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How is a photon emission energy balance maintained ?

If an electron is in motion across space at 260,000 kilometers per second, and it releases a photon in that same direction of travel, the photon will be in motion across space at the speed of 300,000 kilometers per second, thus there will be a difference of 40,000 kilometers per second.

If another electron that is traveling along side the first electron at 260,000 kilometers per second, and it releases a photon in the reverse direction, the photon will be in motion across space at the speed of 300,000 kilometers per second, thus there will be a difference of 560,000 kilometers per second.

In both cases each electron will lose energy when releasing a photon. Is the loss of energy equal in both of these cases ? ( assuming each photon is is identical in its properties of course )

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The easiest way to analyse this process is in the rest frame of the electron. In that frame, the photon emission is symmetrical and the photons have the same energy. If we then boost to back to the lab frame then since energy is not Lorentz invariant, one photon would indeed increase in energy and the other would decrease in energy. They of course move at the same speed, but now they have different frequencies. This is just the relativistic Doppler effect.

Explicitly we have (we denote the electron frame by $e$ and a the lab frame by $lab$), \begin{equation} f_{lab}=\sqrt{\frac{1\mp v/c}{1\pm v/c}}f_e \end{equation} where $v$ is the speed of the electron and $f $ is the frequency of each photon. The top sign is for the photon moving in the direction of the electron and the bottom is for the photon moving opposite to the direction of the electron.

Suppose for example $v=0.8c$. In that case the photon moving in the direction of the electron has a frequency 6 times the frequency of the photon moving in the direction opposite to the electron.

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  • $\begingroup$ How can we bring frequency into the picture when in both cases we are dealing with only one photon? $\endgroup$ – Sean Feb 4 '14 at 10:33
  • $\begingroup$ By frequency here, we don't mean frequency of emission, but frequency of oscillation. If it is more intuitive you can just as well think of wavelength, $\lambda=c/f $. $\endgroup$ – JeffDror Feb 4 '14 at 10:53
  • $\begingroup$ I assume you mean the oscillation of the electron ? $\endgroup$ – Sean Feb 4 '14 at 11:53
  • $\begingroup$ No, I mean of the photon. If frequency bothers you, just think of the photon wavelength (however a photon can be described by either) $\endgroup$ – JeffDror Feb 4 '14 at 12:03
  • $\begingroup$ If I had two identical lasers of equal frequency and these lasers were pointing in opposite directions, and my frame of reference is also moving across space at 260,000 kps in one of those two directions, from my point of view the light frequency is still the same for both lasers. To external observers, the light frequencies would not be the same due to an increase in energy in one direction, and a decrease in energy when emitted in the opposite direction. If I rotate around the lasers I somehow still see the frequencies as equal. Not to drag on, but I was just wondering how this is so. $\endgroup$ – Sean Feb 4 '14 at 12:50

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