PX282 - D4 - virial theorem - energy in stars

the virial theorem relates thermal $(U)$ and gravitational potential $(Ω) :$

Ω = - 2 U

derivation

d Ω_{s h e l l} = - \frac{G M_{r} d m_{s h e l l}}{r}

d m_{s h e l l} = 4 π r^{2} ρ d r

\begin{aligned} \int_{V} d Ω_{s h e l l} & = - \int_{0}^{R} 4 π r^{2} ρ \frac{G M_{r}}{r} d r \\ (1) & \int_{V} Ω d V & = - 4 π \int_{0}^{R} G M_{r} ρ r d r \end{aligned}

\frac{d P}{d r} = - \frac{G M_{r} ρ}{r^{2}}

\begin{aligned} \frac{4}{3} π r^{3} \frac{d P}{d r} & = - \frac{4}{3} π r^{3} \frac{G M_{r} ρ}{r^{2}} \\ (2) & 4 π \int_{0}^{R} r^{3} \frac{d P}{d r} d r & = - 4 π \int_{0}^{R} G M_{r} ρ r d r \end{aligned}

\begin{matrix} 4 π {[P r^{3} - 3 \int P r^{2} d r]}_{0}^{R} = 4 π (0 - 3 \int_{0}^{R} P r^{2} d r) & [∵ P (r = R) = 0] \\ = - 3 \int P d V \end{matrix}

∴ \int_{V} Ω d V = - 3 \int P d V ⟹ Ω = - 3 P

P = n k_{B} T; U = \frac{3}{2} k_{B} T (per particle)

U = \frac{3}{2} n k_{B} T = \frac{3}{2} P

∴ Ω = - 2 U

E = Ω + U

E = - U = \frac{Ω}{2}

for a relativistic gas, $P \neq n k_{B} T$ , so: $Ω = - U$ and $E = 0$
if $E$ increases (ie: from nuclear fusion):
- $Ω$ increases, and star expands
- $U$ decreases, and temperature drops
- if the temperature drops, the fusion decreases, therefore it is self-regulating
if $E$ decreases (ie: from radiation):
- $Ω$ decreases, and star gets smaller
- $U$ increases, and temperature increases
- again, it is self-regulating
self-regulation does not apply to degenerate matter