Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 46 (Accretion Discs)

Write the gravitational potential with spherical symmetry as $\Phi (r)$ and let $\mathbf{n}$ be the unit vector normal to a circular orbit, oriented in the direction of its angular velocity. Put $\mathbf{\Omega }=\Omega \mathbf{n}$ and denote the specific angular momentum by $\mathbf{h}=h\mathbf{n}$. Circular force balance and the specific orbital energy giveDifferentiating along the family of circular orbits,A change in orbital orientation contributes $hd\mathbf{n}$ to $d\mathbf{h}$, but $\mathbf{n}\cdot d\mathbf{n}=0$. ConsequentlyThis circular-orbit energy gradient accounts for radial and inclination changes together; spherical symmetry makes the orbital energy independent of the plane.

For either variable-mass particle, $\mathbf{H}=M\mathbf{h}$ and $E=M\epsilon$. Thus $d\mathbf{H}=Md\mathbf{h}+\mathbf{h}dM$, soConservation of mass and conservation of angular momentum impose $d{M}_2=-d{M}_1$, $d{\mathbf{H}}_2=-d{\mathbf{H}}_1$. Adding the two particle contributions provesTo interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\epsilon -\Omega h$ the first identity gives $f^{\prime }=-h{\Omega }^{\prime }$. The assumed outward decrease of angular velocity therefore makes $f_1<f_2$, so the mass-exchange part lowers the energy when $d{M}_1>0$: mass is transferred inward. In a coplanar, co-rotating pair, ${\Omega }_1>{\Omega }_2$ and an outward angular momentum transfer has $d{\mathbf{H}}_1=-dH\mathbf{n}$ with $dH>0$. Its contribution is $-({\Omega }_1-{\Omega }_2)dH<0$.

The vector difference also identifies the inclination-reducing exchange. For a fixed small magnitude of admissible angular momentum exchange, its energy contribution is most negative whenby the Cauchy-Schwarz inequality. This gives $d{E}_{\mathrm{angular}}=-s|{\mathbf{\Omega }}_1-{\mathbf{\Omega }}_2{|}^2$. Let $q={\mathbf{n}}_1\cdot {\mathbf{n}}_2=\cos \iota$ and $H_i=|{\mathbf{H}}_i|$, taking fixed particle masses for this exchange. Projecting $d{\mathbf{H}}_i$ perpendicular to its own direction givesThereforeThe downhill exchange reduces the relative orbital inclination. Its inner angular momentum component is ${\mathbf{n}}_1\cdot d{\mathbf{H}}_1=-s({\Omega }_1-{\Omega }_{2}q)<0$; in the aligned limit the outer orbit receives that angular momentum directly. For strongly inclined orbits, part of the exchange instead cancels differently directed angular momenta as the planes align. Together these gradients explain the tendencies toward inward mass transfer, outward angular momentum transport and coplanarity.

Let $M_{\ast }$ be the central mass, so ${\Omega }^2=G{M}_{\ast }/r^3$ and the specific orbital energy is $\epsilon =-G{M}_{\ast }/(2r)$. Define the inward accretion rate $\dot{M}>0$. The rate at which the gas acquires binding energy in moving from $b$ to $a$ isThe vertically integrated viscous heating per unit horizontal area, including the entire column, is $\bar{\nu }\Sigma (r{\Omega }^{\prime }{)}^2$. Since $r{\Omega }^{\prime }=-3\Omega /2$, the dissipated power in the annulus isHere the column integral already includes dissipation associated with both disk faces; adding another factor of two would double-count it. Performing the elementary integrals givesFor the whole disk, take $a=r_{\mathrm{in}}$ and $b\to \infty$. Both expressions giveIf $a\gg r_{\mathrm{in}}$, the bracket $1-\sqrt{r_{\mathrm{in}}/r}$ is uniformly close to one throughout the annulus. Comparing the positive integrals showswith the correction negative. Thus far from the inner edge the viscous dissipation is approximately three times the local binding-energy acquisition.

There is no violation of conservation of energy because the viscous torque in an accretion disk also transports mechanical energy. The outward viscous torque in an accretion disk isand its outward energy flux is $\Omega \mathcal{G}=G{M}_{\ast }\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$. The explicit expressions satisfyEnergy transported from the inner disk supplies the extra dissipation at larger radii. The torque power vanishes at the zero-torque inner boundary condition and tends to zero at infinity, leaving global balance. This is torque transport of energy in a steady accretion disk.

