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!

## Best Answer

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 prioriwe 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,

youare 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

cannotshow 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.