Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 75 (Magnetohydrodynamics and Turbulence)

Use Gaussian units, uniform number density $n$, and ion mass density $\rho =n{m}_i$. Here $e>0$ is the magnitude of the elementary charge, as fixed by the printed current relation. Neglect electron inertia and displacement current. ThenFreezing the magnetic flux into this electron velocity, while retaining the ion momentum equation, gives the closed Hall magnetohydrodynamics systemThe pressure enforces the incompressible flow constraint; taking the divergence of momentum determines it up to the usual boundary-dependent constant. The induction equation preserves the solenoidal magnetic-field constraint. One can absorb $B^2/(8\pi )$ into magnetohydrodynamic total pressure, replacing the Lorentz term by magnetic tension. Keeping ion inertia here is important: electron magnetohydrodynamics with fixed ions will be a later limit, not the complete two-field system.

Linearize about zero bulk flow and the uniform guide field. Put $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$, and take plane waves proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$. Let $k_{\parallel }=\mathbf{k}\cdot \hat{\mathbf{z}}$, $k=|\mathbf{k}|$, $q=k_{\parallel }{v}_A$, andBoth $\mathbf{u}$ and $\mathbf{b}$ are perpendicular to $\mathbf{k}$. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term givesIn the induction equation, the Hall contribution is $-c(B_0{\partial }_z)(\nabla \times \delta \mathbf{B})/(4\pi en)$; its two derivative factors of $i$ in the Fourier mode produce a positive coefficient multiplying $\mathbf{k}\times \mathbf{b}$. ThusEliminating $\mathbf{u}$ givesOn the transverse plane, $(i\mathbf{k}\times {)}^2=k^2$, so applying that operator again proves the incompressible Hall-MHD wave dispersion:Equivalently, in a helicity decomposition of a transverse Fourier mode with $i\mathbf{k}\times {\mathbf{b}}_h=hk{\mathbf{b}}_h$, $h=\pm 1$, each polarization obeys ${\omega }^2-hqk{d}_i\omega -q^2=0$. For $k_{\parallel }=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_{\parallel }\ne 0$.

Solving the polarization quadratic gives all four signed roots:A sign-independent description is the pair of positive frequency magnitudesFor wavelengths long compared with the ion skin depth, $k{d}_i\ll 1$, both magnitudes approach $|k_{\parallel }|v_A$. The Hall correction is then small and the usual Alfvén wave dispersion ${\omega }^2=k_{\parallel }^2{v}_A^2$ is recovered. In an incompressible system the two transverse polarizations include the shear and pseudo-Alfvén wave sectors; away from this limit Hall coupling selects circular helicity combinations. A small parallel wavenumber alone does not remove the relative Hall splitting: the expansion parameter is the full $k{d}_i$.

When $k{d}_i\gg 1$, expand the square root in the exact frequencies. The fast branch haswhile the slow branch has ${\Omega }_{\mathrm{slow}}=|q|[(k{d}_i{)}^{-1}+O((k{d}_i{)}^{-3})]$. HenceThe fast dispersive wave is the mode sought in the question's subsequent fast-time limit. In a strongly oblique wave, $k\simeq k_{\perp }$, so its frequency scales as $|\omega |\simeq |k_{\parallel }|v_A{k}_{\perp }{d}_i$. The slow branch instead tends to $(|k_{\parallel }|/k){\Omega }_i$, where the ion cyclotron frequency is ${\Omega }_i=v_A/d_i=e{B}_0/(m_{i}c)$.

There is a terminology qualification: in the pressure-free incompressible Hall equations used here, the fast branch is conventionally a whistler wave. The question calls the requested small-scale waves kinetic Alfvén waves, but a general kinetic-Alfvén dispersion includes pressure/kinetic ordering absent from these equations. The boxed dispersion is the result for the printed model; no extra pressure response or kinetic term has been silently added.

