Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 75 (Magnetohydrodynamics and Turbulence)

Let ${\rho }_i={\rho }_n=\rho$ and $\mu ={\mu }_{in}={\mu }_{ni}$. Use normal modes proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, so $\mathrm{Im}\omega <0$ means damping. The linear displacements satisfyFor a nonzero-frequency perturbation, linearizing the ideal magnetohydrodynamic induction equation givesProject both momentum equations with $\mathsf{P}=I-\mathbf{k}{\mathbf{k}}^T/k^2$. This removes the pressure gradients. The magnetic tension is already transverse to $\mathbf{k}$, and, with $a=k_{\parallel }{v}_A$, the projected equations areHere $v_A=B_0/\sqrt{4\pi \rho }$ uses the ion density, as stipulated. For either of the two independent transverse polarizations, the equal-density ion-neutral Alfvén dispersion relation follows fromIts determinant is $\omega [{\omega }^3+2i\mu {\omega }^2-a^2\omega -i\mu a^2]$. Removing the static neutral-displacement factor givesThe removed factor is a time-independent neutral displacement with no velocity or magnetic perturbation, a relabeling of the homogeneous neutral fluid rather than a fourth physical wave. Direct elimination in the first-order system for $(u_i,u_n,\delta B)$ gives exactly the same cubic. This also handles the zero-frequency limits without introducing a displacement-label mode.

For the strong-collision ion-neutral Alfvén modes, put $a=k_{\parallel }{v}_A$ and assume $0<|a|/\mu \ll 1$. Rewrite the cubic asThe two slow roots obey $\omega =O(a)$, so the first approximation is ${\omega }_0^2=a^2/2$. Set $\omega ={\omega }_0+\delta \omega$ and retain the next order, of size $a^3$:ConsequentlyThese are weakly damped Alfvén waves. Their phase speed is $v_A/\sqrt{2}=B_0/\sqrt{4\pi ({\rho }_i+{\rho }_n)}$, since collisions make the neutral mass participate in the oscillation. Their ambipolar damping rate is ${\Gamma }_A=a^2/(8\mu )$, with ${\Gamma }_A/|\mathrm{Re}{\omega }_{\pm }|\ll 1$.

For the remaining root, at $a=0$ the cubic is ${\omega }^2(\omega +2i\mu )$, so start from ${\omega }_{d0}=-2i\mu$. Denote the cubic by $F(\omega ,a)$. Its value and derivative at this root areThe first correction is therefore $\delta {\omega }_d=i{a}^2/(4\mu )$, givingThis branch is purely damped. Indeed, the real growth-rate polynomial for $s=-i\omega$ is $s^3+2\mu s^2+a^{2}s+\mu a^2$, and its simple real root near $-2\mu$ stays real for sufficiently small real $a^2$. It represents rapid decay of ion-neutral relative motion. In the $a\to 0$ limit, the other two frequencies tend to zero; all three branches are accounted for.

The neutral displacement equation gives, for each nonzero-frequency root,The vector displacements are parallel as complex polarization vectors, with this scalar amplitude and phase ratio.

For the rapidly damped root, $\mu -i{\omega }_d=-\mu +a^2/(4\mu )+O(a^4/{\mu }^3)$, and henceThe two fluids move almost equally and oppositely; collisions remove the counterflow at a rate close to $2\mu$.

For either weakly damped Alfvén wave, expand the ratio around unity:Equivalently, the exact slippage relation is ${\mathbf{\xi }}_i-{\mathbf{\xi }}_n=-(i\omega /\mu ){\mathbf{\xi }}_n$. Thus the relative slippage has size $|a|/(\sqrt{2}\mu )\ll 1$ and is primarily a small phase difference. The fluids almost share a common displacement, but magnetic tension acts directly on the ions and drag accelerates the neutrals. This imperfect locking causes ambipolar damping.

The sign of the damping is also checked by the perturbation energy balance. With periodic or vanishing-flux boundaries,It follows by dotting each linear momentum equation with its velocity and the induction equation with $\delta \mathbf{B}/(4\pi )$; magnetic-tension work cancels magnetic-energy change, pressure contributes only boundary flux, and the two drag terms combine into the negative square. Even though spatial diffusion was neglected, ion-neutral friction still dissipates the wave energy.

