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Stupid Special Relativity

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data
    The question goes: A meter stick with a speed of 0.8c moves past an observer. In the observer’s reference frame, how long does it take the stick to pass the observer ?

    2. Relevant equations
    Special Relativity: where T and L = proper time/length, t, l are not:

    t = T(gamma)
    l(gamma) = L


    3. The attempt at a solution
    So I calculate gamma = 5/3. Here's where I got messed:

    I figured there are two ways to approach the time, one way is to do
    v = l/T = l(gamma)/t. If you plug and chug with this though, you get the wrong answer, gamma is on the top.

    If you use v = L/t = l/[(gamma)t], you get the right answer, since gamma is in the denominator here but in the numerator with the other equation. What makes you decide to use this one over the other one?? Is there something I'm messing up and they should yield the same answer??

    Thanks,
    Ari
     
  2. jcsd
  3. Oct 10, 2011 #2
    You messed up in the first part since T = (gamma)*t. You get the same answer in both cases.
     
  4. Oct 10, 2011 #3
    Sorry for the non-latex. I'm looking at a definition right now for special relativity, and it says

    [ tex ] t = \frac{ \tau}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma \tau [ /tex ] where tau is proper time.

    Thus, by using this method, [tex]\gamma = 5/3[/tex], [tex]t = 5/3 \tau[/tex], or [tex]\tau = 3/5 t[/tex].

    Thus, (where L = proper length) in [tex]v = \frac{d}{t} = \frac{L}{\tau} = \frac{1}{(3/5) t}[/tex],
    we're going to end up with:
    [tex]t = \frac{5}{3v}[/tex], which will give you the wrong answer...what am I doing wrong??

    Why isn't my latex working??
     
    Last edited: Oct 10, 2011
  5. Oct 10, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Let's be clear about which frame is measuring a proper time.

    Let the rest frame of the meterstick be the unprimed frame. The length in that frame is L (of course). Let's call the time Δt. Note that measurement of Δt requires readings from two synchronized clocks (one at each end of the stick) so it is not a proper time.

    Let the frame of the observer be the primed frame. The length of the stick in that frame is L/γ. The time in that frame is Δt'. Note that Δt' is a measurement made on a single clock, so it is a proper time.

    The relationship between the times is Δt = γΔt'.

    You can use either frame to solve for Δt' and you should get the same answer.
     
  6. Oct 10, 2011 #5
    [tex] t = \frac{ \tau}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma \tau [/tex]

    There it is... the spaces inside the brackets were the problem.

    Can't help you with relativity :P
     
  7. Oct 10, 2011 #6
    Ahhh crap. Yeah makes sense. I stupidly assumed that L and T must correspond to the same frame, but whoever is making the measurement is T....

    Thanks.
     
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