In steady-state, when radiative equilibrium holds and the luminosity as a function of radius is constant, we may compute the radiation force per volume $f_\mathrm{rad}$ as $$f_\mathrm{rad} = \frac{L}{4 \pi \, r^2 \, c}\, \rho(r) \, \kappa(r) \; .$$

The total outward force exerted by the radiation is $$\int _V f_\mathrm{rad}\, dV = 4 \pi \left(\frac{L}{4 \pi \, c}\right) \int_{0}^{\infty} \rho(r) \, \kappa(r) \, dr = \tau \frac{L}{c}\; ,$$ where $\tau$ is the radial optical depth for the infrared photons.

In a gas rich galactic nucleus with a column density of $10^{25}$ cm$^{-2}$, a mean mass per particle of 1.5 times the proton mass, and an infrared dust opacity of 10 cm$^{2}$ per gram of gas, an initial guess for the optical depth would be approximately 250. There are two primary effects that will reduce the actual radiation force from such a high value. The first is the lack of spherical symmetry: a torus obscures only a fraction of the solid angle surrounding the accretion disk, and the presence of clumps and voids in the torus can increase the photon mean free path for certain sightlines. The second effect is dust sublimation: dust will be absent from the innermost regions 