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Classical Theory: -

April 27, 20101

1J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Section 1-5 : Electrostatics-Magnetostatics Electrostatics

The behavior of an electrostatic field can be described by two differential equations: ∇~ · E~ = 4πρ (1) (Gauss law) and ∇~ × E~ = 0 (2) the latter equation being equivalent to the statement that E~ is the of a scalar , the scalar potentialΦ: E~ = −∇~ Φ (3) Eqns (1) and (3) can be combined into one differential equation for a single scalar functionΦ( ~x): ∇2Φ = −4πρ (4) This equation is called Poisson equation. In the regions of where there is no , the satisfies the Laplace equation: ∇2Φ = 0 (5)

Classical Field Theory: Electrostatics-Magnetostatics For a general distribution ρ(~x 0), the potential is expected to be the sum over all increments of charge d 3x 0ρ(~x 0), i.e.,

Z ρ(~x 0) Φ(~x) = d 3x 0 (6) |~x − ~x 0|

This potential should satisfy Poisson’s equation. But does it? If we operate with ∇2 on both sides of (6) we get (on ~x not on ~x 0)

Z  1  ∇2Φ(~x) = ρ(~x 0)d 3x 0∇2 (7) |~x − ~x 0|

But ∇2(1/|~x − ~x 0|) = 0 as long as ~x 6= ~x 0 !(Why ?) The singular nature of ∇2(1/|~x − ~x 0|) = ∇2(1/r) can be best expressed in terms of the Dirac δ-function. Since ∇2(1/r) = 0 for r 6= 0 and its volume is −4π (Why?) we can write  1  ∇2 = −4πδ(|~x − ~x 0|) (8) |~x − ~x 0|

Classical Field Theory: Electrostatics-Magnetostatics By definition, if the integration volume contains the point ~x = ~x 0 Z δ3(~x − ~x 0)d 3x = 1 otherwise is zero. This way we recover Poisson’ s equation

2 0 ∇ Φ(~x) = −4πρ(~x )~x0=~x (9)

Thus, we have not only shown that the potential from ’s law satisfies Poisson’s eqn, but we have established (through the solution of Poisson’s eqn) the important result that : the potential from a distributed source is the superposition of the individual potentials from the constituent parcels of charge. We may consider situations in which ρ is comprised of N discrete charges 0 qi , positioned at ~xi so that

N 0 X 3 0 0 ρ(~x ) = qi δ (~x − ~xi ) (10) i=1 In this case the solution for the potential is a combination of terms proportional to1 /|~x − ~x 0|.

Classical Field Theory: Electrostatics-Magnetostatics Green Theorem

If in the electrostatic problem involved localized discrete or continuous distributions of charge with no boundary surfaces, the general solution (6) would be the most convenient and straight forward solution to any problem. To handle the boundary conditions it is necessary to develop some new mathematical tools, namely, the identities or theorems due to George Green (1824). These follow as simple applications of the theorem Z I ∇~ · A~ d 3x = A~ · ~n da (11) V S which applies to any well-behaved vector field A~ defined in the volume V bounded by the closed surface S. Let A~ = Φ∇~ ψ,(Φ and ψ arbitrary scalar fields). Then

∇~ · (Φ∇~ ψ) = Φ∇2ψ + ∇~ Φ · ∇~ ψ (12) ∂ψ Φ∇~ ψ · ~n = Φ (13) ∂n where ∂/∂n is the normal at the surface S.

Classical Field Theory: Electrostatics-Magnetostatics When (12) and (13) substituted into the (8) produces the so-called Green’s 1st identity Z   I ∂ψ Φ∇2ψ + ∇~ Φ · ∇~ ψ d 3x = Φ da . (14) V S ∂n If we rewrite (14) withΦ and ψ interchanged, and then subtract it from (14) we obtain Green’s 2nd identity or Green’s Theorem:

Z I  ∂ψ ∂Φ Φ∇2ψ − ψ∇2Φ d 3x = Φ − ψ da (15) V S ∂n ∂n Now we can apply Poisson’s equation (8) for discrete charge, substituting for ψ = 1/|~x − ~x 0|

Z  4πρ(~x 0) 3 ~ ~ 0 ~ 0 3 −4πδ (x − x )Φ(x ) + 0 d x V |~x − ~x | I     ∂ 1 1 ∂Φ 0 = Φ 0 0 − 0 0 da (16) S ∂n |~x − ~x | |~x − ~x | ∂n

Classical Field Theory: Electrostatics-Magnetostatics Integrating the over all values of ~x 0 within V and for ~x within the volume V yields a nonzero result

Z ρ(~x 0) 1 I  1 ∂Φ ∂  1  ~ 3 0 0 Φ(x) = 0 d x + 0 0 − Φ 0 0 da V |~x − ~x | 4π S |~x − ~x | ∂n ∂n |~x − ~x | (17) • The (blue) correction term goes to zero as the surface S goes to infinity (because S falls of faster than1 /|~x − ~x 0|) • If the integration volume is free of charges, then the first term of equation (17) becomes zero, and the potential is determined only by the values of the potential and the values of its at the boundary of the integration region (the surface S).

