# Relativity on a moving train

I've been given the following scenario:

Observer $B$ is in the center of a train carriage which is moving at velocity $v$ with respect to an observer $A$. Two light signals are emitted from sources L at the left end and R at the right end of the carriage, such that they reach $B$ simultaneously. $B$ observes that the light signals were emitted at the same time. Show that $A$ doesn't agree.

I've no idea how to do this mathematically. I understand the concept; that as the carriage is moving to the right, $A$ will see that the light from the left will have to travel further than light traveling from the right, as $B$ is moving towards the light coming from the right.

What I've done so far (which I feel is wrong) is as follows:

Let the carriage be of length $2d$ from $A$'s point of view. Then the distance that light has travelled from the left is $d+v\Delta t_L=c\Delta t_L$. Similarly, the distance that light has travelled from the right to B is $d-v\Delta t_R=c\Delta t_R$. Can solve these to get $\Delta t_L=\dfrac{d}{c-v}$ and $\Delta t_R=\dfrac{d}{c+v}$.

If this is correct then I'm ok, as if $\Delta t_R \neq \Delta t_L$ then the light can't have been emmitted simultaneously according to $A$. However, I feel like having the denominator $c+v$ is wrong, as in some sense I'm adding velocity $v$ to the speed of light which is not allowed in SR.

Secondly, in B's frame of refence, the two times $\Delta t'_L=\Delta t'_R$. Surely two equal time differences will dilate the same and we get that $\gamma \Delta t_L= \gamma \Delta t_R$ and thus $\Delta t_L=\Delta t_R$?

Thanks for any replies!

• Is there any way to flag your own question as still requiring an answer? I'm really stuck here at the moment and I can't move forward without this understanding! I should add I'm really greatful for all the help I've recieved so far though! – James Machin Feb 14 '14 at 17:23

## 2 Answers

However, I feel like having the denominator c+v is wrong, as in some sense I'm adding velocity v to the speed of light which is not allowed in SR.

It is permissible, in SR, to have 'non-physical speeds' in excess of c.

For example, you observe two trains speeding towards on another. You measure the speed of each train to be say, $0.9c$ relative to the track.

Then, according to you, their closing speed is $1.8c$, i.e., if at some time, the trains are separated by a distance $d$, the time to impact is

$$\frac{d}{0.9c + 0.9c}=\frac{d}{1.8c}$$

This is fine because there is no physical object with speed $1.8c$.

However, the speed of one train, according to an observer on the other train, must be less than $c$ and is given by the relativistic velocity addition formula.

So, the closing speeds, according to A, of B and the light beams are $c + v$ and $c - v$.

UPDATE to address the comments:

Can I just say that B's clocks are synchronized with A's clocks?

No. Regarding your second question, you seem to be trying to use time dilation inappropriately to understand the relativity of simultaneity.

First, it is true that, according to B's clocks, (one in the rear of the train, one in the center, and one in the front), the time it takes for light to travel from the rear to the center and from the front to the center are equal.

In fact, it's true by definition because this how B confirms that his clocks are synchronized; see Einstein synchronization.

But, according to A's (Einstein synchronized) clocks, B's clocks are not synchronized. In fact, according to A, the clock 'in front' is behind the clock 'in back' (though they both run at the same rate).

This is precisely the reason that $\Delta t'_L = \Delta t'_R$ while $\Delta t_L \gt \Delta t_R$; no time dilation is required to explain this - only the constancy of the speed of light and the Einstein synchronization convention.

• Oh I see, it's similar to the "beam of light moving on the cloud" scenario then? Thank you very much! – James Machin Feb 14 '14 at 15:44
• Sorry to hassle you but is there any chance you could take a look at my second question? I've been working on it for a while and not got anywhere. – James Machin Feb 14 '14 at 16:32
• @JamesMachin, regarding your 2nd question: it isn't clear to me what you're trying to do. Don't forget, not only do B's clocks run slower, according to A's clocks (time dilation), B's clocks are not synchronized according to A's clocks. I don't think you're taking this into account. As always, draw a spacetime diagram to gain insight. – Alfred Centauri Feb 14 '14 at 18:12
• Can I just say that B's clocks are synchronized with A's clocks? Or under which conditions can they be synchronized? I guess I'm misunderstanding how time dilation works. I thought that any two frames moving relative to each other regard each other's time as "going slowly." – James Machin Feb 14 '14 at 18:19
• @JamesMachin, you're correct that each regards the other clocks as "going slowly". But, more than that, (spatially separated) events that are simultaneous according to A are not according to B and vica versa and this means that (spatially separated) clocks synchronized according to B are unsynchronized according to A. – Alfred Centauri Feb 14 '14 at 18:50

Observer $B$ is in the center of a train carriage [...] Two light signals are emitted from sources $L$ at the left end and $R$ at the right end of the carriage, such that they reach $B$ simultaneously.

