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Photon Energy when Doppler Shifted!

  1. Apr 13, 2005 #1
    I don't know how to understand doppler shifts and yet conserve energy.

    Consider a red photon is emitted by a torch. This photon is let free to travel through space. Any observer at rest relative to the torch (which by now may be millions of miles away) will see it as a red photon.

    Another observer aproaching the torch from a distance at a high speed will see this photon as blue shifted; let's say as a UV ray. This observer will conclude that this photon has enough energy to cause a photo-electric emission of a Sodium metal electron.

    Still another observer receeding form the torch at high speed will conclude that this photon, severely shifted into the red, say an IR photon, cannot bring about the photo-emission of an electron from the surface of a piece of Sodium.

    Questions:
    1) It seems that whether or not a photo-electron is emitted depends on which frame of reference you are in....i.e. whether the freed electron exists outside from the metal or not depends on how the incident photon looks...UV..ok!. IR no effect. Well, the electron can and can't exist freely simply by virtue of what frame of reference we look at a photon from.

    Worse still; what about using that damned photon for electron pair creation? If I view it as a gamma ray, I'll get the pair creation...not so if I view it as IR!! We must all agree, no matter what our inertial frames of references are to each other, that the electron-positron pair either does or doesn't exist!!!

    2) The photon energy will differ from one observer to another because of the relative motion between observers. This is due to the Doppler effect for light. Where does the photon get it's extra energy from if it's seen as UV; or where does the photon loose it's energy to if it absorbed as IR?
     
  2. jcsd
  3. Apr 14, 2005 #2
    ..or more simply stated..

    If the photon's energy depends on it's frequency; then we can turn a red photon in to a UV, gamma, IR or radio wave photon merely by changing our frame of reference to a long departed photon source which can be considered, to all intents and purposes, out of the picture.

    This means we can make the energy content of a photon anything we like just by viewing it in different reference frames. So much for the conservation of energy??

    I'm obviously missing some essential element in my understanding; but for the life of me I can't see what it is.
     
  4. Apr 14, 2005 #3

    Ich

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    Energy is not conserved when changing reference frames; neither is momentum.
    The trick is to do your calculations in only one frame; you will find that the change of energy adds up to the same amount in every system - enough to set an electron free.
     
  5. Apr 14, 2005 #4

    Chronos

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    You need to adjust all the parameters to get the right result. A redshifted photon is both distance and time dilated. When you view a redshifted light source, you are viewing it in slow motion. The individual photons have less energy, but you receive more of them.
     
  6. Apr 14, 2005 #5

    jcsd

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    No, the explanation is simpler (photon number AFAIK is Lorentz invariant, plus as a photn of a given frequency has a fixed energy you couldn't use this scheme to make energy Lorentz invaraint), as Ich indicates energy is a frame-dependent property, so the total energy of an isolated system is not going to be the same in two frames. The conservation of energy only demands that the total enrgy of an isolated system be constant in all inertial frames (i.e. the energy of an isolated system some inertial frame is not a function of time), not that it should be the same value in all inertial frames).

    The frame dependcence of energy is not peculair to relativty either, infact if you could of asked the same question about Doppler shift in a non-relativstic context (well actually looking at your post you didn't mention relativty specifcally, but I assume it was relatvity you were thinking of as this is the relativty forum) and you would of gotten exactly the same answer.
     
    Last edited: Apr 14, 2005
  7. Apr 14, 2005 #6

    Chronos

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    Let's forget about photons for the moment. Stretching space is the same thing as stretching time. One minute of light energy emitted by a Z=1 redshifted object will take two minutes to pass an observer at Z=0. The emitting object will only appear to be half as bright, but the image will persist twice as long.
     
  8. Apr 14, 2005 #7

    jcsd

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    Ignore power, as that needlessly complicates things, the total enrgy received by the two obsrevers from the source is different.
     
