A. Suppose a moving train. The train has two wheels(front and back) and each wheel is connected to light source inside the train. The light source is triggered(light is emitted) when train passes certain point on track and both light sources are triggered at the same time(the gap between certain points is identical to the gap between two wheels).

A-1. Suppose an observer at the middle of the train(same distance from two light sources). He observes two light at the same time, since in his frame light sources and he is stationary.

A-2. Suppose an observer outside the train, who can observe light reaching the middle of train(by mirror or similar). He observes front light reaching the middle first, since in his frame train is moving.

So the event of both light reaching the middle is not simultaneous for both observers.

Am I understanding and thinking right?

  • 2
    $\begingroup$ When you say "the gap between certain points is identical to the gap between two wheels" do you mean that those to distances are equal when the train is at rest with respect to the rails (i.e, in their own rest frames)? $\endgroup$ – garyp Mar 16 '16 at 17:29

Although the two lights are emitted simultaneous in the train's frame, they do not appear to be emitted simultaneously in the track's frame. The observer on track sees the rear light emitted earlier than the front light, by exactly $\Delta t = \frac{v/c}{\sqrt{1 - (v/c)^2}} \frac{L}{c}$, where $L$ is the rest length of the train and $v$ is its velocity on the track. This is just relativity of simultaneity at work, and the main source of confusion in the understanding of this typical train setup.

So the observer on the track does observe the front light taking a shorter time to reach the middle of the train then the rear light, but both him and the observer on the train do see the two lights reaching the middle of the train simultaneously.

What is seen from the track is that by the time the front light is emitted, the rear light has already traveled a distance $d = c\Delta t = \frac{v/c}{\sqrt{1 - (v/c)^2}} L$, and from that moment both lights travel equal distances until they meet in the middle of the train.

The reason why the latter simultaneity is allowed is because the event occurs at a single point in space-time. The emission events, on the other hand, occur at two distinct locations in any frame and are always spacelike separated. This means that although they are simultaneous in the train's frame, they are not simultaneous in any other frame. Moreover, different frames observe them in different order: they are not causally related.

Another source of confusion: in the track's frame the distance between the locations at which the two lights are seen emitted is neither the contracted length of the train $L\sqrt{1-(v/c)^2}$, nor the rest length of the train, but actually a dilated length $L/\sqrt{1-(v/c)^2} > L > L\sqrt{1-(v/c)^2}$. Please keep in mind though that this is a distance between events that are observed at different moments in the track's time. The phenomenon occurs for any two events that are simultaneous in a given frame.

Derivation of the delay $\Delta t$:

Let the train's frame have primed coordinates $(x', ct')$, and the track frame have unprimed coordinates $(x, ct)$. If the frames are synchronized such that for $ct = 0$ on the track then $ct' = 0$ on the train, the Lorentz transformation from train coordinates to track coordinates reads $$ x = \gamma(x' + \beta ct')\\ ct = \gamma(ct' + \beta x') $$ for $\beta = v/c$ and $\gamma = 1/\sqrt{1 - \beta^2}$. Suppose the two lights are emitted in the train's frame at coordinates $(x'_1 = -L/2, ct'_0 = 0$) and ($x'_2 = L/2, ct'_0 = 0$). Then in the track frame these events have coordinates $$ x_1 = \gamma(-L/2 + \beta \cdot 0) = -\gamma L/2\\ ct_1 = \gamma(0 + \beta (-L/2)) = -\beta \gamma L/2 $$ for the rear light and $$ x_2 = \gamma ( L/2 + \beta \cdot 0) = \gamma L/2\\ ct_2 = \gamma(0 + \beta L/2) = \beta \gamma L/2 $$ for the front light. From these we have that the track observer sees the rear light being emitted first, since $ct_2 > ct_1$, and so the front light appears to be delayed by $$ \Delta t = t_2 - t_1 = \beta\gamma \frac{L}{c} = \frac{v/c}{\sqrt{1-(v/c)^2}} \frac{L}{c} $$ Likewise, the distance between the locations at which the track observer sees the lights being emitted is $$ \Delta x = x_2 - x_1 = \gamma L = \frac{L}{\sqrt{1-(v/c)^2}} $$

Fun exercise: Apply the same kind of reasoning to calculate where and when the meeting of the two lights in the middle is observed in the track frame. Then calculate the distances traveled by the two lights until they meet, as seen on the track :)

  • $\begingroup$ Thank you for so kind answer. Could you please explain how $\Delta t$ is derived? $\endgroup$ – ylem Mar 17 '16 at 22:57
  • $\begingroup$ Welcome. I added the details on $\Delta t$ to the answer. $\endgroup$ – udrv Mar 18 '16 at 3:18

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