At fixed number density $n$, the electron current givesThe second equality uses Ampère's law without displacement current and stationary ions. A perturbation with characteristic perpendicular wave number $k_{\perp }$ therefore produces $\delta u_e\sim \alpha k_{\perp }\delta B$. Its nonlinear magnetic-advection rate isFor the anisotropic electron magnetohydrodynamics wave, $k\simeq k_{\perp }$ and $|\omega |\sim v_A{d}_i|k_{\parallel }|k_{\perp }$. Equating this to the nonlinear rate by critical balance givesThis is small-amplitude but strong turbulence: a small perturbation can have order-one nonlinear distortion over one wave period because its perpendicular gradients are much larger than its parallel gradients.

The solenoidal magnetic-field constraint impliesWith $\delta B/B_0=O(ϵ)$ and ${\partial }_z=O(ϵ{\nabla }_{\perp })$, the right-hand side is $O({ϵ}^2{k}_{\perp }{B}_0)$. Thus the leading, order-$ϵ$ perpendicular field is divergence-free. On a simply connected perpendicular patch, or for the nonzero perpendicular Fourier modes of a periodic domain, it has a stream function. Choose its normalization byConsequently the leading reduced electron magnetohydrodynamics representation isThe error denotes an order-${ϵ}^2$ field relative to $B_0$. The parallel perturbation is an independent scalar and is not removed by leading perpendicular solenoidality.

For clarity, the representation is asymptotic rather than an exact decomposition of an arbitrary three-dimensional solenoidal field. If $\delta B_{\parallel }$ depends on $z$, exact solenoidality can be restored by adding ${\nabla }_{\perp }\chi$ to $\delta {\mathbf{B}}_{\perp }$, where ${\nabla }_{\perp }^2\chi =-{\partial }_z\delta B_{\parallel }$. Its amplitude is $O({ϵ}^2{B}_0)$, so it lies beyond the retained field order. A perpendicular harmonic component can be absorbed into the guide field or fixed separately by boundary conditions.

Put $b=\delta B_{\parallel }/B_0$ and ${\mathbf{b}}_{\perp }=\delta {\mathbf{B}}_{\perp }/B_0=\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$. Define the two-dimensional Poisson bracket byThen, to the retained anisotropic order, the derivative along the perturbed magnetic field isThe omitted $b{\partial }_z$ term is smaller by $ϵ$ than either retained term. The nonlinear field-line-bending term is of the same order as ${\partial }_z$, which is why it must remain.

Use ${\mathbf{u}}_e=-\alpha \nabla \times \mathbf{B}$ with $\alpha B_0=v_A{d}_i$. The leading parallel and perpendicular electron velocities areIn detail, $\nabla \times (B_{0}b\hat{\mathbf{z}})=-B_0\hat{\mathbf{z}}\times {\nabla }_{\perp }b$, while the parallel curl of $B_0\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$ is $B_0{\nabla }_{\perp }^2\Psi /v_A$. The perpendicular curl involving ${\partial }_z\delta {\mathbf{B}}_{\perp }$ is order-smaller. Exact electron incompressibility follows from the divergence of a curl; its small velocity correction supplies any divergence omitted by this leading representation.

The electron magnetohydrodynamics equation is frozen-in induction,because both $\mathbf{B}$ and ${\mathbf{u}}_e$ are solenoidal. Keeping its leading order-${ϵ}^2$ perpendicular terms givesLet $X_f=\hat{\mathbf{z}}\times {\nabla }_{\perp }f=(-f_y,f_x)$. Expanding the two components gives $(X_f\cdot \nabla )X_g-(X_g\cdot \nabla )X_f=X_{\{f,g\}}$. Substitution therefore reduces the perpendicular equation toThe perpendicular-constant freedom in $\Psi$ has no magnetic effect and can be fixed so that

The parallel induction component at the same order isThe advection term vanishes since ${\mathbf{u}}_{e\perp }\cdot {\nabla }_{\perp }b=v_A{d}_i\{b,b\}=0$. ThusTogether these are both required reduced electron magnetohydrodynamics equations. Terms such as $u_{e\parallel }{\partial }_{z}b$ and $b{\partial }_z{\mathbf{u}}_e$ are order-${ϵ}^3$ and consistently absent.

