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Lengths perpendicular to relative motion are unchanged

  1. Jul 9, 2009 #1
    This has been bothering me for a long time. I can see that it's "obviously" true but I'm having trouble arguing it rigorously from what I know about special relativity (i.e. I "believe it" but when I try to (hypothetically) explain it to someone, I can't do so convincingly.)

    I first found this problem in Hartle's book, in which he writes:

    Imagine two meter sticks, one at rest and the other moving along an axis perpendicular to the first and perpendicular to its own length. There is an observer riding at the center of each meter stick.

    (a) Argue that the symmetry about the x-axis implies that both observers will see the ends of the meter sticks cross simultaneously and that both observers will therefore agree if one meter stick is longer than the other.

    (b) Argue that the lengths cannot be different without violating the principle of relativity.

    So assuming that the observers will agree on which stick is shorter, then the lengths cannot be different. Suppose the observer "at rest" sees a length contraction of the other stick. Then both observers agree that the "moving" stick is contracted. However, the observer in motion is equally entitled to consider himself "at rest," in which case he should say (by the principle of relativity) that the other stick is contracted - contradicting the fact that both observers agree.

    The part I'm having trouble with is why the meter sticks crossing simultaneously imply both observers must agree on which stick is shorter. I mean, in the case where the sticks are moving parallel to one another, the relativity of simultaneity in the measurements of lengths is precisely what gives length contraction with each observer claiming that the other stick is contracted, and in this case it's precisely the observers agreeing on which stick is "really" shorter that allows us to conclude that the lengths are not contracted. How does this difference come into the argument, though? What am I missing?
  2. jcsd
  3. Jul 9, 2009 #2


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    Endpoints of stick A: (0,0,0) and (1,0,0).
    Endpoints of stick B: (0,0,vt) and (0,b,vt) (...where b is 1 m, but we're not supposed to assume that)

    This is what you described, right?

    This situation is perfectly symmetrical. The observers sitting on the sticks both see another stick, oriented in the up direction, moving to the left (if we imagine that both observers are facing the other stick at t=0 and that A considers "up" to be in the positive y direction while B considers "up" to be in the positive x direction).
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