Electrostatic Field Problems:
Spherical Symmetry
EE 141 Lecture Notes
Topic 7
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014
Motivation
Spherical Coordinates
Coordinate transformations
x = r sin θ cos φ,
y = r sin θ sin φ, z = r cos θ
(1)
!
p
y p
x2 + y2
, φ = arctan
(2)
r = x 2 + y 2 + z 2 , θ = arctan
z
x
with 0 ≤ r ≤ ∞, 0 ≤ θ ≤ π, and 0 ≤ φ < 2π.
Unit Basis Vectors in Spherical Coordinates
Unit basis vectors
1̂r = 1̂r (θ, φ) = 1̂x sin θ cos φ + 1̂y sin θ sin φ + 1̂z cos φ,
1̂θ = 1̂θ (θ, φ) = 1̂x cos θ cos φ + 1̂y cos θ sin φ − 1̂z sin θ,
(3)
1̂φ = 1̂φ (θ, φ) = −1̂x sin φ + 1̂y cos φ,
where (Orthogonality Relations)
1̂r × 1̂θ = 1̂φ ,
1̂θ × 1̂φ = 1̂r ,
1̂φ × 1̂r = 1̂θ .
(4)
Problem 15. Using the relations in Eq. (3), determine expressions for
the unit basis vectors 1̂x , 1̂y , and 1̂z in terms of the unit basis
vectors 1̂r , 1̂θ , and 1̂φ .
Problem 16. Determine the direct coordinate transformation in
matrix notation from cylindrical to spherical coordinates.
Vectors in Spherical Coordinates
Any vector V may be expressed in spherical polar coordinates as
V = 1̂V V = 1̂r Vr + 1̂θ Vθ + 1̂φ Vφ ,
(5)
where
Vr = 1̂r · V
= (1̂x sin θ cos φ + 1̂y sin θ sin φ + 1̂z cos θ)·(1̂x Vx + 1̂y Vy + 1̂z Vz )
= Vx sin θ cos φ + Vy sin θ sin φ + Vz cos θ,
Vθ = 1̂θ · V
= (1̂x cos θ cos φ + 1̂y cos θ sin φ + 1̂z sin θ)·(1̂x Vx + 1̂y Vy + 1̂z Vz )
= Vx cos θ cos φ + Vy cos θ sin φ + Vz sin θ,
(6)
Vφ = 1̂φ · V = (−1̂x sin φ + 1̂y cos φ) · (1̂x Vx + 1̂y Vy + 1̂z Vz )
= −Vx sin φ + Vy cos φ,
with magnitude
V = |V| =
√
V·V=
q
V 2 + V 2 + V 2.
(7)
Position Vector in Spherical Coordinates
Position Vector of a point P = P(r , θ, φ) is given by
~ = 1̂r (θ, φ)r ,
R = OP
where r =
p
(8)
x 2 + y 2 + z 2.
Notice that this position vector does not have either a 1̂θ or 1̂φ
component. Nevertheless, the orientation of the radial unit vector
1̂r = 1̂r (θ, φ) depends upon the θ and φ coordinates of the point P.
Spherical Coordinates - Differential Elements
Differential elements of length along the 1̂r , 1̂θ , & 1̂φ -directions:
d ℓr = dr ,
d ℓθ = rd θ,
d ℓφ = r sin θd φ.
Differential Length, Surface Area, & Volume
The vector differential element of length:
d ~ℓ = 1̂r dr + 1̂θ rd θ + 1̂φ r sin θd φ.
(9)
Fundamental quadratic form or metric form
d ℓ2 = dr 2 + r 2 d θ2 + r 2 sin2 θd φ2 .
(10)
Differential elements of surface area:
θφ-Spherical Surface: d sr = 1̂r (d ℓθ d ℓφ ) = 1̂r r 2 sin θd θd φ.
r φ-Conical Surface: d sθ = 1̂θ (d ℓr d ℓφ ) = 1̂θ r sin θdrd φ.
r θ-Planar Surface: d sφ = 1̂φ (d ℓr d ℓθ ) = 1̂φ rdrd θ.
Differential element of volume:
dV = d ℓr d ℓθ d ℓφ = r 2 sin θdrd θd φ.
(11)
Distance Between Two Points
Coordinates of P1 (x1 , y1 , z1 ) = P1 (r1 , θ1 , φ1 ):
x1 = r1 sin θ1 cos φ1 ,
y1 = r1 sin θ1 sin φ1 ,
z1 = r1 cos θ1 .
