Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 74 (Stellar and Planetary Magnetic Fields)

Adding $i$ times the equation for $B$ to the equation for $A$ combines the two real amplitudes into $P=A+iB$. With the fluctuating coupling absent, the derivative terms satisfy $-r{B}_x+ir{A}_x=ir{P}_x$, so the complex evolution isBoth real amplitudes satisfy the zero Dirichlet boundary condition, hence $P(0)=P(\ell )=0$. A unit-modulus change of dependent variable removes the first derivative: put $P=e^{-irx/2}F$. Direct differentiation givesThe eigenfunctions satisfying the Dirichlet boundary condition are therefore $e^{-irx/2}\sin (n\pi x/\ell )$, with temporal eigenvaluesThe first nonzero stationary mode appears when ${\sigma }_1=0$, givingThe complex constant $C$ supplies the phase freedom of this unperturbed mean-field dynamo model.

To display the dipole and quadrupole parity in a mean-field dynamo, center the coordinate at the midpoint: $y=x-\ell /2$, $F(y)=\cos (\pi y/\ell )$, $h=r_0/2$. Choose a real amplitude $C_0$. The dipole choice is $P_D=C_{0}F(y)e^{-ihy}$, soHere $A_D$ is an even function and $B_D$ an odd function of $y$, and the cosine envelope vanishes at both boundaries. The quadrupole choice is $P_Q=i{P}_D$, which givesNow $A_Q$ is an odd function and $B_Q$ an even function. Both choices have the same stationary threshold because the unperturbed complex equation is invariant under a constant phase rotation.

Write $P=P_0+d{P}_1+O(d^2)$ and $r=r_0+d{r}_1+O(d^2)$, with each coefficient satisfying the zero Dirichlet boundary condition. The steady complex equation, including the fluctuating coupling, is $0={\mathcal{L}}_{r}P-dB\sin (2\pi x/\ell )$. Since $B_0=(P_0-P_0^{\ast })/(2i)$, its order-$d$ terms giveThis is a regular perturbation equation for the threshold shift. The right projection uses the complex inner product, not a bilinear product without conjugation.

The linear operator ${\mathcal{L}}_{r_0}$ is self-adjoint on the domain $H^2(0,\ell )\cap H_0^1(0,\ell )$ in complex L2 space. For functions $v,w$ in that domain, two integration by parts steps giveThe boundary term vanishes. In particular $i{r}_0{\partial }_x$ is formally self-adjoint: both complex conjugation of $i$ and integration by parts contribute a minus sign. Taking $v=P_0$, $w=P_1$ and using ${\mathcal{L}}_{r_0}{P}_0=0$ provesThis is the Fredholm solvability condition for a self-adjoint operator. Define $s(x)=\sin (2\pi x/\ell )$ and $N={\int }_0^{\ell }|P_0{|}^{2}dx$. The gauged sine mode hassince the real envelope contributes an integral of its own derivative, whose endpoint term is zero. In particular the denominator in the solvability condition is nonzero. ThereforeFor either parity choice this expression is real. Indeed $|P_0{|}^2$, $A_0^2$ and $B_0^2$ are even about the midpoint, while $s$ is odd. Hence $\int |P_0{|}^{2}s=\int B_0^{2}s=0$, and the last numerator reduces to $\int A_0{B}_{0}s$.

No integral evaluation is needed to compare the two shifts. The phase change $P_Q=i{P}_D$ gives $A_Q=-B_D$, $B_Q=A_D$, and leaves $N$ and $I$ unchanged. Thus $A_Q{B}_Q=-A_D{B}_D$, andEquivalently, the term involving $|P_0{|}^2$ vanishes by parity, while $(P_Q^{\ast }{)}^2=-(P_D^{\ast }{)}^2$ reverses the remaining perturbation numerator. This is parity splitting of a mean-field dynamo threshold: the perturbation breaks the phase degeneracy and lowers one onset threshold by the amount it raises the other.

Let $\eta >0$ denote the uniform magnetic diffusivity. The resistive induction equation for a solenoidal vector field in an incompressible flow isHere the radial compression and axial stretching have zero total divergence: $r^{-1}{\partial }_r(-\omega r^2)+{\partial }_z(2\omega z)=0$. A purely axial field independent of $z$ is automatically solenoidal, and the stretching term is $2\omega B_z\hat{\mathbf{z}}$. Thus the ansatz closes under the equation. For its angular mode, with $m$ a nonnegative integer,This is magnetic concentration by an incompressible stagnation flow.

