# 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,

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