Coordinates of P2 (x2 , y2 , z2 ) = P2 (r2 , θ2 , φ2 ):
x2 = r2 sin θ2 cos φ2 ,
y2 = r2 sin θ2 sin φ2 ,
z2 = r1 sin θ1 sin φ1 .
Distance between P1 & P2 is then given by Pythagorean’s theorem as
h
d = (r2 sin θ2 cos φ2 − r1 sin θ1 cos φ1 )2
i1/2
2
2
+(r2 sin θ2 sin φ2 − r1 sin θ1 sin φ1 ) + (r2 cos θ2 − r1 cos θ1 )
h
i1/2
2
2
= r1 + r2 − 2r1 r2 cos θ1 cos θ2 + sin θ1 sin θ2 cos (φ2 − φ1 )
.
(12)
Vector Differential Operators in Spherical Polar
Coordinates
Gradient Operator
∇ = 1̂r
∂
1 ∂
1 ∂
+ 1̂θ
+ 1̂φ
∂r
r ∂θ
r sin θ ∂φ
(13)
The gradient of a scalar function of position f (r) = f (r , θ, φ) is then
given by
1 ∂f
1 ∂f
∂f
+ 1̂θ
+ 1̂φ
.
(14)
∇f = 1̂r
∂r
r ∂θ
r sin θ ∂φ
Laplacian Operator
1 ∂
∇ = 2
r ∂r
2
1
∂
1
∂
∂2
2 ∂
r
+ 2
sin θ
+ 2 2
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
(15)
Vector Differential Operators in Spherical Polar
Coordinates
Vector function of position F(r) = F(r , θ, φ) = 1̂r Fr + 1̂θ Fθ + 1̂φ Fφ
has divergence
∇·F=
1 ∂ 2
1 ∂
1 ∂Fφ
(r Fr ) +
(Fθ sin θ) +
2
r ∂r
r sin θ ∂θ
r sin θ ∂φ
(16)
and curl
∂
∂Fθ
(Fφ sin θ) −
∂θ
∂φ
∂Fr
∂
1
− sin θ (rFφ )
+1̂θ
r sin θ ∂φ
∂r
1 ∂
∂Fr
+1̂φ
.
(rFθ ) −
r ∂r
∂θ
1
∇ × F = 1̂r
r sin θ
(17)
Field of a Point Charge (Symmetry about a Point)
Consider a point charge Q located at the origin of coordinates (if
not, the origin of coordinates can always be shifted to that point).
The spherical symmetry of the problem then requires that
E(r) = 1̂r E (r ).
Field of a Point Charge (Symmetry about a Point)
Application of Gauss’ law to a concentric spherical surface S with
radius r > 0 surrounding the point charge Q at the origin O yields
I
I
Q
=
E (r )1̂r · 1̂r da = E (r ) da = 4πr 2 E (r ),
ǫ0
S
S
so that
E(r) = 1̂r E (r ) = 1̂r
Q
,
4πǫ0 r 2
r >0
The absolute potential due to the point charge is then given by
Z ∞
Q
V (r ) =
1̂r E (r ) · 1̂r dr =
, r > 0.
4πǫ0 r
r
Field of a Point Charge (Symmetry about a Point)
E ~ 1/r2
Q/4πε0
V ~ 1/r
0
0
1
2
r
3
4
Field of a Uniform Spherical Charge Distribution
Consider determining the field due to a uniform spherical charge
distribution of radius r0 > 0 with volume charge density ̺(r ) = ̺0 for
r ≤ r0 centered on the origin, where ̺(r ) = 0 for r > r0 .
The spherical symmetry of the problem then requires that
E(r) = 1̂r E (r ).
Application of Gauss’ law to a concentric sphere of radius r centered
at the origin then yields
(
ZZZ
4
1
1
πr 3 ̺0 , r ≤ r0
3
̺(r )d 3 r =
4πr 2 E (r ) =
ǫ0
ǫ0 43 πr03 ̺0 , r ≥ r0
V
Notice that if ̺0 is positive, then E is directed radially outward from
the origin, whereas if ̺0 is negative, then E is directed radially inward
toward the origin.
Field of a Uniform Spherical Charge Distribution
The total charge Q contained in the spherical charge distribution is
4
Q = πr03 ̺0 ,
3
so that
3Q
.
