# Angular Diameter Distance to the last scattering surface

The definition of the angular diameter distance is the ratio of an object's physical transverse size to its angular size. However when I was reading my textbook, Astrophysics in a Nutshell by Dan Maoz pp.220-221, I am having some trouble trying to understand the notion of angular diameter distance to the last scattering surface. The text calculates the angular diameter distance to the last scattering surface $D_A$:

Consider flat cosmology (k=0) with no cosmological constant. We wish to calculate the angular size on the sky, as it appears today of a region of physical size $$D_s=\frac{2ct_{rec}}{\sqrt{3}}=140 kpc$$ from which light was emitted at time $t_{rec}$. Between recombination and the present time, the Universal expansion is matter-dominated, with $R\propto t^{2/3}$ for this model $$\frac{R}{R_0}=\left( \frac{t}{t_0} \right)^{2/3} = \frac{1}{1+z}$$ and hence we can write $$D_s=\frac{2ct_0}{\sqrt{3}}(1+z_{rec})^{-3/2}$$ The angle subtended by the region equals its size, divided by its distance to us at the time of emission (since that is when the angle between rays emanating from two sides of the region was set).

I'm not sure what does the last line actually mean..Can someone please elaborate more on this? I just simply take the $D_s$ as the "physical transverse size".

As we are concerned with observed angles, the type of distance we are interested in is the distance which, when squared and multiplied by 4π, will give the area of the sphere centered on us and passing through the said region. If the comoving radial coordinate of the surface of last scattering is r, the required distance is currently just $r\times R_0$ and is called the proper motion distance. The proper motion distance can be solved using null geodesic in the FRW metric $$\int_{t_{rec}}^{t_0} \frac{c dt}{R(t)} = \int_{0}^{r}\frac{dr}{\sqrt{1-kr^2}}$$ Setting k = 0, and substituting $$R(t)=R_0 \left( \frac{t}{t_0} \right)^{2/3}$$ and integrate $$rR_0=3ct_0[1-(1+z_{rec})^{-1/2}]$$

So I take this as the physical distance of the region from us. The next part is what confuse me:

However, at the time of emission, the scale factor of the Universe was 1 + z times smaller. The so-called angular diameter distance to the last scattering surface is therefore $$D_A=\frac{rR_0}{1+z}=3ct_0[(1+z_{rec})^{-1}-(1+z_{rec})^{-3/2}]$$

How does a physical distance $rR_0$ comes into play in the angular diameter distance, because from its definition it is just $$D_A=\frac{\text{physical transverse size}}{\text{angular size}}$$??

• There's a prior duplicate question with one answer at Astronomy.SE.
– Rob
Jun 17, 2018 at 17:13
• The angle subtended by the region equals its size, divided by its distance to us at the time of emission (since that is when the angle between rays emanating from two sides of the region was set). At the time of emissions, then angle would have been 90 degrees. That sentence makes no sense. The whole idea of the Sound Horizon is that we have an idea of how far the sound wave travels, and that becomes our cosmic yardstick when we look at it from a great distance away. The further away, the smaller the angle.
– user32023
Dec 26, 2018 at 15:59