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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\November 14, 2006}
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Last time we went over electrostatics.  Now we'll allow $E$ and $B$ to vary over time.  The new phenomena here were discovered by Faraday, and this is Faraday's law of induction.  At first I'll write this in terms of the integral calculus.
$$
\frac{d}{dt}\int_S B(t)=-\int_{\boundary S}E.
$$
We can derive a more infinitessimal statement.  This is only true if you integrate over very small things.  Let's integrate both sides over $t\in [t_0,t_1].$  We get
$$
\int_S B(t_1)-\int_S B(t_0)+\int_{\boundary S\times [t_0,t_1]}E(t)\wedge dt =0.
$$
I'm dealing with some cylinder $C=[t_0,t_1]\times S$ sitting in $\aA\times N.$  What this, well, set $F=B+E\wedge dt.$  This is suggested to us by this equation.  With these definitions, 
what I've basically got written down is that 
$$
\int_{\boundary C} F = 0.
$$
This is what happens when we have a cylinder like this one.  It would be nice to see this for other cylinders.  What if you had a spatial cylinder $C\subset \{t_0\}\times N$ be a three dimensional region at fixed $t_0.$  Then
$$
 \int_{\boundary C}F = \int_{\boundary C} B(t_0),
$$
which is zero because $dB=0.$  So all together we conclude that $dF=0.$

All right, so that gives us one of Maxwell's equations.  Now we need something else to derive the other.  This is Maxwell's discovery.  Other than writing down the equations, he discovered a physical phenomenon, which was a modification of Amp\`ere's Law.  This said
$$
\int_S 4\pi J=\int_{\boundary S}*B
$$
This is fine for static fields.  Maxwell noticed that if everything depends on time, you need to put inside the integral on the left hand side
$$\frac{\partial *E}{\partial t}.$$
We can play the same game now to make this exression local to get, well, let $$G=*E-(*B)\wedge dt.$$ Then these equations tell me that
$$
\int_{\boundary C} G = -\int_C 4\pi J\wedge dt
$$
That was a temporal cylinder.  We can also consider the spatial patch $C\subset \{t_0\}\times N,$ so that $\int_{\boundary C}G=\int_C 4\pi\rho$ by electrostatics.  We can collate these two pieces by calling $\mathscr{j}$ to be $\rho-J\wedge d,$ and then we can say that $dG=4\pi\mathscr{j}.$  We show that the integral over any three dimensional block of these two things are equal, so they are the same form.

What is the relation of $F$ and $G?$  We can extend $*:\Omega(N)\to \Omega(N)$ to $\bigstar:\Omega(\aA^1\times N)\to \Omega(\aA^1\times N)$ via
$$\bigstar(\alpha)=*(\alpha)\wedge dt
$$$$
\bigstar(dt\wedge \beta)=-(*\beta)$$
for $\alpha,\beta\in \Omega(N).$

This encodes a Lorentzian metric of the form $dt^2-g_N.$  East coast rules.

Electromagnetism then motivates special relativity.  The argument I just made was very mathematical, there is also a physical motivation, but this is easier for me.

So we end up with Maxwell's equations, $dF=0$ and $d\bigstar F=4\pi \mathscr{J}.$  As I said, relativity is embedded in Maxwell's equations.

We've come full circle and from empirical information derived Maxwell's equations.

Suppose that $A_0$ satisfies $F=dA.$  We want to get $F=dA$ and $d\bigstar A=0$  So we can write $A=A_0-d\psi.$  Then $0=d\bigstar A=d\bigstar A_0 - d\bigstar d\psi.$  So this is the same as 
$$
d\bigstar A_0=d\bigstar d\psi,
$$
which can be solved.  I want this other requirement to ``choose a gauge.''  I am picking a particular gauge that satisfies this requirement.  Given that $A$ satisfies this requirement, then, just like the magnetostatics, I can show that $\bigstar d\bigstar d+d\bigstar d\bigstar$ applied to $A$ is $4\pi \bigstar \mathscr{j}.$

If you actually did the computation, this sum of operators is $\frac{\partial^2}{\partial t^2}+\Delta_N.$  If you've ever seen anything on, if you've seen anything on Hodge theory, this, for the usual Laplacian, is the Laplacian on forms.  I want an $A$ that satisfies the wave equation with a source, or the inhomogeneous wave equation.

Let's talk about wave equations now.  Today I want to talk about the homogeneous equation where we have no source.  Let $N=\eE^1.$  Here $$\frac{\partial^2 f}{\partial t^2}-\frac{\partial^2 f}{\partial x^2}=0.$$I can write this as 
$$(\frac{\partial}{\partial t}-\frac{\partial}{\partial x})(\frac{\partial}{\partial t}+\frac{\partial}{\partial x})f=0.$$
This means $f(x,t)=h(x+t)+g(x-t).$  What I'm more interested in, well, you can imagine I have these functions $h$ and $g.$  They can be viewed as functions of $x$ with graphical ``profiles'' that move in time.  They are called left and right moving solutions.

We need more information to tell us the solution.  That information is initial conditions.  I want given initial conditions, initial data, often called Cauchy data, which is $f(x,0)=u(x)$ and $\frac{\partial f}{\partial t}(x,0)=v(x).$

Given this, this isn't too hard to show, there exists a unique solution to the homogeneous wave equation on $\eE^1$ with given Cauchy data.  That solution is
$$f(x,t)=\frac{1}{2}(u(x+t)+u(x-t))+\int_{x-t}^{x+t} v(s)ds.$$
\begin{exercise}Prove this.
\end{exercise}
So remarks are in order.  You'll notice that the solution is linear in the Cauchy data.  So you can think of this solution as a linear map.  Then the value of $f(x,t)$ only depends on certain parts of space time, in particular the past light cone.  That's called causality.
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1 = n/4 