Write $ϵ=l/d_i\ll 1$ and $\delta =v_A\tau /l\ll 1$. For magnetic amplitudes of order $B_0$, ion acceleration is of order $v_A^2/l$. Thus its change over the fast time isThe ions are consequently fixed to leading order on this timescale. Their first nonzero response is driven by the momentum equation, with negligible self-advection for an initially resting ion flow; one need not assert that the dimensional Lorentz acceleration literally vanishes. In fast-time, Alfvén-velocity dimensionless variables, the ion equation reads ${\partial }_{\hat{t}}\hat{\mathbf{u}}=O(\delta )$, giving ${\partial }_{\hat{t}}\hat{\mathbf{u}}=0$ at leading order.

Meanwhile the electron-current velocity is $j/(en)\sim v_A{d}_i/l=v_A/ϵ$, larger than a bulk flow of order $v_A$. The ion-advection term in induction is smaller than Hall induction by $O(ϵ)$ even if that slow flow is retained. The leading magnetic equation is thereforewith stationary ions at leading order. This is the fast-time electron-MHD limit of Hall magnetohydrodynamics and is precisely electron magnetohydrodynamics.

The already obtained fast-wave eigenvector has $u/b=|q/\omega |\sim 1/(k{d}_i)\ll 1$, confirming that ion motion cannot affect its leading frequency. The closed magnetic equation therefore suffices to obtain that branch without using momentum to close it. Its timescale is ${\tau }_w\sim 1/(v_A{d}_i|k_{\parallel }|k)$; compatibility with ${\tau }_w\ll l/v_A$ requires $|k_{\parallel }|d_i\gg 1$ when $l\sim k^{-1}$. Arbitrarily small $k_{\parallel }$ can violate this fast-time assumption even though $k{d}_i\gg 1$. The slow branch does not obey the stated ordering and is not retained by the fixed-ion reduction.

For the theta pinch, the electric current is azimuthal:Its Lorentz force is ${\mathbf{j}}_0\times {\mathbf{B}}_0/c=-B_0{B}_0^{\prime }\hat{\mathbf{r}}/(4\pi )$. Static radial balance with the pressure gradient therefore givesThe constant is the equilibrium magnetohydrodynamic total pressure.

Suppress the common Fourier mode factor and define the transverse and full divergences of the fluid Lagrangian displacement byThe perturbed magnetic field is $\mathbf{Q}=({\mathbf{B}}_0\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla ){\mathbf{B}}_0-{\mathbf{B}}_{0}D$. Its components areThus the magnetic part of the energy integrand isUse $p_0^{\prime }=-B_0{B}_0^{\prime }/(4\pi )$ from force balance. The pressure-gradient cross term isThe current term has the azimuthal cross-product component ${\xi }_z^{\ast }{Q}_r-{\xi }_r^{\ast }{Q}_z$, givingEvery term involving $B_0^{\prime }$ cancels in the sum, including the possibly complex ${\xi }_z^{\ast }{\xi }_r$ terms. The magnetohydrodynamic energy principle therefore reduces exactly toFor physical $p_0\ge 0$ and positive adiabatic index $\gamma$, every surviving term is nonnegative. On the admissible displacement domain for the supplied functional, squared ideal-mode frequencies cannot be negative. This is the positive ideal energy of a theta pinch, proving absence of exponentially growing ideal modes. The qualification is stable or marginal, not strictly positive energy for every motion: at $k=0$, suitable incompressible displacements can have zero energy. As usual, the energy principle presumes regularity and boundary conditions eliminating unwanted surface terms; additional external systems are not part of the given functional.

Interpret the multiplicative Gaussian white noise as the limit of a smooth short-correlation-time velocity field, hence in the Stratonovich integral convention. Put ${\partial }_i=\partial /\partial B_i$. For a realization, differentiating the field-space Dirac delta gives its field-space continuity equationThe velocity-gradient covariance has zero trace variance: contracting $T_{mn}^{ij}$ with $m=i,n=j$ gives $3-(9+3)/4=0$. Thus the gradient is trace free, consistently with the incompressible ideal induction equation.

