The Equation of Transfer in Integral Form

Let $L$ be the streaming-collision operator, and $S$ is scattering operator, we have
$$LI=\Omega \cdot \nabla I(r,\Omega )+\sigma (r,\Omega )I(r,\Omega )$$
and
$$SI={\int }_{4\pi }{\sigma }_s(r,{\Omega }^{\prime },\Omega )I(r,{\Omega }^{\prime })d{\Omega }^{\prime }$$
and using $R$ to denote the scattering operator on the boundary $\delta V$ for the intensity $I^{+}$ of medium leaving radiation is introduced as
$$R{I}^{+}=\frac{1}{\pi }d{r}_b^{\prime }{\int }_{2\pi -}{\rho }_b{\mu }^{\prime }I(r,{\Omega }^{\prime })d{\Omega }^{\prime }$$
Using this notions, we can wirte the stationaty radiative transfer equation as
$$LI=SI+q,I^{-}=R{I}^{+}+q_b$$
If $R$ = 0 and $q_b=0$, then the boundary value problem is called standard problem. In this case, we set use $L_0$ to denote the streaming-collision operator.

For standard problem, the integral need to find the $L_0^{-1}$. Let $J=SI+q$, and $u$ represetns either $SI$ or $J$, the function $v=L_0^{-1}u$ satisfies the equation
$$\Omega \cdot \nabla v(r,\Omega )+\sigma (r,\Omega )v(r,\Omega )=u(r,\Omega )$$
with zero boundary condition, $i.e.,u(r_b,\Omega )=0,n(r_b)\cdot \Omega <0$.

Consider a stright line $r_b+\eta \Omega$, along an incoming direction $\Omega$, $n(r_b)\cdot \Omega <0$, this equation takes the following form
$$\frac{dv(r_b+\xi \Omega ,\Omega )}{d\xi }+\sigma (r_b+\xi \Omega ,\Omega )v(r_b+\xi \Omega ,\Omega )=u(r_b+\xi \Omega ,\Omega ),v(r_b,\Omega )=0.$$
This is an ODE w.r.t $\xi$. The integral yields
$$v(r_b+\xi \Omega ,\Omega )={\int }_0^{\xi }{e}^{{\int }_{\xi }^{{\xi }^{\prime }}\sigma (r_b+{\xi }^{\prime \prime }\Omega ,\Omega )d{\xi }^{\prime \prime }}u(r_b+{\xi }^{\prime }\Omega ,\Omega )d{\xi }^{\prime }$$
which is equivelent to
$$v(r_b+\xi \Omega ,\Omega )={\int }_{4\pi }{\int }_0^{\xi }{e}^{{\int }_{\xi }^{{\xi }^{\prime }}\sigma (r_b+{\xi }^{\prime \prime }{\Omega }^{\prime },{\Omega }^{\prime })d{\xi }^{\prime \prime }}u(r_b+{\xi }^{\prime }{\Omega }^{\prime },{\Omega }^{\prime })\delta (\Omega ,{\Omega }^{\prime })d{\Omega }^{\prime }d{\xi }^{\prime }$$
Now, let $r$ and $r^{\prime }=r-{\xi }^{\prime }{\Omega }^{\prime }$ be two points on line $r_b+\eta {\Omega }^{\prime }$. The volumn elements in this point is ${\xi }^{2}d\Omega d\xi$ , and $\Vert r-r^{\prime }\Vert ={\xi }^{\prime }$, so ${\Omega }^{\prime }=\frac{r-r^{\prime }}{\Vert r-r^{\prime }\Vert }$. Then, Eq. (9) can be convert to
$$L_0^{-1}u=v(r,\Omega )={\int }_V\frac{e^{-\tau (r,r^{\prime },\Omega )}}{\Vert r-r^{\prime }{\Vert }^2}u(r^{\prime },\Omega )\delta (\Omega ,\frac{r-r^{\prime }}{\Vert r-r^{\prime }\Vert })d{r}^{\prime }$$
Here, $\tau (r,r^{\prime },\Omega )$ is the optical distance between $r$ and $r^{\prime }$ along $\Omega$, which is defined as
$$\tau (r,r^{\prime },\Omega )={\int }_0^{{\xi }^{\prime }}d{\xi }^{\prime \prime }\sigma (r-{\xi }^{\prime \prime }\Omega ,\Omega ).$$
Here, noting that original representation in Eq. (8) is from $\xi$ to ${\xi }^{\prime }$ as dummy variable, which is discribe the integral from $r$ to $r^{\prime }$. So here in Eq. (11), we integral from $0$ (means r) to ${\xi }^{\prime }$ (means $r-{\xi }^{\prime }\Omega$), which is defined to be $r^{\prime }$​. The Eq. (10) describe the 3-D distribution $v(r,\Omega )$ of photons from the source $u$ arrive at point $r$ along $\Omega$ without suffering a collision. Substitude $u=SI$ into Eq. (10) we have
$$I(r,\Omega )={\int }_V{\mathcal{K}}_I(r^{\prime },{\Omega }^{\prime },\Omega )I(r^{\prime },{\Omega }^{\prime })d{\Omega }^{\prime }d{r}^{\prime }+Q(r,\Omega )$$
where
$${\mathcal{K}}_I(r^{\prime },{\Omega }^{\prime },\Omega )=\frac{e^{-\tau (r,r^{\prime },\Omega )}}{\Vert r-r^{\prime }{\Vert }^2}{\sigma }_s(r^{\prime },{\Omega }^{\prime },\Omega )\delta (\Omega ,\frac{r-r^{\prime }}{\Vert r-r^{\prime }\Vert })$$
and $Q=L_0^{-1}q$ is calculated using Eq. (10). ${\mathcal{K}}_I$ is transition density, means that ${\mathcal{K}}_{I}d{r}^{\prime }d\Omega$ is the probability photons which have undergone interactions at $r^{\prime }$ in the direction ${\Omega }^{\prime }$ will have their next interaction at $r$ along $\Omega$.

