The Equation of Transfer in Integral Form
Let L L L be the streaming-collision operator, and S S S is scattering operator, we have
L I = 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
S I = ∫ 4 π σ s ( r , Ω ′ , Ω ) I ( r , Ω ′ ) d Ω ′ 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 R R R to denote the scattering operator on the boundary δ V \delta V δV for the intensity I + I^+ I+ of medium leaving radiation is introduced as
R I + = 1 π d r b ′ ∫ 2 π − ρ b μ ′ I ( r , Ω ′ ) d Ω ′ 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
L I = S I + q , I − = R I + + q b LI=SI+q,\ I^-=RI^++q_b LI=SI+q, I−=RI++qb
If R R R = 0 and q b = 0 q_b=0 qb=0, then the boundary value problem is called standard problem. In this case, we set use L 0 L_0 L0 to denote the streaming-collision operator.
For standard problem, the integral need to find the L 0 − 1 L_0^{-1} L0−1. Let J = S I + q J=SI+q J=SI+q, and u u u represetns either S I SI SI or J J J, the function v = L 0 − 1 u v=L_0^{-1}u v=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 ( r b , Ω ) = 0 , n ( r b ) ⋅ Ω < 0 i.e., u(r_b,\Omega)=0, n(r_b)\cdot\Omega<0 i.e.,u(rb,Ω)=0,n(rb)⋅Ω<0.
Consider a stright line r b + η Ω r_b+\eta\Omega rb+ηΩ, along an incoming direction Ω \Omega Ω, n ( r b ) ⋅ Ω < 0 n(r_b)\cdot\Omega <0 n(rb)⋅Ω<0, this equation takes the following form
d v ( r b + ξ Ω , Ω ) d ξ + σ ( r b + ξ Ω , Ω ) v ( r b + ξ Ω , Ω ) = u ( r b + ξ Ω , Ω ) , v ( r b , 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 \xi ξ. The integral yields
v ( r b + ξ Ω , Ω ) = ∫ 0 ξ e ∫ ξ ξ ′ σ ( r b + ξ ′ ′ Ω , Ω ) d ξ ′ ′ u ( r b + ξ ′ Ω , Ω ) d ξ ′ 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 ( r b + ξ Ω , Ω ) = ∫ 4 π ∫ 0 ξ e ∫ ξ ξ ′ σ ( r b + ξ ′ ′ Ω ′ , Ω ′ ) d ξ ′ ′ u ( r b + ξ ′ Ω ′ , Ω ′ ) δ ( Ω , Ω ′ ) d Ω ′ d ξ ′ 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 r r r and r ′ = r − ξ ′ Ω ′ r'=r-\xi'\Omega' r′=r−ξ′Ω′ be two points on line r b + η Ω ′ r_b+\eta\Omega' rb+ηΩ′. The volumn elements in this point is ξ 2 d Ω d ξ \xi^2d\Omega d\xi ξ2dΩdξ , 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
L 0 − 1 u = v ( r , Ω ) = ∫ V e − τ ( r , r ′ , Ω ) ∥ r − r ′ ∥ 2 u ( r ′ , Ω ) δ ( Ω , r − r ′ ∥ r − r ′ ∥ ) d r ′ 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 r r r and r ′ r' r′ along Ω \Omega Ω, which is defined as
τ ( r , r ′ , Ω ) = ∫ 0 ξ ′ d ξ ′ ′ 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 \xi ξ to ξ ′ \xi' ξ′ as dummy variable, which is discribe the integral from r r r to r ′ r' r′. So here in Eq. (11), we integral from 0 0 0 (means r) to ξ ′ \xi' ξ′ (means r − ξ ′ Ω r-\xi'\Omega r−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 u u u arrive at point r r r along Ω \Omega Ω without suffering a collision. Substitude u = S I u=SI u=SI into Eq. (10) we have
I ( r , Ω ) = ∫ V K I ( r ′ , Ω ′ , Ω ) I ( r ′ , Ω ′ ) d Ω ′ d r ′ + 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
K I ( r ′ , Ω ′ , Ω ) = e − τ ( r , r ′ , Ω ) ∥ r − r ′ ∥ 2 σ s ( 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 = L 0 − 1 q Q=L^{-1}_0q Q=L0−1q is calculated using Eq. (10). K I \mathcal{K}_I KI is transition density, means that K I d r ′ d Ω \mathcal{K}_Idr'd\Omega KIdr′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 r r r along Ω \Omega Ω.
We can imagine that I ( r ′ , Ω ′ ) I(r',\Omega') I(r′,Ω′) scattered to Ω \Omega Ω direction and then extincted to r r r.
Multiplying Eq. (10) using differential cattering coefficient σ s \sigma_s σs,
σ s L 0 − 1 u = σ s ∫ V e − τ ( r , r ′ , Ω ) ∥ r − r ′ ∥ 2 u ( r ′ , Ω ) δ ( Ω , r − r ′ ∥ r − r ′ ∥ ) d r ′ \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
∫ 4 π σ s ( r , Ω ′ , Ω ) L 0 − 1 u d Ω ′ = ∫ 4 π σ s ( r , Ω ′ , Ω ) ∫ V e − τ ( r , r ′ , Ω ) ∥ r − r ′ ∥ 2 u ( r ′ , Ω ) δ ( Ω , r − r ′ ∥ r − r ′ ∥ ) d r ′ d Ω ′ = S L 0 − 1 u \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 K s \mathcal{K_s} Ks of integral operator S L 0 − 1 SL_0^{-1} SL0−1 is
K S = ∫ 4 π ∫ V e − τ ( r , r ′ , Ω ) ∥ r − r ′ ∥ 2 σ s ( 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 J J J satisfies the following integral equation
J ( r , Ω ) = S L 0 − 1 J + q = ∫ 4 π ∫ V K S J ( r ′ , Ω ′ ) d r ′ d Ω ′ + 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 I I I could be expressed via J J J as I = L 0 − 1 J I=L_0^{-1}J I=L0−1J, where L 0 − 1 L_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 = L 0 − 1 J I=L_0^{-1}J I=L0−1J is a better solution.