Classical Field Theory: Electrostatics-Magnetostatics • Physical experience leads us to believe that specification of the potential on a closed surface defines a unique potential problem. This is called Dirichlet problem or Dirichlet boundary conditions. • Similarly it is plausible that specification of the electric field (normal derivative of the potential) everywhere on the surface (corresponding to a given surface-charge density) also defines a unique problem. The specification of the normal derivative is known as the Newmann boundary condition. • As it turns out either one of the two conditions results in unique solution. It should be clear that a solution to the Poisson eqn with both Φ and ∂Φ/∂n specified arbitrarily on a closed boundary (Cauchy boundary conditions) does not exist since there are unique solutions for Dirichlet and Newmann condition separately.

Classical Field Theory: Electrostatics-Magnetostatics Green Functions

The solution of the Poisson or Laplace eqn in a finite volume V with either Dirichlet or Neumann boundary conditions on the boundary surface S can be obtained by means of Green’s theorem (15) and the so-called Green functions. In obtaining the result (17) we have chosen ψ = 1/|~x − ~x 0| satisfying  1  ∇2 = −4πδ(|~x − ~x 0|) (18) |~x − ~x 0| The function1 /|~x − ~x 0| is only one of a class of functions depending on the variables ~x and ~x 0 and called Green functions. In general ∇02G (~x,~x 0) = −4πδ(|~x − ~x 0|) (19) where 1 G (~x,~x 0) = + F (~x,~x 0) (20) |~x − ~x 0| and F satisfying the Laplace equation inside the volume V ∇02F (~x,~x 0) = 0 (21)

Classical Field Theory: Electrostatics-Magnetostatics If we substitute G(~x,~x 0) in eqn (17) we get Z 1 I  ∂Φ ∂  ~ ~ 0 ~ ~ 0 3 0 ~ ~ 0 ~ 0 ~ ~ 0 0 Φ(x) = ρ(x )G(x, x )d x + G(x, x ) 0 − Φ(x ) 0 G(x, x ) da V 4π S ∂n ∂n (22) The freedom in the definition of G means that we can make the surface integral depend only on the chosen type of BC. • For the Dirichlet BC we demand: 0 0 GD (~x,~x ) = 0 for ~x ∈ S (23) Then the 1st term on the surface integral of (22) vanishes Z 1 I ∂ ~ ~ 0 ~ ~ 0 3 0 ~ 0 ~ ~ 0 0 Φ(x) = ρ(x )GD (x, x )d x − Φ(x ) 0 GD (x, x )da (24) V 4π S ∂n • For Neumann BC the simplest choice of BC on G is ∂G N (~x,~x 0) = 0 for ~x 0 ∈ S (25) ∂n0 but application o the Gauss theorem on (19) shows that (how?) I ∂GN 0 0 da = −4π 6= 0 S ∂n which is incosistent with ∇02G(~x,~x 0) = −4πδ3(~x − ~x 0).

Classical Field Theory: Electrostatics-Magnetostatics This will mean that the outflux of G cannot be zero when there is a source enclosed by S. Then the simplest boundary condition that we can use is ∂G 4π N (~x,~x 0) = − for ~x 0 ∈ S (26) ∂n0 S S is the total area of the boundary surface. Then the solution will be: Z 1 I ∂Φ ~ ~ 0 ~ ~ 0 3 0 ~ ~ 0 0 Φ(x) = ρ(x )GN (x, x )d x + hΦiS + 0 GN (x, x )da (27) V 4π S ∂n where hΦiS is the average value of the potential over the whole surface I 1 0 0 hΦiS ≡ Φ(~x )da (28) S S In most cases S is extremely large (or even infinite), in which case hΦiS → 0. The physical meaning of F (~x,~x 0) : it is a solution of the Laplace eqn inside V and so represents the potential of charges external to the volume V chosen as to satisfy the homogeneous BC of zero potential on the surface S.

Classical Field Theory: Electrostatics-Magnetostatics Green Function

• It is important to understand that no matter how the source is distributed, finding the Green function is completely independent or ρ(~x 0). • G(~x,~x 0) depends exclusively on the geometry of the problem, is a “template”, potential and not the actual potential for a given physical problem. • In other words G(~x,~x 0) is the potential due to a unit charge, positioned arbitrarily within the surface S consistent either with GD = 0 or ∂GN /∂n = −4π/S on the surface. • The “true potential” is a convolution of this template with the given ρ(~x 0).

