Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 59 (Astrophysical Fluid Dynamics)

Take the equilibrium mass density to be constant and absorb any conservative equilibrium force into the pressure. The linearized ideal magnetohydrodynamic induction equation, using the solenoidal vector field conditions, isFor $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$, differentiation of the cylindrical basis is important. Acting on a vector whose cylindrical components have azimuthal dependence $e^{im\varphi }$ gives $(\mathbf{B}\cdot \nabla )\mathbf{u}=imJ\mathbf{u}+J{\mathbf{e}}_z\times \mathbf{u}$. Also $(\mathbf{u}\cdot \nabla )\mathbf{B}=J{\mathbf{e}}_z\times \mathbf{u}$. These basis contributions cancel in induction, giving $i\omega \mathbf{b}=imJ\mathbf{u}$.

The perturbation of the Lorentz force isBoth of the first two terms now contribute a basis-rotation term. With $\tilde{p}=p_1/\rho +\mathbf{B}\cdot \mathbf{b}/({\mu }_0\rho )$, the linearized ideal magnetohydrodynamic momentum equation is thereforeThe magnetic part of $\tilde{p}$ is the perturbation of magnetic pressure, not an extra restoring force to retain after projecting out pressure.

Put $D={\omega }^2-m^2{\tilde{J}}^2$, where $\tilde{J}=J/\sqrt{{\mu }_0\rho }$. For $mJ\ne 0$, substitute $\mathbf{u}=\omega \mathbf{b}/(mJ)$ into the momentum equation and multiply by $mJ$. Taking a curl removes the modified pressure. Sincewe obtain the uniform-current toroidal-field curl reductionA second curl, using $\nabla \times \nabla \times \mathbf{b}=-{\nabla }^2\mathbf{b}$, givesThe divided expression assumes $D\ne 0$. If $mkJ\ne 0$, the undivided curl equation shows that $D=0$ would force $\mathbf{b}=0$, so this restriction loses no such magnetic mode. The $k=0$ degenerate limit can instead be taken in the undivided equations; it has no growing branch of the kind considered below. Axisymmetric $m=0$ disturbances are likewise excluded by the velocity elimination, and induction gives no nonzero-frequency magnetic disturbance in this particular equilibrium.

For a vector Laplacian eigenmode with ${\alpha }^2+k^2>0$, elimination givesThus both squared frequencies are real. The lower branch satisfies ${\omega }_{-}^2\ge {\tilde{J}}^2(|m{|}^2-2|m|)$, so it is positive for $|m|>2$. For $|m|=2$ it is positive when $\alpha \ne 0$, but the printed strict-positivity statement needs an endpoint qualification: $|m|=2$, $\alpha =0$ permits a neutral mode unless boundary conditions exclude it.

This is a genuine counterexample to the unqualified endpoint, not merely a loose algebraic bound. The neutral quadrupolar perturbation of a uniform-current toroidal fieldis regular at the axis, is solenoidal, and has $\nabla \times \mathbf{b}=k\mathbf{b}$, hence ${\nabla }^2\mathbf{b}=-k^2\mathbf{b}$. Also ${\mathbf{e}}_z\times \mathbf{b}=-i\mathbf{b}$, so the non-gradient part of the linearized Lorentz force vanishes for $m=2$. The gas-pressure perturbation $p_1=-\mathbf{B}\cdot \mathbf{b}/{\mu }_0$ cancels its magnetic pressure gradient, as required by $\tilde{p}=0$. Taking its real part gives a real disturbance. Decay at radial infinity or an appropriate homogeneous radial boundary condition would exclude it, restoring strict positivity for the allowed eigenmodes; no such condition is printed.

For $|m|=1$ the lower branch is unstable precisely whenAt equality it is marginal. With the time convention $e^{i\omega t}$, the growing member has $\omega =-is$ andThe other branch remains oscillatory. Actual admissibility of a chosen $\alpha$ depends on the radial domain and its boundary conditions.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

The magnetorotational instability extracts energy from differential rotation through magnetic tension. It can destabilize a disk whose specific angular momentum increases outward, even though such a disk satisfies Rayleigh's circulation criterion. The distinction is that an unmagnetized displaced parcel approximately conserves its own angular momentum, whereas a field line transfers angular momentum between parcels. If angular velocity decreases outward, an outward-displaced parcel lags its inner partner. Magnetic tension transfers angular momentum to the outer parcel and removes it from the inner one, making their radial separation increase. A sufficiently stiff magnetic connection suppresses the separation instead. The instability therefore requires a finite range of field-line-bending strengths.

