Classical Field Theory:
Electrostatics-Magnetostatics
April 27, 20101
1
J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Section 1-5
Classical Field Theory: Electrostatics-Magnetostatics
Electrostatics
The behavior of an electrostatic field can be described by two differential
equations:
~ = 4πρ
~ ·E
∇
(1)
(Gauss law) and
~ =0
~ ×E
∇
(2)
~ is the
the latter equation being equivalent to the statement that E
gradient of a scalar function, the scalar potential Φ:
~ = −∇Φ
~
E
(3)
Eqns (1) and (3) can be combined into one differential equation for a
single scalar function Φ(~x ):
∇2 Φ = −4πρ
(4)
This equation is called Poisson equation.
In the regions of space where there is no charge density, the scalar
potential satisfies the Laplace equation:
∇2 Φ = 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 3 x 0 ρ(~x 0 ), i.e.,
Z
ρ(~x 0 ) 3 0
d x
(6)
Φ(~x ) =
|~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
(7)
∇2 Φ(~x ) = ρ(~x 0 )d 3 x 0 ∇2
|~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 integral 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
δ 3 (~x − ~x 0 )d 3 x = 1
otherwise is zero. This way we recover Poisson’ s equation
∇2 Φ(~x ) = −4πρ(~x 0 )~x 0 =~x
(9)
Thus, we have not only shown that the potential from Coulomb’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
qi , positioned at ~xi0 so that
ρ(~x 0 ) =
N
X
qi δ 3 (~x 0 − ~xi0 )
(10)
i=1
In this case the solution for the potential is a combination of terms
proportional to 1/|~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 divergence
theorem
Z
I
3
~
~ · ~n da
~
∇ · Ad x =
A
(11)
V
S
~ defined in the volume V
which applies to any well-behaved vector field A
bounded by the closed surface S.
~ = Φ∇ψ,
~
Let A
(Φ and ψ arbitrary scalar fields). Then
~ · (Φ∇ψ)
~
∇
=
~ · ~n
Φ∇ψ
=
~ · ∇ψ
~
Φ∇2 ψ + ∇Φ
∂ψ
Φ
∂n
(12)
(13)
where ∂/∂n is the normal derivative at the surface S.
Classical Field Theory: Electrostatics-Magnetostatics
When (12) and (13) substituted into the divergence theorem (8)
produces the so-called Green’s 1st identity
Z I
∂ψ
2
3
~
~
da .
Φ∇ ψ + ∇Φ · ∇ψ d x =
Φ
∂n
V
S
(14)
If we rewrite (14) with Φ and ψ interchanged, and then subtract it from
(14) we obtain Green’s 2nd identity or Green’s Theorem:
Z
I ∂Φ
∂ψ
−ψ
da
(15)
Φ∇2 ψ − ψ∇2 Φ d 3 x =
Φ
∂n
∂n
V
S
Now we can apply Poisson’s equation (8) for discrete charge, substituting
for ψ = 1/|~x − ~x 0 |
Z 4πρ(~x 0 ) 3
−4πδ 3 (~x − ~x 0 )Φ(~x 0 ) +
d x
|~x − ~x 0 |
V
I ∂
1
1
∂Φ
=
Φ 0
−
da0
(16)
0|
0 | ∂n0
~
~
∂n
|~
x
−
x
|~
x
−
x
S
Classical Field Theory: Electrostatics-Magnetostatics
Integrating the Dirac delta function over all values of ~x 0 within V and for
~x within the volume V yields a nonzero result
