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The spatially flat FRW metric in Cartesian co-ordinates is given by: $$ds^2 = -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2)$$

As I understand it, since the metric does not depend on the spatial co-moving co-ordinates $x,y,z$ then there are Killing vectors in the $x,y,z$ directions.

Does this imply that the 3-momentum of a free particle is conserved when measured with respect to the $x,y,z$ co-ordinates? (In terms of expanding proper distances I presume that the particle would seem to lose velocity)

Does this also imply that the 3-momentum of a photon is conserved when measured with respect to the $x,y,z$ co-ordinates?

If the 3-momentum of the photon is conserved then, as $E=pc$ for photons, does this imply that its energy is conserved as well?

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    $\begingroup$ I'm guessing this is related to Valter's answer to your previous question, but it seems to me that Valter's answer answers most if not all of this question too. Can you clarify what you're asking that isn't covered by the previous question? $\endgroup$ – John Rennie May 17 '15 at 15:44
  • $\begingroup$ Does a Killing vector in the x direction imply that x-momentum is conserved for a particle travelling on a geodesic? $\endgroup$ – John Eastmond May 17 '15 at 18:42
  • $\begingroup$ For your last question you have to distinguish between global frames and locally inertial ones. In relativity Energy is generally not a frame-invariant quantity. In a FRW spacetime which expands, photons will lose energy from redshift unless you construct a local approximately inertial frame. $\endgroup$ – Thatpotatoisaspy Jun 8 at 4:58
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Canonical momentum is conserved, but proper momentum density is proportional to $\frac{1}{a}$ where $a$ is the scale factor.

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