It is claimed that redshift due to cosmological expansion doesn't conserve energy. Does this exception also apply to gravitational redshift? Why or why not?
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Gravitational redshift is considered conservative in the field of General Relativity. If you have a photon that climbs up and out of a well it loses energy, and manifests as a redshift. If it falls, it gains energy, and manifests as a blueshift. However, in a gravitational field, that energy lost or grained is balanced by the work done against or by the gravitational field. That in itself respects the entirety of the conservation of energy. We know this MUST be true, because local conservation laws, within General Relativity, hold true.
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$\begingroup$ That certainly seems logical but why would cosmological redshift be different. The photon is coming from a denser place so wouldn't this be equivelent to climbing? $\endgroup$ Commented Jun 14 at 23:15
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$\begingroup$ @DerekSeabrooke, the best way I can describe this is that locality makes things much easier. The density of the universe and the gravitational potential are not static in an expanding universe (this is key). The “climb” in gravitational redshift is through a static or quasi-static potential, whereas in cosmological redshift, the photon travels through a spacetime that itself is expanding and evolving, making energy conservation non-trivial. $\endgroup$– StaticoCommented Jun 14 at 23:20