Z
I 1
∂Φ
∂
1
ρ(~x 0 ) 3 0
1
Φ(~x ) =
d x +
da0
−Φ 0
x − ~x 0 |
4π S |~x − ~x 0 | ∂n0
∂n |~x − ~x 0 |
V |~
(17)
• The (blue) correction term goes to zero as the surface S goes to
infinity (because S falls of faster than 1/|~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 derivatives 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
= −4πδ(|~x − ~x 0 |)
(18)
∇2
|~x − ~x 0 |
The function 1/|~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
∇02 G (~x , ~x 0 ) = −4πδ(|~x − ~x 0 |)
(19)
where
1
+ F (~x , ~x 0 )
|~x − ~x 0 |
and F satisfying the Laplace equation inside the volume V
G (~x , ~x 0 ) =
∇02 F (~x , ~x 0 ) = 0
(20)
(21)
Classical Field Theory: Electrostatics-Magnetostatics
If we substitute G (~x , ~x 0 ) in eqn (17) we get
Z
I 1
∂Φ
∂
Φ(~x ) =
ρ(~x 0 )G (~x , ~x 0 )d 3 x 0 +
G (~x , ~x 0 ) 0 − Φ(~x 0 ) 0 G (~x , ~x 0 ) da0
4π S
∂n
∂n
V
(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:
GD (~x , ~x 0 ) = 0
for ~x 0 ∈ S
Then the 1st term on the surface integral of (22) vanishes
Z
I
1
∂
0
0
3 0
Φ(~x ) =
ρ(~x )GD (~x , ~x )d x −
Φ(~x 0 ) 0 GD (~x , ~x 0 )da0
4π S
∂n
V
(23)
(24)
• For Neumann BC the simplest choice of BC on G is
∂GN
(~x , ~x 0 ) = 0 for ~x 0 ∈ S
∂n0
but application o the Gauss theorem on (19) shows that (how?)
I
∂GN 0
da = −4π 6= 0
0
S ∂n
(25)
which is incosistent with ∇02 G (~x , ~x 0 ) = −4πδ 3 (~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
4π
∂GN
for ~x 0 ∈ S
(26)
(~x , ~x 0 ) = −
∂n0
S
S is the total area of the boundary surface. Then the solution will be:
Z
I
1
∂Φ
Φ(~x ) =
ρ(~x 0 )GN (~x , ~x 0 )d 3 x 0 + hΦiS +
GN (~x , ~x 0 )da0 (27)
4π S ∂n0
V
where hΦiS is the average value of the potential over the whole surface
I
1
Φ(~x 0 )da0
(28)
hΦiS ≡
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
∂
∂Φ
1
∂2Φ
1 ∂2
1
sin
θ
+
(r
Φ)
+
=0
(29)
2
r 2 ∂r 2
r 2 sin θ ∂θ
∂θ
r 2 sin θ ∂φ2
If we assume
1
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
2
2
r sin θ
+
sin
θ
+
=0
(31)
U dr 2
r 2 sin θP dθ
dθ
Q dφ2
Φ=
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
P
sin θ
+ l(l + 1) − 2
r 2 sin θ dθ
dθ
sin θ
d 2U
l(l + 1)
−
U
dr 2
r2
=
0
(34)
=
0
(35)
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 Legendre Polynomials
The equation (34) for P(θ) can be expressed in terms of x = cos θ
m2
d
2 dP
(1 − x )
+ l(l + 1) −
P=0
dx
dx
1 − x2
This is the generalized Legendre equation and its solutions are the
associated Legendre functions.
We will consider the solution of (54) for m2 = 0
dP
d
(1 − x 2 )
+ l(l + 1)P = 0
dx
dx
(37)
(38)
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 series
P(x) = x k
∞
X
aj x j
(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?)
