Pole and Barn paradox - close the doors for good!

1. Jul 30, 2008

jtaravens

As I was reading Fabric of the Cosmos, I got stumped at the relativity of simultaneity section. This led me to Google for some additional explanation. I stumbled upon http://www.phys.unsw.edu.au/einsteinlight/jw/module4_pole_paradox.htm which made me think of this question regarding the pole and barn paradox.

I searched around the forum and I didn’t find an answer to my specific question – here goes:

If the person running with the pole appears to the spectator (at a distance) to be inside the barn completely with the doors shut, how can the person running with the pole think that the pole is hanging out of the barn?

I understand how only if the doors are simultaneously shut (according to the spectator) and immediately opened b/c the person running with the pole would not agree with the simultaneous shutting and opening. So to the runner, the first door shuts while the back door is still open allowing the pole to hang out, and then the front door opens while the front of the pole moves out of the barn and then back door shuts. But if both doors are shut, and shut for good, doesn’t it come down to whether the pole is in the barn, or whether the back door came crashing down on the end of the pole? One of my problems is right here: if the doors are closed forever, is the pole in the barn or not? Shouldn’t this be the same for both observers? The pole would either be crushed by the door or not.

If the simultaneity of the doors closing for good is still relative, couldn’t you time the closing so it appears simultaneous to the runner (i.e. have the back door close a bit earlier than the front door)? In this case it wouldn’t be simultaneous to the spectator. But we can still ask if the pole is in the barn or did it get crushed by a door?

I feel like to the runner, the pole is at a static length so it can’t ever be in the barn because it’s too big. This leads me to ask is this just an illusion based on perspectives of when doors are opening and closing and how the pole appears to the spectator (based on length contraction) to pass through the barn. I must be missing something here.

John

2. Jul 30, 2008

MikeLizzi

True, the runner never observes the pole being completely in the barn. But this fact does not alter anything that must be invariant between reference frames. If the pole is observed to get crushed by the door in one reference frame it will be observed to get crushed by the door in any other reference frame. Likewise if the pole clears the door in one reference frame, it will clear the door in every other.

May I suggets my tutorial on the subject at

http://mysite.verizon.net/mikelizzi/TutorialPoleInBarn.htm

If you have a PC you can actually run the 3d graphics simulation.

3. Jul 30, 2008

JesseM

If the doors shut simultaneously in the runner's rest frame, in the barn's own rest frame the back door must close later than the front door, not earlier. It'd have to work out so that if the back end of the rod has passed through the front end of the barn before the front door shut (in the barn's rest frame), then that would always mean that by the time the back door shut the front end of the rod would have already had time to pass partway through the front door (assuming the rod continues to move inertially) so the door would get closed on it. Should I try to do the math to prove this or would you rather just take my word for it?

4. Jul 30, 2008

jtaravens

Well that confuses me a bit - why would the barn's rest frame mean that the back door must close later than the front? While the runner is in the barn, couldn't the back door close before the front door?

I believe my question wasn't so clear. (but no need for mathematics - i'm only a microbiologist so extra math will make my head spin even harder!)

I think this website answers my question
http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html
The site states:
What if the doors are left shut?
If the doors are kept shut the rod will obviously smash into the barn door at one end. If the door withstands this the leading end of the rod will come to rest in the frame of reference of the stationary observer. There can be no such thing as a rigid rod in relativity so the trailing end will not stop immediately and the rod will be compressed beyond the amount it was Lorentz contracted. If it does not explode under the strain and it is sufficiently elastic it will come to rest and start to spring back to its natural shape but since it is too big for the barn the other end is now going to crash into the back door and the rod will be trapped in a compressed state inside the barn.

My interpretation of this is that the pole actually fits in that barn. The spectator sees it in the barn and by shutting the doors for good, you can squeeze the compressed pole in the barn (probably destroyed by strain) for the runner to see too.

MikeLizze- I believe I have a java issue here at work, so I will try later. Thanks!

Thanks for your help.

5. Jul 30, 2008

JesseM

Because you specified that the doors close simultaneously in the frame of the pole. In general, if two clocks are a distance L apart in their rest frame, and in your frame they are moving at some speed v in a direction parallel to the axis between them, and the clocks are synchronized in their own rest frame, then in your frame they will always be out-of-sync by vL/c^2, with the leading clock showing a time that's behind the trailing clock. So if we have clocks at rest in the barn's frame and placed at the front and back of the barn, and they are synchronized in the barn's frame, then in the pole's frame the clock at the front of the barn is leading and the clock at the back is trailing, so in the pole's frame the clock at the back must be ahead of the clock at the front by vL/c^2, where L is the length of the barn in its own rest frame. So if the front door closes when the clock there reads some time t, and the back door closes simultaneously in the pole's frame, that means the clock at the back must read t + vL/c^2 when it closes. And since these two clocks are synchronized in the barn's own rest frame, that means the back door must close later than the front door in this frame (exactly vL/c^2 later).
The site does not specify exactly how the doors are closed, but I believe they are discussing the situation where the doors are closed simultaneously in the barn's frame, in which case the pole will briefly fit entirely inside in this frame (while in the other frame the front end of the pole will hit the back end of the barn, and the pole will begin to compress on that end, before the back end of the pole passes through the front end of the barn, after which the front door shuts). You were discussing a different situation where the doors close simultaneously in the pole's frame--in this case it is obvious that the doors must shut on some section of the pole, just by looking at things in the pole's frame where the pole is longer than the barn. I did write a quick proof that what I said earlier was correct, the only math needed is the formula for how out-of-sync moving clocks are which I wrote above, as well as the formula for length contraction which says a rod of rest length L will be shrunk to length L*squareroot(1 - v^2/c^2) in a frame where it's moving at speed v...so, if you change your mind about not wanting to see the math let me know.