Arthur Eddington studied the structure of stars, not black holes. He knew that the radiation produced by stars exerts a force on their atmospheres. In massive stars this radiation pressure can be the dominant component of the pressure supporting the star against collapse. Eddington assumed that if the radiation pressure was larger than the gravitational attraction in a particular region of a star, then that region would be blown off the star by its own luminosity. He reasoned that for the star to be stable, the force exerted by radiation on a particle in the atmosphere must be less than the force the particle feels from the gravity of the star: Frad < Fgrav
The radiation force, Frad, is the product of the radiation flux (f) from the star and a constant known as the Thomson scattering cross section (σT) divided by the speed of light (c). A scattering cross section is related to the probability that one particle interacts with another. Both a star and the accreting material around a black hole are ionized hydrogen (the electron has been separated from the proton). In this case, the Thomson cross section is related to the likelihood that a photon interacts with a free electron. And of course the flux is related to how many photons are present to interact, so the following expression gives the total force exerted by photons (the radiation force) in the star’s atmosphere.
\begin{equation} F_{rad} = \frac{f \sigma_T}{c} \end{equation}
We can rewrite the flux using the luminosity of the star. The flux drops as the distance from the center of the star squared (inverse square law): f = L/4πr2, and thus
\begin{equation} F_{rad} = \frac{L \sigma_T}{4 \pi c r^2} \end{equation}
The gravitational force felt by a particle that is part of a star of mass M is given by Newton’s law of gravity.
\begin{equation} F_{grav} = \frac{GMm_p}{r^2} \end{equation}
Here we have explicitly used the mass of a proton, mp, because protons are much more massive than the electrons in the ionized hydrogen gas.
If we now set the radiation force and the gravitational force equal, we get:
\begin{equation} \frac{GMm_p}{r^2} = \frac{L \sigma_T}{4 \pi c r^2} \end{equation}
Now we cancel the common factor of r2 from each side of the equation and then solve for L. We obtain the following expression for the Eddington luminosity:
\begin{equation} L=\left(\frac{4\pi cGm_p}{\sigma_T}\right)M_{BH} \end{equation}
Substituting in the values for the constants, we get:
\begin{equation} L=\left[\frac{4\pi (3 \times 10^8)(6.67\times 10^{–11})(1.67 \times 10^{–27})} {(6.65\times 10^{–29})}\right]M_{BH} = (6.3) M_{BH} \end{equation}
where the luminosity (L) is in W, the mass of the black hole (MBH) is in kg, and the constant 6.3 is in W/kg.
Eddington had stars in mind when he derived this expression, but it works for any object that is luminous and accreting (or supporting) gaseous material composed of hydrogen.
What happens when a system reaches this Eddington luminosity? In the case of a star, strong winds will eject matter outward. For a black hole, all the energy released by accretion does not have to appear as outgoing luminosity, since energy can be lost through the event horizon.