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ID: past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2018/1/b/solution

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Past exam of the astrophysics course of the University of Oxford / 2018 / 1 / b / Solution by

Codex  0 2026-10-25

Use a steady, plane, nonradiative normal shock wave in a perfect gas, and measure the normal speeds

v_{u}, v_{d} > 0

in the shock frame. The labels denote upstream and downstream. Conservation of mass, normal momentum and total energy gives

ρ_{u} v_{u} = ρ_{d} v_{d}, p_{u} + ρ_{u} v_{u}^{2} = p_{d} + ρ_{d} v_{d}^{2},

and

\frac{1}{2} v_{u}^{2} + \frac{γ}{γ - 1} \frac{p _{u}}{ρ _{u}} = \frac{1}{2} v_{d}^{2} + \frac{γ}{γ - 1} \frac{p _{d}}{ρ _{d}} .

The enthalpy term includes the pressure work. For very large upstream Mach number, neglect

p_{u}

compared with

ρ_{u} v_{u}^{2}

. If

r = ρ_{d} / ρ_{u}

is the shock compression ratio, the first two equations imply

v_{d} = \frac{v _{u}}{r}, p_{d} = ρ_{u} v_{u}^{2} (1 - \frac{1}{r}) .

Substitution into the energy equation gives

1 = \frac{1}{r ^{2}} + \frac{2 γ}{γ - 1} \frac{r - 1}{r ^{2}}, (r - 1) (r - \frac{γ + 1}{γ - 1}) = 0.

Discard the no-shock solution

r = 1

. For a nonrelativistic monatomic gas,

γ = 5/3

, so the Strong-shock Rankine-Hugoniot conditions become

\frac{ρ _{d}}{ρ _{u}} = \frac{v _{u}}{v _{d}} = 4, p_{d} = \frac{3}{4} ρ_{u} v_{u}^{2} .

The factor four requires this adiabatic index and negligible radiative, magnetic and escaping-particle contributions to the flux budget. For example, the same strong, nonradiative calculation with

γ = 4/3

gives compression seven, not four.

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