The clocks are all in sync in the ground frame, but they are not in sync in the train's frame. An observer on the train would think that the clocks in the front of the train are ahead, while those to the rear are behind.

Measuring the forward traveling beam against the nearby clocks will show a long time difference, while measuring the rearward traveling beam shows a shorter elapsed time. The observer would think the actual time for the forward and rearward beams was the same and that the difference is because the clocks are not synchronized (in the observer's frame).

Why is the door that is farther away from the train observed to close
first?

By observe, in SR, we don't mean *see*, we mean essentially to record the time and place of events according to rods and (synchronized) clocks at rest. For example,

- Assume that, at the front and back of the train, there are identical
clocks that are synchronized in the frame of the
*train*.
- Further assume that, at the ends of the tunnel, there are identical
clocks that are synchronized in the frame of the
*tunnel*.

By the relativity of simultaneity, the clocks on the train are *not* synchronized in the frame of the tunnel where it is observed that the clock at the front is *behind* the clock at the back.

Symmetrically, the clocks in the tunnel are *not* synchronized in the frame of the train where it is observed that the clock at the entrance of the tunnel is *behind* the clock at the exit.

- Finally, assume that the length contracted train in the frame of the
tunnel just fits within the tunnel.

Thus, there is a moment, *according to the tunnel clocks*, that the contracted train is completely within the tunnel.

But remember, in the frame of the tunnel, the train clocks are not synchronized. In particular, since the clock at the front of the train is observed to be *behind* the clock at the back, it must be the case that, as *recorded by the clocks on the train*, the door at the exit of the tunnel closes *earlier* than the door at the entrance.

That is, in the frame of the train, the front of the train just reaches the exit, the door there closes for an instant without hitting the train *and there is still a trailing portion of the train that is yet to enter the tunnel*.

When the back of the train just clears the entrance of the tunnel, the door at the entrance closes for an instant without hitting the train and there is a leading portion of the train that has exited the tunnel.

As always, I recommend that you draw a spacetime diagram of this sequence of events to get a better 'picture' of how this works.

## Best Answer

Rely on the invariance of the interval $\tau$ between the ball being dropped and the ball hitting the floor. Also using $c=1$ units to keep the typing down.

The moving frame is un-primed and the stationary frame primed.

$$ t^2 - h^2 = \tau^2 = {t'}^2 - [{h'}^2 + (vt')^2] \,.$$

Because the velocity $v$ is horizontal both parties measure the same $h$, allowing us to write

$$\begin{align*} {t'}^2 - [h^2 + (vt')^2] - \tau^2 &= 0 \\ {t'}^2(1 - v^2) - h^2 - \tau^2 &= 0 \\ {t'}^2(1 - v^2) - h^2 - (t^2 - h^2) &= 0 \\ {t'}^2(1 - v^2) - t^2 &= 0 \\ t' = \frac{t}{\sqrt{1 - v^2}} \,, \end{align*}$$

so I owe Mew an apology for my comments in the earlier thread, but it does

notviolate the equivalence principle because we didn't use any information about the value of $h$ in the above work, so it works fine when $h=0$ as in the elevator case.