 Solution

ID: past-exam-of-the-radiative-processes-and-high-energy-astrophysics-course-of-the-university-of-oxford/2024/1/e/solution

ρ_{d} = 4 ρ_{u}

and

v_{d} = v_{u} /4

P_{d} = ρ_{d} k_{B} T_{d} / μ

, with

μ

5 k_{B} T / (2 μ)

. Dividing the energy flux equation in d by the common mass flux gives

\frac{v _{u}^{2}}{2} = \frac{v _{u}^{2}}{32} + \frac{5 k _{B} T _{d}}{2 μ} .

Therefore

\frac{5 k _{B} T _{d}}{2 μ} = \frac{15 v _{u}^{2}}{32}, k_{B} T_{d} = \frac{3}{16} μ v_{u}^{2} .

As an independent check, conservation of momentum gives

P_{d} = ρ_{u} v_{u}^{2} - 4 ρ_{u} (v_{u} /4)^{2} = 3 ρ_{u} v_{u}^{2} /4

, and division by

ρ_{d} = 4 ρ_{u}

gives exactly the same temperature. The numerical coefficient relies on

γ = 5/3

and the absence of dynamically important radiative or magnetic stresses.

 New to topics? Read the docs here!