Methods of Solution for the Electrostatic Potential

Methods of solution for the electrostatic potential

November 20, 2013

1 Method of Images

By placing additional charges outside the region of interest we can build an equivalent system.

1.1 Exterior potential for a grounded sphere in the presence of a point charge Consider a grounded sphere (potential V (R) = 0) of radius R centered on the origin, and a charge q brought to within a distance a > R. We ﬁnd the potential using the method of images. Let the z-axis pass through the charge q. The system then has azimuthal symmetry, so if the problem can be solved using images, any image charge must also lie on the z-axis. Let there be a charge q0 at position b < R. This is allowed since we are interested in the potential outside the sphere. The question is whether we can choose q0 and b so that the sphere is an equipotential. The potential for the charge pair is   1 q q0 V =  +  4π0 ˆ ˆ x − ak x − bk Now,

r ˆ ˆ ˆ x − ak = x − ak · x − ak p = r2 + a2 − 2ar cos θ so that 1 q q0 V = √ + √ 4π0 r2 + a2 − 2ar cos θ r2 + b2 − 2br cos θ For the sphere to be at zero potential, the term in parentheses must vanish when r = R for all θ. This means that p q p R2 + a2 − 2aR cos θ = − R2 + b2 − 2bR cos θ q0 r q b R2 = − (R2 + b2 − 2bR cos θ) q0 R b2 r q b R4 R2 = − + R2 − 2R cos θ q0 R b2 b

R2 The two roots will be equal if b = a, so that R2 b = a

1 and the charges must be related by q b 1 = − q0 R b q0 = −q R qR = − a The potential is then zero on the sphere, and the potential anywhere outside the sphere is given by   q 1 R V =  −  4π0 x − akˆ a x − R2 kˆ a 2 Separation of variables

The separation of variables technique is more powerful than the methods we have studied so far. The approach begins with a simplifying assumption, that the potential may be written as a product (or in some cases, the sum) of some simpler functions.

2.1 Cartesian: The Laplace equation in Cartesian coordiates is ∇2V (x, y, z) = 0 ∂2V ∂2V ∂2V + + = 0 ∂x2 ∂y2 ∂z2 We assume a solution of the form V (x, y, z) = X (x) Y (y) Z (z) Substitution gives d2X d2Y d2Z YZ + XZ + XY = 0 dx2 dy2 dz2 where the partial derivatives now become ordinary. Dividing by V = XYZ gives 1 d2X 1 d2Y 1 d2Z + + = 0 X dx2 Y dy2 Z dz2 and each terms is now a function of only one variable. Taking the partial derivative of the whole expression with respect to x gives ∂ 1 d2X ∂ 1 d2Y ∂ 1 d2Z 0 = + + ∂x X dx2 ∂x Y dy2 ∂x Z dz2 ∂ 1 d2X = + 0 + 0 ∂x X dx2

1 d2X 2 showing that X dx2 must be independent of x, hence constant. Choosing this constant to be −α (either sign is allowed, we choose positive here), gives 1 d2X = −α2 X dx2 d2X + α2X = 0 dx2

2 with the familiar solution, X = A sin αx + B cos αx The remaining equation becomes 1 d2Y 1 d2Z −α2 + + = 0 Y dy2 Z dz2 1 d2Y 1 d2Z + − α2 = 0 Y dy2 Z dz2 and differentiating with respect to y shows that the y term is constant. Choosing −β2 for the constant, 1 d2Y = −β2 Y dy2 d2Y + β2Y = 0 dy2 we again find an oscillating solution, Y = C sin βy + D cos βy Finally, we have d2Z − α2 + β2 Z = 0 dz2 For convenience, define γ2 = α2 + β2, so that d2Z − γ2Z = 0 dz2 It is easy to see that e±γz both give solutions, so the general solution is the arbitrary linear combination Z = E0eγz + F 0e−γz It is often more useful to use the symmetric and antisymmetric combinations, eγz + e−γz cosh γz = 2 eγz − e−γz sinh γz = 2 Then eγz = cosh γz + sinh γz e−γz = cosh γz − sinh γz and our solution is Z = E0 (cosh γz + sinh γz) + F 0 (cosh γz − sinh γz) = (E0 + F 0) cosh γz + (E0 − F 0) sinh γz Renaming the constants, E = E0 + F 0 and F = E0 − F 0, this is just Z = E cosh γz + F sinh γz The solution for our ansatz is therefore

Vα,β = (A sin αx + B cos αx)(C sin βy + D cos βy)(E cosh γz + F sinh γz) and the general solution for V is a sum of such solutions for diﬀerent α and β, where the leading constants may be diﬀerent for each choice of α and β: X V = (Aα,β sin αx + Bα,β cos αx)(Cα,β sin βy + Dα,β cos βy)(Eα,β cosh γz + Fα,β sinh γz) α,β

