Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 321 (Dynamics of Astrophysical Discs)

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives

Here $\beta =p_{\mathrm{rad}}/p_{\mathrm{gas}}$ is the radiation-to-gas pressure ratio, not plasma beta. Write the opacity as ${\kappa }_{\mathrm{op}}$ to distinguish it from the radial epicyclic frequency. The radiation pressure is $p_{\mathrm{rad}}=4\sigma T^4/(3c)$, so radiative diffusion can be written asSince $\beta$ is independent of height, $p_{\mathrm{rad}}=\beta p/(1+\beta )$. Substitution of hydrostatic equilibrium, $p^{\prime }=-\rho {\Omega }^{2}z$, givesDifferentiating and using the viscous heating equation, together with $r{\Omega }^{\prime }=-q\Omega$, yieldsThus the effective dynamic viscosity $\rho \nu$ is independent of height. If the surfaces are at $z=\pm H$, the vertically integrated kinematic viscosity satisfiesUse the previous result and the specified Eddington accretion rate convention,to obtain the full thicknessThe constant-pressure-ratio vertical disk model therefore has thickness proportional to the inward accretion rate. This is a full surface-to-surface thickness, rather than a density disk scale height. Its use as a thin disk requires $2H\ll r$.

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.

Expanding the inertial specific angular momentum gives $h=h_0+r_0{p}_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference circular orbit, and $x_0$ specifies the radius of its associated epicyclic guiding center.

The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical harmonic oscillator has conserved energy

Initially every particle has $A=B=0$, so its Jacobi energy in a shearing sheet is $-3{\Omega }_0^2{x}_0^2/8$. During an inelastic collision, positions are fixed at the instant of impact, while momentum conservation preserves the sum of the tangential velocities. Hence the sum of $p_y=\dot{y}+2{\Omega }_{0}x$, and therefore the sum of the epicyclic guiding center positions $x_0=2p_y/{\Omega }_0$, is unchanged. Between collisions, these quantities are individually conserved.

The collision dissipates kinetic energy without changing the instantaneous tidal potential. Consequently the total Jacobi energy in a shearing sheet decreases. At any later time, writing $x_{0j}$ for the particles' current epicyclic guiding center positions givesIf $D\ge 0$ is the accumulated energy dissipated in the inelastic collisions, comparison with the initial circular orbits givesThus the mean guiding-center position stays fixed, while its variance grows. The initial ensemble has no preference for positive or negative $x$, and the local equations and collision law preserve the symmetry $(x,y,z)\mapsto (-x,-y,z)$. The spreading is therefore symmetric in the ensemble average: angular momentum transport moves some particles inward and others outward. A particular finite random realization need not be exactly symmetric. This is the microscopic energy argument for dissipative spreading of a planetary ring.

In the rotating shearing sheet, write the background orbital shear as ${\mathbf{U}}_0=-q\Omega x{\mathbf{e}}_y$. The rotating momentum equation contains Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{u}$ and the shearing-sheet tidal potential; the background shear balances the radial tidal acceleration.

SetGauss's law for magnetism gives ${\partial }_z{B}_z=0$, and the vertical ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_z=0$. Horizontal invariance and $v_z=0$ remove the nonlinear horizontal advection. The remaining background-shear term is $(\mathbf{v}\cdot \nabla ){\mathbf{U}}_0=-q\Omega v_x{\mathbf{e}}_y$. The horizontal magnetic tension is $B_z{\partial }_z{\mathbf{B}}_h/{\mu }_0$, while horizontal pressure gradients vanish. Subtracting background balance yieldsFor ideal magnetohydrodynamics, the ideal magnetohydrodynamic induction equation is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}$. Its horizontal components giveThese horizontally invariant magnetized shearing-sheet equations are exact within the stated local, incompressible ansatz, even for finite horizontal amplitudes. The vertical equation determines the pressure needed to balance vertical gravity and magnetic pressure; it does not add another horizontal evolution equation.

Write $H=z^{+}$ and $B_{+}=B_x^{+}$, and assume $B_z\ne 0$ and $q>0$. In a steady state, the horizontally invariant magnetized shearing-sheet equations giveSince $B_y$ vanishes at both boundaries, $v_x=0$ for $q\ne 2$, and then $B_y=0$ throughout. The other two equations areEliminating $v_y$ gives the harmonic oscillator equationThe midplane-symmetric magnetic bending in an incompressible disk has odd $B_x$ and even $v_y$. Applying $B_x(\pm H)=\pm B_{+}$ givesFor nonzero imposed inclination $B_{+}$, this equilibrium exists only when $\sin (KH)\ne 0$. More generally, $B_x=A\sin (Kz)+D\cos (Kz)$; the boundary conditions require $A\sin (KH)=B_{+}$ and $D\cos (KH)=0$. Thus the displayed solution is unique away from these resonances. At $\cos (KH)=0$, an additional even homogeneous solution is possible unless midplane symmetry is imposed. At $q=2$, an arbitrary constant $v_x$ is also allowed because its coefficient in the azimuthal equation vanishes; choosing $v_x=0$ gives the same symmetric equilibrium.

The horizontally invariant magnetized shearing-sheet equations are linear in the horizontal fields, so perturbations about the equilibrium satisfy the same equations. The fixed surface boundary conditions require $\delta B_x=\delta B_y=0$ at $z=\pm H$. Choose a normal mode withwhere the vertical wavenumber is $k=m\pi /(2H)$, $m=1,2,\dots$. With the Alfvén frequencythe four amplitude equations becomeEliminating the velocities leavesA nonzero amplitude requires the determinant of this system to vanish, yielding the ideal magnetorotational dispersion relationEquivalently, with ${\kappa }_r^2=2(2-q){\Omega }^2$,As a quadratic equation for $s^2$, its discriminant is ${\kappa }_r^4+16{\Omega }^2{\omega }_a^2>0$. If $0<{\omega }_a^2<2q{\Omega }^2$, its constant term is negative, so one root $s^2$ is positive and there is an exponentially growing mode. If ${\omega }_a^2>2q{\Omega }^2$, both the constant term and the coefficient of $s^2$ are positive, giving two negative roots and only oscillatory modes. Equality is marginal.

The lowest allowed vertical wavenumber, $k=\pi /(2H)$, is the last to be stabilized as $|B_z|$ increases. Therefore the finite-thickness magnetorotational instability criterion isThere is also a vertically uniform velocity mode with zero magnetic perturbation; for the usual orbitally stable regime $q<2$, it is just stable epicyclic motion. For $q>2$, that uniform mode is already hydrodynamically unstable, independently of the magnetic criterion. At $q=2$ it is marginal. The criterion above concerns the magnetic modes.

Using $K^2=2q{\mu }_0\rho {\Omega }^2/B_z^2$, the finite-thickness magnetorotational instability criterion is preciselyFor the midplane-symmetric equilibrium with $B_{+}\ne 0$, the absolute radial magnetic field isIts first maximum occurs at $z_{\ast }=\pi /(2K)$. The instability condition puts $z_{\ast }$ strictly inside the disc: $z_{\ast }<H$. The magnitude rises from zero to this maximum, then decreases on a nonempty interval before the surface. Thus the unstable equilibrium exhibits nonmonotonic magnetic bending, as illustrated below.

The link between bending and instability concerns a nonzero imposed surface inclination. If $B_{+}=0$, the unbent equilibrium $B_x=0$ can still have magnetorotational instability; the literal nonmonotonic-bending conclusion does not apply to that degenerate case. Resonances with $\sin (KH)=0$ also require the separate equilibrium analysis in part (b).