Car-garage paradox with just one door [closed]

Question

Special relativity implies the possibility of some apparently paradoxical situations, which can ususally be made sense of if one applies the theory rigorously. One of these is the car-garage paradox: a car speeds towards a garage which, at rest, is slightly shorter than the car. From the reference frame of the garage, the car appears shorter, so that it will fit into the garage (if the speed is high enough, which we assume it is). From the reference frame of the car, the garage appears shorter, so that the car will not fit. I have seen some very nice solutions, like this one, where the garage has a front door and a back door and the caveat is that their opening and closing times depend on the reference frame, so that in the end the car gets through in both cases, although in one case it is longer than the garage.

But what happens if there is only a front door? Of course the car will eventually crash, so that there are some non-inertial computations involved, if one wants to do things rigorously. But in any case, if the garage door closes immediately (let it be almost as fast as light and very small) after the rear of the car has passed it, then in the garage's frame the door will close and then the car will crash, whereas in the car's frame the door will not be able to close, and since after the car crashes (decelerates to $0$) it is still longer than the garage (neglect shortening due to the accident), the door will not close at all, ever. But this is impossible, since the fact that the door is closed or open for all future times should not depend on the reference frame.

Is there a way to make sense of this applying the theory of special relativity and without waving hands and just saying "this situation is not physically possible"? I know it isn't, but one can certainly modify it appropriately so that it is (e.g. transforming the car into a particle and the door into a sensor,...).

Answer 1

it is still longer than the garage (neglect shortening due to the accident)

You cannot neglect the shortening. The rear of the car is moving forward. The front of the car is having a collision.

The crucial point is that information about this collision cannot reach the back of the car faster than the speed of light. So the rear of the car will always continue forward (at the same speed) until information about the collision reaches it.

The situation is set up so that the rear of the car will not get information about the collision until after it has entered the garage. Only at that point could it be slowed.

this relativistic shortening of the car is not the same as the one we observe in car accidents, is it?

No, not exactly. The "shortening" is really just a different point of view. But it does mean that everyday concepts of rigid objects break down at relativistic speeds.

It's not that the collision is necessary to shorten the car. It's that whether the collision happens or not (maybe you have a choice of whether there's an open door), the rear of the car is going to enter the garage at a particular time in the garage frame. There's nothing you can do to the front of the car to prevent it.

What if it is not a car but a bar of a very hard material, and at the end of the garage there is a very hard elastic wall that can make the bar decelerate in a negligible length. So at the end the bar fits inside the garage, it has not broken, it has not bent. But it is shorter. Have the particles composing it got closer together?

Relativity tells us that the concept of a perfectly rigid material must be impossible. If you push on one end of a material, the other end will not react until after sufficient time for a signal has reached it. That signal must be no faster than the speed of light.

It doesn't tell us exactly what the behavior of the material is. It could compress, it could fail/explode. Real materials tend to either flow or shatter. But if you suppose that you can stop the front of the object, it will necessarily compress (in some manner) because the rear of the object will not stop simultaneously.