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 Incoming links: Central pressure of a constant-density relativistic star

Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 54 / 1 / b / ii / Solution Created 2026-10-03 Updated 2026-10-06

For a perfect fluid in general relativity with nonnegative energy density and pressure, the dominant energy condition is

p \leq ρ

. Pressure decreases outward, so it is enough to test the central pressure. Put

s = 1 - 2 M / R

; the regular Interior Schwarzschild metric requires

3 s - 1 > 0

. The central pressure of a constant-density relativistic star gives

\frac{p _{c}}{ρ} = \frac{1 - s}{3 s - 1} \leq 1 ⟺ 1 - s \leq 3 s - 1 ⟺ s \geq \frac{1}{2} .

Therefore the dominant-energy compactness bound for an incompressible star is

\frac{M}{R} \leq \frac{3}{8} .

Equality has

p_{c} = ρ

and a finite central pressure; it is allowed by the non-strict dominant energy condition. This is stronger than the regular-star Buchdahl's theorem limit

M / R < 4/9

.

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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 54 / 1 / b / i / Solution Created 2026-10-03 Updated 2026-10-06

For the constant-density model,

m (r) = 4 π ρ r^{3} /3 = M r^{3} / R^{3}

. This is the incompressible Interior Schwarzschild metric, replacing the earlier finite-slope barotropic stellar equation of state. The Tolman–Oppenheimer–Volkoff equation separates as

\frac{d p}{( p + ρ ) ( p + ρ /3 )} = - \frac{4 π r d r}{1 - 8 π ρ r ^{2} /3} .

On the left, partial fractions give

\int \frac{d p}{( p + ρ ) ( p + ρ /3 )} = \frac{3}{2 ρ} lo g \frac{p + ρ /3}{p + ρ} .

On the right the antiderivative is

\frac{3}{4 ρ} lo g (1 - 8 π ρ r^{2} /3)

. Applying the surface condition

p (R) = 0

s (r) = 1 - 2 M r^{2} / R^{3}

,

s_{R} = 1 - 2 M / R

, gives

\frac{p + ρ /3}{p + ρ} = \frac{s ( r )}{3 s _{R}} .

Solving for the pressure recovers the Interior Schwarzschild solution:

p (r) = ρ \frac{s ( r ) - s _{R}}{3 s _{R} - s ( r )} .

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

p_{c} = ρ \frac{1 - 1 - 2 M / R}{3 1 - 2 M / R - 1} .

The denominator must be positive for a regular positive-pressure centre, requiring

M / R < 4/9

. As that limiting compactness is approached, the central pressure diverges, in agreement with Buchdahl's theorem.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 36D / b / Solution Created 2026-09-24 Updated 2026-10-03

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} .

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} .

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}) .

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.

a = \frac{8 π ρ _{0}}{3} .

Separating variables gives

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

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}}),

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}} .

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

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}} .

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} .

It diverges when

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

or equivalently

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

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

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