Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 36D / b

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- Solution b

Solution

For the constant density

ρ_{0}

m (r) = 4 π \int_{0}^{r} ρ_{0} s^{2} d s = \frac{4 π}{3} ρ_{0} r^{3} .

(1)

The definition of

m

then gives

\frac{1}{μ ^{2}} = 1 - \frac{2 m ( r )}{r} = 1 - \frac{8 π}{3} ρ_{0} r^{2} .

(2)

Substitution in the Tolman–Oppenheimer–Volkoff equation yields

\frac{d P}{d r} = - \frac{4 π r}{1 - \frac{8 π}{3} ρ _{0} r ^{2}} (P + \frac{ρ _{0}}{3}) (P + ρ_{0}) .

(3)

The minus sign follows both from the stated TOV equation and from the outward decrease of pressure; a version of the requested intermediate formula without that sign is a typographical error.

Set

a = \frac{8 π ρ _{0}}{3} .

(4)

Separating variables gives

\frac{d P}{( P + ρ _{0} /3 ) ( P + ρ _{0} )} = - \frac{4 π r d r}{1 - a r ^{2}} .

(5)

Since

\frac{1}{( P + ρ _{0} /3 ) ( P + ρ _{0} )} = \frac{3}{2 ρ _{0}} (\frac{1}{P + ρ _{0} /3} - \frac{1}{P + ρ _{0}}),

(6)

integration and the surface condition

P (R) = 0

give

\frac{P ( r ) + ρ _{0} /3}{P ( r ) + ρ _{0}} = \frac{1}{3} \frac{1 - a r ^{2}}{1 - a R ^{2}} .

(7)

Writing

M = m (R) = \frac{4 π}{3} ρ_{0} R^{3}, a = \frac{2 M}{R ^{3}},

(8)

one obtains the pressure of the Interior Schwarzschild metric:

P (r) = ρ_{0} \frac{1 - 2 M r ^{2} / R ^{3} - 1 - 2 M / R}{3 1 - 2 M / R - 1 - 2 M r ^{2} / R ^{3}} .

(9)

In particular, the central pressure of a constant-density relativistic star is

P (0) = ρ_{0} \frac{1 - 1 - 2 M / R}{3 1 - 2 M / R - 1} .

(10)

It diverges when

3 1 - \frac{2 M}{R} = 1,

(11)

or equivalently

M = \frac{4 R}{9} .

(12)

This is the limiting equality case of Buchdahl's theorem.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 36D / b

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