Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2016 / Paper 1 / 4

Write the polytropic law as $p=K{\rho }^{1+1/n}$, with constant $K$ and $n>0$. Its pressure potential, or polytropic enthalpy relative to zero density, isHydrostatic balance is $dH/dr=-d\psi /dr$. If the surface has zero density and potential ${\psi }_0$, integration gives $H={\psi }_0-\psi$. At the centre, $\Delta \psi \equiv {\psi }_0-{\psi }_c=(n+1)K{\rho }_c^{1/n}>0$. ThereforeInsert $\psi ={\psi }_0-\Delta \psi \theta$ and $\rho ={\rho }_c{\theta }^n$ into the spherical Poisson equation for Newtonian gravity:The central conditions are $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$; the latter follows from a regular spherically symmetric centre. This reproduces the supplied coordinate and potential scaling rather than introducing an unrelated normalization.

For the sound-speed task, the adiabatic sound speed of a perfect polytropic gas isA diatomic gas with active rotational but frozen vibrational degrees of freedom has $\gamma =7/5$ and $n=5/2$. With the requested surface-potential gauge ${\psi }_0=0$,The interior potential is then negative. Choosing zero potential at the surface is legitimate; for a finite isolated sphere it is a different additive gauge from setting potential to zero at infinity. Both choices cannot generally be imposed simultaneously.

For the mass-radius task, let ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. A finite-radius polytrope hasThusAt fixed central temperature, fixed composition and fixed polytropic index, $M\propto R$. These are homologous polytropes at fixed central temperature. They do not in general have constant $K$; imposing constant entropy instead gives a different polytropic mass-radius sequence. If $n$ or $\mu$ varies between stars, the displayed dimensionless structure factor and composition factor must be retained: equal $T_c$ alone is not a universal exact proportionality for arbitrary stellar models.

Treat the gas as a smooth quasi-one-dimensional flow through a rigid passage, neglecting axial body forces and heat input. It is the confining medium that is incompressible; the flowing gas must remain compressible. With area $\mathcal{A}(x)$ and axial speed $u$, continuity and inviscid momentum balance giveCombining their logarithmic differentials yields the nozzle area-velocity relationA smooth crossing of $\mathcal{M}=1$ requires an area extremum. The given radius is positive and has derivative $r^{\prime }(x)=16r_{0}x\exp [-16x^2/L^2]/L^2$, which vanishes only at $x=0$ within the slab and changes from negative to positive there. Thus this station is the unique throat, with $r_s=r_0/2$ and ${\mathcal{A}}_s=\pi r_0^2/4$. The specified subsonic inlet and supersonic outlet select the smooth accelerating branch, soA throat by itself does not guarantee transonic flow for arbitrary reservoir and back pressures; that branch is specified here. Before the throat a converging section accelerates subsonic flow; beyond it a diverging section accelerates supersonic flow.^[ref]

For the temperature task, evaluate the mass flux and sound speed at the throat using $p=K{\rho }^{\gamma }$:The ideal-gas relation then givesThis expresses the sonic temperature in terms of the entropy constant, geometry, mass flux and composition; no numerical value of $K$ or $\dot{M}$ is provided. Equivalently $c_s^{\gamma +1}=\gamma K(\dot{M}/{\mathcal{A}}_s{)}^{\gamma -1}$ and $T_s=\mu c_s^2/(\gamma R^{\ast })$. The slab length drops out because the sonic thermodynamic state depends on the minimum area, not on the axial distance over which a lossless smooth nozzle contracts. As a check, conserved stagnation enthalpy gives $T_s=2T_{\mathrm{stag}}/(\gamma +1)$.^[ref]

The result assumes cross-sectional thermodynamic uniformity and negligible transverse kinetic energy. Exact multidimensional flow through a sharply varying passage can have a curved sonic surface; the radius profile alone does not solve that boundary-value problem. The original numerical plot below computes the standard quasi-one-dimensional branch, rather than reproducing the supplied passage sketch.