Classical Field Theory: Electrostatics-Magnetostatics Laplace Equation in Spherical Coordinates

In spherical coordinates( r, θ, φ) the Laplace equation can be written in the form 1 ∂2 1 ∂  ∂Φ 1 ∂2Φ (rΦ) + sin θ + = 0 (29) r 2 ∂r 2 r 2 sin θ ∂θ ∂θ r 2 sin2 θ ∂φ2 If we assume 1 Φ = U(r)P(θ)Q(φ) (30) r Then by substituting in (29) and multiplying with r 2 sin2 θ/UPQ we obtain  1 d 2U 1 d  dP  1 d 2Q r 2 sin2 θ + sin θ + = 0 (31) U dr 2 r 2 sin θP dθ dθ Q dφ2 We see that the therms depending on φ have been isolated and we can set 1 d 2Q = −m2 (32) Q dφ2 with solution Q = e±imφ (33)

Classical Field Theory: Electrostatics-Magnetostatics Similarly the remaining terms can be separated as:

1 d  dP   m2  sin θ + l(l + 1) − P = 0 (34) r 2 sin θ dθ dθ sin2 θ d 2U l(l + 1) − U = 0 (35) dr 2 r 2 The radial equation will have a solution

U = Ar l+1 + Br −l (36) while l is still undetermined.

Classical Field Theory: Electrostatics-Magnetostatics Legendre Equation and

The equation (34) for P(θ) can be expressed in terms of x = cos θ

d  dP   m2  (1 − x 2) + l(l + 1) − P = 0 (37) dx dx 1 − x 2 This is the generalized Legendre equation and its solutions are the associated Legendre functions. We will consider the solution of (54) for m2 = 0

d  dP  (1 − x 2) + l(l + 1)P = 0 (38) dx dx The solution should be single valued, finite, and continuous on the interval −1 ≤ x ≤ 1 in order that it represents a physical potential. The solution can be found in the form of a power

∞ k X j P(x) = x aj x (39) j=0 where k is a parameter to be determined.

Classical Field Theory: Electrostatics-Magnetostatics By substitution in (54) we get the recurrence relation (how?)

(k + j)(k + j + 1) − l(l + 1) a = a (40) j+2 (k + j + 1)(k + j + 2) j while for j = 0, 1 we find that

if a0 6= 0 then k(k − 1) = 0 (41)

if a1 6= 0 then k(k + 1) = 0 (42)

These two relations are equivalent and it is sufficient to choose either a0 or a1 different from zero but not both( why?). We also see that the is either only odd or only on even powers of x. By choosing either k = 0 or k = 1 it is possible to prove the following properties: 2 I The series converges for x < 1, independent of the value of l

I The series diverges for x = ±1, unless it terminates. Since we want solution that is finite at x = ±1, as well as for x 2 < 1, we demand that the series terminates. Since k and j are positive integers or zero, the recurrence relation (40) will terminate only if l is zero or a positive integer.

Classical Field Theory: Electrostatics-Magnetostatics If l is even(odd), then only the k = 0( k = 1) series terminates. The polynomials in each case have x l as their highest power. By convention these polynomials are normalized to have the value unity for x = +1 and are called the Legendre polynomials of order l. The first few are:

P0(x) = 1

P1(x) = x 1 P (x) = (3x 2 − 1) (43) 2 2 1 P (x) = (5x 3 − 3x) 3 2 1 P (x) = (35x 4 − 30x 2 + 3) 4 8 The Legendre polynomials can be taken from Rodrigue’s formula:

1 d l P (x) = (x 2 − 1)l (44) l 2l l! dx l

Classical Field Theory: Electrostatics-Magnetostatics The Legendre polynomials form a complete orthonormal set of functions on the interval −1 ≤ x ≤ 1( prove it)

Z 1 2 Pl 0 (x)Pl (x) = δl 0l (45) −1 2l + 1 Since the Legendre polynomials form a complete set of orthonormal functions, any function f (x) on the interval −1 ≤ x ≤ 1 can be expanded in terms of them i.e. ∞ X f (x) = Al Pl (x) (46) l=0 where (how?) 2l + 1 Z 1 Al = f (x)Pl (x)dx (47) 2 −1 Thus for problems with azimuthal i.e. m = 0 the general solution is: ∞ X h l −(l+1)i Φ(r, θ) = Al r + Bl r Pl (cos θ) (48) l=0 where the coefficients Al [it is not the same as in eqn (47)] and Bl can be determined from the boundary conditions.

