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In Einstein's famous thought example involving a fast-moving train (say with a velocity of $c/2$), we assume that lightning strikes occur simultaneously at the front and rear end of the train as viewed from an observer at a platform outside of the moving train. For an observer sitting at the center of the train, however, one of the strikes will then appear to happen before the other as the lightning strike which occurs at the front of the train will have a shorter distance to travel before it "catches up" with the observer. This I fully understand.

Now on to the part which confuses me. For the observer on the train, the two observed lightning strikes may be registered as two separate events, say event A and event B. The observer on the train could then, hypothetically, use a clock to estimate the time difference between the two events. For the observer on the platform, however, due to time dilation effects and the Lorentz factor, I would assume that a clock used by this external observer would register the time between the two events as longer than what the clock used by the observer on the train would show.

All this makes me a bit confused. The observer on the platform views the two lightning strikes as occuring simultaneously, but, at the same time, a clock used by this observer would show that longer time would pass between events A and B compared to the time registered by the observer on the train. This seems paradoxical to me, so if someone could clear up my confusion, then I would be extremely grateful!

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    $\begingroup$ In your first paragraph, you assume that for an observer on the platform, the two lightning strikes are simultaneous --- that is, the time difference between them is zero. In your second, you say that you expect that for an observer on the platform, the time difference between them is longer than it is for an observer on the train, which means in particular that the time difference is not zero. You've assumed it's zero, so how can you expect it to not be zero? $\endgroup$
    – WillO
    Commented Jul 10 at 21:53
  • $\begingroup$ Well, that's what I find paradoxical. Yes, the time difference should be zero for the observer on the platform. For the observer on the train the difference is not zero. So the observer on the train can use a clock to estimate the time difference between the two events. However, the theory of time dilation then tells us that a clock used by the observer on the platform should show a larger time difference. This is where I am confused, as the time is supposed to be zero for the platform observer. $\endgroup$
    – user12277
    Commented Jul 10 at 22:18
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    $\begingroup$ You set up the experiment such that "lightning strikes occur simultaneously at the front and rear end of the train as viewed from an observer at a platform". This means that the single clock at the observer's location (conveniently situated halfway between the lightning strike locations) observes a zero length time interval. Dilate that zero any way you like, it stays zero. (As an aside, I wonder why you say "estimate" when you talk about the train observer's measurements. Those are not estimates.) $\endgroup$
    – Peter - Reinstate Monica
    Commented Jul 11 at 8:24
  • $\begingroup$ Yes, from the answers below, I now understand why my reasoning was incorrect. Also, you are right that "estimate" is not the best word - "calculate" would, of course, be more suitable. Sorry about that. English is not my first language. $\endgroup$
    – user12277
    Commented Jul 11 at 13:26

2 Answers 2

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The time dilation formula applies to measurements by a single clock that is at rest in the "moving" frame. In this case there are two clocks involved, at the front and back of the train. An observer on the train will regard these clocks as synchronized. In a frame in which the train is moving, however, the clocks will not be synchronized, precisely because of the relativity of simultaneity. They tick at the same (dilated) rate, but one of the clocks will appear to be set ahead of the other.

EDIT: To clarify, I'm not necessarily talking about physical clocks, but about time recorded at a given place relative to the frame. The full Lorentz transformation includes a $\Delta x$ term. If you use measurements that happen in the same place then $\Delta x = 0$ and that term disappears, giving the familiar time dilation expression. But if the measurements happen in different places (as they do here!), the simple time dilation formula does not apply.

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  • $\begingroup$ Perhaps my question was not clear, but I am here only assuming one clock on board the train, and that is a clock in the hands of the observer located at the center of the train. Then, I am also assumint, tthat he observer at the platform has a clock. $\endgroup$
    – user12277
    Commented Jul 10 at 22:28
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    $\begingroup$ For purposes of the thought experiment you could hang clocks at the front and back of the train (where the events happen). In the rest frame of the train these are synchronized with that of the observer; in that of the platform all 3 show different times. In other words time dilation is only part of the story; the full story is given by the Lorentz transformation, which has a spatial factor too. $\endgroup$
    – Eric Smith
    Commented Jul 10 at 22:38
  • $\begingroup$ Yes, I am just starting to dig into these concepts, so I still have a lot to learn! $\endgroup$
    – user12277
    Commented Jul 10 at 22:40
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Consider the situation :

Say the lightning stike occurred at $t=0$ and when both the observers ( on platform, $O_p$ and on train , $O_t$ ) were at $x=0$ , the train being traveling along positive $x$-axis.

For more simplicity, say the observer $O_t$ is in middle of the train.

What happened for $O_p$ ? Some time must have taken by the ' light ' of the two lightning strike to reach him simultaneously, say it to be $t_p$ ( in the frame of $O_p$ ): $$t_p=\frac{d}{c}$$ $2d$ , being the length of the train.

Let's change the frame , we are in the frame of $O_t$ . After the lighting strike occurred at $t=0$ , he was approaching the light of the front strike at $c/2$ ( given ) , let's say he saw it time $t_f$: $$t_f=\frac{d}{3c/2}$$

And , he was going away from the rear light at speed $c/2$ , say he saw the rear strike at time $t_r$ ( both the time being measured in his own frame ): $$t_r=\frac{d}{c/2}$$

Clearly, for $O_t$ , $t_f<t_r$ . Hence, he observes the lighting strikes as two different events.

What's the total time for the completion of the events in $O_t's$ frame ? $$t_{total}= t_r-t_f=\frac{4d}{3c} $$

So, we can say that this total time would have been longer in the frame of $O_p$ but we can't say that the two strikes seems to be longer apart for $O_p$ than that to $O_t$ as we have already seen it happened for $O_p$ for just an instant at $t_p$.

That is to say, we can see and measure time dilation from one frame to other but we can't see the events as it is seen in any other frame.

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  • $\begingroup$ Thanks a lot! This is very helpful and I am really starting to see the logic now. The only remaining question I have is in regards to the time dilation effect. From your calculations we have that $t_{total} > t_{p}$. But is it correct that what time dilation tells us is that if the observer on the platform had started a clock at $t_f$, then that clock would have shown that more time would pass until $t_r$ compared to the clock on the train? $\endgroup$
    – user12277
    Commented Jul 10 at 22:32
  • $\begingroup$ @user12277 You are right , I was just comparing the numerical values of the two . Just to ensure if I would have used it later but it doesn't get used ...you can ignore that !! $\endgroup$
    – CP of Physics
    Commented Jul 10 at 22:34
  • $\begingroup$ Fantastic! Thanks a lot for your very helpful comments! $\endgroup$
    – user12277
    Commented Jul 10 at 22:35

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