I'm studying electromagnetic propagation in the ionosphere. The starting point is Maxwell's Equations in an inhomogeneous medium, plus the assumption that these inhomogeneities occur at length scales much larger than the scale of interest (WKB or eikonal approximation). For a time-harmonic source with angular frequency $\omega$, you get a Helmholtz equation,

$$\nabla^2 \vec{E} + k^2n^2\vec{E} = 0,$$

where $n = n(x,y,z)$ is the spatially-dependent index of refraction, and $k^2$ is the magnitude of the wavevector in vacuum, $k=\omega/c$. I'm interested in frequencies higher than the plasma frequency, such that the index is given by

$$n^2 = 1 - \frac{\omega_p^2}{\omega^2} \hspace{10pt} \big(\omega_p^2 \equiv \frac{N e^2}{m_e \epsilon_0}\big)$$

where $N(x,y,z)$ is the number of free electrons per unit volume. The original Helmholtz equation now looks like

$$\nabla^2 \vec{E} + k^2 \vec{E} = \xi N \vec{E},$$

with $\xi$ being a symbol of my own devising. You can prove to yourself or take my word that I've defined $\xi \equiv \frac{\mu_0 e^2}{m_e}$. Since $N$ is a number density, the dimensions of $N$ are $1/L^3$, which implies that $\xi$ is a length scale (you could also see this by simply expanding the units of the physical constants involved).

Usually, whenever I see a dimensionful quantity constructed from physical constants pop out of an equation, I try to interpret it physically. I'm bothered by this one ($\xi$), because I have no clue what the significance of it is. Anyone wanna take a stab at interpreting this quantity?

For your convenience, I have inserted the values of $e$, $\epsilon_0$ and $m_e$ to determine that $\xi = 3.53 \times 10^{-14}$ m.

  • 1
    $\begingroup$ Try converting $\mu_{o} = 1/\varepsilon_{o} c^{2}$ to recover the plasma frequency squared. Then I would recommend looking up something called the inertial length (or skin depth). $\endgroup$ – honeste_vivere Sep 10 '17 at 18:33
  • $\begingroup$ as said by @honeste_vivere, the length scale you might be looking for is the plasma skin depth $\delta$, given by $\delta = c/\omega_{pe}$ $\endgroup$ – Alf May 6 '18 at 21:28

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