Classical Field Theory: Electrostatics-Magnetostatics Example

Boundary Value Problems with Azimuthal Symmetry

Let’s specify as V (θ) the potential on the surface of a sphere of R, and try to find the potential inside the sphere. If there are no charges at the origin( r = 0) the potential must be finite there. Consequently Bl = 0 for all l. Then on the surface of the sphere

∞ X l V (r = R, θ) = Al R Pl (cos θ) (49) l=0

and the coefficients Al will be taken via eqn (47)

2l + 1 Z π Al = l V (θ)Pl (cos θ) sin θdθ (50) 2R 0

Classical Field Theory: Electrostatics-Magnetostatics If, for example V (θ) = ±V on the two hemispheres then the coefficients can be derived easily and the potential inside the sphere is (how?):

3  r  7  r 3 11  r 5  Φ(r, θ) = V P (cos θ) − P (cos θ) + P (cos θ) − ... 2 R 1 8 R 3 16 R 5 (51) To find the potential outside the sphere we merely replace( r/R)l by (R/r)l+1 and the resulting potential will be (how?): " # 3 R 2 7 R 2 Φ(r, θ) = V P (cos θ) − P (cos θ) + ... (52) 2 r 1 12 r 3

Classical Field Theory: Electrostatics-Magnetostatics Legendre Polynomials

An important expansion is that of the potential at ~x due to a unit point charge at ~x0 ∞ 1 X r l = < P (cos γ) (53) |~x − ~x0| l+1 l l=0 r>

where r<(r>) is the smaller (larger) of |~x| and |~x0| and γ is the between |~x| and |~x0|. Show that the potential is : ∞ 1 X   = A r l + B r −(l+1) P (cos γ) on the z-axis |~x − ~x0| l l l l=0 ∞  l 1 1 X r< 0 = on the x-axis |~x − ~x | r> r> l=0

Classical Field Theory: Electrostatics-Magnetostatics Associated Legendre Functions and

For problems without azimuthal (axial) symmetry, we need the generalization of Pl (cos θ), namely, the solutions of d  dP   m2  (1 − x 2) + l(l + 1) − P = 0 (54) dx dx 1 − x 2 for arbitrary l and m. It can be shown that in order to have finite solutions on the interval −1 ≤ x ≤ 1 the parameter l must be zero or a positive integer and that the integer m can take on only the values −l, −(l − 1), ..., 0 , ... ,(l − 1), l (why?). The solution having these properties is called an associated Legendre m function Pl (x) . For positive m it is defined as d m Pm(x) = (−1)m(1 − x 2)m/2 P (x) (55) l dx m l If Rodrigues’ formula is used an expression valid for both positive and negative m is obtained: (−1)m d m Pm(x) = (1 − x 2)m/2 P (x) (56) l 2l l! dx m l

Classical Field Theory: Electrostatics-Magnetostatics −m m There is a simple relation between Pl (x) and Pl (x): (l − m)! P−m(x) = (−1)m Pm(x) (57) l (l + m)! l m For fixed m the functions Pl (x) form an orthonormal set in the index l on the interval −1 ≤ x ≤ 1. The relation is Z 1 m m 2 (l + m)! Pl 0 (x)Pl (x)dx = δl 0l (58) −1 2l + 1 (l − m)! imφ We have found that Qm(φ) = e , this function forms a complete set of orthogonal functions in the index m on the interval0 ≤ φ ≤ 2π. m The product Pl Qm forms also a complete orthonormal set on the surface of the unit sphere in the two indices l, m. From the normalization condition (58) we can conclude that the suitably normalized functions, denoted by Ylm(θ, φ), are : s 2l + 1 (l − m)! Y (θ, φ) = Pm(cos θ)eimφ (59) lm 4π (l + m)! l and also m ∗ Yl,−m(θ, φ) = (−1) Ylm(θ, φ) (60)

Classical Field Theory: Electrostatics-Magnetostatics The normalization and orthogonality conditions are: Z 2π Z π ∗ dφ sin θdθYl 0m0 (θ, φ)Ylm(θ, φ) = δl 0l δm0m (61) 0 0 An arbitrary function g(θ, φ) can be expanded in spherical harmonics

∞ l X X g(θ, φ) = AlmYlm(θ, φ) (62) l=0 m=−l where the coefficients are Z ∗ Alm = dΩYlm(θ, φ)g(θ, φ) . (63)

The general solution for a boundary-value problem in spherical coordinates can be written in terms of spherical harmonics and powers of r in a generalization of (48) :

∞ l X X h l −(l+1)i Φ(r, θ, φ) = Almr + Blmr Ylm(θ, φ) (64) l=0 m=−l If the potential is specified on a spherical surface, the coefficients can be determined by evaluating (64) on the surface and using (63).

