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A photon emitted by a distant source billions of light years away arrives here with a lower frequency hence less energy than it started with. What happened to the energy?

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marked as duplicate by Brandon Enright, DavePhD, Kyle, Kyle Kanos, Chris White Jun 10 at 0:58

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Excellent question! My first guess is that the energy-momentum four vector is conserved and that the redshift is just an indication of the fact that the photon is reaching us with lower momentum since we're moving away from the source. I'll be interested to see what the correct answer is. –  dolphus333 Jun 9 at 19:23
Similar (same?) question: physics.stackexchange.com/questions/13577/… –  Joshua Jun 9 at 20:08
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4 Answers 4

The energy of a particle is an observer-dependent quantity in General Relativity. For a particle with four-momentum $ P^\mu $, measured by an observer with four velocity $ u^\mu $, is defined as:

\begin{equation} \mathcal{E}^{(u)}=-g_{\mu \nu}u^{\mu} P^{\nu} >0 \end{equation}

For instance, for a static observer $ u^\mu_{st} =(1,0,0,0) $ in Minkowski space-time, we have: \begin{equation} \mathcal{E}^{(u_{st})}=-P_0 \end{equation}

That is constant, and the energy is conserved. But this is not true in general. If the four velocity is time dependent, like in an expanding universe, the energy is not a conserved quantity. You can find from the geodesic equation (using the Robertson-Walker metric) that the velocity is inverse proportional to the cosmic scale factor, so decrease with time. From another point of view, you can say that is the time dependence of the metric that breaks conservation of energy.

At the end it really depends on the definition of energy you want to use. Very often in the definition of energy you need a time-like Killing vector field to have a constant energy. But the Robertson-Walker metric doesn't admit such a vector field.

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I have heard that even in expanding spacetime, there is an energy-like quantity that is still conserved. Can you comment on this? –  rob Jun 9 at 23:53
Not to my knowledge. Unfortunately you need a time-like killing vector field (that means you have symmetry in the time direction) to define an energy-like quantity. Maybe you are talking about the energy-momentum pseudotensor that includes the contribute of gravitational field, but this is not a good quantity for a conservation law (you need a tensor with vanishing divergence). Anyway, locally (that is when the curvature and expansion of the universe is negligible) energy is conserved even in an expanding universe. –  Rexcirus Jun 10 at 10:30
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Let us take a simple redshift of a spectral line from a moving galaxy. This means we are dealing with special relativity equations.


In the center of mass system of the excited atom ("deexcited-atom and photon" )the spectral line is fixed if our rest frame coincides with the rest frame of the atom . We observe it as at the star level on the left of the image.

The next line from the bottom is a nearby galaxy, this galaxy is moving, and so the rest system of the atom is moving with respect to us. We see the photon with less energy and if we could measure the deexcited atom we would see it balance the energy. In relativistic speeds one should use the special relativity equations.That is how we find that the galaxy is moving after all!

It is similar to shooting from a moving train: if ahead the bullet will gain energy from the train, if behind, the train will gain energy from the bullet and the bullet will be slower. For an observer at rest on the ground the two bullets will have different energies even though the gun shoots with the same energy at the rest frame of the train.

These redshift observations led to Hubbles law,

Hubble's law is the name for the observation in physical cosmology that: (1) objects observed in deep space (extragalactic space, ~10 megaparsecs or more) are found to have a Doppler shift interpretable as relative velocity away from the Earth; and (2) that this Doppler-shift-measured velocity, of various galaxies receding from the Earth, is approximately proportional to their distance from the Earth for galaxies up to a few hundred megaparsecs away. This is normally interpreted as a direct, physical observation of the expansion of the spatial volume of the observable universe.

My argument above relates to reference frames and does not discuss expansion of space itself as in cosmological models. As far as measurements go I do not see why it would not hold, that the atom moving away from us and the photon arriving in our detectors have to balance the energy , after all each photon signals a single interaction. It is all a matter of reference frames, imo.

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anna :may be you would like to read my version here –  Helder Velez Jun 10 at 22:11
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No energy is lost. The photon does not change, we just perceive it differently because of our relative velocities. The doppler effect is not a change in a wave, just the change in the apparent frequency of the wave. Technically speaking, the doppler effect changes the wavelength of the wave, which alters the apparent frequency, which is why waves get red- or blue-shifted. $E = hf$, so since the frequency of the photon does not change (again, only the frequency we observe changes), the energy does not change.

Another way to imagine it is that, for an observer at any distance from the star but moving with the same velocity we do, the light will always be redshifted no matter how far it is from the star/galaxy. Since the observed frequency of the light does not change, the energy cannot have changed.

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May I ask why I was downvoted in an effort to improve my answers and thus avoid future downvotes? –  Dave Coffman Jun 9 at 20:07
While I have no idea how to answer the question, your answer is wrong because a photon doesn't have kinetic energy in the $E = \frac12 mv^2$ sense. Its energy is given by $E = hf$, so if its frequency drecreases, its energy will too. –  Javier Badia Jun 9 at 20:15
OK...I think I'll remove that part. (I was thinking $0.5mv^2$ because photons can have mass) But I'm willing to bet that the doppler part is right - the change is in the frequency observed, not in the frequency of the photon itself –  Dave Coffman Jun 9 at 20:20
"photons can have mass" - what? This is new information for me. What makes you think photons have mass? –  Hunter Jun 9 at 20:27
@hunter I probably shouldn't have said that. You're right in a practical sense, and part of my brain must have shut off to let me say something like that. Interestingly enough, though, scientists are trying to figure out whether or not photons do have mass - energy is the same as mass, and photons definitely have energy. Wikipedia has an interesting discussion about the "Experimental Mass of a Photon". –  Dave Coffman Jun 9 at 20:38
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I think Dave's answer is correct. My I add as an alternative approach. We can speak of small volume elements $\delta V$ then energy is conserved because the reference frame is almost inertial. Since at long distance scales this inertial reference frame approach fails and energy appears to have been lost.

Edit: Joshua's link gives a better explanation of what I just said.

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