A noise impulse in component ${\sigma }_{jn}$ changes ${\tilde{B}}_j$ by ${\tilde{B}}_n$ times that impulse. Its instantaneous retarded action on the Dirac delta is thereforeOnly earlier times affect the present field. In the white-noise limit the integral over this causal interval has ${\int }_0^t\delta (t-t^{\prime })d{t}^{\prime }=1/2$, corresponding to the symmetric short-time regularization. Applying the Furutsu–Novikov formula and then averaging givesSubstitute into the averaged field-space continuity equation:The differentiated $B_n$ term vanishes because ${\sum }_j{T}_{mj}^{ij}=0$, and the remaining contraction isConsequently the closed vector Fokker-Planck equation for Gaussian white-noise magnetic stretching isThis is a positive diffusion operator: the displayed matrix has eigenvalues $B^2$ on the two transverse directions and $B^2/2$ along $\mathbf{B}$. Replacing the endpoint half-weight by one would double the diffusion and fail to reproduce the later rate.

For isotropic initial statistics, isotropic noise preserves $P=P(B,t)$. Isotropy of the noise alone does not turn an arbitrary anisotropic initial vector distribution into an isotropic one; alternatively take the angularly averaged magnitude distribution. The radial gradient satisfies ${\partial }_{j}P=P_B{B}_j/B$, so the diffusion-vector contraction in the preceding equation is $B{B}_i{P}_B/2$. Its divergence givesNow $F=4\pi B^{2}P$ is the magnitude probability density function, normalized by ${\int }_0^{\infty }FdB=1$. Sincewe obtain the radial magnetic-field Fokker-Planck equationThe outward Fokker-Planck probability current is $J=-D(B^2{F}_B-2BF)$. Zero probability flux at zero and infinity preserves normalization. For positive initial magnitude the solution below remains on $B>0$; an atom initially at $B=0$ instead remains an absorbing zero-field component.

In Gaussian units, the ensemble-averaged magnetic energy density isLet $M_q={\int }_0^{\infty }{B}^{q}FdB$, and assume the relevant moment exists and endpoint terms vanish. Twice applying integration by parts to the conservative magnitude equation givesIn particular, the magnetic stretching moment growth for $q=2$ isThis is a kinematic magnetic dynamo result with prescribed velocity statistics and no Ohmic loss. Once the Lorentz force appreciably changes the flow, this indefinitely exponential kinematic approximation is no longer self-consistent.

For $B_0>0$, change to logarithmic magnitude and its correctly transformed probability density function:Using ${\partial }_B=B^{-1}{\partial }_x$, the radial equation becomesThis is diffusion with constant drift. Set $y=x-3Dt$ and $g(y,t)=f(y+3Dt,t)$; then $g_t=D{g}_{yy}$. The initial Dirac delta transforms to $f(x,0)=\delta (x)$, so the heat-kernel solution givesDividing by $B$ yields the lognormal magnetic-field amplification lawExpanding the square and using $B=B_0{e}^x$ gives the equivalent requested formThe logarithm is a Gaussian random variable with mean $3Dt$ and variance $2Dt$, which proves normalization and spreading directly. The exponential moments of that Gaussian distribution give $⟨B^q⟩=B_0^q{e}^{Dq(q+3)t}$, independently checking the energy growth. The limit at $t\downarrow 0$ is the initial Dirac delta distribution, not a pointwise finite function.

Use incompressible ideal MHD at scales above the ion skin depth, a strong uniform mean field, and a statistically steady local energy cascade with mean flux $ϵ$ per unit mass. Assume balanced counterpropagating wave populations, comparable velocity and magnetic amplitudes in velocity units, and no additional scale-dependent alignment or intermittency correction. Write $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$ and introduce the fluctuation Elsässer variables ${\mathbf{z}}^{\pm }=\mathbf{u}\pm \mathbf{b}$. Their equations areOnly opposite populations distort one another. A single population has no leading nonlinear self-interaction, so the balance assumption is essential to a one-amplitude cascade estimate.

