Perpendicular Wave Propagation

One obvious way of solving this equation is to have , or

with the eigenvector . Since the wave-vector now points in the -direction, this is clearly a transverse wave polarized with its electric field parallel to the equilibrium magnetic field. Particle motions are along the magnetic field, so the mode dynamics are completely unaffected by this field. Thus, the wave is identical to the

The other solution to Eq. (545) is obtained by setting the determinant
involving the - and - components of the electric field to zero. The
dispersion relation reduces to

Let us, first of all, search for the cutoff frequencies, at which goes to zero. According to Eq. (547), these frequencies are the roots of and . In fact, we have already solved these equations (recall that cutoff frequencies do not depend on ). There are two cutoff frequencies, and , which are specified by Eqs. (541) and (544), respectively.

Let us, next, search for the resonant frequencies, at which goes to
infinity. According to Eq. (547), the
resonant frequencies are solutions of

(549) |

(550) |

(551) |

The first resonant frequency,
, is greater than the
electron cyclotron or plasma frequencies, and is called the *upper hybrid
frequency*. The second resonant frequency,
, lies between the
electron and ion cyclotron frequencies, and is called the
*lower hybrid frequency*.

Unfortunately, there is no simple explanation of the origins of the two hybrid resonances in terms of the motions of individual particles.

At low frequencies, the mode in question reverts to the compressional-Alfvén wave discussed previously. Note that the shear-Alfvén wave does not propagate perpendicular to the magnetic field.

Using the above information, and the easily demonstrated fact that

(552) |

Wave propagation at oblique angles is generally more complicated than propagation parallel or perpendicular to the equilibrium magnetic field, but does not involve any new physical effects.