The vertical hydrostatic equilibrium equation balances the gradient of the total pressure against gravity. In a thin disk with Keplerian rotation, expanding the central gravitational field about the midplane gives vertical acceleration $-{\Omega }^{2}z$, so $p_z=-\rho {\Omega }^{2}z$. Disk self-gravity is neglected in this approximation.

The flux-gradient equation is local thermal balance: the upward radiative flux increases with height because differential rotation dissipates energy. For a Newtonian effective dynamic viscosity $\mu$, the volumetric heating rate is $\mu (r{\Omega }^{\prime }{)}^2$, which becomes $9\mu {\Omega }^2/4$ for Keplerian rotation. The symbol $\mu$ here is dynamic viscosity, not kinematic viscosity; the latter is $\nu =\mu /\rho$.

The radiative diffusion equation applies to an optically thick medium in local thermodynamic equilibrium. Its diffusion coefficient is $c/(3\kappa \rho )$, and the equilibrium radiation energy density is $u_{\mathrm{rad}}=4\sigma T^4/c$. HenceThe minus sign means energy flows down the temperature gradient. Here $\sigma$ is the Stefan-Boltzmann constant, $\kappa$ the opacity per unit mass, and $c$ the speed of light. Constant Thomson scattering opacity is the adopted transport approximation; local thermodynamic equilibrium is an additional assumption behind the radiation equation of state.

Finally the equation of state adds ideal gas gas pressure $p_g=\rho kT/({\mu }_m{m}_H)$ and isotropic radiation pressure $p_r=u_{\mathrm{rad}}/3=4\sigma T^4/(3c)$. The Boltzmann constant is $k$, and ${\mu }_m{m}_H$ is the mean mass per gas particle. Both contributions supply the total pressure entering vertical hydrostatic equilibrium.

In the limiting model with negligible gas pressure, $p=p_r=4\sigma T^4/(3c)$. Its gradient is $p_z=16\sigma T^3{T}_z/(3c)$. Combining radiative diffusion and hydrostatic equilibrium givesBecause $\kappa$ and $\Omega$ are constant with height at the fixed disk radius,Equating this to the viscous heating rate $9\mu {\Omega }^2/4$ forcesThis proves constant viscosity in radiation-supported vertical balance. It fixes the local effective dynamic viscosity uniquely, independent of height, surface density and radius within this ideal limiting model. It does not fix a height-independent kinematic viscosity: $\nu (z)=4c/(9\kappa \rho (z))$ still depends on the local mass density.

Set $\mathcal{R}=k/({\mu }_m{m}_H)$, so $p_g=\mathcal{R}\rho T$ and the prescribed dynamic viscosity is $\mu =\alpha \mathcal{R}\rho T/\Omega$. Keplerian shear and radiative diffusion then giveDifferentiating the second expression, equating it to the first and dividing by $\rho$ givesThis reduction uses the gas-pressure viscosity closure, not an assumption that gas pressure dominates the hydrostatic support.

For $z\ge 0$, introduce the vertical mass coordinate of a diskLet $\Sigma ={\int }_{-z_s}^{z_s}\rho dz$ be the full surface density of a reflection-symmetric disk. The upper surface is at $m_s=\Sigma /2$. With $\zeta =2m/\Sigma$, the temperature equation becomesWrite $T=T_{\ast }t(\zeta )$ and chooseThe equation reduces exactly to $(t^4{)}^{\prime \prime }+t=0$. Midplane symmetry gives $t^{\prime }(0)=0$, and the zero surface temperature gives $t(1)=0$. The stipulated unique positive nontrivial dimensionless solution therefore applies at every radius and surface density, with all physical dependence contained in $T_{\ast }$ and the mass-coordinate scaling. A zero temperature at the ideal surface does not mean zero escaping flux: $F=-(4\sigma /(3\kappa )){\partial }_m{T}^4$ can have a finite nonzero surface value.

The required viscosity is the density-weighted mean kinematic viscosity, not an unweighted vertical mean. Since $\rho \nu =\mu$,Substituting the dimensionless solution gives, with $I_t={\int }_0^{1}t(\zeta )d\zeta$,For central mass $M_{\ast }$, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$, so the requested explicit radius and surface-density dependence isThe universal integral need not be evaluated. No separate solution for $\rho (z)$ is needed to find this mean: the mass-coordinate transformation removes it from both the temperature equation and the viscosity integral. The result is the gas-pressure viscosity closure with mixed pressure support.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.