# 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.