Difference between revisions of "CMB anisotropy"

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=== Problem 1 ===
 
=== Problem 1 ===
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<p style= "color: #999;font-size: 11px">problem id: </p>
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<p style= "color: #999;font-size: 11px">problem id: per25</p>
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Obtain the equation of motion for photon in metrics ($ds^2=a^2(\eta)[(1+2\Phi)d\eta^2-(1-2\Phi)\delta_{ij}dx^idx^j]$) in linear approximation in $\Phi$.
 
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     <p style="text-align: left;">Equation of geodesics describing the propagation of radiation in space with arbitrary curvature can be represented as:
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$$
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\frac{dx^\alpha }
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{d\lambda } = p^\alpha,\; \frac{dp_\alpha }
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{d\lambda} = \frac{1}
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{2}\frac{\partial g_{\gamma \delta }}
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{\partial x^\alpha }p^\gamma p^\delta,
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$$
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where $\lambda$ is an arbitrary affine parameter, which is taken along geodesic. Since photon has zero mass, the first integral of this equation is:
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$$
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p^\alpha p_\alpha = g^{\mu \nu }p_\mu p_\nu  = 0.
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$$
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Using this relation, zero component can eliminated from the equations of geodesics:
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$$
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p^0 = \frac{1}
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{a^2}\left( \sum\limits_{i = 1}^3 p_i^2 \right)^{1/2} = \frac{p}{a^2}
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$$
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and
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$$
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p_0 = \left( 1 + 2\Phi\right)p.
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$$
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Hence,
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$$
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\frac{dx^i}
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{d\eta } = \frac{p^i}
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{p^0} = \frac{ - \frac{1}
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{a^2}\left( 1 + 2\Phi  \right){p_i}}
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{p^0} = r^i\left( 1 + 2\Phi\right),
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$$
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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:
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$$\frac{dp_\alpha }
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{d\eta} = \frac{1}
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{2}\frac{\partial g_{\gamma \delta }}
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{\partial x^\alpha }\frac{p^\gamma p^\delta }
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{p^0} = 2p\frac{\partial \Phi }
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{\partial x^\alpha }.
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$$</p>
 
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=== Problem 1 ===
 
=== Problem 1 ===
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<p style= "color: #999;font-size: 11px">problem id: </p>
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<p style= "color: #999;font-size: 11px">problem id: per26</p>
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For the Universe, dominated by a substance with equation of state $ p = w \rho $, connect in the first approximation the fluctuations of the gravitational potential of the CMB with $ \Phi$.
 
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     <p style="text-align: left;">Using the invariant $aT = const$:
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$$
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\frac{\delta T}{T} =-\frac{\delta a}{a}.
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$$
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In case of $p = w\rho ,\; a\left( t \right) = t^{\frac{2}
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{3\left( 1 + w \right)}}$ and, hence,
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$$
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\frac{\delta a}
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{a} = \frac{2}
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{3\left(1 + w\right)}\frac{\delta t}{t}.
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$$
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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
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$$
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d\tau=\sqrt{g_{00}}dt\approx\sqrt{1 + 2\Phi } dt \approx \left(1+\Phi\right)dt,
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$$
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where $\Phi$ is a Newtonian gravitational potential and, thus, ${\delta t}/t \simeq \Phi$:
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$$
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\left(\frac{\delta T}{T} \right)_e =  - \frac{2}{3\left( 1 + w\right)}{\Phi_e}
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$$
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where subscript $e$ denotes the moment of emission.
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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
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$$
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\left( \frac{\delta T}
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{T} \right)_0 = \left(\frac{\delta T}
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{T}\right)_e + \Phi _e
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$$
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where subscript $0$ denotesthe moment of detection. Finally,
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$$
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\left(\frac{\delta T}
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{T} \right)_0 = \left(\frac{1 + 3w}
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{3 + 3w} \right)\Phi _e
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$$</p>
 
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=== Problem 1 ===
 
=== Problem 1 ===
 
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Estimate the spatial scale of Silk effect. Assume that at the temperatures we are interested in, photon changes direction randomly, and  its energy does not change when scattering on electrons.
 
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     <p style="text-align: left;">In this approximation, the oscillations of the photon component
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smooth on scales smaller than the distance at which the photon diffuses
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in characteristic period of evolution. If considering the before recombination era this time is the Hubble time.
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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
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$$H(z_r)=H_0\sqrt{\Omega_{m0}(1+z_r)^3+\Omega_{r0}(1+z_r)^4}\simeq 5\, \mbox{Mpc}^{-1}.$$
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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
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$$n_e(z)=0.75 \frac{\rho_b(z)}{m_p}=6\cdot 10^{-4}\Omega_b (1+z)^3 \mbox{sm}^{-3}.$$
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At the beginning of recombination the number density of free electrons is
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$n_e(\eta_r)=230\,\mbox{sm}^{-3}.$
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Hence,
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$$\lambda_S(\eta) \simeq \sqrt{\frac{1}{\sigma_{_T}}n_e(\eta) H(\eta)},$$
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which gives $\lambda_S(\eta_r) \simeq 0.02 \,\mbox{Mpc.}$</p>
 
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=== Problem 1 ===
 
=== Problem 1 ===
 
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Estimate the angular scale of the CMB anisotropy due to the Silk effect.
 
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     <p style="text-align: left;">Physical distance $R(z)$  is connected to the angular diameter distance $R_a(z)$ as
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$$R_a(z)=\frac{1}{1+z}R(z),$$ 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.
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Using the relation $\theta \simeq \frac{\lambda_S}{R_a(z_r)}$ one can obtain $\theta \simeq 10^{-3}$.</p>
 
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Revision as of 00:51, 25 February 2014

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