Classical Electrodynamics
Lecture 3
Sven Nordebo
Department of Physics and Electrical Engineering
Linnæus University, Sweden
4FY826 Classical Electrodynamics, October 1, 2013
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
Consider the following partial differential equation in spherical coordinates
∇2 Φ(x) + A(r)Φ(x)
=
1 ∂ 2 ∂Φ
1
∂
∂Φ
1
∂2Φ
r
+ 2
sin θ
+ 2
+ A(r)Φ = 0
r2 ∂r ∂r
r sin θ ∂θ
∂θ
r sin2 θ ∂φ2
I A(r) = 0 ⇒ Laplace equation in electrostatics
I A(r) = k2 =
ω2
c2
0
⇒ Helmholtz equation in electrodynamics
I A(r) = 2m
(E − V (r)) ⇒ Time-independent Schrödinger equation for a single
~2
non-relativistic particle, e.g., the Hydrogen atom
The radial part of the solution will be different (potens functions, spherical Bessel
functions and Laguerre functions, etc) but the angular part will be the same spherical
harmonics Ylm (θ, φ)
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 2(48)
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
Consider the following partial differential equation in spherical coordinates
∇2 Φ(x) + A(r)Φ(x)
1 ∂ 2 ∂Φ
1
∂
∂Φ
1
∂2Φ
r
+ 2
sin θ
+ 2
+ A(r)Φ
2
2
r ∂r ∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
1 ∂ 2 ∂Φ
1
1 ∂
∂Φ
1 ∂2Φ
r
+ A(r)Φ + 2
sin θ
+
2
2
2
r ∂r ∂r
r
sin θ ∂θ
∂θ
sin θ ∂φ
1 ∂ 2 ∂Φ
1
= 2
r
+ A(r)Φ + 2 ∇2Ω Φ = 0
r ∂r ∂r
r
=
where
∇2Ω =
1 ∂
∂
1
∂2
sin θ
+
sin θ ∂θ
∂θ
sin2 θ ∂φ2
is the (angular) Laplace operator on the unit sphere
In quantum mechanics the orbital angular momentum operator is L = −i(r × ∇) and
L2 = L · L = −∇2Ω
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 3(48)
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
Consider the following partial differential equation in spherical coordinates
1 ∂ 2 ∂Φ
1
r
+ A(r)Φ + 2 ∇2Ω Φ = 0
r2 ∂r ∂r
r
Separate the variables in spherical coordinates
Φ(r, θ, φ) = R(r)P (θ)Q(φ)
1 ∂ 2 ∂
1
r
R(r) + A(r)R(r)P (θ)Q(φ) + R(r) 2 ∇2Ω P (θ)Q(φ) = 0
r2 ∂r ∂r
r
1 1 ∂ 2 ∂
1
1
r
R(r) + A(r) + 2
∇2 P (θ)Q(φ) = 0
R(r) r2 ∂r ∂r
r P (θ)Q(φ) Ω
P (θ)Q(φ)
1 ∂ 2 ∂
1
r
R(r) + r2 A(r) +
∇2 P (θ)Q(φ) = 0
R(r) ∂r ∂r
P (θ)Q(φ) Ω
|
{z
} |
{z
}
=l(l+1)
=−l(l+1)
where l(l + 1) is the separation constant.
It can be shown that if the whole angular interval 0 ≤ θ ≤ π is allowed, then the only
possible values of l are l = 0, 1, . . .
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 4(48)
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
The separation of variables yields



∂ 2 ∂
r
R(r) + r2 A(r) = l(l + 1)R(r)
∂r ∂r

 ∇2 P (θ)Q(φ) = −l(l + 1)P (θ)Q(φ)
Ω
For the Laplace equation in electrostatics, we have A(r) = 0 and the radial function
R(r) satisfies
∂ 2 ∂
r
R(r) = l(l + 1)R(r)
∂r ∂r
with the general solution
R(r) = Arl + Br−(l+1)
where A and B are constants. Physical solutions in source free regions must be finite.
Hence, if r = 0 is included in the solution domain where there are no sources then
B = 0, and if r = ∞ is included in the solution domain where there are no sources
then A = 0.
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 5(48)
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
For the angular part, the separation of variables yields
∇2Ω P (θ)Q(φ) = −l(l + 1)P (θ)Q(φ)
Hence, with the definition of ∇2Ω
Q(φ)
1 ∂
∂P (θ)
1 ∂ 2 Q(φ)
sin θ
+ P (θ) 2
= −l(l + 1)P (θ)Q(φ)
sin θ ∂θ
∂θ
sin θ ∂φ2
1 ∂
∂P (θ)
1
1 ∂ 2 Q(φ)
1
sin θ
+
+ l(l + 1) = 0
P (θ) sin θ ∂θ
∂θ
Q(φ) sin2 θ ∂φ2
1 ∂ 2 Q(φ)
1
∂
∂P (θ)
sin θ
sin θ
+ l(l + 1) sin2 θ +
=0
P (θ)
∂θ
∂θ
Q(φ) ∂φ2
|
{z
} |
{z
}
=m2
=−m2
where m2 is the separation constant
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 6(48)
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
The separation of variables yields

∂
∂P (θ)

2
2


 sin θ ∂θ sin θ ∂θ + l(l + 1) sin θP (θ) = m P (θ)
 ∂ 2 Q(φ)


= −m2 Q(φ)

∂φ2
or

m2
∂P (θ)
1 ∂



 sin θ ∂θ sin θ ∂θ + l(l + 1) − sin2 θ P (θ) = 0
 ∂ 2 Q(φ)



