Contract $\nabla_aT^{ab}=0$ for the <perfect fluid in general relativity> with $u_b$. Using $u_bu^b=-1$ and $u_b\nabla_au^b=0$ gives the <relativistic perfect-fluid energy equation>
$$
\boxed{u^a\nabla_a\rho+(\rho+p)\nabla_au^a=0},
$$
the local <first law of thermodynamics>. Project instead with $g^a{}_b+u^au_b$ to obtain the <relativistic Euler equation>
$$
\boxed{(\rho+p)u^b\nabla_bu^a
=-(g^{ab}+u^au^b)\nabla_bp}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 4 / a / Solution

In the orthonormal frame, the comoving <perfect fluid in general relativity> has $T_{00}=\rho$ and $T_{ij}=p\delta_{ij}$. The <Einstein field equations> with <cosmological constant> therefore give
$$
\frac34-\Lambda=8\pi\rho,
\qquad
-\frac14+\Lambda=8\pi p.
$$
Thus
$$
\boxed{\rho=\frac{3/4-\Lambda}{8\pi},
\qquad p=\frac{\Lambda-1/4}{8\pi}}.
$$
Choosing
$$
\boxed{\Lambda=\frac14}
$$
gives $p=0$ and $\rho=1/(16\pi)>0$. The metric is then the <Einstein static universe> supported by pressureless matter and positive vacuum energy.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 4 / d / Solution

{c}
{title2=TOV equation}
{wiki}

The Tolman–Oppenheimer–Volkoff equation governs a static, spherically symmetric <perfect fluid in general relativity>. In geometrized units, if $m(r)$ is the mass inside areal radius $r$, then
$$
\frac{dm}{dr}=4\pi r^2\rho,
\qquad
\frac{dp}{dr}=-\frac{(\rho+p)(m+4\pi r^3p)}{r(r-2m)}.
$$

Tolman–Oppenheimer–Volkoff equation

{title2=$T_{ab}=(\rho+p)u_au_b+pg_{ab}$}
{wiki=Perfect_fluid}

A relativistic perfect fluid has energy density $\rho$, isotropic pressure $p$ and unit timelike four-velocity $u$. Projecting $\nabla_aT^{ab}=0$ along and orthogonally to $u$ gives its energy and Euler equations.


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 Incoming links: Perfect fluid in general relativity