aj+2 =
(k + j)(k + j + 1) − l(l + 1)
aj
(k + j + 1)(k + j + 2)
(40)
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 series
expansion 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:
I
The series converges for x 2 < 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)
=
P2 (x)
=
P3 (x)
=
P4 (x)
=
x
1
(3x 2 − 1)
2
1
(5x 3 − 3x)
2
1
(35x 4 − 30x 2 + 3)
8
(43)
The Legendre polynomials can be taken from Rodrigue’s formula:
Pl (x) =
1 dl 2
(x − 1)l
2l l! dx l
(44)
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 0 l
(45)
2l
+1
−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
Al =
2
Z
1
f (x)Pl (x)dx
(47)
−1
Thus for problems with azimuthal symmetry i.e. m = 0 the general
solution is:
∞ h
i
X
Φ(r , θ) =
Al r l + Bl r −(l+1) 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 radius 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
V (r = R, θ) =
∞
X
Al R l Pl (cos θ)
(49)
l=0
and the coefficients Al will be taken via eqn
(47)
Z
2l + 1 π
Al =
V (θ)Pl (cos θ) sin θdθ (50)
2R l 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?):
7 r 3
11 r 5
3 r
P1 (cos θ) −
P3 (cos θ) +
P5 (cos θ) − . . .
Φ(r , θ) = V
2 R
8 R
16 R
(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?):
#
2 "
2
7 R
3 R
Φ(r , θ) =
V P1 (cos θ) −
P3 (cos θ) + . . .
(52)
2 r
12 r
Classical Field Theory: Electrostatics-Magnetostatics
Legendre Polynomials
An important expansion is that of the
potential at ~x due to a unit point charge
at ~x 0
∞
X rl
1
<
=
Pl (cos γ)
|~x − ~x 0 |
r>l+1
(53)
l=0
where r< (r> ) is the smaller (larger) of |~x |
and |~x 0 | and γ is the angle between |~x |
and |~x 0 |.
Show that the potential is :
∞ X
1
l
−(l+1)
=
A
r
+
B
r
Pl (cos γ) on the z-axis
l
l
|~x − ~x 0 |
l=0
∞ 1
1 X r< l
=
on the x-axis
|~x − ~x 0 |
r>
r>
l=0
Classical Field Theory: Electrostatics-Magnetostatics
Associated Legendre Functions and Spherical Harmonics
For problems without azimuthal (axial) symmetry, we need the
generalization of Pl (cos θ), namely, the solutions of
m2
d
2 dP
(1 − x )
+ l(l + 1) −
P=0
dx
dx
1 − x2
(54)
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
function Plm (x) . For positive m it is defined as
dm
Pl (x)
(55)
dx m
If Rodrigues’ formula is used an expression valid for both positive and
negative m is obtained:
Plm (x) = (−1)m (1 − x 2 )m/2
Plm (x) =
(−1)m
dm
(1 − x 2 )m/2 m Pl (x)
l
2 l!
dx
(56)
Classical Field Theory: Electrostatics-Magnetostatics
There is a simple relation between Pl−m (x) and Plm (x) :
Pl−m (x) = (−1)m
(l − m)! m
P (x)
(l + m)! l
(57)
For fixed m the functions Plm (x) form an orthonormal set in the index l
on the interval −1 ≤ x ≤ 1. The orthogonality relation is
Z 1
2 (l + m)!
δl 0 l
(58)
Plm0 (x)Plm (x)dx =
2l + 1 (l − m)!
−1
We have found that Qm (φ) = e imφ , this function forms a complete set of
orthogonal functions in the index m on the interval 0 ≤ φ ≤ 2π.