̺0 =
4πr03
The radial component of the electric field vector E(r) = 1̂r E (r ) is
then given by
(
Q
r , r ≤ r0
4πǫ0 r03
E (r ) =
Q
, r ≥ r0
4πǫ0 r 2
Notice that measurements of the electric field due to a spherically
symmetric charge distribution localized in space (r ≤ r0 ) for r > r0
are independent of the radius of the charge distribution.
In the limit as r0 → 0, the field due to a point charge Q at the origin
is obtained.
Field of a Uniform Spherical Charge Distribution
The absolute potential [Eq. (4.8)] due to the uniform spherical
charge distribution is then given by (with d ~ℓ = 1̂r dr )
Z ∞
V (r ) =
E (r )dr .
r
For r ≥ r0 ,
Z ∞
Q
dr
Q
=
V (r ) =
2
4πǫ0 r r
4πǫ0 r
which is the same as that due to a point charge Q at the origin.
For r ≤ r0 ,
Z r0
Z ∞ Q
1
dr
V (r ) =
rdr +
3
4πǫ0 r0 r
r2
r0
Q
Q
2
2
+
.
=
r
−
r
0
3
8πǫ0 r0
4πǫ0 r0
{z
}
|
|
{z
}
increase in potential above the surface value
potential at the sphere surface
Field of a Uniform Spherical Charge Distribution
3Q/8πε0r0
Q/4πε0r0
V(r)
Q/4πε0r02
0
E(r)
0
r0
2r0
r
Field of a Uniform Spherical Charge Distribution
The potential V (r) may also be determined using Poisson’s &
Laplace’s equations directly, taking advantage of the spherical
symmetry of the source charge distribution.
Inside the uniform charge distribution (r
∇2 V (r ) = −̺0 /ǫ0 becomes
1 ∂
̺0
2 ∂V
r
=−
=⇒
2
r ∂r
∂r
ǫ0
̺0
∂V
= − r 3 + C =⇒
∴ r2
∂r
3ǫ0
̺0
C
+D =⇒
∴ V (r ) = − r 2 −
6ǫ0
r
|{z}
C=0
≤ r0 ), Poisson’s equation
̺0
∂
2 ∂V
r
= − r2
∂r
∂r
ǫ0
̺0
C
∂V
=− r+ 2
∂r
3ǫ0
r
Q
V (r ) = −
r 2 + D.
3
8πǫ0 r0
Field of a Uniform Spherical Charge Distribution
Outside the uniform charge distribution
∇2 V (r ) = 0 becomes
1 ∂
2 ∂V
r
=0
r 2 ∂r
∂r
∂V
∴ r2
=A
∂r
A
∴ V (r ) = − + B; V (∞) ≡ 0
r
A
∴ V (r ) = − .
r
(r ≥ r0 ), Laplace’s equation
=⇒
=⇒
=⇒
∂
2 ∂V
r
=0
∂r
∂r
∂V
A
= 2
∂r
r
B=0
Field of a Uniform Spherical Charge Distribution
Continuity of V (r ) at r = r0 requires that
−
A
Q
=−
+ D,
r0
8πǫ0 r0
while continuity of ∂V /∂r [i.e., continuity of E (r )] at r = r0 requires
Q
Q
A
=−
=⇒ A = −
2
2
r0
8πǫ0 r0
8πǫ0
so that
D=
Q
A
Q
Q
− =
+
.
8πǫ0 r0 r0
8πǫ0 r0 4πǫ0 r0
Field of a Uniform Spherical Charge Distribution
The electrostatic potential is then given by
V (r ) =
Q
Q
r02 − r 2 +
; r ≤ r0
3
8πǫ0 r0
4πǫ0 r0
Q
V (r ) =
; r ≥ r0
4πǫ0 r
so that the electric field intensity E(r ) = −∇V (r ) = −1̂r ∂V /∂r is
Q
r ; r ≤ r0
4πǫ0 r03
Q
E(r ) = 1̂r
; r ≥ r0
4πǫ0 r 2
E(r ) = 1̂r
in agreement with the result obtained using Gauss’ law.
Take Home Exam Problem 2
A spherical region of radius a > 0 situated in free space contains a
volume charge density given by
̺(r ) = ̺0 1 + αr 2 ; r ≤ a,
with ̺(r ) = 0 for r > a, where ̺0 and α are constants.
