Why does the Doppler-shifted photon frequency change the observed energy loss?

[Pandemonium]

Consider a situation in which a light source is stationary with respect to an observer, A. This source emits n photons of frequency f, each of energy E = hf, towards A. Hence, A will be able to detect the energy loss in the source, which is E = nhf.

Now, this light source is moving with respect to A, and emits n photons of frequency f just like before. However, due to the Doppler Effect, the apparent frequency in which A detects is different, say v. As a result, the energy loss in the source detected at A will be E = nhv, and not E = nhf, which should be the actual energy loss.

How can this be explained? And is this consistent with the wave theory of light?

 

[selfAdjoint]

You have done this all in a Newtonian framework. But because photons move at c you have to do it with the relativistic contractions/dilations. When you do this (there are many sites on the web that do the calculations, or you could read Spacetime Physics by Wheeler and Thorne), it all works out. Relativity provides a consistent theory of the doppler effect.

 

[dhris]

Well, what you are pointing out is simply that the energy a particle has depends upon the reference frame in which it is measured. There is no "actual" energy loss of the emitter just as there is no "actual" time that an event occurs, but rather these quantities must be related to a particular reference frame.

dhris

 

[Pandemonium]

Thanks for the help, guys!