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Lecture 5 Notes, 95.657 Electromagnetic Theory I, Fall 2011
Dr. Christopher S. Baird, UMass Lowell
1. The Laplace Equation in Spherical Coordinates
 In this coordinate system,
r
is the radial distance from the origin to the observation point,
θ
is the
polar angle that the point makes with the
z
axis, and
ϕ
is the azimuthal angle in the
x

y
plane
relative to the
x
axis.
 Spherical coordinates are useful when the boundary conditions have a spherical shape or
symmetry.
 The Laplace equation in spherical coordinates:
∇
2
=
0
1
r
∂
2
∂
r
2
r
1
r
2
sin
∂
∂
sin
∂
∂
1
r
2
sin
2
∂
2
∂
2
=
0
 Use the method of separation of variables by trying a solution of the form:
r ,
,
=
R
r
r
P
Q
Here an extra factor (1/
r
) is included to anticipate that the mathematics will be simplified if each
factor has the same dimensionality.
 Substitute this into the Laplace equation:
1
r
∂
2
∂
r
2
r
[
R
r
r
P
Q
]
1
r
2
sin
∂
∂
sin
∂
∂
[
R
r
r
P
Q
]
1
r
2
sin
2
∂
2
∂
2
[
R
r
r
P
Q
]
=
0
P
Q
1
r
∂
2
R
r
∂
r
2
R
r
r
Q
1
r
2
sin
∂
∂
sin
∂
P
∂
R
r
r
P
1
r
2
sin
2
∂
2
Q
∂
2
=
0
 This equation is complex enough that we can not make each term independent all at once. First,
get
Q
in a form to show it is independent by multiplying by
r
3
sin
2
/
R
r
P
Q
:
r
2
sin
2
R
r
d
2
R
r
d r
2
sin
P
d
d
sin
d P
d
1
Q
d
2
Q
d
2
=
0
 Here the partial derivatives have become total derivatives because the functions they operate on
are now functions of only one variable.
 The last term is now independent of
ρ
and
θ
, and must hold for all
ρ
and
θ
, so that it must equal a
constant:
r
2
sin
2
R
r
d
2
R
r
d r
2
sin
P
d
d
sin
d P
d
−
m
2
=
0
and
−
m
2
=
1
Q
d
2
Q
d
2
 We can solve the second equation. First put it in a more intuitive form:
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View Full Document d
2
Q
d
2
=−
m
2
Q
 Now the general solution is clearly:
Q
(ϕ)=
A
m
e
i m
ϕ
+
B
m
e
−
im
ϕ
if
m
≠ 0 and
Q
(ϕ)=
A
m
=
0
+
B
m
=
0
ϕ
if
m
= 0.
 The constant
m
is in general not necessarily an integer. If the region of interest includes the full
azimuthal sweep of values, then
m
must be an integer to keep the solution singlevalued and the
case of
m
= 0 reduces to
Q
(
φ
) =
A
m
=0
. From here on, we are dealing with this special case, which is
still quite general.
 We now turn to the rest of the equation:
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This note was uploaded on 02/13/2012 for the course PHYSICS 95.657 taught by Professor Staff during the Fall '11 term at UMass Lowell.
 Fall '11
 Staff
 Mass

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