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Momentum of photons in CM frame.

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Two photons in the laboratory system have frequencies [itex]\nu_1[/itex] and [itex]\nu_2[/itex]. The angle between the propagation directions is [itex]\theta[/itex].

    a) Write down the expressions for the total energy and momentum of the photons in the laboratory system.

    b) Find the photons’ frequency in the center of mass system.

    c) Is it allways possible to find a center of mass system for the photons?


    2. Relevant equations
    Four momentum
    [tex] \underline P = (E/c, \vec p)[/tex]

    Lorentz transformation
    [tex] P'^\mu = \Lambda^\mu_{\ \ \nu} x^\nu [/tex]

    Duality equations
    [tex] p = h/\lambda \ \ \ E = h\nu [/tex].

    3. The attempt at a solution
    a)
    I've taken the x-axis in the laboratory system to be parallel with photon 2 such that

    [tex] \vec p_1 = \frac{h\nu_1}c(\cos \theta, \sin \theta, 0) \ \ \vec p_1 = \frac{h\nu_2}c(1,0,0)[/tex]

    The total four momentum would be

    [tex] \underline P = (E_{tot}/c, \vec p_{tot})[/tex]

    where
    [tex] E_{tot} = \frac{h(\nu_1 + \nu_2}c[/tex].

    b) To find the momentum in the CM system I would have to do a lorentz boost into that system which is defined so that the spatial total momentum is zero. I do however have problems in finding the correct [itex]\beta = v/c[/itex] for this transformation.
    I know the velocity of this frame would have to satisfy
    [tex] \vec v = \frac{(\vec p_1 + \vec p_2)c^2}E[/tex]

    but that frame would not have it's x-axis parallel with my original one, so I do not
    know how to find the correct transformation. Have I chosen my axis badly in this problem
    or is there a clever way out?

    Thanks for any help.
     
  2. jcsd
  3. Mar 16, 2012 #2

    vela

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    Try aligning the x-axis to point along ##\vec{p}_1 + \vec{p}_2##. You'll have to find the angle each photon makes with the axis, but that should be straightforward.
     
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