+ m2 Q(φ) = 0
∂φ2
If the whole azimuthal range 0 ≤ φ ≤ 2π is allowed, then due to the periodicity
(uniqueness) the general solution for Q(φ) is given by
Q(φ) = Ceimφ + De−imφ
where m is an integer
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 7(48)
Boundary-value problems in electrostatics: II
Laplace equation in spherical coordinates (3.1)
The equation for P (θ) is given by
∂P (θ)
m2
1 ∂
sin θ
+ l(l + 1) −
P (θ) = 0
sin θ ∂θ
∂θ
sin2 θ
Introduce the substitution x = cos θ where 0 ≤ θ ≤ π and −1 ≤ x ≤ 1.
We have that
and hence
∂
∂x ∂
∂
=
= − sin θ
∂θ
∂θ ∂x
∂x
∂
1 ∂
=−
sin θ ∂θ
∂x


 sin θ ∂ = − sin2 θ ∂ = −(1 − x2 ) ∂
∂θ
∂x
∂x
which yields the generalized Legendre equation
∂
∂
m2
(1 − x2 )
P (x) + l(l + 1) −
P (x) = 0, −1 ≤ x ≤ 1
∂x
∂x
1 − x2




The solutions Plm (x) are called the associated Legendre functions. In order to have
regular (finite) solutions on the interval −1 ≤ x ≤ 1 it is required that l and m are the
integers l = 0, 1, . . ., and m = −l, . . . , l
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 8(48)
Boundary-value problems in electrostatics: II
Legendre equation and Legendre polynomials (3.2)
The ordinary Legendre differential equation (m2 = 0)
∂
∂
(1 − x2 )
P (x) + l(l + 1)P (x) = 0,
∂x
∂x
−1 ≤ x ≤ 1
To find a solution, assume that
P (x) = xα
∞
X
aj xj
j=0
It is straightforward to show that
∞
X
(α + j)(α + j − 1)aj xα+j−2 − [(α + j)(α + j + 1) − l(l + 1)] aj xα+j = 0
j=0
Each power of x must vanish
 α−2
⇒

 x


 α−1
x
⇒




 xα+j ⇒
separately. Hence
α(α − 1)a0 = 0
α(α + 1)a1 = 0
aj+2 =
(α + j)(α + j + 1) − l(l + 1)
aj
(α + j + 2)(α + j + 1)
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 9(48)
Boundary-value problems in electrostatics: II
Legendre equation and Legendre polynomials (3.2)
Each power of x in P (x) = xα
∞
X
aj xj must vanish separately. Hence
j=0
 α−2
x




 α−1
x




 xα+j
⇒
α(α − 1)a0 = 0
⇒
α(α + 1)a1 = 0
⇒
aj+2 =
(α + j)(α + j + 1) − l(l + 1)
aj
(α + j + 2)(α + j + 1)
Every second coefficient aj will be zero, starting either from a0 6= 0 or from a1 6= 0.
Choosing a0 6= 0 ⇒ α = 0 or α = 1 ⇒ a1 = 0 (the other option a1 6= 0 is equivalent)
If α = 0 then P (x) is an even function
If α = 1 then P (x) is an odd function
It can be shown that the series diverges at x = ±1, unless it terminates
The series terminates at α + j = l, but only if l is zero or a positive integer,
l = 0, 1, . . .
(
l even α = 0 j = 0, 2, . . . , l
l odd
α=1
j = 0, 2, . . . , l − 1
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 10(48)
Boundary-value problems in electrostatics: II
Legendre equation and Legendre polynomials (3.2)
The Legendre polynomials of order l, Pl (x), satisfy the ordinary Legendre differential
equation
∂
∂
(1 − x2 )
Pl (x) + l(l + 1)Pl (x) = 0, −1 ≤ x ≤ 1
∂x
∂x
and are normalized such that Pl (1) = 1. The first polynomials are

