Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 75 / 2

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.