We can imagine that $I(r^{\prime },{\Omega }^{\prime })$ scattered to $\Omega$ direction and then extincted to $r$.

Multiplying Eq. (10) using differential cattering coefficient ${\sigma }_s$​,
$${\sigma }_s{L}_0^{-1}u={\sigma }_s{\int }_V\frac{e^{-\tau (r,r^{\prime },\Omega )}}{\Vert r-r^{\prime }{\Vert }^2}u(r^{\prime },\Omega )\delta (\Omega ,\frac{r-r^{\prime }}{\Vert r-r^{\prime }\Vert })d{r}^{\prime }$$
and integral
$$\begin{array}{rl}& {\int }_{4\pi }{\sigma }_s(r,{\Omega }^{\prime },\Omega )L_0^{-1}ud{\Omega }^{\prime }=\\ & {\int }_{4\pi }{\sigma }_s(r,{\Omega }^{\prime },\Omega ){\int }_V\frac{e^{-\tau (r,r^{\prime },\Omega )}}{\Vert r-r^{\prime }{\Vert }^2}u(r^{\prime },\Omega )\delta (\Omega ,\frac{r-r^{\prime }}{\Vert r-r^{\prime }\Vert })d{r}^{\prime }d{\Omega }^{\prime }\\ & =S{L}_0^{-1}u\end{array}$$
and the kernel ${\mathcal{K}}_s$ of integral operator $S{L}_0^{-1}$ is
$${\mathcal{K}}_S={\int }_{4\pi }{\int }_V\frac{e^{-\tau (r,r^{\prime },\Omega )}}{\Vert r-r^{\prime }{\Vert }^2}{\sigma }_s(r,{\Omega }^{\prime },\Omega )\delta (\Omega ,\frac{r-r^{\prime }}{\Vert r-r^{\prime }\Vert })$$
And the source function $J$ satisfies the following integral equation
$$\begin{array}{rl}J(r,\Omega )& =S{L}_0^{-1}J+q\\ & ={\int }_{4\pi }{\int }_V{\mathcal{K}}_{S}J(r^{\prime },{\Omega }^{\prime })d{r}^{\prime }d{\Omega }^{\prime }+q(r,\Omega )\end{array}$$
The $I$ could be expressed via $J$ as $I=L_0^{-1}J$, where $L_0^{-1}$ is in Eq. (10). In many cases, the solution to Eq. (16) is easier for Eq. (40), so using Eq. (16) and using $I=L_0^{-1}J$ is a better solution.