1
Utilize Gauss’ law together with the inherent symmetry of the
problem to derive the electrostatic field vector E(r) both inside
and outside the spherical charge region.
2
Use both Poisson’s and Laplace’s equations to directly determine
the electrostatic potential V (r) both inside and outside the
spherical region. From this potential function, determine the
electrostatic field vector E(r).
3
Determine the value of the parameter α for which the
electrostatic field vanishes everywhere in the region outside the
spherical charge region (r > a). Plot Er (r ) and V (r ) as a
function of r for this value of α.
Boundary Value Problems in Spherical Coordinates
In spherical coordinates (r , θ, ϕ) defined by the set of transformation
equations x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ with
r ∈ [0, ∞), ϕ ∈ [0, 2π), and θ ∈ [0, π], Laplace’s equation ∇2 φ = 0
assumes the form
∂φ
1
1
∂2φ
1 ∂ 2 (r φ)
∂
sin
θ
+
+
= 0.
(18)
r ∂r 2
r 2 sin θ ∂θ
∂θ
r 2 sin2 θ ∂ϕ2
This equation then admits separated solutions of the form
1
φ(r , θ, ϕ) = U(r )P(θ)Q(ϕ),
r
(19)
where the factor r −1 is explicitly displayed in order to reflect the form
of the electrostatic potential for a spherical charge distribution.
Boundary Value Problems in Spherical Coordinates
With this substitution, Laplace’s equation (18) becomes
dP
1 d 2Q
1 d 2U
1
d
2
2
r sin θ
sin
θ
=
−
+
= m2
U dr 2
Pr 2 sin θ d θ
dθ
Q d ϕ2
(20)
where m2 is the separation constant. The ode for Q(ϕ) then has the
elementary solutions Q(ϕ) = e ±imϕ . In order that Q(ϕ) be singlevalued when the full azimuthal range ϕ = 0 → 2π is allowed, the
separation constant m must be an integer or zero.
The remaining part of Eq. (20) may then be written as
dP
m2
1
d
r 2 d 2U
sin
θ
+
= ℓ(ℓ + 1),
(21)
=
−
U dr 2
P sin θ d θ
dθ
sin2 θ
where ℓ(ℓ + 1) is another separation constant. The ode for the radial
part of this separated equation then has the elementary solution
U(r ) = Ar ℓ+1 + Br −ℓ ,
where A and B are constants of integration.
(22)
Boundary Value Problems in Spherical Coordinates
The remaining angular part of Eq. (21) is then
dP(θ)
m2
1 d
sin θ
+ ℓ(ℓ + 1) − 2 P(θ) = 0,
sin θ d θ
dθ
sin θ
(23)
where ℓ is as yet undetermined. With the change of variable
ζ = cos θ, for which d ζ = − sin (θ)d θ so that d /d θ = − sin (θ)d /d ζ,
this ode assumes the standard form
m2
d
2 dP
+ ℓ(ℓ + 1) −
1−ζ
P=0
(24)
dζ
dζ
1 − ζ2
known as the generalized or associated Legendre equation, whose
solutions are the associated Legendre functions. In order that its
solutions represent a physically realizable potential, they must be
single-valued, finite, and continuous on the interval ζ ∈ [−1, 1].
Adrien-Marie Legendre (1752–1833)
Boundary Value Problems Spherical Coordinates I
Case I: Legendre’s Equation and the Legendre Polynomials:
Special case when the problem possesses azimuthal symmetry so that
there is no ϕ-dependence and m = 0. The generalized Legendre
equation (24) then simplifies to the ordinary Legendre differential
equation
d
2 dP
(25)
+ ℓ(ℓ + 1)P = 0
1−ζ
dζ
dζ
Solutions of this equation are the Legendre polynomials of order ℓ
with series representation (obtained by the method of Frobenius)
Pℓ (ζ) =
[ℓ/2]
X
j=0
(−1)j
(2ℓ − 2j)!
ζ ℓ−2j
ℓ
2 j!(ℓ − j)!(ℓ − 2j)!
(26)
which are normalized such that Pℓ (1) = 1. Here [ℓ/2] denotes the
greatest integer value of ℓ/2, where [ℓ/2] = ℓ/2 if ℓ is even and
[ℓ/2] = (ℓ − 1)/2 if ℓ is odd.
