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20. Green's Functions in Spherical
Coordinates: Constructing an Image
Preliminaries: Single Point Charge
As an example of using spherical harmonics in electrostatics,
we’ll take another look at the old favorite of a point charge outside
a grounded conducting sphere, and show how simple
considerations of symmetry lead automatically to the image.
Let's begin with the single point charge. The Green's function
→ →′
G( r , r ) =
1
→′ ∣
∣→
r − r
∣
∣
,
is the solution of
→ →′
→
→′
2
∇ G ( r , r ) = −4πδ ( r − r ),
that is, it's the potential φ (
→
r )
from a unit charge at
→′
r .
To translate to spherical coordinates, recall from the Spherical
Harmonics lecture that, in terms of the angle between the two
vectors and of their spatial orientation,
∞
1
′
→ ∣
∣→
∣ r − r ∣
∣
∣
= ∑
ℓ=0
ℓ
r<
ℓ+1
r>
∞
P ℓ (cos γ) = 4π ∑
ℓ=0
1
2ℓ + 1
ℓ
r<
ℓ+1
r>
ℓ
∑ Y
m=−ℓ
m
ℓ
where r> , r< are the
greater and lesser of
r,
′
r .
It’s important to realize that
the last term is symmetric
in r, r′ (as it must be)
even though it’s not
symmetric in
r> , r<
because swapping
changes which one is
′
r, r
r> .
But how can we see the expression on the right is a solution to
→ →′
→
→′
2
∇ G ( r , r ) = −4πδ ( r − r )
?
First we need the delta function itself in spherical coordinates.
Now
→
→′
3
d rδ ( r − r ) = ∫
1 = ∫
and δ (cos θ ′
→
→′
2
r dr sin θdθdϕδ ( r − r ),
′
− cos θ) = (1/ sin θ)δ (θ − θ ),
′
→
→
δ( r − r ) =
1
r
2
′
δ (r − r )δ (cos θ
′
so
− cos θ)δ (ϕ
′
− ϕ).
Furthermore,
∞
ℓ
∑ ∑ Y
ℓ=0 m=−ℓ
m
ℓ
′
′
* (θ , ϕ )Y
m
ℓ
′
(θ, ϕ) = δ (ϕ − ϕ )δ (cos θ − cos θ
′
(This completeness is nontrivial to prove: the functions are all the
eigenstates of the Laplacian on the spherical surface,
Y
m
ℓ
having
eigenvalue −ℓ (ℓ + 1).)
Except on the shell r
2
∇
the radial component of the
′
= r ,
operator multiplies the ℓ
+ℓ (ℓ + 1),
th
term in the expression by
giving the (full operator) result ∇2 φ
= 0,
as we
expect.
However,
at
′
r = r
the
slope of the
ℓ
th
radial
function
(putting
r< = r>
after differentiating) changes by
1
2ℓ+1
(− (ℓ + 1) − ℓ)
1
r
2
= −
1
r
2
.
Notice this is independent of ℓ. The radial part of the ∇2 operator
can be written
2
d R
dr
differentiation −
2
1
r
2
+
2
dR
r
dr
and this change in slope yields on
′
δ (r − r ).
So ∇2 φ has the same radial shell of delta function in each term,
but this is to be multiplied by the angular term, then summing the
series gives the angular delta function, leaving nonzero only the
point on the shell (θ, ϕ). This recovers the full three-dimensional
delta function in spherical polars:
→
→′
−4πδ ( r − r ) = −
4π
r
2
′
δ (r − r )δ (cos θ
′
− cos θ)δ (ϕ
′
− ϕ)
Adding the Grounded Spherical Conductor
The trivial example described above had no boundary conditions.
Consider the same problem, a single point charge at distance
′
r
from the origin, but now there is a grounded conducting sphere
of radius
′
a < r
centered at the origin.
The angular structure will be the same as for the single point
charge analyzed above, but the function in a
will have
′
< r < r
both positive and negative powers of r, cancelling at r
= a.
The Green’s function must have the spherical harmonic series
from the point charge we’ve already found, but each term in the
−(ℓ+1)
series must now have an r<
term added to the rℓ< term, to
ensure the potential vanishes at the grounded conducting surface,
r = a.
Furthermore, these new terms must also vanish if
a = 0,
and finally they must have the Green’s function symmetry
in r, r
′
.
Putting all these requirements together, and adjusting the power of
a
to get the dimensions correct, we find
∞
→ →′
G ( r , r ) = 4π ∑
ℓ=0
ℓ
1
(
2ℓ + 1
r<
ℓ+1
r>
a
−
2ℓ+1
ℓ+1
r<
ℓ+1
r>
ℓ
) ∑ Y
m=−ℓ
m
ℓ
Note now that since the second term is symmetric in r> , r< we
can replace them with r, r′ , and slightly rearrange to find
∞
→ →′
G ( r , r ) = 4π ∑
(
2ℓ + 1
ℓ=0
2
ℓ
1
r<
ℓ+1
ℓ
′
ℓ
a (a /r )
−
) ∑ Y
′
(ℓ+1)
r
r>
r
m=−ℓ
The first term is identical to the potential from a single charge
found in the previous section, the second term has the same
form: but now the charge is
position is
2
′
a /r
′
−a/r
instead of one, and the
instead of r , but along the same axis. In other
′
words, we've found the image charge the hard way!
