So I know that the Eddington luminosity is given by:

$L_{Eddington} = 4\pi GMc/\kappa$.

I want to calculate this luminosity for a plasma of purely ionized helium, as well as for an electron-positron plasma.

How do I find $\kappa$ in each of these cases? I know sometimes this is approximated as $m/\sigma_T$, but I'm not sure when to invoke this approximation, and what the respective mass would be.

  • $\begingroup$ electron-positron plasma Do you really mean an electron-proton plasma? $\endgroup$
    – PM 2Ring
    May 22 '19 at 14:57
  • $\begingroup$ No, I meant electron-proton, so that's been fixed. $\endgroup$
    – Ken G
    May 22 '19 at 23:26

I think you mean that kappa = sigma / m. The sigma is the "cross section," like the area you expose when you slice a watermelon in half. The m is the mass of the particle that has the cross section in question (it's not literally the cross section of the physical particle, but an effective cross section for interaction with light), or more often, the mass of all the particles that come along with the cross section in question (often there is one particle contributing most of the cross section, but several particles that come with it which provide the mass). For example, if you mix protons and electrons, the electrons have a larger cross section because their low mass makes them interact more easily with light, but that lower mass also means it is not the electrons, but the protons, that provide the "m" in the above formula.

So these quantities depend on the composition of the gas you have in mind. It's easy for an electron-proton [edited] plasma, because if you don't worry about resonant interactions in the plasma, you can just consider the cross sections of the individual particles, and you get "Thomson opacity" of 0.4 square centimeters per gram (that's sigma/m for the sigma of an electron and the m of a proton, rounded to one significant figure because nothing is ever really a pure electron-proton plasma). Purely ionized helium is also easy, because if you again ignore any resonant interactions between the electrons and the nuclei, it's just the same Thomson opacity except reduced by the extra neutrons [edited] in the heavy helium nuclei. Since there is one neutron per proton, that reduces 0.4 by factor of 2, yielding 0.2. [edited for forgetting that the 0.4 already includes the protons.]


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