P0 (x) = 1






P1 (x) = x



P2 (x) = (3x2 − 1)/2





P3 (x) = (5x3 − 3x)/2




P4 (x) = (35x4 − 30x2 + 3)/8
Rodrigues’ formula
1 dl
(x2 − 1)l
2l l! dxl
The Legendre polynomials form a complete orthogonal set of functions on the interval
−1 ≤ x ≤ 1 with orthogonality relationship
Z 1
Z π
2
Pl (x)Pl0 (x) dx =
Pl (cos θ)Pl0 (cos θ) sin θ dθ =
δll0
2l + 1
−1
0
Pl (x) =
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Boundary-value problems in electrostatics: II
Boundary-value problems with azimuthal symmetry (3.3)
General solution for the electrostatic potential with azimuthal symmetry
Φ(r, θ) =
∞ X
Al rl + Bl r−(l+1) Pl (cos θ)
l=0
Exercise 3.1: Suppose that all sources are strictly outside a sphere of radius a, and
that the potential on the surface of the sphere is given by V (θ). Determine the
potential inside the sphere.
Solution: For r ≤ a
Φ(r, θ) =
∞
X
Al rl Pl (cos θ)
l=0
and for r = a
V (θ) = Φ(a, θ) =
∞
X
Al al Pl (cos θ)
l=0
Use the orthogonality of the Legendre polynomials to find that
Z
2l + 1 −l π
a
V (θ)Pl (cos θ) sin θ dθ
Al =
2
0
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Boundary-value problems in electrostatics: II
Exercise 3.2: Suppose that all sources are strictly inside a sphere of radius a, and that
the potential on the surface of the sphere is given by V (θ). Determine the potential
outside the sphere.
Solution: For r ≥ a we have that ∇2 Φ = 0 and hence
Φ(r, θ) =
∞
X
Bl r−(l+1) Pl (cos θ)
l=0
and for r = a
V (θ) = Φ(a, θ) =
∞
X
Bl a−(l+1) Pl (cos θ)
l=0
Use the orthogonality of the Legendre polynomials to find that
Bl =
2l + 1 l+1
a
2
π
Z
V (θ)Pl (cos θ) sin θ dθ
0
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Boundary-value problems in electrostatics: II
Exercise 3.3: Find an expansion of the Green’s function in terms of the Legendre
polynomials, where
1
G(x − x0 ) =
4π|x − x0 |
Solution: Introduce a spherical coordinate system where γ is the angle between x and
x0 . Let x0 be oriented along the positive z-axis. The Green’s function G(x − x0 ) is
now azimuthally symmetric and can generally be expanded as
G(x − x0 ) =
∞ X
Al rl + Bl r−(l+1) Pl (cos γ)
l=0
For r < r0 the expansion reads
G(x − x0 ) =
∞
X
Al rl Pl (cos γ)
l=0
For r > r0 the expansion reads
G(x − x0 ) =
∞
X
Bl r−(l+1) Pl (cos γ)
l=0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 14(48)
Boundary-value problems in electrostatics: II
Exercise 3.3 (solution):
For r < r0 the expansion reads
G(x − x0 ) =
∞
X
Al rl Pl (cos γ)
l=0
and for γ = 0 (use Pl (1) = 1)
∞
X
Al r l =
l=0
∞
∞
1
1 X r l
1
1
1 X 0 −(l+1) l
r
r
=
=
r =
0
0
0
0
4π(r − r)
4πr 1 − r0
4πr l=0 r
4π l=0
and hence
Al =
1 0 −(l+1)
r
4π
Finally, the solution is
G(x − x0 ) =
∞
1 X 0 −(l+1) l
r
r Pl (cos γ)
4π l=0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 15(48)
Boundary-value problems in electrostatics: II
Exercise 3.3 (solution):
For r > r0 the expansion reads
G(x − x0 ) =
∞
X
Bl r−(l+1) Pl (cos γ)
l=0
and for γ = 0 (use Pl (1) = 1)
∞
X
Bl r−(l+1) =
l=0
∞ ∞
1
1
1
1 X r0 l
1 X 0 l −(l+1)
r r
=
=
0 =
0
r
4π(r − r )
4πr 1 −
4πr l=0 r
4π l=0
r
and hence
Bl =
1 0l
r
4π
Finally, the solution is
G(x − x0 ) =
∞
1 X 0 l −(l+1)
r r
Pl (cos γ)
4π l=0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 16(48)
Boundary-value problems in electrostatics: II
Exercise 3.3 (solution):
The two solutions can be written jointly as
G(x − x0 ) =

∞
1 X 0 −(l+1) l



r
r Pl (cos γ)


 4π l=0
1
=
∞

4π|x − x0 |

1 X 0 l −(l+1)


r r
Pl (cos γ)

