Difference between revisions of "Non relativistic small perturbation theory"

General equations of hydrodynamics and gravity for ideal gas in Newtonian approximation are $$ \begin{gathered} \frac{\partial \rho} {\partial t} + \nabla \cdot \left(\rho \vec v \right) = 0, \hfill \\ \frac{\partial \vec v} {\partial t} + \left(\vec v\nabla \right)\vec v + \frac{1} {\rho }\nabla P + \nabla \Phi = 0, \hfill \\ \Delta \Phi = \nabla \cdot \left(\nabla \Phi\right) = 4\pi G\rho , \hfill \\ \frac{\partial S} {\partial t} + \left(\vec v\nabla \right)S = 0, \hfill \\ \end{gathered} $$ where $\rho$ is density, $\vec v$ is velocity, $S$ is unit entropy of matter and $\Phi$ is gravitational potential. Perturbed solution could be expressed as follows: $$\begin{gathered} \rho (\vec x,t) = \rho _0\left[ 1 + \delta (t)e^{i\vec k\vec x}\right];\hfill \\ \vec v(\vec x,t) = \vec w(t)e^{i\vec k\vec x};\hfill \\ p (\vec x,t) = p_0 + f(t)e^{i\vec k\vec x};\hfill \\ S = S_0 + \sigma (t)e^{i\vec k\vec x};\hfill \\ p = p_0 + \frac{\partial p} {\partial \rho}\left( \rho - \rho _0 \right) + \frac{\partial p} {\partial S}\left( S - S_0\right) = p_0+ c_S^2\rho _0\delta e^{i\vec k\vec x} + h^2\sigma e^{i\vec k\vec x}. \hfill \\ \end{gathered} $$ In latter expression we use notation $$ \frac{\partial p} {\partial \rho} = h^2,\;\frac{\partial p} {\partial S} = c_S^2 $$ where $c_S^2$ is adiabatic speed of sound. Substituting these expressions into hydrodynamic equations, we obtain the foolowing system: $$ \begin{gathered} \frac{d\delta} {dt} + i\vec k\vec w = 0; \hfill \\ \frac{d\vec w} {dt} + i\vec kf + i\vec kc_S^2\delta+i\vec kh^2\rho _0^{-1}\sigma =0; \hfill \\ {k^2}f = - 4\pi {\rho _0}\delta ; \hfill \\ \frac{d\sigma } {dt} = 0. \hfill \\ \end{gathered} $$

Let's denote the solution as $\varphi(t)$, where $\varphi$ is a set of quantities $\delta ,w,f,\sigma$, defined in the previous problem. Since equations for small perturbations are linear and homogeneous, $A\varphi (t),\; A=const,$ is also a solution. At the same time, since coefficients do not depend on time, solution could be shifted in time, so that $A\varphi (t+\tau)$ is a solution too. Assuming that there exist such solutions (eigensolutions), that both groups of transformations yield the same result, i.e. $A\varphi (t) = \varphi (t + \tau ),$ and each $\tau$ has corresponding (one or more) value of $A$. Such functional relationship takes place if $\varphi = \varphi _0e^{\omega t},A = e^{\omega t}$. Thus, the dependence of perturbations on time has the form: $$ \begin{gathered} \delta = \delta _0e^{\omega t}, \hfill \\ \vec w =\vec w_0e^{\omega t}, \hfill \\ f = f_0e^{\omega t},\hfill \\ \sigma = \sigma _0e^{\omega t}. \hfill \\ \end{gathered} $$ The equations for perturbations in this case (see. problem \ref{per2}) have the form: $$ \omega \delta _0 = - ikw_0; $$ $$ \omega \vec w_0 = - i\vec kf_0 - i\vec kc_S^2\delta _0 - i\vec k\rho _0^{ - 1}h^2\sigma_0; $$ $$ \; - k^2f_0 = 4\pi G\rho _0\delta _0;\;\omega \sigma _0 = 0. $$

