# Calculating distance from the FRW metric

My question arises from the book Cosmology by Daniel Baumann, specifically from equation (2.81), where I don't understand how the expression for the distance is calculated. I will start by providing some context, which you can skip if you are familiar with the FRW metric.

Context

First of all, we consider the Friedmann-Robertson-Walker metric, which is given by this expression in spherical spatial coordinates:

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

where $$a(t)$$ is the scale factor, which measures the expansion of the universe, and $$d\Omega^2\equiv d\theta^2+\sin^2\theta d\phi^2$$.

This metric can be rewritten as:

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

by redefining the radial coordinate in the following way, where it is easy to prove that $$\chi$$ is the comoving distance:

$$d\chi=\dfrac{dr}{\sqrt{1-k(r/R_0)^2}}$$

The result of integrating this expression is denoted by $$r=S_k(\chi)$$.

Question

We consider a light source of transverse physical size $$D$$ and an observer, at a comoving distance $$\chi$$. If the spacetime is considered static, the angular size $$\delta\theta$$ measured by the observer will be, assuming $$\sin\delta\theta\simeq\delta\theta$$:

$$\delta\theta=\dfrac{D}{\chi}$$

But the book states that, in an expanding universe (where we consider the FRW metric described above), this formula becomes:

$$\delta\theta=\dfrac{D}{a(t_1)S_k(\chi)}$$

where $$t_1$$ is the time at which the light is emitted by the source. First, I don't understand why this point in time is the one that should be considered, instead of integrating in the time variable between the times of emission and detection, since the universe is expanding and therefore the distance is changing. And second, I don't see where $$S_k(\chi)$$ comes from in this expression.

What I have attempted is to restrict the metric to $$t=t_1$$ (although I can't see why this should be done), $$\theta=constant$$ and $$\phi=constant$$, which yields:

$$ds^2=a^2(t_1)\dfrac{dr^2}{1-k(r/R_0)^2}=a^2(t_1)d\chi^2$$

and therefore the absolute value of the determinant of the metric tensor is, in this case:

$$g=a^2(t_1)$$

Integrating, the distance we are looking for would be:

$$L=\int_0^\chi \sqrt{g}d\chi'=\int_0^\chi a(t_1)d\chi'=a(t_1)\chi$$

Why is this incorrect?

Edit: in my calculations, I am following the guidance provided in the very clear answer I received to a previous question here, but I am not sure of whether that is valid in this context.

I think the answer is simple if you just set $$dt=0$$, $$d\chi=0$$ and $$d\phi=0$$. In this case the metric

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

becomes

$$ds = a(t)S_k(\chi)d\theta$$

or

$$d\theta = \frac{ds}{a(t)S_k(\chi)}$$

or

$$\delta \theta = \frac{D}{a(t)S_k(\chi)}$$