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Dopler Shift and Photon Number

  1. Jun 14, 2008 #1
    Here's something that I recently thought about:

    If we assume a Doppler shifted EM wave retains the amplitude of the unshifted wave, then we must be creating/annihilating photons. This seems consistent with the conservation of energy; the removal of several photons removes energy from the system but this energy is conserved by the increase in energy of the remaining photons, as would be the case of a blue shift.

    Does this seam reasonable?

    I realize that number conservation of photons is never a requirement, but it originally bothered me that photons can just come in and out of existence in the manner described above. My explanation for this is that during a Doppler shift, a single photon is either joining with several others or splitting into several others in a manner that conserves energy.

    Does this seem reasonable?
  2. jcsd
  3. Jun 14, 2008 #2
    Probably a good idea to look into coherent states and bogoliubov transformations.
  4. Jun 14, 2008 #3


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    The Doppler shift does not change the number of photons.
    The frequency, energy and momentum of each photon changes.
  5. Jun 14, 2008 #4
    If the number of photons does not change, then a change in wavelength will induce a change in amplitude. For the amplitude to remain the same, the number of photons must change - hence my original post.

    lbrits, I will be looking into it.....
  6. Jun 14, 2008 #5


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  7. Jun 15, 2008 #6


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    Why do you assume the amplitude won't change? The E and B fields change in a Lorentz transformation.
    Consider light so weak that there is only one photon.
    Could a small change in frequency add or subtract a photon?
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