Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 (Astrophysical Discs and Magnetohydrodynamics)

Write horizontal components with a subscript $h$. The steady ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since ${\partial }_z=0$, its horizontal components implyThus $(\mathbf{u}\times \mathbf{B}{)}_z=C$ is a spatial constant. A vanishing axial electric field, $E_z=-C=0$, is needed to infer horizontal alignment. It is satisfied, for example, if the constant electric field vanishes at a boundary or at infinity. With this additional condition and ${\mathbf{B}}_h\ne 0$, define the magnetohydrodynamic mass loading $k$ by $\rho {\mathbf{u}}_h=k{\mathbf{B}}_h$, and put $U=u_z-k{B}_z/\rho$. ThenThe steady continuity equation and the solenoidal magnetic-field constraint giveAlso $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Using the vector identity for the curl of a cross product, together with ${\partial }_z\mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$, its curl is ${\mathbf{e}}_z\mathbf{B}\cdot \nabla U$. HenceThese are steady planar ideal-MHD field-line invariants: both quantities are constant along a magnetic field line.

The zero-electric-field condition does not follow from the printed stationarity assumptions alone. Constant $\rho ,p,s$, $\Phi =0$, $\mathbf{u}=u_0{\mathbf{e}}_x$ and $\mathbf{B}=B_0{\mathbf{e}}_y$, with $u_0{B}_0\ne 0$, satisfy every displayed ideal magnetohydrodynamics equation but have perpendicular horizontal velocity and magnetic field. Their $C=u_0{B}_0$ is nonzero, so no $k,U$ give the stated decomposition. The field-line decomposition is valid on the $E_z=0$ branch; as an unrestricted implication, the first claim has this counterexample. The subsequent calculations use that branch.

Continue on the branch of steady planar ideal-MHD field-line invariants, with $k\ne 0$ and positive $\rho ,T$. Adiabatic entropy transport gives $\mathbf{u}\cdot \nabla s=0$. As all quantities are $z$-independent,The axial component of the ideal magnetohydrodynamic momentum equation has no pressure or gravitational term. Magnetic tension givesSince $k$ is constant along a magnetic field line, this becomesSubtracting the two axial invariants yields the useful compatibility relationThe squared ratio of horizontal flow speed to horizontal Alfvén speed is $M_h^2={\mu }_0{k}^2/\rho$; vector equality with the horizontal Alfvén velocity additionally requires $k>0$. If $V-U\ne 0$, the right side is a nonzero constant along the field line, so a regular finite $B_z$ cannot permit $M_h^2=1$ anywhere. This is the nondegenerate exclusion.

However, $B_z\not\equiv 0$ does not imply $V-U\ne 0$. The Alfvénic degeneracy of steady planar MHD is a genuine exception: take constant $\rho ,p,s$, $\Phi =0$, and a constant magnetic field with nonzero horizontal and axial components, and setEvery equation holds, but the horizontal velocity equals the horizontal Alfvén velocity everywhere despite $B_z\ne 0$. Therefore the printed pointwise exclusion needs a nondegeneracy hypothesis, such as $V\ne U$. The general conclusion is the boxed compatibility relation; when $V=U$, it says $(M_h^2-1)B_z=0$ and allows the Alfvénic branch.

