# 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 '20 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. – Carl Witthoft Mar 18 '20 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 '20 at 9:24
• @CarlWitthoft ditto. – uhoh Mar 19 '20 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? – B--rian Apr 15 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). – Thomas Apr 20 at 21:35