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Reference frames for photon collisions

  1. Sep 26, 2007 #1
    Hi,

    My question probably has a simple answer, but I've been scratching my
    head over it a little too long so I thought I would ask it here. I
    have three initial photons involved in a collision with 4-momenta k1,
    k2 and k3. I have two reference frames:

    frame 1: the centre of mass frame of photons k1 and k2 (so that 3-momenta k1+k2=0)

    frame 2: the centre of mass frame of all photons k1, k2 and k3 (sothat 3-momenta k1+k2+k3=0)

    I want to find out what the relativistic beta v/c and gamma 1/Sqrt(1-(v/c)^2 are. So there seems no way to directly find the relative velocity of the two frames is as only photons are involved. I thought
    to use the expression for the relativistic doppler shift, but that doesn't seem appropriate.

    Can anyone help me out here?

    cheers.
     
  2. jcsd
  3. Sep 26, 2007 #2

    jtbell

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    Staff: Mentor

    Try using the Lorentz transformation for energy and momentum. That is, if an object (including a photon) has energy E and momentum p in one frame, then in another frame with relative velocity v (with respect to the first frame):

    [tex]p^{\prime}c = \gamma (pc - \beta E)[/tex]

    [tex]E^{\prime} = \gamma (E - \beta pc)[/tex]
     
  4. Sep 26, 2007 #3
    thanks, but [tex]\beta[/tex] and [tex]\delta[/tex] are what I'm trying to determine
     
  5. Sep 26, 2007 #4
    err I meant beta and gamma
     
  6. Oct 2, 2007 #5

    clem

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    Science Advisor

    Use: P=k3
    E=k1+k2+k3
    LT with \beta=p/E in the direction of k3.
     
  7. Oct 2, 2007 #6

    pervect

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    Staff Emeritus
    Science Advisor

    Note that you have only one variable to solve for, [itex]\beta[/itex]. By definition,

    [tex]\gamma = \frac{1}{\sqrt{1-\beta^2}}[/tex]

    You can (and will have to) invert this to solve for beta as a function of gamma.

    If I understand the problem correctly, you have 6 3-vectors as input? I.e. you have the components of k1 in frame1 and frame2, the components of k2 in frame1 and frame2, and the components of k3 in frame 1 and frame 2.

    I suppose you could call these k1 and k1', k2 and k2', k3 and k3'?

    First step: replace the 3-vectors with 4-vectors. The "missing" component, which is energy, can be found from |k1| = |k2| = |k3| = 0, i.e. E^2 - |p|^2 = 0.

    k1 + k2 must be related by a Lorentz transform to (k1'+k2'), because k1' is the Lorentz transform of k1, and k2' is the Lorentz transform of k2.

    In geometric units, we know that the components of (k1+k2) are (1,0,0,0). In standard units, this would be (c,0,0,0).

    The zeroeth. component of (k1'+k2') will be gamma in geometric units, or c * gamma in standard units.


    You apparently are not interested in solving for the direction of the Lorentz boost, just it's absolute value?
     
    Last edited: Oct 2, 2007
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