right.

There is no need to have two photons going in each direction. The meeting between S and S' is a single event, i.e., a single point in spacetime. Emitting a single photon in a given direction from a single event is sufficient. This follows from the postulate of the constancy of the speed of light in any inertial frame, which is being assumed for the problem. Anything you say about a given pair of photons propagating together in your version, can also be said about a single photon in the standard version.

 

[Peter Strohmayer]

That is correct. The difference is in didactics. That seems to have been the problem in this thread.
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.
If, on the other hand, each of the observers emits a photon at origin coverage, which propagate together, then it becomes clear why the coordinates of an event also differ in time. Then one can derive the L-T instead of presupposing it.

 

[Ibix]

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.

Huh? It's just an intercept caculation in either frame (albeit a trivial one in one frame). You need to use the invariance of the speed of light, sure, but that's how you derive the Lorentz transforms.

 

[Peter Strohmayer]

I do not dispute the possibility of a purely mathematical derivation of the L-T in the standard situation. However, I think that the intuitive access, which Einstein's thought experiment shall open, is better served with two complementary photons. The connection with the finiteness of the causal propagation becomes clear. The identity of signal path and (signal)time between two events becomes conscious.

 

[Ibix]

I do not dispute the possibility of a purely mathematical derivation of the L-T in the standard situation. However, I think that the intuitive access, which Einstein's thought experiment shall open, is better served with two complementary photons. The connection with the finiteness of the causal propagation becomes clear. The identity of signal path and (signal)time between two events becomes conscious.

Seems like an unnecessary complication to me, but if it works for you that's fine. It's a rather different claim from "[t]hen one can derive the L-T instead of presupposing it", which you made earlier.

 

[Peter Strohmayer]

Yes, that was a bit inaccurate, I should have written "instead of presupposing it mathematically."

 

[Ibix]

Yes, that was a bit inaccurate, I should have written "instead of presupposing it mathematically."

That's still wrong. You aren't pre-supposing the Lorentz transforms whether you use light pulses emitted only by the train observer or pulses emitted by both train and embankment observers. All the latter does is reinforce the consequence of the invariance of the speed of light by making the case that the parallel pulses should travel side-by-side.

 

[Peter Strohmayer]

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. I doubt like DmitryS that Einstein's thought experiment (the "observation" of a light pulse) is helpful for it. But you are welcome to see it differently.

 

[PeterDonis]

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.

This is not correct. Whether the photon is emitted by S' or S is irrelevant, since, as I have already pointed out, the emission event is a single event (a single point in spacetime), namely, the event at which the worldlines of S and S' cross. So each photon is traveling at $c$ in a known direction from a known starting point, and the ends of the train are traveling at a known speed $v$ in a known direction from known starting points. This is true in both frames, so the coordinates of the reception events can be calculated in both frames independently, without having to make use of any transformation whatsoever.

 

[Peter Strohmayer]

The positions of the ends of the train (the starting points) are by no means known to both observers. Only if a starting point of one end of the train is fixed for one of the observers, the starting point for the other observer results from the L-T. The L-T is the precondition for this.
In other words: the event 2 of the arrival of the light pulse must be defined for one of the observers, only then it can be said for the second observer by means of L-T when and where it occurs for him.

 

[PeterDonis]

The positions of the ends of the train (the starting points) are by no means known to both observers.

They don't need to be. Each observer only needs to know that, in his frame, the two reception points are in opposite directions, at equal distances, from the emission point. The emission event alone is sufficient to link things together in both frames because, by stipulation, that event occurs when the two worldlines of the observers (one on the embankment and one at the center of the train) cross.

 

[PeterDonis]

Einstein's thought experiment is confusing because S is supposed to "observe" two light pulses at the platform, which S' has emitted in the middle of the train.

Btw, this is not Einstein's original thought experiment. In Einstein's original thought experiment, two light pulses are emitted from opposite ends of the embankment (when lightning strikes occur at those places), and are received at the same event by the embankment observer who is halfway between the emission locations. Einstein takes this as a definition of what "simultaneous" means for the embankment observer. He then shows that the same two events (the emission events of the two light pulses) cannot meet the same definition of simultaneity for a train observer who is spatially co-located with the embankment observer at the same instant in the embankment frame that the light pulses are emitted.

You are turning all this around and having light pulses be emitted from the center towards the ends, instead of from the ends towards the center. This still illustrates relativity of simultaneity, but in a different way.

 

[Peter Strohmayer]

PeterDonis #102:
If the receiving points are at the same distance for one observer, then under no circumstances they can be at the same distance for the other observer.
In other words: If for each of the observers receiving points are at the same distance, they are not talking about the same receiving events.
The emission event is fixed for both observers (0/0), but the reception events are fixed for only one of them. The other one must derive them in theory (in the thought experiment) from the L-T or measure them in practice.

 

[PeterDonis]

If the receiving points are at the same distance for one observer, then under no circumstances they can be at the same distance for the other observer.

Yes, they can, if the receiving points and their distances are different for each observer, and relative to each observer's frame.

If for each of the observers receiving points are at the same distance, they are not talking about the same receiving events.

Of course not. That is an obvious consequence of your version of the scenario: the train reception events are a different pair of events from the embankment reception events.

Note that in Einstein's original version, there is only one pair of events because the pair is a pair of emission events, which are simultaneous in the embankment frame by construction. But you could also add to his version a second pair of emission events that were simultaneous in the train frame; these events would of course have to be different from the embankment pair (and this second pair would not be simultaneous in the embankment frame).

The emission event is fixed for both observers (0/0), but the reception events are fixed for only one of them.

Nope. Each one can compute his own reception events simply from the knowledge of the emission event's location in their frame (that it is exactly halfway between the two reception points). Each pair of reception events will be simultaneous in the corresponding observer's frame, but will not be simultaneous in the other frame.