Difference between revisions of "CMB anisotropy"

Equation of geodesics describing the propagation of radiation in space with arbitrary curvature can be represented as: $$ \frac{dx^\alpha } {d\lambda } = p^\alpha,\; \frac{dp_\alpha } {d\lambda} = \frac{1} {2}\frac{\partial g_{\gamma \delta }} {\partial x^\alpha }p^\gamma p^\delta, $$ where $\lambda$ is an arbitrary affine parameter, which is taken along geodesic. Since photon has zero mass, the first integral of this equation is: $$ p^\alpha p_\alpha = g^{\mu \nu }p_\mu p_\nu = 0. $$ Using this relation, zero component can eliminated from the equations of geodesics: $$ p^0 = \frac{1} {a^2}\left( \sum\limits_{i = 1}^3 p_i^2 \right)^{1/2} = \frac{p}{a^2} $$ and $$ p_0 = \left( 1 + 2\Phi\right)p. $$ Hence, $$ \frac{dx^i} {d\eta } = \frac{p^i} {p^0} = \frac{ - \frac{1} {a^2}\left( 1 + 2\Phi \right){p_i}} {p^0} = r^i\left( 1 + 2\Phi\right), $$ where $r^i = - \frac{p_i}{p^0}$. Expressing $p^i$ and $p^0$ through $p$ and substituting metrics into the second equation while keeping only linear terms in $\Phi$ one obtains: $$\frac{dp_\alpha } {d\eta} = \frac{1} {2}\frac{\partial g_{\gamma \delta }} {\partial x^\alpha }\frac{p^\gamma p^\delta } {p^0} = 2p\frac{\partial \Phi } {\partial x^\alpha }. $$

Using the invariant $aT = const$: $$ \frac{\delta T}{T} =-\frac{\delta a}{a}. $$ In case of $p = w\rho ,\; a\left( t \right) = t^{\frac{2} {3\left( 1 + w \right)}}$ and, hence, $$ \frac{\delta a} {a} = \frac{2} {3\left(1 + w\right)}\frac{\delta t}{t}. $$ In the case of a weak gravitational field (in this approximation, we assume that the density perturbations generate a small perturbation of the metric, which is valid for the time at which we see these disturbances), the true time $\tau$ is connected to coordinate time $t$ as $$ d\tau=\sqrt{g_{00}}dt\approx\sqrt{1 + 2\Phi } dt \approx \left(1+\Phi\right)dt, $$ where $\Phi$ is a Newtonian gravitational potential and, thus, ${\delta t}/t \simeq \Phi$: $$ \left(\frac{\delta T}{T} \right)_e = - \frac{2}{3\left( 1 + w\right)}{\Phi_e} $$ where subscript $e$ denotes the moment of emission. When light is propagating in expanding Universe, the relation $\omega (t) \propto a(t)^{-1}$ holds. Using the same arguments in Newtonian approximation, one could write $$ \left( \frac{\delta T} {T} \right)_0 = \left(\frac{\delta T} {T}\right)_e + \Phi _e $$ where subscript $0$ denotesthe moment of detection. Finally, $$ \left(\frac{\delta T} {T} \right)_0 = \left(\frac{1 + 3w} {3 + 3w} \right)\Phi _e $$

In this approximation, the oscillations of the photon component smooth on scales smaller than the distance at which the photon diffuses in characteristic period of evolution. If considering the before recombination era this time is the Hubble time. In this random walk approximation Silk scale $\lambda_S$ can be estimated as geometric average of mean free path of a photon $\lambda_\gamma$ and horizon scale $l_H$ (or Hubble time $t_H\sim H^{-1}$). The number collisions between photon and electrons during the Hubble time is estimated as $t_H/\lambda_\gamma$ and distance between collisions is of order of $\lambda_\gamma$, so that a photon diffuses at distance $$\lambda_S \sim \sqrt{N}\lambda_\gamma =\sqrt{\frac{t_H}{\lambda_\gamma}}\lambda_\gamma = \sqrt{\lambda_\gamma l_H}$$ during the Hubble time. Hubble parameter at recombination era is $$H(z_r)=H_0\sqrt{\Omega_{m0}(1+z_r)^3+\Omega_{r0}(1+z_r)^4}\simeq 5\, \mbox{Mpc}^{-1}.$$ Densities of free electrons and protons coincide and the latter before the resombination is about 75 \% of barion density: other barions (and electrons) are contained in helium atoms, which formed in the Universe somewhat earlier. Thus, before the start of resombination the electrons density satisfies $$n_e(z)=0.75 \frac{\rho_b(z)}{m_p}=6\cdot 10^{-4}\Omega_b (1+z)^3 \mbox{sm}^{-3}.$$ At the beginning of recombination the number density of free electrons is $n_e(\eta_r)=230\,\mbox{sm}^{-3}.$ Hence, $$\lambda_S(\eta) \simeq \sqrt{\frac{1}{\sigma_{_T}}n_e(\eta) H(\eta)},$$ which gives $\lambda_S(\eta_r) \simeq 0.02 \,\mbox{Mpc.}$

Physical distance $R(z)$ is connected to the angular diameter distance $R_a(z)$ as $$R_a(z)=\frac{1}{1+z}R(z),$$ where $$R(z)=R_H\int_0^{z}\frac{dz}{\sqrt{\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}}},$$ and $R_H$ is Hubble radius. Using the relation $\theta \simeq \frac{\lambda_S}{R_a(z_r)}$ one can obtain $\theta \simeq 10^{-3}$.