Post by Bruchpilot AkiInto which form of energy is the energy actually converted, which
photons lose through the cosmic redshift?
Nothing "happens" to the energy of the EM radiation known as the CMBR,
because energy is not a "thing", it is a quantity used for bookkeeping.
A) Remember that the energy of a given object depends on the coordinates
used to measure it.
B) Remember also that Noether's theorem relates conservation of energy
to time-translation invariance, which in GR requires that a timelike
Killing vector is present
If there is no such invariance (no timelike Killing vector) one should
not expect energy to be conserved, and in the FLRW manifolds used to
model cosmology there is neither such a Killing vector nor conservation
of energy. [#]
[#] In a "small" local 4-volume in which the metric can be
considered to be constant within measurement resolutions,
one can construct locally-inertial coordinates in which
the time coordinate is indistinguishable from a timelike
Killing vector. So in this region, using such coordinates,
energy is conserved to within measurement resolutions.
This applies to every classical experiment that was used
to establish conservation of energy. And it most definitely
does not apply to the CMBR between emission and now.
Similarly for conservation of momentum and angular momentum.
(I have omitted some details and caveats for clarity.)
IOW: To calculate the energy of the CMBR when it was emitted requires
one to use locally-inertial coordinates at rest relative to the emitting
particles, and the dot product used to calculate it involves the metric
at that place and time. To calculate the energy of that same radiation
now requires one to use locally-inertial coordinates at rest relative to
the earth, and the dot product involves the metric here and now. Both
the coordinates and metric involved are different, which explains why
different values are calculated.
Tom Roberts