Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 70 / 3

Let $v=-u_r>0$, $c=v_s$ and $c_0=v_{s0}$. Steady smooth adiabatic flow has $p=K{\rho }^{\gamma }$ with the entropy parameter fixed by infinity, andThe second expression follows by integrating the radial momentum equation with $dp/\rho =d(c^2/(\gamma -1))$ and imposing the outer rest condition. Differentiate continuity to give ${\rho }^{\prime }/\rho =-2/r-v^{\prime }/v$, and substitute into momentum. This yields the Bondi accretion wind equationAt a smooth positive-radius sonic point, $v=c=c_s$ and both sides' vanishing factors require $GM/r_s=2c_s^2$. Evaluate the Bernoulli relation there:Both the squared sound speed and sonic radius must be positive and finite. HenceFor $\gamma =5/3$ no finite-radius regular sonic crossing is possible; the critical solution discussed next approaches sonic speed only at the central limit.

At $\gamma =5/3$, the sound-speed relation becomes $\rho /{\rho }_0=(c/c_0{)}^3$. With $\mathcal{M}=v/c$, mass conservation gives $c^4=\dot{M}{c}_0^3/(4\pi {\rho }_0{r}^2\mathcal{M})$. ThusMultiply $v^2/2+3c^2/2=3c_0^2/2+GM/r$ by $r$ and substitute. This provesFor the curves, define $x=r{c}_0^2/(GM)$, $\lambda =\dot{M}{c}_0^3/(4\pi G^2{M}^2{\rho }_0)$ and $h(\mathcal{M})={\mathcal{M}}^{3/2}/2+3{\mathcal{M}}^{-1/2}/2$. ThenThe function decreases up to Mach one, increases thereafter, and has minimum $h(1)=2$. The original sketch uses rates relative to ${\dot{M}}_{\max }$, that is $\eta =4\lambda$:

For $0<\lambda <1/4$, the minimum of the formal radius curve lies at negative radius. Its physical parts are separate subsonic and supersonic branches extending to $r=0$. Only the subsonic branch satisfies the specified outer state: as $r\to \infty$, it has $c\to c_0$, $\rho \to {\rho }_0$ and $v\sim \dot{M}/(4\pi {\rho }_0{r}^2)\to 0$. The supersonic branch instead has $\rho \to 0$, $c\to 0$ and $v^2\to 3c_0^2$, so it is incompatible with that boundary condition.

At $\lambda =1/4$, the two branches meet at $(r,\mathcal{M})=(0,1)$. The outer-rest branch remains subsonic at every positive radius and approaches Mach one at the centre. For $\lambda >1/4$, they meet at a positive minimum radius $r_{\min }=(2GM/(3c_0^2))(2\sqrt{\lambda }-1)$, and no real solution exists below it. Such a curve cannot describe a global smooth inflow to a point mass.

Thus globally defined positive-rate solutions satisfying the given outer condition are the subsonic branches with $0<\lambda \le 1/4$. These constitute monatomic Bondi accretion. The outer condition alone does not select a unique member; the conventional maximal Bondi solution is the limiting sonic one, while a specific inner accretor condition is additional physical information. Zero rate is the separate hydrostatic limit, not a positive accretion flow.

A solution reaching arbitrarily small radii must obey $\sqrt{\lambda }h(\mathcal{M})=1+3x/2$ with $h\ge 2$. Taking $x\downarrow 0$ therefore requires $2\sqrt{\lambda }\le 1$, or $\lambda \le 1/4$. This upper bound is attained: at $\lambda =1/4$ the decreasing branch $0<\mathcal{M}<1$ has one solution for every $x>0$ and approaches Mach one as $x\to 0$. HenceThe argument establishes the maximum allowed by a globally continuing steady flow, without incorrectly claiming a finite-radius sonic point at $\gamma =5/3$.