Electromagnetic Theory I
Problem Set 7
Due: 10 October 2011
25.
J, Problem 3.23.
Note that you are actually obtaining various representations of the
Dirichlet Green function (just set
q
=
ǫ
0
) for the interior of a cylinder when you do this
problem.
In particular, the form in part c) is an expansion of the Green function in a
complete set of functions in all three coordinates as we discussed in class.
Solution
: The potentials of this problem are just
q/ǫ
0
, (
q/
4
πǫ
0
) times my (Jackson’s) Green
functions for this geometry. The first two expansions a) and b) are adaptations, to a finite
range in
ρ
: 0
< ρ < a
, of the Green functions for the region between two parallel planes,
that we obtained in problems 22a) and 23c) in Problem Set 6 respectively. The easiest one
to obtain is the second form which differs from that in 23c) by the following extra term in
the radial factor
−
1
πL
I
m
(
nπρ
<
/L
)
I
m
(
nπρ
>
/L
)
K
m
(
nπa/L
)
I
m
(
nπa/L
)
=
−
1
πL
I
m
(
nπρ/L
)
I
m
(
nπρ
′
/L
)
K
m
(
nπa/L
)
I
m
(
nπa/L
)
.
(1)
The equality follows because the left side is symmetric in
ρ
<
↔
ρ
>
.
This term has no
singularity at
ρ
=
ρ
′
and satisfies Laplace’s equation throughout the region. It’s role is to
make the potential vanish at
ρ
>
=
a
. Since the new term has no singularity at
ρ
=
ρ
′
, it
doesn’t contribute to the delta function. The first (discontinuous) term is exactly the same
as in problem 23c) so it produces the proper delta function.
Thus the correctness of the
second expansion is proved.
The first expansion differs from that in 22a) by the replacement of the integral over
continuous
k
by a sum over the discrete values
k
mn
=
x
mn
/a
, so that the Bessel functions
in the expansion vanish at
ρ, ρ
′
=
a
. We must only check that the new (constant) factors
are such as to produce the proper delta function. From the orthogonality condition, Jackson
(3.95). it follows that the completeness condition for discrete
k
is
2
a
2
∞
summationdisplay
n
=1
J
m
(
x
mn
ρ/a
)
J
m
(
x
mn
ρ
′
/a
)
J
2
m
+1
(
x
mn
)
=
1
ρ
δ
(
ρ
−
ρ
′
)
(2)
Comparing this to the continuous
k
completeness relation,
1
ρ
δ
(
ρ
−
ρ
′
)
=
integraldisplay
∞
0
kdkJ

m

(
kρ
)
J

m

(
kρ
′
)
(3)
we see that the extra constant factors 2
/
(
a
2
k
mn
J
2
m
+1
(
x
mn
)) = 2
/
(
ax
mn
J
2
m
+1
(
x
mn
)) in the
first expansion, compared to those in 22a), are precisely what is needed to give the correct
normalization of the delta function.
1
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Finally the third expansion involves a triple sum of eigenfunctions of
−∇
2
with eigenval
ues (
x
mn
/a
)
2
+ (
kπ/L
)
2
. Each term satisfies the boundary conditions in
ρ, z
and periodicity
in
ϕ
.
Applying
−∇
2
to the expansion leaves the representation for the product of delta
functions
1
ρ
δ
(
ρ
−
ρ
′
)
δ
(
ϕ
−
ϕ
′
)
δ
(
z
−
z
′
)
already discussed. To obtain either of the first two expansions from this triple sum, either
the sum over
k
or the sum over
n
must be explicitly performed. Since all forms have been
proved, we can infer the result of that summation by comparison.
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 Spring '08
 Staff
 real axis, Upper Half, Dirichlet Green function

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