ROUGH DRAFT authorea.com/116890

# Boundary condition for a semitransparent particle under diffuse irradiation using the $$\mathrm{P_{1}}$$ approximation

Here we go. I am going to include most of the piddly mathematical steps to make sure it is clear.

Consider a semitransparent particle which we will denote region 1 surrounded by a transparent medium denoted region 2. The derivation starts with a statement that says when radiation at a particular frequency passes from region 1 to region 2, energy is conserved. So at the interface, we can write,

$$dQ_{1}^{+}=dQ_{2}^{+}\\$$

where the positive sign indicates the radiation is moving in the positive direction which we define to mean going from the particle to the surrounding medium. Then using the definition of spectral intensity $$I^{+}_{1,\nu}$$ and spectral, directional reflectivity, $$\rho^{\prime}_{1\rightarrow 2,\nu}$$

$$\label{eq:plusbalance}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta_{1})]I^{+}_{1,\nu}(\hat{s}_{1}\cdot\hat{n})\,\mathrm{d}\Omega_{1}\,\mathrm{d}A_{1}\,\mathrm{d}\nu_{1}=I^{+}_{2,\nu}(\hat{s}_{2}\cdot\hat{n})\,\mathrm{d}\Omega_{2}\,\mathrm{d}A_{2}\,\mathrm{d}\nu_{2}\\$$

where $$\nu$$ is the frequency of radiation traveling in the direction $$\hat{s}$$ impinging upon a surface with a unit normal vector $$\hat{n}$$ and area $$\,\mathrm{d}A$$. This is precisely Eq. (17.39) in (Howell 2011). I will expound on the exact definition of the reflectivity and arrow notation I have adopted later. We have chosen to define intensity in terms of frequency. Since frequency does not change with refractive index, we can say

$$\label{eq:nu}\nu_{1}=\nu_{2}.\\$$

To simplify (\ref{eq:plusbalance}), we can introduce the definition of the solid angle $$\,\mathrm{d}\Omega_{i}=\sin\theta_{i}\,\mathrm{d}\theta_{i}\,\mathrm{d}\psi_{i}$$ and the dot product of two unit length vectors, $$(\hat{s}_{i}\cdot\hat{n})=\cos\theta_{i}$$. We are using Modest’s notation; $$\theta$$ is the polar and $$\psi$$ is the azimuthal angle. Substituting gives

$$\label{eq:plusbalancesub}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta_{1})]I^{+}_{1,\nu}\cos\theta_{1}\sin\theta_{1}\,\mathrm{d}\theta_{1}\,\mathrm{d}\psi_{1}\,\mathrm{d}A_{1}\,\mathrm{d}\nu_{1}=I^{+}_{2,\nu}\cos\theta_{2}\sin\theta_{2}\,\mathrm{d}\theta_{2}\,\mathrm{d}\psi_{2}\,\mathrm{d}A_{2}\,\mathrm{d}\nu_{2}\\$$

Now we can bring in some results from electromagnetic theory. We assume we have a planar interface—a fine assumption since the particle is so much bigger than the wavelengths of light under consideration. We know the azimuthal angle does not change when a wave passes between media with different refractive indices. So

$$\label{eq:azi}\,\mathrm{d}\psi_{1}=\,\mathrm{d}\psi_{2}.\\$$

However, the polar angle does change in a very well known way according to Snell’s law (more appropriately called Sahl’s law),

$$\label{eq:sahl}n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2},\\$$

where $$n_{i}$$ is the complex index of refraction of medium $$i$$. Further, we can differentiate both sides by $$\theta_{1}$$ to get,

$$\label{eq:dsahl}\frac{\mathrm{d}}{\mathrm{d}\theta_{1}}\left[n_{1}\sin\theta_{1}\right]=\frac{\mathrm{d}\theta_{2}}{\mathrm{d}\theta_{1}}\frac{\mathrm{d}}{\mathrm{d}\theta_{2}}\left[n_{2}\sin\theta_{2}\right]\rightarrow n_{1}\cos\theta_{1}\,\mathrm{d}\theta_{1}=n_{2}\cos\theta_{2}\,\mathrm{d}\theta_{2}.\\$$

Substituting (\ref{eq:nu}), (\ref{eq:azi}), (\ref{eq:sahl}) and (\ref{eq:dsahl}) into (\ref{eq:plusbalancesub}) gives

$$\label{eq:plusbalancesimp}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta_{1})]I^{+}_{1,\nu}\cos\theta_{1}\sin\theta_{1}\,\mathrm{d}\theta_{1}\,\mathrm{d}\psi_{1}\,\mathrm{d}\nu_{1}=I^{+}_{2,\nu}\left(\frac{n_{1}}{n_{2}}\cos\theta_{1}\right)\left(\frac{n_{1}}{n_{2}}\sin\theta_{1}\,\mathrm{d}\theta_{1}\right)\,\mathrm{d}\psi_{1}\,\mathrm{d}\nu_{1}\\$$

