# Special relativity thought experiment using light

I have a clock using lights bouncing off two mirrors in a moving train as shown below. The train is moving in the left direction as shown.

In this situation, the light in direction 1 takes more distance to travel for an observer on the ground level. The light in direction 2 takes less time to travel because it's against the direction of the train movement (and it travels shorter distance) which should compensate for the direction 1 case. Doesn't it mean the clock is the same for both moving and ground observers? What is that I am doing wrong here?

• Thanks for the answers. I see the confusion now. Since the time that light travels in direction 2 is smaller, the distance moved by the mirror due to the train velocity is not the same in both directions. Commented Aug 3 at 14:39

The light in direction 2 takes less time to travel because it's against the direction of the train movement (and it travels shorter distance) which should compensate for the direction 1 case.

Not quite. You have to do some maths to see exactly how the round trip time is different for the two observers.

For the observer on the train the mirrors are stationary and are a distance, say, $$d$$ apart, so the round trip travel time for light from one mirror to the other and back again is $$t=2 \frac d c$$.

For the observer on the ground, on the first leg the mirror is moving away from the point where the light was emitted at speed $$v$$, which is the speed of the train. After a time $$t$$ the mirror has moved a distance $$vt$$ and the light has travelled a distance $$ct$$. So the light will reach the second mirror at time $$t_1$$ where

$$d + vt_1 = ct_1 \\ \displaystyle \Rightarrow t_1 = \frac d {c-v}$$

On the return leg the first mirror is now moving towards the point where the light was reflect at speed $$v$$, so the return leg takes time $$t_2$$ where

$$d - vt_2 = ct_2 \\ \displaystyle \Rightarrow t_2 = \frac d {c+v}$$

Adding these two times together we see that for the observer on the ground the round trip time is

$$\displaystyle t' = t_1 + t_2 = \frac d {c-v} + \frac d {c+v} = 2 \frac {cd} {c^2-v^2}$$

(If the shorter time for the return leg exactly cancelled out the longer time for the first leg then we would have $$t_1+t_2 = 2 \frac d c$$, but this is not quite correct.)

The ratio of the round trip time measured by the observer on the train to the round trip time measured by the observer on the ground is

$$\displaystyle \frac t {t'} = \frac {c^2-v^2} {c^2} = 1 - \frac {v^2}{c^2}$$

• I can't comment so I am adding my reply here. The answer that has been accepted is NOT correct. @gandalf61 has forgotten to take into account length contraction. The length in the frame of the ground observer is $\frac{d}{\gamma}$ The times in the two frames only differ by a factor of gamma not gamma squared as has been suggested in that answer. Commented Aug 3 at 15:30
• Someone mentioned there is a length contraction factor. In that case, the formula needs to be updated? Commented Aug 3 at 15:33
• @iVenky Since the light is travelling in the same direction as the train we are seeing the combined effect of length contraction and time dilation here. If you wanted to calculate the time dilation effect on its own you would have to place the mirrors on opposite sides of a carriage so that the light (from the point of view of the train observer) travels at right angles to the train's motion. Commented Aug 4 at 7:23
• I don't intend to digress, but is it possible to explain how the slowness in time of a moving frame of reference can result in a simultaneous lightning strike that a person sees at the midpoint of a moving train intuitively ? Commented Aug 4 at 7:56
• @iVenky Try reading en.wikipedia.org/wiki/Relativity_of_simultaneity. If that does not help then post another question. Commented Aug 4 at 8:00

A couple of points.

1. For someone observing from the ground the clocks on the train move slower than his clocks.

2. However, imagine there are physical clocks on every carriage of the train. These clocks would be showing the same time in the frame of the train, but would show different times for the observer on the ground. This is the relativity of simultaneity. The rear clocks are ahead.

3. For the observer on the ground the lengths on the train have contracted, so the distance between the mirrors has shrunk by a factor of gamma.

4. let the distance between the mirrors be L be in the frame of the train and let the speed of light be c. Then the time taken $$t$$ = $$\frac{2L}{c}$$

5. In the frame of the observer on the ground: $$t'$$ = $$\frac{L}{\gamma(c-v)}$$+$$\frac{L}{\gamma(c+v)}$$ = $$\frac{2L\gamma}{c}$$

6. The answer we get is consistent with time dilation. For the observer on the ground, time got dilated by a factor of gamma.

So the train is at rest in $$S$$, and there are 3 events (labeled by $$[t, x]$$):

Emission:

$$E_1 = [0, 0]$$

Reflection:

$$E_2 = [L, L]$$

Final Detection:

$$E_3 = [2L, 0]$$

I can transform those to a frame $$S'$$ moving at $$\beta$$:

$$E_1 = [0, 0]'$$

$$E_2 = [\gamma(1-\beta)L, \gamma(1-\beta)L]'$$

$$E_3 = [2\gamma L, -2\beta\gamma L]'$$

from which it is easy to see that:

$$(E_2-E_1)^2 = (E_3-E_2)^2 = 0$$

in $$S'$$ (That's just an algebra check on null-world lines).

Now for the question about how long the moving observer sees between events, we need to flip the sign of $$\beta$$ (because you set the problem up backwards). With $$L=1$$, we can put it back latter:

$$\Delta t'_{2,1} = \gamma(1+\beta)$$

$$\Delta t'_{3,2} = 2\gamma - \gamma(1+\beta) = \gamma(1-\beta)$$

which should make sense, and the asymmetry is:

$$\Delta t' \equiv \Delta t'_{2,1}-\Delta t'_{3,2}= 2\beta\gamma L$$

or we do the ratio:

$$R = \frac{1+\beta}{1-\beta}$$

which is the square of the Doppler shift factor, which is has to be, since the number of nodes in-between the mirrors is a Lorentz invariant.