Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 1

Define the surface density of a disk and outward radial mass flux byVertical integration of mass conservation eliminates the surface term because $\rho u_z\to 0$, and axisymmetry eliminates the azimuthal derivative. Hence

For $u_{\varphi }\simeq r\Omega (r)$, the specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the viscous torque in an accretion diskSubtracting $h$ times the mass equation from the integrated angular-momentum equation givesFor an axisymmetric circular flow, ${\Pi }_{r\varphi }=\mu rd\Omega /dr$. Ifthen

Combining the two conservation laws givesThe first term is the local rate of change of angular momentum per radial interval, $\mathcal{F}h$ is outward advective angular-momentum flux, and $\mathcal{G}$ is outward stress-carried angular-momentum flux.

Solved by gpt-5.6-sol high.

In steady state, mass conservation makes the outward flux constant, $\mathcal{F}=\dot{M}$. Angular-momentum conservation then givesThe zero-torque condition at $r_{\mathrm{out}}$ fixes the constant to $\dot{M}{h}_{\mathrm{out}}$, soFor a Kepler orbit, $h=\sqrt{GMr}$ and $d\Omega /dr=-3\Omega /(2r)$. The viscous torque is consequently $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Equating the two expressions yields the steady decretion disk structure

Solved by gpt-5.6-sol high.

At the inner edge, the imposed torque isThereforeThe angular-momentum conservation law says that $\mathcal{F}h+\mathcal{G}$ is constant. The magnetic process supplies angular momentum $\Theta$ at the inner boundary; viscous stress passes it outward, and the mass leaving at $r_{\mathrm{out}}$ carries the injected angular momentum together with the angular momentum $\dot{M}{h}_{\mathrm{in}}$ that entered with the mass. A larger torque must therefore move the removal radius outward so that each unit mass can carry more specific angular momentum.

Solved by gpt-5.6-sol high.

The luminosity from both disk faces is the radial integral of the given viscous heating rate:Substituting the steady profile and ${\Omega }^2=GM/r^3$ givesUsing $\Theta =\dot{M}\sqrt{GM}(\sqrt{r_{\mathrm{out}}}-\sqrt{r_{\mathrm{in}}})$ turns this intoThe first term is mechanical power injected by the stellar magnetic torque. The second is $\dot{M}(E_{\mathrm{in}}-E_{\mathrm{out}})$, where $E=-GM/(2r)$ is the specific energy of a circular orbit; it is negative because outward motion makes the material less tightly bound and consumes some torque power. The remainder is radiated by viscous dissipation.

Matter reaching the outer edge still has binding energy $GM/(2r_{\mathrm{out}})$ per unit mass. A wind that reaches infinity with negligible terminal energy must therefore receivefrom the stellar radiation field. In a signed disk-energy balance, the material removed at the edge carries orbital energy flux $\dot{M}{E}_{\mathrm{out}}=-GM\dot{M}/(2r_{\mathrm{out}})$.

Solved by gpt-5.6-sol high.

For $r_{\mathrm{in}}\ll r\ll r_{\mathrm{out}}$, part (i) gives $\bar{\nu }\Sigma \propto r^{-1/2}$. Since ${\Omega }^2\propto r^{-3}$, the dissipative flux obeysIn the spectrum, introduce the dimensionless variableThen $rdr$ is proportional to ${\omega }^{-16/7}{x}^{9/7}dx$, soAt intermediate frequencies the inner limit is much smaller than one and the outer limit much larger than one, allowing them to be replaced by zero and infinity. Thus

Solved by gpt-5.6-sol high.