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Asking this question I assume the example photons travel in vacuum. Once a source emits a photon they get separated by some space. Doesn't that space expand with the Hubble constant like the rest of the universe? If it does (even very slightly when the distance is small) can we assume that every photon travels with superluminal speed away from its source after any meaningful time has passed from emission?

As a secondary question: would the total relative speed of the photon be a simple addition of the speed of light and the product of the Hubble constant and the proper distance between the source and the photon?

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    no, photons always travel at speed c. they get a red shift due to the expansion. see https://en.wikipedia.org/wiki/Gravitational_redshift see answer https://physics.stackexchange.com/questions/332100/does-observed-red-shift-indicate-expanding-universe and https://physics.stackexchange.com/questions/113939/differentiating-the-gravitational-redshift-and-the-cosmological-redshift/113941#113941 – anna v Mar 24 '18 at 16:09
  • In summary, when space itself is expanding it can increase separation without altering velocity. – JMLCarter Mar 24 '18 at 17:43
  • @annav this fails to explain to the confused how we are able to detect light from objects with an apparent distance of greater than 14.7 billion light-years when the universe is only 14.7 billion years old. – Asher Mar 24 '18 at 18:03
  • @Asher we cannot detect them, where did you get that idea. see https://en.wikipedia.org/wiki/Observable_universe – anna v Mar 24 '18 at 18:52
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    General relativity doesn't have any unambiguous way to define the velocity of A relative to B, when A and B are at at cosmological distances from one another. For more on this, see this answer: https://physics.stackexchange.com/a/141219/4552 –  Mar 24 '18 at 19:00
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    @annav: no, photons always travel at speed c. That's misleading in the present context. The speed of a photon to a local observer is always c. The speed of a photon (or of any object) according to a distant observer is simply not defined in general relativity. The question is asking about the global idea, whereas your claim is only valid for the local case. –  Mar 24 '18 at 19:02
  • @BenCrowell Hubble constant is in the frame of the observable universe, and I am am commenting within the context of the observable universe,( not all possible general relativity solutions) – anna v Mar 25 '18 at 06:15

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