Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 60 (Magnetohydrodynamics)

In SI units, magnetostatic equilibrium without gravity is $\nabla p={\mu }_0^{-1}(\nabla \times \mathbf{B})\times \mathbf{B}$. For a purely axial magnetic field, the magnetic tension term $(\mathbf{B}\cdot \nabla )\mathbf{B}$ vanishes. Thus the radial balance isMatching the magnetohydrodynamic total pressure to the field-free exterior gives the gas pressure and axial field bound:The inequality follows directly from nonnegative gas pressure. It is the magnetic pressure limit for an untwisted flux tube, without inward hoop tension.

A toroidal magnetic field introduces an inward hoop force. In cylindrical coordinates, the radial magnetic tension is $-B_{\varphi }^2/({\mu }_{0}s)$, so cylindrical magnetostatic pressure balance in SI units becomesAt the field-free boundary $P(s_0)=p_e$. HenceRegularity on the axis gives $B_{\varphi }=O(s)$, which makes the integral finite and gives $B_{\varphi }(0)=0$. Evaluating at the axis proves the central-field identity:The positive hoop-tension contribution allows the central axial field to exceed $B_p$ without negative pressure. For an explicit regular example, put $x=s^2/s_0^2$, choose a constant $a>0$ and $0<\epsilon <1$, and take inside the flux tubeSubstitution verifies the radial balance, with continuous zero field at $s=s_0$ and positive gas pressure. For $a=3$, $\epsilon =1/4$, the central value is $B_z(0{)}^2=3B_p^2/2$.

To obtain the cross-sectional average, multiply the radial balance by $s^2$ and integrate by parts:The toroidal contributions cancel. The resulting flux-tube axial field virial identity isThis is at most $B_p^2$, and is strictly smaller whenever the weighted pressure integral is positive. In particular, continuous matching with $p_e>0$ guarantees strictness. If only $p\ge 0$ is assumed without a nondegenerate gas or continuous positive boundary pressure, the justified conclusion is the non-strict bound; the identity specifies exactly when equality occurs.

Use the central-field identity and substitute the prescribed relation between the toroidal magnetic field and axial magnetic field:The flux-tube axial field virial identity bounds the last average by $B_p^2$, while $p(0)\ge 0$. Therefore the uniform-twist central axial field bound isThe stronger exact expression is $B_z(0{)}^2=(1+{\gamma }^2)B_p^2-2{\mu }_{0}p(0)-(4{\mu }_0{\gamma }^2/s_0^2){\int }_0^{s_0}sp(s)ds$.

The azimuthal derivative of a cylindrical vector Fourier mode must include basis rotation: the azimuthal dependence refers to the components in the rotating cylindrical basis. Since ${\partial }_{\varphi }\hat{\mathbf{s}}=\hat{\mathbf{\varphi }}$ and ${\partial }_{\varphi }\hat{\mathbf{\varphi }}=-\hat{\mathbf{s}}$, a vector amplitude $\mathbf{v}\propto e^{im\varphi }$ obeysAlso $\mathbf{B}=(J/2)(-y,x,0)$ in Cartesian coordinates, so $(\mathbf{v}\cdot \nabla )\mathbf{B}=(J/2)\hat{\mathbf{z}}\times \mathbf{v}$. Consequentlyand the basis-rotation terms cancel in the ideal magnetohydrodynamic induction equation:Here the printed $J$ is the coefficient in the specified magnetic field: $\nabla \times \mathbf{B}=J\hat{\mathbf{z}}$. If $J_{\mathrm{phys}}$ denotes SI electric current density, then $J={\mu }_0{J}_{\mathrm{phys}}$. The field and all subsequent printed coefficients are mutually consistent with this normalization.

For a normal mode with $\sigma \ne 0$, induction gives $\mathbf{b}=imJ\mathbf{u}/(2\sigma )$. Substitution into the momentum equation, including the Coriolis force, gives the reduced velocity equationThus the coefficients are the printed $\lambda$ and $\nu$. Let $C=J^2/({\mu }_0\rho )$. Inserting the allowed relation $\lambda =i\delta \nu$ and multiplying by $\sigma$ yieldsHence the uniform-current rotating magnetohydrodynamic wave dispersion isThe radicand is real. For $|m|\ge 2$, $2m\delta -m^2<0$ because $|\delta |<1$, so both roots are purely imaginary. For $m=0$ the roots of the quadratic are also imaginary; any zero-frequency case must be checked in the original equations, since the elimination divided by $\sigma$. A positive real part requiresFor integer $m$ this is possible only when $|m|=1$ and $m\delta >1/2$. Taking the conventional representative $m\ge 0$ gives the printed exception $m=1$, with $\delta >1/2$ and sufficiently large $J$. Literally allowing negative integers also gives $m=-1$, $\delta <-1/2$; complex conjugation maps $(m,\delta ,\sigma )$ to $(-m,-\delta ,\overline{\sigma })$, so these describe the conjugate real disturbance. Purely imaginary roots describe neutral oscillatory modes; a repeated root at threshold does not itself prove boundedness of arbitrary initial perturbations.