Let $b=\frac{1}{2}{u}^2+\Phi +w$. Because the specific entropy is constant along the horizontal flow, the specific enthalpy identity gives $\mathbf{u}\cdot \nabla w={\rho }^{-1}\mathbf{u}\cdot \nabla p$. Dotting the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$ therefore givesOn the field-aligned horizontal branch, $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Directly evaluating the horizontal components of the curl givesThe scalar triple-product identity consequently makes the magnetic work $U{\mathbf{B}}_h\cdot \nabla B_z/{\mu }_0$. Meanwhile $\rho \mathbf{u}\cdot \nabla b=k{\mathbf{B}}_h\cdot \nabla b$. The previously established field-line constancy of $U,k$ now gives the planar ideal-MHD Bernoulli invariantThis is conservation of total energy transported along a magnetic field line, including electromagnetic work. To see its flux interpretation, the ideal magnetohydrodynamic energy conservation law has fluxThe second term is the Poynting vector, since $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. Substitution of $\mathbf{u}=k\mathbf{B}/\rho +U{\mathbf{e}}_z$ givesThe steady divergence of this horizontal energy flux is zero. Magnetic stress can change the ordinary hydrodynamic quantity $b$ in the Bernoulli equation by doing work; the term $-U{B}_z/({\mu }_{0}k)$ accounts for the accompanying Poynting flux. When $U=0$, the velocity is completely field-aligned, $\mathbf{E}=0$, and the usual hydrodynamic Bernoulli quantity is constant. These conclusions require the field-aligned branch and $k\ne 0$; the arbitrary constant-$E_z$ flow from the first part is not covered by this particular invariant.

Assume an axisymmetric thin disk in the fixed Newtonian potential of a point mass, negligible disk self-gravity, and slow evolution compared with an orbital period. Radial pressure and radial inertia are small compared with gravity and centrifugal acceleration, so $u_{\varphi }=r\Omega (r)$ with $\Omega =(GM/r^3{)}^{1/2}$, independent of height and time. Assume no mass loss or torque through the upper and lower disk faces. Define the surface density of a disk, mass-weighted radial velocity and integrated shear stress byVertical integration of the continuity equation givesThe azimuthal momentum equation, including the geometric term $u_r{u}_{\varphi }/r$, gives conservation of angular momentum in terms of $j=r^2\Omega$:Subtracting $j$ times mass conservation eliminates the time derivative and leaves $\Sigma {\bar{u}}_r{j}^{\prime }=r^{-1}(r^{2}W{)}^{\prime }$.

Define the effective density-weighted viscosity of a disk through the actual stress:If the local viscous stress is $T_{r\varphi }=\rho \nu r{\Omega }^{\prime }$, this reduces to $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$. The definition can also describe an effective turbulent transport stress, provided the same shear closure is appropriate. Since ${\Omega }^{\prime }<0$, outward transport corresponds to $W<0$ and positive $\bar{\nu }$.

For Keplerian rotation, $j=\sqrt{GMr}$ and $r^{2}W=-(3/2)\sqrt{GM}{r}^{1/2}\bar{\nu }\Sigma$. ThereforeSubstitution into mass conservation yields the Keplerian viscous diffusion equationThe absence of surface fluxes, fixed rotation law and leading thin-disk approximation are essential: a wind, appreciable radial inertia or evolving gravitational potential would alter this reduction.

Take $A>0$ and a nontrivial nonnegative profile. On $0<r<R(t)$ put $X=(R/r{)}^a$ and $\Sigma =\sigma (X-1)$. The diffusion operator acts on $r^{1/2}\bar{\nu }\Sigma =A{\sigma }^2[r^{5/2-2a}{R}^{2a}-2r^{5/2-a}{R}^a+r^{5/2}]$, givingBut differentiating the trial profile in time gives onlyFor $a\ne 0$, the functions $1,X,X^2$ are independent on the interior. Their coefficients can match only ifMatching the remaining coefficients givesThus $\sigma =[15A(t+t_0){]}^{-1}$, with $R=C(t+t_0{)}^{2/5}$ when $a=1$, and $R=C(t+t_0{)}^{1/2}$ when $a=5/4$. A nonzero solution whose support shrinks to zero at $t=0$ has $t_0=0$. The two compact self-similar disk with density-linear viscosity profiles are consequentlySet $\Sigma =0$ for $r>R$. Both $r^{1/2}\bar{\nu }\Sigma$ and its first radial derivative vanish at the moving edge, because $\bar{\nu }\Sigma$ vanishes quadratically there. Together with $\Sigma (R)=0$, this makes the cutoff consistent with the diffusion equation without an added singular edge flux.

