To this point, we have made not said anything of the \(\mathrm{P_{1}}\) approximation so the previous equations holds for any solution method. Now we have to depart upon the final act of the derivation requires an argument I find a bit too delicate for my taste. Perhaps there is something more convincing that can be said. The first part is easy, and I doubt anyone will have a problem with it: Any intensity existing in the particle (medium 1) we have chosen to represent with the \(\mathrm{P_{1}}\) approximation. Since the \(\mathrm{P_{1}}\) approximation only considers location and pays no mind to direction when assigning an approximate intensity, we know the intensities within the particle (medium 1) are represented by the same incident irradiation, \(G\),
\begin{equation} I^{+}_{1,\nu}=I^{-}_{1,\nu}\approx\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right],\\ \end{equation}where \(A_{1}\) is the linear coefficient in the linear anisotropic scattering phase function approximation, and \(\tau\) is the optical depth. Further—and this is where I think the ice we stand on is a bit thin—we can say any intensity leaving the particle from this infinitely thin interface must also have the same intensity since it originated from the particle. So,
\begin{equation} I^{+}_{2,\nu}\approx\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right],\\ \end{equation}These last two statements seem strange because they are far from true in general. In the \(\mathrm{P_{1}}\) approximation, however, I think they stand (please tell me why it is wrong!). Substituting we get,
\begin{align} \label{eq:longmess} \int_{0}^{\infty} & \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta)]\frac{1}{n_{1}^{2}}\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right]\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu \\ + & \int_{0}^{\infty}\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}\frac{1}{n^{2}_{1}}\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right]\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu\notag \\ = & \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\frac{1}{n^{2}_{2}}\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right]\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu\notag \\ + & \label{eq:longmess}\int_{0}^{\infty}\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}[1-\rho^{\prime}_{2\rightarrow 1,\nu}(\theta)]\frac{I^{-}_{2,\nu}}{n_{2}^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu\notag \\ \end{align}Icky.
You probably recognize these integrals as the ones from Modest we were fussing over to decide which way the signs should go. It is Eq. (15.43) in my second edition. For completeness, I am going to fill in the details to make sure I am not ”pretending to understand.” I find reading too many books and not doing enough math/programming yourself can lead to this ”condition.” Anyway, the ugly part is the integral over \(\nabla_{\tau}G\cdot\hat{s}\). Lets look at what this quantity looks when we pull it apart,
\begin{equation} \label{eq:nablaGs} \label{eq:nablaGs}\nabla_{\tau}G\cdot\hat{s}=[\nabla_{\tau}G]_{t1}\sin\theta\cos\psi+[\nabla_{\tau}G]_{t2}\sin\theta\sin\psi+[\nabla_{\tau}G]_{n}\cos\theta.\\ \end{equation}The funny bracket notation corresponds to which component of the gradient we are after. Since we are working with gradient in terms of the optical depth, there is no \(x,y,z\), just a coordinate system along the path of travel, so we can define \(t1\) and \(t2\) to be the components tangent to the surface and \(n\) to be the normal. The components of \(\hat{s}\) have been chosen accordingly. So. I would substitute this stuff into (\ref{eq:longmess}) but that would take up a ton of space. Instead, note that the only \(\psi\) depenedence in the integrations appear in the two tangent components as integrals of \(\sin\psi\) and \(\cos\psi\) over the full \(0\) to \(2\pi\). These are both zero. Thus the tangent components die and the normal component lives to fight another day. Let’s substitute (\ref{eq:nablaGs}) and carry out the integrals over \(\psi\) to get,
\begin{align} \frac{1}{2}\int_{0}^{\infty} & \int_{0}^{\frac{\pi}{2}}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta)]\frac{1}{n_{1}^{2}}\left[G-\frac{3}{3-A_{1}\omega}\frac{\partial G}{\partial n}\cos\theta\right]\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\nu \\ + & \frac{1}{2}\int_{0}^{\infty}\int_{\frac{\pi}{2}}^{\pi}\frac{1}{n^{2}_{1}}\left[G-\frac{3}{3-A_{1}\omega}\frac{\partial G}{\partial n}\cos\theta\right]\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\nu\notag \\ = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}\frac{1}{n^{2}_{2}}\left[G-\frac{3}{3-A_{1}\omega}\frac{\partial G}{\partial n}\cos\theta\right]\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\nu\notag \\ + & \int_{0}^{\infty}\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}[1-\rho^{\prime}_{2\rightarrow 1,\nu}(\theta)]\frac{I^{-}_{2,\nu}}{n_{2}^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu.\notag \\ \end{align}We are left with a bunch of integrals that we can evaluate exactly:
\begin{equation} \int_{0}^{\pi/2}\cos\theta\sin\theta\,\mathrm{d}\theta=\frac{1}{2},\quad\int_{0}^{\pi/2}\cos^{2}\theta\sin\theta\,\mathrm{d}\theta=\frac{1}{3}\\ \end{equation}and
\begin{equation} \int_{\pi/2}^{\pi}\cos\theta\sin\theta\,\mathrm{d}\theta=-\frac{1}{2},\quad\int_{\pi/2}^{\pi}\cos^{2}\theta\sin\theta\,\mathrm{d}\theta=\frac{1}{3}\\ \end{equation}Thanks Mathematica. After introducing the hemispherical reflectance \(\rho_{i\rightarrow j,\nu}\),
\begin{equation} \rho_{i\rightarrow j,\nu}=\frac{\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\rho^{\prime}_{i\rightarrow j,\nu}I\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi}{\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}I\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi},\\ \end{equation}we are left with,
\begin{align} \frac{1}{2}\int_{0}^{\infty} & [1-\rho_{1\rightarrow 2,\nu}]\frac{1}{n_{1}^{2}}\left[\left(\frac{1}{2}\right)G-\left(\frac{1}{3}\right)\frac{3}{3-A_{1}\omega}\frac{\partial G}{\partial n}\right]\,\mathrm{d}\nu \\ + & \frac{1}{2}\int_{0}^{\infty}\frac{1}{n^{2}_{1}}\left[\left(-\frac{1}{2}\right)G-\left(\frac{1}{3}\right)\frac{3}{3-A_{1}\omega}\frac{\partial G}{\partial n}\right]\,\mathrm{d}\nu\notag \\ = & \frac{1}{2}\int_{0}^{\infty}\frac{1}{n^{2}_{2}}\left[\left(\frac{1}{2}\right)G-\left(\frac{1}{3}\right)\frac{3}{3-A_{1}\omega}\frac{\partial G}{\partial n}\right]\,\mathrm{d}\nu\notag \\ + & \int_{0}^{\infty}\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}[1-\rho_{2\rightarrow 1,\nu}]\frac{I^{-}_{2,\nu}}{n_{2}^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu.\notag \\ \end{align}To simplify, let’s remove the integration over frequency and do some rearranging. Note that since we are assuming linear scattering, removing the frequency integrals is absolutely allowed.