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I am reading a book "The Early Universe " by Kolb and Turner. On P.48, it says

For $T^\mu\,_\nu=\operatorname{diag} (\rho, -p,-p,-p)$, the $\mu=0$ component of the conservation of stress energy $(\partial_\nu T^{\mu\nu}=0)$ gives the first law of thermodynamics in the familiar form $$d(\rho R^3)=-pd(R^3).$$

Unfortunately I can't see how the latter follows from the conservation equation. Can anyone help me?

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The conservation law is that the divergence of the covariant derivative vanishes: $$ T^{0 \nu}_{\ \ \ \ \ \ ;\nu}=0.\qquad (*) $$ To compute what this means, you need a metric. I'm going to assume that Kolb and Turner are using the Robertson-Walker metric, since this is very common in cosmology. This metric is $$ ds^2=g_{\mu\nu} dx^{\mu} dx^{\nu}= dt^2-R(t)^2 \left( \frac{dr^2}{1-kr^2}+r^2 d\theta^2+r^2\sin^2 \theta d\phi^2\right), $$ where $k\in\{0,\pm 1\}$ is constant and determines whether the Universe is spatially flat, spherical, or hyperbolic, and $R=R(t)$ gives the evolution of the scale of the Universe with time. Then, you just have to use the formulae for the connection, $$ \Gamma^{\rho}_{\mu\nu}=\frac12 g^{\rho\sigma}(\partial_\mu g_{\nu\sigma}+\partial_\nu g_{\mu\sigma}-\partial_\sigma g_{\mu\nu}),$$ where the $\partial_\alpha$s mean coordinate differentiation, $\partial_\alpha=\frac{\partial}{\partial x^\alpha}$, the covariant derivative, $$ T^{\mu\nu}_{\ \ \ \ \ \ ;\omega}=\partial_\omega T^{\mu\nu}+\Gamma^\mu_{\alpha\omega} T^{\alpha\nu} +\Gamma^\nu_{\beta\omega} T^{\mu \beta}, $$ and the form of the stress-energy tensor given above in the question to simplify (*). This gives $$ \frac{\partial \rho}{\partial t}+\frac{3p}{R} \frac{dR}{dt} + \frac{3\rho}{R} \frac{dR}{dt}=0 $$ which, after it is multiplied by $R^3$, is equivalent to the equation Kolb and Turner give.

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