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This question already has an answer here:

Imagine a galaxy millions of lightyears away and, obeying Hubble's law, moving very quickly away from us.

Now imagine the same galaxy emitted a green photon in our direction (a photon with a frequency of $f_g$.) That photon reached earth, due to the redshift, with a frequency $f_r$, with $f_g>f_r$. That means photon $f_g$ left the distant galaxy with an energy of $H_g=hf_g$, but when it reached Earth, it had an energy of $H_r=hf_r$.

But $H_r < H_g$, so where did the difference between the energies go?

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marked as duplicate by ACuriousMind, Kyle Kanos, John Rennie, user10851, Qmechanic Jul 19 '15 at 11:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ possible duplicate of Redshifting of Light and the expansion of the universe $\endgroup$ – ACuriousMind Jul 19 '15 at 1:51
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    $\begingroup$ I'm going to throw this out there, but to the observer, it's never a green Photon, it's always red. To the object that generates the photon, it's always green, The energy changes if the frame of reference changes, but that's true for Newtonian physics as well, like throwing baseballs off a train. $\endgroup$ – userLTK Jul 19 '15 at 3:10
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Where does the energy go when light is redshifted?

It doesn't go anywhere. Let's say we're motionless with respect to some source which is emitting a stream of photons. We agree that the photons have some energy E=hf. Now let's say I push you such that you're moving away from the source. You now claim that the photons are redshifted, but those photons haven't changed one jot. Instead you have. I did work on you, I added energy to you. The photon energy hasn't changed, but yours has, so your measurement of the photon energy has reduced. You might say you're going faster than you were, so you measure the selfsame photon energy as reduced. Even though it hasn't, because conservation of energy applies.

It's similar for gravitational redshift$^*$. Let's say we're down on the ground near some source which is emitting a stream of vertical photons. We agree that the photons have some energy E=hf. Now let's say I lift you up away from the source. You now claim that the photons are redshifted, but those photons haven't changed one jot. Instead you have. I did work on you, I added energy to you. The photon energy hasn't changed, but yours has, so your measurement of the photon energy has reduced. At the higher elevation there's less gravitational time dilation. Your clocks are now going faster than they were, so you measure the selfsame photon energy as reduced. Even though it hasn't, because conservation of energy applies.

Now imagine the same galaxy emitted a green photon in our direction (a photon with a frequency of fg). That photon reached earth, due to the redshift, with a frequency fr... where did the difference between the energies go?

It didn't go anywhere. Your and your clocks are now going faster than they were, so you measure the selfsame photon energy as reduced. Even though it hasn't, because conservation of energy applies.

$*$ And blue shift. Throw a 511keV photon into a black hole and the mass increases by 511keV/c². Conservation of energy applies.

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  • $\begingroup$ This should be the accepted answer. $\endgroup$ – Jossie Calderon Aug 11 '18 at 20:39
  • $\begingroup$ @JossieCalderon This answer basically says the opposite of the accepted answer in the second linked question. $\endgroup$ – Michael Aug 14 at 16:22
  • $\begingroup$ @Michael then that answer is wrong. $\endgroup$ – Jossie Calderon Aug 14 at 16:35
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Nobody ever said that different observers have to agree on the energy of a photon (or anything else). The invariant quantity is energy minus momentum (i.e. rest mass), which is equal to zero whether the photon is red or green. (Edited to add: I see now that userLTK already said as much in a comment.)

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