More appropriate terminology is to say that "the two light signals reached $B$ in coincidence"; or that "the two light signals were observed by $B$ in coincidence".

$B$ observes that the light signals were emitted at the same time.

More appropriate terminology is to say that "due to $B$ having observed the signal from $L$ and the signal from $R$ in conincidence, and due to $B$ having been in the center of (or: the middle between) $L$ and $R$, it is concluded (or measured) that these two light signals were emitted simultaneously".

[...] train carriage which is moving at velocity $v$ with respect to an observer $A$.
[...] Show that $A$ doesn't agree [with the indicated conclusion that $L$'s emission and $R$'s emission had been simultaneous].

This cannot be shown.
Since it is given as a factual setup prescription that $B$ was in the center (or: the middle between) $L$ and $R$, it is not possible to "show" otherwise.

Since it is given as a factual setup prescription that $B$ had observed in coincidence that $L$ had emitted a signal and that $R$ had emitted a signal, it is not possible to "show" otherwise.

Finally, as far as the given setup description implies that the indicated conclusion had been drawn correctly, following Einstein's definition, i.e. especially as far as the refractive index $n$ throughout the region containing $L$, $B$, and $R$ had the value $n = 1$, it is not possible to conclude otherwise; everyone, including anyone (such as $A$) moving wrt. $L$, $B$, and $R$, should be able to comprehend and agree with the conclusion as indicated above.

However:
it should be instructive to suitably amend the given scenario as follows: In addition to the particpants $L$, $B$, $R$, and $A$ as they were already given above consider participants $K$ and $Q$ such that

• $K$ and $A$ and $Q$ were (pairwise) at rest to each other,

• $L$ observed $K$'s passage in coincidence with emitting the signal, and
$K$ observed $L$'s passage in coincidence with observing the signal emitted by $L$,

• $R$ observed $Q$'s passage in coincidence with emitting the signal, and
$Q$ observed $R$'s passage in coincidence with observing the signal emitted by $R$.

Now (only now) it may be asked, for instance, whether $K$'s indication of observing $L$'s passage (as well as $L$'s signalling) and $Q$'s indication of observing $R$'s passage (as well as $R$'s signalling) had been simultaneous to each other, or not.

Based on the scenario described so far this question can still not be answered; but it depends on whether $A$'s, $K$'s, and $Q$'s motion wrt. $L$, $B$, and $R$ (who make up the "carriage axis"), is "perpendicular to the carriage axis", or not.

Of course, the subsequent considerations stated in the question suggest that $A$'s (and $K$'s, and $Q$'s) motion is supposed to be "along the carriage axis", and "from $L$, past $B$, towards $R$" (not the other way around).

In any case, it would again be decisive to identify the participant who is the middle between $K$ and $Q$ and to consider whether this participant had observed $K$'s indication of observing $L$'s passage (as well as $L$'s signalling) and $Q$'s indication of observing $R$'s passage (as well as $R$'s signalling) in coincidence, or not.

It is not obvious from the question, whether or not $A$ was supposed to be the middle between $K$ and $Q$; so let's call the middle between $K$ and $Q$ explicitly $M$, instead.

Then the briefest argument I know goes like this:
by the given scenario and the added suggestions, $L$'s signal indication and $R$'s signal indication and $B$'s indication of $M$'s passage were all simultaneous. But $B$ observed $L$'s signal and $R$'s signal only afterwards (namely $LB / c \equiv d / c$ later), when $M$ has already moved away again, towards $R$. Consequently $M$ first observed $Q$'s indication of observing $R$'s passage (as well as $R$'s signalling) and only afterwards $K$'s indication of observing $L$'s passage (as well as $L$'s signalling); $M$ did not make these two observations in concidence.

In analogy to the argument made above, therefore $K$'s indication of observing $L$'s passage and $Q$'s indication of observing $R$'s passage were not simultaneous.

It might also be interesting, or even required, to quantify $M$'s duration $\tau_M[ \,_{\circledR}^{Q \circledast R}, \,_{\circledR}^{K \circledast L} ]$ from having made the former observation until having made the latter observation; for instance in comparison to the value $LR / c$.

Using what's sketched in the question already I'd calculate $\tau_M[ \,_{\circledR}^{Q \circledast R}, \,_{\circledR}^{K \circledast L} ] = (\Delta t_L - \Delta t_R) / \gamma = (\frac{d}{c - v} - \frac{d}{c + v}) / \gamma = 2 \, d \, \frac{v}{c} / \sqrt{ c^2 - v^2} \equiv \gamma \, \beta \, LR / c$.