Put $q=r/g(t)$ and $D_{m}b=b^{\prime \prime }+q^{-1}{b}^{\prime }-m^2{q}^{-2}b$. Substitution and division by $f$ giveA convenient self-similar magnetic mode in a stagnation flow is obtained by making the two dimensionless separation coefficients constant. ChooseThe constant $\lambda$ labels the radial modes. More generally the first right side can be $\beta \eta$ and the radial drift term $\beta q{b}^{\prime }$; the displayed equations choose the positive-width normalization $\beta =1$.

If $g(0)=g_0>0$, the width isFor $\omega =0$ its continuous limiting expression is $g^2=g_0^2+2\eta t$. Since $\eta /g^2=\omega +g^{\prime }/g$, the amplitude isFor the compressive case $\omega >0$, the width tends to the diffusion-strain balance scale $\sqrt{\eta /\omega }$ rather than collapsing to zero.

Let $I_b={\int }_0^{\infty }b(q{)}^{2}qdq$ be finite, and let $C_m={\int }_0^{2\pi }{\cos }^2(m\varphi )d\varphi$, equal to $2\pi$ for $m=0$ and $\pi$ otherwise. The magnetic energy per unit axial length isso the complete time dependence isThe angular integral is only a constant; zero signed magnetic flux does not force $I_b$ or $E$ to vanish.

For $m=0$, smoothness on the axis requires $b(0)$ finite and $b^{\prime }(0)=0$. Set $s=-q^2/2$ and $b=h(s)$. The radial equation becomes the Kummer differential equationIts regular solution iswhere $M$ is the confluent hypergeometric function of the first kind. The second independent solution has a logarithmic singularity at the axis and is inadmissible for a smooth field. For real $\lambda >0$, the regular solution vanishes at infinity: generically it has a $q^{-\lambda }$ tail; at positive even integer $\lambda$ it instead has a Gaussian times a polynomial. Thus decay at infinity alone does not select a unique separation parameter. Generically finite magnetic energy further requires $\lambda >1$.

The simplest finite-flux solution is $\lambda =2$, for which $M(1,1,s)=e^s$ andDirect differentiation gives $b^{\prime \prime }+q^{-1}{b}^{\prime }=(q^2-2)b$ and $q{b}^{\prime }=-q^{2}b$, verifying the radial equation. Its axial magnetic flux is $2\pi b_{0}f{g}^2=2\pi b_0{f}_0{g}_0^2$, while its magnetic energy grows by $g_0^2/g^2$ if the original width is larger than $\sqrt{\eta /\omega }$. For positive $\eta$ this Gaussian amplification saturates, rather than giving indefinite exponential growth. The other rapidly decaying regular axisymmetric modes have $\lambda =2(n+1)$ and profiles $e^{-q^2/2}{L}_n(q^2/2)$, where $L_n$ solves $y{L}_n^{\prime \prime }+(1-y)L_n^{\prime }+n{L}_n=0$.

The energy mechanism is transparent from a separate integral calculation. For any sufficiently decaying field of this axial form, integration by parts in the transverse plane givesThe positive term represents work by the axial stretching, with radial compression accounted for in the area integral. In the ideal limit, $g=g_0{e}^{-\omega t}$ and $f=f_0{e}^{2\omega t}$ for any smooth profile, so $E=E_0{e}^{2\omega t}$ even when $m\ne 0$ and the angular mean is zero. For sufficiently weak initial diffusion, this also gives genuine finite-time energy amplification at positive $\eta$.

Even sustained energy growth is possible in this unbounded model if slowly decaying profiles are admitted. For example, with $\omega >0$, take the constant width $g=\sqrt{\eta /\omega }$, $m=1$ and $1<\lambda <2$. The regular radial solution $b=qM((1+\lambda )/2,2,-q^2/2)$ has a $q^{-\lambda }$ tail and finite $I_b$. It gives $E\propto e^{2(2-\lambda )\omega t}$ and zero signed flux through every centered disk. Its absolute flux integral over the whole plane diverges, so the zero-flux statement here means the limit of these disk integrals. This example is not an isolated bounded-body dynamo: the velocity is unbounded at infinity and the field has an algebraic tail. The dynamo property concerns conversion of fluid energy into magnetic energy, not production of a nonzero signed magnetic flux; growth and its boundary assumptions must be specified.