Linearization removes the Poisson bracket from ${\nabla }_{\parallel }$. For Fourier perturbations proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, the two reduced equations becomeEliminating either amplitude givesSince $k=(k_{\perp }^2+k_{\parallel }^2{)}^{1/2}=k_{\perp }[1+O({ϵ}^2)]$, this is the specified dispersion relation to the accuracy of the anisotropic reduction. The reduced equations cannot retain the subleading correction replacing $k_{\perp }$ by the full $k$.

The full electron magnetohydrodynamics equation checks that correction directly. Its linearization is ${\partial }_t\delta \mathbf{B}=-\alpha B_0{\partial }_z\nabla \times \delta \mathbf{B}$. HenceApplying the cross-product operator twice and using $\mathbf{k}\cdot \delta \mathbf{B}=0$ givesThus the full model gives the stated relation exactly and the reduced model gives its leading anisotropic limit, with all signs and normalization factors consistent.

Use the Goldreich–Sridhar turbulence model in its balanced form. Assume a uniform strong guide field; anisotropic fluctuations $k_{\parallel }\ll k_{\perp }$; comparable counterpropagating Alfvén wave amplitudes; a statistically steady inertial range with negligible forcing and dissipation within it; transfer mainly between comparable perpendicular scales; and a scale-independent energy flux per unit mass $\mathcal{E}$. The estimate also assumes no extra scale-dependent alignment factor in the nonlinear interaction and neglects intermittency corrections.

Let ${\ell }_{\perp }$ and ${\ell }_{\parallel }$ be an eddy's perpendicular and parallel scales, and $\delta u_{\ell }\sim \delta B_{\ell }/\sqrt{4\pi {\rho }_0}$ its Alfvénic velocity amplitude. The nonlinear interaction is supplied by the oppositely propagating Elsässer variable, so comparable populations giveThe Alfvén propagation time is ${\tau }_A\sim {\ell }_{\parallel }/v_A$. For strong turbulence, critical balance makes these times comparable; the decorrelation and cascade time is therefore of order ${\tau }_{\mathrm{nl}}$, not a much longer time obtained by adding many weak encounters.

A constant perpendicular energy flux givesDefine the one-dimensional perpendicular turbulent energy spectrum so that $\int E(k_{\perp })d{k}_{\perp }$ measures fluctuation energy per unit mass. Locality of the spectrum then givesHencewhere $C$ is a dimensionless normalization constant not determined by the scaling argument. The magnetic and kinetic spectra have the same leading scaling under the assumed Alfvénic amplitude relation.

Critical balance also determines the scale-dependent anisotropy:Thus the eddies become relatively more elongated along the guide field at smaller perpendicular scales. The $-5/3$ exponent follows under the listed cascade assumptions; it is not established for every imbalanced or aligned MHD state solely by invoking critical balance.

At leading anisotropic order, perpendicular Alfvénic turbulence evolves independently of the compressive fluctuations. The slow modes and the entropy mode are transported by that turbulence. The fast magnetosonic frequency is of order $k_{\perp }(v_A^2+c_s^2{)}^{1/2}$, much larger than the reduced Alfvén frequency; the low-frequency reduced magnetohydrodynamics ordering filters out that fast-wave sector.

The passive evolution can be made explicit for a uniform ideal adiabatic plasma. Let $c_s^2=\gamma p_0/{\rho }_0$, $\delta =\delta \rho /{\rho }_0$, $b=\delta B_{\parallel }/B_0$, and define the normalized entropy perturbationUse $D={\partial }_t+{\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }$ and ${\nabla }_{\parallel }={\partial }_z+(\delta {\mathbf{B}}_{\perp }/B_0)\cdot {\nabla }_{\perp }$. Perpendicular force balance makes the leading total-pressure perturbation vanish:The ideal entropy equation gives $Ds=0$. The leading continuity and parallel-induction equations have the same compressive divergence term; subtracting them gives $D(\delta -b)=-{\nabla }_{\parallel }{u}_{\parallel }$. Parallel momentum gives $D{u}_{\parallel }=-{\nabla }_{\parallel }\delta p/{\rho }_0$. PutThe two compressive equations therefore becomeLet $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ be the tube speed and $h=c_s^{2}r/c_T$. ThenThus the slow and entropy fluctuations in reduced magnetohydrodynamics have characteristic amplitudes $f^{\pm }=u_{\parallel }\pm h$ obeyingThe two slow-wave populations propagate in opposite directions along the bent field lines while being mixed across them by ${\mathbf{u}}_{\perp }$. The entropy perturbation is carried by the fluid without a restoring-wave frequency. A pure entropy perturbation has $r=b=u_{\parallel }=0$ and $\delta =-s/\gamma$, so its density variation is accompanied by a temperature change with no leading pressure variation.

