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Given that there is no change in relative kinetic energy attributable to the expansion of the universe, what accounts for the difference in energy emitted vs energy absorbed (i.e. the red shift component) by two very distant bodies of mass whose relative velocities only exist due to the expansion of the universe and are therefore effectively zero?

Consider a photon that got emitted some time back in a far-off galaxy and absorbed today on earth. Also assume that there is no peculiar velocity involved, and that it was emitted due to an energy change in a hydrogen atom. The wavelength we would see were we able to observe the photon would be greater than that of a similar photon emitted from earth, and we might put that red-shift down to the space stretching effect of the expansion of the universe that had taken place between emission and absorption. All well and good in terms of observation, but not so good in terms of energy conservation. The photon energy, being only dependent on its frequency, is lower than expected. But unless there is another energy transfer taking place, the energy emitted must equal the energy absorbed right? How can this be explained?

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marked as duplicate by Rob Jeffries, John Rennie cosmology Dec 2 '18 at 20:44

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Given that there is no change in relative kinetic energy attributable to the expansion of the universe

This assumption does not hold, if one comoving observer throws a projectile it will reach the second observer with a lower local velocity than its original initial velocity.

The photon energy, being only dependent on its frequency, is lower than expected. But unless there is another energy transfer taking place, the energy emitted must equal the energy absorbed right?

Not at all, energy is only conserved in static spacetimes, which is not the case in an expanding universe, see Sean Carroll: energy is not conserved

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  • $\begingroup$ Thanks for this. My point about KE was intended to show that the loss of mass due to the photon emission did not account for the apparent energy loss. Do you agree? $\endgroup$ – Alan Gee Dec 2 '18 at 18:30
  • $\begingroup$ As for conservation of energy. The article seems to suggest that if you accept GR then you must also accept breaking of the COE law. Without necessarily knowing why. Am I right? $\endgroup$ – Alan Gee Dec 2 '18 at 18:34
  • $\begingroup$ The conservation of energy does not need to be broken since it never applied for dynamic spacetimes in the first place. $\endgroup$ – Yukterez Dec 3 '18 at 0:17
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Imagine you're measuring the distance to a galaxy ${\bf r} = a {\bf x}$, where ${\bf x}$ is its comoving distance and $a$ the scale factor. So, even if the galaxy is fixed in location, it still moves because the scale factor is a function of cosmic time $a = a(t)$.

You can actually calculate the velocity

$$ \dot{\bf r} = a\dot{\bf x} + \dot{a}{\bf x} = a\dot{\bf x} + \frac{\dot{a}}{a}{\bf r} = \underbrace{a\dot{\bf x}}_{\text{peculiar velocity}} + \underbrace{H(a){\bf r}}_{\text{Hubble flow}} $$

So the velocity of the galaxy you observe is the result of the Hubble flow and the its peculiar velocity. Again, if the later is zero, you still have the flow induced by expansion

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    $\begingroup$ In answers to other questions, there seems to be some disagreement about whether the velocity due to the Hubble flow is of the same nature as the peculiar velocity. For example, see the accepted answer to the following question physics.stackexchange.com/questions/442986/… (and my dissenting answer) $\endgroup$ – D. Halsey Dec 1 '18 at 13:44
  • $\begingroup$ @caverac This is useful information, but it doesn't appear to answer my question. If you ignore the peculiar velocity bit, there will be a red shift due to the Hubble flow, which means there is an imbalance between the energy that leaves one place and arrives at a distant other. How does that imbalance occur? $\endgroup$ – Alan Gee Dec 1 '18 at 14:03
  • $\begingroup$ @AlanGee Energy density is decreasing as the universe expands, as a matter of fact for radiation the energy decreases as $a^{-4}$. Note that this is not the same as saying the energy is being lost, it is just that it gets diluted $\endgroup$ – caverac Dec 1 '18 at 14:09
  • $\begingroup$ @caverac How do you dilute a photon? $\endgroup$ – Alan Gee Dec 1 '18 at 14:11

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