 4π
l=0
or, as
G(x − x0 ) =
r < r0
r > r0
∞
l
1 X r<
1
=
P (cos γ)
l+1 l
4π|x − x0 |
4π l=0 r>
where
r< = min r, r0
and
r> = max r, r0
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Boundary-value problems in electrostatics: II
Exercise 3.4: Find the potential due to a total charge q uniformly distributed around a
circular ring of radius a with its axis
√ the z axis and its center at z = b. The distance
from the origin to the ring is c = a2 + b2 where the angle is α (see Fig. 3.4 in
Jackson).
The problem has azimuthal symmetry and the solution can hence be written generally
as
∞ X
Al rl + Bl r−(l+1) Pl (cos θ)
Φ(r, θ) =
l=0
For r < c the expansion reads
Φ(r, θ) =
∞
X
Al rl Pl (cos θ)
l=0
For r > c the expansion reads
Φ(r, θ) =
∞
X
Bl r−(l+1) Pl (cos θ)
l=0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 18(48)
Boundary-value problems in electrostatics: II
Exercise 3.4 (solution):
For r < c the expansion reads
Φ(r, θ) =
∞
X
Al rl Pl (cos θ)
l=0
and for a point x = rẑ on the symmetry axis z (use Pl (1) = 1)
Φ(r, 0) =
∞
X
Al r l =
l=0
1
4π0
Z
∞
1
q X −(l+1) l
dq
q
c
r Pl (cos α)
=
=
|x − c|
4π0 |x − c|
4π0 l=0
and hence
Al =
q −(l+1)
c
Pl (cos α)
4π0
and the solution is
Φ(r, θ) =
∞
q X −(l+1)
c
Pl (cos α)rl Pl (cos θ)
4π0 l=0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 19(48)
Boundary-value problems in electrostatics: II
Exercise 3.4 (solution):
For r > c the expansion reads
Φ(r, θ) =
∞
X
Bl r−(l+1) Pl (cos θ)
l=0
and for a point x = rẑ on the symmetry axis z (use Pl (1) = 1)
Φ(r, 0) =
∞
X
Bl r−(l+1) =
l=0
1
4π0
Z
∞
1
q X −(l+1) l
dq
q
r
c Pl (cos α)
=
=
|x − c|
4π0 |x − c|
4π0 l=0
and hence
Bl =
q l
c Pl (cos α)
4π0
and the solution is
Φ(r, θ) =
∞
q X l
c Pl (cos α)r−(l+1) Pl (cos θ)
4π0 l=0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 20(48)
Boundary-value problems in electrostatics: II
Associated Legendre functions and the spherical harmonics (3.5)
The associated Legendre functions Plm (x) are regular (finite) solutions of the
generalized Legendre equation
∂
m2
∂
(1 − x2 )
P (x) + l(l + 1) −
P (x) = 0, −1 ≤ x ≤ 1
∂x
∂x
1 − x2
where l = 0, 1, . . ., and m = −l, . . . , l. Note that Pl0 (x) = Pl (x)
For m ≥ 0 the associated Legendre functions are defined by
Plm (x) = (−1)m (1 − x2 )m/2
dm
Pl (x),
dxm
−1 ≤ x ≤ 1
By using Rodrigues’ formula a representation of Plm (x) is obtained that is valid for
both positive and negative m
Plm (x) =
(−1)m
dl+m
(1 − x2 )m/2 l+m (x2 − 1)l ,
l
2 l!
dx
−1 ≤ x ≤ 1
It can be shown that
Pl−m (x) = (−1)m
(l − m)! m
P (x),
(l + m)! l
−1 ≤ x ≤ 1
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Boundary-value problems in electrostatics: II
Associated Legendre functions and the spherical harmonics (3.5)
The associated Legendre functions form a complete orthogonal set on the interval
−1 ≤ x ≤ 1 with orthogonality relationship
Z 1
2 (l + m)!
Plm (x)Plm
δll0
0 (x) dx =
2l + 1 (l − m)!
−1
The spherical harmonics are defined by
s
2l + 1 (l − m)! m
P (cos θ)eimφ
Ylm (θ, φ) =
4π (l + m)! l
∗ (θ, φ)
where l = 0, 1, . . ., and m = −l, . . . , l, and Yl,−m (θ, φ) = (−1)m Ylm
The spherical harmonics form a complete orthonormal set on the unit sphere with
orthogonality relationship
Z 2π Z π
Ylm (θ, φ)Yl∗0 m0 (θ, φ) sin θ dθ dφ = δll0 δmm0
φ=0
θ=0
The spherical harmonics are eigenfunctions of the Laplace operator on the unit sphere
∇2Ω Ylm (θ, φ) = −l(l + 1)Ylm (θ, φ)
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Boundary-value problems in electrostatics: II
Associated Legendre functions and the spherical harmonics (3.5)
Note that
Plm (cos θ) = (−1)m (sin θ)m
dm
Pl (cos θ),
d(cos θ)m
0≤m≤l
and Ylm (θ, φ) are polynomials in {cos θ, sin θ} and in {cos φ, sin φ} of order l
q

