Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 54 / 1

Assume an axisymmetric thin disc rotating in the fixed potential of a dominant central mass, with $\Omega =(G{M}_{\ast }{)}^{1/2}{r}^{-3/2}$ independent of time and height. Neglect vertical mass loss and vertical angular-momentum flux at the two faces, as well as self-gravity and radial pressure corrections to the rotation law. Define the surface density and density-weighted kinematic viscosity byand let $v_r={\Sigma }^{-1}\int \rho u_{r}dz$. These assumptions give the vertically averaged viscous disk equationswhere $l=r^2\Omega$ is specific angular momentum. Subtract $l$ times conservation of mass from conservation of angular momentum. Since $l$ is fixed in time,Substitution into conservation of mass proves the Keplerian viscous diffusion equationNo assumption of height-independent kinematic viscosity is needed; its density-weighted average is the one appearing in the integrated stress. A wind or surface magnetic stress would add terms and must not be silently discarded.

Put $F=\bar{\nu }\Sigma$. In steady state, $r^{1/2}{\partial }_r(r^{1/2}F)$ is constant, soThe inward mass accretion rate and outward viscous torque in an accretion disk are

For a zero-torque inner boundary condition at finite $r_{\mathrm{in}}$,With positive finite inner kinematic viscosity, $\Sigma$ tends to zero there. At fixed nonzero inward mass flux, the corresponding $v_r=-\dot{M}/(2\pi r\Sigma )$ becomes large, signaling the breakdown of the nearly circular thin disc approximation in the inner transition region.

For no central accretion, $A=0$ andThis nonaccreting constant-torque disk has a finite inner stress and normally a nonzero inner surface density. An external inner torque supplies the angular momentum transported to an outer sink, although mass does not flow. If both this zero-mass-flux condition and a zero inner torque are imposed, $B=0$; there is no nontrivial positive-viscosity steady disk satisfying both.

The instantaneous kinematic viscosity law is $\bar{\nu }(r,\Sigma )$, with no explicit constitutive memory. Expand its transported quantity at the possibly evolving background:The viscous transport response exponent satisfies ${\partial }_{\Sigma }(\bar{\nu }\Sigma ){|}_0=q{\bar{\nu }}_0$. Subtract the background equation and retain only linear terms to obtainThis remains a linear equation with space- and time-dependent background coefficients; neither a steady background nor constant $q$ is required for this step. The sign of the response coefficient is the negative-diffusion criterion for viscous disk instability.

For $\bar{\nu }=C{r}^a{\Sigma }^b$, the response is $q=b+1$. A nonaccreting background has ${\bar{\nu }}_0{\Sigma }_0\propto r^{-1/2}$, while ${\Sigma }_0\propto r^{-p}$, soConsequently ${\bar{\nu }}_0\propto r^{p-1/2}$. On this steady background, introduce the square-root-radius diffusion transformSince $q$ is constant and ${\partial }_r=(2x{)}^{-1}{\partial }_x$, the linear equation becomesThus $A(x)\propto x^{2p-3}$, and the required choice isFor a Fourier mode $g\propto e^{\lambda t+i{k}_{x}x}$, $\lambda =-A{k}_x^2$. Positive kinematic viscosity makes $A$ have the sign of $q$, so $q<0$ gives growing modes whose rate increases with $k_x^2$. The formal equation is a backward heat equation and predicts arbitrarily rapid small-scale amplification. Physically the thin disc diffusion closure applies only to wavelengths sufficiently larger than the thickness and stress-relaxation scales; this formal limit identifies the need for a cutoff, rather than a finite fastest wavelength absent from the model. The boundary case $q=0$ has vanishing linear transport response.