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I think the title says it. Did expansion of the universe steal the energy somehow?

Qmechanic
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sonardude
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2 Answers2

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Energy isn't a nice concept in GR, so all I'm giving is an intuitive way of looking at it.

For gravitationally redshifted stuff: A photon has energy, thus it gravitates (as energy can gravitate analogous to mass from $E=mc^2$), thus it has some (negative) gravitational potential energy when on the surface of a planet. If it's emitted, its GPE eventually becomes 0. So, this increase in GPE had to come from somewhere: the photon's redshift gave the energy. It's pretty much the same thing that happens when you throw a ball up. It loses kinetic energy (slows down).

The GPE in relativity is basically related to the energy stored in spacetime curvature; in a complicated way that I don't know.

For a normally redshifted photon from a moving body: Energy need not be conserved if you swith frames. Energy is different from each reference frame.

See the answers to the question provided by Qmechanic above as well. Over there, they're talking about the entire universe, though, which leads to additional issues.

Manishearth
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  • You know, it's like why's it expanding? No one say's oh yes, of course, why wouldn't it be. – sonardude Feb 29 '12 at 02:28
  • You can write GR in so many ways that there's no rightish way to look at it. – sonardude Feb 29 '12 at 02:32
  • @sonardude That's why every way of looking at it is wrongish :P. – Manishearth Feb 29 '12 at 02:40
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    This is not wrongish, but correct. Still, many people will tell you that it is wrong, even though it is correct, because they are uncomfortable with the energy in GR, because it is a pseudotensor, which is only globally interesting, and only uncontroversial for asymptotically flat spaces. But whatever. +1. – Ron Maimon Feb 29 '12 at 04:58
  • @RonMaimon Didn't know that, thanks . You've managed to to point out that I'm wrong when I think I'm right, and right when I think I'm wrong. Wierd. :P – Manishearth Feb 29 '12 at 12:58
  • @RonMaimon No way, so, over cosmic times, energy gets tired. Is the packet going to spread on out? And what, the space-time is sort of just going back to flat? That is just weird. – sonardude Mar 01 '12 at 01:12
  • @RonMaimon I mean, the EM energy (distortion) is going back to flat as the wavep-acket stretches out. And, turns out it's giving it up to the space-time which is also going flat as the photon looses mass. Strange. – sonardude Mar 01 '12 at 01:19
  • _".. same thing that happens when you throw a ball up.." , right, but what if the energy of the photon is completely drained out? it can never fall again, so energy is lost. No PE, no conservation of energy? where did energy go? Thanks – bobie Nov 01 '14 at 11:31
  • @bobie It can't be completely drained out until it reaches infinity (or if it is beyond an event horizon). General relativity doesn't have a nice concept of "potential energy", however remember that a gravitational well has negative PE, so throwing an energetic photon up (and having it redshift into oblivion) conserves energy. – Manishearth Nov 01 '14 at 14:23
  • @bobie Asymptotic is asymptotic, regardless of the size. I never didn't accept that energy can be drained out. It can. However, again, the photon started with a net negative "potential energy", so everything evens out. – Manishearth Nov 01 '14 at 15:30
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The short answer is "yes". The energy lost from the photons is taken up by the energy in the gravitational field. Of course energy is a relative concept but if you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is

$E = Mc^2 + \frac{P}{a} - \frac{3a}{\kappa} (\frac{da}{dt})^2 = 0$

$M$ is the fixed mass of cold matter in the volume, $\frac{P}{a}$ is the decreasing radiation energy in the volume with $P$ constant, and the third term is the gravitational energy in the volume which is negative. The rate of expansion $\frac{da}{dt}$ will evolve in such a way that the (negative) gravitational energy increases to keep the total constant and zero.

For a more general discussion of energy conservation in general relativity see my paper http://vixra.org/abs/1305.0034

Philip Gibbs - inactive
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