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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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Related: and links therein. – Qmechanic Mar 29 '15 at 21:17
up vote 12 down vote accepted

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

If it is a gravitational redshift, to a first, non-rigorous order, the energy loss is due to the fact that it is moving in a gravitational field, and thus is gaining potential energy while losing kinetic energy.

If it is a redshift due to the actual motion of the object, then the energy lost in the redshift is imparted to the object doing the emitting since energy and momentum are conserved in the emission process--it is an energy transfer due to recoil.

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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
@space_cadet: classically, that's true. In GR, however, it is not--you get a perfectly well-defined "gravitational potential" in, for example, the Schwarzschild solution for null geodesics--the energy lost by photons in a gravitational field was one of Einstein's starting thought experiments for GR, actually. And the recoil effect completely applies, because if you ignore the recoil effect, neither energy nor momentum are conserved in the emission process. – Jerry Schirmer Feb 8 '11 at 19:58
My apologies @Jerry. I was hasty in judging your answer. Also, my own answer was way off target. – user346 Feb 8 '11 at 20:21
@space_cadet: no worries. :) – Jerry Schirmer Feb 9 '11 at 5:28
The first paragraph is not quite right. In a homogeneous cosmological spacetime, for an observer at rest relative to the Hubble flow, the gravitational field vanishes by symmetry. The second paragraph is true but not relevant. We all agree that energy is conserved in the small patch of spacetime surrounding the emission of the photon. The issue is what happens between that time and the time when the photon is received at a cosmological distance from the source. – Ben Crowell Jul 21 '11 at 17:21

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