Why is the density of photons in the Eddington Limit derived this way?

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SUMMARY

The discussion centers on the derivation of the photon density formula in relation to the Eddington Limit, as presented in "Astrophysics in a Nutshell" by Dan Maoz. The formula for photon density at a radius r in an ionized gas is given by nph = Lν / (4πr2chν). The participants clarify that while Lν represents energy flux, dividing by hν converts this into a photon flux, and further division by c yields a density of photons per volume. The conversation emphasizes understanding the Eddington limit through momentum flux rather than photon density.

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Aleolomorfo
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I am studyng accretion process on "Astrophysics in a nutshell" by Dan Maoz and I have some doubts about the derivation of the formula for the eddington limit. I understand what the edding limit is. The accretion rate cannot be arbitrarly large. The starting point is to consider an electron at a radius ##r## in an ionized gas that is taking part in an accretion flow towards some compact object. The accretion flow produces a luminosity per frequency interval ##L_\nu##, and therefore the density of photons with energy ##h\nu## at ##r## is:
$$n_{ph}=\frac{L_\nu}{4\pi r^2 ch\nu}$$
I do not understand why the density of photons is written in this way. I see that it is dimensionally correct but I do not see the reason.
##\frac{L_\nu}{4\pi r^2}## is the flux of photons with frequency ##\nu## but I do not understand why it is divided by ##ch\nu##.
 
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L_{\nu} is not the flux of photons, it is the flux of energy. So you have to divide by the energy per photon to get the flux of photons.
 
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Yes, you first have to divide by the energy per photon, h*nu, to get it into a number flux (and I believe you mean flux per area), but then you also have to divide by the particle speed to get it into a density of particles per volume. You'd have to do the same thing with a flux of bullets. If you still don't see it, it might help to take the c up onto the left side of the equation, and think about what a number density times a speed is.

However, I would also point out that the easiest way to understand the Eddington limit is to think in terms of the momentum flux per area, not the photon flux and not the photon density. This is because to get the radiative force per gram, you simply take the momentum flux per area, and multiply by the cross section per gram. That's the simplest way to see what is going on.
 
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Aleolomorfo said:
therefore the density of photons with energy hνhνh\nu at rrr is:
Is there any special point in restating the Eddington limit in terms of photons?
 

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