Wednesday, June 19, 2013

CMB recombination

the CMB photosphere has a thickness of $\Delta z\simeq 100$ in redshift. how much time (in years) does it take for the atoms to recombine?

maths bonus: can you show that the Gauss-distribution is shape-invariant under convolution? what about the Cauchy-distribution?

5 comments:

  1. What's photospere? Just the duration of the recombination period in redshift?

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    Replies
    1. the photosphere is the surface from which the CMB photons are emanating... so the question aims at how long recombination takes in units of time corresponding to 100 units in redshift.

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  2. Hmm, I am going to present a solution attempt, but I don't trust the outcome too much.

    I am assuming the CMB is sitting at a redshift of roughly 1100.
    The relation between redshift and scale factor is
    $a = \frac{1}{1+z}\,$
    so the scale factor changed by a factor of about 1.09.
    I am in the matter dominated epoch, so the time behaves as:
    $t = a^{\frac{3}{2}}\.$
    Accordingly, the time changed by the factor of 1.13.
    The age of the universe at CMB creation should be something like 372000years. Thirteen percente of that are a bit more than 50.000years, which strikes me as bit long...

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    Replies
    1. EDIT: The formula I mistyped was:
      $t = a^{\frac{3}{2}}\,.$

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  3. Yes, quite so: You can estimate the time passing during recombination by writing
    \begin{equation}
    \Delta t = \frac{\partial t}{\partial z}\Delta z
    \end{equation}
    in which one can substitute the Hubble function by rewriting
    \begin{equation}
    \Delta t = \frac{\partial t}{\partial a}\frac{\partial a}{\partial z}\Delta z
    \end{equation}
    and by using $\partial a/\partial t = a H$ as well as $a=1/(1+z)$:
    \begin{equation}
    \Delta t = \frac{1}{aH} \times a^2 = \frac{a}{H}\Delta z
    \end{equation}
    Substituting the Hubble function $H^2 = H_0^2(\Omega_\gamma/a^4+\Omega_m/a^3+\Omega_\Lambda)$ yields then the numerical value of a few hundred thousand years.

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