The Equation of Transfer in Integral Form
Let LLL be the streaming-collision operator, and SSS is scattering operator, we have
LI=Ω⋅∇I(r,Ω)+σ(r,Ω)I(r,Ω)
LI=\Omega\cdot\nabla I(r,\Omega)+\sigma(r,\Omega)I(r,\Omega)
LI=Ω⋅∇I(r,Ω)+σ(r,Ω)I(r,Ω)
and
SI=∫4πσs(r,Ω′,Ω)I(r,Ω′)dΩ′
SI=\int_{4\pi}\sigma_s(r,\Omega',\Omega)I(r,\Omega')d\Omega'
SI=∫4πσs(r,Ω′,Ω)I(r,Ω′)dΩ′
and using RRR to denote the scattering operator on the boundary δV\delta VδV for the intensity I+I^+I+ of medium leaving radiation is introduced as
RI+=1πdrb′∫2π−ρbμ′I(r,Ω′)dΩ′
RI^+=\frac{1}{\pi}dr_b'\int_{2\pi-}\rho_b\mu'I(r,\Omega')d\Omega'
RI+=π1drb′∫2π−ρbμ′I(r,Ω′)dΩ′
Using this notions, we can wirte the stationaty radiative transfer equation as
LI=SI+q, I−=RI++qb
LI=SI+q,\ I^-=RI^++q_b
LI=SI+q, I−=RI++qb
If RRR = 0 and qb=0q_b=0qb=0, then the boundary value problem is called standard problem. In this case, we set use L0L_0L0 to denote the streaming-collision operator.
For standard problem, the integral need to find the L0−1L_0^{-1}L0−1. Let J=SI+qJ=SI+qJ=SI+q, and uuu represetns either SISISI or JJJ, the function v=L0−1uv=L_0^{-1}uv=L0−1u satisfies the equation
Ω⋅∇v(r,Ω)+σ(r,Ω)v(r,Ω)=u(r,Ω)
\Omega\cdot\nabla v(r,\Omega)+\sigma(r,\Omega)v(r,\Omega)=u(r,\Omega)
Ω⋅∇v(r,Ω)+σ(r,Ω)v(r,Ω)=u(r,Ω)
with zero boundary condition, i.e.,u(rb,Ω)=0,n(rb)⋅Ω<0i.e., u(r_b,\Omega)=0, n(r_b)\cdot\Omega<0i.e.,u(rb,Ω)=0,n(rb)⋅Ω<0.
Consider a stright line rb+ηΩr_b+\eta\Omegarb+ηΩ, along an incoming direction Ω\OmegaΩ, n(rb)⋅Ω<0n(r_b)\cdot\Omega <0n(rb)⋅Ω<0, this equation takes the following form
dv(rb+ξΩ,Ω)dξ+σ(rb+ξΩ,Ω)v(rb+ξΩ,Ω)=u(rb+ξΩ,Ω),v(rb,Ω)=0.
\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.
dξdv(rb+ξΩ,Ω)+σ(rb+ξΩ,Ω)v(rb+ξΩ,Ω)=u(rb+ξΩ,Ω),v(rb,Ω)=0.
