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In Feynman's Lectures on Physics , chapter 15, page 6 he writes about 2 identical, synchronized light signal clocks. These are clocks that consist of a rod (meter stick) with a mirror at each end, and the light goes up and down between the mirrors, making a ,click each time it goes down. He describes giving one of these clocks to a man flying out in space, while the other remains stationary. The man in the space ship mounts the clock perpendicular to the motion of the spaceship. Feynman then writes:

"the length of the rod will not change. How do we know that perpendicular lengths do not change? The men can agree to make marks on each other's y-meter stick as they pass each other. By symmetry, the two marks must come a the same y- and y' coordinates, since otherwise, when they get together to compare results, one mark will be above or below the other, and so we could tell who was really moving."

what exactly is the "test" with the marking of meter-sticks that Feynman is describing? why would it violate relativity, since it seems the person in the space ship would be looking outside? why would a change in a perpendicular length violate relativity, but not a change in parallel length--couldn't the men also make marks on each other's sticks, in the case of parallel lengths? Thank you!

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3 Answers 3

It is because the observers will always claim, that as long as they are moving in uniform velocity, they are always moving with respect to each other in only one dimension. There will be no relativistic effects in the perpendicular direction because there is no motion in that direction. And as for comparing, each person will be correct to say that the other person's parallel rods, or anything parallel to their relative motion, have contracted. To see who "really" has contracted, they must come together to see their markings once they have made them while they were passing by, but if they do so, the symmetry will no longer be valid.

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You're right, the symmetry in Feynman's argument doesn't show why there can't be a symmetrical Lorentz contraction perpendicular to the direction of motion. This is because he's looking at the two frames where each person does their marking.

Let the two moving frames have their times set to zero when their origins cross. Feynman is saying that if each person is positioned at the same location in their local frame and marks the other frame at the same local time, then the marks will appear at the same position in one another's frame. This is right for the reason he gave - any difference would mean one frame was in "motion". But this doesn't mean the marks have to appear at the perpendicular position of each person, and so he's wrong to conclude that there isn't a Lorentz contraction perpendicular to the direction of motion from his argument using these two frames.

He needs to bring into his argument the frame where the two rods approach one another with the same velocity. Therefore if the rods are contracted perpendicular to their movement, it must done in the same way which follows from the isotropy of space; it shouldn't matter in which direction a rod moves. This means each person must mark the other rod at the location of the other person. Hence there is no Lorentz contraction perpendicular to the motion of each rod.

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Sorry, this is completely incorrect---- imagine two pipes colliding and bouncing off each other. If there is any perpenduicular contraction, the two pipes don't touch if you boost to the rest frame of either one. This is a paraphrase of Feynman. –  Ron Maimon Jun 18 '12 at 5:41
    
@ron Feynman's argument uses the local frame of each person, not the centre of mass frame which is the key to your correct argument. I've updated my answer using yours which is much nicer - I hope you don't mind ;) –  John McVirgo Jun 18 '12 at 12:41
    
The argument I gave is the same as Feynmans, don't be fooled by superficial differences in intuition. –  Ron Maimon Jun 18 '12 at 20:06
    
@RonMaimon have you read the section? Perhaps you're allowing your admiration of Feynman to cloud your impartial judgement. It's clear to me his argument makes no mention of the centre of mass frame. –  John McVirgo Jun 18 '12 at 21:50
    
I read the section, the argument I gave is just an elaboration of Feynman's, Feynman's explanation is fine and sufficient, and I am not clouding anything. Feynman doesn't give the velocities of the two people, he expects you to think about it. By considering one moving, the other stationary, or both moving, etc the invariance of the marking procedure is the same argument as the cylinder collision, and is correct. He just expects you to fill in this obvious part of the reasoning, so he doesn't go into detail. This is best done by each person alone, I just gave the way I fill in the details. –  Ron Maimon Jun 19 '12 at 3:16

Not sure what happened to Ron's response, since it was probably the best one, so I'll try to shed some more accurate light on this old question.

The idea is suppose you are holding a rod of a certain rest length, and someone else is holding an identical rod but is moving toward you. Both rods have the same orientation, perpendicular to the line of motion. A priori we can't be sure whether the other person's rod will seem longer, shorter, or the same length as yours. (The result of such a measurement is well defined at the moment you pass each other, since then either the rods will coincide or one will stick out past the other.)

Assume the other person's rod will be shorter. Then when you coincide, your rod will stick out past his (you can mark your rod at the points coinciding with the endpoints of his rod), and this is a qualitative, frame-invariant result. He will have to agree. But then look at the reverse situation. From his reference frame, you are the one moving, and since we assert that direction doesn't matter, just relative speed, we would have to conclude that he thinks your rod is shorter than his. This is a contradiction, proving there can be no perpendicular length contraction. The exact same argument proves there can be no perpendicular length expansion.

This line of argument can be made to work with any two frames in relative motion, but these two symmetric frames in which one or the other rod is at rest make it easiest to grasp. Be careful, as if you consider things "objectively" from only a central frame wherein the two rods are moving at the same speed toward one another, you cannot show anything at all, since any perpendicular changes would be equal and the rods would always appear the same length.

As Ron seems to have hinted at, there are variations on this by considering whether or not two equal-rest-size cylinders can pass through each other, and then asking, if they can, which one passed on the inside. My favorite variation involves a train moving very fast. If there were perpendicular contraction, it would see the tracks get closer together and fall inside the wheels, while a person standing still on the ground would see the train's wheels come together and fall inside the tracks. These two outcomes are incompatible, and so we know that there is no change in the perpendicular direction.

Parallel length is a bit trickier, since there are different times to consider. The times at which you compare rods become important.

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