Difference between revisions of "Introduction to relativistic theory of small perturbations"

Substituting the expressions $\rho_w = \rho _{w0} + \delta \rho_w;\; a = a_0 + \delta a$ into second Friedman equation (here subscript 0 denotes the unperturbed quantities) and keeping only first order terms, we obtain $$\delta \ddot a = - \frac{8\pi G}{3} a_0\delta \rho _w.$$ Using $$ \frac{\delta \rho _w}{\rho _w} = - 3(1 + w)\frac{\delta a}{a},$$ one could obtain $$ \delta \ddot a - 4\pi G\rho _{w0}(1 + w)(1 + 3w)\delta a = 0. $$ In the general case the solution of this equation has the form: $$ \delta a=C_1e^{\sqrt{4\pi G\rho _{w0}(1+w)(1+3w)}\cdot t}+C_2e^{-\sqrt{4\pi G\rho _{w0}(1+w)(1+3w)}\cdot t}. $$ In the region $-1<w<1/3$, thus, Universe is stable to small perturbations and $\delta \propto \sin(t)$. Outside of this region of $w$, perturbations exponentially increase in time. Such Universe is unstable. The case of Einstein's closed stationary Universe must be considered separately (see Time-dependent_Cosmological_Constant#DE19 problem 1 in chapter 9).

$$ \begin{array}{l} ds^{2} =\left[g_{\mu \nu }^{(0)} +\delta g_{\mu \nu } \right]dx^{\mu } dx^{\nu } \\ {\left|\delta g_{\mu \nu } \right|\ll \left|g_{\mu \nu }^{(0)} \right|} \end{array} $$ Using the conformal time the unperturbed part of the metrics can be expressed in the following form: $$ g_{\mu \nu }^{(0)} dx^{\mu } dx^{\nu } =a^{2} (\eta )\left(d\eta ^{2} -\delta _{ij} dx^{i} dx^{j} \right). $$ Perturbations of the metrics $\delta _{\mu \nu }$ can be scalar, vector and tensor. This classification is based on the invarience of unperturbed metrics under spatial rotations and translations. Scalar component of perturbation is $$ \delta g_{00} =2a^{2} \varphi , $$ where $\varphi (\vec{r},t)$ is a scalar. Vector component $\delta g_{0i}$ can be expressed as a sum of a gradient of scalar function $B$ and vector $S_{i} $ with zero divergence. This component transforms as $$ \delta g_{0i} =a^{2} \left(\frac{\partial B}{\partial x^{i} } +S_{i} \right). $$ Due to the condition $S_{,i}^{i} =0$ this vector has only two independent components. Component $\delta g_{ij} $ transform as a tensor under three--dimensional rotations and can be expressed as $$ \delta g_{ij} =a^{2} \left(2\psi \delta _{ij} +2\frac{\partial E}{\partial x^{i} \partial x^{j} } +\frac{\partial F_{i} }{\partial x^{j} } +\frac{\partial F_{j} }{\partial x^{i} } +h_{ij} \right). $$ Here $\psi $ and $E$ are scalar functions, $F_{i} $ is a vector with zero divergence and $h_{ij} $ is a tensor, which satisfies four conditions: $$ h_{i}^{i} =0;\; h_{j,i}^{i} =0. $$ Scalar perturbations describe inhomogeneities in energy (and mass) distribution. Due to Jeans instability these components lead to the formation of structures in the Universe. Vector perturbations are are associated with rotational motions in fluid. Tensot perturbations have no classical analogs and describe the gravitational degrees of freedom and gravitational waves in particular.

Scalar perturbations describe inhomogeneities in energy distribution and are determined by four functions $\varphi ,\psi ,B,E$. Vector (rotational) perturbations are described by the vectors $S_{i} ,F_{i} $ with corresponding conditions of zero divergence, so that four independent functions describe this class of perturbations. Tensor perturbations are described by two independent functions, since the number of independent components of symmetric order--3 tensor is 6 and four conditions apply: $$ h_{i}^{i} =0;\quad h_{j,i}^{i} =0. $$ Thus, ten independent functions is needed in order to describe the whole variety of considered perturbations.

