Time dilation in Einstein's train

Question

a clock that is moving relative to [observer] will be measured to tick slower than a clock that is at rest in their frame of reference

and they explain it using two pictures:

Left: Observer at rest measures time 2L/c between co-local events of light signal generation at A and arrival at A.

Right: Events according to an observer moving to the left of the setup: bottom mirror A when signal is generated at time t'=0, top mirror B when signal gets reflected at time t'=D/c, bottom mirror A when signal returns at time t'=2D/c

Let's assume that the second picture is a train going from left to right. Signal travels perpendicular (90°) to the motion of the train.

Now instead of positioning the mirrors on the floor and ceiling, what if I put mirror A to the back and mirror B to the front of the train (still facing each other). The signal now runs parallel (0° or 180°) to the motion of the train. It travels long way from A to B and short way back to A, effectively cancelling the time dilation effect at the end of the A-B-A trip.

It seems to me that I can tweak the angle between perpendicular (90°) and parallel (0° or 180°) to increase or decrease the time dilation effect.

Now if above is true does it mean that, contrary to what Wiki says, it is possible to position the clock in a way that it will be measured to tick at the same rate (not slower) than a clock that is at rest in observer's frame of reference?

Does this answer your question? https://physics.stackexchange.com/a/276603/4993 — WillO, Jul 30 '20 at 23:17

score 0 · Accepted Answer · answered Jul 30 '20 at 22:57

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If you have a horizontal signal, you will get that $\Delta t=2L/c$, and $\Delta t'=2L'/c=2L/c\gamma=\Delta t'/\gamma$. The reason is that now you have a component of the light moving in the same direction than S', and so length contraction plays a role.

However, to use such clock you need to know and correct for length contraction. In the case the light goes perpendicular to the motion you do not need to consider this effect.

answered Jul 30 '20 at 22:57

Makes sense but consider modified setup: what if the train passenger fires the signal under an angle between 90° and 180° to the motion of the train (in other words, not directly to the ceiling but slightly against the direction of motion). He'll see the signal reflect from ceiling and hit the floor at some distance from where he stands. But people in a train station will see it hit the floor faster because the motion of the train shortened the signal path for them. Length contraction doesn't seem to help me in this case, seems like the signal traveled faster for the people in the station. – elendir Aug 04 '20 at 09:06
imagine that on the train at any given position you have many clocks with many different orientations and lengths. These clocks will always be synchronized with each other in the train, and must also look synchronized from the platform. So the orientation or mechanism of the clock will not matter. they will all experience the same time dilation. – Aug 04 '20 at 14:06
I know that must be true and I'm trying to find error in my reasoning. So let's simplify: forget clocks, let's just talk about the length the signal has to travel in my previous comment. It seems obvious to me that it is possible to fire the signal in a direction which will make it's path (before hitting a barrier of ceiling/floor) in the frame of a train passenger longer than in the frame of a stationary train platform. Or am I wrong? – elendir Aug 04 '20 at 21:28
yes, but it will be a combination of the vertical and the horizontal clock, which both give the same result. But being more precise, yes, in that example the length seems to be smaller in the platform. But to calculate the exact dilation you do need to consider length contraction or you will get the wrong result – Aug 04 '20 at 21:34
I think I begin to understand. Although it's possible to have the signal travel shorter path in the frame of the platform, it's not enough to build a clock. Because to build a clock we need the signal to come back to source and that will somehow make the path longer again (I must learn some more theory and math to do the precise calculation). – elendir Aug 06 '20 at 08:34

score 0 · Answer 2 · answered Jul 31 '20 at 19:38

It travels long way from A to B and short way back to A, effectively cancelling the time dilation effect at the end of the A-B-A trip.

If we could write the longer path as $L'+\Delta L'$, and the shorter one as $L'-\Delta L'$, and if there was not anything such as length contraction ($L=L'$) , your claim would be correct and the clock would be measured to tick at the same rate.

However, calculations show that the longer path complies with $L/(1-\sqrt{\beta})$, and the shorter one does with $L/(1+\sqrt{\beta})$, where $\sqrt{\beta}=v/c$. On the other hand, the Lorentz contraction implies $L=L'\sqrt{1-\beta^2}$. These mean that the shorter and longer paths do not compensate for each other to maintain the proper time measured in the clock's rest frame.

Time dilation in Einstein's train

2 Answers2