3

cos θ
 Y10 (θ, φ) =
4π
1
Y00 (θ, φ) = √
q
4π 
 Y (θ, φ) = − 3 sin θeiφ
11
8π
q

5
3

cos2 θ − 12
Y20 (θ, φ) =

4π
2




q
15
Y21 (θ, φ) = − 8π
sin θ cos θeiφ



q


 Y (θ, φ) = 1 15 sin2 θei2φ
22
4
2π
q

7
5

cos3 θ − 23 cos θ
Y30 (θ, φ) =

4π 2



q



21

sin θ 5 cos2 θ − 1 eiφ
 Y31 (θ, φ) = − 14 4π
q

1
105


Y
(θ,
φ)
=
sin2 θ cos θei2φ
32

4
2π




q


35
Y33 (θ, φ) = − 14 4π
sin3 θei3φ
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Boundary-value problems in electrostatics: II
Associated Legendre functions and the spherical harmonics (3.5)
From the completeness of the spherical harmonics follows that the general solution to
Laplace equation ∇2 Φ = 0 can be written as the series
Φ(r, θ, φ) =
∞ X
m X
Alm rl + Blm r−(l+1) Ylm (θ, φ)
l=0 l=−m
where Alm and Blm are constants. Physical solutions in source free regions must be
finite. Hence, if r = 0 is included in the solution domain where there are no sources
then Blm = 0, and if r = ∞ is included in the solution domain where there are no
sources then Alm = 0.
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 24(48)
Boundary-value problems in electrostatics: II
Associated Legendre functions and the spherical harmonics (3.5)
From the completeness of the spherical harmonics follows that an arbitrary function
g(θ, φ) can be expanded in spherical harmonics
g(θ, φ) =
∞ X
m
X
Alm Ylm (θ, φ)
l=0 l=−m
where the coefficients are
Z 2π Z π
Z
∗
∗
Alm =
Ylm
(θ, φ)g(θ, φ) sin θ dθ dφ =
Ylm
(θ, φ)g(θ, φ) dΩ
0
0
Note that
r
Yl0 (θ, φ) =
2l + 1
Pl (cos θ)
4π
and for θ = 0 (cos θ = 1)
Plm (1) =
Pl (1) = 1
0
r
Ylm (0, φ) =
[g(θ, φ)]θ=0 =
∞
X
l=0
r
Al0
2l + 1
4π
where
m=0
m 6= 0
= δm0
2l + 1
δm0
4π
r
Z
2l + 1
Al0 =
Pl (cos θ)g(θ, φ) dΩ
4π
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 25(48)
Boundary-value problems in electrostatics: II
Addition theorem for spherical harmonics (3.6)
Let x and x0 be two points on the unit sphere with spherical angles (θ, φ) and
(θ0 , φ0 ), respectively. Let γ be the angle between the two vectors (see Fig. 3.7 in
Jackson). The addition theorem for spherical harmonics is given by
Pl (cos γ) =
l
X
4π
∗
Ylm (θ, φ)Ylm
(θ0 , φ0 )
2l + 1 m=−l
Proof: (3.6 in Jackson and 12.8 in Arfken)
Let x0 be fixed and consider a rotated coordinate system with x0 on the positive
z-axis and with spherical angles (γ, β). We have
θ = θ(γ, β)
φ = φ(γ, β)
γ = γ(θ, φ)
β = β(θ, φ)
γ = 0 ⇔ (θ, φ) = (θ0 , φ0 )
The spherical harmonics are eigenfunctions of ∇2Ω with eigenvalue −l(l + 1)
∇2Ω Ylm (θ, φ) = −l(l + 1)Ylm (θ, φ)
2
∇0 Ω Ylm (γ, β) = −l(l + 1)Ylm (γ, β)
where ∇0 refers to the coordinate system with spherical angles (γ, β)
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 26(48)
Boundary-value problems in electrostatics: II
Proof: (3.6 in Jackson and 12.8 in Arfken)
The Laplace operator is invariant to a rotation of the coordinate system. Hence
∇2Ω = ∇0 2Ω and
2
∇0 Ω Ylm (θ, φ) = −l(l + 1)Ylm (θ, φ)
Hence, Ylm (θ, φ) is an eigenfunction of the operator ∇0 2Ω with the same eigenvalue
−l(l + 1), and can hence be expanded as
Ylm (θ, φ) =
l
X
Alm0 Ylm0 (γ, β)
m0 =−l
where
Z
∗
Ylm
0 (γ, β)Ylm (θ, φ) dΩ(γ, β)
Alm0 =
Z
∗
Yl0
(γ, β)Ylm (θ, φ) dΩ(γ, β)
Al0 =
Let g(γ, β) = Ylm (θ, φ). Then
0
0
[g(γ, β)]|γ=0 = Ylm (θ , φ ) =
l
X
m0 =−l
r
