# How does apparent brightness (or stellar magnitude) change with distance in an expanding universe

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?

 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.

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 $$1{+}z$$.
• @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. Commented Sep 13, 2022 at 21:21