The Eddington limit can be shown as:
Momentum of a photon: $ p = E/c = h\nu / c $
Force due to this radiation is change in momentum. Supposedly then,
$$ F_{rad} = \int^\inf_0 \frac{L_\nu}{h\nu}\frac{h\nu}{c}d\nu$$
Please explain how this integral encapsulates this - force is the first derivative?
Looking at a small bandwidth $\delta\nu$, the integral captures:
$$ \frac{L_{\nu}\delta\nu}{h\nu} $$
The numerator is total power radiated in this bandwidth, and the denominator normalizes that to a total number of photons in the bandwidth $(\nu, nu+\delta\nu)$ radiated per unit time.
The second factor converts that to a momentum per unit time so that the momentum change caused by photons in the bandwidth over a small time interval is:
$$ F_{\nu} = \frac{\Delta p_{\nu}}{\Delta t} $$
which conforms with the Newtonian idea that:
$$ F = ma = m\frac{dv}{dt} = \frac{dp}{dt} $$
Here I've assumed $L$ is total luminosity. You can now proceed by dividing by the total area at a given radius to get a radiation pressure (force per unit area).