In this case we have $\sigma_0 =0$ and vector $\vec w_0$ (see. problem 2) is parallel to wavevector $\vec k$. Denoting \[\vec w_0 = w_0 \frac{\vec k}{k},\] we obtain the equations for time dependent perturbations: $$ \begin{gathered} \omega \delta _0 = - ikw_0, \hfill \\ \omega \vec w_0 = - i\vec kf_0 - i\vec kc_S^2\delta _0, \hfill \\ - k^2f_0 = 4\pi G\rho _0\delta _0,\hfill \\ \omega \sigma _0 = 0. \hfill \\ \end{gathered} $$ This system has non--trivial solution $$ \omega = \pm \sqrt {4\pi G\rho _0 - c_S^2k^2}. $$ The properties of this solution critically depend on the sign of radicand. Given the characteristics of matter $(\rho_0$ and $c_S^2)$, the critical value of $k_J$ is determined by condition $\omega = 0:$ $$ \begin{gathered} 4\pi G\rho _0 - c_S^2k_J^2 = 0, \hfill \\ k_J = \frac{1} {c_S}\sqrt {4\pi G\rho _0},\hfill \\ \lambda _J = \frac{2\pi} {k_J} = c_S\sqrt {\frac{\pi } {G\rho _0}} . \hfill \\ \end{gathered} $$

a) $$ \begin{gathered} 4\pi G{\rho _0} - c_S^2{k_J}^2 > 0,\hfill \\ \omega _1 = - \omega _2 = \omega \operatorname{Im}\delta _0 = 0,\hfill \\ \rho = \rho _0\operatorname{Re}(1 + \delta_0e^{\omega _{1,2}t+i\vec k\vec x}) = \rho _0(1 + \delta _0e^{\omega _{1,2}t}\cos \vec k\vec x), \hfill \\ \vec v = \operatorname{Re}(i\frac{\vec k} {k}e^{i\vec k\vec x}\frac{\omega } {k}\delta _0e^{\omega _{1,2}t}) = - \frac{\vec k\omega} {k}e^{\omega _{1,2}t}\sin \vec k\vec x .\hfill \\ \end{gathered} $$ The solution with positive $\omega,$, which increases as $e^{\omega t}(\omega > 0),$ corresponds to gravitational instability of homogeneous distribution of matter. At $\lambda \to \infty $ increment is $\omega \to \sqrt {4\pi G\rho _0}.$ The time scale $\tau$ of perturbations $e$--fold increase $(\tau = 1/{\omega} = 1/{\sqrt {4\pi G\rho _0}})$ is of order of cosmological time, during which density increses from $\rho =\infty$ to $\rho=\rho_0$ in Friedmann model.
b)\[ \begin{gathered} \omega = \pm i\sqrt {c_S^2k^2 - 4\pi G\rho _0}, \hfill \\ \rho = \rho _0\left[ 1 - \delta _0\cos \left( \vec k\vec x - \left| \omega \right|t \right) \right],\hfill\\ \vec v = \frac{\vec k\omega } {k}\cos \left( \vec k\vec x - \left| \omega \right|t \right),\hfill \\ \left| \omega \right| = \sqrt {c_S^2k^2 - 4\pi G\rho _0}. \hfill \\ \end{gathered} \] Long wave perturbations are mostly influenced by gravity, which cause the instability of homogeneous distribution of matter and exponentially increasing density perturbations. Pressure in this case could be neglected. Consider the case of short waves: $$ k^2 \gg k_J^2 = \frac{4\pi G\rho _0}{c_S^2}. $$ In this case the solution is just sound waves: $$ \begin{gathered} \rho =\rho _0\left[1 -\delta _0\cos \left(\vec k\vec x \right)\right], \hfill \\ \vec v = \frac{\vec k\omega } {k}\cos \left( \vec k\vec x\right). \hfill \\ \end{gathered} $$ Pressure plays a main role for short wave perturbations: perturbations of density cause the pressure perturbations, which leads to propagation of acoustic waves with constant amplitude. Gravity can be neglected for short wave perturbations, wince it once changes speed of sound in higher approximations, while the amplitude remains the same and instability does not arise.

