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GR0877 Problem 23

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) Imagine the earth (S frame, x-axis is to the right and passes through the spaceships) and two spaceships approaching the earth from the left and right with equal speeds v. Let’s move to the reference frame of the left spaceship (system S', two systems are oriented the same way). Then, according to the velocity addition formula, the velocity of the right spaceship in this frame of reference is: \displaystyle u_r' = \frac{u_r - V}{1 - u_r V/c^2}, where u_r = -v is the velocity of the right spaceship in the S frame and V denotes the speed of S' relative to S, V = v. Thus, \displaystyle u_r' = \frac{-2v}{1+v^2/c^2}. But this is exactly the speed with which two spaceships approach one another (as seen from the reference frames associated with them). And we know from the length contraction formula that, if the relative speed between two frames is u, a stick at rest with respect to one reference frame is observed from the other reference frame to be contracted by the factor of \gamma = 1/\sqrt{1 - u^2/c^2} = l_0/l. The problem statement suggests \gamma = l_0/l = 1/0.6 = 5/3. Substituting u_r' into the equation for \gamma and working with units in which c = 1:
\displaystyle \frac{5}{3} = \frac{1}{\sqrt{1- \frac{4v^2}{(1+v^2)^2}}} = \frac{1+v^2}{1-v^2},
5 - 5v^2 = 3 + 3v^2, 2 = 8v^2, v = 0.5 or v = 0.5c as in choice (B).

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  1. oliver
    05.11.2012 at 21:33

    You can also do this with the Lorentz transformation in matrix form. So if g = gamma and B = beta are associated with the relative velocity between a spaceship and earth, then composing two matrices gives the transformation from one ship frame to the other. In other words, defining g_eff as the gamma associated with the relative motion of one ship to another ship, we have [g -Bg; -Bg g]^2 = [g^2 + (B*g)^2 -2Bg^2; -2Bg^2 g^2 + (B*g)^2] so that g_eff = g^2 + (B*g)^2. Now you know that g_eff = 5/3 (as in the original solution). So just solve g^2 + (B*g)^2 = g^2(1+B^2) = (1+B^2)/(1-B^2) = 5/3 to get B^2 = 1/4 => B = 1/2 (working in units of c = 1). Sorry for the mess. Not saying this way is better; it’s just another method for solving the problem.

  2. caspernorth
    30.04.2013 at 19:48

    S1, S2 = Spaceship 1 frame, Spaceship 2 frame
    fix the frame in s1 (thus earth and the other spaceship are to the right +ve x axis).

    By law of addition of velocities :
    u’ = (u-V) / (1-uV/c^2)

    { u’ = velocity of S2 in Earth frame : -ve
    u = velocity of S2 in S1 : -ve
    V = velocity of earth in S1 : -ve }

    u = 0.8c (from length contraction formula)
    u’ = V = v (same sign for v since both are approaching S1)
    .:. v = (0.8c – v) / (1 – 0.8v/c)
    solve quadratic for v (one solution gives 2c which is impossible, so the other one is correct which is 0.5c)

    v = 0.5 c ( put c=1 during calculations for simplicity)

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