Boundary Value Problems Spherical Coordinates I
Explicit expressions for the first few Legendre polynomials are:
P0 (ζ) = 1,
P1 (ζ) = ζ,
1
3ζ 2 − 1 ,
P2 (ζ) =
2
1
P3 (ζ) =
5ζ 3 − 3ζ ,
2
1
35ζ 4 − 30ζ 2 + 3 ,
P4 (ζ) =
8
1
63ζ 5 − 70ζ 3 + 15ζ ,
P5 (ζ) =
8
1
P6 (ζ) =
231ζ 6 − 315ζ 4 + 105ζ 2 − 5 ,
16
1
P7 (ζ) =
429ζ 7 − 693ζ 5 + 315ζ 3 − 35ζ ,
16
:
Legendre Polynomials
Boundary Value Problems Spherical Coordinates I
Series representation of the Legendre polynomials may be written as
[ℓ/2]
ℓ!
1 dℓ X
ζ 2(ℓ−j) .
(−1)j
Pℓ (ζ) = ℓ
ℓ
2 ℓ! d ζ j=0
j!(ℓ − j)!
The summation limit of [ℓ/2] can now be extended to ℓ because the
power of ζ 2(ℓ−j) is less than ℓ after the [ℓ/2] term and so its ℓth -order
derivative vanishes. The resulting summation is the expansion of
ℓ
(ζ 2 − 1) , so that
Pℓ (ζ) =
ℓ
1 dℓ
2
ζ
−
1
2ℓ ℓ! d ζ ℓ
which is known as Rodrigues’ formula.
(27)
Boundary Value Problems Spherical Coordinates I
The Legendre polynomials Pℓ (ζ) form a complete orthogonal set of
functions on the interval ζ ∈ [−1, 1], satisfying the orthogonality
relation
Z 1
2
δℓ′ ℓ
Pℓ′ (ζ)Pℓ (ζ)d ζ =
(28)
2ℓ + 1
−1
Any sufficiently well-behaved function f (ζ) on the interval ζ ∈ [−1, 1]
can then be expanded in a Legendre series representation as
f (ζ) =
∞
X
Aℓ Pℓ (ζ),
ℓ=0
−1 ≤ ζ ≤ 1,
(29)
with expansion coefficients
2ℓ + 1
Aℓ =
2
Z
1
−1
f (ζ)Pℓ (ζ)d ζ.
(30)
Boundary Value Problems Spherical Coordinates I
For a given boundary value problem for Laplace’s equation in
spherical coordinates which possesses azimuthal symmetry (i.e., is
independent of the azimuthal angle ϕ), Eq. (20) demands that
m = 0. The general solution for the potential is then given by
φ(r , θ) =
∞
X
ℓ=0
Aℓ r ℓ + Bℓ r −(ℓ+1) Pℓ (cos θ).
The coefficients Aℓ and Bℓ are then determined by the imposed
boundary conditions on some prescribed surface.
(31)
Boundary Value Problems Spherical Coordinates I
As an example, suppose that the electrostatic potential is specified as
V (θ) on the surface of a sphere of radius a, and it is required to
determine the potential within the spherical region bounded by that
surface. If there is no charge at the origin, then the potential must be
finite there and Bℓ = 0 for all ℓ. The expansion given in Eq. (31)
then becomes
∞
X
φ(r , θ) =
Aℓ r ℓ Pℓ (cos θ), r ≤ a.
(32)
ℓ=0
The coefficients Aℓ are then determined byP
evaluating Eq. (32) on
ℓ
the surface of the sphere, so that V (θ) = ∞
ℓ=0 Aℓ a Pℓ (cos θ). This
is just a Legendre series representation with ζ = cos θ, so that the
coefficients are given by
Z
2ℓ + 1 π
V (θ)Pℓ (cos θ) sin θd θ.
(33)
Aℓ =
2aℓ
0
Boundary Value Problems Spherical Coordinates I
Suppose next that the electrostatic potential is to be determined in
the region external to the sphere. Because the potential must now be
finite at r = ∞, it is required that Aℓ = 0 for all ℓ. The expansion
given in Eq. (31) now becomes
∞
X
Bℓ
φ(r , θ) =
Pℓ (cos θ),
r ℓ+1
ℓ=0
r ≥ a.
As in the previous case, the coefficients Bℓ are determined by
evaluating
this expression on the surface of the sphere as
P∞
ℓ
V (θ) = ℓ=0 aBℓ+1
Pℓ (cos θ), so that
Z
2ℓ + 1 ℓ+1 π
Bℓ =
a
V (θ)Pℓ (cos θ) sin θd θ.