Expressing small perturbations in the form $$ \begin{gathered} \tilde{\rho} = \rho + \delta \rho \left(\vec x,t \right);\hfill \\ \vec{\tilde {V}} = \vec V + \delta \vec v;\hfill \\ \tilde \Phi = \Phi + \delta \Phi; \hfill \\ \tilde p = p + \delta p. \hfill \\ \end{gathered} $$ Substituting these expressions into hydrodynamic equations, we obtain the following linearized equations $$ \begin{gathered} \frac{\partial \delta \rho} {\partial t} + \rho \nabla \delta \vec v + \nabla \left( \delta \rho \cdot \vec V \right) = 0; \hfill \\ \frac{\partial \delta \vec v} {\partial t} + \left( \vec V \cdot \nabla \right)\delta \vec v + \left( \delta \vec v \cdot \nabla \right)\vec V + \frac{c_s^2} \rho\nabla \delta \rho + \nabla \delta \Phi = 0;\hfill \\ \Delta \delta \Phi = 4\pi G\delta \rho . \hfill \\ \end{gathered} $$ Transforming Euler coordinates $\vec X$ to Lagrangian coordinates $\vec Y$ according to expression $\vec X\left(\vec x,t \right) = a\left( t \right)\vec Y$ and accounting for $$ \left( \frac{\partial } {\partial t} \right)_{\vec X} = \left( \frac{\partial } {\partial t} \right)_{\vec Y} - \left( \vec V \cdot \nabla _{\vec Y}\right), $$ we obtain the transformation for space coordinates $$\nabla _{\vec X} = \frac{1} {a}\nabla _{\vec Y}. $$ Denoting $\delta = \delta \rho /\rho$, one could rewrite equations in the form $$\begin{gathered} \frac{\partial \delta }{\partial t} + \frac{1}{a}\nabla \delta \vec v = 0;\hfill\\ \frac{\partial \delta }{\partial t} + H\delta \vec v + \frac{c_s^2}{a}\nabla \delta + \frac{1}{a}\nabla \Phi = 0;\hfill \\ \Delta \delta \Phi = 4\pi G\rho \delta , \hfill \\ \end{gathered} $$ where $\nabla = \nabla _{\vec Y}$. Applying divergence to both sides of second equation and using the continuity equation and Poisson equation, we can express $\nabla \delta \vec v$ è $\Delta \delta \Phi $ in terms of $\delta$. As a result, we obtain closed equation with respect to $\delta$, which describes gravitational instability in expanding Universe: $$ \ddot \delta + 2H\dot \delta - \frac{c_s^2} {a^2}\Delta \delta - 4\pi G\rho \delta = 0. $$

Expanding the perturbation in plane waves $$ \delta \left( t,\vec Y \right) = \frac{1} {\left( 2\pi \right)^{2/3}}\int\limits_{ - \infty }^\infty \delta \left( t \right)\exp \left( {i\vec k\vec Y} \right)d\vec k , $$ and substituting into the equation from previous problem we obtain $$ \ddot \delta _{\vec k} + 2H\dot \delta _{\vec k} + \left( \frac{k^2c_s^2} {a^2} - 4\pi G\rho \right)\delta _{\vec k} = 0. $$ Evolution of each perturbation critically depends on quantity with dimensionality of length: $$ \lambda _J^{ph} = \frac{2\pi a} {k_J} = c_S\sqrt {\frac{\pi } {G\rho }}. $$ In matter dominated Universe the density is $$ \rho = \frac{1}{6\pi Gt^2} $$ and thus $$\lambda _J^{ph} = 2\pi c_St \sim c_St.$$ This means that in comoving coordinates Jeans wavelength is of order of magnitude of sound horizon.

