I'm studying from these lecture notes (Author: Wayne Hu, title: Lecture Notes on CMB Theory: From Nucleosynthesis to Recombination). At the very end of page 10 and beginning of page 11 he says:

Let us begin with the simple approximation that the temperature field at recombination is isotropic but inhomogeneous and the anisotropy viewed at the present is due to the observer seeing different portions of the recombination surface in different directions (see Fig. 8).

Now I'm not really understanding this argument. If, before last scattering the fluid is only inhomogeneous but isotropic I still expect CMB to be isotropic.

For example, let's suppose that with respect to an observer, before last scattering, at a distance $d$ in a direction $\hat{n}$ there is a positive fluctuation of temperature. I expect that, due to isotropy, there will be the same positive fluctuation in a direction $\hat{n}'$ at a distance $d$. After last scattering, let's say approximately after a time $t=d/c$ (not considering expansion of the universe here) the photons from that surface of radius $d$ arrive to the observer and it can measure the positive fluctuation in temperature in both the directions. Apply this to all the directions and we don't have any kind of anisotropy.

Am I missing something and there's a different way to look at it and understand this?

  • $\begingroup$ I'm pretty sure Hu means that the sky was/is isotropic on large scales. That is, the small inhomogeneities makes the Universe anisotropic on small scales, but statistically you'd see the same whether you look up or down. I think it's a bit badly phrased, because on the scales that the Universe is anisotropic, it is also inhomogeneous. This is true both before and after recombination. $\endgroup$
    – pela
    Jan 21, 2020 at 14:59

1 Answer 1


I think the paragraph you quote must be using the terms somewhat loosely.

To unpack this, I will first comment on a more strict usage, then I will comment on what I think the paragraph is saying (i.e. a less strict usage).

The word "homogeneous" means "same at all locations" or "independent of position"; the word "isotropic" means "same in all directions" or "independent of angle". A uniform electric field is homogeneous but not isotropic. The wavefunction of the hydrogen ground state is isotropic from the centre, but not homogeneous. And there we note a possible ambiguity about the meaning of the word 'isotropic'. A distribution can be isotropic when viewed from one place, but not when viewed from another.

(homogeneous) and (isotropic from one place) implies (isotropic from all places)

(inhomogeneous) and (isotropic from one place) does not imply (isotropic from all places)

(isotropic from all places) seems to me very likely to imply (homogeneous) but I have not tried to prove it or look up a mathematical proof---perhaps someone would like to comment on this.

Now coming back to the text you quoted, if it means "is isotropic as viewed from Earth" then isotropic it is (viewed from Earth) and no amount of inhomogeneity will change this. Therefore I think the author can't mean that. Very likely they mean "on the large scale, no direction is preferred to any other; we don't need to put explicitly a function of angle into our model of the density and temperature. On the other hand, if we allow that the density and temperature is inhomogeneous, but not such as to result in a privileged position for Earth, then when viewed from Earth the scattering surface will cut through these inhomogeneities and therefore be anisotropic, even though we put no explicit anisotropy into our model."

  • $\begingroup$ I'd like to answer, but after reading, I think that's been done. $\endgroup$ Jan 24, 2020 at 21:18

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