3 2.2 Spherical coordinates In spherical coordinates the Laplace equation takes the form

1 ∂ ∂V 1 ∂ ∂V 1 ∂2V r2 + sin θ + = 0 r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂ϕ2 Let V = R (r)Θ(θ)Φ(ϕ) and substitute. Dividing by RΘΦ, we have

1 d dR 1 1 d dΘ 1 1 d2Φ r2 + sin θ + = 0 r2R dr dr r2 sin θ Θ dθ dθ r2 sin2 θ Φ dϕ2

This time, nothing separates until we multiply by r2 sin2 θ,

r2 sin2 θ d dR sin θ d dΘ 1 d2Φ r2 + sin θ + = 0 r2R dr dr Θ dθ dθ Φ dϕ2 and now we see that, enforcing periodicity of Φ by our choice of sign,

d2Φ + m2Φ = 0 dϕ2 Φ = C sin mϕ + D cos mϕ

leaving sin2 θ d dR sin θ d dΘ r2 + sin θ − m2 = 0 R dr dr Θ dθ dθ Now divide by sin2 θ, 1 d dR 1 1 d dΘ m2 r2 + sin θ − = 0 R dr dr Θ sin θ dθ dθ sin2 θ This time, we choose the constant to be l (l + 1), obviously with hindsight. This results in

d dR r2 − l (l + 1) R = 0 dr dr 1 d dΘ m2 sin θ + l (l + 1) − Θ = 0 sin θ dθ dθ sin2 θ The ﬁrst of these is not bad. Let R = rα. Then d dR r2 − l (l + 1) R = 0 dr dr d drα r2 − l (l + 1) rα = 0 dr dr d αrα+1 − l (l + 1) rα = 0 dr α (α + 1) rα − l (l + 1) rα = 0

Being careful to get all solutions, we solve the quadratic,

α2 + α − l (l + 1) = 0 −1 ± p1 + 4l (l + 1) α = 2

4 √ −1 ± 4l2 + 4l + 1 = 2 −1 ± (2l + 1) = 2 l = − (l + 1) giving two solutions as we should expect for a second order equation. Taking a general linear combination, B R (r) = Arl + rl+1 The solutions to the θ equation with m = 0 are called the Legendre polynomials. Solutions with general m may be found by diﬀerentiating the Legendre polynomials, giving the associated Legendre polynomials. We consider only the m = 0 case. To start, we deﬁne a new variable, x = cos θ Then d dx d = dθ dθ dx d = − sin θ dx and therefore d 1 d = − dx sin θ dθ Thus, with m = 0, d dΘ sin2 θ + l (l + 1) Θ = 0 dx dx d dΘ 1 − x2 + l (l + 1) Θ = 0 dx dx This is the Legendre equation, with solutions

Θ(θ) = Pl (x) where x = cos θ. Like the harmonic functions, there are many relationships among the Legendre polynomials. The most important for us now is the orthogonality relationship,

π 2 2 P (cos θ) P 0 (cos θ) sin θdθ = δ 0 ˆ l l 2l + 1 ll 0 2 2 0 Notice that the right side may be written as either 2l+1 δll0 or as 2l0+1 δll0 since it is only nonzero when l = l . We may also write this in terms of x = cos θ. With dx = − sin θdθ, we have

1 2 P (x) P 0 (x) dx = δ 0 ˆ l l 2l + 1 ll −1 The general solution for a potential depending only on θ and r, but not ϕ, is ∞ X Bl V (r, θ) = A rl + P (cos θ) l rl+1 l l=0

5 Example 1: Potential on a sphere Let a sphere of radius R have potential

4 V (r = R, θ, ϕ) = V0 sin θ

Find the potential everywhere inside and outside the sphere.

Inside: Since we have spherical boundary conditions, it is easiest to use the spherical separation, and since the problem has azimuthal symmetry, we may use the solution to the Laplace equation in the form ∞ X Bl V (r, θ) = A rl + P (cos θ) l rl+1 l l=0 Our boundary conditions are

V (0) = finite 4 V (R, θ, ϕ) = V0 sin θ

The ﬁrst of these conditions shows that Bl = 0 for all l. The outer boundary condition becomes

∞ 4 X l V0 sin θ = AlR Pl (x) l=0

4 Essentially, we must express V0 sin θ in terms of Legendre polynomials. Let x = cos θ. Then noting that we may write

4 2 2 V0 sin θ = V0 1 − cos θ 22 = V0 1 − x 2 4 = V0 1 − 2x + x we should be able to write the potential on the sphere in terms of Legendre polynomials of even order less than or equal to x4. Using higher order polynomials would introduce undesired higher powers of x. The three relevant polynomials are:

P0 (x) = 1 1 P (x) = 3x2 − 1 2 2 1 P (x) = 35x4 − 30x2 + 3 4 8 Rearranging and combining to produce x4 − 2x2 + 1, we start with 1 x4 = 8P (x) + 30x2 − 3 35 4 1 x2 = (2P (x) + 1) 3 2 1 x4 − 2x2 + 1 = 8P (x) + 30x2 − 3 − 2x2 + 1 35 4 8 30 3 = P (x) + − 2 x2 + 1 − 35 4 35 35

6 8 6 14 32 = P (x) + − x2 + 35 4 7 7 35 8 8 1 32 = P (x) − (2P (x) + 1) + 35 4 7 3 2 35 8 16 8 32 = P (x) − P (x) − + 35 4 21 2 21 35 8 16 8 4 1 = P (x) − P (x) + − 35 4 21 2 7 5 3 8 16 8 7 = P (x) − P (x) + 35 4 21 2 7 15 8 16 8 = P (x) − P (x) + P (x) 35 4 21 2 15 0 Therefore, we must solve ∞ 8 16 8 X V P (x) − P (x) + P (x) = A RlP (x) 0 35 4 21 2 15 0 l l l=0

for the coeﬃcients Al. To do this, we use the orthogonality relation, 1 2 Pl (x) Pl0 (x) dx = δll0 ˆ−1 2l + 1

Multiply our equation by an arbitrary Pk (x) for any fixed k, and integrate over all x, ∞ 1 8 16 8 X 1 V P (x) − P (x) + P (x) P (x) dx = A Rl P (x) P (x) dx 0 ˆ 35 4 21 2 15 0 k l ˆ l k −1 l=0 −1 1 1 1 ∞ 8V0 16V0 8V0 X 2 P (x) P (x) dx − P (x) P (x) dx + P (x) P (x) dx = A Rl δ 35 ˆ 4 k 21 ˆ 2 k 15 ˆ 0 k l 2l + 1 lk −1 −1 −1 l=0 8V 2 16V 2 8V 2 2 0 δ − 0 δ + 0 δ = A Rk 35 2k + 1 k4 21 2k + 1 k2 15 1 0k k 2k + 1 8V 2 16V 2 8V 2 2 0 δ − 0 δ + 0 δ = A Rk 35 9 k4 21 5 k2 15 1 0k k 2k + 1 The left side vanishes unless k = 0, 2 or 4, and in these cases we have 8V A = 0 0 15 16V A = − 0 2 21R2 8V A = 0 4 35R4 The full solution everywhere inside the sphere is found by putting these coefficients back into the general form, giving the final potential as ∞ X Bl V (r, θ) = A rl + P (cos θ) l rl+1 l l=0 0 2 4 = A0r P0 (cos θ) + A2r P2 (cos θ) + A4r P4 (cos θ) 8V 16V 8V = 0 r0P (cos θ) − 0 r2P (cos θ) + 0 r4P (cos θ) 15 0 21R2 2 35R4 4 1 2r2 r4 = 8V − P (cos θ) + P (cos θ) 0 15 21R2 2 35R4 4

7 We immediately verify that if we set r = R and use the form of sin4 θ in terms of Legendre polynomials, we recover the boundary condition. At the center of the sphere, we ﬁnd (setting r = 0) that the potential is 8 V (0) = 15 V0.

Aside: Verify Theorem 3.1.4 If we integrate the potential over the surface of the sphere (using Wolfram integrator),

π 2π π sin θdθ dϕV sin4 θ = 2πV sin5 θdθ ˆ ˆ 0 0 ˆ 0 0 0 π 5 5 3 1 = 2πV0 − cos θ + cos θ − cos 5θ 8 48 80 0 5 5 1 5 5 1 = 2πV − + − − + − 0 8 48 80 8 48 80 5 5 1 5 5 1 = 2πV − + + − + 0 8 48 80 8 48 80 5 5 1 = 2πV − + 0 4 24 40 150 25 3 = 2πV − + 0 120 120 120 128 = 2πV 0 120 32π = V 15 0 so that the potential at the center of the sphere is

π 2π 8 1 V = V = V (R, θ, ϕ) R2 sin θdθdϕ center 15 0 4πR2 ˆ ˆ 0 0 as required by the theorem of section 3.1.4 in Griﬃths.

