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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\December 5, 2006}
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Last time we had Maxwell's equations in medio.  Today we can take our time.  After this lecture, we're moving on to more general theories, connections on bundles.  Next semester we'll be doing general relativity, thermo and statistical mechanics, non-relativistic quantum mechanics, and quantum field theory.  That's roughly what we need to cover next semester.

To recall, these equations were
$$
\nabla \times\vec{E}+\frac{\partial \vec{B}}{\partial t}=0
$$$$
\nabla\cdot \vec{B}=0
$$$$
\nabla\times \vec{H}-\frac{\partial \vec D}{\partial t} = \vec{J}
$$$$
\nabla\cdot \vec{D}=\rho
$$
Here $\vec{D}=\epsilon\vec{E}$ and $\vec{H}=\frac{1}{\mu}\vec{B}.$
We were using a linear ohmic uncharged medium, meaning $\epsilon$ and $\mu$ were constant, $\vec{J}$ was $g\vec{E},$ where $g$ is called the conductance, and $\rho$ is zero.  Then the wave equations became 
$$\box \vec{E}-g\mu\frac{\partial \vec{E}}{\partial t}=0$$ and a similar equation for $\vec{B}.$
Here $\box$ is $-\mu\epsilon \frac{\partial ^2}{\partial t^2}+\nabla^2.$

Now we can take the Fourier transform 
$$
\vec{E}(\vec{x},t)=\int\frac{d\omega}{2\pi}\frac{d^2\vec{k}}{(2\pi)^3}\vec{E}_{\omega,\vec{k}}e^{i(\omega t-\vec{k}\cdot\vec{x})}
$$
If you plug this in, you can write down an equation with frequency and wave number instead of differential operators.  $\vec{k}$ is in the direction of propagation and counts the number of waves per unit length.

So these are basic and motivate the definition of a plane wave, which are these pieces.  there's a constant amplitude.  Oh, everywhere you're supposed to take the real part.  A plane wave is $$\vec{E}_0e^{i(\omega t-\vec{k}\cdot\vec{x})}$$ where $\omega$ is real.  Plugging a plane wave into this equation, you replace time derivatives with $i\omega$ and $\nabla$s with $-i\vec{k}.$  Then the wave equation becomes 
$$\frac{\omega^2}{v^2}-k\cdot\vec{k}+ig\mu\omega=0$$
If we're in vacuum, the $v$ is $c$ and the first two terms are $k_\mu k^\mu,$ the four-momentum squared.  This is the dispersion relation.

These things were still not solutions to all of Maxwell's equations.  The equations can't actually be decoupled, which is what we have now.  Do the replacement again and you get $\nabla\cdot \vec{D}=0$ gives $\vec{k}\cdot\vec{E}_0=0.$  We get $\nabla\cdot\vec{B}=0$ implies $\vec{k}\cdot \vec{B}_0=0.$  $\nabla\times E=-\frac{\partial B}{\partial t}$ gives $\vec{k}\times \vec{E}_0=\omega \vec{B}_0,$ and the last equation gives $\vec{k}\times \vec{B}_0=-\frac{\omega}{c^2}(K+\frac{ig}{\epsilon_0\omega})\vec{E}_0.$  Here $K$ is $\epsilon/\epsilon_0.$

You have a space-time and erect a bundle over it, and this will have a potential which is a section of that bundle.  It would be consistent to have no electromagnetic field.

So what is the point of these equations?  If $\omega$ is real and $g$ is not zero, then $\vec{k}$ is complex.  So write it in its real and imaginary components and get $$
\vec{E}(\vec{x},t)=\vec{E}_0e^{-\vec{k}_i\cdot\vec{x}}e^{i(\omega t-\vec{k}_r\cdot x)}.$$  We want the phase $\omega t-\vec{k}_r \cdot x$ to be real, and the other part is the amplitude.  

So $g\ne 0$ means that you have an imaginary part, these are closely related.  So as the wave propagates through the medium, its amplitude moves to zero, fastest along $\hat{\vec{k}}_i.$  As the wave hits medium with conductance, this thing is damped, it's suppressed.

Let me back up.  A homogeneous wave is one in which the imaginary part is parallel to the real part.  A homogeneous medium would have solutions with this property.  So you could write $\vec{k}=(\vec{k}_r-i\vec{k} i)\vec{u}$ with $\vec{u}\cdot\vec{u}=1.$

So we can write
$$
\vec{E}(\vec{x},t)=\vec{E}_0e^{-ki(\vec{\mu}\cdot\vec{x})}\cos (\omega t-k_r(\vec{u}\cdot \vec{x}))
$$
Take $\vec{u}=\hat{z}.$  Then this becomes 
$$
\vec{E}(\vec{x},t)=\vec{E}_0e^{-k_iz}\cos(\omega t-k_0z).$$

Let me draw a picture.



\end{document}
q = sqrt(m l) l/t

sqrt(m/l)/t
1 = n/4 