Make this mechanism quantitative in the local shearing sheet of a cylindrical rotation profile $\Omega (r)$. Put $q=-d\log \Omega /d\log r$, and take $0<q<2$ so the unmagnetized flow is stable. Its radial epicyclic frequency is ${\kappa }^2=2(2-q){\Omega }^2$. Let the uniform field be vertical, $B_z{\mathbf{e}}_z$, with constant mass density, and use an axisymmetric horizontal disturbance proportional to $e^{st+ikz}$. Write $\mathbf{h}=\mathbf{b}/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{B}_z/\sqrt{{\mu }_0\rho }$. The linearized ideal magnetohydrodynamics equations areThe first pair contains Coriolis acceleration and the epicyclic response; the last equation is the shear-winding of a radial field perturbation into an azimuthal one. Omitting that winding term would remove the driving mechanism.

Eliminating $h_r,h_{\varphi }$ gives $(s^2+{\omega }_A^2)u_r=2\Omega s{u}_{\varphi }$ and $s(s^2+{\omega }_A^2)u_{\varphi }+\Omega [(2-q)s^2-q{\omega }_A^2]u_r=0$. Their determinant is the ideal magnetorotational dispersion relationIts discriminant as a quadratic in $s^2$ is ${\kappa }^4+16{\Omega }^2{\omega }_A^2>0$. Since $2{\omega }_A^2+{\kappa }^2>0$, a positive $s^2$ exists precisely when its constant term is negative:Thus angular velocity, rather than specific angular momentum, must decrease outward. A Keplerian disk has $q=3/2$, ${\kappa }^2={\Omega }^2$ and is hydrodynamically stable, yet magnetic modes with $0<k^2{v}_A^2<3{\Omega }^2$ grow.

The growing root is $s^2=-{\omega }_A^2-{\kappa }^2/2+\frac{1}{2}\sqrt{{\kappa }^4+16{\Omega }^2{\omega }_A^2}$. Differentiation with respect to ${\omega }_A^2$ gives the maximum growth rate of ideal magnetorotational instability:For Keplerian shear these are $3\Omega /4$ and $15{\Omega }^2/16$. The ideal maximum is set by shear; reducing field strength reduces the fastest wavelength instead of reducing this maximum growth rate. This is why the exactly zero-field problem differs from a sequence of weak-field problems with progressively larger wave number. At fixed wave number, growth does tend to zero as the field vanishes. The wavelength must fit the actual disk, and diffusion eventually matters at sufficiently short scales.

The instability has the required angular-momentum-transfer sign. In a growing mode, $u_{\varphi }/u_r=(s^2+{\omega }_A^2)/(2\Omega s)>0$. At the fastest mode this ratio is one, while induction gives $h_{\varphi }=-h_r$ after allowing for their common Fourier phase. Thus the averaged Reynolds stress minus Maxwell stress tensor, $\rho ⟨u_r{u}_{\varphi }-h_r{h}_{\varphi }⟩$, is positive: angular momentum moves outward. Its product with $q\Omega$ is the energy supplied by the background shear. These local mechanisms and coefficient conventions are also checked in the primary MRI stability calculation.

