# photons in expanding space: how is energy conserved?

If a photon (wave package) redshifts (streches) traveling in our expanding universe, is it's energy reduced? If so, where does it go?

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The same question has been asked several times, in different guises, on this website. Please see: physics.stackexchange.com/q/1327/829 physics.stackexchange.com/q/2597/829 physics.stackexchange.com/q/10309/829 physics.stackexchange.com/q/296/829 The answer basically is: you need to be careful what you mean by "energy conservation" in the context of general relativity. –  Willie Wong Aug 15 '11 at 15:34
The questions are mostly about energy conservation on cosmological scale, I was wandering about the local phenomenon - though it seems to be the matter of quantum-mechanic and gravitational unification :/ –  troyaner Aug 15 '11 at 17:24
This should be closed. It's an exact duplicate of the questions Willie Wong has linked to. –  Ben Crowell Aug 16 '11 at 1:09
Since Marek's post in 1327 covers at least 95% of this question, second the move to close. Short addendum: there is no local coordinate-independent measure of gravitational energy, nor can there be one, because its existence would violate the equivalence principle. So the answer is just "it isn't!" –  Stan Liou Aug 16 '11 at 5:06
Ted's answer below seems to address the matter of local definitions more squarely that the answers I read on 1327, and I am tempted to leave this open just for that reason. However, it is more the statement of the answer rather than the content which is different so I could still be talked into using my diamond power close. Thought? –  dmckee Aug 16 '11 at 16:58

Since you say you're talking about what happens locally (in a small volume), I'll answer from that point of view. The usual formulation of energy conservation in such a volume is that energy is conserved in an inertial reference frame. In general relativity, there are no truly inertial frames, but in a sufficiently small volume, there are reference frames that are approximately inertial to any desired level of precision. If you restrict your attention to such a frame, there is no cosmological redshift. The photon's energy when it enters one side of the frame is the same as the energy when it exits the other side. So there's no problem with energy conservation.

The (apparent) failure of energy conservation arises only when you consider volumes that are too large to be encompassed by a single inertial reference frame.

To be slightly more precise, in some small volume $V=L^3$ of a generic expanding Universe, imagine constructing the best possible approximation to an inertial reference frame. In that frame, observers near one edge will be moving with respect to observers near the other edge, at a speed given by Hubble's Law (to leading order in $L$). That is, in such a frame, the observed redshift is an ordinary Doppler shift, which causes no problems with energy conservation.

If you want more detail, David Hogg and I wrote about this at considerable (perhaps even excessive!) length in an AJP paper.

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I see, so photon did not actually lost any energy, but it is measured in a different reference frame. –  troyaner Aug 16 '11 at 16:01
@troyaner: Saying that the photon "did not actually lost any energy" is not really right. It assumes there is some way to talk about conservation of energy nonlocally, and there just isn't. –  Ben Crowell Aug 17 '11 at 20:31
If measured in comoving coordinates the photon actually lost its energy. –  Anixx Aug 19 '11 at 7:35
That's true. The quantity energy-as-measured-in-comoving-coordinates is not a conserved quantity. Personally, I'd rather not interpret that sentence as meaning that energy is not conserved, but rather as meaning that energy-as-measured-in-comoving-coordinates is not a concept that rightfully deserves to be considered as "energy." –  Ted Bunn Aug 19 '11 at 13:48