# Redshifted Photon Energy

A photon emitted from a receding source (Doppler redshift) has less energy when detected at an observer's location. Please explain the energy loss from the perspective of energy conservation.

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Consider the following scenario: I am on a train moving away from you. I throw a ball to you. The speed of the ball as measured by you when you catch it, is less than the speed of the ball as measured by me when I threw it. Where did the energy go?

This situation is precisely the same as the Doppler shift situation you describe. In both cases, there's no problem with energy conservation, because the energies in question are measured in two different reference frames. Energy conservation says that, in any given reference frame, the amount of energy doesn't change. It says nothing about how the energy in one frame is related to the energy in another frame.

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Thanks, I was just going to try to provide the voice of reason. Different observers measure different energies/frequencies/whatever when looking at the same phenomena. There is no physics in this, just accounting. –  user566 Feb 8 '11 at 19:32
Plus one, Ted. Best answer, without unnecessary stuff that only creates extra fog. This thought experiment is the photon counterpart of the very basic toy models where we may study what energy conservation means - so not understanding this situation means to understand nothing about energy conservation. The energy is conserved when one carefully uses a consistent inertial frame to measure it. Claiming that energy of a photon - or anything else - has to be the same in two different inertial frames isn't a disproof of energy conservation; it's a misunderstanding of the relativity of energy etc. –  Luboš Motl Feb 8 '11 at 19:49
+1 for that. Some time back I posted an answer to a related question that might be of interest here (with respect to the idea that conservation of energy doesn't apply between different reference frames). –  David Z Feb 8 '11 at 21:38
I'm not sure this analogy is helpful. The speed of light is a constant regardless of the frame of reference so the calculation of a photon's total energy (via wavelength) should be the same regardless of the frame of reference. Oh how I wish I'd taken math and physics in college. I'm very interested in cosmology but can't do the math myself. –  Kelly S. French Jun 29 '11 at 15:51
This answer is somewhat misleading. In the train example, there is the logical possibility that we could cover both the throw and the catch with a single frame of reference. In the case of a cosmological redshift, there is no such possibility; there is no Lorentzian frame of reference that can encompass both the emission and the detection of the photon. It's also misleading because it would tend to lead the OP to believe that energy is conserved in GR. It isn't. We have conserved scalar measures of mass-energy only in certain special types of spacetimes (static, asymptotically flat). –  Ben Crowell Jul 21 '11 at 17:29
Your reasoning about the recoil is completely incorrect in regards to what the questioner is asking. The statement about gravitational redshift is also incorrect. There is no such thing as the potential energy for a photon where $m=0$. –  user346 Feb 8 '11 at 19:30