polarizabilityProof.tex

\noindent $\textrm{Knowing } \frac{\epsilon-1}{\epsilon+2}=\frac{4\pi\alpha}{3d^3}$, \\

\noindent Let $\alpha'=\frac{4\pi\alpha}{3d^3}$ where $\alpha$ is the polarizability \\ and $d$ is the dipole distance in micrometers in DDSCAT.\\\\
\noindent$\implies\frac{\epsilon-1}{\epsilon+2}=\alpha'$
$\implies\epsilon-1=\alpha'(\epsilon+2)$
$\implies\epsilon-1=\alpha'\epsilon+2\alpha'$\\
\noindent$\implies\epsilon=\alpha'\epsilon+2\alpha'+1$
$\implies\epsilon-\alpha'\epsilon=2\alpha'+1$\\
\noindent$\implies\epsilon(1-\alpha')=2\alpha'+1$
$\implies\epsilon=\frac{2\alpha'+1}{1-\alpha'}$\\

\noindent Knowing that $\alpha'$ is complex, let $\alpha'=\textrm{a}+i\textrm{b}$\\
\noindent$\implies\epsilon=\frac{2(\textrm{a}+i\textrm{b})+1}{1-(\textrm{a}+i\textrm{b})}$
$\implies\epsilon=\frac{2(\textrm{a}+i\textrm{b})+1}{1-(\textrm{a}+i\textrm{b})}\frac{1-\textrm{a}+i\textrm{b}}{1-\textrm{a}+i\textrm{b}}$\\
\noindent$\implies\epsilon=\frac{(2(\textrm{a}+i\textrm{b})+1)(1-\textrm{a}+i\textrm{b})}{(1-\textrm{a}+i\textrm{b})(1-\textrm{a}+i\textrm{b})}$
$\implies\epsilon=\frac{(2(\textrm{a}+i\textrm{b})+1)(1-\textrm{a}+i\textrm{b})}{(1-\textrm{a})^2+\textrm{b}^2}$\\
\noindent$\implies\epsilon=\frac{2(\textrm{a}+i\textrm{b})-2\textrm{a}(\textrm{a}+i\textrm{b})+2i\textrm{b}(\textrm{a}+i\textrm{b})+1-\textrm{a}+i\textrm{b}}{(1-\textrm{a})^2+\textrm{b}^2}$\\
\noindent$\implies\epsilon=\frac{2\textrm{a}+2i\textrm{b}-2\textrm{a}^2-2i\textrm{a}\textrm{b}+2i\textrm{a}\textrm{b}-2\textrm{b}^2+1-\textrm{a}+i\textrm{b}}{(1-\textrm{a})^2+\textrm{b}^2}$\\
\noindent$\implies\epsilon=\frac{\textrm{a}-2\textrm{a}^2-2\textrm{b}^2+1+3i\textrm{b}}{(1-\textrm{a})^2+\textrm{b}^2}$\\

\noindent Knowing that $\epsilon$ is complex, let $\epsilon=\textrm{c}+i\textrm{d}$\\
\noindent$\implies\textrm{c}+i\textrm{d}=\frac{\textrm{a}-2\textrm{a}^2-2\textrm{b}^2+1+3i\textrm{b}}{(1-\textrm{a})^2+\textrm{b}^2}$\\
\noindent$\implies\textrm{c}=\frac{\textrm{a}-2\textrm{a}^2-2\textrm{b}^2+1}{(1-\textrm{a})^2+\textrm{b}^2}$ and $\textrm{d}=\frac{3\textrm{b}}{(1-\textrm{a})^2+\textrm{b}^2}$\\
\noindent$\implies\textrm{c}=\frac{(2\textrm{a}+1)(1-\textrm{a})-2\textrm{b}^2}{(1-\textrm{a})^2+\textrm{b}^2}$ and $\textrm{d}=\frac{3\textrm{b}}{(1-\textrm{a})^2+\textrm{b}^2}$