For the centrally condensed point-source model,

M_{r} = M

L_{r} = L

, and gas pressure dominates. The equations are

\frac{d P}{d r} = - \frac{GMρ}{r ^{2}}, \frac{d T}{d r} = - \frac{3 κ _{0} ρ ^{2} L}{16 π a _{rad} c r ^{2} T ^{13/2}}, P = R ρT,

where

R = k_{B} / (μ H)

. Dividing the first equation by the second and eliminating

ρ = P / (R T)

gives

P \frac{d P}{d T} = \frac{16 π a _{rad} c GM R}{3 κ _{0} L} T^{15/2} .

The surface conditions

P = T = 0

then yield

P = (\frac{64 π a _{rad} c GM R}{51 κ _{0} L})^{1/2} T^{17/4},

ρ = \frac{1}{R} (\frac{64 π a _{rad} c GM R}{51 κ _{0} L})^{1/2} T^{13/4} .

Substitution into hydrostatic equilibrium cancels the common power of

T

and gives

\frac{d T}{d r} = - \frac{4 GM}{17 R r ^{2}} .

Using

T (R) = 0

T (r) = \frac{4 GM}{17 R} (\frac{1}{r} - \frac{1}{R}) .

ii

Solution

At the core-envelope interface, continuity at temperature

T_{c}

requires the envelope pressure

P = C_{0} (\frac{M}{L})^{1/2} T_{c}^{17/4}, C_{0}^{2} = \frac{64 π a _{rad} c G R}{51 κ _{0}},

to equal the nonrelativistic electron degeneracy pressure

P = K (\frac{ρ}{μ _{e}})^{5/3}, ρ = \frac{P}{R T _{c}} .

Eliminating

ρ

gives

P = (R μ_{e} T_{c})^{5/2} K^{- 3/2} .

Equating the two pressures and solving for luminosity gives Mestel's cooling law

\frac{L}{M} = \frac{64 π a _{rad} c G}{51 κ _{0}} \frac{K ^{3}}{R ^{4} μ _{e}^{5}} T_{c}^{7/2} .

In particular,

L / M \propto T_{c}^{7/2}

There are no discussions about this article yet.

There are currently no matching articles.