  9. Apr 15, 2005 #8

    Ich

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    The question was about a single photon.
    When you use m²=E²-p², and calculate the change in m (preferably a small dm), it will always be equal to hf, with f as seen from the absorbing body.
     
  10. Apr 15, 2005 #9

    jcsd

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    m is zero for photons anyway, so E = p in natural units, but I don't see your point as f is not Lorentz invariant. Energy is not a Lorentz scalar (infact it's a component of the momentum 4-vector for which the associated Lorentz scalar is mass)
     
  11. Apr 16, 2005 #10

    Ich

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    m, E, and p refer to the absorbing body. It´s (rest)mass increases by dm=h*f0, with f0 being the photon´s frequency in the frame of the absorber. dm is invariant.
     
  12. Apr 16, 2005 #11

    jcsd

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    Why do you think that is correct? Remember momentum as well as energy is conserved.
     
  13. Apr 16, 2005 #12
    Red shift and blue shift arise from time dilation and conservation of momentum considerations.

    juju
     
  14. Apr 17, 2005 #13
    I guess I still need to know that if this photon in one frame seems to have enough energy to create a proton-antiproton pair; where did this energy come from if the original photon was emitted by a mere red torch?
     
  15. Apr 18, 2005 #14

    jcsd

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    This is precisely why single free photons cannot create a proton-antiproton pair.
     
  16. Apr 18, 2005 #15

    Ich

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    I think it´s correct because
    1 I did the calculation and I think I did not mix up all those signs
    2 it is the only answer consistent with the principle of relativity: it is allowed to view things from the absorber´s frame, and any calculation in a different frame must yield the same result.
     
  17. Apr 18, 2005 #16

    Ich

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    As jcsd said, that´s impossible. You need a second particle to absorb the excess momentum. And then the required energy is defined in the frame of this particle (at least if it´s big enough).
     
  18. Apr 18, 2005 #17

    jcsd

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    1. It can't be true, the actual formula is: dm = sqrt(2Mh*f), (where M is the intial mass of the object, f the frequency of the photon in the frame which the absorber is initally at rest in and h is Planck's constant, all in units such that c = 1).

    1) The reason your formula is not correct is that your fromual cosnerves energy, but it doesn't conserve momentum (remember the photon has a momentum of h*f).

    2) As enegry and momentum are the tiem and spatial compenents repsectively of the same four-vector you find that not only is moamntum not conserved, but enrgy is not conserved in all frames either if your formula was correct.
     
    Last edited: Apr 18, 2005
  19. Apr 18, 2005 #18
    Then the total energy is conserved. That means that the sum of the energy of the torch plus the energy of the photon remains unchanged.
    It appears to mean that you're assuming that the potential energy of the sodium atoms remains constant. Why?
    Do not dismiss that which is emitting both photons. There must be something which emitted them. In this case you chose sodium (torch).
    That's called "pair production." To create two photons one has an atom around to take up the difference in the momentum and energy that is required to balance the energy/momentum books. The atoms do this and thus act as a sort of catalyst for pair production. So the atom starts moving and two photons are created.

    Pete
     
    Last edited by a moderator: May 2, 2005
  20. Apr 19, 2005 #19
    jcsd...are you saying one photon would not conserve together the momentum-energy? What about a pair of photons? Can they do it?

    Do these two photons then need to be viewed from a frame in which the newly created pair would have a zero COM? That would overcome my problems as together they would need the required total energy; even if the two photons viewed in a frame seeing them with different frequencies would have one red shifted and the other blue shifted.
     
  21. Apr 19, 2005 #20

    Ich

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    Your formula dm = sqrt(2Mh*f) is incomplete, it must read:
    M+dm = sqrt(M²+2Mhf (+(hf)²-(hf)²) ); <-- only in M´s rest frame!
    using M>>hf (and sqrt(1+x)~=1+x/2) yields
    dm = hf.
    1) Momentum is conserved, but in M´s frame does not contribute to the increase in rest mass.
    2) In different frames neither E nor p stay the same, but mass (which is the length of the four vector) does.
     
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