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An electrostatic plasma wave has $\delta \mathbf{E}=-i\mathbf{k}\varphi$. In the absence of externally imposed charge density, the Gauss law for the magnetized cold-plasma dielectric tensor is $\mathbf{k}\cdot \mathbf{ϵ}\cdot \delta \mathbf{E}=0$. Consequently,The antisymmetric terms of the magnetized cold-plasma dielectric tensor cancel in this scalar contraction. For oblique propagation this is the leading longitudinal condition in the large-refractive index approximation; a small transverse electric field is generally required by the full Maxwell equations at finite $n$. For parallel propagation an exactly longitudinal solution exists: $P=0$, hence ${\omega }^2\simeq {\omega }_{pe}^2$. It coincides with the $\theta \to 0$ limit of the finite-coefficient plasma wave resonance condition in c. At exactly $\theta =0$, $P$ factors out of the transverse dispersion relation, leaving the independent plasma oscillation rather than an isolated divergence of the transverse branches.

For the ordinary plasma wave, choose $\mathbf{k}=k\hat{\mathbf{x}}$ and $\delta \mathbf{E}=E_z\hat{\mathbf{z}}$. The equations in a give $u_{xs}=u_{ys}=0$ andThe driven particle motion is entirely parallel to ${\mathbf{B}}_0$: the parallel Lorentz force has no magnetic contribution. Nevertheless, the ordinary plasma wave is electromagnetic, since $\delta \mathbf{E}\perp \mathbf{k}$ and the Faraday law givesAn electrostatic plasma wave instead has $\delta \mathbf{E}\parallel \mathbf{k}$ and zero perturbed magnetic field. Alignment with the equilibrium magnetic field does not make a wave electrostatic. At the exact plasma wave cutoff $k=0$, this propagating-wave distinction becomes degenerate.