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I am trying to follow some calculations in the book Cosmology by Daniel Baumann, where we have a light source and an observer, and we need to compute the area of a sphere centered at the source and so that the observer lays on its surface, in order to determine the expression for the observed flux in an expanding universe.

My question is the following. The expression that the book provides for this area, at the time $t=t_0$ when the photons from the source are detected by the observer, is:

$$A(t_0)=4\pi a^2(t_0)S_k^2(\chi)$$

where $\chi$ is the comoving distance between the source and the observer and $S_k(\chi)$ coincides with the radial coordinate $r$ in comoving spherical coordinates, that is:

$$S_k(\chi)=\begin{cases}R_0\sinh(\chi/R_0) & \text{ if }k=-1 \\ \chi & \text{ if } k=0 \\ R_0\sin(\chi/R_0) & \text{ if }k=-1\end{cases}$$

since it is used to rewrite the Friedmann-Robertson-Walker (FRW) metric, from

$$ds^2=-c^2dt^2+a^2(t)\bigg[\dfrac{dr^2}{1-kr^2/R_0^2}+r^2d\Omega^2\bigg]$$

to the simplified expression

$$ds^2=-c^2dt^2+a^2(t)\big[d\chi^2+S_k^2d\Omega^2\big]$$

by using the following redefinition of the radial coordinate:

$$d\chi=\dfrac{dr}{\sqrt{1-kr^2/R_0^2}}$$

I have already proven that $\chi$, defined in this way, coincides with the notion of the comoving distance. But I don't understand why the radius of the sphere that is used to compute its area has to be $a(t_0)r$ instead of $a(t_0)\chi$, which would be the physical distance between the source and the observer at the time of detection of the arriving photons. Shouldn't the physical distance between two points be the distance that is considered to calculate the area of a sphere defined as I indicated at the beginning? Why then is the radial comoving coordinate $r=S_k(\chi)$ present in the expression of the area instead of $\chi$? What I would write is:

$$A(t_0)=4\pi d^2_{phys}(t_0)=4\pi \big[a(t_0)\chi\big]^2$$

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To compute the area of the sphere, you restrict to the surface of the sphere $t={\rm const.}, r={\rm const.}, \chi={\rm const}$, and integrate over the angular coordinates weighted by the volume element for the angular coordinates.

When written in terms of $r$, the volume element for the angular coordinates is the same as what you get in Euclidean space \begin{equation} A = \int_{S^2} d\theta d\phi \sqrt{g} = a^2(t) r^2 \int_0^{\pi} d\theta \sin \theta \int_0^{2\pi} d\phi = 4 \pi a^2(t) r^2 \end{equation}

When written in terms of $\chi$, the volume element is modified, so you get \begin{equation} A = \int_{S^2} d\theta d\phi \sqrt{g} = a^2(t) S_k(\chi)^2 \int_0^{\pi} d\theta \sin \theta \int_0^{2\pi} d\phi = 4 \pi a^2(t) S_k(\chi)^2 \end{equation} Of course for a spatially flat Universe, $k=0$, $S_k(\chi)=r$ and these two expressions are the same. However, for a Universe with spatial curvature, the correct expression in terms of $\chi$ will involve a factor $R_0^2 \sin^2(r/r_0)$ or $R_0^2 \sinh^2(r/r_0)$.

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  • $\begingroup$ Just to be sure, in your notation, $g$ corresponds to what I call $ds^2$ in my question, correct? So, restricting it to the surface of the sphere, that is, setting $t,r,\chi$ as constants, we get $g\equiv ds^2=a^2(t)r^2d\Omega^2=a^2(t)S_k^2(\chi)d\Omega^2$, right? Thank you very much for your answer! $\endgroup$ Jan 13, 2023 at 19:08
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    $\begingroup$ @WildFeather $g$ is the (absolute value of the) matrix determinant of the metric tensor, which in your case is $g = c^2 a(t)^2 S_k^2$. It is also common to let $g$ simply be the determinant (which is negative) and to explicitly negate the minus sign by writing $\sqrt{-g}$ or $\sqrt{|g|}$ in the integral. $\endgroup$
    – J. Murray
    Jan 13, 2023 at 21:09

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