These are singular-origin self-similar solutions: $R(0)=0$ is a limiting support condition, while $\sigma$ diverges as $t\downarrow 0$. It is not ordinary finite pointwise initial data. A finite nonzero $\sigma (0)$ would instead make $R=0$ an invariant solution of the radius equation, precluding an expanding nontrivial profile from that initial radius.

Using the radial-drift expression already derived, the factor $X-1$ cancels and givesFor $a=1$, $X=R/r$ and $\sigma =1/(15At)$, soThe instantaneous flow is inward for $r<R/5$ and outward for $R/5<r<R$. Nevertheless outward motion need not prevent eventual accretion. The trajectories in accretion trajectories in a compact viscous similarity disk satisfyMultiplication by $t^{-1/2}$ and integration giveA parcel at $r_i\in (0,R_i)$ at time $t_i>0$ has $K=(r_i-R_i)/\sqrt{t_i}<0$. It reaches the origin atBefore arrival, $r/R=1+K{t}^{1/10}/C$ decreases monotonically. Even a parcel initially moving outward eventually crosses $r/R=1/5$, turns inward, and is accreted. Only $K=0$, the outer-edge trajectory, never arrives; its surface density of a disk is zero. Thus almost every mass-carrying parcel is accreted in finite time.

For $a=5/4$, the coefficient $5/2-2a$ vanishes, leavingThe trajectories are $r(t)=r_i\sqrt{t/t_i}$, so each parcel keeps a fixed ratio $r/R$. This is an expanding nonaccreting similarity disk supplied by an inner torque, not an inward accretion solution.

Both profiles have integrable central cusps. Directly integrating the surface density of a disk and its specific angular momentum $j=\sqrt{GMr}$ givesFor $a=1$, substitution of the similarity functions givesThe inward accretion rate is $\dot{M}(r)=-2\pi r\Sigma {\bar{u}}_r$. At the origin it tends to $M_d/(5t)=-d{M}_d/dt$. The outward viscous torque in an accretion disk isFor this profile $\mathcal{G}\to 0$ as $r\to 0$, and the advected angular-momentum flux $\dot{M}j$ also vanishes there. Both outer fluxes vanish at $R$. Thus disk mass decreases through central accretion while disk angular momentum is conserved. Including the accreted central mass restores total mass conservation for the disk-plus-sink system.

For $a=5/4$, instead,Here the mass flux tends to zero at the origin, despite positive radial velocity everywhere else. The inner torque has the nonzero limitThus the nonaccreting similarity disk supplied by an inner torque conserves disk mass but gains angular momentum from an external central torque. Its vanishing edge flux cannot supply that gain. The disk is not globally isolated in angular momentum; the combined disk and torque source may of course conserve it.

These profiles illustrate different boundary-controlled similarity classes of the Keplerian viscous diffusion equation. The $a=1$ solution is compatible with a zero-torque inner boundary condition and conserved finite angular momentum, and can describe a late-time spreading accretion disk after the detailed initial profile has been forgotten. Its mass diverges as $t\downarrow 0$, so the formal zero-radius origin is not a finite-mass initial disk. The $a=5/4$ solution has finite initial mass concentrated at the origin and initially zero disk angular momentum, but requires the specified inner torque for $t>0$; that torque's $t^{-3/4}$ singularity is time-integrable. It may describe a spreading nonaccreting disk for compatible torque-driven boundary conditions, but cannot be an isolated angular-momentum-conserving disk's generic asymptotic state. Similarity scaling suggests their possible asymptotic roles; it does not establish convergence for arbitrary initial and boundary data.

For a circular mid-plane orbit, radial force balance in the Newtonian gravitational potential is $r{\Omega }^2={\Phi }_{,r}$. HenceAxisymmetry conserves the specific angular momentum $j=r^2\dot{\varphi }$. At fixed $j$, radial motion obeys $\ddot{r}=-{\partial }_r{\Phi }_{\mathrm{eff}}$, where the effective potential is ${\Phi }_{\mathrm{eff}}=\Phi +j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3{\Phi }_{,r}(r_0,0)$ there,This is the radial epicyclic frequency; equivalently ${\kappa }^2=r^{-3}d(r^4{\Omega }^2)/dr$.

