I asked a question earlier, why ions are not included in the refractive index of plasma .

What I do not understand is how a refractive index of plasma can be formed mathematically for plasma that is just partially ionized? Let's say 10% of atoms are ionized, so it is a mixture of ions, electrons and neutral atoms.

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    $\begingroup$ The effective electron concentration would be lower by an order of magnitude, but there will be optical activity since there are still a lot of free charges around. $\endgroup$
    – CuriousOne
    Commented Jun 24, 2016 at 19:51

1 Answer 1


The refractive index comes, mostly, from the electric polarizability of the medium: that is, the amount of electric dipole induced in the medium by a unit driving electric field. This is a macroscopic quantity, calculated by adding up the contributions of each microscopic components, and if you have multiple species in the medium (such as neutral atoms and free electrons) then you simply add together their contributions to the total induced dipole moment in that unit of volume.

If you have a single species, then you look at the dipole moment induced in each microscopic component, $$ \mathbf p=\varepsilon_0 \alpha \mathbf E, $$ where $\alpha$ is the electric polarizability of the species, a dimensionless constant, and you multiply it by the number of systems per unit volume $N$ to form the electric polarization (i.e. dipole moment per unit volume) $$ \mathbf P=N\mathbf p = \varepsilon N \alpha \mathbf E=\varepsilon \chi_e\mathbf E, $$ where $\chi_e=N\alpha$ is the electric susceptibility of the medium. This is the constitutive relation of the (linear) medium, and it gets fed into the (relative) electric permittivity $\varepsilon_r=1+\chi_e$ that then goes to make the refractive index $n=\sqrt{\varepsilon_r\mu_r}$.

If you have multiple species, the game is similar. Each species will have its own linear response of the form $$ \mathbf p_j=\varepsilon_0 \alpha_j \mathbf E, $$ with its own volumetric number density $N_j$, giving them a contribution $$ \mathbf P_j=N_j\mathbf p_j = \varepsilon N_j \alpha_j \mathbf E $$ to the total polarizability, which comes out as $$ \mathbf P=\sum_j \mathbf P_j = \varepsilon_0 \left(\sum_j N_j\alpha_j\right) \mathbf E = \varepsilon_0 \chi_e \mathbf E, $$ so the electric susceptibility gets modified to $$ \chi_e=\sum_j N_j\alpha_j. $$ This then goes into $\varepsilon_r=1+\chi_e$ and $n=\sqrt{ \varepsilon_r \mu_r}$ as before.

For a partially ionized gas, it sort of depends on how strong the response is from each of the components. If you are near the plasma frequency of free electrons in a thin gas, and the response of the neutrals is relatively flat there, then decreasing the plasma to an ionized fraction $r$ will decrease the optical activity by $r$. If you're in a more complicated situation (such as resonances of the neutral, or a thicker plasma), then you need to do the full calculation as above.

  • $\begingroup$ Thanks my question is more simpler , how the plasma index $1-w^2/w^2_p$ changes in partially ionized gas. $\endgroup$
    – Anonymous
    Commented Jun 24, 2016 at 23:07
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    $\begingroup$ And, as was the case in the question you linked, the answer is there. You do actually have to do some work to understand the answers you solicit. $\endgroup$ Commented Jun 24, 2016 at 23:41

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