# How is a photon emission energy balance maintained?

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 )

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.
• 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, $\lambda=c/f$. – JeffDror Feb 4 '14 at 10:53