Alm0 Ylm0 (0, β) =
| {z }
2l + 1
Al0
4π
q
= 2l+1
δm0 0
4π
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 27(48)
Boundary-value problems in electrostatics: II
Proof: (3.6 in Jackson and 12.8 in Arfken)
We have also
r
Yl0 (γ, β) =
l
X
2l + 1
Pl (cos γ) =
Am (θ0 , φ0 )Ylm (θ, φ)
4π
m=−l
where
Am (θ0 , φ0 ) =
Z
∗
Ylm
(θ, φ)Yl0 (γ, β) dΩ(θ, φ) =
Z
∗
Ylm
(θ, φ)Yl0 (γ, β) dΩ(γ, β) = A∗l0
and hence
Pl (cos γ) =
l
X
m=−l
r
4π
2l + 1
Am (θ0 , φ0 )
|
{z
}
Ylm (θ, φ)
q
4π Y ∗ (θ 0 ,φ0 )
=A∗
= 2l+1
l0
lm
=
l
X
m=−l
4π
∗
Ylm (θ, φ)Ylm
(θ0 , φ0 )
2l + 1
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 28(48)
Boundary-value problems in electrostatics: II
Addition theorem for spherical harmonics (3.6)
Expansion of the free space Green’s function in terms of the spherical harmonics
G(x − x0 ) =
∞
l
1
1 X r<
=
P (cos γ)
l+1 l
4π|x − x0 |
4π l=0 r>
=
∞
l
l
X
1 X r<
4π
∗
Ylm (θ, φ)Ylm
(θ0 , φ0 )
l+1
4π l=0 r> m=−l 2l + 1
=
∞ X
l
X
l=0 m=−l
l
r<
1
∗
Y (θ, φ)Ylm
(θ0 , φ0 )
l+1 2l + 1 lm
r>
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 29(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Consider the following Helmholtz equation (Laplace equation when k = ω/c0 = 0) in
cylindrical coordinates
∇2 Φ(x) + k2 Φ(x) = 0
1 ∂ ∂
1 ∂2
∂2
ρ Φ(x) + 2
Φ(x) +
Φ(x) + k2 Φ(x) = 0
2
ρ ∂ρ ∂ρ
ρ ∂φ
∂z 2
Introduce the separation Φ(x) = R(ρ)Q(φ)Z(z), yielding
QZ
1 ∂2
∂2
1 ∂ ∂
ρ R + RZ 2
Q + RQ 2 Z + k2 RQZ = 0
ρ ∂ρ ∂ρ
ρ ∂φ2
∂z
1 1 ∂ ∂
1 1 ∂2
1 ∂2
ρ R + k2 +
Q+
Z=0
2
2
R ρ ∂ρ ∂ρ
Q ρ ∂φ
Z ∂z 2
1 ∂ ∂
1 ∂2
1 ∂2
ρ ρ R + k 2 ρ2 +
Q +ρ2
Z=0
2
R ∂ρ ∂ρ
Q ∂φ
Z ∂z 2
| {z }
=−ν 2
Solution for the azimuthal part
∂2
Q + ν2Q = 0
∂φ2
⇒
Q(φ) = e±iνφ
If 0 ≤ φ ≤ 2π is allowed ⇒ v = m (integer)
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 30(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Continue to separate the radial and longitudinal part
1 ∂ ∂
1 ∂2
ρ ρ R + k2 ρ2 − ν 2 + ρ2
Z=0
R ∂ρ ∂ρ
Z ∂z 2
ν2
1 ∂2
1 1 ∂ ∂
ρ R + k2 − 2 +
Z=0
R ρ ∂ρ ∂ρ
ρ
Z ∂z 2
| {z }
=γ 2
Solution for the longitudinal part
∂2
Z − γ2Z = 0
∂z 2
⇒
Z(z) = e±γz
(γ is the propagation constant)
Solution for the radial part
1 1 ∂ ∂
ν2
ρ R + γ 2 + k2 − 2 = 0
| {z } ρ
R ρ ∂ρ ∂ρ
=κ2
1 ∂ ∂
ν2
ρ R + κ2 − 2 R = 0
ρ ∂ρ ∂ρ
ρ
where κ2 = γ 2 + k2 (For the Laplace equation κ = γ)
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 31(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Solution for the radial part
1 ∂ ∂
ν2
ρ R + κ2 − 2 R = 0
ρ ∂ρ ∂ρ
ρ
Substitution: x = κρ ⇒
∂
∂ρ
∂
= κ ∂x
The Bessel equation
∂
∂
x
R(x) + (x2 − ν 2 )R(x) = 0
∂x ∂x
Solutions are called Bessel functions of order ν
x
The Bessel functions of the first kind and order ν (real or complex) are given by the
series expansion
Jν (x) =
∞
X
j=0
x 2j+ν
(−1)j
,
j!Γ(j + ν + 1) 2
x, ν ∈ C
where Γ is the Gamma function. For integer order m ≥ 0 the series expansion becomes
Jm (x) =
∞
X
j=0
(−1)j x 2j+m
j!(j + m)! 2
and where
J−m (x) = (−1)m Jm (x)
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 32(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Derivation of the series expansion
Define ∇ν = x
∞
X
∂
∂
x
+ (x2 − ν 2 ) and make the ansatz R(x) =
aj xα+j , a0 6= 0
∂x ∂x
j=0
It is straightforward to show that
∇ν R(x) =
∞
X
∞
X
aj xα+j (α + j)2 − ν 2 +
aj xα+j+2 = 0
j=0
Identify powers
 α
x :