2
0
(34)
(35)
Boundary Value Problems Spherical Coordinates I
The uniqueness of the potential φ(r , θ) provides a convenient means
by which the solution of some potential problem may be obtained
from a knowledge of the potential along the axis of symmetry.
Along the positive z-axis, z = r and cos θ = 1, so that
∞ X
Bℓ
ℓ
Aℓ r + ℓ+1 , z ≥ 0,
φ(z) =
(36)
r
ℓ=0
while along the negative z-axis, z = r and cos θ = −1, so that
∞
X
Bℓ
ℓ
ℓ
φ(z) =
(−1) Aℓ r + ℓ+1 , z ≤ 0.
r
(37)
ℓ=0
If φ(z) can be determined at any point z on the symmetry axis, and
if this potential can be expressed in a power series in z = r of the
form given above with known coefficients, then the solution for the
potential at any point in space is obtained simply by multiplying each
power of r ℓ and r −(ℓ+1) by the Legendre polynomial Pℓ (cos θ).
Boundary Value Problems Spherical Coordinates I
An expansion of considerable practical importance is that of the
potential φ(r, r′) = |r − r′ |−1 at a field point r due to a unit point
charge at r′ . Its Legendre series representation is obtained by first
rotating the coordinate axes so that the z-axis lies along the position
vector r′ to the unit point source. The potential φ(r, r′), which
satisfies Laplace’s equation everywhere except at the point r = r′ ,
then possesses azimuthal symmetry and can therefore be expanded as
∞ X
Bℓ
1
ℓ
Aℓ r + ℓ+1 Pℓ (cos γ), r 6= r′ ,
=
(38)
|r − r′ |
r
ℓ=0
where γ is the angle between the vectors r and r′ . If the point r is on
the positive z-axis, then the right-hand side of Eq. (38) reduces to
the form given in Eq. (36) while the left-hand side becomes
1
1
1
−→
=
1/2
′
|r − r |
|r − r ′ |
(r 2 + r ′2 − 2rr ′ cos γ)
as γ → 0.
Boundary Value Problems Spherical Coordinates I
Let r< denote the smaller of r and r ′ , and let r> denote the larger of
r and r ′ . One then has the expansion
1
1
=
=
′
|r − r |
r> − r<
1
r>
1
1 − r< /r>
ℓ
∞ 1 X r<
=
.
r>
r>
ℓ=0
For points off of the z-axis it is then only necessary to multiply each
term in this series expansion by the Legendre polynomial Pℓ (cos γ),
resulting in the general Legendre series representation
∞
X rℓ
1
1
<
=
=
P (cos γ)
ℓ+1 ℓ
1/2
2
′2
′
|r − r′ |
r
(r + r − 2rr cos γ)
>
ℓ=0
(39)
which is a useful expansion of Green’s function φ(r, r′) = |r − r′ |−1 in
spherical coordinates.
Boundary Value Problems Spherical Coordinates II
Case II: Associated Legendre Functions & The Spherical Harmonics:
The general potential problem in spherical coordinates possesses
azimuthal dependency. In order that the elementary solution
Q(ϕ) = e ±imϕ of Laplace’s equation (20) be single-valued in this
general situation, m must be an integer. It is then necessary to
determine the solution of the associated Legendre equation
2
dP
m2
2 d P
P = 0,
(40)
+ ℓ(ℓ + 1) −
(1 − ζ ) 2 − 2ζ
dζ
dζ
1 − ζ2
for arbitrary values of ℓ and arbitrary integer values of m. For its
solution, one defines
d
P(ζ) ≡ (−1)m (1 − ζ 2 )m/2 m u(ζ),
dζ
in which case the associated Legendre equation becomes
2
dm
du
2 d u
(1 − ζ ) 2 − 2ζ
+ ℓ(ℓ + 1)u = 0.
