1
$\begingroup$

I guess most of you are familiar with the "paradox" of the train passing through a tunnel smaller than itself on a speed close to the speed of light and 2 guillotines (1 in the exit and 1 in the entrance of the tunnel) going down/up when the train is "fully inside" the tunnel ( http://www.pcbheaven.com/opendir/index.php?show=3550vf5099tla8512e2c ).

I think I fully understand the explanation of this paradox, and I think it's ok for a person in the frame of reference of the train to think that the events are not simultaneous.
But what would happen if, instead of both guillotines being activated at the same time, the entrance guillotine would be activated first and this would cause the activation of the exit guillotine. For instance, when the entrance guillotine is activated, it starts to fall uncovering a light signal that goes to exit guillotine and activate itself. (Remember that the train is running in a speed close to light, let's assume 0.5c, so the signal is faster.)

How would the explanation be in that case in both frames of reference?

$\endgroup$
3
  • $\begingroup$ The explanation is the same. If the train can still be fully enclosed in the tunnel from an observer's point of view, then either the tunnel is long enough for the train or the same condition applies. For a short tunnel though, you'll likely have the train hit the front before the back is closed $\endgroup$
    – Jim
    May 14, 2015 at 18:04
  • $\begingroup$ Yes, cause-effect is always preserved in relativity. However I think there are some issues with this "paradox". IMHO there aren't any real paradoxes. When people think there's a paradox, it's because they don't understand something. And this one, which is often called the pole-and-barn paradox, comes up time and time again. $\endgroup$ May 14, 2015 at 20:19
  • $\begingroup$ If event A and event B are in each others light cones for some inertial frame of reference, then there is no inertial frame of reference where they are not in each others light cones (future or past). If they are, then they are possibly but not necessarily related by cause-effect. But if they are not, then there will be some inertial observer that sees them as simultaneous, while other frames of reference will see A before B or B before A depending on their motion (but they would all agree A and B couldn't have caused the other to happen). $\endgroup$ May 14, 2015 at 22:50

1 Answer 1

0
$\begingroup$

In the original paradox, there are two events of note: the front of the train exits the tunnel, and the rear of the train enters the tunnel. Call these events A and B, respectively. Because these two events are spacelike separated, the two observers (tunnel-based and train-based can disagree on the order in which they occurred. According to the tunnel observer, A and B are simultaneous; according to the train observer, A happens before B.

Now we want to introduce event C, which occurs when the receiver at the end of the tunnel gets a light signal emitted at A. Both observers will agree that event A happens before event C: the two events are light-like separated. Moreover, both observers will agree on the relative order of B and C, since they are timelike separated. But whether or not B happens before C or vice versa is going to depend on the relative rest lengths of the train and the tunnel, along with the speed at which the train is traveling relative to the tunnel. If the train's length in the tunnel frame is longer than this, it gets chopped in half; if the train's length is shorter, it's trapped inside the tunnel for some period of time (according to the tunnel observer.) If you want to have $v = 0.5 c$, then it's not too hard to see that the events B & C will be simultaneous when the train's length in the tunnel frame is exactly 1/2 the length of the tunnel; this then translates to the train's rest length being about 57.7% ($=\gamma/2$) the length of the tunnel.

Note, in particular, that if the train is too short, then C happens before B in both reference frames; and since A happens before C in both frames, it must be that A happens before B in both frames. So for sufficiently short trains (at a given speed), there's no paradox in the first place.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.