Green’s Function Inside Grounded Hollow
Conducting Sphere
It is straightforward to adjust the above analysis to this scenario:
consider first the region 0
∞
→ →′
G ( r , r ) = 4π ∑
ℓ=0
clearly going to zero at r
′
< r
< r < a.
Here
′ℓ
1
′ℓ
r
(
2ℓ + 1
ℓ+1
r
−
r
a
ℓ
ℓ
r
2ℓ+1
) ∑ Y
m
ℓ
*
m=−ℓ
= a.
Rearranging:
∞
→ →′
G ( r , r ) = 4π ∑
ℓ=0
′ℓ
1
r
(
2ℓ + 1
ℓ+1
r
−
′
r
ℓ
ℓ
a
r
2
′
ℓ+1
(a /r )
The second term is the potential from a charge equal to
′
− (a/r )
at radius (a2 /r′ ) on the radial line of the original
charge—the image is bigger, it’s a concave mirror.
) ∑
m=−
Poisson’s Formula and the Green’s Function
Suppose there is an electric potential, but no electric charge, in a
sphere of radius
a
centered at the origin. Then, inside the sphere,
we can expand the potential in spherical harmonics, but only
those nonsingular at the origin:
∞
ℓ
→
ℓ
φ ( r ) = 4π ∑ r
∑ a ℓm Y
ℓ=0
m
ℓ
(θ, ϕ).
m=−ℓ
Poisson showed that if we know the potential everywhere on the
spherical surface r
= a,
call it V
(a, θ, ϕ),
then we can find it
anywhere within the sphere.
The essential point is that this knowledge uniquely determines the
coefficients
a ℓm :
from the orthonormality relation we find
2π
π
1
a ℓm=
4πa
ℓ
∫
dϕ ∫
0
sin θdθY ℓm* (θ, ϕ)V (a, θ, ϕ).
0
So Poisson’s formula for the potential anywhere inside the sphere
is:
∞
r
φ (r, θ, ϕ) = ∑ (
ℓ=0
ℓ
)
a
2π
ℓ
∑ ∫
m=−ℓ
0
π
′
dϕ ∫
′
′
′
′
sin θ dθ Y ℓm* (θ , ϕ )V
0
How does this relate to the Green’s function in this space? Recall
that from the Reciprocation Theorem, if there is zero charge within
the sphere, and you know the potential everywhere on the
spherical surface, then the potential at any inside point is given by
integrating over the spherical surface the normal derivative of the
Green’s function from that point, that is (remembering that the
normal points into the space)
1
φ (r, θ, ϕ) = −
∫
2
′
′
d
′
a dϕ sin θ dθ (
4π
dr
Now, the Green’s function in the region r′
> r,
′
)
′
G (r, θ, ϕ; r ,
′
r =a
(true for r′
= a
)
is
∞
→ →′
G ( r , r ) = 4π ∑
ℓ=0
′ℓ
ℓ
1
r
(
2ℓ + 1
′
ℓ+1
r
−
a
(r )
so, differentiating with respect to r′ , then putting r′
→ →′
⎛ ∂G ( r , r ) ⎞
⎜
⎜
′
∂r
⎝
∞
⎟
⎟
= −4π ∑ (
⎠
ℓ
r
a
ℓ=0
2ℓ+1
a
2
) ∑ Y
m
ℓ
m=−ℓ
= a,
ℓ
1
)
ℓ
ℓ
r
∑ Y
m
ℓ
′
* (θ , ϕ
m=−ℓ
′
r =a
and putting this in the expression for the potential,
∞
φ (r, θ, ϕ) = ∫
′
′
′
dϕ sin θ dθ ∑ (
ℓ=0
ℓ
r
)
a
ℓ
∑ Y
m
ℓ
′
′
* (θ , ϕ )Y
m
ℓ
m=−ℓ
identical to Poisson’s result.
More General Spherical Green's Function Problems
This method will work for situations where the image technique is
much messier. For example, suppose the charge is between two
grounded conducting concentric spheres, so
′
a < r, r
< b.
This will need an infinite series of images. But by
the present method, it is straightforward. In the two regions, both
positive and negative powers of r are present. The boundary
conditions are simple: the radial Green’s function must vanish
when r<
= a
or r>
= b,
′
ℓ
so (Jackson 3.122)
a
g ℓ (r, r ) = C (r< −
2ℓ+1
ℓ+1
ℓ
r>
1
)(
r<
ℓ+1
−
b
r>
2ℓ+1
),
and the constant C is determined by the radial Green's function
change of slope at r
′
= r
′
(
dg ℓ (r, r )
′
)
dr
− (
′
r=r +ε
dg ℓ (r, r )
4π
)
dr
= −
′
r=r −ε
′
r
2
.
(Note: Jackson works with rg rather than just g for some reason,
but the equations are equivalent, the result is the same.)
This gives
4π
C =
.
(2ℓ + 1) (1 − (a/b)
Exercise: Prove it.
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2ℓ+1
)