(41)
dζm
dζ
dζ
Boundary Value Problems Spherical Coordinates II
The expression appearing inside the square brackets of this equation
is precisely the ordinary Legendre differential equation [cf. Eq. (25)]
with ℓ a nonnegative integer, and so its solution is the Legendre
polynomial Pℓ (ζ). The appropriate solution of the associated
Legendre differential equation (41) is then given by
Pℓm (ζ) = (−1)m (1 − ζ 2 )m/2
dm
Pℓ (ζ)
dζm
(42)
for positive integer values of m. If Rodrigues’ formula (27) is used to
represent Pℓ (ζ), an expression for the associated Legendre functions
Pℓm (ζ) that is valid for both positive and negative integer values of m
is obtained as
Pℓm (ζ) = (−1)m (1 − ζ 2)m/2
d ℓ+m 2
(ζ − 1)ℓ
ℓ+m
dζ
from which it is seen that Pℓ0 (ζ) = Pℓ (ζ).
(43)
Boundary Value Problems Spherical Coordinates II
The solutions given by Eq. (43) will be finite on the closed interval
−1 ≤ ζ ≤ 1 provided that (1) ℓ is either zero or a positive integer
and (2) that m can only take on the values
m = −ℓ, −ℓ + 1, −ℓ + 2, . . . , −1, 0, 1, . . . , ℓ − 2, ℓ − 1, ℓ.
Because the defining differential equation (40) depends only upon m2
and m can only take on positive or negative integer values, it is seen
that Pℓ−m (ζ) and Pℓm (ζ) are proportional. From Rodrigues’ formula,
it is found that
Pℓ−m (ζ) = (−1)m
(ℓ − m)! m
P (ζ)
(ℓ + m)! ℓ
(44)
Boundary Value Problems Spherical Coordinates II
For a given fixed value of m, the associated Legendre functions
Pℓm (ζ) form an orthogonal set on the closed interval −1 ≤ ζ ≤ 1,
satisfying the orthogonality relation
Z
1
−1
Pℓm′ (ζ)Pℓm (ζ)d ζ =
2 (ℓ + m)!
δℓ′ ℓ
2ℓ + 1 (ℓ − m)!
(45)
Because of Eq. (44), this orthogonality relation holds for both
positive and negative integer values of m. In addition, it reduces to
the orthogonality relation given in Eq. (28) for the Legendre
polynomials when m = 0.
Boundary Value Problems Spherical Coordinates II
The solution of Laplace’s equation (18) in spherical coordinates by
the separation of variables method assumed a product of single
variable functions of the three coordinate variables r , θ, ϕ of the form
given in Eq. (19). It is convenient to combine the angular functions
Qm (ϕ) and Pℓm (θ) of this solution in such a way so as to construct a
set of orthogonal functions over the unit sphere. Such a set of
functions is called the set of spherical harmonics
The exponential functions Qm (ϕ) = e imϕ form a complete set of
orthogonal functions in the index m on the angular interval
0 ≤ ϕ < 2π, and the associated Legendre functions Pℓm (θ) form a
complete set of orthogonal functions in the index ℓ for each allowed
value of m on the interval 0 ≤ θ ≤ π.
Their product Pℓm (θ)Qm (ϕ) thus forms a complete orthogonal set of
functions on the surface of a unit sphere in the two indices ℓ and m.
Boundary Value Problems Spherical Coordinates II
From the orthogonality relation (45), this normalized set of functions
is
s
(2ℓ + 1)(ℓ − m)! m
Pℓ (cos θ)e imϕ
(46)
Yℓm (θ, ϕ) ≡
4π(ℓ + m)!
The spherical harmonics satisfy the symmetry relation
∗
Yℓ,−m (θ, ϕ) = (−1)m Yℓm
(θ, ϕ).
the orthonormalization condition
Z 2π
Z π
dϕ
sin θd θ · Yℓ∗′ m′ (θ, ϕ)Yℓm (θ, ϕ) = δℓ′ ℓ δm′ m
0
(47)
(48)
0
and the completeness relation
∞ X
ℓ
X
ℓ=0 m=−ℓ
∗
Yℓm
(θ′ , ϕ′ )Yℓm (θ, ϕ) = δ(ϕ − ϕ′ )δ(cos θ − cos θ′ )
(49)
Boundary Value Problems Spherical Coordinates II
Any sufficiently well-behaved function g (θ, ϕ) that is defined on the
unit sphere can be expanded in terms of spherical harmonics as
g (θ, ϕ) =
∞ X
ℓ
X
Aℓm Yℓm (θ, ϕ),
(50)
where [from the orthonormalization condition (48)]
I
∗
Aℓm = g (θ, ϕ)Yℓm
(θ, ϕ)d Ω,
(51)
ℓ=0 m=−ℓ
where d Ω = sin θd θd ϕ is the differential element of solid angle, the
integration being taken over the entire surface of the unit sphere.