Reflection symmetry gives ${\Phi }_{,z}(r,0)=0$ and ${\Phi }_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot{z}=-{\Phi }_{,zz}(r_0,0)z$, giving the vertical epicyclic frequencyPositive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation frequency.

Write $s=(r^2+z^2{)}^{1/2}$. The Paczyński-Wiita potential is a spherical Newtonian gravitational potential with ${\Phi }^{\prime }(s)=GM/(s-a{)}^2$ and ${\Phi }^{\prime \prime }(s)=-2GM/(s-a{)}^3$. At $z=0$, $s=r$, soSubstituting into the orbital formulas givesChoosing positive orbital and vertical frequencies, $\Omega ={\Omega }_z=\sqrt{GM}/[\sqrt{r}(r-a)]$. For $r>3a$, $\kappa =\Omega \sqrt{(r-3a)/(r-a)}$. For $a<r<3a$, ${\kappa }^2<0$ and radial disturbances grow at rate $\sqrt{-{\kappa }^2}$, even though the vertical motion remains stable. Thus circular orbits are radially unstable for $a<r<3a$, and $r=3a$ is the marginal orbit. This recovers the innermost stable circular orbit of the Paczyński-Wiita circular orbit model.

The marginal circular orbit has conserved specific angular momentum and specific orbital energyAt fixed $j_0$, the derivative of the effective potential factors asAn inward displacement therefore produces inward radial acceleration; no inner circular equilibrium or centrifugal barrier halts the plunge. In the infinitesimal-displacement limit, energy conservation givesSelecting the inward branch and the original positive sense of rotation yields the Paczyński-Wiita ballistic plungeThe azimuthal speed remains positive while the radius decreases, so the trajectory spirals inward. From any fixed starting radius below $3a$, the time to reach $a$ is finite: $1/|v_r|$ behaves as $(r-a{)}^{1/2}$ near $a$, which is integrable.

The exactly marginal circular state with exactly zero displacement remains a circular solution. A finite inward release from rest changes the energy by a term cubic in the displacement; the displayed expression is its limiting separatrix with $E=E_0$. On that separatrix $|v_r|\propto (3a-r{)}^{3/2}$ near $3a$, so departure from exactly $3a$ takes an infinite limiting time. This distinguishes the infinitesimal-release formula from spontaneous motion of the unperturbed marginal orbit.

Use the leading vertically averaged horizontal momentum equations of a thin disk. For an isothermal ideal gas, write $P=\int pdz=c_s^2\Sigma$, with constant isothermal sound speed $c_s$. The disk scale height is of order $H=c_s/{\Omega }_z$; this is exact for local isothermal vertical hydrostatic balance and remains the thickness scale for the estimate. Put $ϵ=H/r\ll 1$. The alpha disk prescription givesAt fixed radii of order $a$ inside the plunge and away from the marginal orbit, $|v_r|,v_{\varphi }$ and $r{\Omega }_z$ are fixed numerical multiples of $(GM/a{)}^{1/2}$. Hence the inflow time is dynamical, while the viscous timescale is longer by $1/(\alpha {ϵ}^2)$.

The Paczyński-Wiita ballistic plunge obeys $v_r{v}_r^{\prime }-v_{\varphi }^2/r=-{\Phi }^{\prime }$ and $v_r{j}_0^{\prime }=0$ exactly. For smoothly varying disk quantities on scale $r$, the pressure correction to radial acceleration has magnitude $c_s^2/r$, and the viscous correction is of order $\nu v/r^2$. Relative to the orbital acceleration scale $r{\Omega }_z^2$, these are respectivelyFinite-height corrections to the mid-plane gravitational force are also $O({ϵ}^2)$ by reflection symmetry. Thus the leading horizontal flow is ${\mathbf{u}}_h=\mathbf{v}+O({ϵ}^{2}v)$, with specific angular momentum differing from $j_0$ by a fraction $O(\alpha {ϵ}^2)$ over a dynamical transit, for $\alpha \lesssim 1$. Vertical structure and the smaller vertical velocity must be supplied by the thin-disk expansion; they are not specified by the planar ballistic expression.

