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Angular diameter distance

  1. Mar 26, 2014 #1
    So the angular diameter distance is given by the angle it extends in the sky and its actual size x
    [itex]
    d_A= \frac{x}{\theta}[/itex]
    This definition is okay. It means that if I consider the universe to be homogeneous and isotropic, I will have a conformal expansion. Therefore the angle should remain the same and only x expands.
    But if I now use the following definition, I am lost.
    [itex]
    d_A = \frac{r(\chi)}{1+z}[/itex]

    with [itex]\chi[/itex] being the comoving distance.
    [itex]
    r(\chi) = \begin{cases}
    \sin \left( \sqrt{-\Omega_k} H_0 \chi \right)/\left(H_0\sqrt{|\Omega_k|}\right) & \Omega_k < 0\\
    \chi & \Omega_k=0 \\
    \sinh \left( \sqrt{\Omega_k} H_0 \chi \right)/\left(H_0\sqrt{|\Omega_k|}\right) & \Omega_k >0
    \end{cases}[/itex]

    I know [itex]r(\chi)[/itex] is called proper motion distance or transverse comoving distance.

    1) But what does it represent in a non flat universe?

    If I had a light ray travelling along some path. Then this quantity would not include the expansion of the universe. So it cannot represent the physical distance the lightray has travelled.

    2) What is then a physical distance in a curved universe?

    Well if I use the definition anyway together with
    [itex]{1+z}=\frac{a_{now}}{a_{past}}[/itex]
    then for an expanding universe [itex]
    d_A = \frac{r(\chi)}{1+z}[/itex], the angular diameter distance would be decreased. So why that? 3) If the angle in the sky remains the same, why should it be necessary to decrease this value?
    4) Does the angle maybe chance for a curved universe?

    I would be very happy about any help.
     
  2. jcsd
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