Roger's Equations

The Monopole Potential

The Electric potential for a point charge is given by the equation

where φ is the electric potential, ε₀ is the permittivity of free space, q is the charge, and r is the distance from the point charge producing the potential.

The electric field is the gradient of the electric potential,

which means that the electric field of a point charge (1) is,

So if you know the electric potential, you can get the Electric Field.

The Dipole Potential

A dipole is a configuration of two charges, one positive and one negative as in the picture below,

The two charges are 2a distance from each other. We are interested in measuring the potential at a distance r from the midpoint of the two charges. The principle of superposition tells us that at any point we can simply add the potentials produced by the two charges above. So the potential of the dipole is,

Φ_dipole= (1/4πε₀) (q/r₁ + (-q)/r₂ = (1/4πε₀) (q/r₁ - q/r₂) = (q/4πε₀)(1/r₁ - 1/r₂)

At the midpoint between q and -q we draw a line r. This creates two irregular triangles. The first one has sides with lengths r, r₁, and (2a)/2, and an angle θ, the other has sides with lengths r, r₂, and (2a)/2, and an angle θ-π. We can use the law of cosines to express r₁ and r₂ in terms of r and θ,

r₁²= r² + a² - racosθ
r₂²= r² + a² - racos(θ-π) = r² + a² + racos(θ)

rearranging we get,

r₁²= r²(1 + (a²/r^2₎ - (acosθ)/r)
r₂²= r²(1 + (a²/r^2₎ + (acosθ)/r)

If we limit ourselves to situations where r >> a the we can neglect the middle term (a²/r²) in both expressions. That leaves,

r₁²= r²(1 - (acosθ)/r)
r₂²= r²(1 + (acosθ)/r)

inverting and taking the square root of both sides gives us,

1/r₁= (1/r)(1 - (acosθ)/r)^-1/2
1/r₂= (1/r)(1 + (acosθ)/r)^-1/2

we can use the binomial expansion to get,

1/r₁= (1/r)(1 + ((a/2)cosθ)/r)
1/r₂= (1/r)(1 - ((a/2)cosθ)/r)

Now we can express our original expression in terms of r and θ,

Φ_dipole= (q/4πε₀)(1/r₁ - 1/r₂) = (q/4πε₀)(1/r - 1/r + (acosθ/r²))

Φ_dipole= (q/4πε₀)(acosθ/r²) = (1/4πε₀)(qacosθ/r²)

Notice that the potential depends only upon the distance between the charges and θ. A monopole has a potential that drops off 1/r, notice the dipole drops off 1/r².

Multipole Expansion

So what if we have a charge configuration? How can we represent the potential of such a configuration. It turns out you can express any charge configuration as the sum of Monopole, Dipole, Quadrupole, Octopole, etc. potentials.

Consider the equation for the potential of a charge configuration:

Φ= (1/4πε₀)∫(1/d) ρ dτ

where ρ is the charge density and d is the distance between an infinitesimal charge and point at which the field is measured. Basically we are taking a charge configuration, chopping it up into little volume elements and calculating the potential produced by each element, then we sum all the elements in order to calculate the total potential due to the charge configuration (see diagram below).

Using the law of cosines, we can express d as:

d²= r² + R² - 2rRcosθ = R² (1 + r²/R² - 2(r/R)cosθ)

If we assume we are far enough from the charge configuration that R>>r, then we can discard the r²/R² term as approximately zero. This leaves,

d²= R² (1 - 2(r/R)cosθ)

which we invert and taking the square root of both sides to give us,

1/d= 1/R (1 - 2(r/R)cosθ)^-1/2

We now can expand the binomial,

(1 - 2(r/R)cosθ)^-1/2 = (1 - 1/2(r/R)(r/R - 2cosθ) + 3/8 (r/R)² (r/R - 2cosθ)² + ......

Noting that expansion of the binomial is of the form,

(1 - 2(r/R)cosθ)^-1/2 = Σ (r/R)ⁿ P_n(cos θ) for n=0 to ∞

In this last step, I've collected like powers of (r/R) and noticed that the coefficients associated with them (the cosine terms) were Legendre Polynomials where x=cosθ(P_n(cos θ)). This last step is simply for convenience, we could just as easily leave it in its less compact form, but but grouping it in this way we gain insight into what the binomial expansion is telling us (to be explained in a moment).

Substituting the summation for the binomial in our original expression, we get:

1/d= 1/R (1 - 2(r/R)cosθ)^-1/2 = (1/R) Σ (r/R)ⁿ P_n(cos θ) for n=0 to ∞

which when we replace 1/d in our expression for the potential,

Φ= (1/4πε₀)∫(1/d) ρ dτ

gives us,

Φ= (1/4πε₀)∫((1/R) Σ (r/R)ⁿ P_n(cos θ)) ρ dτ for n=0 to ∞

or more explicitly if we expand the summation for the first couple of terms we get,

Φ= 1/4πε₀ [1/R ∫ ρ dτ + 1/R² ∫ rcosθ ρ dτ + 1/R³ ∫ (r)² (3/2 cos²θ - 1/2) ρ dτ + ...]

So what does that mean? Lets take a closer look at the equation above.

The Multipole Potential

Φ= 1/4πε₀ [1/R ∫ ρ dτ + 1/R² ∫ rcosθ ρ dτ + 1/R³ ∫ (r)² (3/2 cos²θ - 1/2) ρ dτ + ...]

The first thing you should notice is that the first term,

1/4πε₀ (1/R) ∫ ρ dτ

is essentially the equation for a monopole and the second term,

1/4πε₀ (1/R²) ∫ rcosθ ρ dτ

is the equation for a dipole. This pattern continues with the third term corresponding to the potential of a quadrupole and the term after that corresponding to the potential of an octopole. By expanding the binomial we found through manipulating the original equation of the potential, we have expressed the potential as a sum of monopole, dipole, ....etc. contributions. Thus the name, Multipole Expansion.

From far away the monopole term, which has a potential proportional to 1/R dominates whereas the closer you get to the configuration of charges, the more the dipole (1/R²), Quadrupole (1/R³), Octopole (1/R⁴), etc. contribute.

Thanks to the following websites:

http://en.wikipedia.org/wiki/Electric_field

http://physics.tamuk.edu/~suson/html/1402/potential.html

And a very sincere thanks to Introduction to Electrodynamics by David J. Griffiths, a gem of a book. This topic is covered in Chapter 3 from page 145 to 148.

I'd also like to thank Classical Electrodynamics by John David Jackson. A book I hope to be smart enough to understand one day.