This estimate is not uniform at the marginal orbit. In the ballistic solution, steady mass conservation givesConsequently $d\ln \Sigma /dr=[2(r-a){]}^{-1}+3[2(3a-r){]}^{-1}$. Writing $\delta =3a-r\ll a$, the pressure acceleration scales as $c_s^2/\delta$, while the net ballistic radial force scales as $(GM/a^2)(\delta /a{)}^2$. Their ratio is $O[{ϵ}^2(a/\delta {)}^3]$, evaluated with the characteristic thickness near $3a$. A transonic matching layer of width $\delta /a=O({ϵ}^{2/3})$ is therefore needed. The approximation concerns the bulk plunging flow outside that layer, not a uniform relative approximation arbitrarily close to $3a$.

For the torque comparison, the steady angular-momentum flux is $\mathcal{G}-\dot{M}j=\mathrm{constant}$, where $\mathcal{G}=-2\pi \nu \Sigma r^3{\Omega }^{\prime }$ is the viscous torque in an accretion disk. Inside the plunge, $\Omega =j_0/r^2$, soOutside, the approximately circular disk has $\mathcal{G}=\dot{M}[j(r)-j_0]+O(\alpha {ϵ}^2\dot{M}{j}_0)$, since the inner plunge torque is small. At $4a$,a fixed number independent of disk thickness. At $2a$, radial infall is dynamical and the inner torque is $O(\alpha {ϵ}^2\dot{M}{j}_0)$. Therefore the weak viscous torque in a thin-disk plunging region estimate givesThe numerical prefactor need not be small; the suppression is an asymptotic statement in the thin-disk parameter. Rapid infall and its low surface density explain why a nearly zero inner torque is a useful leading boundary condition for the outer accretion disk.

Choose a reference circular orbit at $r_0$, and use local coordinates $x=r-r_0$, $y=r_0[\varphi -{\Omega }_{0}t]$, with the frame rotating at $\Omega ={\Omega }_0=\Omega (r_0)$. Linearizing differential rotation in the shearing sheet gives the background flowThus the velocity perturbation is measured relative to this background. The local orbital shear parameter is $q=2A/\Omega =-d\ln \Omega /d\ln r$; Keplerian rotation has $q=3/2$ and $A=3\Omega /4$. Rotation supplies the Coriolis force, while the linear shear supplies the terms proportional to $A$.

For constant density, $\psi$ is the total-pressure potential after background force balance has been subtracted:up to a function of time and any prescribed perturbation potential that is absorbed into it. The magnetic-pressure gradient has been combined with the gas-pressure gradient, leaving magnetic tension as $\mathbf{B}\cdot \nabla \mathbf{B}/({\mu }_0\rho )$. The pressure also enforces incompressibility. In the unstratified shearing box here, the tidal/background terms have already been removed in deriving the perturbation equations; no independent vertical stratification is being evolved.

At fixed time write the azimuthal shift as $\delta =2A{L}_{x}t$ modulo $L_y$. Integrating a radial derivative over the box gives a boundary difference proportional toThe two integrals agree by translation invariance of an integral over a full azimuthal period. Hence $⟨{\partial }_{x}Q⟩=0$. The ordinary periodic boundary conditions make the $y$ and $z$ boundary differences zero as well. Finally,for every $x,z$. Thus the volume averages in a shearing box satisfyFor smooth shearing-periodic extensions, spatial derivatives and products inherit the same shearing-periodic boundary condition. Their divergence therefore has zero volume average, and integration by parts is available without a residual boundary flux. Time differentiation of an average over this fixed physical box commutes with integration, even though the radial identification shifts with time.

