Angular Diameter Distance

  • #1
36
2

Main Question or Discussion Point

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 $$D_A=\frac{\text{physical transverse size}}{\text{angular size}}$$??
 

Answers and Replies

  • #2
stefan r
Science Advisor
Gold Member
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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".
Yes. It is easier if you say 2 stars instead of "region". That was the distance between the two stars when they each radiated a photon. The stars may have moved while the photons were on their way here.

The textbook wording is more correct and it applies to any source.
 
  • #3
36
2
I found the derivation in "Gravitation and Cosmology" by Weinberg at pp.421-424
 

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