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Schutz First Course in GR Problem 15b, chapter 1. Mistake?

  1. May 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that the velocity of an observer O' relative to O is nearly that of light, |v|=1-ε, 0<ε<<1. Show that the Lorentz contraction formula can by approximated by:
    ∆x≈∆x'/√(2ε)


    2. Relevant equations
    Lorentz contraction, ∆x=∆x'/γ


    3. The attempt at a solution
    I think it should be ∆x≈∆x'(√(2ε)). (As opposed to divided by the square root of 2ε). Is this a mistake in the book, or am I just being stupid? Don't tell me how to solve it or anything- just if it's a mistake or not; if not, I'll keep trying but I don't want to waste my time if the problem is stated incorrectly. Thanks!


    Ps. Anybody who likes SR- try out problem 12 from that same chapter, it's very fun :).
    PPs. Just thinking intuitively, the approximation given by the problem is incorrect because it'd give a longer length measured by observer O, which just makes no sense. The famous effect is a contraction, after all!
     
    Last edited: May 14, 2014
  2. jcsd
  3. May 14, 2014 #2

    ehild

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    The length is maximum in the frame in which the object is at rest. Usually x' refers to the moving frame of reference. Imagine you stand on the ground and determine the length of a stick in a spaceship. The length of a stick is measured Lo by the astronaut, the observer who moves together with the stick: Δx' = Lo. You measure L=Δx, a shorter length, L=Lo√(1-(v/c2), but that means Δx'=Δx/√(1-(v/c)2).

    ehild
     
  4. May 15, 2014 #3
    I agree, and I don't believe I said anything that contradicted any of that. What are your thoughts on the problem? I think it has a typo, and it should say that ∆x is approximately ∆x' *times* sqrt(2 epsilon).
     
  5. May 15, 2014 #4

    ehild

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    Yes, you are right if Lo=∆x' and L=∆x, as L should be shorter than Lo :) . It is a typo in the book, or it defined x' and x in the opposite way.

    ehild
     
  6. May 15, 2014 #5
    Cool, thanks!
     
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