The product Plm 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)! m
P (cos θ)e imφ
(59)
Ylm (θ, φ) =
4π (l + m)! l
and also
∗
Yl,−m (θ, φ) = (−1)m Ylm
(θ, φ)
(60)
Classical Field Theory: Electrostatics-Magnetostatics
The normalization and orthogonality conditions are:
Z 2π
Z π
dφ
sin θdθYl∗0 m0 (θ, φ)Ylm (θ, φ) = δl 0 l δm0 m
0
(61)
0
An arbitrary function g (θ, φ) can be expanded in spherical harmonics
g (θ, φ) =
∞ X
l
X
Alm Ylm (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) :
Φ(r , θ, φ) =
∞ X
l
h
i
X
Alm r l + Blm r −(l+1) 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
l =0
l =1
l =2
Y00 =
1
4π
r
3
Y11 = −
sin θe iφ
8π
r
3
cos θ
Y10 =
4π
r
1 15 2 2iφ
Y22 =
sin θe
4 2π
r
15
Y21 = −
sin θ cos θ e iφ
8π
r
5 3
1
2
Y20 =
cos θ −
4π 2
2
Classical Field Theory: Electrostatics-Magnetostatics
Spherical Harmonics Ylm (θ, φ)
l =3
Y33
Y32
Y31
Y30
r
1 35 3 3iφ
sin θ e
=−
4 4π
r
1 105 2
=
sin θ cos θe 2iφ
4 2π
r
1 21
=−
sin θ(5 cos2 θ − 1) e iφ
4 4π
r
1
7
=
sin θ 5 cos3 θ − 3 cos θ
2 4π
(65)
Classical Field Theory: Electrostatics-Magnetostatics
Figure: Schematic representation of
Ylm on the unit sphere. Ylm is equal
to 0 along m great circles passing
through the poles, and along l − m
circles of equal latitude. The
function changes sign each time 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
angles θ, φ and θ0 , φ0 :
Pl (cos γ) =
l
X
4π
∗
Ylm
(θ0 , φ0 )Ylm (θ, φ)
l(l + 1)
(66)
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 ).
Pl (cos γ) is a function of the angles θ, φ with the angles θ0 , φ0 as
parameters and it maybe expanded in a series (63) :
0
Pl (cos γ) =
∞ X
l
X
Al 0 m0 (θ0 , φ0 )Yl 0 m0 (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
∞ X
l
l
X
r<
1
1
∗
=
4π
Ylm
(θ0 , φ0 )Ylm (θ, φ)
l+1
0
|~x − ~x |
2l + 1 r>
(68)
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
Multipole Expansion
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
Φ(~x ) =
∞ X
l
X
l=0 m=−l
4π
1
qlm l+1 Ylm (θ, φ) (69)
2l + 1
r
This type of expansion is called multipole expansion;
The l = 0 term is called monopole term,
The l = 1 are called dipole 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 ) 3 0
Φ(~x ) =
d x
(70)
|~x − ~x 0 |
Classical Field Theory: Electrostatics-Magnetostatics
Using the expansion (68) for 1/|~x − ~x 0 | i.e.
Φ(~x ) =
∞ X
l
X
l=0 m=−l
1
4π
Ylm (θ, φ)
l+1
2l + 1 r
Z
∗
Ylm
(θ0 , φ0 )ρ(~x 0 )r 0l d 3 x 0
Consequently the coefficients in (69) are :
Z
∗
qlm = Ylm
(θ0 , φ0 )ρ(~x 0 )r 0l d 3 x 0
(71)
(72)
and called the multipole moments of the charge distribution ρ(~x 0 )
Classical Field Theory: Electrostatics-Magnetostatics
Monopole moment l = 0
Here, the only component is
Z
Z
1
q
0 00 ∗
0
0
3 0
q00 = ρ(~x )r Y00 (θ , φ )d x = √
ρ(~x 0 )d 3 x 0 = √
4π
4π
(73)
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)
q
∗ 1
Y00 + · · · = + . . .
