# Monte Carlo Simulations of Radiative Transfer: Basics of Radiative Transfer Theory (Part I)

SOURCE FOR CONTENT: Chandrasekhar, S., 1960. Radiative Transfer. 1.

In this post, I will be discussing the basics of radiative transfer theory necessary to understand the methods used in this project. I will start with some definitions, then I will look at the radiative transfer equation and consider two simple cases of scattering.

The first definition we require is the specific intensity, which is the amount of energy associated with a specific frequency $dE_{\nu}$ passing through an area $dA$ constrained to a solid angle $d\Omega$ in a time $dt$. We may write this mathematically as

$dE_{\nu}=I_{\nu}\cos{\theta}d\nu d\Sigma d\Omega dt. (1)$

We must also consider the net flux given by

$\displaystyle d\nu d\Sigma dt \int I_{\nu}\cos{\theta}d\Omega, (2)$

where if we integrate over all solid angles $\Omega$ we get

$\pi F_{\nu}=\displaystyle \int I_{\nu}\cos{\theta}d\Omega. (3)$

Let $d\Lambda$ be an element of the surface $\Lambda$ in a volume $V$ through which radiation passes. Further let $\Theta$ and $\theta$ denote the angles which form normals with respect to elements $d\Lambda$ and $d\Sigma$. These surfaces are joined by these normals and hence we have the surface across which energy flows  includes the elements $d\Lambda$ and $d\Sigma$, given by the following:

$I_{\nu}\cos{\Theta}d\Sigma d\Omega^{\prime}d\nu = I_{\nu}d\nu \frac{\cos{\Theta}\cos{\theta}d\Sigma d\Lambda}{r^{2}} (4),$

where $d\Omega^{\prime}=d\Lambda \cos{\Theta}/r^{2}$ is the solid angle subtended by the surface element $d\Lambda$ at a point $P$ and volume element $dV=ld\Sigma \cos{\theta}$ is the volume that is intercepted in volume $V$. If we take this further, and integrate over all $V$ and $\Omega$ we arrive at

$\displaystyle \frac{d\nu}{c}\int dV \int I_{\nu} d\Omega=\frac{V}{c}d\nu \int I_{\nu}d\Omega, (5)$

where if the radiation travels some distance $L$ in the volume, then we must multiply Eq.(5) by $l/c$, where $c$ is the speed of light.

We now define the integrated energy density as being

$U_{\nu}=\displaystyle \frac{1}{c}\int I_{\nu}d\Omega, (6.1)$

while the average intensity is

$J_{\nu}=\displaystyle \frac{1}{4\pi}\int I_{\nu}d\nu, (6.2)$

and the relation between these two equations is

$U_{\nu}=\frac{4\pi}{c}J_{\nu}. (6.3)$

I will now introduce the radiative transfer equation. This equation is a balance between the amount of radiation absorbed and the radiation that is emitted. The equation is,

$\frac{dI_{\nu}}{ds}=-\epsilon \rho I_{\nu}+h_{\nu}\rho, (7)$

where if we divide by $\epsilon \rho$ we get

$-\frac{1}{\epsilon_{\nu}\rho}\frac{dI_{\nu}}{ds}=I_{\nu}+U_{\nu}(\theta, \phi), (8)$

where $U(\theta,\phi)$ represents the source function given by

$U_{\nu}(\theta,\phi)=\displaystyle \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}p(\theta,\phi;\theta^{\prime},\phi^{\prime})I_{\nu}\sin{\theta^{\prime}}d\theta^{\prime}d\phi^{\prime}. (9)$

The source function is typically the ratio between the absorption and emission coefficients. One of the terms in the source function is the phase function which varies according to the specific scattering geometry. In its most general form, we can represent the phase function as an expansion of Legendre polynomials:

$p(\theta, \phi; \theta^{\prime},\phi^{\prime})=\displaystyle \sum_{j=0}^{\infty}\gamma_{j}P_{j}(\mu), (10)$

where we have let $\mu = \cos{\theta}$ (in keeping with our notation in previous posts).

In Part II, we will discuss a few simple cases of scattering and their corresponding phase functions, as well as obtaining the formal solution of the radiative transfer equation. (DISCLAIMER: While this solution will be consistent in a mathematical sense, it is not exactly an insightful solution since much of the more interesting and complex cases involve the solution of either integro-differential equations or pure integral equations (a possible new topic).)