LECTURE NOTES FOR PHYSICS 308: ELECTRICITY AND MAGNETISM, Spring 2007
TEXT: GRIFFITHS' INTRODUCTION TO ELECTRODYNAMICS
Last updated: 1/16/2007

SOME USEFUL REFERENCE STUFF: Greek alphabet, metric prefixes, conversion factors

ASSIGNMENTS: (Subject to change: check back often.)
HW #4: Due Friday, March 2 at noon: Problem 2:39, 3:3, plus numerical spreadsheet problem, TBA.

HW #5: Due Friday, March 9 at noon: Problems 3.6, 27, 32, plus, solve for V(r,θ) in all space, if V(R,θ) = C cos³θ on the boundary, R of a sphere.
"Dailies":
No daily due Tuesday, Feb. 27.
Due Thursday, March 2: Calculate P₆(x), using either the Rodrigues or Iterative formula. Which is easier?
Due Tuesday, March 6: For Figure 3.34B, calculate the dipole moment for each of the following choices of origin: the leftmost charge and the topmost charge.

GO TO LECTURE 11, 12
LECTURE 11:
SECTION 3.3: SEPARATION OF VARIABLES
Skip Section 3.3.1. We will not do this in class. In class we will focus on the spherically symmetric case. (Section 3.3.2) Still, you should be aware of the existence of the rectangular case, and be able to look it up if you ever need to use it yourself.

SECTION 3.3.2: SPHERICALLY SYMMETRIC CASE AND LEGENDRE POLYNOMIALS
Consider a boundary value problems in which the electric potential, V, is defined on the surface of some sphere. Furthermore, for the sake of simplicity, let's assume that the potential has no azimuthal dependence: the potential depends only on the angle θ, not on φ. What can we say about the electric potential in the rest of space?

Because this problem has spherical symmetry, we will consider a the electric potential to be a function of r and θ only: V=V(r,θ). In spherical coordinates the Laplacian is given by,...well ... (this is where the inside front cover of your text is very very useful: open it up and take a look at all of the useful reference material in it)
The method of separation of variables begins by hoping that we can write the electric potential as the product of two functions: We want a function of r only and another function of θ only: V(r,θ)=R(r)θ(θ). If we plug this formed into Laplace's equation -- -- we get
where P_l(cosθ) is a polynomial in the family known as the Legendre polynomials. We'll come back to this in a second. In general, the electric potential is the sum of an infinite series, where only integer values of l work.

Now, depending on whether not we are looking for the electric potential outside of or inside of the spherical boundary, we will throw out either all of the A_l terms or all of the B_l terms, whichever set of terms would lead to infinite potential, V, in the region of space we are considering. (r< R implies all B_l=0, r>R implies all A_l=0)

LEGENDRE POLYNOMIALS
You have seen Legendre polynomials before if you applied Schrödinger's Equation solution to the hydrogen atom in Modern Physics. Why do these polynomials come up again here in classical electrostatics? The reason is that both mathematical problems involve (a) the Laplacian operator and (b) spherical asymmetric boundary values. Since these polynomials come up in a number of places in physics, we should talk a little bit about them. First of all, the first few Legendre polynomials are given by
Second, it would be useful to know how to derive the rest of the terms in this expansion. There are at least two useful ways of doing this. The first of these is known as the Rodrigues formula, after Benjamin Olinde Rodrigues. The Rodrigues formula for the Legendre polynomial P_l is
This equation is a bit of a bother. Why? Because if we want to calculate P₄, it doesn't help that we already know P₀ through P₃: we have to start from scratch again. A more convenient equation (one which the book doesn't provide you with) is the iterative formula
We'll do some examples either on the board or as "daily" problems to give you some practice with this one.

Now how do we actually use these polynomials? Let's assume that we are given the electric potential on the sphere of radius R. The first task is to write this function in terms of the appropriate Legendre polynomials. Then we match this function to the form of the general solution to the spherically symmetric boundary value problem. That's it. For really nasty problems, we would have to calculate an infinite number of coefficients in the expansion, very much like calculating the coefficients in a Fourier expansion. But, since I'm such a nice guy, we will only look at boundary value problems which include a handful of finite terms. (By the way, check out 1, 2, or 3, or search the web using "fourier series java applet" to find some cool Fourier series simulations.)