Classical Field Theory: Electrostatics-Magnetostatics Spherical Harmonics Ylm(θ, φ)

r 1 l = 0 Y = 00 4π r 3 l = 1 Y = − sin θeiφ 11 8π r 3 Y = cos θ 10 4π 1r 15 l = 2 Y = sin2 θe2iφ 22 4 2π r 15 Y = − sin θ cos θ eiφ 21 8π r 5 3 1 Y = cos2 θ − 20 4π 2 2

Classical Field Theory: Electrostatics-Magnetostatics Spherical Harmonics Ylm(θ, φ)

1r 35 l = 3 Y = − sin3 θ e3iφ 33 4 4π 1r105 Y = sin2 θ cos θe2iφ 32 4 2π 1r 21 Y = − sin θ(5 cos2 θ − 1) eiφ 31 4 4π 1r 7 Y = sin θ 5 cos3 θ − 3 cos θ 30 2 4π (65)

Classical Field Theory: Electrostatics-Magnetostatics Figure: Schematic representation of Ylm on the unit sphere. Ylm is equal to0 along m great circles passing through the poles, and along l − m circles of equal latitude. The function changes sign each it crosses one of these lines. Classical Field Theory: Electrostatics-Magnetostatics Addition Theorem for Spherical Harmonics

The spherical harmonics are related to Legendre polynomials Pl by a relation known as the addition theorem. The addition theorem expresses a Legendre polynomial of order l in the angle γ in terms of products of the spherical harmonics of the θ, φ and θ0, φ0:

l 4π X P (cos γ) = Y ∗ (θ0, φ0)Y (θ, φ) (66) l l(l + 1) lm lm m=−l where γ is the angle between the vectors ~x and ~x 0, ~x · ~x 0 = x · x 0 cos γ and cos γ = cos θ cos θ0 + sin θ sin θ0 cos(φ − φ0). 0 0 Pl (cos γ) is a function of the angles θ, φ with the angles θ , φ as parameters and it maybe expanded in a series (63) :

∞ l 0 X X 0 0 Pl (cos γ) = Al 0m0 (θ , φ )Yl 0m0 (θ, φ) (67) l 0=0 m=−l 0

Classical Field Theory: Electrostatics-Magnetostatics The addition theorem offer the possibility to extend the expansion valid for a point charge (axially symmetric distribution) to an arbitrary charge distribution. By substituting (66) into (53) we obtain

∞ l l 1 X X 1 r< ∗ 0 0 = 4π Y (θ , φ )Ylm(θ, φ) (68) |~x − ~x 0| 2l + 1 r l+1 lm l=0 m=−l >

This equation gives the potential in a completely factorized form in the coordinates ~x and x~0. This is useful in any integration over charge densities, when one is the variable of integration and the other the observation point.

Classical Field Theory: Electrostatics-Magnetostatics A localized distribution of charge is described by the charge density ρ(~x 0), which is nonvanishing only inside a sphere a around some origin. The potential outside the sphere can be written as an expansion in spherical harmonics

∞ l X X 4π 1 Φ(~x) = q Y (θ, φ) (69) 2l + 1 lm r l+1 lm l=0 m=−l

This type of expansion is called multipole expansion; The l = 0 term is called monopole term, The l = 1 are called terms etc. The problem to be solved is the determination of the constants qlm in terms of the properties of the charge distribution ρ(~x 0). The solution is very easily obtained from the integral for the potential Z ρ(~x 0) Φ(~x) = d 3x 0 (70) |~x − ~x 0|

Classical Field Theory: Electrostatics-Magnetostatics Using the expansion (68) for1 /|~x − ~x 0| i.e.

∞ l X X 4π 1 Z Φ(~x) = Y (θ, φ) Y ∗ (θ0, φ0)ρ(~x 0)r 0l d 3x 0 (71) 2l + 1 r l+1 lm lm l=0 m=−l

Consequently the coefficients in (69) are : Z ∗ 0 0 0 0l 3 0 qlm = Ylm(θ , φ )ρ(~x )r d x (72) and called the multipole moments of the charge distribution ρ(~x 0)

Classical Field Theory: Electrostatics-Magnetostatics Monopole l = 0

Here, the only component is Z Z 0 00 ∗ 0 0 3 0 1 0 3 0 q q00 = ρ(~x )r Y (θ , φ )d x = √ ρ(~x )d x = √ (73) 00 4π 4π Observed from a large distance r, any charge distribution acts approximately as if the total charge q (monopole moment) would be concentrated at one point since the dominant term in (68) 1 q Φ(~x)=4 πq∗ Y + ··· = + ... (74) 00 r 00 r

NOTE: The moments with m ≥ 0 are connected (for real charge density) too the moments with m < 0 through

m ∗ ql,−m = (−1) qlm (75)

