\frac{d M _{r}}{d r} = 4 π r^{2} ρ, \frac{d P}{d r} = - \frac{G M _{r} ρ}{r ^{2}},

\frac{d L _{r}}{d r} = 4 π r^{2} ρ ϵ, \frac{d T}{d r} = - \frac{3 κ ρ L _{r}}{16 π a _{rad} c r ^{2} T ^{3}},

with

P = P_{g} + P_{rad}, P_{g} = \frac{ρ k _{B} T}{μ H}, P_{rad} = \frac{a _{rad} T ^{4}}{3}, κ = κ_{0} ρ T^{- 7/2} .

The last relation is the Kramers opacity law. Uniform energy release per unit mass means

ϵ

is constant, so integration of the mass and luminosity equations gives

L_{r} = ϵ M_{r}

. At the surface,

ϵ = \frac{L}{M} = M^{- 1} L .

Thus

ϵ = M^{a} L^{b}

has

a = - 1, b = 1 .

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