xα+1 :



 α+2
x
:


..



.



 α+m
x
:
j=0
a0 (α2 − ν 2 ) = 0
⇒
α = ±ν
a1 (α + 1)2 − ν 2 = 0
a2 (α + 2)2 − ν 2 + a0 = 0
⇒
a1 = 0
..
.
am (α + m)2 − ν 2 + am−2 = 0
General solution am = −
1
am−2 for m ≥ 2
(α + m)2 − ν 2
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 33(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Derivation of the series expansion
General solution for m = 2, 4, . . . (use α2 = ν 2 )
am = −
1
1
am−2 = −
am−2
(α + m)2 − ν 2
m(m + 2α)
or for j = 1, 2, . . .
a2j = −
1
a2j−2
4j(j + α)
Iterate to obtain
a2j =
(−1)j
(−1)j Γ(α + 1)
a0 = 2j
a0
22j j!(α + j)(α + j − 1) · · · (α + 1)
2 j! Γ(α + j + 1)
and choose
a0 =
1
2α Γ(α + 1)
to obtain (with α = ν)
Jν (x) =
∞
X
j=0
a2j x2j+ν =
∞
X
j=0
x 2j+ν
(−1)j
j!Γ(j + ν + 1) 2
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 34(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
The Gamma function is defined by Euler’s integral
Z ∞
e−t tz−1 dt
Γ(z) =
0
which is holomorphic in the half-plane Re z > 0. A single integration by parts yields
the recurrence formula
Γ(z + 1) = zΓ(z)
It is seen that Γ(1) = 1 and Γ(n + 1) = n! for n = 0, 1, 2, . . .
Γ(z) is defined by analytic continuation for Re z ≤ 0. It is a meromorphic function
1
at z = −n for
with no zeros and with simple poles with residues (−1)n n!
n = 0, 1, 2, . . .. The function 1/Γ(z) is entire with simple zeros at z = −n.
Proof (residues): A Taylor series expansion at z = 0 gives Γ(z + 1) = 1 + zf (z) where
f (z) is holomorphic in a neighborhood of z = 0. It follows that
Γ(z − n) =
Γ(z)
1 + zf (z)
=
(z − 1)(z − 2) · · · (z − n)
z(z − 1)(z − 2) · · · (z − n)
1
1
1 + zf (z)
1 (−1)n
=
+ ···
z
z =
n
z (−1) n! (1 − z)(1 − 2 ) · · · (1 − n )
z n!
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 35(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
The Gamma function
Note that
Γ(α + j + 1) = (α + j)Γ(α + j) = (α + j)(α + j − 1)Γ(α + j − 1) = · · ·
= (α + j)(α + j − 1) · · · (α + 1)Γ(α + 1)
so that
Note also that
1
Γ(α + 1)
=
(α + j)(α + j − 1) · · · (α + 1)
Γ(α + j + 1)
1
∂ 1 =
= (−1)n n!
∂z Γ(z) z=−n
Res Γ(z)|z=−n
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 36(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Exercise 3.5: Use the properties of the Gamma function to show that
J−m (x) = (−1)m Jm (x) where m is a non-negative integer (m ≥ 0).
J−m (x) =
∞
X
j=0
x 2j−m (−1)j
1
= 0 for j − m + 1 ≤ 0
=
j!Γ(j − m + 1) 2
Γ(j − m + 1)
=
∞
X
j=m
=
x 2j−m
(−1)j
= {j = k + m}
j!Γ(j − m + 1) 2
∞
X
k=0
x 2(k+m)−m
(−1)k+m
(k + m)!Γ(k + 1) 2
(−1)m
∞
X
k=0
(−1)k x 2k+m
= (−1)m Jm (x)
(k + m)!k! 2
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 37(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
The Wronskian
If f1 and f2 are two solutions of a linear differential equation of the second order, and
if f10 and f20 denote their derivatives, then f1 and f2 are linearly independent if and
only if the Wronskian determinant does not vanish identically, i.e.,
f
W (f1 , f2 ) = 10
f1
f2 = f1 f20 − f2 f10 6= 0
0
f2 Proof (sketch): f1 and f2 are linearly dependent ⇔
f 0 f2 − f1 f20
f1
=C⇔ 1
= 0 ⇔ f1 f20 − f2 f10 = 0
f1 − Cf2 = 0 ⇔
f2
f22
It can be shown that
W (Jν (x), J−ν (x)) = −
2 sin νπ
πx
Hence, Jν (x) and J−ν (x) are linearly independent if ν is not an integer, and they are
linearly dependent if ν is an integer ν = m. Note that J−m (x) = (−1)m Jm (x) for all
integers m.
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 38(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
The Neumann function
The Neumann function (or Bessel function of the second kind) is defined by
Nν (x) =
Jν (x) cos νπ − J−ν (x)
sin νπ
when ν is not an integer. The Neumann function and the Bessel function of the first
2
kind are linearly independent and their Wronskian is W (Jν (x), Nν (x)) = πx
For integer order m the Neumann function is defined by the limit
Nm (x) = lim
ν→m
Jν (x) cos νπ − J−ν (x)
sin νπ
It can be shown that Nm (x) and Jm (x) are linearly independent, their Wronskian is
2
W (Jm (x), Nm (x)) =
, and that N−m (x) = (−1)m Nm (x).
πx
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 39(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Exercise 3.6: Show that
Nm (x) =
1 ∂
∂
1
Jν (x)
Jν (x)
+ (−1)m
π ∂ν
π
∂ν
ν=m
ν=−m
Solution:
Jν (x) cos νπ − J−ν (x)
sin νπ
Jν (x) cos νπ − Jm (x)(−1)m − (J−ν (x) − J−m (x))
= lim
ν→m
sin νπ
Jν (x) cos νπ − Jm (x)(−1)m
J−ν (x) − J−m (x)
−
lim
ν→m
(−1)m sin(ν − m)π
−(−1)m sin(−ν + m)π
Jν (x)(−1)m − Jm (x)(−1)m
J−ν (x) − J−m (x)
lim
−
ν→m
(−1)m (ν − m)π
−(−1)m (−ν + m)π
1 ∂
1
∂
=
Jν (x)
+ (−1)m
Jν (x)
π ∂ν
π
∂ν
ν=m
ν=−m
Nm (x) = lim
ν→m
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 40(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Exercise 3.7: Find the dominating singularity behavior of Nm (x) for m ≥ 0
1 ∂
∂
1
Solution: Use Nm (x) =
Jν (x)
Jν (x)
+ (−1)m
and the series
π ∂ν
π
∂ν
ν=m
ν=−m
∞
∞
j
j
X
X
x
(−1)
x 2j+ν
(−1)
e(2j+ν) ln 2
expansion Jν (x) =
=
j!Γ(j
+
ν
+
1)
2
j!Γ(j
+
ν
+
1)
j=0
j=0
Nm (x) =
∞
x 2j+ν
1 X (−1)j ∂
1
∂ x 2j+ν
1
+
π j=0 j!
∂ν Γ(j + ν + 1) 2
Γ(j + ν + 1) ∂ν 2
ν=m
∞
x 2j+ν
X
1
(−1)j ∂
1
1
∂ x 2j+ν
+ (−1)m
+
π
j!
∂ν Γ(j + ν + 1) 2
Γ(j + ν + 1) ∂ν 2
ν=−m
j=0
−m
1
∂
1
x
2
x
= (−1)m
+ · · · + Jm (x) ln + · · ·
π
∂ν Γ(ν + 1) ν=−m 2
π
2
|
{z
}
=(−1)m−1 (m−1)!
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 41(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Summary of Bessel functions of integer order
Bessel function of the first kind
∞
X
(−1)j x 2j+m
1 x m
Jm (x) =
=
+ O{xm+2 }
j!(j
+
m)!
2
m!
2
j=0
2
Bessel function of the second kind (Neumann function), W (Jm , Nm ) = πx

(m − 1)! x −m
2
x


+ · · · + Jm (x) ln + · · ·
 Nm (x) = −
π
2
π
2

2
x

 N0 (x) = J0 (x) ln + · · ·
π
2
(1)
Bessel function of the third kind (Hankel functions), W (Jm , Hm ) =
(1)
(2)
W (Hm , Hm )
2i
,
πx
4i
− πx
=
r

2 i(x−mπ/2−π/4)

(1)


H
(x)
=
J
(x)
+
iN
(x)
∼
e
as x → ∞
m
m
m

πx
r


2 −i(x−mπ/2−π/4)
(2)

 Hm
(x) = Jm (x) − iNm (x) ∼
e
as x → ∞
πx
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 42(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Recursion formulas for Bessel functions. Here Cm (x) is a Bessel function of the first,
second or third kind