Outside: The exterior solution is similar, but this time the boundary conditions are

V (∞) = 0 4 V (R, θ, ϕ) = V0 sin θ

which means that we must have Al = 0 for all l. The second condition now reads ∞ 8 16 8 X Bl V P (x) − P (x) + P (x) = P (x) 0 35 4 21 2 15 0 Rl+1 l l=0

l Bl so we need only replace AlR of the interior solution by Rl+1 . This gives R 2R3 R5 V (r, θ) = 8V − P (cos θ) + P (cos θ) 0 15r 21r3 2 35r5 4

8 Problem 3.22: Extending the solution for a disk We have found previously that the potential on the z-axis above a circular disk of radius R lying in the xy-plane is σ p V (z) = z2 + R2 − z 20 Since we have axial symmetry, this must equal V (r, θ = 0) for our general solution to the Laplace equation, ∞ X Bl V (r, θ) = A rl + P (cos θ) l rl+1 l l=0

Since, on the positive z axis we have cos θ = cos 0 = 1, we use Pl (1) = 1 for all l. Furthermore, we have r = z, and the equality requires

∞ σ p 2 2 X l Bl r + R − r = Alr + l+1 Pl (cos θ) 20 r l=0 Although this is not a boundary condition in the usual sense, it is still enough information to ﬁnd all of the coeﬃcients and construct the potential everywhere. To do this, we need to expand the left side of the r R equation as a power series in r. For the series to converge, we need to expand in powers of either R or r , whichever is less than one. This means we divide the problem in to two regions, r < R and r > R. For r < R, we subdivide into two regions since we must match the charge density on the disk, using

dV dV 1 − = − σ dn above dn below 0

dV 1 ∂V π On the disk, the normal derivative is dn = − r ∂θ θ = 2 and σ is a given constant. We must consider the series solutions for positive and negative z separately. For the negative z axis, we have cos θ = cos π = −1, and because the Legendre polynomials are either l even or odd polynomials, we have Pl (−1) = (−1) . Therefore, we will equate

∞ 0 σ p 2 2 X l 0 l Bl r + R + r = (−1) Alr + l+1 20 r l=0

0 0 Notice that this is a distinct region, so the constants Al and Bl are distinct from those above the disk.

π Case 1a: r < R, θ < 2 In this case we expand r p r2 r2 + R2 = R 1 + R2 r2 √ Let x = R2 so that we need the Taylor series for f (x) = 1 + x. Taking derivatives,

f (x) = (1 + x)1/2 1 f (1) (x) = (1 + x)−1/2 2 1 1 f (2) (x) = − (1 + x)−3/2 2 2 1 1 3 f (3) (x) = (1 + x)−5/2 2 2 2 1 1 3 5 f (4) (x) = − (1 + x)−7/2 2 2 2 2

9 so evaluating f (n) (0), the ﬁrst few deriatives are

f (0) = 1 1 f (1) (0) = 2 1 f (2) (0) = − 4 3 f (3) (0) = 8 15 f (4) (0) = − 16 and we have ∞ X 1 f (x) = f (n) (0) xn n! n=0 1 1 1 5 = 1 + x − x2 + x3 − x4 − · · · 2 8 16 128 The potential on the z-axis for r < R therefore has the expansion σ p V (r, 0) = r2 + R2 − r 20 σR r2 r = f 2 − 20 R R σR 1 r2 1 r4 1 r6 5 r8 r = 1 + 2 − 4 + 6 − 8 − · · · − 20 2 R 8 R 16 R 128 R R σR r 1 r2 1 r4 1 r6 5 r8 = 1 − + 2 − 4 + 6 − 8 + ··· 20 R 2 R 8 R 16 R 128 R and this must satisfy

2 4 6 8 ∞ σR r 1 r 1 r 1 r 5 r X l Bl 1 − + 2 − 4 + 6 − 8 + ··· = Alr + l+1 Pl (cos 0) 20 R 2 R 8 R 16 R 128 R r l=0

σR 1 Clearly, with Pl (1) = 1, this can only be satisﬁed if Bl = 0 for all l and Al = 0 for l odd except A1 = − , 20 R while for l even we equate σR A0 = 20 σR 1 A2 = 2 20 2R σR 1 A4 = − 4 20 8R σR 1 A6 = 6 20 16R σR 5 A8 = − 8 20 128R and so on, as far as we need to go. The potential everywhere with r < R is therefore σR r 1 r2 1 r4 1 r6 5 r8 V (r, θ) = 1 − P1 (cos θ) + 2 P2 (cos θ) − 4 P4 (cos θ) + 6 P6 (cos θ) − 8 P8 (cos θ) − · · · 20 R 2 R 8 R 16 R 128 R

10 π Case 1b: r < R, θ > 2 This time we expand require

∞ 0 σ p 2 2 X l 0 l Bl r + R + r = (−1) Alr + l+1 20 r l=0 so we still need the expansion for r p r2 r2 + R2 = R 1 + R2 The potential on the z-axis for r < R therefore has the expansion σ p V (r, 0) = r2 + R2 + r 20 σR 1 r2 1 r4 1 r6 5 r8 r = 1 + 2 − 4 + 6 − 8 − · · · + 20 2 R 8 R 16 R 128 R R σR r 1 r2 1 r4 1 r6 5 r8 = 1 + + 2 − 4 + 6 − 8 + ··· 20 R 2 R 8 R 16 R 128 R and this must satisfy