The background and perturbed parts of four--velocity are \begin{equation}\label{dalta_u} u^0=\frac{1}{a}(1+\delta u^0),~u^i=\frac{1}{a} v^i. \end{equation} Using the general expression $$g_{\mu\nu}u^\mu u^\nu=1.$$ one obtains $$(1+h_{00})(1+2\delta u^0)=1$$ and finally $$\delta u^0=-\frac{1}{2}h_{00}.$$ It's also useful to calculate $u_0$: $$u_0=a^2(1+h_{00})u^0=a\left( 1+h_{00}\right)$$ Spatial components of $v^i$ remain undetermined. The explanation is that local motions of matter in linear approximation do not influence metrics.

The four--velocity is expressed as $$ \delta u^i=\frac{dx^i}{ds}\approx\frac{dx^i}{dt}, $$ which is exact in the first order. Physical velocity is, in turn, $$ v^i=\frac{\delta l^i}{\delta t}=\frac{a dx^i}{dt}, $$ where $\delta l^i=a dx^i$ is physical length. Let's note that only in formulation \eqref{dalta_u} $v^i$ is a physical velocity.

By definition \begin{equation}\label{delta_T} \delta T^\mu_\nu =\widetilde{T}^\mu_\nu -T^\mu_\nu = T^\mu_\nu \left(\widetilde{\rho},\widetilde{p}\right)-T^\mu_\nu \left(\rho,p\right). \end{equation} Substituting the components of four--velocity from previous problem into this expressions and using the expression for unperturbed energy-momentum tensor (see chapter 2), we obtain in linear approximation \begin{eqnarray} \delta T^0_0 &=& \delta \rho \\ \delta T^0_i &=& -(\rho+p)v_i \\ \delta T^i_j &=& - \delta^i_j \delta p. \end{eqnarray}

Due to Lorentz covariance of energy--momentum tensor the four non--trivial equations exist: one for zero component of energy--momentum tensor $\nabla_\mu T^\mu_{0}:$ \begin{equation}\label{cov_ten_00} \delta \rho '+3\mathcal{H}\left( \delta \rho +\delta p\right)+(p+\rho)\left( \partial_i v_i - \frac 12 h'\right)=0 \end{equation} and three for spatial components $\nabla_\mu T^\mu_{i}:$ \begin{equation}\label{cov_ten_ij} \partial_i \delta p+ \left( \rho + p\right)\left(4 \mathcal{H} v_i - \frac 12 \partial_i h_{00}\right)+\left[ (\rho + p)v \right ]'=0 \end{equation} where $h=h_{ii}$, and dash denotes derivative with respect to conformal time.

As is well known (see problem \ref{equ25} in Chapter 2), the following relation holds: $$ h^{\mu \nu } \equiv \delta g^{\mu \nu } = - g^{\lambda \nu }g^{\mu \rho }\delta g_{\rho \lambda}, $$ with components $h^{ij} = - a^{ - 4}h_{ij};\;h^{i0} = a^{ - 2}h_{i0}; \;h^{00} = - h_{00}.$ Energy--momentum tensor for ideal fluid has the form: $$ T_{\mu \nu } = \left( p + \rho \right)u_\mu u_\nu - pg_{\mu \nu }, $$ where $g^{\mu \nu }u_\mu u_\nu = 1$. Thus, $$ u_\mu u_\nu \delta g^{\mu \nu } = - 2g^{\mu \nu }u_\mu\delta u_\nu = u^\nu \delta u_\nu. $$ For ideal fluid filling the Universe ${u_0} = 1,\:{u_i} = 0$, which immediately gives $$ \delta {u_0} = \delta {u^0} = \frac{1}{2}h_{00}, $$ $$\delta T_{\mu \nu } = u_\mu u_\nu \delta \left( p + \rho \right) + \left( u_\mu \delta u_\nu + u_\nu\delta u_\mu \right)\left( p + \rho\right) - p\delta g_{\mu \nu } - g_{\mu \nu }\delta p. $$ $$ \begin{gathered} \delta T_{ij} = - w\left(\rho h_{ij} - a^2\left( t \right)\delta _{ij}\delta \rho \right); \\ \delta T_{i0} = \left( 1 + w \right)\rho \delta u_i - w\rho h_{i0};\; \\ \delta T_{00} = \delta \rho + w\rho h_{00}.\\ \end{gathered} $$

