 Solution

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.