2 4 6 8 ∞ σR r 1 r 1 r 1 r 5 r X l l Bl 1 + + 2 − 4 + 6 − 8 + ··· = (−1) Alr + l+1 20 R 2 R 8 R 16 R 128 R r l=0

σR 1 Again this can only be satisﬁed if Bl = 0 for all l and Al = 0 for l odd except A1 = + 2 R . For l even we π 0 have the same series as in Case 1a so the potential everywhere with r < R, θ > 2 is σR r 1 r2 1 r4 1 r6 5 r8 V (r, θ) = 1 + P1 (cos θ) + 2 P2 (cos θ) − 4 P4 (cos θ) + 6 P6 (cos θ) − 8 P8 (cos θ) − · · · 20 R 2 R 8 R 16 R 128 R To evaluate this on the disk, we need 1 dV dV − σ = − 0 dn above dn below π θ= 2 d = (Vabove − Vbelow) dn π θ= 2 1 d σr σr = − − P1 (cos θ) − P1 (cos θ) r dθ 20 20 π θ= 2 1 d σr = − − cos θ r dθ 0 π θ= 2 σ = − sin θ 0 π θ= 2 σ = − 0 which miraculously checks.

Case 2: r > R When r > R, we need to expand r p R2 r2 + R2 = r 1 + r2

11 R2 so we use the same Taylor series but this time with x = r2 , σ p V (r, 0) = r2 + R2 − r 20 r ! σr R2 = 1 + 2 − 1 20 r σR r 1 R2 1 R4 1 R6 5 R8 = 1 + 2 − 4 + 6 − 8 + · · · − 1 20 R 2 r 8 r 16 r 128 r σR 1 R 1 R3 1 R5 5 R7 = − 3 + 5 − 7 + ··· 20 2 r 8 r 16 r 128 r so equating coeﬃcients we now see that all Al vanish as well as the odd-indexed Bl so that B2k+1 = 0. We then solve for the even Bl

3 5 7 ∞ σR 1 R 1 R 1 R 5 R X Bl − 3 + 5 − 7 + ··· = l+1 20 2 r 8 r 16 r 128 r r l=0

σR R B0 = 20 2 σR R3 B2 = − 20 8 σR R5 B4 = 20 16 σR 5R7 B6 = − 20 128 and reconstruct the series,

σR R 1 R3 1 R5 1 R7 V (r, θ) = − 3 P2 (cos θ) + 5 P4 (cos θ) − 7 P6 (cos θ) + ··· 20 2r 8 r 16 r 128 r 3 Multipole moments 3.1 The multipole expansion Suppose we have a localized charge distribution, confined to a region near the origin with r < R. Then for values of r > R, the electric field must be described by solutions to the Laplace equation with vanishing boundary condition at infinity. This requires Al = 0 for an expansion in Legendre polynomials and the potential for any azimuthally symmetric source is

∞ X 1 V (r, θ) = B P (cos θ) rl+1 l l l=0 For large values of r, the lowest nonvanishing term will dominate this approximation, giving an excellent approximation for the potential Now consider the full solution for any localized source near the origin. Let the charge density be ρ (r0), and suppose we with to know the potential at a point r far from the source. The general solution for the potential is 1 ρ (r0) d3r0 V (r) = 0 4π0 ˆ |r − r |

12 Let the angle between r and r0 be α, so the denominator is p |r − r0| = r2 + r02 − 2rr0 cos α r r02 − 2rr0 cos α = r 1 + r2 √ Since r r0, the square root has the form 1 + x with x 1. To expand the integrand, we need 1 1 3 5 √ = 1 − x + x2 − x3 + ··· 1 + x 2 8 16 where r02 − 2rr0 cos α r02 2r0 x = = 1 − cos α r2 r2 r

1 1 0 3 0 V (r) = 0 ρ (r ) d r 4π0 ˆ |r − r |

1 Compute the Legendre series for |r−r0| :

∞ 1 X 1 = B P (cos α) |r − r0| rl+1 l l l=0

To ﬁnd the coeﬃcients, multiply by Pk (cos α) and integrate over all angles.

