Trains in a Tunnel - Solution

1. The front end of train 1 enters the light field at Q. Q is simultaneous with and P and R in the train 1 frame. Therefore, the front end of train 1 is at Q, and the rear end is at P.
2. The rear end of train 1 exits the light field at F. F is simultaneous with and E and G in the train 1 frame. Therefore, the front end of train 1 is at G, and the rear end is at F.
3. The rear end of train 2 enters the light field at V. V is simultaneous with and S, T, and U in the train 1 frame. Therefore, the front end of train 2 is at U, and the rear end is at V.
4. The front end of train 2 exits the light field at A. A is simultaneous with and B, C, and D in the train 1 frame. Therefore, the front end of train 2 is at A, and the rear end is at B.
5. Even though train 2 is shorter than train 1 in the train 1 frame of reference, as can be seen by comparing the lengths of AB and CD, it spends more time in the light field. The rear end of train 2 starts entering the light field at V, and is completely immersed in light (QR) by the time the front end of train 1 starts entering the light field at Q. The rear end of train 1 exits the light field at F, but at that time train 2 is still completely immersed in light (EF). So the effect of Lorentz contraction is completely cancelled out by the fact that train 2 spends a longer time in the light than train 1.
   The important thing to notice is that the lights inside the tunnel do not turn on and off simultaneously in either of the trains' frames. In the train 1 frame, what you observe is a band of light moving from the right to the left. Since train 1 is moving in the opposite direction as the band of light while train 2 is moving in the same direction, train 1 spends less time in the light than train 2. By the way, the band of light travels at a speed faster than the speed of light, but that does not violate causality since nothing is really moving, nor is any information being transmitted.