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Homework Help: Spec Rel: Proof concerning IF through Minkowski diag

  1. Nov 14, 2008 #1

    phz

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    1. The problem statement, all variables and given/known data
    This problem is from "Relativity" by Rindler, second edition, problem 3.4:
    Use a Minkowski diagram to establish the following result: Given two rods of equal length [tex]l_1[/tex] and [tex]l_2[/tex] ([tex]l_2 < l_1[/tex]), moving along a common line with relative velocity [tex]v[/tex], there exists a unique inertial frame [tex]S'[/tex] moving along the same line with velocity [tex]c^2 \frac{l_1-l_2/\gamma (v)}{l_1 v} [/tex] relative to the longer rod, in which the two rods have equal lengths, provided [tex] l^2_1 (c-v) < l^2_2(c+v) [/tex].


    2. Relevant equations
    [tex] \tan \Theta = \frac{v}{c} [/tex] : (1)


    3. The attempt at a solution
    I have denoted the velocity of [tex]S'[/tex] relative the rod [tex]l_1[/tex] as [tex]v'[/tex]. By use of (1) I identify that [tex]tan \Theta = \frac{v'}{c} = c \frac{l_1-l_2/\gamma (v)}{l_1 v} [/tex] : (2). When I saw this I thought I should be able to find a trigonometric relation in my diagram satisfying that relation.

    2dt0vlv.jpg
    I drew the diagram with [tex]l_1[/tex] along the principal x-axis, identified a side with length [tex]l_1-l_2/\gamma(v) [/tex] in the figure, found through similarity that [tex] \Theta [/tex] was in that triangle and calculated through Pyth. theorem the cathetus remaining (denoted [tex]X[/tex] in the figure) to be [tex]X = l_2 \sqrt{ 1 - \frac{1}{\gamma (v) ^ 2} } = l_2 \sqrt{ 1-(1-v^2/c^2) } = l_2 \frac{v}{c} [/tex]. Had it been [tex] l_1 \frac{v}{c} [/tex] I would have been done I believe since that would have shown (2). What am I missing?

    I also don't see the what effect the last constraint [tex] l^2_1 (c-v) < l^2_2(c+v) [/tex] has on the problem.

    EDIT: Right now I'm thinking that I am using the Minkowski diagram incorrectly. When trying to find the length of [tex]l_2[/tex] on the principal x-axis I shouldn't draw a line perpendicular to the x-axis but instead parallell to the time axis corresponding to the inertial frame in which [tex]l_2[/tex] is at rest, right? That makes my geometry wrong, and more complicated I guess.

    EDIT 2: But when I'm looking at the figures in the book it seems like I interpretated the diagrams correctly the first time, and I'm back at my previous problem again. Or more precisely, I'm more lost than before. Is it the calibrating hyperbolas I am missing? How do they affect my drawing?

    EDIT 3: If I calculate my [tex]X[/tex] another way, by denoting the angle between [tex]l_1[/tex] and [tex]l_2[/tex] as [tex]\alpha[/tex] and using that [tex]\tan \alpha = \frac{v}{c} [/tex] I get [tex]X = \frac{v l_2}{c \gamma(v)}[/tex] which gives [tex] v' = c^2 ( l_1 - \frac{l_2}{\gamma(v)} ) \frac{\gamma(v)}{v l_2 } [/tex], a different result. In this case, if [tex]\frac{l_2 }{\gamma (v) } = l_1 [/tex] the given expression for [tex]v'[/tex] would have been fulfilled. The fact that I get different results depending on which relation I use to calculate [tex]X[/tex] clearly indicates that I have misunderstood something.
     
    Last edited: Nov 15, 2008
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  3. Nov 16, 2008 #2

    phz

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    I used the Minkowski diagrams incorrectly. The projection of [tex]l_2[/tex] onto [tex]l_1[/tex] wasn't perpendicular but with an angle corresponding to the time-axis for [tex]l_2[/tex], not included in my image above. At first I thought this would lead to a contradiction when I tried to project [tex]l_2[/tex] onto [tex]x'[/tex], but that was because I didn't consider the calibrating hyperbola.

    I believe the last constraint is just an expression that says that the relativistic speed limit should be obeyed.
     
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