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Special Relativity 4-vector problem - Doing my head in!

  1. Aug 12, 2009 #1
    1. The problem statement, all variables and given/known data

    OK - the problem is thus:

    In an inertial frame two observers (called a & b) travel along the positive x-axis with velocities Va and Vb. They encounter a photon traveling in the opposite x-direction. Without using the Lorentz transformations, show that the ratio of the energies of the photon observed by observers a & b is given by:

    [tex]

    \frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}

    [/tex]


    2. Relevant equations

    [tex]

    E_o = -p_o.u_o

    [/tex]

    where E_o is the observed energy of a photon with 4-momentum p_0, by a given observer moving with 4-velocity u_o.

    I set up the 4-velocities of observers a & b as:

    [tex]

    u^\alpha_a = (\gamma_a,\gamma__a V_a ,0,0)

    [/tex]
    [tex]

    u^\alpha_b = (\gamma_b,\gamma__b V_b ,0,0)

    [/tex]

    and the 4-momentum of the photon as:

    [tex]

    p^\alpha = (p^t,p^x, 0,0)

    [/tex]


    3. The attempt at a solution

    With these 4-vectors set up, the energies of the photon for each observer should just be the dot product of each 4-velocity with the negative of the 4-momentum of the photon - i.e:

    [tex]

    E_o = -p_\alpha .u^\alpha = - \eta_\alpha_\beta p^\alpha .u^\alpha
    [/tex]

    where [tex] \eta_\alpha_\beta [/tex] is the metric.

    so

    [tex]
    E_a = \gamma_a p_t + \gamma_a V_a p_x \\
    [/tex]

    [tex]
    E_b = \gamma_b p_t + \gamma_b V_b p_x
    [/tex]

    Then, I used the fact that for a photon,

    [tex]
    p^\alpha.p^\alpha = 0 \\
    [/tex]

    [tex]
    i.e. -p_t^2+ p_x^2 = 0 \\
    [/tex]

    so

    [tex]
    p_t = p_x
    [/tex]

    and we get:

    [tex]


    E_a = \gamma_a p_t + \gamma_a V_a p_t = \gamma_a p_t (1+V_a) \\
    [/tex]

    [tex]
    E_b = \gamma_b p_t + \gamma_b V_b p_t = \gamma_b p_t (1+V_b) \\
    [/tex]

    so

    [tex]
    \frac{E_a}{E_b} = \frac{\sqrt{1-V_b^2}(1+V_a)}{\sqrt{1-V_a^2}(1+V_b)}
    [/tex]

    ...which is where I'm falling down. As I said before, I'm after the relation

    [tex]

    \frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}

    [/tex]

    Can anyone see what I'm doing wrong? Have I made a mistake somewhere, or is there some mathematical trick to take me further from where I am to the required answer? I've stared at this for a while now and can't work out why it's falling down.

    Any help would be greatly appreciated.
     
    Last edited: Aug 12, 2009
  2. jcsd
  3. Aug 12, 2009 #2
    Actually, I think I botched the dot product earlier between the 4-vels and the 4-mom. I think it should give me:

    [tex]
    E_a = \gamma_a p_t - \gamma_a V_a p_t = \gamma_a p_t (1-V_a) \\
    [/tex]

    [tex]
    E_b = \gamma_b p_t - \gamma_b V_b p_t = \gamma_b p_t (1-V_b) \\
    [/tex]

    as opposed to:

    [tex]
    E_a = \gamma_a p_t + \gamma_a V_a p_t = \gamma_a p_t (1+V_a) \\
    [/tex]

    [tex]
    E_b = \gamma_b p_t + \gamma_b V_b p_t = \gamma_b p_t (1+V_b) \\
    [/tex]

    i.e. with minus signs, not the plusses I have before. But this seems to take me even further away from where I want to be. Arrg!

    Help!
     
  4. Aug 12, 2009 #3

    Avodyne

    User Avatar
    Science Advisor

    Nothing is wrong. These two expressions are equal. It's easier to see this by squaring each expression, then using [itex]1-V^2 = (1-V)(1+V)[/itex].
     
  5. Aug 12, 2009 #4
    Ah - thanks heaps. I'll work through it!

    Cheers.
     
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