2m

Cm−1 (x) + Cm+1 (x) =
Cm (x)



x

0
Cm−1 (x) − Cm+1 (x) = 2Cm
(x)




m
 C 0 (x) =
Cm (x) − Cm+1 (x)
m
x
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 43(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
The Fourier-Bessel series
The zeros xmn of the Bessel function of the first kind Jm (x) satisfy
Jm (xmn ) = 0, for n = 1, 2, . . .
and are given asymptotically as xmn ∼ nπ + (m − 1/2)π/2 as n → ∞
It can be shown that the following orthogonality relationship holds
a
Z
0
ρ
ρ
a2 2
ρJm (xmn )Jm (xmn0 ) dρ =
J
(xmn )δnn0
a
a
2 m+1
and hence that a function f (ρ) with f (a) = 0 can be expanded in the Fourier-Bessel
series
∞
X
ρ
f (ρ) =
Amn Jm (xmn )
a
n=1
and where
Amn =
2
2
a2 Jm+1
(xmn )
a
Z
0
ρ
ρf (ρ)Jm (xmn ) dρ
a
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 44(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Z
Exercise 3.8: Show that
0
a
ρ
ρ
ρJm (xmn )Jm (xmn0 ) dρ = 0 when n 6= n0
a
a
Solution: with x = κρ, the Bessel equation is
1 ∂ ∂
ν2
ρ R(ρ) + κ2 − 2 R(ρ) = 0
ρ ∂ρ ∂ρ
ρ
Now, put κ =
xmn
a
⇒ x = xmn aρ , so that
2
ρ
1 ∂ ∂
ρ
xmn
ν2
Jm (xmn ) = 0
ρ Jm (xmn ) +
−
ρ ∂ρ ∂ρ
a
a2
ρ2
a
Multiply the equation with ρJm (xmn0 aρ ) and integrate
a
Z
0
∂ ∂
ρ
ρ
ρ Jm (xmn )Jm (xmn0 ) dρ+
∂ρ ∂ρ
a
a
Z
0
a
x2mn
ν2
ρ
ρ
−
ρJm (xmn )Jm (xmn0 ) dρ = 0
a2
ρ2
a
a
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 45(48)
Boundary-value problems in electrostatics: II
Laplace equation in cylindrical coordinates; Bessel functions (3.7)
Exercise 3.8: Integrate by parts
Z a
∂
ρ a
∂
ρ x 0
ρ
ρ 0
Jm (xmn )Jm (xmn0 ) −
ρ Jm (xmn )Jm
(xmn0 ) mn dρ
∂ρ
a
a 0
∂ρ
a
a
a
0
|
{z
}
ρ
=0
a
Z
+
0
a
Z
−
0
x2mn
ν2
ρ
ρ
−
ρJm (xmn )Jm (xmn0 ) dρ = 0
a2
ρ2
a
a
ρ 0
ρ xmn xmn0
0
ρJm
(xmn )Jm
(xmn0 )
dρ
a
a a
a
Z a 2
ρ
xmn
ρ
ν2
ρJm (xmn )Jm (xmn0 ) dρ = 0
+
−
a2
ρ2
a
a
0
Interchange n and n0 and subtract, to obtain
x2mn − x2mn0
a2
a
Z
0
ρ
ρ
ρJm (xmn )Jm (xmn0 ) dρ = 0
a
a
showing that the integral vanishes when n 6= n0
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 46(48)
Boundary-value problems in electrostatics: II
Boundary-value problems in cylindrical coordinates (3.8)
Exercise 3.9: Find an analytical solution to Laplace equation ∇2 Φ = 0 for a finite
cylinder as depicted in Fig. 3.9 in Jackson. The radius of the cylinder is a and the
hight is L. The boundary conditions are Φ = 0 at all surfaces except at the top at
z = L where Φ(ρ, φ, L) = V (ρ, φ)
Solution: The separation of variables gives solutions of the form
Φ(ρ, φ, z) = R(ρ)Q(φ)Z(z) = (Am Jm (κρ) + Bm Nm (κρ))eimφ (Cm eκz + Dm e−κz )
where γ = κ and x = κρ and m = 0, ±1, ±2, . . .
Since the origin ρ = 0 is included in the problem we have that Bm = 0
Since Φ = 0 for z = 0 we have that Z(0) = 0 yielding Cm + Dm = 0 or
Z(z) = Cm (eκz − e−κz ) = 2Cm sinh(κz)
Since Φ = 0 for ρ = a, we have that R(a) = Jm (κa) = 0 yielding κa = xmn or
κ = κmn = xmn
for n = 1, 2 . . .
a
The following general expansion is obtained
Φ(ρ, φ, z) =
∞
X
∞
X
Amn Jm (κmn ρ)eimφ sinh(κmn z)
m=−∞ n=1
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 47(48)
Boundary-value problems in electrostatics: II
Boundary-value problems in cylindrical coordinates (3.8)
The boundary condition at the top of the cylinder is
Φ(ρ, φ, L) = V (ρ, φ) =
∞
X
∞
X
Amn Jm (κmn ρ)eimφ sinh(κmn L)
m=−∞ n=1
which is a Fourier-Bessel series in the radial variable and an ordinary Fourier series in
the azimuthal variable. We have the following orthogonality relationships
 Z




Z




a
0
ρ
a2 2
ρ
J
ρJm (xmn )Jm (xmn0 ) dρ =
(xmn )δnn0
a
a
2 m+1
2π
ei(m
0
−m)φ
dφ = 2πδmm0
0
and it follows that
Amn =
1
2
πa2 sinh(κmn L)Jm+1
(xmn )
Z
0
2π
a
Z
0
ρ
ρV (ρ, φ)Jm (xmn )e−imφ dρ dφ
a
Sven Nordebo, Department of Physics and Electrical Engineering, Linnæus University, Sweden. 48(48)