Difference between revisions of "Bianchi I Model"

If we admit the energy-momentum tensor of a perfect fluid, then the field equations of the BI universe are found as, \begin{eqnarray} \label{feforgm}\frac{{\dot{a}_{1}}{\dot{a}_{2}}}{a_{1} a_{2}}+\frac{{\dot{a}_{1}}{\dot{a}_{3}}}{a_{1} a_{3}}+\frac{{\dot{a}_{2}}{\dot{a}_{3}}}{a_{2} a_{3}}&=&\rho,\\ \frac{{\ddot{a}_{1}}}{a_{1}}+\frac{{\ddot{a}_{3}}}{a_{3}}+\frac{{\dot{a}_{1}}{\dot{a}_{3}}}{a_{1} a_{3}}&=& -p,\\ \frac{{\ddot{a}_{2}}}{a_{2}}+\frac{{\ddot{a}_{1}}}{a_{1}}+\frac{{\dot{a}_{2}}{\dot{a}_{1}}}{a_{2} a_{1}}&=&-p,\\ \frac{{\ddot{a}_{3}}}{a_{3}}+\frac{{\ddot{a}_{2}}}{a_{2}}+\frac{{\dot{a}_{3}}{\dot{a}_{2}}}{a_{3} a_{2}}&=&-p. \end{eqnarray}

Inserting the directional Hubble parameters and their time derivatives \[\dot H_1=\frac{\ddot a_1}{a_1}-\left(\frac{\dot a_1}{a_1}\right)^2,\ \dot H_2=\frac{\ddot a_2}{a_2}-\left(\frac{\dot a_2}{a_2}\right)^2,\ \dot H_3=\frac{\ddot a_3}{a_3}-\left(\frac{\dot a_3}{a_3}\right)^2\] into the modified Friedmann equations we obtain \begin{align} \nonumber H_1H_2+H_1H_3+H_2H_3 & =\rho,\\ \nonumber \dot H_1+ H_1^2 +\dot H_3+ H_3^2 +H_1H_3& =-p,\\ \nonumber \dot H_1+ H_1^2 +\dot H_2+ H_2^2 +H_1H_2& =-p,\\ \nonumber \dot H_2+ H_2^2 +\dot H_3+ H_3^2 +H_2H_3& =-p. \end{align}

\[\rho_{cr}=H_1H_2+H_1H_3+H_2H_3.\]

The energy conservation equation $T^\mu_{\nu;\mu}=0$ yields \[\dot\rho+3\bar H(\rho+p)=0,\quad \bar H\equiv \frac13(H_1+H_2+H_3)=\frac13\left(\frac{\dot a_1}{a_1} +\frac{\dot a_2}{a_2} +\frac{\dot a_3}{a_3}\right),\] where $\bar H$ represents the mean of the three directional Hubble parameters in the BI Universe.

Adding the three latter Friedmann equations (see Problem \ref{bi_2}) one obtains \begin{equation}\label{bi_5_1} 2\frac{d}{dt}\sum\limits_{i=1}^3 H_i+2(H_1^2+H_2^2+H_3^2)+H_1H_2+H_1H_3+H_2H_3=-3p. \end{equation} where $\bar H$ represents the mean of the three directional Hubble parameters in the BI Universe. Substituting \[\sum\limits_{i=1}^3 H_i^2=\left(\sum\limits_{i=1}^3 H_i\right)^2-2(H_1H_2+H_1H_3+H_2H_3)\] and \[H_1H_2+H_1H_3+H_2H_3=\rho\] into equation (\ref{bi_5_1}), we then obtain \[\frac{d}{dt}\sum\limits_{i=1}^3 H_i+\left(\sum\limits_{i=1}^3 H_i\right)^2=\frac32(\rho-p).\] Using the mean of the three directional Hubble parameters $\bar H$ we obtain a nonlinear first order differential equation \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p).\]

\[\bar H=\frac13\frac{d}{dt}\ln(a_1a_2a_3)=\frac13\frac{d}{dt}\ln V=\frac13\frac{\dot V}{V}.\]

Using the relation between volume $V$ and the mean Hubble parameter $\bar H$, obtained in the previous problem, one finds \[\dot{\bar H}=\frac13\frac{\ddot V}{V}-3\bar H^2.\] As \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p),\] we obtain \[\ddot V-\frac32(\rho-p)V=0.\]

The equations \begin{align} \nonumber \dot H_1+ 3H_1\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_2+ 3H_2\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_3+ 3H_3\bar H & =\frac12(\rho-p), \end{align} allow us to write the generic solution of the directional Hubble parameters, \[H_i(t)=\frac1{\mu(t)}\left[K_i+\frac12\int\mu(t)(\rho(t)-p(t))dt\right],\quad i=1,2,3,\] where $K_i$s are the integration constants. The integration factor $\mu$ is defined as, \[\mu(t)=\exp\left(3\int\bar H(t)dt\right).\] As can be seen, the initial values (integration constants) determine the solution of each directional Hubble parameter. These values are the origin of the anisotropy. Note that the generic solution of the directional Hubble parameters is incomplete. To obtain exact solutions of the Hubble parameters and therefore the Einstein equations, one has to know the state equation for the component which fills the Universe.