Φ(~x ) = 4πq00
r
r
(74)
NOTE:
The moments with m ≥ 0 are connected (for real charge density) too the
moments with m < 0 through
∗
ql,−m = (−1)m qlm
(75)
Classical Field Theory: Electrostatics-Magnetostatics
Dipole moment l = 1
In Cartesian coordinates the dipole moment is given by
Z
~p = ~x 0 ρ(~x 0 )d 3 x 0
(76)
In Spherical representation one obtains
Z
∗
∗
q11
=
ρ(~x 0 )r 0 Y11
(θ0 , φ0 )d 3 x 0
r
Z
3
= −
ρ(~x 0 )r 0 (sin θ0 cos φ0 − i sin θ0 sin φ0 ) d 3 x 0
8π
in Cartesian represantation
r
r
Z
3
3
0
0
0
3 0
= −
ρ(~x )(x − iy )d x = −
(px − ipy ) (77)
8π
8π
also
r
Z
Z
3
∗
0 0 ∗
0
0
3 0
q10 =
ρ(~x )r Y10 (θ , φ )d x =
ρ(~x 0 )r 0 cos θ0 d 3 x 0
4π
r
r
Z
3
3
0
0
3
z ρ(~x )d x =
pz
(78)
=
4π
4π
Classical Field Theory: Electrostatics-Magnetostatics
Quadrupole moment l = 2
∗
q22
Z
∗
ρ(~x )r 02 Y22
(θ0 , φ0 )d 3 x 0
r
Z
1 15
2
=
ρ(~x ) [r 0 sin θ0 (cos φ0 − i sin φ0 )] d 3 x 0
4 2π
r
Z
1 15
=
ρ(~x )(x 0 − iy 0 )2 d 3 x 0
4 2π
1 02
because (x 0 − iy 0 )2 =
(3x − r 02 ) − 6ix 0 y 0 − (3y 02 − r 02 )
3
r
15
1
=
(Q11 − 2iQ12 − Q22 )
(79)
12 2π
=
where Qij is the traceless (why?) quadrupole moment tensor:
Z
Qij =
3xi0 xj0 − r 2 δij ρ(x~0 )d 3 x 0
(80)
Classical Field Theory: Electrostatics-Magnetostatics
Quadrupole moment l = 2
Analogously
∗
q21
Z
=
r
0
02
∗
Y21
(θ0 , φ0 )d 3 x 0
ρ(~x )r
r
1 15
= −
(Q13 − iQ23 )
3 8π
=−
15
8π
Z
ρ(~x 0 )z 0 (x 0 − iy 0 )d 3 x 0
(81)
and
∗
q20
Z
=
=
0
02
∗
Y20
(θ0 , φ0 )d 3 x 0
ρ(~x )r
r
1
5
Q33
2 4π
1
=
2
r
5
4π
Z
ρ(~x 0 )(3z 0 − r 02 )d 3 x 0
(82)
From eqn (60) we can get the moments with m < 0 through
∗
ql,−m = (−1)m qlm
(83)
Classical Field Theory: Electrostatics-Magnetostatics
Multipole Expansion
By direct Taylor expansion of 1/|~x − ~x 0 | the expansion of Φ(~x ) in
rectangular coordinates is:
1X
xi xj
q ~p · ~x
Qij 5 + . . .
Φ(~x ) = + 3 +
r
r
2
r
(84)
i,j
The electric field components for a given multipole can be expressed
~ = −∇Φ
~ and (69)
most easily in terms of spherical coordinates. From E
for fixed (l, m) we get(how?):
Er
Eθ
Eφ
1
4π(l + 1)
qlm l+2 Ylm
2l + 1
r
4π
1 ∂
= −
qlm l+2 Ylm
2l + 1
r
∂θ
4π
1 1 ∂
= −
qlm l+2
Ylm
2l + 1
r
sin θ ∂φ
=
(85)
For a dipole ~p along the z-axis they reduce to:
Er =
2p cos θ
,
r3
Eθ =
p sin θ
,
r3
Eφ = 0
(86)
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
~ (~x ) = 3~n (~p · ~n) − ~p
E
|~x − ~x0 |3
with ~n =
~r
r
(87)
because Φ(~x ) = ~p · ~x /r 3 and
~ (~x ) = −∇Φ
~ = − ∇(~p · ~x ) − (~p · ~x )∇
E
r3
1
r3
=
3~n (~p · ~n) − ~p
|~x − ~x0 |3
(88)
Classical Field Theory: Electrostatics-Magnetostatics
Energy 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 3 x
(89)
V
The external field (if its is slowly varying over the region where ρ(~x ) is
non-negligible) may be expanded in a Taylor series:
3
3
1 XX
∂2Φ
~
Φ(~x ) = Φ(0) + ~x · ∇Φ(0)
+
xi xj
(0) + . . .
