Why do we ignore the Flux - Luminosity relation when deriving the Eddington Luminosity? I was just doing the derivation for the Eddington Luminosity by considering the momentum equation,
$$ -\frac{Gm}{4\pi r^4} = \frac{dP}{dm} $$
and the radiative transport pressure
$$ \frac{dT}{dr} = - \frac{3 \kappa \rho }{4ac T^3} \frac{F}{4 \pi r^2} $$
paired with the radiation pressure
$$ P_{rad} = \frac{1}{3} a T^4 $$
Taking the ratio of the hydrostatic pressure and the radiation pressure we get
$$ dP_{rad}/dP \leq 1 $$
Which gives
$$ F = \frac{4\pi c Gm}{\kappa} $$
We then immediately jump to the Eddington luminosity
$$ L_{Edd} = \frac{4\pi c Gm}{\kappa} $$
without taking into consideration that
$$ F = \frac{L}{4 \pi r^2} $$
Why? We have carried the constants through all this way so why do just now only consider the proportionality of F with L and not the complete relationship?
 A: I think there is at least one mistake in what you've written above. The radiative diffusion equation is
$$ \frac{d(aT^4)}{dr} = -\frac{3 \kappa \rho}{c} F_{rad}$$
$$\frac{dT}{dr} = -\frac{3 \kappa \rho}{4acT^3} F_{rad}$$
(I think you have mixed up $F$ and $L$ in your second equation, and the answer may be as simple as that).
Radiation pressure is
$$P_{rad} = \frac{a}{3}T^4 $$
Hydrostatic equilibrium (assuming only radiation pressure)
$$ \frac{dP_{rad}}{dr} = - \rho g$$
$$ \frac{4aT^{3}}{3}\frac{dT}{dr} = -\rho g$$
$$ \frac{4a T^3}{3} \frac{3\kappa \rho}{4acT^3} F_{rad} = \rho g = \rho G \frac{M}{r^2}$$
Therefore, assuming that all of the luminosity is due to radiation
$$ 4\pi r^2 F_{rad} = L_{rad} = \frac{4\pi GM c}{\kappa}$$
EDIT: After the exchange about the Prialnik book you used, it seems there is confusion over the definition of $F$. I have used $F_{rad}$ to be a flux, that is power per unit area. Prialnik's definition of $F$ is a power. Thus using Prialnik's definition then indeed $F= L_{edd}$ at the surface and the final equation in your question is incorrect.
