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Photon Doppler Shift?

  1. Mar 31, 2004 #1
    Photon Doppler Shift???

    I am a practicing EE who has used the standard Doppler shift equation for thirty years in radar design; therefore, I am not disputing the correctness of the equation but the "explanation" that is so often given in text books. The general explanation always discusses relative velocity between source and receiver. I have always wondered how the "photon" that was emitted or absorbed "knew" the relative velocity in question. The pictorials always describe the wavefronts being bunched closer or further apart as the source/receiver moves. But at the time of emission or absorption how does the photon know its velocity?? And relative to what?? The effect is such that on the relative velocity is important but the explanation of how this info is imparted to the photon leaves a lot to be desired. If anyone on this forum has a good reference to this I would greatly appreciate hearing about it. My own thoughts are that there is a background FOR to which all source/receiver velocities are referenced. It is the difference in the effects that the velocity relative to this FOR has on the photon that is eventually made known at the receiver. The FOR maybe unnecessary from the point of view of SR but the explanantion of how this relative velocity info is imparted to the photon is surly lacking. Also could this be a case for the quantum mechanical argument that it is only observables that are important in a theory since one can really only measure energy of the photon directly??
  2. jcsd
  3. Mar 31, 2004 #2


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    Light travels at c in both the transmitter's and the receiver's rest frames. It is the transmitter that "knows" to emit at c, and it is the receiver that "knows" to receive at c. Don't blame the poor innocent photon. :wink:

    There are hundreds of treatments by hundreds of authors on common theoretical applications of SR. Try Wolfgang Rindler's "Relativity: Special, General, and Cosmological" (2001) I'm pretty sure he says something about the doppler shift.

    Basically, an ether? This is not believed to be the case. And I'm pretty sure that the doppler shift formula that you have is not compatable with that notion. Check it against the first semester physics doppler shift for sound, which does propagate through an absolute FOR.

    Actually, the way you described it, not only is it unecessary, but completely imcompatable with the fundamental premise of SR.

    I don't think so, but that is just MHO. In the QM context, observable indicates a Hermitian operator, in other words, having a measurement that results in purely real vs. complex/imaginary numbers. I don't think that this is the same kind of "observable" that you are talking about regarding the transmission and reception of E&M radiation.

    Try this:

    I'm assuming that you're familiar with Maxwell's equations. Well, you can get a wave equation from them that has a propagation constant that shows up. These equations are believed to be valid regardless of how fast you are traveling WRT some arbitrary FOR, with the condition that you are approximately in an inertial frame. Since this is the case, that constant is assumed to be just that, a constant.

    So, applying this idea to the transmitter, it uses Maxwell's equations with that constant to transmit the E&M radiation. The receiver is sitting there and all of a sudden starts experiencing E&M waves. These satisfy Maxwell's equations at the receiver, so, they also have that same constant.

    That propagation constant is c.
    Last edited: Mar 31, 2004
  4. Apr 1, 2004 #3
    I believe you're right. For a full set of Doppler equations that explain this in terms of an ether frame of reference, see section 9.3 / equation set 9.6

    If the motion of the source relative to the ether is 0, then the equations become those of SR. I would be interested to know if using these equations in your work gives better results or the same results as SR?
    If you need help with the time dilation terms, let me know.
    Last edited by a moderator: Apr 20, 2017
  5. May 15, 2004 #4

    You are absolutely correct.

    Are you aware that under the Doppler Law, there are TWO causes of redshifts and blueshifts?

    1) “Stretched out” or “compressed” waves due to the motion of the emitter.


    2) A faster or slower relative speed of the wave, relative to the observer, due to a moving observer.
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