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I am writing an article about the Hubble Tension and when I was looking through the angular diameter distance I get confused over something.
In many articles the angular diamater distance to the LSS defined as the
$$D_A^* = \frac{r_s^*}{\theta_s^*}$$ where ##r_s^*## is the comoving sound horizon distance to the LSS and ##\theta_s^*## is the angular size of the LSS.
The angular diameter distance can be defined as
$$d_A = \frac{D}{\theta}~~~~Eqn. (1)$$
By using the FLRW metric for ##d\chi = d\phi = 0## we can write
$$ds = a(t)S_k(\chi)d\theta$$
By taking the integral
$$D \equiv \int ds = a(t)S_k(\chi)\theta~~~~Eqn. (2)$$
In this case by combining (1) and (2) we can write
$$d_A = \frac{D}{\theta}= \frac{a(t)S_k(\chi) \theta}{\theta} = \frac{S_k(\chi)}{(1+z)}$$
where ##1+z = a^{-1}##
$$r \equiv S_k(\chi) =
\begin{cases}
sinh(\chi) & k= -1 \\
\chi & k = 0 \\
sin(\chi) & k = +1
\end{cases}$$
From these definitions, ##r_s^* = D## must be satisfied. But that also does not make sense. So my question is, Am I mixing the notations or these are different definitions ? Beacause in the article its claimed that
$$r_s^* = \int_0^{t_*} c_s(t)\frac{dt}{a(t)} = \int_ {z_*}^{\infty} c_s(t)\frac{dz}{H(z)}$$
but ##D = a(t)S_k(\chi)\theta = r##, which does not make sense
Note: I am using the \begin{equation} ds^2 = -c^2dt^2 + a^2(t)[d\chi^2 + S_k^2(\chi)d\Omega^2]\end{equation} as my metric where
$$\chi = \int_0^r \frac{dr}{\sqrt{1-kr^2}}\equiv \begin{cases}
sinh^{-1}(r) & k= -1 \\
r & k = 0 \\
sin^{-1}(r) & k = +1
\end{cases}$$
For the reference at the sound horizon etc see https://arxiv.org/abs/1908.03663 Page 3
For the derivation of the angular diameter distance see
Daniel Baumann. Cosmology Part III Mathematical Tripos, pages 14–16.
http://theory.uchicago.edu/~liantaow/my-teaching/dark-matter-472/lectures.pdf
Last accessed on 2020-10-1.
In many articles the angular diamater distance to the LSS defined as the
$$D_A^* = \frac{r_s^*}{\theta_s^*}$$ where ##r_s^*## is the comoving sound horizon distance to the LSS and ##\theta_s^*## is the angular size of the LSS.
The angular diameter distance can be defined as
$$d_A = \frac{D}{\theta}~~~~Eqn. (1)$$
By using the FLRW metric for ##d\chi = d\phi = 0## we can write
$$ds = a(t)S_k(\chi)d\theta$$
By taking the integral
$$D \equiv \int ds = a(t)S_k(\chi)\theta~~~~Eqn. (2)$$
In this case by combining (1) and (2) we can write
$$d_A = \frac{D}{\theta}= \frac{a(t)S_k(\chi) \theta}{\theta} = \frac{S_k(\chi)}{(1+z)}$$
where ##1+z = a^{-1}##
$$r \equiv S_k(\chi) =
\begin{cases}
sinh(\chi) & k= -1 \\
\chi & k = 0 \\
sin(\chi) & k = +1
\end{cases}$$
From these definitions, ##r_s^* = D## must be satisfied. But that also does not make sense. So my question is, Am I mixing the notations or these are different definitions ? Beacause in the article its claimed that
$$r_s^* = \int_0^{t_*} c_s(t)\frac{dt}{a(t)} = \int_ {z_*}^{\infty} c_s(t)\frac{dz}{H(z)}$$
but ##D = a(t)S_k(\chi)\theta = r##, which does not make sense
Note: I am using the \begin{equation} ds^2 = -c^2dt^2 + a^2(t)[d\chi^2 + S_k^2(\chi)d\Omega^2]\end{equation} as my metric where
$$\chi = \int_0^r \frac{dr}{\sqrt{1-kr^2}}\equiv \begin{cases}
sinh^{-1}(r) & k= -1 \\
r & k = 0 \\
sin^{-1}(r) & k = +1
\end{cases}$$
For the reference at the sound horizon etc see https://arxiv.org/abs/1908.03663 Page 3
For the derivation of the angular diameter distance see
Daniel Baumann. Cosmology Part III Mathematical Tripos, pages 14–16.
http://theory.uchicago.edu/~liantaow/my-teaching/dark-matter-472/lectures.pdf
Last accessed on 2020-10-1.
Last edited: