Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2017 / 3

Work in the rest frame of a thin planar shock wave, with normal upstream and downstream velocities $u_1,u_2>0$. Assume a steady flow, negligible gravitational change across its thickness, no radiative energy loss, and negligible viscous stress and heat flux at the two remote faces of a control volume. Dissipation within the shock is permitted and is essential to the physical discontinuity.

Integrating mass conservation through that volume gives equal mass flux per unit area:The dimensions of $j$ are $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. The normal momentum entering minus that leaving must balance the pressure forces. Equivalently, the conserved normal momentum flux is $p+\rho u^2$, givingFor conservation of energy, matter advects kinetic and internal energy while pressure performs work at rate $pu$ per unit area. Writing $ϵ$ for specific internal energy, the energy flux is $\rho u(u^2/2+ϵ)+pu$. With specific enthalpy $h=ϵ+p/\rho$, its conservation givesThe division by $j$ is valid for a genuine through-flow shock. Together these are the three Rankine-Hugoniot conditions. The conserved energy flux has units $\mathrm{W}{\mathrm{m}}^{-2}$, and both momentum fluxes have units of pressure.

For a calorically perfect ideal gas, $ϵ=p/[(\gamma -1)\rho ]$ and $h=\gamma p/[(\gamma -1)\rho ]$. The three jump conditions can be combined explicitly. Put $r={\rho }_2/{\rho }_1$, so $u_2=u_1/r$ andSubstitute these into the enthalpy equation and use ${\mathcal{M}}_1^2={\rho }_1{u}_1^2/(\gamma p_1)$, with the upstream adiabatic sound speed. Besides the trivial no-shock root $r=1$, the physical compression root isFor unchanged composition, $T_2/T_1=(p_2/p_1)/r$. A compressive shock has ${\mathcal{M}}_1>1$ and increases the specific entropy. Thus an adiabatic shock conserves total energy but is not an isentropic process.

Interpret $v$ as the normal upstream speed in the shock rest frame and $\rho$ as its upstream mass density. If the quoted jet speed is in another frame, the shock speed must first be supplied to obtain this relative velocity. Use a calorically perfect gas of mean particle mass $\bar{m}=\mu m_u$.

In the strong-shock limit, ${\mathcal{M}}_1\gg 1$. The Strong-shock Rankine-Hugoniot conditions giveThe pressure expression also follows from the supplied ratio by inserting ${\mathcal{M}}_1^2=\rho v^2/(\gamma p_1)$. The ideal gas law then gives the immediate postshock temperatureThe upstream thermal contribution is negligible at this order. This is a temperature in kelvin, rather than a temperature change per unit time.

To turn the postshock thermal energy into a rate, the processed mass per unit time is $\dot{m}=\rho vS$. Its downstream specific internal energy isThus the thermal internal energy carried into the shocked gas per unit time isPer unit area, divide this expression by $S$. Its dimensions are watts; the mean molecular weight cancels because the heat capacity per unit mass is inversely proportional to it, while $T_2$ is proportional to it.

There is a second common convention for shock heating power: the loss of directed bulk kinetic energy flux. The energy fluxes of a strong adiabatic shock giveThis equals the downstream enthalpy flux, and includes pressure work as well as the advected internal energy. For $\gamma =5/3$, the compression ratio is $4$, $T_2=3\mu m_u{v}^2/(16k_B)$, ${\dot{E}}_{\mathrm{int}}=9\rho S{v}^3/32$, and ${\dot{E}}_{\mathrm{kin},\mathrm{loss}}=15\rho S{v}^3/32$. Specifying the energy convention avoids conflating these rates. A rate $dT/dt$ for a fixed slab cannot be inferred without its participating mass and thermal evolution; the supplied data fix the postshock temperature and the processed energy fluxes.

Let $a_T^2=k_{B}T/(\mu m_u)$ be the constant isothermal sound speed squared. The stationary galactic gas obeys hydrostatic equilibrium in the specified gravitational potential:Integrating from the pressure-matching radius $r_0$ gives $\rho (r)=\rho (r_0)\exp [-(\psi (r)-\psi (r_0))/a_T^2]$. Since $\rho (r_0)=p_0/a_T^2$ and $\psi =-A{r}^{-1/2}$,The exponent is dimensionless: the quantity in parentheses times $A$ is a difference of gravitational potential, of units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Density increases inward, as expected. The idealized potential is singular at the centre, so this formal profile should not be extrapolated into a region where the prescribed potential or the neglect of gas self-gravity fails.

The smaller galaxy is treated as a test particle released from rest at $r_0$. Conservation of energy in the fixed gravitational potential givesThe galactic gas is stationary in its hydrostatic model, so this is the relative speed squared. Write $\Delta (d)=d^{-1/2}-r_0^{-1/2}\ge 0$ on the inward path. The astrophysical ram pressure is the incident normal momentum flux $\rho (d){\dot{d}}^2$, givingIt vanishes at release and increases during infall. This ram pressure convention is $\rho v^2$, not the $\rho v^2/2$ sometimes called dynamic pressure. Gas drag and the object's response are neglected consistently with the prescribed ballistic motion.