2
$\begingroup$

In order to obtain Friedmann equations, we use the metric $g_{\mu \nu}$ and some metric dependent component, i.e Ricci (curvature) tensor and Ricci scalar. Plugging these and energy-momentum tensor $T_{\mu \nu}$ into Einstein field equations yields two equations.

$\textbf{Temporal Part:}$ \begin{align*} R_{00}- \frac{1}{2} R g_{0 0} = \frac{8 \pi G}{c^4} T_{00} \\ -3 \frac{\ddot{a}}{a} + 3c^2 \left[ \frac{\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 + \frac{k}{a^2} \right] &= \frac{8 \pi G }{c^2} \rho \end{align*}

\begin{align} \label{eq15} \boxed{\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \rho - \frac{kc^2}{a^2(t)}} \end{align}

$\textbf{Space part:}$

\begin{align*} R_{ii}- \frac{1}{2} R g_{ii} = 8 \pi G (-p)g_{ii} \end{align*} \begin{align} \label{eq16} \boxed{\frac{2\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 = -\frac{8 \pi G }{c^2}p - \frac{kc^2}{a^2(t)}} \end{align}

However, I do not understand that how can time component have a corresponding element ($\rho$) in energy-momentum tensor. By the way I derived $T_{\mu \nu}$:

\begin{align} \label{eq13} T_{\mu \nu}=(\rho + p)u_\mu u_\nu +pg_{\mu \nu} \end{align} where $g_{\mu \nu}$ is the metric of the manifold and $u_\alpha$ is the 4-velocity of the medium.

I mean it is just "time", how can it determines the evolution of the universe? Although I am using first Friedmann equation, I can not understand the issue discussed above.

$\endgroup$
  • $\begingroup$ Perhaps you should read up on Wikipedia the Stress Energy Tensor. I confess it is a little confusing and it will tae some time to get it, but the concept can only be understood if you mull over it and research various explanations from everywhere. $\endgroup$ – Horus Sep 2 '16 at 18:19
1
$\begingroup$

For the Friedman equations one is interessted in the "expansion of space in homogeneous and isotropic models of the universe". The demand for the universe to be homogeneous and isotropic does only allow a time dependence in the metric. In this framework one describes a universe filled with a perfect fluid with a given energy density $\rho$ and pressure $p$. The fluid itself is static.

The expression you gave for the EM-tensor is correct but you need to put in the four velocities: $$u^{\alpha }\equiv \frac{\text{dq}^{\alpha }}{\text{d$\tau $}}=\gamma \left(1,v^1,v^2,v^3\right).$$ The spatial components $v^i$ are zero since the fluid is static and therefore the Lorentz factor $\gamma$ is one. So we have \begin{align} u^\mu & = (1,0,0,0)\\\\ u_\mu &= g_{\mu\nu}u^\nu=(-1,0,0,0)\end{align} and the standart normalization of $u^\mu u_\mu=-1$ is satisfied.

With that the EM-tensor can be expressed by $\rho$, $p$ and $g_{\mu\nu}$ only and the only time dependence comes from the metric potential $a$. If you plug in the EM-Tensor with this four velocities into the field equations you get the two equations you presented in your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.