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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\November 30, 2006}
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We haven't really discussed the physics of Maxwell's equations, polarization and so on.  We thought that maybe we'd do electric and magnetic fields to give some idea of what was going on.  As usual we start with Maxwell's equations.  This time we'll deal with mediums.  We'll take simple cases of high interest.  The Bianchi identity $dF=0$ splits into
$$\nabla\times E+\frac{\partial B}{\partial t}=0, \quad \nabla\cdot B=0.$$
Then we get $*d*F=k$ which splits into
$$\nabla\times D=\rho, \quad \nabla\times H - \frac{\partial D}{\partial t}=J$$
Here $D=\epsilon E$ is the electric displacement, and $\epsilon$ is a constant, not $\epsilon_0$ which is for free space.  We also have $H=B/\mu,$ the magnetic intensity, and we'll normally take it to be $\mu_0,$ the numerical value for the vacuum, so that the medium is non-magnetic.

We want this to be uncharged, so $\rho$ is zero, and conducting, meaning that $J=gE$ where $g$ is a constant, the conductivity.  One special case is vacuum, where $\epsilon=\epsilon_0,\ \mu=\mu_0,\rho=0,$ and $g=0.$

Okay, so take the curl of the first equation, and use the equation $\nabla\times\nabla\times=\nabla(\nabla\cdot)-\nabla^2,$ we get
$$0=-\nabla^2 E+\frac{\partial}{\partial t}(\mu J+\frac{1}{v^2}\frac{\partial}{\partial t}E).$$
But define $\box$ as $-\frac{1}{v^2}\frac{\partial^2}{\partial t^2}+\nabla^2.$  Then we have the equation $$\box E=\frac{\partial}{\partial t} \mu J= g\mu \frac{\partial E}{\partial t}.$$  The box is the d'Alembertian.
So we want to see some implications from this equation.  Because these are linear we can complexify.  Then you do all of your manipulations and eventually take the real part.  The 

I want to complexify all of the fields, and at the end of the day, I'll just take the real part.  All this does is allow me to use complex exponentials.

Now we get $$-\frac{1}{v^2}\frac{\partial^2}{\partial t^2} E +\nabla^2 E -g\mu \frac{\partial E}{\partial t}=0.$$
Now I claim that there is a constant $A$ which will make $\nabla^2 R + AR=0$ and $\frac{1}{v^2}\frac{d^2}{dt^2}T+g\mu \frac{dT}{dt}+AT=0.$  Here $E(r,t)=R(r)T(t).$  Now we solve these.  The first is the Helmholtz equation and gives
$$R(r)=R_+ e^{ikr}+R_-e^{-ikr}$$ where $k\cdot k=A.$
The second one gives
$$T=T_1e^{-i\omega_1 t}+T_2 e^{-i\omega_2 t}$$ where $$\frac{1}{v^2} \omega^2+ig\mu \omega-A=0.$$  
The dispersion relation is then
$$\frac{1}{v^2}\omega^2+igu\omega-k\cdot k=0.$$
Now I'm going to simplify my life by assuming $\omega\in \rR,$ the frequency.  We'll define a monochromatic wave as one in which $E(r,t)=E_0e^{-i(\omega t-k\cdot r)}.$  The $E_0$ is the amplitude, and the $\omega t-k\cdot r$ is the phase.

The most general form would just be an integral over $\omega$ and $k$ of such $E_{\omega,k}.$  So I can find out all the physics by means of this case.

How did we get this?  Clearly a solution, well, this does not yet satisfy Maxwell's equations, we need to satisfy the other equation or equations.  Since $E_0$ is a constant, and all of the $r$ and $t$ dependence is in the exponent, we can just replace $\frac{\partial}{\partial t}$ with $-i\omega$ and $nabla$ with $ik.$

So what do we get?  Let $\epsilon=K\epsilon_0.$  Then I get $k\cdot E_0=0,k\cdot B_0=0,$ and $$k\times E_0=\omega B_0, k\times B_0=\frac{-\omega}{c^2}(K+\frac{ig}{\epsilon_0 \omega})E_0.$$

Now let's decompose this big thing.  Let $k=k_r+ik_i.$  Well, in vacuum, $g=0$ which means that $k$ is real, so that the first two equations are transversality equations.  Suppose the phase is constant.  Then you have a wave propagating in the $r$ direction.

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1 = n/4 