# Photon energy & Energy lost

1. Apr 26, 2010

### Bjarne

Light (photons) that are reaching Earth after travelling billion of years, - the EM-spectre have become redshift.

This means that photons must be losing energy.

But energy can’t be lost, - so where is that “lost of photon-energy”?

2. Apr 26, 2010

### Filip Larsen

A short answer in the context of classical physics would be, that the red- and blue-shift of photons is due to the Doppler effect, which again is due to the relative speed between source and sink of the photon. If you in this context look at the energy and momentum transferred by a photon between two reference frames in relative motion it actually fits with the conservation law of energy and momentum.

Same can be said about gravitational red-shift (i.e. that no energy is lost), so that only leaves cosmological red-shift which is due to the expansion of space on cosmological distances, which I hope someone else can give a nice short answer on

3. Apr 26, 2010

### Ich

Oh, you can break cosmological redshift down to doppler and gravitational redshift, too. But the procedure becomes ambiguous if spacetime changes strongly with time - you can neither define what "at rest" means nor what the gravitational potential should be.

Maybe it's safer to say that energy conservation is a provable mathematical theorem that relies on the time invariance of the description of your system. If your description changes with time (as in cosmology), so may energy. Energy conservation is no longer valid.

A more comforting viewpoint seems to be that the lost energy goes into the gravitational field. It's true in a sense, except that you can't unambiguously define the energy of gravitation in GR.

4. Apr 26, 2010

### bapowell

I've also heard this explanation, but I've always been confused about just what the gravitational field is supposed to be in a homogeneous and isotropic cosmology.

5. Apr 26, 2010

### Ich

"Field" not a vector field like in Newtonian standard terminology. It's tensorial in GR, so not necessarily directional, thus isotropy is not a problem.

6. Apr 26, 2010

### bapowell

Thanks Ich, but that's not what I mean. I understand GR fine. But the gravitational field in GR is the gradient of the metric. This vanishes in a homogeneous and isotropic spacetime.

7. Apr 27, 2010

### Bjarne

OK, - this is what I thought, - it’s a dead end ?
I was speculated in the possibility: what when also distances are relative. This could solve the problem, - but of course also only speculation.

8. Apr 27, 2010

### Ich

If by "gradient of the metric" you mean the Christoffel symbols: they do not all vanish in a FRW metric.

9. Apr 27, 2010

### Ich

Distances are relative, as are velocities. That's why I could give three different "explanations" for the phenomenon.

10. Apr 27, 2010

### bapowell

Right. I guess my point is that there are no non-vanishing spatial gradients of the metric. The nonzero connection terms are all functions of time ($$\dot{a}/a$$ to be exact). I've always identified the Newtonian perturbation, $$\Phi({\bf x},t)$$ (eg from the definition $$g_{00} = 1 + 2\Phi$$) as giving rise to the gravitational field. Are you suggesting that we identify the nonzero connection terms with the gravitational field?

11. Apr 27, 2010

### Ich

Yes, that's what I meant with "Newtonian standard terminology". It's a 3D vector field, as a gradient of a scalar potential. Such things vanish necessarily in the homogeneous case.
I was struggling a bit with your terminology, and this was the closest thing to a "gradient of the metric". I think it's not unusual to think of it as the "gravitational field", analoguous to the Newtonian case. The important point is that you can't use spatial symmetry to argue that the "field", and with it the energy, vanishes.
Whatever, they appear in the http://en.wikipedia.org/wiki/Landau-Lifgarbagez_pseudotensor" [Broken], and that is what I had in mind with my statement concerning the "energy of the gravitational field".

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