∞ 1 1 X 1 Pk (x) dx Bl Pl (x) Pk (x) dx = √ rl+1 ˆ ˆ r2 + r02 − 2rr0x l=0 −1 −1 1 1 2 Pk (x) dx k+1 Bk = r 2k + 1 ˆ q r02 r0 −1 r 1 + r2 − 2x r 1 ! 1 1 r02 r0 3 r02 r0 2 5 r02 r0 3 = P (x) dx 1 − − 2x + − 2x − − 2x + ··· r ˆ k 2 r2 r 8 r2 r 16 r2 r −1 1 1 r02 r0 3 r04 3 r03 3 r02 = P (x) dx 1 − + x + − x + x2 r ˆ k 2r2 r 8 r4 2 r3 2 r2 −1 1 1 5 r06 15 r05 15 r04 5 r03 + P (x) dx − + x − x2 + x3 + ··· r ˆ k 16 r6 8 r5 4 r4 2 r3 −1 1 1 r0 r02 3 r02 3 r03 5 r03 r04 = P (x) dx 1 + x − + x2 − x + x3 + O r ˆ k r 2r2 2 r2 2 r3 2 r3 r4 −1 1 1 r0 r02 1 r03 1 r04 = P (x) dx 1 + x + 3x2 − 2 + 5x3 − 3x + O r ˆ k r r2 2 r3 2 r4 −1 1 1 r0 r02 r03 r04 = P (x) dx P (x) + P (x) + P (x) + P (x) + O r ˆ k 0 r 1 r2 2 r3 3 r4 −1

13 1 ∞ n 1 X r0 ⇒ P (x) dxP (x) r r ˆ k n n=0 −1 1 r0 k 2 = r r 2k + 1 0k Bk = r

While showing the ﬁrst three terms of an inﬁnite series is not a proof, this pattern does continue, and we end up with ∞ l 1 1 X r0 = P (cos α) |r − r0| r r l l=0 A complete proof makes use of the addition theorem for spherical harmonics. Now, returning to the potential,

1 1 0 3 0 V (r) = 0 ρ (r ) d r 4π0 ˆ |r − r | ∞ 0 l 1 1 X r 0 3 0 = Pl (cos α) ρ (r ) d r 4π0 ˆ r r l=0 ∞ 1 X 1 0 l 0 3 0 = l+1 (r ) ρ (r ) Pl (cos α) d r 4π0 r ˆ l=0 Notice that the integrals, 1 0 l 0 3 0 Ql = (r ) ρ (r ) Pl (cos α) d r 4π0 ˆ are independent of r, and therefore only characterize the source. These are the multipole moments of the charge distribution. The potential is then

∞ X Ql V (r) = rl+1 l=0 It is important to be able to recognize and understand the properties of the ﬁrst few multipole moments. They are extremely useful for describing arbitrary charge distributions.

3.2 Total charge The zeroth moment is proportional to the total charge,

1 0 0 0 3 0 Q0 = (r ) ρ (r ) P0 (cos α) d r 4π0 ˆ 1 = ρ (r0) d3r0 4π0 ˆ 1 = Qtotal 4π0 The potential of a pure monopole is given by Coulomb’s law for a single point charge, Q V (r) = 4π0r

14 and the resulting electric ﬁeld is

E = −∇V ∂ Q = −ˆr ∂r 4π0r Q ˆ = 2 r 4π0r as we know.

3.3 Dipole moment The next term depends on the dipole moment,

1 0 0 3 0 Q1 = r ρ (r ) P1 (cos α) d r 4π0 ˆ 1 = r0ρ (r0) cos αd3r0 4π0 ˆ 1 = ˆr · r0ρ (r0) d3r0 4π0 ˆ 1 = ˆr · r0ρ (r0) d3r0 4π0 ˆ We deﬁne the dipole moment to be p ≡ r0ρ (r0) d3r0 ˆ A dipole moment may be thought of as a pair of equal and opposite charges, slightly displaced. If we take charges ±q displaced by a vector dkˆ, then the charge density is

d d ρ (r0) = q δ3 r0 − kˆ − δ3 r0 + kˆ 2 2

so the dipole moment is

p ≡ r0ρ (r0) d3r0 ˆ d d = r0q δ3 r0 − kˆ − δ3 r0 + kˆ d3r0 ˆ 2 2 d d = q r0δ3 r0 − kˆ d3r0 − r0δ3 r0 + kˆ d3r0 ˆ 2 ˆ 2 d d = q kˆ + kˆ 2 2 = qdkˆ that is, the vector displacement between the two charges times one charge. To see if this is a pure dipole, we check the higher moments of the distribution,

1 0 l 0 3 0 Ql = (r ) ρ (r ) Pl (cos α) d r 4π0 ˆ q 0 l 3 0 d ˆ 3 0 d ˆ 3 0 = (r ) δ r − k − δ r + k Pl (cos α) d r 4π0 ˆ 2 2

15 q 0 l 3 0 d ˆ 3 0 d ˆ 0 02 0 0 0 0 = (r ) δ r − k − δ r + k Pl (cos θ ) r dr sin θ dθ dϕ 4π0 ˆ 2 2 q 3 0 d ˆ 0 0l+2 0 0 0 0 3 0 d ˆ 0 0l+2 0 0 0 0 = δ r − k Pl (cos θ ) r dr sin θ dθ dϕ − δ r + k Pl (cos θ ) r dr sin θ dθ dϕ 4π0 ˆ 2 ˆ 2

