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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\December 7, 2006}
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Fields have moved from being verbs to being nouns.  So now I want to consider, we want to study this as your first field theory.  Let $M=\aA^1_{\text{time}}\times N,$ for $N$ Riemannian or $\eE^3.$  This all works for $N$ any dimension Riemannian.  Here $M$ has metric $(dt)^2-g_N.$

The players were $E(t)\in \Omega^1(N), B(t)\in \Omega^2(N),$ and $\rho(t)\in \Omega^3_c(N), J(t)\in \Omega^2_c(N).$

We found it convenient to collate this information via $F=B+E\wedge dt,$ and we brought together the charge density and current via $j=\rho-J\wedge dt.$

Maxwell's equations have a nice form when you collate, $dF=0$ and $d*F=4\pi j.$  That's more or less where we've been.

For now, let's consider what this tells us.  Let's make an extra assumption so that we can, we essentially did this when we did our previous classes.  Let's make the assumption, we'll add this to Maxwell's equations, which is that $[F]=0$ in $H^2_{dR}(M).$  In that case we can write $F=dA$ for some $A\in \Omega^1(M).$  So some remarks, 
\begin{enumerate}
\item $A$ is only determined up to closed forms. So the space of fields is $\Omega^1(M)/\Omega^1_{cl}(M).$
\item notice that $dj=0.$  In particular, $[j]\in H^3_c(M).$  Then the determinant is an element in there.  There is a natural map to ordinary cohomology by inclusion. In ordinary homology it represents a trivial class.  It could be a current around a loop, which would be nontrivial compactly.
\end{enumerate}
One of the things about field theories is that their equations of motion can be derived from Lagrangians.  Remember we can derive the equations of motion from a Lagrangian formulation of the field theory.  I'll take a slightly different approach than William.  Remember what I called a Lagrangian.  This is a map from a space of fields into volume forms on $M.$  This is like what we talked about in mechanics.  William wrote this down already.  This was $A\mapsto \frac{1}{2}dA\wedge *dA+A\wedge j.$  The second part is not well-defined on $\Ff,$ the space of fields.  This is local on $M,$ only depending on the $1$-jet of $A.$

This is not well defined as written on $\Ff=\Omega^1/\Omega^1_{cl}.$  But Lagrangians are only defined up to exact terms.  So $A\wedge j=A\wedge dG$ for some $G.$  So this is $dA\wedge G -d(A\wedge G).$  So this is well-defined on $A$ up to an exact term.

We obtain that up to exact terms, $L(A)$ depends on $dA.$  Now we can derive the Euler-Lagrange equations.    We consider the action of this theory $S(A)=\int_M L(A).$  But $M$ is not compact.  So this is not necessarily well-defined for a lot of fields.  But variation with respect to fields with compact support is well-defined and derives the Euler Lagrange equations.  If I can show that this is true on every neighborhood, I can show it everywhere.  I can then show that something is zero everywhere by showing it is true on every neighborhood.

You don't need to consider the action if you don't want to.  You can avoid the action completely by using the Lagrangian and other data, but that would require an exposition that I'm not sure you would find fruitful.  This is a non-local derivation of a local equation.  There is a completely local derivation, but I think the explanation and definitions required would not be worth the effort.

Recently people have had insight into what a non-local field theory would look like.  All field theories we'll deal with will be local, will come from compactly supported variations of the action.

$T_A\Ff,$ in classical mechanics this took more work because we were on a manifold, but here we're in a vector space, so it's easy, I know what its tangent field looks like.  Let $\xi\in \Omega^1_c(M)$ and let $\delta$ denote the exterior differential on $\Ff.$

This is the notation physicists use.  Then $$
\delta L(\xi)=-d\xi\wedge *dA+\xi\wedge j.$$

This is $$\frac{d}{ds}|_{s=0}L(A+s\xi)=\frac{d}{ds}|_{s=0}
(-\frac{1}{2}(sd\xi\wedge *dA + sdA\wedge *d\xi) +s\xi\wedge j +\ldots)
=-d\xi\wedge * dA+\xi\wedge j$$
To derive the Euler Lagrange equations I need something linear in the variation, so then I write $\delta L(\xi)=\xi\wedge \wedge (-d*dA+j)-d(\xi\wedge *dA).$  Now $$\delta S(\xi)=\int_M \xi \wedge (-d*dA+j)$$ so that $\delta S|_A=0$ if and only if $d*dA=j.$

I could have done this in any direction, finding the part that was linear in $xi$ and then having the rest being the exact part.

[I have a general question about this process.  Why do we make up Lagrangians to give us what we were looking for?]

Everything in this course was known before Lagrangian principles.  Then it was noticed that all these principles had a Lagrangian interpretation.  This made people sit up and take notice.  There's a lot of appeal to that, all the information was zipped up in the Lagrangian.

Go to the turn of the century when people were discovering other theories.  People thought it was prudent to explore other theories using Lagrangian principles.

Once I've finished packaging this up I'll have an infinite number of generalizations.

$A$ is what we call a classical field because it satisfies the classical equations of motion.  Here $A\in \Mm,$ meaning it's solutions to Maxwell's equations.   This is similar to what we did with particle mechanics.

In order to get a Lagrangian, I had to use $A,$ I couldn't use $F.$ You'll never get two derivatives.

So one of the things that was nice about the Lagrangian formulation of classical mechanics, we got a bunch of structure on the space of solutions.  We'll have a symplectic structure on the space of solutions.  We'll use that structure, because that's the only way we know how to quantize.  Here's an heuristic argument.  For a moment we only consider fields with compact spatial support along $\{t\}\times N$ for all $t\in \aA^1_t.$  Consider the region $[t_0,t_1]\times N\subset M.$  Define $$S_{[t_0,t_1]}(A)=\int_{[t_0,t_1]}\int_N L(A).$$
Restrict $S$ to the space of classical solutinos, and look at $\delta S_{[t_0,t_1]}(\xi).$  On classical solutions the main term vanishes and we get the exact term integrated over the boundary,
$$\int_{\{t_1\}\times N} \xi\wedge *dA - \int_{\{t_0\}\times N}\xi\wedge *dA.$$

Now we see that there is a variational one-form.  We can assign to each time $t\in \aA_t^1$ a one-form $\gamma_t$ on $\Mm.$  Now $\gamma_t(\xi)=\int_{\{t\}\times N} \xi \wedge *dA$ 
for $\xi\in T_A(\Mm).$  Then the $2$-form $\omega=\delta\gamma_t$ is independent of $t.$  So $\omega$ is a symplectic form on $\Mm.$


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q = sqrt(m l) l/t

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