Let $L$ be a core-scale length, $B_{\ast }$ a representative strength of the prescribed poloidal magnetic field, and $v_A=B_{\ast }/\sqrt{{\mu }_0\rho }$ the Alfvén speed. A consistent weakly damped, small-amplitude ordering isThe first ratio is the Lehnert number. It makes the wave slow relative to rotation and allows almost geostrophic flow, with the azimuthal velocity approximately constant along a rotation-axis cylinder. The small wave amplitude gives $|B_{\varphi }|\sim \sqrt{{\mu }_0\rho }|U|\ll B_{\ast }$. Poloidal advection then changes neither the background field nor the wave substantially over one wave period. If there are no-slip boundary layers, also require their damping rate $\sqrt{\nu \Omega }/L\ll \sigma$, or otherwise retain their torque; a small bulk viscosity alone does not remove that boundary torque.

In the rotating magnetohydrodynamic momentum equation, axisymmetry removes the azimuthal pressure derivative. The neglected nonlinear azimuthal advection has size $|{\mathbf{u}}_p|U/L$, and the neglected viscous term has size $\nu U/L^2$. The retained azimuthal Coriolis force is $2\Omega u_r$. A typical compatible local balance has $|u_r|\sim \sigma U/\Omega$, small compared with $U$ when the Lehnert number is small. The azimuthal Lorentz force density can be written using $\nabla \cdot {\mathbf{B}}_p=0$:The physical prefactor is $1/({\mu }_0\rho )$: the extra $i$ printed inside the first reduced momentum equation is a typographical error, also contradicted by the subsequent real-valued system.

In the resistive induction equation, neglecting poloidal advection of the perturbation and magnetic diffusion leaves the winding of the prescribed poloidal magnetic field by differential rotation. ThusThis is the small-amplitude torsional Alfvén wave approximation, with a time-independent background and no external azimuthal forcing.

We now average over the full height of a rotation-axis cylinder. The sphere is impenetrable. Integrating incompressible flow over the volume inside any such cylinder gives zero flux through its cylindrical side, since there is no flux through its spherical caps. ThereforeThis is why the Coriolis force disappears from the cylinder average; it need not vanish locally. An axisymmetric field in the insulating exterior is a potential field with $B_{\varphi }=0$, so continuity gives $B_{\varphi }=0$ on the sphere. Consequently both the end flux terms and the moving-endpoint terms in the magnetic-stress integral vanish. Expanding the divergence givesSince $U$ is independent of $z$, its integrated time derivative is $2z_b{U}_t$. Defining $T={\int }_{-z_b}^{z_b}{B}_r{B}_{\varphi }dz$ givesThe stationary background and the fact that $U/r$ depends only on $r,t$ also givesoTogether these form the cylinder-averaged torsional Alfvén wave system.

Assume $H>0$ on the cylinders carrying the wave, with regular endpoint limits. If $H$ vanishes on an interior cylinder, then $B_r$ and $T$ vanish there and that cylinder has no radial magnetic coupling; the displayed $H^{-1}$ energy term must be interpreted on the coupled region rather than literally dividing by zero. Since $H$ is time independent, differentiation of the proposed conserved energy of a cylinder-averaged torsional wave givesRegularity at the rotation axis makes the lower endpoint vanish, while vanishing magnetic torque at the collapsed outer cylinder makes the upper endpoint vanish. This provesThe magnetic term has a direct interpretation. Decompose $B_{\varphi }=(T/H)B_r+B_{\varphi }^{\perp }$, with $\int B_r{B}_{\varphi }^{\perp }dz=0$. Then $\int B_{\varphi }^{2}dz=T^2/H+\int (B_{\varphi }^{\perp }{)}^{2}dz$, and the induction equation gives $(B_{\varphi }^{\perp }{)}_t=0$. Thus, apart from a common dimensional factor and a stationary orthogonal magnetic component, the conserved quantity is the physical kinetic plus magnetic energy.