2
∂xi ∂xj
(90)
i=1 j=1
~ = −∇Φ
~ for the external field
Since E
3
~ (0) −
Φ(~x ) = Φ(0) − ~x · E
3
1 XX
∂Ej
xi xj
(0) + . . .
2
∂xi
(91)
i=1 j=1
Classical Field Theory: Electrostatics-Magnetostatics
~ = 0 for the external field we can substract
~ ·E
Since ∇
1 2~ ~
r ∇ · E (0)
6
from the last term to obtain finally the expansion:
3
~ (0) −
Φ(~x ) = Φ(0) − ~x · E
3
∂Ej
1 XX
3xi xj − r 2 δij
(0) + . . .
6
∂xi
(92)
i=1 j=1
When this inserted into (89) the energy takes the form:
3
~ (0) −
W = qΦ(0) − ~p · E
3
1 XX
∂Ej
Qij
(0) + . . .
6
∂xi
(93)
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 except q00 .
If the sphere has a radius R0 and constant charge density ρ(~x ) = ρ then
∗
qlm
Z
Z
R0
=ρ
r
Ω0
0l
∗
Ylm
(θ0 , φ0 )r 02 dr 0 dΩ0
0
R l+3
=ρ 0
l +3
Z
Ω0
∗
Ylm
(θ0 , φ0 )dΩ0
√
but since Y00 = 1/ 4π we have from the orthogonality relation
∗
qlm
=ρ
R0l+3 √
4π
l +3
Z
Ω0
∗
Ylm
Y00 dΩ0 = ρ
R0l+3 √
4πδl0 δm0
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 +
a2m Y2m (θ, φ) , |a2m | << 1
m=−2
The multipole moments are
Z Z R(θ0 ,φ0 )
∗
∗ 02 0
qlm =
ρr 0l Ylm
r dr dΩ0
Ω0
= ρ
0
R0l+3
l +3
Z
Ω0
∗
Ylm
(θ0 , φ0 )
1+
2
X
!l+3
∗
∗
a2m
Y2m
(θ0 , φ0 )
dΩ0
m=−2
Classical Field Theory: Electrostatics-Magnetostatics
∗
qlm
R l+3
≈ ρ 0
l +3
Z
Ω0
∗
Ylm
(θ0 , φ0 ) 1 + (l + 3)
2
X
!
∗
∗
a2m
Y2m
(θ0 , φ0 ) dΩ0
m=−2
Apart from the monopole moment q00 of the previous example, we find
∗
qlm
=ρ
Z
2
X
X
R0l+3
∗
∗
∗
(l+3)
a2µ
Ylm
Y2µ dΩ0 = ρR0l+3
a2µ
δmµ δl2 (l > 0)
l +3
0
Ω
µ
µ=−2
∗
∗
Thus the nonvanishing multipole moments are : q2m
= ρR05 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
~ (also called magnetic induction).
flux density , denoted by B
• The magnitude of the flux density can be defined by the magnetic
~ exerted on the magnetic dipole:
torque N
~ =µ
~
N
~ ×B
(94)
where µ
~ is the magnetic moment of the dipole.
