This is what Feynman had to say about transverse length contraction in special relativity.

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 at 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.

I understand other arguments used to explain why we can't have transverse length contraction (e.g. an event in spacetime not being the same in all reference frames when there is transversal contraction). However, I don't understand this one.

  1. If they're moving relative to each other, the marks will be the same height on each one's meter stick, so when they come together they both "decontract" so there is no contradiction and one could not tell "who was really moving".
  2. Even if one could tell who was really moving, feynman's own explanation of the "twin paradox" could be applied here. When they come together one of them accelerates back to the other and he felt this so this would mean that he "absolutely" could tell he was moving away from the other person staying on earth. The one who felt the acceleration has property X when he comes back to compare; X in the twin paradox being difference in age and here being difference in position of the mark.

Maybe I totally misunderstand this argument so any help would be appreciated. Also, I would like for someone to clarify what he means by "by symmetry", symmetry in the sense that nothing should change when switching (inertial) reference frames?

  • $\begingroup$ related: physics.stackexchange.com/questions/399281/… $\endgroup$
    – user4552
    May 9 '18 at 21:05
  • $\begingroup$ Thank you. Like I said, I understand the arguments provided there, I just don't understand Feynman's. But of course Feynman being Feynman I'm the one not getting something and that bothers me. $\endgroup$
    – delivosa
    May 9 '18 at 21:09
  • $\begingroup$ The argument in the top answer in Ben's link is exactly the same as Feynman's. Feynman just uses meter sticks instead of rings. $\endgroup$
    – Pulsar
    May 9 '18 at 21:29
  • $\begingroup$ Please elaborate on how it's the same. I get that it's the same in the sense that the physics is not the same in all reference frames (each stick sees itself having a mark and the other having no mark so the event here is the mark) but how is that related to "knowing who's moving"? $\endgroup$
    – delivosa
    May 9 '18 at 21:56

I’m stationary with my stick. If there was transverse contraction/dilation, the mark on my stick will be in a different place than the moving one: there’s no need to stop the moving stick to see that. For simplicity, say the moving stick is shorter.

But you’re holding that stick. You don’t think you’re moving. But at the crossing, you see your stick marked as contracted! (You can’t see it as not; that’s not what would have happened in this hypothetical-transverse case)

So there’s the break in symmetry: you both should be able to think of the other as moving. And you can’t, because only one of the two “moving” sticks can be shorter.

Note: this is different from measuring length because that involves simultaneity along the motion. This involves simultaneity at a point, the marking-the-stick point, which is (almost) automatic.

  • $\begingroup$ Why can't they both see their sticks as being contracted, so when they meet there is symmetry? So you're saying that the contraction only goes one way. But it could go both ways and then one would see the other as moving and the two marks will be at the same place on the respective sticks. When they meet the sticks would both "decontract". $\endgroup$
    – delivosa
    May 11 '18 at 7:53
  • $\begingroup$ What do you mean by only one of the two moving sticks can be shorter? $\endgroup$
    – delivosa
    May 11 '18 at 8:51
  • $\begingroup$ @delivosa you never see your own stick as “contracted”. That’s the point of a meter stick. So if the moving observer’s meter stick has the same transverse length as yours, there’s no transverse length contraction. QED. $\endgroup$ May 11 '18 at 10:50
  • $\begingroup$ Oh now I see, my reasoning about decontraction and stuff was completely wrong. Thanks $\endgroup$
    – delivosa
    May 11 '18 at 10:59
  • $\begingroup$ @BobJacobsen I don't really understand your answer here. You say you are stationary and he is moving, but as you said, each observer sees himself as stationary. When you pass each other, you see your own stick as a meter long and the other guy's stick as shorter than a meter. Say you mark each other's sticks 0.5 meters above ground according to your reference. Each of you will see a mark on your stick, made by the other guy, a bit above the half-way line. When you stop moving relative to him, his stick returns to a meter and the marks are in the same place, identical. What am I missing here? $\endgroup$
    – Adgorn
    Jul 31 '19 at 14:36

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