 Solution

Put $a=c/(4\pi en)$ and retain terms linear in $\delta \mathbf{B}$. The electron magnetohydrodynamics induction equation becomesHere the curl of $(\nabla \times \delta \mathbf{B})\times B_0\hat{\mathbf{z}}$ reduces to $B_0{\partial }_z\nabla \times \delta \mathbf{B}$, since the background magnetic field is constant and the divergence of a curl vanishes. For a plane wave $\delta \mathbf{B}=\mathbf{b}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the linear equation and the solenoidal constraint areOn the transverse plane, the operator $i\mathbf{k}\times$ has eigenvalues $\pm k$ by the helicity decomposition of a transverse Fourier mode. Equivalently, squaring the equation and using the vector triple product gives ${\omega }^2=a^2{B}_0^2{k}_{\parallel }^2{k}^2$. Since $a{B}_0=v_A{d}_i$, where $v_A$ is the Alfvén speed and $d_i$ the ion skin depth,The two branches have opposite circular polarizations. The specified electron-only induction model describes a whistler wave; the paper calls these branches kinetic Alfvén waves. The following cascade calculation uses the specified model and its dispersion relation, without adding the pressure response needed for the usual kinetic Alfvén interpretation.