# What is the dielectric constant of a star?

1. What is the dielectric constant of a star, especially its corona? Is it of order 1 or is it quite large?

2. How much do "impurities" (elements other than H and He) affect the dielectric constant?

• Thanks for the edit, I've adjusted the wording a bit to better fit the site. Just fyi the answer will be different at radio frequencies and optical frequencies (visible light).
– uhoh
Mar 18, 2020 at 1:19
• Trying for the world's biggest capacitor here? :-) . Now, as pointed out, plasmas conduct, but depending on the frequency band you're interested in, you may find some ranges where the plasma does hold energy rather than conduct it. Mar 18, 2020 at 19:28
• @JamesK there's nothing wrong with talking about the dielectric constant of conductors or other lossy materials; they are simply complex valued rather than real: en.m.wikipedia.org/wiki/Permittivity#Complex_permittivity
– uhoh
Mar 19, 2020 at 9:24
• @CarlWitthoft ditto.
– uhoh
Mar 19, 2020 at 9:25
• @JamesK I found Determination of the Dielectric Constant of a Plasma which gives a formula and might be helpful for the corona - I have to think of it. Do you think we could estimate all the parameters? Apr 15, 2021 at 8:14

### Disclaimer

This not (yet) a full answer, since I only focus on approaching part 1 of the question, namely the calculation of the dielectric constant of the corona.

### Dielectric constant of the corona

The Sun's corona

... extends millions of kilometres into outer space and is most easily seen during a total solar eclipse [...] measurements indicate strong ionization in the corona and a plasma temperature in excess of $$1\,000\,000 {\rm K}$$ much hotter than the surface of the Sun.

So let's dig a bit into plasma physics.

According to I. M. Podgornyi's article Determination of the Dielectric Constant of a Plasma, in the absence of a magnetic field, the dielectric constant $$\varepsilon$$ of a plasma can be written

$$\varepsilon(\omega) = 1 - {{4\pi {e^2}n} \over {{m_e}{\omega ^2}}} \cdot {1 \over {1 - i{{{v_{col}}} \over \omega }}}$$

where $$\omega$$ is the angular frequency of the electromagnetic wave, $$e = 1.60217662 \cdot 10^{-19} {\rm C}$$ is the charge of an electron, $$m_e = 511 {\rm keV} = 9.10938356 \cdot 10^{-31} {\rm kg}$$ is the mass of an electron, $$i$$ the imaginary unit.

The frequency of collisions $$ν_{col}$$ with charged particles is given by

$${v_{col}} = {{{{2.10}^{ - 5}}{n_i}Z} \over {T_e^{3/2}\left( {\rm eV} \right)}}.$$

Here, we have $$T_e \approx 80\ldots90 {\rm eV} = 10^6 {\rm K}$$ the temperature of a plasma given in electron volt, and (average?) $$Z$$ the charge number.

The next step towards an answer would be to figure out all parameters in this equation, in particular the average composition of the plasma (in terms $$n$$ and $$Z$$).

I did not yet manage that (yet), see also my question Physical properties of the stellar corona?

### References

The following publications might be helpful for determining $$n_i$$ and $$Z$$ for some or all consitutents of the coronal plasma:

### Related

• The solar corona is practically a collisionless plasma, so you can safely neglect the collision term here (even in the photosphere it should be negligible unless $\omega$ corresponds to radio waves). Apr 20, 2021 at 21:35