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Old 07-27-2005, 01:47 PM
fnord_too fnord_too is offline
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Join Date: May 2004
Location: Norfolk, VA
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Default Fun relatavisic frame of reference problem

This is an oldie, but I don't know how many people have heard it.

The set up information is consistent with current theory, you can treat it as fact for the purposes of the question.

Set up:

All frames of reference are equally valid, and all frames of reference must observe the same physical events.

An observer in a frame of reference observing an object moving in relation to this frame of referene will see the object as being shorter along the axis of relative motion. This is called length contraction.

Any object can consider itself as an object at rest if it is not accelerating. (That is, there is a valid frame of reference that it is at rest in, its rest frame.)

The speed of light (photons) is constant to all frames of reference. (In the case about to be presented, the barn and tube would both see the photons traveling at exactly the same speed.)

The problem:

There is a hollow tube closed at both ends that has photons bouncing back and forth (axially) between the ends (in the tubes rest frame exactly 50% are always traveling in each direction), and a barn with doors on each end.

The tube is tube is traveling (relative to the barn) such that it will pass through the two doors perpendicularly.

From the barns perspective, taking the length contraction into account, the tube is barely too long to fit in the barn, such that if it shuts both its doors at once when the tubes center is at the center of the barn, it will chop off the end caps and release the photons. There are detectors outside the barn on each end. So the tube pass through the barn and the barn chops off both ends at exactly the same time and the detectors detect the photons that escape.

From the tubes perpesctive, though, the barn is contracted, so it does not fit almost exactly in the barn, but rather hangs out a bit. Since both frames need to agree on the events, the tube needs to see its ends just shaved off. It sees the far door cutting off the lead end then the back door cutting off the back end slightly later.

One might hypothesise that from the barn's perspective an equal number of photons exit from each end of the tube, but from the tube's perspective more exit from the front since that end was cut off before the back end. Both have to agree on what the detectors say, however, so who is correct and how is this apparent paradox resolved.
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