 Solution

ID: past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2013/paper-1/1/c/solution

Work in the rest frame of the shock. The normal momentum flux is conserved, so

p_{1} + ρ_{1} u_{1}^{2} = p_{2} + ρ_{2} u_{2}^{2}

. In the strong-shock limit

p_{1} ≪ ρ_{1} u_{1}^{2}

ρ_{2} u_{2} = ρ_{1} u_{1}

p_{2} ≃ ρ_{1} u_{1}^{2} (1 - \frac{u _{2}}{u _{1}}) = \frac{2 ρ _{1} u _{1}^{2}}{γ + 1}, ρ_{2} = ρ_{1} \frac{γ + 1}{γ - 1} .

Using the downstream ideal gas law

p_{2} / ρ_{2} = k_{B} T_{2} / m

gives

\frac{k _{B} T _{2}}{m} = \frac{2 ( γ - 1 )}{( γ + 1 ) ^{2}} u_{1}^{2} .

For a monatomic perfect gas,

γ = 5/3

and the coefficient is

3/16

. This is the immediate post-shock temperature, before substantial radiative cooling; ionization, dissociation or unequal electron and ion temperatures require additional physical modeling.

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