Letting x0 = sin θ0, the delta functions may be written as

d 4 d δ3 r0 − kˆ = δ r0 − δ (x0 − 1) 2 2πd2 2 d 4 d δ3 r0 + kˆ = δ r0 − δ (x0 + 1) 2 2πd2 2 so that

 ∞ 1 2π ∞ 1 2π  q 4 0 d 0 0 0l+2 0 0 0 0 d 0 0 0l+2 0 0 0 Ql = 2  δ r − δ (x − 1) Pl (x ) r dr dx dϕ − δ r − δ (x + 1) Pl (x ) r dr dx dϕ  4π0 2πd ˆ ˆ ˆ 2 ˆ ˆ ˆ 2 0 −1 0 0 −1 0  ∞ 1 ∞ 1  q 0l+2 0 d 0 0 0 0 0 d 0l+2 0 0 0 0 = 2  r δ r − dr δ (x − 1) Pl (x ) dx − δ r − r dr δ (x + 1) Pl (x ) dx  π0d ˆ 2 ˆ ˆ 2 ˆ 0 −1 0 −1 ! q dl+2 dl+2 = 2 Pl (1) − Pl (−1) π0d 2 2 q dl = 1 − (−1)l 4π0 2 ( l 2q d l odd = 4π0 2 0 l even

This means that the distribution is not a “pure” dipole, but contains higher moments as well. To obtain a pure dipole, we take the limit as the separation of charges vanishes, d → 0, in such a way that the product p = qd remains constant, d d ρ (r0) = lim q δ3 r0 − kˆ − δ3 r0 + kˆ d→0 2 2 In this case, the odd moments become

2q d2k+1 Q2k+1 = lim d→0 4π0 2 qd d2k = limd→0 4π0 2 p d2k = lim 4π0 d→0 2 p k = 0 = 4π0 0 k > 0

and we have a pure dipole with dipole moment p = pkˆ For an arbitrary pure dipole with dipole moment p, the potential is therefore

ˆr · p V (r) = 2 4π0r

16 The electric field of a dipole lying along the z-axis, p = pkˆ, is found by taking the gradient, E = −∇V (r) ˆr · p = −∇ 2 4π0r ∂ 1 ∂ 1 ∂ p cos θ ˆ ˆ ˆ = − r + θ + ϕ 2 ∂r r ∂θ r sin θ ∂ϕ 4π0r ∂ p cos θ 1 ∂ p cos θ ˆ ˆ = − r 2 + θ 2 ∂r 4π0r r ∂θ 4π0r 2p cos θ p sin θ ˆ ˆ = − r − 3 − θ 3 4π0r 4π0r 1 ˆ ˆ ˆ = 3 2 (r · p) r + p sin θθ 4π0r We may rewrite the second term using ˆr × (ˆr × p) = (ˆr · p) ˆr − p (ˆr · ˆr) = (ˆr · p) ˆr − p together with ˆr × (ˆr × p) = −ˆr × ϕˆp sin θ = θˆp sin θ The electric field becomes 1 ˆ ˆ ˆ ˆ E = 3 (2 (r · p) r + (r · p) r − p) 4π0r 3 (ˆr · p) ˆr − p E = 3 4π0r 3.4 Quadrupole For the quadrupole moment, we need to compute ∞ 1 2π 1 02 0 0 0 0 2 0 Q2 = r dr dx dϕ (r ) ρ (r ) P2 (cos α) 4π0 ˆ ˆ ˆ 0 −1 0 Choose the observation point r along the z-axis, so that cos α = cos θ0 = x0. Then we have ∞ 1 2π 1 02 0 0 0 0 02 02 Q2 = r dr dx dϕ ρ (r ) 3x − 1 r 8π0 ˆ ˆ ˆ 0 −1 0 We need to rewrite this to separate the direction ˆr at which we wish to know the field, from the primed integration variables. Using index notation, this is not hard: 3x02 − 1 r02 = 3r02 cos2 θ0 − r02 = 3 (ˆr · r0)2 − r0 · r0 3 3 3 X 0 X 0 02 X = 3 rîri rˆjrj − r δijrîrˆj i=1 j=1 i,j=1 3 X 0 0 02 = rîrˆj 3rirj − r δij i,j=1

17 P3 This uses the dot product of the unit vector, 1 = ˆr · ˆr = i,j=1 δijrˆirˆj and the summation form of the dot 0 P3 0 product, ˆr · r = i=1 rˆiri. Now we may write the quadrupole moment in a way that is independent of where we choose the direction of ˆr:

3  ∞ 1 2π  1 X 02 0 0 0 0 0 0 02 Q2 = rˆirˆj  r dr dx dϕ ρ (r ) 3rirj − r δij  8π0 ˆ ˆ ˆ i,j=1 0 −1 0 3 1 X = rˆ Q rˆ 8π i ij j 0 i,j=1 where we have deﬁned the quadrupole moment tensor,

∞ 1 2π Q ≡ r02dr0 dx0 dϕ0ρ (r0) 3r0r0 − r02δ ij ˆ ˆ ˆ i j ij 0 −1 0

We can develop a pure quadrupole moment tensor in the same way as we built a dipole moment vector, d by placing four charges at the corners of a square of side 2 in the xy plane. Then the charge density is d d d d d d d d ρ (r0) = lim q δ3 r0 − î − ˆj − δ3 r0 − î + ˆj + δ3 r0 + î + ˆj − δ3 r0 + î − ˆj d→0 2 2 2 2 2 2 2 2 and the quadrupole moment may be evaluated easily in Cartesian coordinates. The Dirac delta functions become, for example, d d d d δ3 r0 − î − ˆj = δ (z0) δ x0 − δ y0 − 2 2 2 2 and so on, changing signs appropriately. Then, noticing that the quadrupole tensor is a symmetric matrix, Qij = Qji, we need to compute Qxx,Qxy,Qxz,Qyy,Qyz,Qzz. We can avoid one of thes by noticing that the trace of the matrix is

3 X tr (Qij) = Qii i=1 3 X 0 0 02 ∝ 3riri − r δii i=1 3 X 0 0 02 = 3riri − 3r i=1 = 3r02 − 3r02 = 0

Therefore, Qzz = −Qxx − Qyy. Computing one at a time. For the diagonal elements,

∞ ∞ ∞ Q = dx0 dy0 dz0ρ (r0) 3x02 − x02 + y02 + z02 δ xx ˆ ˆ ˆ 11 −∞ −∞ −∞ ∞ ∞ d d d d = q dx0 dy0 δ x0 − δ y0 − − δ x0 − δ y0 + 2x02 − y02 ˆ ˆ 2 2 2 2 −∞ −∞

18 ∞ ∞ d d d d +q dx0 dy0 δ x0 + δ y0 + − δ x0 + δ y0 − 2x02 − y02 ˆ ˆ 2 2 2 2 −∞ −∞ ! d2 d2 d2 d2 = q − + − 2 2 2 2 = 0 and ∞ ∞ ∞ Q = dx0 dy0 dz0ρ (r0) 3y02 − x02 + y02 + z02 δ yy ˆ ˆ ˆ 22 −∞ −∞ −∞ ∞ ∞ d d d d = q dx0 dy0 δ x0 − δ y0 − − δ x0 − δ y0 + 2y02 − x02 ˆ ˆ 2 2 2 2 −∞ −∞ ∞ ∞ d d d d +q dx0 dy0 δ x0 + δ y0 + − δ x0 + δ y0 − 2y02 − x02 ˆ ˆ 2 2 2 2 −∞ −∞ ! d2 d2 d2 d2 = q − + − 2 2 2 2 = 0 and therefore Qzz = 0 as well. For the oﬀ-diagonal components, ∞ ∞ ∞ Q = dx0 dy0 dz0ρ (r0) 3x0y0 − x02 + y02 + z02 δ xy ˆ ˆ ˆ 12 −∞ −∞ −∞ ∞ ∞ ∞ = dx0 dy0 dz03x0y0ρ (r0) ˆ ˆ ˆ −∞ −∞ −∞ ∞ ∞ d d d d = 3q dx0 dy0 δ x0 − δ y0 − − δ x0 − δ y0 + x0y0 ˆ ˆ 2 2 2 2 −∞ −∞ ∞ ∞ d d d d +3q dx0 dy0 δ x0 + δ y0 + − δ x0 + δ y0 − x0y0 ˆ ˆ 2 2 2 2 −∞ −∞ ! d2 d2 d2 d2 = 3q + + + 2 2 2 2 3 = qd2 4 and ﬁnally ∞ ∞ ∞ Q = dx0 dy0 dz0ρ (r0) 3x0z0 − x02 + y02 + z02 δ xz ˆ ˆ ˆ 13 −∞ −∞ −∞ ∞ ∞ ∞ = dx0 dy0 dz0ρ (r0) 3x0z0 ˆ ˆ ˆ −∞ −∞ −∞

19 = 0

0 because of the δ (z ) in the density. Qyz vanishes in the same way. The limit as d → 0 in such a way that Q = qd2 remains constant gives a pure quadrupole. The quadrupole matrix is therefore simply

 0 1 0  3 Q = qd2 1 0 0 ij 4   0 0 0 and the potential at an arbitrary point r is

1 3 2 V (r) = 5 qd Qijrirj 8π0r 4 3qd2xy = 5 16π0r