This is an ODE w.r.t ξ\xiξ. The integral yields
v(rb+ξΩ,Ω)=∫0ξe∫ξξ′σ(rb+ξ′′Ω,Ω)dξ′′u(rb+ξ′Ω,Ω)dξ′
v(r_b+\xi\Omega,\Omega)=\int_0^{\xi}e^{\int_{\xi}^{\xi'}\sigma(r_b+\xi''\Omega,\Omega)d\xi''}u(r_b+\xi'\Omega,\Omega)d\xi'
v(rb+ξΩ,Ω)=∫0ξe∫ξξ′σ(rb+ξ′′Ω,Ω)dξ′′u(rb+ξ′Ω,Ω)dξ′
which is equivelent to
v(rb+ξΩ,Ω)=∫4π∫0ξe∫ξξ′σ(rb+ξ′′Ω′,Ω′)dξ′′u(rb+ξ′Ω′,Ω′)δ(Ω,Ω′)dΩ′dξ′
v(r_b+\xi\Omega,\Omega)=\int_{4\pi}\int_0^{\xi}e^{\int_{\xi}^{\xi'}\sigma(r_b+\xi''\Omega',\Omega')d\xi''}u(r_b+\xi'\Omega',\Omega')\delta(\Omega,\Omega')d\Omega'd\xi'
v(rb+ξΩ,Ω)=∫4π∫0ξe∫ξξ′σ(rb+ξ′′Ω′,Ω′)dξ′′u(rb+ξ′Ω′,Ω′)δ(Ω,Ω′)dΩ′dξ′
Now, let rrr and r′=r−ξ′Ω′r'=r-\xi'\Omega'r′=r−ξ′Ω′ be two points on line rb+ηΩ′r_b+\eta\Omega'rb+ηΩ′. The volumn elements in this point is ξ2dΩdξ\xi^2d\Omega d\xiξ2dΩdξ , and ∥r−r′∥=ξ′\|r-r'\|=\xi'∥r−r′∥=ξ′, so Ω′=r−r′∥r−r′∥\Omega'=\frac{r-r'}{\|r-r'\|}Ω′=∥r−r′∥r−r′. Then, Eq. (9) can be convert to
L0−1u=v(r,Ω)=∫Ve−τ(r,r′,Ω)∥r−r′∥2u(r′,Ω)δ(Ω,r−r′∥r−r′∥)dr′
L_0^{-1}u=v(r,\Omega)=\int_{V}\frac{e^{-\tau(r,r',\Omega)}}{\|r-r'\|^2}u(r',\Omega)\delta(\Omega,\frac{r-r'}{\|r-r'\|})dr'
L0−1u=v(r,Ω)=∫V∥r−r′∥2e−τ(r,r′,Ω)u(r′,Ω)δ(Ω,∥r−r′∥r−r′)dr′
Here, τ(r,r′,Ω)\tau(r,r',\Omega)τ(r,r′,Ω) is the optical distance between rrr and r′r'r′ along Ω\OmegaΩ, which is defined as
τ(r,r′,Ω)=∫0ξ′dξ′′σ(r−ξ′′Ω,Ω).
\tau(r,r',\Omega)=\int_{0}^{\xi'}d\xi''\sigma(r-\xi''\Omega,\Omega).
τ(r,r′,Ω)=∫0ξ′dξ′′σ(r−ξ′′Ω,Ω).
Here, noting that original representation in Eq. (8) is from ξ\xiξ to ξ′\xi'ξ′ as dummy variable, which is discribe the integral from rrr to r′r'r′. So here in Eq. (11), we integral from 000 (means r) to ξ′\xi'ξ′ (means r−ξ′Ωr-\xi'\Omegar−ξ′Ω), which is defined to be r′r'r′. The Eq. (10) describe the 3-D distribution v(r,Ω)v(r,\Omega)v(r,Ω) of photons from the source uuu arrive at point rrr along Ω\OmegaΩ without suffering a collision. Substitude u=SIu=SIu=SI into Eq. (10) we have
I(r,Ω)=∫VKI(r′,Ω′,Ω)I(r′,Ω′)dΩ′dr′+Q(r,Ω)
I(r,\Omega)=\int_{V}\mathcal{K}_I(r',\Omega',\Omega)I(r',\Omega')d\Omega'dr' + Q(r,\Omega)
I(r,Ω)=∫VKI(r′,Ω′,Ω)I(r′,Ω′)dΩ′dr′+Q(r,Ω)
where
KI(r′,Ω′,Ω)=e−τ(r,r′,Ω)∥r−r′∥2σs(r′,Ω′,Ω)δ(Ω,r−r′∥r−r′∥)
\mathcal{K}_I(r',\Omega',\Omega)=\frac{e^{-\tau(r,r',\Omega)}}{\|r-r'\|^2}\sigma_s(r',\Omega',\Omega)\delta(\Omega,\frac{r-r'}{\|r-r'\|})
KI(r′,Ω′,Ω)=∥r−r′∥2e−τ(r,r′,Ω)σs(r′,Ω′,Ω)δ(Ω,∥r−r′∥r−r′)
and Q=L0−1qQ=L^{-1}_0qQ=L0−1q is calculated using Eq. (10). KI\mathcal{K}_IKI is transition density, means that KIdr′dΩ\mathcal{K}_Idr'd\OmegaKIdr′dΩ is the probability photons which have undergone interactions at r′r'r′ in the direction Ω′\Omega'Ω′ will have their next interaction at rrr along Ω\OmegaΩ.
