6
JACOB LEWIS BOURJAILY
Problem 3.9
Let us consider a right circular cylinder with radius b whose central axis is coincident with the z-axis.
The ends of the cylinder are grounded and the potential on the face is given by ϕ(b, φ, z). We are to
determine a series solution for the potential everywhere inside of the cylinder.
Almost canonically, let us assume that the potential ϕ can be completely separated into
ϕ(ρ, φ, z) = R(ρ)Φ(φ)Z(z).
We know that the Laplacian of ϕ must vanish—the interior of the cylinder is assumed to be free of
charge. Therefore, taking the Laplacian of ϕ and dividing by ϕ, we see that
µ
¶
1 ∂
∂ϕ
1 ∂2ϕ ∂2ϕ
∇2 ϕ(ρ, φ, z) = 0 =
ρ
+ 2
+
,
ρ ∂ρ
∂ρ
ρ ∂φ2
∂z 2
1 ∂R
1 ∂2Φ
∂2Z
∂2R
ΦZ
+
ΦZ
+
RZ
+
RΦ
,
=
∂ρ2
ρ ∂ρ
ρ2 ∂φ2
∂z 2
½
¾
1 ∂2R
1 ∂R
1 ∂2Φ
1 ∂2Z
=
+
+
+
.
2
2
2
R ∂ρ
Rρ ∂ρ
Φρ ∂φ
Z ∂z 2
Now, because the expression in brackets explicitly depends on ρ while the other term does not, each
part must be separately constant. Without any loss of generality we can say that the term in brackets
is equal to λ2 —if this causes problems, which it won’t, we could choose λ to be purely imaginary. In
particular, this implies that
1 ∂2Z
= −λ2 ,
Z ∂z 2
and hence, we see that
Z(z) = A cos(λz) + B sin(λz).
From our boundary conditions, Z(0) = Z(L) = 0, we know that A = 0 and λn = πn
L for n ∈ Z.
Therefore, in more generality, the term in brackets that we saw earlier can be equal to λ2n for any n ∈ Z.
Using our work above for λn , we can derive similar relations for Φ. Specifically, we see that
1 ∂2R
1 ∂R
1 ∂2Φ
+
+
− λ2n ,
R ∂ρ2
Rρ ∂ρ
Φρ2 ∂φ2
ρ2 ∂ 2 R
ρ ∂R
1 ∂2Φ
=
+
+
− ρ2 λ2n ,
R ∂ρ2
R ∂ρ
Φ ∂φ2
½ 2 2
¾
ρ ∂ R
ρ ∂R
1 ∂2Φ
2 2
=
+
−
ρ
λ
+
.
n
R ∂ρ2
R ∂ρ
Φ ∂φ2
0=
Again, because the expression in brackets is clearly dependent on ρ whereas the other term is not, each
must be separately constant. We will lose no generality by simply demanding that the expression in
brackets be equal to m2 . In almost exact analogy to the case for Z, we have
1 ∂2Φ
= −m2 ,
Φ ∂φ2
which immediately implies that
Φ(φ) = C eimφ .
Because of the periodicity of the cylindrical coordinate φ, we have that Φ(φ) = Φ(φ + 2π) which
demands that m ∈ Z.
Let us now determine the function R. Using our work above, we have arrived at the following
differential equation for R.
ρ ∂R
ρ2 ∂ 2 R
+
− ρ2 λ2n − m2 ,
R ∂ρ2
R ∂ρ
µ
¶
∂ 2 R 1 ∂R
m2
2
=
+
−
λ
+
R.
n
∂ρ2
ρ ∂ρ
ρ2
0=
Using the clarity of hindsight from similar problems in Jackson, we will define the variable η ≡ λn ρ and
we see that the above equation becomes
µ
¶
m2
∂ 2 R 1 ∂R
+
− 1 + 2 R = 0.
∂η 2
η ∂η
η
PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 4
7
Because we have spent many hours roaming the bowels of mathworld, we notice that the above
expression is simply the modified Bessel differential equation whose solution is a modified Bessel function
of the first kind, Im (λn ρ). To be exceedingly pedantic, this is explicitly,
³
´2k+m
λn ρ
∞
X
2
R(ρ) ∝ Im (λn ρ) =
.
k!Γ (m + k + 1)
k=0
Therefore, returning to our original separated form of the potential ϕ, we see that
³ πnρ ´
³ πnz ´
X
Anm eimφ Im
sin
.
ϕ(ρ, φ, z) =
L
L
n,m
In a way that has become so canonical that it is of little worth mentioning, we can use the orthogonality
of the harmonic functions in the expression above to see that
Z 2π Z L
³ πnz ´
2
Anm =
ϕ(b, φ, z)eimφ sin
dzdφ.
πLIm (πnb/L) 0
L
0
‘
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óπ²ρ
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