 Solution

Use positive angular frequencies and abbreviate $p={\omega }_{pe}$, $b={\omega }_{ce}$. The left-handed index is $n_L^2=1-p^2/[\omega (\omega +b)]$. It is negative below its positive zero,For the right-handed mode, $n_R^2=1-p^2/[\omega (\omega -b)]$. Below the cyclotron resonance $0<\omega <b$ its denominator is negative and the index square is positive. Just above the resonance it is negative untilThis is the lossless electron-only cold-plasma model. Ions, collisions and resonance absorption matter outside its applicable range.

For a whistler with $\omega \ll b$ and $p^2/(\omega b)\gg 1$,The travel time through a uniform slab is $t-t_s=Lp{\omega }^{-1/2}/(2c\sqrt{b})$. In terms of ordinary frequency,An arbitrary emission time changes the intercept, not the slope. The original figure labels a dispersion constant $D=81\pm 0.5\mathrm{H}{\mathrm{z}}^{1/2}\mathrm{s}$: it is the reciprocal of the plotted slope, not the slope itself. ThereforeFor $B=80$ nT and $L=6.0\times 10^7$ m this gives $b\simeq 1.41\times 10^4{\mathrm{s}}^{-1}$, $p\simeq 2.41\times 10^5{\mathrm{s}}^{-1}$, andThe resulting forbidden right-handed frequency width isThe lower edge is about $2.24$ kHz and the upper edge about $39.5$ kHz. In the real nonuniform plasma, the travel-time coefficient measures a path integral proportional to $\int {\omega }_{pe}/\sqrt{{\omega }_{ce}}ds$. The inferred density is consequently an effective uniform-slab estimate, not a unique local density at the spacecraft. Uncertainty in the path and magnetic field likely exceeds the formal error in $D$.