# What is the refractive index for partially ionized plasma (mixture of electrons, ions and atoms)?

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.

• 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. Commented Jun 24, 2016 at 19:51

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.
• Thanks my question is more simpler , how the plasma index $1-w^2/w^2_p$ changes in partially ionized gas. Commented Jun 24, 2016 at 23:07