# Equilibrium State¶

In the static equilibrium state the fluid velocity \(\vv\) vanishes. The momentum equation then becomes the hydrostatic equilibrium equation

\[\nabla P = - \rho \nabla \Phi.\]

Assuming the equilibrium is spherically symmetric, this simplifies to

\[\deriv{P}{r} = - \rho \deriv{\Phi}{r}.\]

Poisson’s equation can be integrated once to yield

\[\deriv{\Phi}{r} = \frac{G}{r^{2}} \int 4 \pi \rho r^{2} \, \diff{r} = \frac{G M_{r}}{r^{2}},\]

where the second equality introduces the interior mass

\[M_{r} \equiv \int 4 \pi \rho r^{2} \, \diff{r}.\]

The hydrostatic equilibrium equation thus becomes

\[\deriv{P}{r} = - \rho \frac{G M_{r}}{r^{2}}.\]

The heat equation in the equilibrium state is

\[\rho T \pderiv{S}{t} = \rho \epsnuc - \nabla \cdot (\vFrad + \vFcon).\]

If the star is in thermal equilibrium then the left-hand side vanishes, and the nuclear heating rate balances the flux divergence term. Again assuming spherical symmetry, this is written

\[\deriv{}{r} \left( \Lrad + \Lcon \right) = 4 \pi r^{2} \rho \epsnuc,\]

where

\[\Lrad \equiv 4 \pi r^{2} \Fradr, \qquad
\Lcon \equiv 4 \pi r^{2} \Fconr\]

are the radiative and convective luminosities, respectively.