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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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3 Answers 3

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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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    $\begingroup$ 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. $\endgroup$
    – user566
    Feb 8, 2011 at 19:32
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    $\begingroup$ 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. $\endgroup$ Feb 8, 2011 at 19:49
  • $\begingroup$ +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). $\endgroup$
    – David Z
    Feb 8, 2011 at 21:38
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    $\begingroup$ 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. $\endgroup$ Jun 29, 2011 at 15:51
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    $\begingroup$ 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). $\endgroup$
    – user4552
    Jul 21, 2011 at 17:29
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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.

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  • $\begingroup$ 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 $m=0$. $\endgroup$
    – user346
    Feb 8, 2011 at 19:30
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    $\begingroup$ @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. $\endgroup$ Feb 8, 2011 at 19:58
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    $\begingroup$ My apologies @Jerry. I was hasty in judging your answer. Also, my own answer was way off target. $\endgroup$
    – user346
    Feb 8, 2011 at 20:21
  • $\begingroup$ @space_cadet: no worries. :) $\endgroup$ Feb 9, 2011 at 5:28
  • $\begingroup$ 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. $\endgroup$
    – user4552
    Jul 21, 2011 at 17:21
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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.

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  • $\begingroup$ The volume stretches, the energy remains the same, but not occupies larger volume, energy density is reduced. $\endgroup$
    – Ziezi
    Nov 16, 2017 at 14:48
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    $\begingroup$ "but now occupies larger volume" . Yes, that's exactly what I meant. $\endgroup$
    – user162182
    Nov 16, 2017 at 16:08

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