Conservation law for FRW metric

  • #1
109
14

Homework Statement:

Given the FRW metric
$$d s^{2}=-d t^{2}+a(t)^{2}\left[\frac{d r^{2}}{1-k r^{2}}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \varphi^{2}\right)\right]$$
and the Stress-energy tensor of an ideal fluid,
$$T_{a b}=(P+\rho) u_{a} u_{b}+P g_{a b}, \quad \text{where}\quad P=\omega \rho\quad \text{and}\quad u^a:=\frac{dx^a}{d\tau},$$
My Prof. claimed during yesterdays lecture that one can show that the zero component of the convervation equation of ##T_{ab}## results in
$$\frac{d}{d t} \log \rho\propto\frac{d}{d t} \log a(t)\tag{1}.$$
I've been trying to show this now for a while and I just can't figure it out...

Relevant Equations:

All given above.
My attempt:
  1. Realize we can work in whatever coordinate system we want, therefore we might as well work in the rest frame of the fluid. In this case ##u^a=(c,\vec{0})##.
  2. The conservation law reads ##\nabla^a T_{ab}=0##. Let us pick the Levi-Civita connection so that we don't have to worry about ##\nabla g##. We then get $$\begin{align*}\nabla^a T_{ab} &= \nabla^a\left((1+\omega)\rho u_{a} u_{b}+\omega\rho g_{a b}\right)\\&= (1+\omega)\nabla^a(\rho u_au_b)+ \omega g_{ab}\nabla^a\rho\\&\overset{(*)}{=} (1+\omega) u_au_b\nabla^a\rho+ \omega g_{ab}\nabla^a\rho\\&= \left((1+\omega) u_au_b+ \omega g_{ab}\right)\nabla^a\rho,\end{align*}$$ where in ##(*)## we use the fact that ##u^a## are constant in the comoving frame.
  3. For the zero component I now get $$\begin{align*}\nabla_a T^{a0} &= \left((1+\omega) u^au^0+ \omega g^{a0}\right)\nabla_a\rho = \left((1+\omega)c u^a+ \omega g^{a0}\right)\nabla_a\rho\\&= -(1+\omega)c^2 \nabla_0\rho -\omega \nabla_0\rho\\&= -(1+\omega)c^2 \partial_t\rho -\omega \partial_t\rho\\&\overset{!}{=}0 \end{align*}$$ I don't really think there is a way to get to the claimed result from here, since I'm completely missing the scale factor ##a##.
Questions
I suspect that my problem lies in the fact that I'm not sure if
  • I'm actually allowed to pick a specific frame of reference and work in it and
  • I understand correctly how the covariant derivative acts on scalars (##\rho##) and constant values (##u^a##, since they are only zero or ##c## in the comoving frame).
The answer to the second question is, I think, that the covariant derivative in this case just acts like a "normal" derivative. But I'm of course not sure and it isn't working... So I'm presumably misunderstanding something. Any hints on how to show this are appreciated!
 

Answers and Replies

  • #2
164
38
Yes. It is better to pick up a specific frame. By Principal of equivalence, it should hold in every frame. In the comoving frame in which fluid elements is at origin, the connection coefficient simplify considerably.

Some Hints:
##\nabla_j T^{0j}=\frac {\partial T^{0j}}{\partial x_j}+ \Gamma^0_{\mu\nu}T^{\mu\nu}+\Gamma^{\mu}_{\mu\nu}T^{0\nu}##

Now: ##T^{00}=\rho, T^{0i}=0, T^{ij}=P##


And near origin ##g_{ij}=a^2 \delta_{ij}##
 

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