Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 3

Steady spherical mass conservation gives $4\pi r^2\rho u=\dot{M}$. For the polytropic equation of state, $c_s^2=dp/d\rho$, soSubstitution into the radial Euler equations for an inviscid fluid gives the Parker wind equationAt a sonic point, the coefficient of $du/dr$ vanishes. A smooth transonic branch can pass it only if the right-hand side vanishes simultaneously. Hence

The steady Bernoulli equation isAt the regular sonic point, part a givesDefineThe isentropic flow relation $c_s^2\propto {\rho }^{\gamma -1}$ and mass conservation between the stellar surface and the sonic point giveEquating the surface and sonic values of the Bernoulli function therefore yields the required relationAn outflow reaching infinity with positive terminal kinetic energy requires $E>0$, henceAt $\gamma =5/3$ the terminal state is marginal with $E=0$; larger $\gamma$ cannot support the stipulated wind to infinity.

At $\alpha =1/2$, setting ${\mathcal{M}}_0=1$ gives $x=1$ in the mass relation, and both sides of the boxed relation in part b equal $\nu$. Moreover,so the stellar surface itself is the sonic point. This conclusion is independent of $\gamma$ throughout $1<\gamma <5/3$.

There is no other value of $\alpha$ giving a smooth solution with ${\mathcal{M}}_0=1$. Indeed, eliminating $\alpha$ in favour of $x$ reduces the Bernoulli relation toThe left side has its unique minimum at $x=1$, where it equals the right side. Thus $x=1$ and $\alpha =1/2$ are forced.

For fixed $0<{\mathcal{M}}_0<1$, eliminate $\alpha$ byThe remaining equation isIts left side diverges at both ends and has one minimum, atFor $0<{\mathcal{M}}_0<1$ this minimum lies strictly below the right side, so there are exactly two positive values of $x$, and hence exactly two values of $\alpha$. The branches merge at the marginal point $(\alpha ,{\mathcal{M}}_0)=(1/2,1)$.

On the very hot branch, $\alpha \to \infty$ and ${\mathcal{M}}_0=O({\alpha }^{-2})$. Dropping $1/\alpha$ and ${\mathcal{M}}_0^2$ in the relation from part b gives