# Length contraction of a train

Suppose a train of length $3x~\rm m$ and a tunnel of length $x~\rm m$. When the velocity of train is $282842712.5~\rm m/s$ (from length contraction equation), then its length contracts to $x~\rm m$. Suppose that when the train is inside the tunnel, two flaps at the ends of tunnel get closed.

My question here is that "Will the train get trapped inside the tunnel?"

If the flap at the end of the tunnel is indestructible, then yes, the train will get trapped in the tunnel. Though it will be completely destroyed.

What happens when the front of the train hits the flap is that it comes to a sudden halt. But the rest of the train is still moving. It has to still be moving, since the fastest any signal can propagate is the speed of light. So the train will crumple like an accordion.

If you do the math in the original frame of the train, you'll see that the back of the train always makes it into the tunnel before that time has elapsed.

The end result of this is that the train ends at rest, with a length of at most $x$ meters.

A natural question to ask at this point is "what if the train is rigid and indestructible?" The real answer to this is "it can't be." There is no such thing as a perfectly rigid object in special relativity- the same reason you can't send faster-than-light signals just by pushing a really long pole.

Yes, as the train comes by at high speed you can close both doors at the same time and the train will be totally inside, then open them up again very quickly as the train continues.

The train will see the front door closing and opening, as the front of the train is almost exiting the tunnel and the back door will close and open some time later after the back of the train had just entered the tunnel.

• The question does not contemplate re-opening the flaps. – WillO Feb 18 '18 at 14:19