Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 57 / 1

Write the radial velocity as $u(r)$, allowing $u<0$ for accretion flow, and let $c^2=\gamma p/\rho$ be the squared adiabatic sound speed. The steady continuity equation, radial Euler equations for an inviscid fluid and entropy advection equation giveOn a nonzero smooth flow branch, the polytropic equation of state is $p=K{\rho }^{\gamma }$ with constant $K$. Thus the two useful first integrals for spherically symmetric adiabatic flow areHere $\mathcal{J}$ is the signed mass flux and $\mathcal{B}$ is the Bernoulli function. The specific enthalpy is $c^2/(\gamma -1)$; the prescribed Newtonian gravitational potential has no contribution from the gas's own gravity. Eliminating the density derivative using mass conservation gives the differential formThis sonic point equation displays both the singular coefficient and the numerator that must vanish for a smooth transonic branch.

At a finite smooth sonic point $r_{\ast }$, the equation from (a) requiresSet $\delta =\gamma -1$ and $q=2/\beta -1/2$. Substitution into the Bernoulli function givesAn inflow from a warm reservoir at infinity has $\mathcal{B}=c_{\infty }^2/\delta >0$; an outflow reaching infinity has $\mathcal{B}=u_{\infty }^2/2+c_{\infty }^2/\delta \ge 0$. Therefore a nondegenerate transonic spherical flow in a power-law potential requiresFor the usual range $0<\beta <4$, the critical adiabatic exponent for spherical power-law flow is consequentlyThe printed assumption $\beta >0$ also permits $\beta \ge 4$. In that range the boxed inequality holds for every finite $\gamma >1$: there is no finite upper bound. Equivalently, take $f(\beta )=+\infty$ for $\beta \ge 4$. Extending the rational expression beyond $\beta =4$ would give an incorrect restriction.

The strict inequality also follows from local sonic-point slope discriminant analysis, including the potentially cold zero-energy endpoint. Put $X=(r_{\ast }/u_{\ast })u_{\ast }^{\prime }$. Since mass conservation implies $(c^2{)}^{\prime }/c^2=-\delta (u^{\prime }/u+2/r)$, differentiating the sonic point equation at $r_{\ast }$ yieldsIts discriminant is $8[2\beta -(4-\beta )\delta ]$. Moreover, the logarithmic Mach number derivative at the sonic point isA genuine crossing has two distinct branches and a nonzero derivative. At equality the crossing degenerates. In particular, for $\gamma =f(\beta )$ and $\mathcal{B}=0$, mass conservation and the polytropic equation of state make the Mach number constant along the scale-invariant solution; a solution that is sonic there is sonic everywhere, rather than crossing an isolated sonic point.

Use the positive mass accretion rate $\dot{M}=-\mathcal{J}$, and retain $\delta =\gamma -1$, $q=2/\beta -1/2$. Define $h=1-q\delta$, which is positive under the condition in (b). Matching the Bernoulli function to the reservoir and using the polytropic equation of state givesThe transonic spherical accretion rate in a power-law potential is thereforeThe combination $A/c_{\infty }^2$ has dimensions of length to the power $\beta$, so this expression has dimensions of mass per time. The sonic point selects the flux that connects the subsonic reservoir to the inward supersonic transonic branch.

For the endpoint limits of power-law spherical accretion, hold $A,\beta ,{\rho }_{\infty },c_{\infty }$ fixed. As $\delta \to 0$, $\ln h=-q\delta +O({\delta }^2)$, so $(q-1/\delta )\ln h\to q$. HenceThis is also obtained directly from an isothermal equation of state: the Bernoulli function becomes $u^2/2+c_{\infty }^2\ln (\rho /{\rho }_{\infty })-A/r^{\beta }=0$, so ${\rho }_{\ast }/{\rho }_{\infty }=e^q$. For $\beta =1$, $A=GM$, it recovers the isothermal Bondi accretion rate.

For $0<\beta <4$, $q>0$ and $\gamma \uparrow f(\beta )$ means $h\downarrow 0$. Since $\delta =(1-h)/q$,ThusThe finite limiting flux accompanies $r_{\ast }\to 0$ and $c_{\ast },{\rho }_{\ast }\to \infty$; it does not assert a finite-radius sonic point at the endpoint. For $\beta =1$ this is the familiar $\gamma \uparrow 5/3$ limit $\pi G^2{M}^2{\rho }_{\infty }/c_{\infty }^3$.

For completeness, the printed range $\beta \ge 4$ has the endpoint $f=+\infty$. At $\beta =4$, $q=0$ and the flux is independent of $\gamma$, namely $4\pi {\rho }_{\infty }(2A{)}^{1/2}$. For $\beta >4$, $q<0$ and $h\sim |q|\delta$, so $\dot{M}\to 0$ as $\gamma \to \infty$. These are the corresponding extended endpoint limits.