For constant alpha tensor and positive magnetic diffusivity $\eta$, the steady anisotropic alpha-squared dynamo equation isFor a nonzero Fourier mode with wavevector $\mathbf{K}=(k,l,m)$, this becomes $\eta K^2\mathbf{B}=i\mathbf{K}\times (\mathbf{\alpha }\mathbf{B})$. Rotational symmetry of the alpha tensor in the horizontal plane lets us set $\mathbf{K}=(q,0,m)$, with $q^2=k^2+l^2$. The component equations areSubstitute the first and third into the second. A nonzero steady amplitude requires ${\eta }^2{K}^4={\alpha }_0^2{m}^2+{\alpha }_0{\alpha }_1{q}^2$. Conversely, this relation supplies a nonzero amplitude through the same component equations, and the solenoidal magnetic-field constraint $q{B}_x+m{B}_z=0$ is automatically satisfied. Thus the steady-mode condition isFor the first form, no division by $m^2+\delta q^2$ is needed. The quotient applies when that denominator is nonzero, in particular for $q>0$ and $0<\delta <1$.

Put $Q=q^2>0$ and $r=m^2\ge 0$. The derivative of $F(r)=(Q+r{)}^2/(r+\delta Q)$ isFor $0<\delta <1/2$, the derivative changes from negative to positive at $r=(1-2\delta )Q$. For $1/2\le \delta <1$ it is nonnegative throughout the allowed half-line and the minimum is at $r=0$. Therefore the optimized uniaxial alpha dynamo threshold isBoth expressions agree at $\delta =1/2$. If the horizontal boundary conditions permit $q=0$, a nonzero mode instead has ${\gamma }^2=m^2$ and the infimum is zero as $m\to 0$; it is not attained by a nonzero wavevector. The exactly uniform mode $q=m=0$ has no diffusive or alpha curl term and must be treated separately.

Let $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ be the rate-of-strain tensor of an incompressible flow, and let $s(t)$ be the supremum over the conductor of its largest eigenvalue. State Backus' necessary condition for dynamo action with its magnetic boundary conditions: an isolated bounded conductor of uniform positive magnetic diffusivity $\eta$, surrounded by an electrical insulator with a decaying potential exterior field, and no imposed magnetic field or boundary energy input. For definiteness take a no-slip boundary condition on the fluid, which eliminates the stretching surface term. If the conductor lies within a sphere of radius $R$ and $S={\sup }_{t}s(t)$, a necessary condition for a nondecaying dynamo isThe constant is the free-decay spectral bound for an insulating exterior, not a universal constant for every magnetic boundary condition. The condition is necessary, not sufficient, and involves maximum stretching rather than an rms velocity.

To see both the condition and the growth-rate bound, include exterior magnetic energy:This follows from the resistive induction equation and integration by parts, with the stated boundary assumptions. The magnetic free-decay spectral bound is ${\int }_D|\nabla \times \mathbf{B}{|}^{2}dV\ge ({\pi }^2/R^2){\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$. For a sphere its lowest mode is the dipolar poloidal free-decay mode; enclosing a smaller conductor gives the same valid lower bound. Since $B_i{e}_{ij}{B}_j\le s(t)|\mathbf{B}{|}^2$, we obtainIntegrating this differential inequality gives decay whenever $S<\eta {\pi }^2/R^2$. More generally the exponential rate of the field norm, rather than of its squared energy, satisfiesSimply discarding the nonnegative resistive dissipation already proves the requested maximum-strain bound. The energy exponent is twice the field-amplitude exponent.

For the alpha-Omega dynamo model, write $a=|{\alpha }_0|$, $w=|\Omega |$, $k\ge 0$ and $d=\eta k^2$. Direct differentiation givesFor $a,kw>0$, use the weighted energy estimate for two coupled modes and form the positive weighted norm $N=kw|A{|}^2+a|B{|}^2$. The inequality $N\ge 2\sqrt{akw}|A||B|$ givesFor each fixed $k$ and model parameters, this norm is equivalent to the amplitude norm; its square-root exponential rate is therefore bounded by $h(k)=\sqrt{awk}-\eta k^2$, uniformly over all admissible $f(t)$. Maximizing over $k$ givesThe exponent is also achievable in order of magnitude. Choose the admissible constant $f=1$. The growing eigenvalue of the two-component system has real part $-\eta k^2+\sqrt{awk/2}$. Its maximum occurs at $k=(aw/(32{\eta }^2){)}^{1/3}$ and equals $32^{-10/3}(a^2{w}^2/\eta {)}^{1/3}$. Thus the bounded-modulation alpha-Omega growth estimate has the scalingThis means the maximum over allowed modulations and wavenumbers, not that every bounded modulation grows; $f=0$ supplies no regenerating alpha coupling.

The Omega effect rapidly makes toroidal field from poloidal field, but exponential dynamo action also requires the slower alpha effect to regenerate the poloidal component. The coupled amplification rate is of order $\sqrt{a\Omega k}$ rather than $\Omega$; shortening the wavelength to increase it also increases magnetic diffusion as $\eta k^2$. Their optimal balance gives $k\sim {\Omega }^{1/3}$ and growth $\sim {\Omega }^{2/3}$. The Backus' necessary condition for dynamo action estimate controls stretching alone and does not incorporate this regeneration bottleneck. A shear without regeneration can give transient amplification but not this sustained exponential feedback.