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An important consideration in the post-recombination epoch is the issue of the optical depth $\tau$ of the Universe due to Compton scattering. This is a dimensionless quantity such that $exp(−\tau)$ (often called the visibility) describes the attenuation of the photon flux as it traverses a certain length. The probability $dP$ that a photon has suffered a scattering event from an electron while travelling a distance $cdt$ is given by

$$dP=-\frac{dN_\gamma}{N_\gamma}=-\frac{dI}{I}=\frac{dt}{\tau_{\gamma e}}=n_e\sigma_Tcdt=-\frac{x\rho_m}{m_p}\sigma_T c\frac{dt}{dz}dz=-d\tau\hspace{1cm} (9.4.2)$$

where $N_\gamma$ is the photon flux, so that

$$I(t_0,z)=I(t)exp\left(-\int_0^z\frac{x\rho_m}{m_p}\sigma_T c\frac{dt}{dz}dz\right)=I(t)exp[-\tau(z)];\hspace{2cm} (9.4.3)$$

$I(t_0,z)$ is the intensity of the background radiation reaching the observer at time $t_0$ with a redshift $z$ if it is incident on a region at a redshift $z$ with intensity $I[t(z)]$; $\tau(z)$ is called the optical depth of such a region. The probability that a photon, which arrives at the observer at the present epoch, suffered its last scattering event between $z$ and $z-dz$ is

$$-\frac{d}{dz}\{1-exp[-\tau(z)]\}dz=exp[-\tau(z)]d\tau=g(z)dz\hspace{4cm}{(9.4.4)}$$

The quantity $g(z)$ is called the differential visibility or effective width of the surface of last scattering;

This text was taken from "Cosmology. The origin and evolution of cosmic structure" by P.Coles, F.Lucchin.

I have some doubt about $(9.4.2)$ and $(9.4.4)$.

I would write for $(9.4.2)$:

$$dP=-\frac{dN_\gamma}{N_\gamma}=-\frac{dI}{I}=...=...=-\frac{x\rho_m}{m_p}\sigma_T c\left|\frac{dt}{dz}\right|dz=-d\tau$$

because $z$ decreases from the Big Bang up to us and $t$ increases, so $\frac{dt}{dz}<0$, $dz<0$ and $dt>0$. (The same correction obviously inside the integral in $(9.4.3)$).

About $(9.4.4)$, what is not clear is the way in which "The probability that a photon, which arrives at the observer at the present epoch, suffered its last scattering event between $z$ and $z-dz$", was derived. How do I get the first expression of $(9.4.4)$?

Someone has any suggestion?

Thank you very much.

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  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ Commented Sep 24, 2018 at 20:44
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    $\begingroup$ Oh sorry. I'll fix it. $\endgroup$
    – MattG88
    Commented Sep 24, 2018 at 20:51
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    $\begingroup$ Nice job with the typesetting! Note that \exp (with a leading backslash) sets that operator in upright type. $\endgroup$
    – rob
    Commented Sep 24, 2018 at 21:26
  • $\begingroup$ @rob Thanks!:-) I know about the \exp, but I wanted to be an amanuensis. $\endgroup$
    – MattG88
    Commented Sep 24, 2018 at 21:35

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Here is my attempt, if someone wants to check.

Thinking about $(9.4.2)$ I made a try:

The probability $dP$ that a photon has suffered a scattering event from an electron while travelling a distance $cdt$ is given by

$$dP=-\frac{dI}{I}$$

so I tought that $\Delta P=-\frac{\Delta I}{I}$ it should be the probability that a photon has suffered a scattering event from an electron while travelling a distance $c\Delta t$; thus if $c\Delta t$ is the distance from us:

$$\Delta P=-\frac{\Delta I}{I}=-\frac{I_f-I_i}{I_i}=-\frac{I(t_0,z)-I(t)}{I(t)}\overbrace{=}^{(9.4.3)}-\frac{I(t)e^{-\tau(z)}-I(t)}{I(t)}=1-e^{-\tau (z)}$$

$\Delta P=1-e^{-\tau (z)}$ is the probability that a photon (at redshift $z$) will undergo at least one more scattering before it reaches us (at $z=0$).

If this is correct, I can say that $1-(1-e^{-\tau(z)})=e^{-\tau(z)}$ is the probability that a photon suffers a scattering event before the redshift drops to $z$.

Thus

The probability that a photon, which arrives at the observer at the present epoch, suffered its last scattering event between $z$ and $z-dz$

is:

$$1-e^{-\tau(z)}-(1-e^{-\tau(z-dz)})=(1-e^{-\tau(z)})-(1-e^{-\tau(z-dz)})=\frac{d}{dz}\{1-e^{-\tau(z)}\}dz$$

and we have done, apart from the minus sign in front of $(9.4.4)$. It could be a typo.

What do you think?

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