The coefficients and advecting fields in these three scalar equations come from the Alfvénic cascade; the scalars do not enter the leading perpendicular Alfvén equations. They therefore develop small perpendicular scales through passive mixing. If a scalar $f$ has constant variance flux ${\mathcal{F}}_f$ and is mixed on the Goldreich–Sridhar time ${\tau }_{\ell }\sim {\mathcal{E}}^{-1/3}{\ell }_{\perp }^{2/3}$, thenThe flux and amplitude need not equal the Alfvénic energy flux. This passive-spectrum statement uses the same locality and stationary inertial-range assumptions as part (a), and ignores additional damping. If a passive mode is absent initially and is not forced, its homogeneous leading equation keeps it absent; the Alfvénic cascade does not have to generate every compressive mode.

In the ideal, leading anisotropic system, there are five separately conserved quadratic cascade quantities: the two oppositely propagating Alfvén populations, the two slow-wave populations, and the entropy fluctuation. With the amplitudes of part (b), one convenient normalization iswhere ${\mathbf{z}}_{\perp }^{\pm }={\mathbf{u}}_{\perp }\pm \delta {\mathbf{B}}_{\perp }/\sqrt{4\pi {\rho }_0}$. Constant background factors convert the velocity variances into modal energies and the entropy variance into an entropy-related free-energy contribution. These are the five quadratic cascade invariants of anisotropic magnetohydrodynamic turbulence.

To verify conservation rather than just count polarizations, the perpendicular Elsässer variables obeyDot each equation with $2{\mathbf{z}}_{\perp }^{\pm }$ and integrate. The propagation and nonlinear terms become surface fluxes of $|{\mathbf{z}}_{\perp }^{\pm }{|}^2$, and the pressure term becomes a surface flux because the corresponding field is solenoidal. For periodic boundaries, or boundaries with vanishing invariant flux, all these terms vanish and $d{I}_A^{\pm }/dt=0$. Their sum is proportional to perpendicular kinetic plus magnetic energy, while their difference is proportional to perpendicular cross-helicity; the two Elsässer energy invariants contain both pieces of information.

Likewise the slow-mode equations giveHere ${\mathbf{u}}_{\perp }$ is divergence-free and ${\nabla }_{\parallel }$ differentiates along the divergence-free vector $\hat{\mathbf{z}}+\delta {\mathbf{B}}_{\perp }/B_0$, so both integrals are boundary fluxes. Finally $Ds=0$ gives $d{I}_e/dt=-\int {\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }{s}^{2}dV=0$. This proves all five quadratic conservation laws under the same boundary assumptions.

The two Alfvén cascades interact dynamically: each population strains the other and is needed for its nonlinear transfer between scales. Their integrated invariants are nevertheless conserved separately, so this interaction does not transfer one population's invariant into the other. The slow and entropy cascades are driven by the Alfvénic mixing and field-line motion, with no feedback on the perpendicular cascade at the retained order. Their injection rates set independent fluxes; those fluxes need not be equal, even when the resulting perpendicular spectra share the $-5/3$ scaling.

The count is a count of independent quadratic modal cascade channels. Ideal passive advection also preserves higher moments, such as $\int F(s)dV$ for arbitrary smooth $F$, and analogous integrals for the slow characteristic scalars. These are not additional quadratic modal energies. Forcing and dissipative terms make the corresponding invariants source-and-sink balances; a stationary inertial range conserves their flux through scales rather than their total content in a continually forced domain.