Faraday Rotation is the rotation of the axis of polarization of radiation while propagating through a magnetized plasma. We'll derive this by examining the equation of motion of an $e^-$ in plasma. A long time ago, we wrote the equation of motion of electrons in plasma when subjected to a plane $\ef$ wave: \def\vvTemplate:\vec v $$m\vv=-e\ef-{e\over c}\vv\times\bfield$$ We had thrown away the $\bfield$ dependency because we hadn't imposed and $\bfield$ on the plasma, but this time, we'll examine what happens if we don't throw it away, but instead say that $\bfield=\bfield_0+\bfield_{rad}$, where $\bfield_0$ is an externally imposed field. Then $$m\dot{\vv}=-e\ef-{e\over c}\vv\times\bfield_0$$ We'll just guess the solution: $$\ef=Re\left(E_0(\^x\mp i\^y)e^{i(k\mp z-\omega t)}\right)$$ where the '-' of $\mp$ corresponds to a right-circularly-polarized wave, and '+' corresponds to left-circular-polarization. For the magnetic field, if $\bfield_0=B_0\^z$, then we need to solve the equation: $$m\dot{\vv}=-eE_0(\^x\mp i\^y)e^{i(k\mp z-\omega t)} -{eB_0\over c}(-v_x\^y+v_y\^x)$$ This leaves us with equations for $\dot v_x$ and $\dot v_y$. Instead of doing all the algebra, we'll just trust that the following is a solution: $$\boxed{\vv={-ie\ef\over m(\omega\pm\omega_{cyc})}}$$ Recall that $\omega_{cyc}={eB_0\over mc}$. Now we'll solve for the current in the plasma: \def\ocyc{\omega_{cyc}} $$\vec j=-n_ee\vv={-n_ee(-ie)\over m(\omega\pm\ocyc)}\ef\equiv\sigma\ef$$ where $\sigma={in_ee^2\over m(\omega\pm\ocyc)}$ is the conductivity. Now let's solve Maxwell's equations for a perturbing wave through the plasma (this is very similar to what we did months ago...). Note that in the following equation, $\bfield=\bfield_0+\bfield_{rad}$, but $\ppt{\bfield_0}=0$: \def\erad{{\ef_{rad}}} \def\brad{{\bfield_{rad}}} \def\jvTemplate:\vec j \def\divTemplate:\vec\nabla \begin{aligned}\div\times\erad&={-1\over c}\ppt{\bfield}\\ \div\times\bfield&={4\pi\over c}\jv+\inv{c}\ppt{\ef}\\ \end{aligned} Note also that $\div\times\bfield_0=0$. Now we can solve for the index of refraction: \begin{aligned}\eta^2=\left({ck_\mp\over\omega}\right)^2 &=1-{4\pi\sigma\over i\omega}\\ &=1-{4\pi in_ee^2\over i\omega m(\omega\pm\ocyc)}\\ &=1-{\omega_p^2\over\omega(\omega\pm\ocyc)}\\ \end{aligned} Thus, $k_-\eval{\omega}\ne k_+\eval{\omega}$. If we put a linearly polarized wave through a magnetized plasma, the left-circularly-polarized component will evolve more quickly than the right-circularly-polarized component, by a factor of $\Delta\phi$, so when the wave emerges from the plasma it will be fixed into a new phase where $\Delta\theta={\Delta\phi\over2}$. If we wanted to solve for this in terms of a distance through the plasma, we'd write: $$\Delta\theta={\Delta\phi\over2}=\hf\int_0^d{(k_--k_+)ds}$$ where $\left({ck_\mp\over\omega}\right)^2=1-{\omega_p^2\over\omega(\omega\pm \ocyc)}$. We'll do two approximations of this equation: \begin{itemize} \item $\ocyc\ll\omega$. Then $\ocyc=3MHz\left({B_0\over ??}\right)$. \item $\omega\gg\omega_p$. Then: \begin{aligned}\left({ck_\mp\over\omega}\right)^2 &=1-{\omega_p^2\over\omega^2}(1\mp{\ocyc \over\omega})\\ k_\mp&={\omega\over c} \sqrt{1-{\omega_p^2\over\omega^2}(1\mp{\ocyc\over\omega})}\\ &={\omega\over c} \left(1-{\omega_p\over2\omega^2}(1\mp{\ocyc\over\omega})\right)\\ k_--k_+&={\omega\over c}{\omega_p^2\ocyc\over\omega^3}\\ &={\omega_p^2\ocyc\over c\omega^2}\\ \end{aligned} \end{itemize} Getting back to $\Delta\theta$: \begin{aligned}\Delta\theta&=\hf\int_0^d{{\omega_p^2\ocyc\over c\omega^2}ds}\\ &={2\pi e^3\over c^2m_e^2\omega^2}\underbrace{\int_0^d{n_eB_\|ds}}_{rotation\ measure}\\ \end{aligned} In practice, if we want to figure out the effect of a magnetized plasma on radiation, we need to look at the phase change for two different frequencies from a linearly polarized source: $$\Delta\theta\eval{\omega_1}-\Delta\theta\eval{\omega_2} ={2\pi e^3\over c^2m_e^2}RM\left(\inv{\omega_1^2}-\inv{\omega_2^2}\right)$$ where $RM$ is the rotation measure. Some sources of linearly polarized radiation are pulsars, AGN, and the galactic radio synchrotron from cosmic rays. A magnetized plasma, if it has different rotation measures over different paths from a source to us, can depolarize a source which was originally polarized. This is {\it Faraday De-Polarization}.
