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Compton scattering; conservation of momentum violated?

  1. Jan 12, 2015 #1
    While reading the derivation of the formula [itex] \lambda' - \lambda = \frac{h}{ m_ec}(1-cos(\theta)) [/itex] on Wikipedia, they point out that the momentum gained by the electron is larger than the momentum lost by the photon:

    $$ p_e=\frac{\sqrt{h^2(\nu-\nu')^2 +2h(\nu-\nu')m_ec^2}}{c} > \frac{h(\nu-\nu')}{c} = p_{\gamma}-p_{\gamma'} $$

    How is this not a violation of conservation of momentum?
     
    Last edited: Jan 12, 2015
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  3. Jan 12, 2015 #2

    ShayanJ

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    I don't think that's the right of way talking about it! When we consider SR in analysing phenomena, we should keep in mind that it is no longer true that energy and momentum are separately conserved, but it is the total 4-momentum [itex] P^\mu=(\frac E c, \vec p) [/itex] which is conserved. So such a comparison has no meaning and violates nothing!
     
  4. Jan 12, 2015 #3

    jtbell

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    Momentum is a vector quantity. The incoming and outgoing photons are in different directions. The difference in the magnitudes of their momenta is not a meaningful physical quantity, as far as I know.
     
  5. Jan 12, 2015 #4

    Vanadium 50

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    Shyan, what you wrote is wrong. It is not necessary to reply to every single question here, particularly if you don't know the answer. If the energy-momentum 4-vector (or any 4-vector) is conserved, each and every component is separately conserved.

    jtbell - and Wikipedia - has it right. Comparing the magnitudes of two vectors is, in general (and in this particular case) comparing them along two different directions. You don't expect equality here.
     
  6. Jan 13, 2015 #5

    jtbell

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    ...but the magnitude of the 3-momentum is not. (just to make this explicit)
     
  7. Jan 13, 2015 #6

    Vanadium 50

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    I don't think that helps. The magnitude of the 3-momentum component is conserved, since each of the three components is conserved separately. What we have here is a comparison between two 3-vectors pointing in two different directions. These are, as you said, not going to be equal. You would only get equality if you compared the total initial 4-momentum with the total final 4-momentum.
     
  8. Jan 13, 2015 #7

    jtbell

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    You're right, that was a bad way to put it. I was thinking of the fact that the magnitude of the 3-momentum is not "additive". That is, the sum of the magnitudes of two momenta does not generally equal the magnitude of the total momentum.

    Right, the magnitude of the total momentum is conserved. Before the interaction, the total momentum is the incoming photon's momentum. Afterwards, the total momentum is the sum of the outgoing photon's momentum and the recoiling electron's momentum. The magnitude of the sum is smaller than the magnitude of the outgoing photon's momentum.
     
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