Notice that all terms in the spherical harmonic expansion (50) with
m 6= 0 vanish at θ = 0. In that case, this expansion q
takes on the
P∞
Aℓ0 with
special limiting form g (0, ϕ) = g (θ, ϕ)θ=0 = ℓ=0 2ℓ+1
4π
q
H
g (θ, ϕ)Pℓ (cos θ)d Ω.
Aℓ0 = 2ℓ+1
4π
Boundary Value Problems Spherical Coordinates II
From Eqs. (19), (20), (50), the general solution for a given boundary
value problem for Laplace’s equation in spherical coordinates is given
by
∞ X
ℓ
X
bℓm
ℓ
aℓm r + ℓ+1 Yℓm (θ, ϕ).
φ(r , θ, ϕ) =
(52)
r
ℓ=0 m=−ℓ
If the potential is specified on a spherical surface, the expansion
coefficients aℓm and bℓm can then be determined by evaluating this
expansion on the spherical surface which reduces it to the form given
in Eq. (50).
• For the interior problem one typically requires that bℓm = 0 ∀ℓ, m
so that the potential remains finite at the origin,
• whereas for the exterior problem one requires that aℓm = 0 ∀ℓ, m
so that the potential vanishes at r = ∞.
In either case, the appropriate expansion coefficients are obtained
through direct application of Eq. (51).
Addition Theorem for the Spherical Harmonics
The addition theorem for the spherical harmonics, which is of
considerable mathematical importance, is now briefly considered.
To begin, consider two position vectors r and r′ from the same origin
O that are separated by an angle γ, the first having spherical
coordinates (r , θ, ϕ) and the second having coordinates (r ′ , θ′ , ϕ′ ).
Because
cos γ = cos θ cos θ′ + sin θ sin θ′ cos (ϕ − ϕ′ ),
it is then of interest to express the Legendre polynomial Pℓ (cos γ) in
terms of independent spherical harmonics of the angles θ, ϕ and
θ′ , ϕ′ . If r′ is considered to be fixed in space, then Pℓ (cos γ) is a
function of the angles θ, ϕ with the angles θ′ , ϕ′ as parameters. It
may then be expanded in a series of the form given in Eq. (50) as
Pℓ′
P
′
′
Pℓ (cos γ) = ∞
m=−ℓ′ Aℓ′ m (θ , ϕ )Yℓ′ m (θ, ϕ).
ℓ′ =0
Addition Theorem for the Spherical Harmonics
However, because Pℓ (cos γ) is a spherical harmonic of order ℓ alone,
this expansion simplifies to
Pℓ (cos γ) =
ℓ
X
Am (θ′ , ϕ′ )Yℓm (θ, ϕ),
(53)
m=−ℓ
with expansion coefficients
I
′
′
∗
(θ, ϕ)d Ω =
Am (θ , ϕ ) = Pℓ (cos γ)Yℓm
4π
Y ∗ (θ′ , ϕ′ ).
2ℓ + 1 ℓm
(54)
Combination of these two results then yields the addition theorem for
the spherical harmonics
ℓ
4π X ∗ ′ ′
Y (θ , ϕ )Yℓm (θ, ϕ)
Pℓ (cos γ) =
2ℓ + 1 m=−ℓ ℓm
where cos γ = cos θ cos θ′ + sin θ sin θ′ cos (ϕ − ϕ′ ).
(55)
Addition Theorem for the Spherical Harmonics
If the angle γ goes to zero so that θ = θ′ and ϕ = ϕ′ , then Eq. (55)
yields the sum rule
ℓ
X
m=−ℓ
|Yℓm (θ, ϕ)|2 =
2ℓ + 1
4π
(56)
The addition theorem (55) can be used to express the expansion (39)
of the Green’s function φ(r, r′) = |r − r′ |−1 into its most general form
as
∞ X
ℓ
X
1
1
r<ℓ ∗ ′ ′
Y (θ , ϕ )Yℓm (θ, ϕ)
=
4π
ℓ+1 ℓm
|r − r′ |
2ℓ
+
1
r
>
ℓ=0 m=−ℓ
(57)
where r< denotes the smaller of r and r ′ and r> denotes the larger of
r and r ′ . This expression gives the potential at the point r due to a
unit point charge at the point r′ in a completely factorized form in
terms of the spherical coordinates of r and r′ .