Rybicki \& Lightman, CH1, gives us the following 3-dimensional integro-differential equation for the evolution of specific intensity through an adsorptive, scattering medium: \def\nabsTemplate:\nu,abs \def\nscatTemplate:\nu,scat \def\nemisTemplate:\nu,emis $${dI_\nu\over ds}=-(K_\nabs+K_\nscat)\rho I_\nu+j_\nu$$ where $j_\nu=j_\nscat+j_\nemis$, and $j_\nscat(\^s)$ is given by: $$j_\nscat(\^s)=\rho K_\nscat\times\int_{4\pi}{I_\nu(\^n)p(\^s,\^n)d\Omega}$$ where $p(\^s,\^n)$ is proportional to the probability that $\^n$-photons get redirected to $\^s$-photons, and $\int_{4\pi}{p(\^s,\^n)d\Omega}=1$. Finally, we have: $$j_\nemis\equiv S_\nemis(\^s)\rho K_\nabs$$ So we have ourselves a 3-D Integro-differential equation. If we are going to be able to actually solve this, we're going to have to make some assumptions. We'll make the following 4 idealizations: \begin{itemize} \item{(A)} We have isotropic scattering. Then $p(\^s,\^n)=\inv{4\pi}$, and the mean intensity is given by: $$J_\nu\equiv\inv{4\pi}\int{I_\nu d\Omega}$$ which gives us: $${dI_\nu\over\rho(K_\nabs+K_\nemis)ds}=-I_\nu+\~\omega_\nu J_\nu+(1- \~\omega_\nu)S_\nu(\^s)$$ where $\~\omega_\nu\equiv{K_\nscat\over K_\nscat+K_\nabs}$ is the {\it single-scattering albedo}, and $0<\~\omega_\nu<1$. \item{(B)} $S_\nu(\^s)=B_\nu(T)$ (i.e. Local Thermal Equilibrium). \item{(C)} We have a plane-parallel atmosphere. We'll say that parallel surfaces of constant temperature. If $\^z$ is the direction straight up through the atmosphere, and $\theta$ is the angle of a vector from the $\^z$ direction, the we define $\mu\equiv\cos\theta$. Then $ds={dz\over\cos\theta} ={dz\over\mu}$. Note that the change in optical depth along $\^z$ is: $$d\tau_\nu=-(K_\nabs+K_\nemis)\rho\,dz$$ \item{(D)} No scattering. That is, $\~\omega_\nu=0$. We'll relax this eventually (otherwise it would have been silly to assume A). \par \end{itemize} So the final form of our equation is: $$\boxed{\mu{dI_\nu\over d\tau_\nu}=I_\nu-B_\nu(T(z))}$$ We'll solve this by applying the {\it Eddington 2-stream Formalism}. We begin by taking a moment $\int{equation\,d\Omega}$ of the above: \def\ddtau#1Template:D $$\ddtau{}\int{\mu I_\nu(\mu)d\Omega}=\int{I_\nu d\Omega}-\int{B_\nu d\Omega}$$ Using that $d\Omega=-2\pi d\mu$: $$\ddtau{}\int_1^{-1}{\mu I_\nu(-2\pi d\mu)}=4\pi J_\nu-4\pi B_\nu$$ Using that $2\pi\int_{-1}^1{\mu I_\nu d\mu\equiv F_\nu}$ (Rybicki \& Lightman 1.3b), we have: $$\ddtau{F_\nu}=4\pi(J_\nu-B_\nu)$$ Now Eddington's 2-stream approximation was to say that for a point in our atmosphere, the only important components of the specific intensity are those from above and those from below. That is, we need only consider the components: \begin{aligned}I_\nu^+(\tau_\nu)&=I_\nu(\tau_\nu,0<\mu<1)\\ I_\nu^-(\tau_\nu)&=I_\nu(\tau_\nu,-1<\mu<0)\\ \end{aligned}