The second order tensor is transformed according to $$ g_{\alpha \beta }\left(\tilde x^\rho \right) = \frac{\partial x^\gamma } {\partial \tilde x^\alpha }\frac{\partial x^\delta } {\partial \tilde x^\beta }{g_{\gamma \delta }}\left( {x^\rho } \right) \approx {}^{(0)}g_{\alpha \beta } + \delta g_{\alpha \beta } - {}^{(0)}g_{\alpha \delta }\xi _{,\beta }^\delta - {}^{(0)}g_{\gamma \beta }\xi _{,\alpha }^\gamma , $$ where only linear terms in $\delta g_{\alpha \beta }$ è ${\xi ^\rho }$ are holded. In new coordinates $\tilde x$ the metrics can be expressed as a sum of background and perturbed parts: $$ \tilde g_{\alpha \beta }\left( \tilde x^\rho \right) = {}^{(0)}g_{\alpha \beta }\left( \tilde x^\rho \right) + \delta g_{\alpha \beta }, $$ where ${}^{(0)}g_{\alpha \beta }$ is an unperturbed metrics, which depends on $\tilde x$.At the same time, it easy to obtainthe following relation: $$ {}^{(0)}g_{\alpha \beta }\left( x^\rho \right) = {}^{(0)}g_{\alpha \beta }\left( \tilde x^\rho \right) - {}^{(0)}g_{\alpha \beta ,\gamma }{\xi ^\gamma }, $$ Finally, the transformation is $$ \delta g_{\alpha \beta } \to \delta \tilde g_{\alpha \beta } = \delta g_{\alpha \beta } - {}^{(0)}g_{\alpha \beta ,\gamma }\xi ^\gamma- {}^{(0)}g_{\gamma \beta }\xi _{,\alpha }^\gamma - {}^{(0)}g_{\alpha \delta }\xi _{,\beta }^\delta . $$

In conformal time the metrics of Friedman universe has the form: $$ g_{00} = a^2\left( \eta \right),\;g_{ij} =-a^2\left( \eta \right)\delta_{ij}. $$ using the results of previous problem: $$ \begin{gathered} \delta \tilde g_{00} = \delta g_{00} - 2a\left( a\xi ^0\right)^\prime,\\ \delta \tilde g_{0i}= \delta g_{0i}+ a^2\left[ {\xi '}_{ \bot i} + \left( \zeta ' - \xi ^0 \right)_{,i} \right],\\ \delta \tilde g_{ij} = \delta g_{ij}+ a^2\left[ 2\frac{a'} {a}\delta _{ij}\xi ^0 + \zeta _{,ij} + \left( \xi _{ \bot i,j} +\xi _{ \bot j,i} \right)\right], \\ \end{gathered} $$ where the relation $\xi _{\bot i}\equiv \xi _ \bot ^i$ is used and dash denoted the derivative with the respect to conformal time $\eta$.

It is easy to demonstrate by direct substitution, that transformations $$ t'=t-\lambda^{-1/2}f(t,\vec{x}),~ \vec{y}=\vec{x}-\mathcal{\nabla B},~ \mathcal{B}=\lambda^{-1/2}\int a^{-2}f(t,\vec{x})dt,~a=a_0(t')=a_0\cdot (1-f(t,\vec{x})), $$ with $f(t,\vec{x}) = k\frac{\vec{x}{\,}^2}{2},$ convert the metrics $$ ds^2= \left(1-\frac{2}{\sqrt{\lambda}}\dot{f}(t,\vec{r})\right)dt^2-a^2(\delta_{ij}-2\mathcal{B}_{,ij})dx^idx^j, $$ into the metrics for homogeneous space: $$ ds'^2=dt'^2-a^2d\vec{y}. $$