Let $\delta z_l$ denote the typical balanced amplitude and let $l_{\perp },l_{\parallel }$ be coherence lengths relative to the local mean field. The Alfvén wave time and eddy turnover time areFor $\chi \ll 1$, one counterpropagating encounter changes the packet fractionally by order $\chi$. Assume phases decorrelate between encounters. After $N$ encounters, random accumulation is of order $\sqrt{N}\chi$; an order-one change requires $N\sim {\chi }^{-2}$. The weak wave cascade time is thereforeWeakness by itself does not prove this random walk estimate; phase decorrelation and the interaction geometry are additional assumptions.

For the Iroshnikov-Kraichnan spectrum, impose isotropy, $l_{\parallel }\sim l_{\perp }\sim l$, in addition to $\delta z_l\ll v_A$. Then ${\tau }_A=l/v_A\ll {\tau }_{\mathrm{nl}}=l/\delta z_l$ andHence $\delta z_l\sim (ϵv_{A}l{)}^{1/4}$. A local shell-integrated turbulent energy spectrum has $\delta z_l^2\sim kE(k)$, $k\sim l^{-1}$, soThe isotropy is imposed phenomenologically; a strong guide field actually distinguishes directions and does not force this isotropic picture. The spectrum is not an automatically valid asymptotic result for every strongly magnetized cascade.

For the anisotropic weak Alfvénic cascade, keep $l_{\parallel }$ approximately fixed while cascading to smaller $l_{\perp }$, and require $\chi \ll 1$. This geometry follows from resonant three-wave Alfvén interactions: frequency conservation together with parallel-wavenumber conservation makes one member of an oppositely propagating resonant triad have zero parallel wavenumber, so the propagating modes retain their parallel wavenumber in the leading resonant transfer. Thus there is no leading parallel cascade in this ideal weak-wave ordering. Using the fixed parallel scale,Therefore $\delta z_l\sim (ϵv_A/l_{\parallel }{)}^{1/4}{l}_{\perp }^{1/2}$. For the one-dimensional perpendicular spectrum, $\delta z_l^2\sim k_{\perp }{E}_{\perp }(k_{\perp })$ givesUnlike the isotropic estimate, the parallel length does not decrease with the perpendicular length. Since $\chi =\delta z_l{l}_{\parallel }/(v_A{l}_{\perp })\propto l_{\perp }^{-1/2}$, the interactions become stronger down the cascade. If ${\chi }_0\ll 1$ at outer perpendicular scale $L_{\perp }$, the weak approximation breaks at $l_{\perp ,\ast }\sim L_{\perp }{\chi }_0^2$. This is the weak-to-strong transition of an Alfvénic cascade.

For Goldreich–Sridhar turbulence, abandon the weak-collision estimate and impose critical balance, ${\tau }_A\sim {\tau }_{\mathrm{nl}}$. The local cascade is strong, with transfer time of order one eddy turnover time. ThenThe parallel scale is set dynamically, not held fixed:At smaller scales $l_{\parallel }/l_{\perp }$ grows, so the eddies are increasingly elongated along the local magnetic field. Strong turbulence does not mean $\delta z_l\sim v_A$: small transverse amplitudes can still be strong when the eddies are sufficiently anisotropic to satisfy critical balance. Imbalance, alignment or nonlocal interactions would require different cascade reasoning.

Finally, in the usual isotropic hydrodynamic Kolmogorov 1941 theory inertial-range argument, the only relevant dimensional quantities are $ϵ$ and $k$. With $[ϵ]=L^2{T}^{-3}$ and $[E]=L^3{T}^{-2}$, writing $E\propto {ϵ}^a{k}^b$ forces $a=2/3$, $b=-5/3$. This uses the physical assumptions that outer-scale, viscous and intermittency effects do not supply another leading dependence; dimensions alone are not a proof of those assumptions.

A magnetic guide field supplies the independent speed $v_A$, and anisotropy adds independent parallel lengths. Even before adding anisotropy, dimensions permitwith arbitrary dimensionless $\Phi$. For example $\Phi (x)\propto x^{1/2}$ reproduces the isotropic Iroshnikov-Kraichnan spectrum estimate. This dimensional freedom of an MHD spectrum explains why propagation, decorrelation, anisotropy and cascade-time assumptions must select among the spectra: purely dimensional considerations do not select a unique MHD exponent.