• In electrostatics the conservation of charge demands that the charge
density at any point in space be related to the current density in that
neighborhood by the continuity equation
~ · ~J = 0
∂ρ/∂t + ∇
(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
~ at the
P, then the elementary flux density d B
point P is given in magnitude and direction by
d ~` × ~x
(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
~ = kI
dB
~ = kq ~v × ~x
B
|~x |3
(98)
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 & Ampere’s Law
The basic law (97) for the magnetic induction can be written down in
general form for a current density ~J(~x ):
Z
0
~ x ) = 1 ~J(~x 0 ) × (~x − ~x ) d 3 x 0
B(~
(99)
0
c
|~x − ~x |3
~ x ) is the magnetic analog of electric field in terms
This expression for B(~
of the charge density:
Z
0
~ (~x ) = 1 ρ(~x 0 ) (~x − ~x ) d 3 x 0
(100)
E
c
|~x − ~x 0 |3
In order to obtain DE equivalent to (99) we use the relation
(~x − ~x 0 )
1
~
~ 1 = − ~r
=
−
∇
as
∇
|~x − ~x 0 |3
|~x − ~x 0 |
r
r3
(101)
and (99) transforms into
~ x) = 1 ∇
~ ×
B(~
c
Z
~J(~x 0 )
d 3x 0
|~x − ~x 0 |
(102)
Classical Field Theory: Electrostatics-Magnetostatics
Classical Field Theory: Electrostatics-Magnetostatics
From (102) follows immediately:
~ =0
~ ·B
∇
(103)
• This is the 1st equation of magnetostatics and corresponds to
~ = 0 in electrostatics.
~ ×E
∇
~
By analogy with electrostatics we now calculate the curl of B
~ = 1∇
~ ×∇
~ ×
~ ×B
∇
c
Z
~J(~x 0 )
d 3x 0
|~x − ~x 0 |
~ · ~J = 0) reduces to (how?)
which for steady-state phenomena (∇
(104)
2
~ = 4π ~J
~ ×B
∇
c
(105)
• This is the 2nd equation of magnetostatics and corresponds to
~ = 4πρ in electrostatics.
~ ·E
∇
2
“
”
“
”
~ =∇
~ − ∇2 A
~
~ × ∇
~ ×A
~ ∇
~ ·A
Remember : ∇
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
Z
~J · ~nda
~ · ~n da = 4π
~ ×B
(106)
∇
c S
S
is transformed into
I
Z
~ · d ~` = 4π
~J · ~nda
B
c S
C
(107)
Since the surface integral of the current density is the total current I
passing through the closed curve C , Ampére’s law can be written in the
form:
I
~ · d ~` = 4π I
B
(108)
c
C
Classical Field Theory: Electrostatics-Magnetostatics
Vector Potential
The basic differential laws in magnetostatics are
~ = 4π ~J
~ ×B
∇
c
~ =0
~ ·B
and ∇
(109)
The problems is how to solve them.
~ =0
~ ×B
• If the current density is zero in the region of interest, ∇
~
permits the expression of the vector magnetic induction B as the gradient
~ = −∇Φ
~ M , then (109) reduces to the
of a magnetic scalar potential B
Laplace equation for ΦM .
~ = 0 everywhere, B
~ must be the curl of some vector field
~ ·B
• If ∇
~
A(~x ), called the vector potential
~ x) = ∇
~ x)
~ × A(~
B(~
(110)
~ is:
and from (102) the general form of A
~ x) = 1
A(~
c
Z
~J(~x 0 )
~ x)
d 3 x 0 + ∇Ψ(~
|~x − ~x 0 |
(111)
Classical Field Theory: Electrostatics-Magnetostatics
The added gradient of an arbitrary scalar function Ψ shows that, for a
~ the vector potential can be freely
given magnetic induction B,
transformed according to
~ →A
~ + ∇Ψ
~
A
(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
~ − ∇2 A
~ = 4π ~J
~ = 4π ~J or ∇
~ ∇
~ ·A
~ × ∇
~ ×A
∇
c
c
(113)
If we exploit the freedom implied by (112) we can make the convenient
~ = 0. Then each component of
~ ·A
choice of gauge (Coulomb gauge) ∇
the vector potential satisfies the Poisson equation
~ =−
∇2 A
4π ~
J
c
(114)
~ in unbounded space is (111) with Ψ =constant:
The solution for A
Z ~ 0
1
J(~x ) 3 0
~
d x
(115)
A(~x ) =
c
|~x − ~x 0 |
Classical Field Theory: Electrostatics-Magnetostatics
Further Reading
I
Green Function
I
I
Greiner W. ”Classical Electrodynamics”, Chapter 1, 2, 3, 4
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