For the frequency estimate, put $V=U/r$ and eliminate $T$:Locally this is a wave equation with speedIt is the cylinder-averaged radial field component that matters, not simply the dipole measured at the Earth's surface. Taking a cylinder height of order $6\times 10^6\mathrm{m}$, $\rho \sim 10^4\mathrm{kg}{\mathrm{m}}^{-3}$, and an assumed internal cylindrical radial field $B_{r,\mathrm{rms}}\sim 0.2$--$2\mathrm{mT}$ givesFor a standing wave of radial length $L_c\sim 2$--$3.5\times 10^6\mathrm{m}$, the cyclic frequency is of order $c_T/(2L_c)$ and the angular frequency is of order $\pi c_T/L_c$. HenceGeometry and the assumed internal field can shift the estimate substantially. As a concrete calibration, the 2010 torsional-wave study inferred a cylindrical radial rms field near $2\mathrm{mT}$ and a roughly four-year crossing time, with an observed six-year recurrence. A much weaker assumed interior field instead gives the older multi-decadal estimate. The paper supplies no numerical interior field strength, so a unique numerical frequency cannot be deduced without such an assumption.

An anti-dynamo theorem identifies a restriction that prevents a conducting fluid from regenerating its magnetic field against resistive losses. The hypotheses matter: throughout the following arguments take positive uniform magnetic diffusivity, smooth fields, no imposed magnetic source, and the stated isolated or energy-closed boundary conditions. An anti-dynamo theorem does not rule out transient magnetic amplification, winding of an imposed field, or an externally supplied magnetic flux. The resistive induction equation is the common starting point:

First consider Backus' necessary condition for dynamo action. For a conductor contained in a sphere of radius $a$, include the potential field in the insulating exterior when defining the total magnetic energy:For an incompressible flow, with boundary conditions eliminating the mechanical surface term, such as a no-slip boundary condition for the fluid, integration by parts giveswhere $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ is the rate-of-strain tensor. Its antisymmetric part does no stretching, since $B_i{B}_j$ is symmetric. If $S$ bounds the largest eigenvalue of $e$ throughout the flow, the first term is at most $2S{E}_B$.

The magnetic free-decay spectral bound for the enclosing insulating sphere isThe exterior energy is essential to this variational estimate. Its constant comes from the slowest freely decaying dipolar poloidal mode: the spherical radial matching condition reduces to $j_0(\kappa a)=0$, whose first positive root is $\kappa a=\pi$. Expanding an admissible field in the orthogonal free-decay modes, or using their variational characterization, gives the bound. A smaller conductor can only increase the minimum dissipation quotient for the enclosing-sphere comparison. Combining the two estimates yieldsThusBelow this threshold the Gronwall inequality gives exponential energy decay. Above it the estimate merely permits growth; it does not prove that the geometry regenerates the field. A time-dependent version bounds the field-amplitude exponent by the time average of the maximum strain minus the diffusive rate.

The geometric obstruction in Cowling anti-dynamo theorem is different: an isolated exactly axisymmetric magnetic field cannot be self-maintained. To see the missing regeneration mechanism, write its poloidal magnetic field in terms of the poloidal magnetic flux function $\psi$:The poloidal part of the resistive induction equation becomesThere is no term containing $B_{\varphi }$ that creates new poloidal flux. Fix the gauge by setting $\psi =0$ on the regular symmetry axis, and match to a decaying exterior field with no imposed flux. At a positive interior maximum away from the axis, the advective term vanishes and the diffusive term is nonpositive. The parabolic comparison principle therefore prevents increase of the maximum; it likewise prevents decrease of a negative minimum. In the steady problem the strong maximum principle, regularity at the axis and decay outside exclude a nonzero maximum altogether. Matching to the insulating exterior does not create a maximum at the interface: the exterior flux is harmonic in the corresponding axisymmetric operator, and the two normal-derivative conditions exclude such an interface extremum. The same homogeneous scalar diffusion problem gives decay in time on a bounded conductor with no replenishing boundary data.

