I'm trying to derive the equation for the cosmological fluid:
$$\dot \rho + 3 \frac{\dot a}{a}(\rho +P)=0$$
by starting from the conservation of the stress-energy tensor:
$$\nabla^\mu T_{\mu \nu} = 0$$
with the stress-energy for a perfect fluid in its own frame being:
$$ T_{\mu \nu} = \text{diag} (\rho, a(t)^2 P,a(t)^2 P,a(t)^2 P) $$
in a spatially flat FLRW metric:
$$g_{\mu \nu} = \text{diag}(1,-a(t)^2,-a(t)^2,-a(t)^2)$$
But I keep getting a bogus answer! Consider the equation you get from $ \nabla^\mu T_{\mu \nu} = 0$ when $\nu =0$:
$$ \begin{align*} \nabla^\mu T_{\mu 0} &= 0 \\ g^{\mu \alpha}\nabla_\alpha T_{\mu 0} &= 0 \end{align*} $$
$T$ is diagonal, so $\mu$ must be zero, but $g$ is diagonal as well, so if $\mu$ is zero, then so is $\alpha$. This gives:
$$ \begin{align*} g^{0 0}\nabla_0 T_{0 0} &= 0 \\ \nabla_0 \rho &= 0 \\ \dot \rho &= 0 \end{align*} $$
Because $\rho$ is just a scalar, so the covariant derivative is the partial derivative. Except this answer is wrong.
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