Classical Field Theory: Electrostatics-Magnetostatics Dipole moment l = 1

In Cartesian coordinates the dipole moment is given by Z ~p = ~x 0ρ(~x 0)d 3x 0 (76)

In Spherical representation one obtains Z ∗ 0 0 ∗ 0 0 3 0 q11 = ρ(~x )r Y11(θ , φ )d x

r 3 Z = − ρ(~x 0)r 0 (sin θ0 cos φ0 − i sin θ0 sin φ0) d 3x 0 8π in Cartesian represantation r 3 Z r 3 = − ρ(~x 0)(x 0 − iy 0)d 3x 0 = − (p − ip ) (77) 8π 8π x y also Z r 3 Z q∗ = ρ(~x 0)r 0Y ∗ (θ0, φ0)d 3x 0 = ρ(~x 0)r 0 cos θ0d 3x 0 10 10 4π r 3 Z r 3 = z0ρ(~x 0)d 3x = p (78) 4π 4π z Classical Field Theory: Electrostatics-Magnetostatics moment l = 2

Z ∗ 02 ∗ 0 0 3 0 q22 = ρ(~x)r Y22(θ , φ )d x

1r 15 Z = ρ(~x)[r 0 sin θ0(cos φ0 − i sin φ0)]2 d 3x 0 4 2π 1r 15 Z = ρ(~x)(x 0 − iy 0)2d 3x 0 4 2π 1 because( x 0 − iy 0)2 = (3x 02 − r 02) − 6ix 0y 0 − (3y 02 − r 02) 3 1 r 15 = (Q − 2iQ − Q ) (79) 12 2π 11 12 22

where Qij is the traceless (why?) quadrupole moment : Z 0 0 2  ~0 3 0 Qij = 3xi xj − r δij ρ(x )d x (80)

Classical Field Theory: Electrostatics-Magnetostatics Quadrupole moment l = 2

Analogously

Z r 15 Z q∗ = ρ(~x 0)r 02Y ∗ (θ0, φ0)d 3x 0 = − ρ(~x 0)z0(x 0 − iy 0)d 3x 0 21 21 8π 1r 15 = − (Q − iQ ) (81) 3 8π 13 23 and

Z 1r 5 Z q∗ = ρ(~x 0)r 02Y ∗ (θ0, φ0)d 3x 0 = ρ(~x 0)(3z0 − r 02)d 3x 0 20 20 2 4π 1r 5 = Q (82) 2 4π 33 From eqn (60) we can get the moments with m < 0 through

m ∗ ql,−m = (−1) qlm (83)

Classical Field Theory: Electrostatics-Magnetostatics Multipole Expansion

By direct Taylor expansion of1 /|~x − ~x 0| the expansion ofΦ( ~x) in rectangular coordinates is:

q ~p · ~x 1 X xi xj Φ(~x) = + + Q + ... (84) r r 3 2 ij r 5 i,j The electric field components for a given multipole can be expressed most easily in terms of spherical coordinates. From E~ = −∇~ Φ and (69) for fixed( l, m) we get(how?): 4π(l + 1) 1 E = q Y r 2l + 1 lm r l+2 lm 4π 1 ∂ E = − q Y (85) θ 2l + 1 lm r l+2 ∂θ lm 4π 1 1 ∂ E = − q Y φ 2l + 1 lm r l+2 sin θ ∂φ lm For a dipole ~p along the z-axis they reduce to: 2p cos θ p sin θ E = , E = , E = 0 (86) r r 3 θ r 3 φ

Classical Field Theory: Electrostatics-Magnetostatics Multipole Expansion

The field intensity at a point ~x due to a dipole ~p at a point x~0 (r = |~x − ~x0|) is

3~n (~p · ~n) − ~p ~r ~ ~ ~ E(x) = 3 with n = (87) |~x − ~x0| r

becauseΦ( ~x) = ~p · ~x/r 3 and

∇(~p · ~x)  1  3~n (~p · ~n) − ~p ~ ~ ~ ~ ~ E(x) = −∇Φ = − 3 − (p · x)∇ 3 = 3 (88) r r |~x − ~x0|

Classical Field Theory: Electrostatics-Magnetostatics of a Charge Distribution in an External Field The multipole expansion of the potential of a charge distribution can also be used to describe the interaction of the charge distribution with an external field. The electrostatic energy of the charge distribution ρ(~x) placed in an external field Φ(~x) is given by Z W = ρ(~x)Φ(~x)d 3x (89) V The external field (if its is slowly varying over the region where ρ(~x) is non-negligible) may be expanded in a : 3 3 1 X X ∂2Φ Φ(~x) = Φ(0) + ~x · ∇~ Φ(0) + xi xj (0) + ... (90) 2 ∂xi ∂xj i=1 j=1 Since E~ = −∇~ Φ for the external field 3 3 ~ 1 X X ∂Ej Φ(~x) = Φ(0) − ~x · E(0) − xi xj (0) + ... (91) 2 ∂xi i=1 j=1 Classical Field Theory: Electrostatics-Magnetostatics Since ∇~ · E~ = 0 for the external field we can substract 1 r 2∇~ · E~ (0) 6 from the last term to obtain finally the expansion:

3 3 ~ 1 X X 2  ∂Ej Φ(~x) = Φ(0) − ~x · E(0) − 3xi xj − r δij (0) + ... (92) 6 ∂xi i=1 j=1

When this inserted into (89) the energy takes the form:

3 3 ~ 1 X X ∂Ej W = qΦ(0) − ~p · E(0) − Qij (0) + ... (93) 6 ∂xi i=1 j=1

Notice, that:

I the total charge interacts with the potential,

I the dipole moment with the electric field,

I the quadrupole with the electric field gradient, etc

Classical Field Theory: Electrostatics-Magnetostatics Examples

EXAMPLE 1 : Show that for a uniform charged sphere all multipole moments vanish exceptq 00.

If the sphere has a radius R0 and constant charge density ρ(~x) = ρ then

Z Z R0 l+3 Z ∗ 0l ∗ 0 0 02 0 0 R0 ∗ 0 0 0 qlm = ρ r Ylm(θ , φ )r dr dΩ = ρ Ylm(θ , φ )dΩ Ω0 0 l + 3 Ω0 √ but since Y00 = 1/ 4π we have from the orthogonality relation

l+3 √ Z l+3 √ ∗ R0 ∗ 0 R0 qlm = ρ 4π YlmY00dΩ = ρ 4πδl0δm0 l + 3 Ω0 l + 3

Classical Field Theory: Electrostatics-Magnetostatics EXAMPLE 2: Perform multipole decomposition of a uniform charge distribution whose surface is a weakly deformed sphere:

2 ! X ∗ R = R0 1 + a2mY2m(θ, φ) , |a2m| << 1 m=−2

The multipole moments are Z Z R(θ0,φ0) ∗ 0l ∗ 02 0 0 qlm = ρr Ylmr dr dΩ Ω0 0 l+3 l+3 Z 2 ! R0 ∗ 0 0 X ∗ ∗ 0 0 0 = ρ Ylm(θ , φ ) 1 + a2mY2m(θ , φ ) dΩ l + 3 0 Ω m=−2

Classical Field Theory: Electrostatics-Magnetostatics l+3 Z 2 ! ∗ R0 ∗ 0 0 X ∗ ∗ 0 0 0 qlm ≈ ρ Ylm(θ , φ ) 1 + (l + 3) a2mY2m(θ , φ ) dΩ l + 3 0 Ω m=−2

Apart from the monopole moment q00 of the previous example, we find

l+3 2 Z ∗ R0 X ∗ ∗ 0 l+3 X ∗ qlm = ρ (l+3) a2µ YlmY2µdΩ = ρR0 a2µδmµδl2 (l > 0) l + 3 0 µ=−2 Ω µ

∗ 5 ∗ Thus the nonvanishing multipole moments are : q2m = ρR0 a2m.

Classical Field Theory: Electrostatics-Magnetostatics Magnetostatics

• There is a radical difference between magnetostatics and electrostatics : there are no free magnetic charges • The basic entity in magnetic studies is the magnetic dipole. • In the presence of magnetic materials the dipole tends to align itself in a certain direction. That is by definition the direction of the magnetic flux density , denoted by B~ (also called magnetic induction). • The magnitude of the flux density can be defined by the magnetic torque N~ exerted on the magnetic dipole: N~ = ~µ × B~ (94) where ~µ is the of the dipole. • In electrostatics the conservation of charge demands that the charge density at any point in space be related to the in that neighborhood by the continuity equation ∂ρ/∂t + ∇~ · ~J = 0 (95) • Steady-state magnetic phenomena are characterized by no change in the net charge density anywhere in space, consequently in magnetostatics ∇~ · ~J = 0 (96) Classical Field Theory: Electrostatics-Magnetostatics Biot & Savart Law

If d~` is an element of length (pointing in the direction of current flow) of a wire which carries a current I and ~x is the coordinate vector from the element of length to an observation point P, then the elementary flux density dB~ at the point P is given in magnitude and direction by

d~` × ~x dB~ = kI (97) |~x|3 NOTE: (97) is an inverse square law, just as is Coulomb’s law of electrostatics. However, the vector character is very different. • If we consider that current is charge in motion and replace Id~` by q~v where q is the charge and ~v the velocity. The flux density for such a charge in motion would be ~v × ~x B~ = kq (98) |~x|3 This expression is time-dependent and valid only for charges with small velocities compared to the speed of light. Classical Field Theory: Electrostatics-Magnetostatics Diff. Equations for Magnetostatics & ’s Law

