Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 10D / b

Solution

Use the uniform-density configuration as a trial configuration for the white dwarf. Write

P = C_{F} h^{2} n^{5/3} / m_{e}

, with

C_{F} > 0

the numerical coefficient suppressed by the approximate equation of state. Its particle density is

n = 3 M / (4 π m_{p} R^{3})

3 P /2

. Hence

E_{kin} = \frac{α}{R ^{2}}, α = \frac{3}{2} C_{F} \frac{h ^{2}}{m _{e}} (\frac{3}{4 π})^{2/3} (\frac{M}{m _{p}})^{5/3} .

For uniform density

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

, so integration of the shell energy gives

E_{grav} = - \frac{β}{R}, β = \frac{3}{5} G M^{2} .

Both constants are positive and independent of

R

. Differentiating

E = α / R^{2} - β / R

gives its unique stationary radius

R = 2 α / β

2 α / R^{4} > 0

, and this is the global minimum since

E \to + \infty

at zero and

E \to 0^{-}

at infinity. Thus

R_{WD} = 5 C_{F} (\frac{3}{4 π})^{2/3} \frac{h ^{2} M ^{- 1/3}}{G m _{e} m _{p}^{5/3}} \sim \frac{h ^{2} M ^{- 1/3}}{G m _{e} m _{p}^{5/3}} .

The uniform-density approximation is used for this energy estimate; it does not require a spatially constant pressure to satisfy the exact hydrostatic equilibrium equation.

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