# Explain relationship between angular diameter distance and luminosity distance, Etherington Theorem

I have a question relating to the Etherington Theorem.

The luminosity distance is defined by the equation for flux, i.e.

$F=\frac{L}{4\pi D_L^2}$

where flux is in units energy per unit time (luminosity) per unit area.

The angular diameter distance is defined by

$D_A=\theta/R$, where $\theta$ is the observed angular size measured by a telescope, and $R$ denotes the proper size of an object.

These two quantities are related by $D_L=(1+z)^2D_A$

I have never read a clear explanation for this relationship, nor have I come across a derivation. Could anyone explain to me where the redshift dependence $(1+z)^2$ comes from?

This relation is quite important, non trivial, and mathematical, and was proved by Etherington along with the other closely related theorem in this paper

I. M. H. Etherington (Philosophical Magazine ser. 7, vol. 15, 761 (1933))

This theorem only depends on photon conservation and the fact that photons only travel in null geodesics in Reimannian geometries. For a more detailed overview, read the original paper.

• Unfortunately, this article is not free for students. Are you aware of an arXiv article which discusses this result? This relationship is used very frequently in the large-scale structure and baryon acoustic oscillations BAO. May 18, 2015 at 17:43
• Unfortunately, there is no free result which explains the result, though there are some which experimentally prove the result. However, Wienbergs General Relativity book apparently explains the result quite nicely. May 19, 2015 at 0:25
• Can I ask you where in Weinberg's GR book? I don't find the Etherington relationship in my 1972 copy. May 19, 2015 at 19:24
• I don't know, which is why I said apparently. It was mentioned in a text discussing etherington. May 19, 2015 at 23:13
• Found it. Weinberg's section on Cosmography. Equations 14.4.22-14.4.23. Page 423. Also, see David Hogg's paper on "Distance measures in cosmology", arxiv.org/abs/astro-ph/9905116, Section 7 "Luminosity Distance". May 19, 2015 at 23:16

See D. Hogg's Distance measures in cosmology, 2000

http://arxiv.org/abs/astro-ph/9905116

Section 7, Luminosity Distance, p. 6

$D_L=(1+z)^2 D_A$

follows because the surface brightness of a receding object is reduced by a factor $(1+z)^{−4}$, and the angular area goes down as $D^{-2}_A$.

• This doesn't explain it, it just states it. I'm still looking for the derivation. It appears it has something to do with the power output and Cosmological Time Dilation, but I still haven't found a cogent explanation of 'why'. Apr 12, 2020 at 16:39

I am not familiar with the derivation of Etherington's reciprocity theorem and the following is my layman's superficial interpretation. If you look at Distance measure#Overview, you see that $$D_A = D_M/(1+z)$$, and $$D_L = (1+z)D_M$$, where $$D_M$$ is the transverse comoving distance (which is equal to $$D_C$$, the comoving distance, in a flat universe; for simplicity, we will treat $$D_M$$ as the comoving distance since our universe is practically flat). So the $$(1+z)^2$$ factor can be broken down into two parts: the $$(1+z)$$ factor between $$D_A$$ and $$D_M$$, and the $$(1+z)$$ factor between $$D_M$$ and $$D_L$$.

When we look at a faraway object, the angle that it takes up in the sky is the same angle that it actually took up compared to our location when the light was emitted (even though we couldn't see it back then). The uniform expansion of a flat universe does not change the angles of the light. Since the angular diameter distance is inferred based on the size of the object when the light was emitted, plus the angle in the sky we see now (which is the angle it took up in reality when the light was emitted), the angular diameter distance would basically equal the distance the object was from us when the light was emitted. This distance is a factor of $$(1+z)$$ smaller than the comoving distance (i.e. the distance where the object would be now if it followed the Hubble flow) simply because the universe (and thus the distance between us and the object) expanded by a factor of $$(1+z)$$ between when the light was emitted and now. This explains why $$D_A = D_M/(1+z)$$.

(An equivalent way to think about this is that faraway objects appear to us where it would be located today, if it simply followed the Hubble flow and had no other interactions. If the parts of the object just followed the Hubble flow, then the diameter of the object would have expanded by a factor of $$(1+z)$$ from when the light was emitted until now, and it appears to us to be at the location it would be if it had expanded in this manner. (Of course, the object is probably bound gravitationally and thus didn't expand as the universe expanded, but the light doesn't know that.) Thus, we see the object as if it were $$(1+z)$$ times the size it would be today if it had stayed the same size, and thus our calculation of the distance based on the angular diameter and the size of the object will be too small by a factor of $$(1+z)$$ compared to where it is currently located today.)

The light that the object emitted when it was emitted is now spread over a sphere with radius equal to the comoving distance (the distance between us and the object today). So at first, you would expect the luminosity distance to be equal to the comoving distance, and that would be true if we were not moving relative to the object. However, since we are moving away from the object, two additional factors come into play.

The first is that events in the object play out in "slow motion" for us. This happens whenever there is a Doppler effect -- a song played moving towards you will sound sped up, and a song played moving away from you will sound slowed down. The redshift is a direct measure of the amount of "slow motion", as redshift is just the wiggling of the electromagnetic fields playing out in slow motion (and thus lower frequency). This means that the rate of photons reaching us will go down by a factor of $$(1+z)$$ -- if it would have been 1 photon per second for an observer not moving with respect to the object, it would now be 1 photon per $$(1+z)$$ seconds for us.

The other effect is that each photon of light has less energy, due to the redshift. The frequency decreased by a factor of $$(1+z)$$, and thus the energy of each photon decreased by a factor of $$(1+z)$$. These two factors combine to decrease the flux by a factor of $$(1+z)^2$$, and since the luminosity distance is proportional to the inverse of the square root of the flux, a decrease in flux by a factor of $$(1+z)^2$$ increases the luminosity distance by a factor of $$(1+z)$$. This explains why $$D_L = (1+z)D_M$$.