Once the poloidal field has decayed, let $b=B_{\varphi }/r$ and let ${\Omega }_f=u_{\varphi }/r$ denote the fluid's angular velocity. Its equation isDifferential rotation can make toroidal field from poloidal field, but it cannot close the reverse loop. When ${\mathbf{B}}_p=0$, this too is a homogeneous advection-diffusion equation, with a regular axis and zero toroidal field at an insulating exterior. Hence it cannot sustain the remaining field. The exactly axisymmetric field, rather than merely an axisymmetric velocity, is excluded. If an exactly axisymmetric field were maintained by a nonaxisymmetric velocity, averaging the linear induction equation in azimuth would replace that velocity by its azimuthal mean and give the same contradiction. An axisymmetric flow can, however, excite a nonaxisymmetric field. Also an axisymmetric averaged field in mean-field dynamo theory has an additional fluctuating electromotive force, so this unmodified scalar proof does not apply to it.

The toroidal-velocity anti-dynamo theorem restricts the flow instead. In a conducting sphere, suppose $\mathbf{u}\cdot \mathbf{x}=0$, so the incompressible flow is tangent to every concentric sphere. No assumption of axisymmetry is made. Define $P=\mathbf{x}\cdot \mathbf{B}$. A direct Cartesian calculation is particularly useful:The last equality uses $\nabla \cdot \mathbf{B}=0$, so ${\nabla }^2(\mathbf{x}\cdot \mathbf{B})=\mathbf{x}\cdot {\nabla }^2\mathbf{B}$. Thus the radial magnetic scalar is a passive diffusing scalar, not a regenerated field. Multiplying by $P$, integrating over the sphere and using tangential incompressible advection givesIn an insulating exterior, a spherical harmonic component of degree $l\ge 1$ has $P\propto r^{-l-1}$, so its boundary contribution is negative: ${\partial }_{n}P=-(l+1)P/a$. The monopole component is absent because the magnetic field is solenoidal and regular. The resulting dissipative scalar problem makes $P$, and hence the poloidal field, decay.

When that radial field is absent, the remaining tangential solenoidal field can be written $\mathbf{B}=\nabla \times (T\mathbf{x})$. Its induction equation reduces to a second passive scalar equation. Indeedand the curl of the last gradient term is zero. Therefore $T_t+\mathbf{u}\cdot \nabla T=\eta {\nabla }^{2}T$, up to an irrelevant radial gauge function, which can be removed by taking the angular mean of $T$ to be zero. Its angular variations vanish at the insulating spherical boundary, so this scalar also decays by its quadratic energy identity. Equivalently, a putative growing magnetic normal mode must first have $P=0$, after which the scalar $T$ equation rules it out. Purely toroidal spherical flow cannot close a dynamo regeneration loop, even if it is nonaxisymmetric. This does not assert that purely poloidal flow is always incapable of a dynamo, or that any azimuthally directed cylindrical flow is governed by the same spherical theorem.

Other restrictions illuminate the same distinction between stretching and regeneration. The planar anti-dynamo theorem has a similar scalar proof in the usual two-dimensional geometry. For planar velocity and magnetic field independent of the third coordinate, write the in-plane field as $\nabla \times (A\hat{\mathbf{z}})$. Its vector potential obeys $A_t+\mathbf{u}\cdot \nabla A=\eta \Delta A$ after fixing the gauge, so $\frac{1}{2}\frac{d}{dt}\int A^2=-\eta \int |\nabla A{|}^2$ under periodic or homogeneous energy-closed boundary conditions. A third field component independent of that coordinate is also a passive scalar. A spatially uniform imposed field may survive in a periodic box but is not regenerated dynamo field. Three-component motions depending on only two coordinates are not covered by this strictly planar formulation.

There are also necessary bounds in terms of velocity amplitude rather than strain. The magnetic work term is bounded by $\Vert \mathbf{u}{\Vert }_{\infty }\Vert \mathbf{B}{\Vert }_2\Vert \nabla \times \mathbf{B}{\Vert }_2$; combining this with the same free-decay inequality gives a velocity-based lower threshold on the magnetic Reynolds number. These bounds, like Backus' necessary condition for dynamo action, are necessary rather than sufficient. Pure motion that only rearranges a scalar magnetic potential cannot evade diffusion merely by becoming faster. Conversely the stagnation-flow energy amplification in Question 2 does not contradict the bounded-body theorems: it uses an unbounded three-dimensional strain field and, for its growing finite-energy modes, a slowly decaying field at infinity. Similarly, the torsional Alfvén waves in Question 3 transfer energy between velocity and a toroidal perturbation about a prescribed background; they are not a construction of that background by dynamo action. The anti-dynamo results identify failed regeneration mechanisms and the assumptions under which those failures are rigorous.