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Cosmological redshift causes wavelengths of a distant object to stretch by a factor $1/(1-Hr/c)$ where H is the Hubble constant, r is distance, and c is the speed of light. Consequently the received power density drops by $1-Hr/c$ from that factor alone. Combined with the inverse square law, this would suggest the apparent brightness dropping as $(1-Hr/c)/r^2$.

However, the Cosmic Microwave Background (CMB) has maintained its characteristic black body radiation density with just a drop in temperature by a factor of (1-Hr/c). Since black-body radiation goes as the fourth power of temperature, it suggests that the apparent brightness of a stellar black-body including the expansion of the universe should go as $(1-Hr/c)^4/r^2$

So the question boils down to this: Does a distant red-shifted stellar object look like a slightly cooler black body consistent with its actual diameter, or does it appear brighter than that?

[Edit] Just to capture @benrg's answer which I marked correct: A red-shifted black body looks cooler by a factor $(1-Hr/c)$ and it's apparent brightness is less by a factor $(1-Hr/c)^4$ consistent with its temperature. This is true whether the redshift is due to simple velocity or cosmological expansion.

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There are two separate issues here, the inverse-square law and redshift.

Inverse-square law: the total power received from an object decreases as $r^2$, but also the area over which it's spread decreases by $r^2$, so the areal power density, and hence blackbody temperature, doesn't depend on the distance. In the case of the CMB, which fills the whole sky, the inverse-square law doesn't change the received power at all. Another way of looking at it is that there's $x^2$ times as much CMB-emitting plasma $x$ times farther away. (Technically, in cosmology, all of these distances should be angular diameter distances.)

Redshift: when an object is moving away from you with a redshift of $1{+}z$, then each photon has $1{+}z$ times less energy; also, the rate at which photons arrive is slower by a factor of $1{+}z$; and also, because of the headlight effect, $(1{+}z)^2$ times fewer photons are emitted in your direction. In total, the received power decreases by a factor of $(1{+}z)^4$, so the temperature decreases by a factor of $1{+}z$.

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  • $\begingroup$ Great, so, if I take z as Hr/c, then the received power density drops as 1/(1+Hr/c)^4/r^2 and the solid angle drops as 1/r^2. So it still appears as a simple black body but cooler by 1/(1+Hr/c). And this applies to cosmological expansion as we as actual velocity? $\endgroup$
    – Roger Wood
    Sep 13, 2022 at 20:55
  • $\begingroup$ @RogerWood It applies to anything redshifted. (There's only one kind of redshift in general relativity; "cosmological", "special-relativistic", and "gravitational" redshifts are just special cases for spacetimes with certain symmetries.) Note that $Hr/c\approx z$ is only an approximation. Also, the power density (watts per steradian) doesn't drop as $1/r^2$, but maybe you mean something else by power density. $\endgroup$
    – benrg
    Sep 13, 2022 at 21:21
  • $\begingroup$ Yes, received power density in watts per square meter corresponding to apparent brightness. I assume that's ok? $\endgroup$
    – Roger Wood
    Sep 13, 2022 at 21:33
  • $\begingroup$ @RogerWood Of course, silly me. $\endgroup$
    – benrg
    Sep 13, 2022 at 22:57
  • $\begingroup$ thanks. - edited my question to sumarize in terms of Hr/c $\endgroup$
    – Roger Wood
    Sep 14, 2022 at 17:42

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