Dropping the subscripts on the angles and frequency then rearranging we get,

$$\label{eq:I12}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta)]\frac{I^{+}_{1,\nu}}{n_{1}^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu=\frac{I^{+}_{2,\nu}}{n^{2}_{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu\\$$

So let’s have a look at the final result. It says: “The amount of energy carried by a ray within the particle in some direction impinging on an interface with a certain reflectivity must be the same as the energy carried away in some other direction as decided by Snell’s law.” This seems reasonable. The exact same statement must must be true for rays coming from medium 2 to medium 1. So if we were to follow the exact same arguments in the other direction, we would get

$$\label{eq:I21}[1-\rho^{\prime}_{2\rightarrow 1,\nu}(\theta)]\frac{I^{-}_{2,\nu}}{n_{2}^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu=\frac{I^{-}_{1,\nu}}{n^{2}_{1}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu\\$$

Note that the reflectivity in this statement is different from the previous one because here we define the reflectivity in terms of an electromagnetic wave going from medium 2 with refractive index $$n_{2}$$ to medium 1 with refractive index $$n_{1}$$. The order of refractive index does matter. The case of a wave traveling from non-absorbing to an absorbing material is covered in Modest. The case of a wave traveling from an absorbing material to a non-absorbing material is not. I am not sure if this result appears anywhere explicitly, but it can be derived with only a tolerable amount of pain by following the development of the other case.

We are now able to write down in detail what it means to conserve the total energy going through the semitransparent–transparent interface. The net heat flux due to all the rays at all frequencies on the semitransparent–transparent interface must equal the net heat flux leaving the same interface. That is,

$$q_{1}^{+}+q_{1}^{-}=q_{2}^{+}+q_{2}^{-}\\$$

This expression can be obtained by the logic we used to introduce it, or by simply adding (\ref{eq:I12}) and (\ref{eq:I21}) and integrating over solid angle and frequency. That is,

\begin{align} \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}[1-\rho^{\prime}_{1\rightarrow 2,\nu}(\theta)]\frac{I^{+}_{1,\nu}}{n_{1}^{2}} & \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{I^{-}_{1,\nu}}{n^{2}_{1}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi\,\mathrm{d}\nu \\ = & \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\frac{I^{+}_{2,\nu}}{n^{2}_{2}}\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}[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}

It should be noted here that the integration limits reflect the two different directions according to polar angle denoted by the $${}^{+}$$ and $${}^{-}$$ superscripts, i.e. the parts of the integral that zero because the intensity must be zero have been removed.

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, $$G$$,

$$I^{+}_{1,\nu}=I^{-}_{1,\nu}\approx\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right],\\$$

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,

$$I^{+}_{2,\nu}\approx\frac{1}{4\pi}\left[G-\frac{3}{3-A_{1}\omega}\nabla_{\tau}G\cdot\hat{s}\right],\\$$

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} \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,

$$\label{eq:nablaGs}\nabla_{\tau}G\cdot\hat{s}=[\nabla_{\tau}G]_{t1}\sin\theta\cos\psi+[\nabla_{\tau}G]_{t2}\sin\theta\sin\psi+[\nabla_{\tau}G]_{n}\cos\theta.\\$$

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:

$$\int_{0}^{\pi/2}\cos\theta\sin\theta\,\mathrm{d}\theta=\frac{1}{2},\quad\int_{0}^{\pi/2}\cos^{2}\theta\sin\theta\,\mathrm{d}\theta=\frac{1}{3}\\$$

and

$$\int_{\pi/2}^{\pi}\cos\theta\sin\theta\,\mathrm{d}\theta=-\frac{1}{2},\quad\int_{\pi/2}^{\pi}\cos^{2}\theta\sin\theta\,\mathrm{d}\theta=\frac{1}{3}\\$$

Thanks Mathematica. After introducing the hemispherical reflectance $$\rho_{i\rightarrow j,\nu}$$,

$$\rho_{i\rightarrow j,\nu}=\frac{\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\rho^{\prime}_{i\rightarrow j,\nu}I\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi}{\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}I\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi},\\$$

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.

\begin{align} \frac{1}{4}[1-\rho_{1\rightarrow 2,\nu}]\frac{1}{n_{1}^{2}} & \left[G-\frac{2}{3-A_{1}\omega}\frac{\partial G}{\partial n}\right]+\frac{1}{4}\frac{1}{n^{2}_{1}}\left[-G-\frac{2}{3-A_{1}\omega}\frac{\partial G}{