Okay, here's a quick example of the kind of problem I might ask. Imagine that we are given a boundary condition that the electric potential on the surface of a sphere of radius R is equal to D cos²θ, where D is some constant. We know that the solution should look like , which means that it should equal some linear combination of Legendre polynomials: E P₀(cosθ)+ F P₁(cosθ)+.... The trick is to figure out what E, F, etc. are. Now, for this particular problem, with V = A cos²θ, the expansion will have no Legendre components higher than 2, since that is the highest-order term in cosθ.

After a little algebra, you should be able to convince yourself that cos²θ = [2P₂(cosθ)+P₀(cosθ)]/3. From this we conclude that
                E = 2/3 = (A₂R²+B₂R^-3)   and
                F = 1/3 = (A₁R+B₁R^-2).
If we are looking for the solution for r>R, then we set all the A_l's equal to zero: B₂=(2/3)R³ and B₀=(1/3)R, or V = 2R³/(3r³)P₂(cosθ)+R/(3r)P₀(cosθ). If we are looking for the solution for r < R, then we set all the B_l's equal to zero: A₂=(2/3)R^-2 and A₀=(1/3), or V = 2r²/(3R²)P₂(cosθ)+(1/3)P₀(cosθ).

LECTURE 12:
SECTION 3.4: MULTIPOLE EXPANSION AND DIPOLES
Now let's look at the the most general expression for the electric potential due to a charge distribution problem. Now, it may look like this has nothing to do with all that Legendre polynomial stuff we have just done, but we will see that it does.

Consider some distribution of charge located at the origin. The formal expression for the electric potential is the following integral:
This is of course a very ugly integral to have to do on one's own. First of all, the curly r in the denominator is the difference between the the position vector at the point where you want to know the potential and the distance vector which is the location of the increment of charge, dq you are integrating over. So not only is varying as you integrate over volume, but so is curly r. The book executes a couple of tricks involving the Law of Cosines which allows us to rewrite this expression as an infinite power series which converges as we move further and further away from the charge distribution. What we get turns out to be
There are those pesky Legendre polynomials again! We can look at this expression term by term and write it as
The first term has the exact same r-dependence that we would expect for the electric potential due to a single point charge. In fact, it turns out to equal k times the total charge of the charge distribution. We will call this the "monopole term". The next term falls off more rapidly. One can show that it is the electric potential that you would see if you have two equal but opposite electric charges located at the origin, but separated from each other by an infinitesimal distance. This is called the "dipole term". The terms following these two are known as the quadrupole term, the octupole term and so on. At large distances from the charge distribution, the monopoles term will swamp out all the other terms, unless the monopole term is zero, or very small. In that case, the dipole term will dominate, unless it, in turn, is zero.

What's the point here? That even uncharged objects can produce electrostatic fields. (In fact, cellophane does a good job of wrapping sandwiches because different parts of it attract via electrostatic forces, even though the entire piece is more or less uncharged.) In the next chapter, we will look at a macroscopic phenomenon created by these dipole fields in uncharged matter. For that reason, it is worth looking a bit closer at electrostatic dipoles.

THE ELECTROSTATIC DIPOLE MOMENT
The second term of the expansion for the electric potential can be written as K₁/r², where
where "r-cap" is a unit vector which points from the position of the dipole to the location in space where you want to know the electric potential. The charge distribution is equivalent, for our purposes, to two finite charges, +Q and -Q, separated by a distance d, and having a dipole moment, :
where the vector points in the direction from the negative charge to the positive one.

THE ELECTRIC FIELD DUE TO A DIPOLE
Let's define the z-axis as the direction along which the dipole, , points, and x be in the plane perpendicular to z. Let's let be the position vector pointing to the location where we want to calculate the electric field due to that dipole, and let θ be the angle between to z-axis and the vector. We can define the components of the electric field in a couple of different ways. In spherical coordinates we have
                  while in rectangular coordinates we have
                  This sounds straightforward enough, but once we start looking at the electric field at a point, say, midway between two dipoles, then it gets interesting because we need to define our z and for each dipole, or r and θ relative to each dipole.

In the next chapter, we will look at the effect that an external electric field has on a dipole.

---

YSBATS FOR CHAPTER 3:
Memorize:
The relaxation formula (memorize it in visual form)

Understand:
When to apply the Method of Images
Why Legendre polynomials keep popping up
Origin and significance of the first two multipole terms

You should be able to:
Find the image charge[s], apply them to solving problems
Solve "easy" spherical BVP's
Calculate K₀, K₁ for collections of point charges
Calculate V, E for simple point-charge problems