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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\November 14, 2006}
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So we talked about the boundary conditions for the electric field in the presence of a conductor.  A conductor is a blob with charges in it free to move in the blob.  If you set the current equal to zero, you can say that the electric field inside the blob is zero.  Then we claim that if there are charges in this thing they reside on the boundary.  The other thing is the boundary condition of the electric field.  You don't know what the electric field is outside the boundary.  The idea there is if you took a small chunk of the surface and blew it up, if you had an electric field line that came in at an angle, it would move the charge around until the force lines were perpendicular.

Let's move on to the electric potential.  This is defined up to a derivative as $-\nabla\varphi,$ which you can say because $\nabla\times E=0.$  So we can solve the Bianchi identity.  We know $E=\frac{q}{4\pi}\frac{\vec{r}}{r^3}=-\frac{q}{4\pi}\nabla \frac{1}{r}.$

So we learn that in terms of electric potential a point charge looks like $\frac{q}{4\pi}\frac{1}{r}.$  If $\varphi$ is constant on the interior, then on the exterior you have equal potential everywhere.  That is useful because you can use this for Liouiville's theorem.

Let me make an aside on delta functions.  I gave you a point charge potential.  We started out getting the electric field using an anzats, we chose a functional form for the charge density, and got the electric field at a point.  We then have the equations $p=q\delta(vec{r})$ and $E=\frac{q}{4\pi}\frac{\vec{r}}{r^3}$ and $\nabla E=\rho$ so that $\nabla \frac{\vec{r}}{r^3}$ is $4\pi \delta(\vec{r}).$  Then $\nabla^2(\frac{1}{r})=-4\pi\delta(r).$

Now I want to move on to multipoles.  Let's quickly go through this.  We had a point charge.  We discussed the monopole just now.  We called it a point charge.  In a minute you'll know why it's called a monopole.  There are also dipoles, quadripoles, and so on.  For a more general charge distribution there will be extra, well, the dipole, if you took what we had last time, took a plus charge and a minus charge, you get a charge distribution that looks like this.

In particular, a surface integral around these I get thet enclosed charge which is zero.  But the configuration is still interesting.  It has this additional symmetry which is rotational around the axis.  I want to see what the field looks like very far away.  Why am I going through this?  I want to take a massive dipole moment and shrink to what's left.  The electric field at the point $\vec{r}$ is just $$\frac{q}{4\pi}(\frac{vec{r}-\vec{r'}-\ell}{|r^2-r'^2-\ell^2|-\frac{\vec{r}-\vec{r}'}).$$
I have $$\frac{1}{|r-r'-\ell|^3}\frac{1}{|r-r'|^3}\right(1+3\frac{vec{r}-\vec{r}'\ell +\mathscr{O}(\ell^2))}.$$

The electric field, I'll just rewrite this thing as
$$
E(\vec{r})=\frac{1}{4\pi}\{\frac{3(\vec{r}-\vec{r}')\vec{p}}{|\vec{r}-\vec{r}'|^5}-\frac{\vec{p}}{|\vec{r}-\vec{r}'|}^3
$$

This is $$\frac{-1}{4\pi}\nabla\left(\frac{\vec{p}(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3\right)+\mathscr{O}(\ell^2)$$

So in the limit when $\vec{r}'$ goes to zero we have $\varphi(\vec{r})=\frac{1}{4\pi}\frac{\vec{p}\hat{r}}{r^2}.$  This is weaker than the monopole part.

In general you have a charge distribution.  Pick a point in this, and go far away, and look at a distance $\vec{r}.$  Then you can coumpute the electric potential at this point by means of a Riemann integral by picking little patches with volume, and $\rho$ is a density.  You do the integral, and all you should try to do is to repeat the calculation of the electric field but keeping one higher term, so that if I compute the potential I get
$$
\frac{4\pi}\varphi = \frac{q}{r}+\frac{\vec{p}\cdot \hat{r}}{r^2}+\frac{1}{2}\frac{\hat{x}_i\hat{x}_j}{r^3}Q_{ij}+\mathscr{O}(r^{-4})
$$
Here you have 
$$
Q_{ij}=\int (d^3 r 3x'_ix'_j - \delta_{ij}r'^2)\rho(r').$$

This is a moment tensor, it takes two coordinates where $\vec{p}$ only takes one.

Now let's talk about the method of images.  I was going to remind you of theorems that Jerry ploved, but basically we're looking at Laplace's equation, and $\nabla^2\varphi=0$ has unique solutions.  So you either specify boundary conditions or fix the normal derivative at the boundary.  So the solution is unique with either.  The first type is called Dirichlet and the second type Neumann.  So what?  If I somehow invent a solution that satisfies Laplace's equation, if the boundary configurations match, I have the boundary condition.  Imagine now this setup:  If we have an infinite half-space of conductor and a point charge with charge $q,$ we want to find out $\vec{E}(\vec{r}).$  This would be terrible.  We can roughly tell you what it looks like.  We know that it comes out radially.  We're trying to give an impression why the uniqueness is powerful.  If we added up charges this would be difficult.  Now instead take a charge and an opposite charge.  Then take the plane in the middle and perpendicular to the seperation vector.  This fictitious point charge (called the image) gives you the solution because this matches the boundary conditions.

In the half space we have a solution to the Laplace equations.  Then subject to these boundary conditions it is unique.  Similarly you can write down the electric field of a line of charges and a plane of conductors.  The potential of this is $\frac{q}{4\pi}(\frac{1}{r_i}-\frac{1}{r_2}).$  That's all of electrostatics except for the Poisson equation.  There the idea was to take the equation $\nabla^2\varphi=-\rho,$ replace this with $\nabla^2 G(\vec{r})=-\delta(\vec{r}).$  This is called Green's function, and if you can solve the second one, you solve the first one by
$$\varphi(\vec{r})=\int \left\{d^3\vec{r}' G(\vec{r}-\vec{r}')\rho(\vec{r}')\right\}+\varphi_0$$
which solves.

First Fourier transform, then solve the algebra and inverse Fourier transform.  Then $$G(\vec{x})=\int \frac{d^3 k}{(2\pi)^3}\frac{e^{i(\vec{k}-\vec{x})}{\vec{k}^2}.$$
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1 = n/4 