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For a uniform guide field, write the transverse perturbation in velocity units as $\mathbf{b}$. Linear momentum gives $\omega \mathbf{u}=-k_{\parallel }{v}_A\mathbf{b}$. Linear induction gives $\omega \mathbf{b}=-k_{\parallel }{v}_A\mathbf{u}+i{k}_{\parallel }{v}_A{d}_i\mathbf{k}\times \mathbf{b}$. The helicity decomposition of a transverse Fourier mode then yields the displayed dispersion relation. Its two positive frequency magnitudes are $|k_{\parallel }|v_A[\sqrt{1+(k{d}_i{)}^2/4}\pm k{d}_i/2]$, approaching the Alfvén wave frequency for $k{d}_i\ll 1$. For $k{d}_i\gg 1$, the fast branch is a whistler wave and the slow branch tends to $|k_{\parallel }|v_A/(k{d}_i)$.

Linearize about zero bulk flow and the uniform guide field. Put $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$, and take plane waves proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$. Let $k_{\parallel }=\mathbf{k}\cdot \hat{\mathbf{z}}$, $k=|\mathbf{k}|$, $q=k_{\parallel }{v}_A$, andBoth $\mathbf{u}$ and $\mathbf{b}$ are perpendicular to $\mathbf{k}$. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term givesIn the induction equation, the Hall contribution is $-c(B_0{\partial }_z)(\nabla \times \delta \mathbf{B})/(4\pi en)$; its two derivative factors of $i$ in the Fourier mode produce a positive coefficient multiplying $\mathbf{k}\times \mathbf{b}$. ThusEliminating $\mathbf{u}$ givesOn the transverse plane, $(i\mathbf{k}\times {)}^2=k^2$, so applying that operator again proves the incompressible Hall-MHD wave dispersion:Equivalently, in a helicity decomposition of a transverse Fourier mode with $i\mathbf{k}\times {\mathbf{b}}_h=hk{\mathbf{b}}_h$, $h=\pm 1$, each polarization obeys ${\omega }^2-hqk{d}_i\omega -q^2=0$. For $k_{\parallel }=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_{\parallel }\ne 0$.

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.

For a nonzero Fourier mode $\mathbf{k}=k\hat{z}$, the divergence-free condition implies $B_z=0$. The complex vectors $m^{(\pm )}=(\hat{x}\pm i\hat{y})/\sqrt{2}$ form a Hermitian orthonormal basis of this transverse plane. Their opposite rotation phases give the helicity decomposition of a transverse Fourier mode; the factor $i/\sqrt{2}$ in the amplitudes is a convention. The baryon velocity in this vector sector has the analogous decomposition.

For helicity $\sigma =\pm 1$, the angular source isThe same formula holds for ${\mathbf{v}}_b^{(\sigma )}$. Thus these sources have only angular order $m=\sigma$. Streaming multiplies by $ik\cos \theta$ and the stated Thomson scattering operator is diagonal in $m$, so neither mixes this sector with other angular orders. Assuming the initial anisotropy is in the same vector-helicity sector, or considering the anisotropy generated from zero initial data, the solution therefore contains only $Y_{\ell \sigma }$ with $\ell \ge 1$. Arbitrary independently imposed anisotropies of other $m$ would instead evolve in their own homogeneous sectors.

There is a genuine normalization inconsistency in the PDF. To retain its printed temperature expansion, writewhere $q_{\ell }$ denotes exactly the multipoles called ${\Theta }_{\ell }$ in that expansion. The spherical-harmonic streaming recurrence gives the coefficient of $Y_{\ell \sigma }$ in $ik\cos \theta \Theta$, divided by $C_{\ell }$, asHere $C_{\ell +1}/C_{\ell }=-i\sqrt{(2\ell +3)/(2\ell +1)}$ and $C_{\ell -1}/C_{\ell }=i\sqrt{(2\ell -1)/(2\ell +1)}$. Both streaming denominators are therefore $2\ell +1$, rather than the PDF's $2\ell +3$ and $2\ell -1$. The dipole source also changes: $C_1=-i\sqrt{3/(8\pi )}$, so $-(\mathbf{e}\cdot \dot{\mathbf{B}}+\dot{\tau }\mathbf{e}\cdot {\mathbf{v}}_b)/C_1$ is $-\sigma (4\pi /3)({\dot{B}}^{(\sigma )}+\dot{\tau }{v}_b^{(\sigma )})Y_{1\sigma }$. The quadrupole collision correction is still $-\dot{\tau }{q}_2{\delta }_{\ell 2}/10$.

Thus the vector photon Boltzmann hierarchy consistent with the literal printed expansion isA concrete counterexample to the printed hierarchy with this expansion is a collisionless instant with $q_2=1$ and every other multipole zero. Direct streaming gives ${\dot{q}}_1=-k\sqrt{3}/3$, whereas the printed hierarchy would give $-k\sqrt{3}/5$.

Alternatively, preserve the PDF's desired hierarchy by making the photon multipole normalization changeThis is the reciprocal angular normalization, not the printed expansion. Multiplying the corrected hierarchy by $(2\ell +1)/(4\pi )$ gives preciselyThis supplies both consistent conventions explicitly. The lower coupling vanishes at $\ell =1$, so no vector monopole is introduced. Cosmological optical depth to the observer has $\dot{\tau }<0$, giving damping of the dipole and the factor $9/10$ quadrupole damping in this temperature-only collision model; no polarization collision term has been added to the supplied equation.

In the inertia-free, constant-density electron magnetohydrodynamics model with uniform guide field $B_0\hat{\mathbf{z}}$, a transverse magnetic plane wave obeys $\omega \mathbf{b}=i{v}_A{d}_i{k}_{\parallel }\mathbf{k}\times \mathbf{b}$. The helicity decomposition of a transverse Fourier mode gives the displayed dispersive branches with opposite circular polarizations. This electron-current mode is called a whistler wave; see Lyutikov's analysis of electron magnetohydrodynamics. Its pressure-free induction equation should not be identified with a general kinetic Alfvén model.