Redshifted Photon Energy

Question

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.

Related: http://physics.stackexchange.com/q/7060/2451 and links therein. — Qmechanic, Mar 29 '15 at 21:17
Does it seem relevant whether you take on one single photon or a wave, a bunch of? Counterintuitively: yes, it does. As the source of the photons as a wave moves away - however, any single photon does not,it does the opposite, by the way. — Peter Bernhard, Oct 29 '22 at 17:35

score 19 · Accepted Answer · answered Feb 08 '11 at 19:27

19

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.

answered Feb 08 '11 at 19:27

Ted Bunn

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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. – Feb 08 '11 at 19:32
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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 08 '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 08 '11 at 21:38
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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
The speed of light is the same in different reference frames, but both a photon's energy and its wavelength are different in different frames. Indeed, that's the whole point! – Ted Bunn Jun 29 '11 at 15:54
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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). – Jul 21 '11 at 17:29
It's true that, for photons that travel cosmological distances, there is no Lorentz frame that encompasses both emission and reception. But for small distances there is, and we can (and do!) talk about the cosmological redshift over small distances. Even for large distances, you can (and, I would claim, you should) understand the redshift as being built up incrementally as many infinitesimal redshifts, each one of which is precisely analogous to the situation I describe. (If you want to know what I think about this in annoying detail, http://arxiv.org/abs/0808.1081 )

Ted Bunn

Jul 21 '11 at 17:49

You're right that energy is not conserved in GR. I don't think that my answer suggests otherwise, but there's no harm in saying so explicitly.

– Ted Bunn Jul 21 '11 at 17:50

score 8 · Answer 2 · answered Feb 08 '11 at 18:59

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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.

answered Feb 08 '11 at 18:59

Zo the Relativist

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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 $m=0$. – Feb 08 '11 at 19:30
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@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. – Zo the Relativist Feb 08 '11 at 19:58
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My apologies @Jerry. I was hasty in judging your answer. Also, my own answer was way off target. – Feb 08 '11 at 20:21
@space_cadet: no worries. :) – Zo the Relativist Feb 09 '11 at 05:28
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. – Jul 21 '11 at 17:21
@Ben Crowell: "the gravitational field vanishes by symmetry"--no. The dust-equation of state FRW solution is identical to the solution for a finite size collapsing dust star, for example. One wouldn't say that the gravitational field vanishes THERE. The FRW solution is not minkowski space, so I dont' see how you can argue that the gravitational field vanishes. – Zo the Relativist Jul 23 '11 at 14:17

score 3 · Answer 3 · answered Nov 16 '17 at 13:56

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Redshift is due to expansion of the universe. Not just the space, but space time. So it's not just the space between the source and the observer which is stretched. The light itself is also stretched. This means a ten minute burst of light will be stretched to a 11 minute burst of light. (Just an example). The extra minute of light means a reduction in intensity, like stretching a piece of putty makes it thinner. The total energy remains constant but spaced over a longer period.

answered Nov 16 '17 at 13:56

The volume stretches, the energy remains the same, but not occupies larger volume, energy density is reduced. – Ziezi Nov 16 '17 at 14:48
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"but now occupies larger volume" . Yes, that's exactly what I meant. – Nov 16 '17 at 16:08

Redshifted Photon Energy

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