We can imagine that I(r′,Ω′)I(r',\Omega')I(r′,Ω′) scattered to Ω\OmegaΩ direction and then extincted to rrr.
Multiplying Eq. (10) using differential cattering coefficient σs\sigma_sσs,
σsL0−1u=σs∫Ve−τ(r,r′,Ω)∥r−r′∥2u(r′,Ω)δ(Ω,r−r′∥r−r′∥)dr′
\sigma_sL_0^{-1}u=\sigma_s \int_{V}\frac{e^{-\tau(r,r',\Omega)}}{\|r-r'\|^2}u(r',\Omega)\delta(\Omega,\frac{r-r'}{\|r-r'\|})dr'
σsL0−1u=σs∫V∥r−r′∥2e−τ(r,r′,Ω)u(r′,Ω)δ(Ω,∥r−r′∥r−r′)dr′
and integral
∫4πσs(r,Ω′,Ω)L0−1udΩ′=∫4πσs(r,Ω′,Ω)∫Ve−τ(r,r′,Ω)∥r−r′∥2u(r′,Ω)δ(Ω,r−r′∥r−r′∥)dr′dΩ′=SL0−1u
\begin{aligned}
&\int_{4\pi}\sigma_s(r,\Omega',\Omega)L_{0}^{-1}ud\Omega'=\\&\int_{4\pi}\sigma_s(r,\Omega',\Omega)\int_{V}\frac{e^{-\tau(r,r',\Omega)}}{\|r-r'\|^2}u(r',\Omega)\delta(\Omega,\frac{r-r'}{\|r-r'\|})dr'd\Omega'\\
&=SL_0^{-1}u
\end{aligned}
∫4πσs(r,Ω′,Ω)L0−1udΩ′=∫4πσs(r,Ω′,Ω)∫V∥r−r′∥2e−τ(r,r′,Ω)u(r′,Ω)δ(Ω,∥r−r′∥r−r′)dr′dΩ′=SL0−1u
and the kernel Ks\mathcal{K_s}Ks of integral operator SL0−1SL_0^{-1}SL0−1 is
KS=∫4π∫Ve−τ(r,r′,Ω)∥r−r′∥2σs(r,Ω′,Ω)δ(Ω,r−r′∥r−r′∥)
\mathcal{K}_S=\int_{4\pi}\int_V\frac{e^{-\tau(r,r',\Omega)}}{\|r-r'\|^2}\sigma_s(r,\Omega',\Omega)\delta(\Omega,\frac{r-r'}{\|r-r'\|})
KS=∫4π∫V∥r−r′∥2e−τ(r,r′,Ω)σs(r,Ω′,Ω)δ(Ω,∥r−r′∥r−r′)
And the source function JJJ satisfies the following integral equation
J(r,Ω)=SL0−1J+q=∫4π∫VKSJ(r′,Ω′)dr′dΩ′+q(r,Ω)
\begin{aligned}
J(r,\Omega)&=SL_0^{-1}J+q\\
&=\int_{4\pi}\int_{V}\mathcal{K}_SJ(r',\Omega')dr'd\Omega'+q(r,\Omega)
\end{aligned}
J(r,Ω)=SL0−1J+q=∫4π∫VKSJ(r′,Ω′)dr′dΩ′+q(r,Ω)
The III could be expressed via JJJ as I=L0−1JI=L_0^{-1}JI=L0−1J, where L0−1L_0^{-1}L0−1 is in Eq. (10). In many cases, the solution to Eq. (16) is easier for Eq. (40), so using Eq. (16) and using I=L0−1JI=L_0^{-1}JI=L0−1J is a better solution.