a) In matter dominated Universe we have $$ \ddot{\delta }+2H\dot{\delta }=\delta \left(4\pi G\rho _{0} -\frac{c_{s}^{2} k^{2} }{a^{2} } \right). $$ This equation can be solved analytically if term $c_{s}^{2} k^{2} /a^{2} $ is small in comparison to $4\pi G\rho _{0} $. In physical terms it means that dynamics is determined by Hubble expansion or, in other words, we neglect Jeans mass in comparison to mass of galaxy.
In case of matter domination we have: $$ a(t)=\left(\frac{t}{t_{0} } \right)^{2/3} ;\quad H=\frac{2}{3t};\quad \rho =\frac{1}{6\pi Gt^{2} } $$ and equation for density fluctuations is $$ \ddot{\delta }+\frac{4}{3t} \dot{\delta }-\frac{2}{3t^{2} } \delta =0. $$ Expressing the as $At^{n}$ we obtain the equation for $n$ $$ n^{2} +\frac{1}{3} n-\frac{2}{3} =0, $$ which has the following solutions \[n=\frac{2}{3} ,-1,\] and, thus, $$ \delta =At^{2/3} +Bt^{-1}. $$ The increasing mode is $\propto t^{2/3} $, while decreasing one is $t^{-1} $. Even if these modes are comparable at the beginning of evolution, only increasing mode matters at later times and we can set $B=0$.
b)In contrast to previous case, the pressure of radiation makes substantial contribution to gravitational interaction. In this case $p=\frac{1}{3} \rho c^{2} $ and effective density is $\rho +3p/c^{2} =2\rho$, so that hydrodynamic equations have to be modified. Modified factor aooears in in the equation for density fluctuations: $$ \ddot{\delta }+2H\dot{\delta }=\delta \left(\frac{32}{3} \pi G\rho _{0} -\frac{c_{s}^{2} k^{2} }{a^{2} } \right); $$ $$ a(t)=\left(\frac{t}{t_{0} } \right)^{1/2} ;\quad H=\frac{1}{2t} ;\quad \rho =\frac{3}{32\pi Gt^{2} }. $$ In the same approximation (i.e., considering that Jeans wavenumber is greater than the scale of any inhomogeneity): $$ \ddot{\delta }+\frac{1}{t} \dot{\delta }-\frac{1}{t^{2} } \delta =0. $$ Thus, the solution is $At^{n};\; n^{2} =1,\; \delta =At+Bt^{-1}$.

As is well known, general solution of a second--order ordinary differential equation is a superposition of two linear independent particular solutions. The following trick could be used to find the second particular solution. By taking the time derivarive of Wronskian of the solutions $\dot W = \ddot \delta _1\delta _2 - \delta _1\ddot \delta _2$ and substituting the second derivatives using equation $$\ddot \delta = - \left(\frac{k^2c_s^2} {a^2} - 4\pi G\rho \right)\delta - 2H\dot \delta $$ one obtains the equation $\dot W = - 2HW$, which can be easily integrated as $W = \frac{A}{a^2}$. Thus, $$ W =\dot\delta_1\delta _2-\delta_1\dot\delta_2=\frac{A}{a^2}. $$ Using the ansatz $\delta _2(t) = \delta _1\left( t \right)f\left( t \right)$: $$ f\left( t \right) = - B\int \frac{dt} {a^2\delta _1^2}. $$ General solution is $$ \delta \left( t \right) = \delta _1\left( t \right) + \delta _1\int \frac{dt} {a^2\delta _1^2}, $$ or, taking the relation $\delta _1\left( t \right) = C_1H\left( t \right)$ into account, $$ \delta \left( t \right) = C_1H\left( t \right) + C_2H\int \frac{dt} {a^2H^2} $$ In flat Universe filled with substance with equation of state $p = w\rho$: $$ \delta \left( t \right) = C_1t^{ - 1} + C_2t^{ - \frac{2}{3(1 + w)}}. $$