The basic law (97) for the magnetic induction can be written down in general form for a current density ~J(~x): 1 Z (~x − ~x 0) B~ (~x) = ~J(~x 0) × d 3x 0 (99) c |~x − ~x 0|3 This expression for B~ (~x) is the magnetic analog of electric field in terms of the charge density: 1 Z (~x − ~x 0) E~ (~x) = ρ(~x 0) d 3x 0 (100) c |~x − ~x 0|3 In order to obtain DE equivalent to (99) we use the relation (~x − ~x 0)  1  1 ~r = −∇~ as ∇~ = − (101) |~x − ~x 0|3 |~x − ~x 0| r r 3 and (99) transforms into 1 Z ~J(~x 0) B~ (~x) = ∇~ × d 3x 0 (102) c |~x − ~x 0|

Classical Field Theory: Electrostatics-Magnetostatics Classical Field Theory: Electrostatics-Magnetostatics From (102) follows immediately:

∇~ · B~ = 0 (103)

• This is the 1st equation of magnetostatics and corresponds to ∇~ × E~ = 0 in electrostatics. By analogy with electrostatics we now calculate the of B~

1 Z ~J(~x 0) ∇~ × B~ = ∇~ × ∇~ × d 3x 0 (104) c |~x − ~x 0| which for steady-state phenomena (∇~ · ~J = 0) reduces to (how?) 2

4π ∇~ × B~ = ~J (105) c • This is the 2nd equation of magnetostatics and corresponds to ∇~ · E~ = 4πρ in electrostatics.

“ ” “ ” 2Remember : ∇~ × ∇~ × A~ = ∇~ ∇~ · A~ − ∇2A~ Classical Field Theory: Electrostatics-Magnetostatics The integral equivalent of (105) is called Ampere’s law It can be obtained by applying Stokes’s theorem to the integral of the normal component of (105) over the open surface S bounded by a closed curve C. Thus Z 4π Z ∇~ × B~ · ~n da = ~J · ~nda (106) S c S is transformed into I 4π Z B~ · d~` = ~J · ~nda (107) C c S Since the surface integral of the current density is the total current I passing through the closed curve C, Amp´ere’slaw can be written in the form: I 4π B~ · d~` = I (108) C c

Classical Field Theory: Electrostatics-Magnetostatics Vector Potential

The basic differential laws in magnetostatics are 4π ∇~ × B~ = ~J and ∇~ · B~ = 0 (109) c The problems is how to solve them. • If the current density is zero in the region of interest, ∇~ × B~ = 0 permits the expression of the vector magnetic induction B~ as the gradient ~ of a magnetic scalar potential B = −∇~ ΦM , then (109) reduces to the Laplace equation forΦ M . • If ∇~ · B~ = 0 everywhere, B~ must be the curl of some vector field A~ (~x), called the vector potential

B~ (~x) = ∇~ × A~ (~x) (110)

and from (102) the general form of A~ is:

1 Z ~J(~x 0) A~ (~x) = d 3x 0 + ∇~ Ψ(~x) (111) c |~x − ~x 0|

Classical Field Theory: Electrostatics-Magnetostatics The added gradient of an arbitrary scalar functionΨ shows that, for a given magnetic induction B~ , the vector potential can be freely transformed according to A~ → A~ + ∇~ Ψ (112) This transformation is called a gauge transformation. Such transformations are possible because (110) specifies only the curl of A~ . • If (110) is substituted into the first equation in (109), we find   4π   4π ∇~ × ∇~ × A~ = ~J or ∇~ ∇~ · A~ − ∇2A~ = ~J (113) c c If we exploit the freedom implied by (112) we can make the convenient choice of gauge (Coulomb gauge) ∇~ · A~ = 0. Then each component of the vector potential satisfies the Poisson equation 4π ∇2A~ = − ~J (114) c The solution for A~ in unbounded space is (111) with Ψ =constant:

1 Z ~J(~x 0) A~ (~x) = d 3x 0 (115) c |~x − ~x 0|

Classical Field Theory: Electrostatics-Magnetostatics Further Reading

I Green Function

I Greiner W. ”Classical Electrodynamics”, Chapter 1, 2, 3, 4 I Jackson J.K ”Classical Electrodynamics”, 2nd Edition, Sections 1.7-1.10, 3.1-3.6, 4.1-4.2, 5.1-5.4

Classical Field Theory: Electrostatics-Magnetostatics