When a star exhaust all its nuclear fuel, it does not collapse immediately, but rather evolves into an isothermal state. This is because there is no longer an energy source to support a temperature gradient, and also it is assumed that that thermal conduction time scale is faster than the Kelvin Helmholtz time scale. The star can be divided into an inner core where all the nuclear fuel was burned, and an outer envelope, that still contains fuel.

## Virial Theorem Edit

The hydrostatic equation for the core is

$ \frac{dP}{dr} = -\frac{GM\left( r \right)}{r} \rho $

Multiplying both sides by $ 4 \pi r^3 $ in integrating over the entire volume of the core yields (starting from the left hand side)

$ \int_0^{R_c} 4 \pi r^3 \frac{dP}{dr} dr = $

$ = \left[ 4 \pi r^3 P \right]^{R_c}_0 - 3 \int_0^R P 4 \pi r^2 dr = $

$ = 4 \pi P_c R_c^3 - 3 \frac{M_c k T_c}{\mu_c} $

where $ \mu_c $ is the average molecular mass of the particles in the core. The right hand side is evaluated using the assumption that the density is uniform throughout the core and equal to the average density $ \rho_c = \frac{M_c}{4 \pi R_c^3/3} $

$ -\int_0^{R_c} 4\pi r^3 \rho_c \frac{G \rho_c \cdot 4\pi r^3/3}{r^2} = -\frac{5}{3}{G M_c^2}{R_c} $

Finally,

$ 4 \pi P_c R_c^3 - 3 \frac{M_c k T_c}{\mu_c} = - \frac{5}{3} \frac{G M_c^2}{R_c} $

$ P_c = \frac{1}{4 \pi R_c^3/3} \left( \frac{M_c k T_c}{\mu} - \frac{1}{5} \frac{G M_c^2}{R_c} \right) $

The maximum pressure is

$ \frac{d P_c}{d R_c} = 0 \Rightarrow R_{crit} = \frac{4 G M_c \mu}{15 k T_c} $

$ P_{crit} = \frac{10125}{1024 \pi G^3 M_c^2} \left( \frac{k T}{\mu} \right)^4 $

The last value is the maximum pressure that the core can support.