Q:

Version of the Relativistic Train Paradox

I am having trouble resolving a version of the well known relativistic train paradox. I would prefer a qualitative explanation although any help is appreciated.
Imagine a train and a tunnel which, when they are at rest relative to each other, are of the same length.
The tunnel has a left opening (entrance) and a right opening (exit). Imagine that the tunnel has a door on its right opening (exit) which is initially closed. 
Assume that there are matching sensors on the tunnel's entrance and on the back of the train, so that when the back of the train enters the tunnel a signal is sent to open the front door so that the train can pass safely through the tunnel.
Now let the train travel from left to right at a speed fast enough, so that the train shrinks to a quarter of the legnth of the tunnel in the tunnel’s frame. 
In the tunnel's frame of reference, the tunnel is 4 times as long as the train. The back of the train enters the tunnel (sending a signal), the front door opens, and the train passes through safely.
But in the train's frame of reference, the train is 4 times as long as the tunnel. So a signal cannot be sent to open the initially closed door before the front of the train reaches it. In the train's frame of reference there is a terrible accident and all of the passengers die.
How can these two scenarios be reconciled? which one is true? I think the main difference between this version of the paradox and the "classic" one is that here the opening of the door depends on the back of the train entering the tunnel. 

A:

But in the train's frame of reference, the train is 4 times as long as the tunnel. So that a signal cannot be sent to open the initially closed door before the front of the train reaches it. In the train's frame of reference there is a terrible accident and all of the passengers die.

Remember that the fastest that a signal can go is c and that c is invariant. What this means is that if a signal cannot be sent at c to the closed door in one frame then it cannot be sent at c to the closed door in any frame.
In this scenario it is easy to see that it cannot be sent in time in the train’s frame. Therefore the signal cannot be sent in time in any other frame either. There is a horrible accident in all frames. 

In the tunnel's frame of reference, the tunnel is 4 times as long as the train. The back of the train enters the tunnel (sending a signal), the front door opens and the train passes through safely.

This is incorrect. To achieve a length contraction factor of 4 means the train is moving at 0.968 c. So suppose the tunnel is length 1000 ft and the train is 250 ft long in the tunnel frame. Since c=1 ft/ns it takes 1000 ns for the signal to reach the closed door. In that time the train has moved 968 ft, so the back of the train is only 32 ft from the door and the front of the train has already smashed into the door. 