Let's consider the small perturbations to physical quantities $\rho ,v,\vec{F},p$, denoting them as $\rho_{1},v_1,\vec{F}_1,p_1$. Since these perturbations are small, the equations can be linearized: \begin{equation}\label{7.7per9} \dot{\rho }_1+\frac{\dot{a}(t)}{a(t)}\left( \vec{r}\cdot \nabla \right)\rho_1+3\frac{\dot{a}(t)}{a(t)}\rho_1+\rho \nabla \cdot \vec{v}_1=0 \end{equation} \begin{equation}\label{7.8per9} \dot{\vec{v}}_1+\frac{\dot{a}(t)}{a(t)}\left[ \left( \vec{r}\cdot \nabla \right)v_1+v_1\right]+\frac{1}{\rho }\nabla p_1+\vec{F}_1=0 \end{equation} \begin{equation}\label{7.9per9} \nabla \times F_1=0,\quad \nabla \cdot \vec{F}_1=-4\pi G\rho_1 \end{equation} Additionaly, for adiabatic fluctuations we have \begin{equation}\label{7.10per9} p_1=c_{s}^{2}\rho_1 \end{equation} where ${{c}_{s}}$ is the speed of sound, $c_s=\sqrt{dp_1/d\rho_1}$. The plane--wave solution of equations \eqref{7.7per9}-\eqref{7.9per9} is \begin{equation}\label{7.11per9} \rho_1(\vec{r},t)=\bar{\rho }_1e^{i\chi },\quad \vec{v}_1(\vec{r},t)=\vec{\bar{v}}_1(t)e^{i\chi} \end{equation} where \begin{equation}\label{7.12per9} \chi =\frac{\vec{r}\cdot \vec{k}}{a(t)} \end{equation} The corresponding equations for Fourier amplitudes are \begin{equation}\label{7.13per9} \dot{\bar{\rho}}_1+\frac{3\dot{a}}{a}\bar{\rho}_1+\frac{i\vec{k}\cdot \bar{v}_1}{a}\rho =0 \end{equation} \begin{equation}\label{7.17per9} \vec{\bar{v}}_1+\frac{\dot{a}}{a}\vec{\bar{v}}_1+\frac{ic_{s}^{2}}{a\rho }\vec{k}\bar{\rho}_{1}-\bar{F}_{1}=0 \end{equation} \begin{equation}\label{7.141per9} \vec{\bar{v}}_1+\frac{\dot{a}}{a}\vec{\bar{v}}_1+\frac{ic_{s}^{2}}{a\rho }\vec{k}\bar{\rho}_1-\bar{F}_1=0 \end{equation} \begin{equation}\label{7.14per9} \vec{k}\times \vec{F}_1=0,\quad i\vec{k}\cdot \bar{F}_1=-4\pi G \bar{\rho}_1a \end{equation} Spliting $\vec{\bar{v}}_1$ on normal and parallel to $\vec{k}$ components: \begin{equation}\label{7.16per9} \vec{\bar{v}}_1=\frac{\vec{k}\cdot \vec{\bar{v}}_1}{k^2}\vec{k}+\frac{\vec{k}\times \left(\vec{\bar{v}}_1\times \vec{k} \right)}{k^2}=\vec{v}_{\parallel}+\vec{v}_{\bot}. \end{equation} taking vector product of \eqref{7.16per9} and $\vec{k}$, we obtain \begin{equation} \left( \dot{\vec{v}}_1+\frac{\dot{a}}{a}\vec{v}_1 \right)\times \vec{k}=0. \end{equation} Thus, \begin{equation}\label{7.16per9} \vec{v}_{\bot} a=const \end{equation} This means that transverse or rotational mode in expanding Universe decreases.