Difference between revisions of "CMB interaction with other components"

Let the index $\gamma$ denotes the photon and primed values correspond to that after the scattering. The $4$-momntum conservation law reads: $$ p_{\gamma i}+p_i= p'_{\gamma i}+p'_i. $$ First of all exclude the $4$-momentum $p'_i$ from the above conservation law to obtain: $$ (p_{\gamma i}+p_i- p'_{\gamma i})^2 = {p'_i}^2=m^2 . $$ As is well known $(p_{\gamma i})^2=p_{\gamma i}p^i_{\gamma}=0,$ then one obtains $$ E(E_{\gamma}-E'_{\gamma})+\vec{p}'(\vec{p}_{\gamma}-\vec{p}'_{\gamma})= E_{\gamma}E'_{\gamma}-\vec{p}_{\gamma}\vec{p}'_{\gamma} . $$ The maximum energy transmission takes place at the scattering angle $180°$. Taking into account that $|\vec{p_\gamma}|= E_\gamma ,$ one obtains $$ E(E_{\gamma}-E'_{\gamma})+p(E_{\gamma}+E'_{\gamma})=2E_{\gamma}E'_{\gamma}, $$ and finally: $$ E'_{\gamma}=\frac{E_\gamma(E+P)}{2E_\gamma+E-p}. $$

The required probability equals to \[ W = \sigma _T Nd, \] where $\sigma _T\simeq6.65 \cdot 10^{ - 25} \mbox{cm}^2 $ is the Thomson cross section, $n$ is the electron density, $d$ is the distance passed by the photon. For the case of flat Universe $\rho = \rho _{cr} $ and as the baryons contribution is of order of $4\% $ of total density then $N \simeq 2.3 \cdot 10^{ - 7} \mbox{cm}^{ - 3}.$ The passed distance is of order of observable part of Universe in the present time: $d \sim 10^{28} \mbox{cm}$ . As the result one obtains the following \[ W \simeq 2 \cdot 10^{ - 3} = 0.2\%. \]

The required expression for $E'_\gamma$ can be obtained using the approximate relation $$ p = (E^2-m^2)^{1/2}\simeq E-\frac{1}{2}\frac{m^2}{E}. $$ which corresponds to the considered case with $(E\gg m).$ Then one finally obtains $$ E'_{\gamma}=\frac{E}{1+\frac{1}{4}\frac{m^2}{EE_\gamma}}. $$ Taking into account that $E_\gamma \simeq 10^{-3} \mbox{ eV} \mbox{ and } m \simeq 10^{9} \mbox{ eV} $ one obtains that $E'_{\gamma} \simeq 3\cdot 10^{19}\mbox{ eV}.$ As the result of the considered effect the number of CMB quanta that come from the direction of a cluster effectively decreases, which manifests in lowering of the CMB temperature. The latter is called the Sunyaev-Zel'dovich effect.

Let $\langle\Delta E\rangle $ is the average energy amount that is lost in the collision. it follows from the problem formulation that $E/mc^2\ll 1$ and $T/mc^2 \ll 1$ (in the units with k =1), so in the double series decomposition $$ \langle\Delta E\rangle = mc^2[a_1+a_2(E/m)+a_3(T/m)+a_4(E^2/m^2)+a_5(ET/m^2)+a_6(T^2/m^2)+ \dots] $$ one should keep only first non-zero terms. At $T = E = 0$ nothing happens, and therefore $a_1=0.$ At $T=0,\; E \neq 0$ one has usual Compton scattering with the following cross-section $$ d\sigma/d\Omega \sim (1- \cos^2\theta) $$ and energy transmitted is equal to $$ \Delta E = (E^2/m)(1-\cos \theta). $$ Because the cross section is symmetric with respect to forward-backward direction, the term with $\cos\theta$ cancels in averaging over all angles and $$ \langle \Delta E\rangle = (E^2/m), ~T=0. $$ Thus $a_2=0$ and $a_4 = 1.$ At $E = 0, T \neq 0$ the photon has zero energy and turns into vacuum, therefore $a_3=a_6 = 0.$ At last one needs the coefficient $a_5.$ Consider a diluted flow of photons (the black-body radiation) with the temperature equal to that of the gas: $$ \frac{dN}{dE} = const \cdot E^2 e^{-E/T}. $$ Substitute the above obtained coefficients into the decomposition for $\langle \Delta E\rangle $ to obtain the following $$ \langle\Delta E\rangle = mc^2[a_4(E^2/m^2)+a_5(ET/m^2)]. $$ The thermal equilibrium condition implies that $$ \int^{\infty}_0\langle\Delta E\rangle E^2 e^{-E/T}dE=0, $$ and therefore $a_5 = -4,$ so one finally obtains the required relation $$ \langle\Delta E\rangle = (E/mc^2)(E-4kT). $$