More general fields require separating field-line bending from field gradients and curvature. At leading local order, for a stationary compatible field in an incompressible flow with axisymmetric wavevector $(k_r,0,k_z)$, the bending frequency isPressure elimination projects the radial force by $f=k_z^2/k^2$. The radial momentum equation becomes $s{u}_r-2f\Omega u_{\varphi }=i{\omega }_A{h}_r$; the azimuthal and induction equations retain the form above. Eliminating the amplitudes gives the oblique axisymmetric magnetorotational dispersion relationHere $k_z\ne 0$. The unstable range is $0<{\omega }_A^2<-fd{\Omega }^2/d\log r$. The poloidal magnetic field therefore determines whether an axisymmetric perturbation bends a field line. A locally uniform toroidal magnetic field does not enter $\mathbf{k}\cdot \mathbf{B}$ for axisymmetry, and can contribute magnetic pressure that is projected out of this incompressible leading-order problem. Thus a toroidal component does not suppress the weak-field MRI merely by being present. Conversely a purely toroidal field has no ordinary axisymmetric local bending MRI in this approximation. This does not imply that every globally curved toroidal equilibrium is stable.

A steady general poloidal magnetic field must also be compatible with the equilibrium rotation. The azimuthal ideal magnetohydrodynamic induction equation contains ${\partial }_t{B}_{\varphi }=r{\mathbf{B}}_p\cdot \nabla \Omega$. A stationary axisymmetric equilibrium with no meridional flow consequently requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$, Ferraro's law of isorotation. A radial field crossing cylindrical differential-rotation surfaces would otherwise wind secularly; it cannot simply be substituted into a stationary normal-mode calculation. Boundary conditions and the along-field eigenvalue problem replace a freely chosen local $k$ in nonuniform or bounded equilibria. Strong enough field-line tension can then stabilize all admissible wavelengths.

An explicit axisymmetric effect of nonuniform field strength and curvature is obtained with a purely toroidal field $B_{\varphi }(r){\mathbf{e}}_{\varphi }$, constant mass density and cylindrical rotation. A radial fluid displacement changes the field byThis follows directly from $\mathbf{b}=\nabla \times (\mathbf{\xi }\times \mathbf{B})$ and incompressibility. Its radial curvature force, after absorbing the magnetic-pressure perturbation, is $-2B_{\varphi }{b}_{\varphi }/({\mu }_0\rho r)$. The axisymmetric azimuthal momentum equation is $s{u}_{\varphi }+(2\Omega +r{\Omega }^{\prime })u_r=0$. Combining it with the radial force and then projecting out pressure givesThis is the local Michael criterion for axisymmetric toroidal-field interchange. The new term can either stabilize or destabilize: for $B_{\varphi }\propto r^p$, $\mathcal{D}={\kappa }^2+2(1-p)v_{A\varphi }^2/r^2$. A current-free $B_{\varphi }\propto r^{-1}$ is stabilizing; $B_{\varphi }\propto r$ leaves this axisymmetric interchange coefficient unchanged; a sufficiently steep increase with $p>1$ can destabilize even a centrifugally stable rotation law. That last instability can persist without differential rotation, so its energy source is the magnetic-current equilibrium rather than the ordinary MRI. Nonaxisymmetric current-driven modes are outside the requested essay scope.

For mixed poloidal and toroidal fields, the full induction perturbation contains $(\mathbf{B}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{B}-\mathbf{B}\nabla \cdot \mathbf{\xi }$, and the linear Lorentz force contains both $(\nabla \times \mathbf{b})\times \mathbf{B}$ and $(\nabla \times \mathbf{B})\times \mathbf{b}$. These terms explain why gradients and curvature cannot be replaced everywhere by a scalar $k^2{v}_A^2$: they can draw on equilibrium-current energy and couple the radial and vertical motions. Compressibility introduces additional magnetic pressure and magnetosonic responses; a strong toroidal field can alter the dispersion even though it disappeared from the leading incompressible weak-field result.

Finally, a vertically or radially stratified general field also supports part of the equilibrium weight through magnetic pressure. A displaced flux tube carries its frozen-in mass-to-flux relation, so its mass density at the surrounding total pressure can differ from its new environment. This introduces magnetic buoyancy instability and competition with ordinary buoyancy. Axisymmetric interchange displacements can access this energy without an azimuthal wave number, while poloidal tension resists variation along the field. Growth that survives when $d\Omega /dr=0$ must not be attributed to shear-driven MRI. The general-field stability problem therefore combines shear, field-line tension, current/curvature forces, compressibility, buoyancy and boundary constraints; the uniform-field result isolates the basic angular-momentum feedback mechanism.