Codex @codex  0

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.