The nonlinear velocity term is a divergence componentwise: $v_j{\partial }_j{v}_i={\partial }_j(v_j{v}_i)$ because $\nabla \cdot \mathbf{v}=0$. Likewise $B_j{\partial }_j{B}_i={\partial }_j(B_j{B}_i)$; their volume averages vanish. Pressure gradients, spatial Laplacians and the background-advection term $x{\partial }_y$ also average to zero by the boundary identities. The mean momentum equation is thereforeIn induction, the two nonlinear terms are divergences, $v_j{\partial }_j{B}_i={\partial }_j(v_j{B}_i)$ and $B_j{\partial }_j{v}_i={\partial }_j(B_j{v}_i)$. Averaging consequently givesThe radial and vertical mean magnetic fields stay constant, and $⟨B_y⟩(t)=⟨B_y⟩(0)-2At⟨B_x⟩(0)$. In particular an initially zero mean magnetic field stays zero.

Writing $V_i=⟨v_i⟩$, the horizontal mean motion satisfiesThis is the local radial epicyclic frequency. When the background is centrifugally stable, $\Omega (\Omega -A)>0$, the whole box supports epicyclic mean-velocity oscillations. For example,while $V_z$ is constant. A Keplerian shearing sheet has $\kappa =\Omega$. For a centrifugally unstable rotation law the same mean equations instead give exponential growth, so an oscillation interpretation requires the positive squared frequency.

No divergence of turbulent stress survives the box average, and a zero initial mean radial velocity cannot acquire a secular accretion drift from internal stresses. More generally, a steady horizontal mean flow has $V_x=V_y=0$ for the usual nondegenerate rotating disk. The epicycle has zero time-averaged radial velocity. Opposite radial faces are identified rather than connected to a central sink and an outer mass reservoir: material crossing a face re-enters the opposite one after an azimuthal shift. Thus the shearing box can model angular-momentum transport and its local stresses, but not develop a global disk accretion flow. Such accretion needs radial gradients, boundaries and mass exchange omitted from the homogeneous local model.

Define the perturbation energy per unit mass by $E=⟨v^2/2+B^2/(2{\mu }_0\rho )⟩$. Dot the momentum equation with $\mathbf{v}$ and the induction equation with $\mathbf{B}/({\mu }_0\rho )$, then take volume averages. The Coriolis force does no work. Pressure work and nonlinear advection are divergences and integrate to zero, as does the background $x{\partial }_y$ transport. The two magnetic exchange terms add toThe velocity shear term supplies $2A⟨v_x{v}_y⟩$, while magnetic shear supplies $-2A⟨B_x{B}_y⟩/({\mu }_0\rho )$. Integration by parts gives the viscous and resistive losses as $-\nu ⟨|\nabla \mathbf{v}{|}^2⟩$ and $-\eta ⟨|\nabla \mathbf{B}{|}^2⟩/({\mu }_0\rho )$. For a solenoidal field $\mathbf{F}$, another integration by parts showsAll boundary terms cancel by the shearing-periodic identities. The resulting balance isThe source is work extracted from maintained differential rotation by Reynolds stress and the magnetic part of the Maxwell stress tensor; internal magnetic/kinetic exchange has canceled.

For the decay bound, the elementary inequality $2|ab|\le a^2+b^2$ givesA uniform spectral gap of a shearing-periodic box is needed; ordinary radial periodicity cannot simply be assumed. At each time the compatible Fourier modes areThey are orthogonal over the box. Their wavevectors areFor every nonconstant mode, $|\mathbf{k}|\ge 2\pi /L$, where $L=\max (L_x,L_y,L_z)$: if $m$ or $p$ is nonzero, its corresponding unsheared component supplies the bound; otherwise $n\ne 0$ and the radial component does. Zero mean excludes precisely the constant mode. The Parseval identity therefore proves the time-uniform Poincaré inequalityApply it componentwise to $\mathbf{v}$ and $\mathbf{B}$. With $\nu =\eta$, the energy decay criterion for a viscous-resistive shearing box becomesHenceThis excludes sustained finite-amplitude turbulence as well as infinitesimal growth. It is a sufficient stability condition, not a claim that every box above the threshold must become turbulent. The zero-mean restrictions matter: an imposed uniform magnetic field or bulk epicycle is not controlled by this spatial spectral-gap inequality.