Special Relativity in 14 Easy (Hyper)steps

12. Paradox explained

This one's easy to sort out if we include the tunnel clocks in the drawing showing the train's frame.  Recall that clocks which are synchronized in one frame will appear out of synch in another frame. In particular, the "forward" clock is set to an earlier time than the "aft" clock.

Recall that  where v is 0.8 c and L is the rest-frame separation between the clocks. Plugging in reveals that the left tunnel clock reads -640 nsec (according to observers on the train) when the right tunnel clock reads zero.  Keep in mind that the train sees the tunnel clocks ticking slowly, so the left clock won't reach zero until  (640 / 0.6) nsec  (about 1007 nsec) after the right clock does.

Here's another diagram, including the tunnel clocks:

The right door slams when its clock reads zero.  The train crashes through it and continues moving. By the time the left tunnel clock finally reads zero (~1007 nsec after the right door slams, in the train's rest frame), the left door of the tunnel will have moved an additional 853.3 feet, past the left end of the train.