# 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

(5)$M_{r} \equiv \int 4 \pi \rho r^{2} \, \diff{r}.$

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.