Engineering and Applied Sciences Applied Physics 216 Lecture 9: Lecture 9
21 Feb 2017
School
Department
Professor
1
AP 216 Lecture 9 Electrostatic Problems with Spherical Symmetry 1
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We now demonstrate the use of the Ylm in electrostatics.
Potential Problems with Spherical Boundary Conditions
Separate variables in
2
2
222
11
ˆ0rL
rr r
2
8
9
(1)
Now separate variables
(, , ) () ( , )rfrY
(2) 10
11
12
solves the LE if
2
2
22
1ˆ
() (, ) 0
() ( , )
rd
rf r L Y
frdr Y
(3) 13
14
15
16
So each term must be constant. Because of the presence of L
ˆ2 we set the first term equal
to l(l+1) and the second to –l(l+1).
2
2
22
() ( 1) ()
ˆ(,) ( 1) (,)
ll
lm lm
d
r rfr ll fr
dr
LY l l Y
17
18
19
20
21
22
(4)
Where solutions are labeled by eigenvalues. Second equation is solved by Ylm. The
solution of the first depending on which l values is chosen in the second.
First equation is solved by rfl = rl+1 and rfl = 1 ¤ rl to yield
1
1
() , () l
ll
l
f
rfr
rr
(5) 23
24 So a general solution to LE is
1
0
0
1
(, , ) ( , )
l
l
l m lm lm
l
lml
l
in out
l m lm lm lm
lml
rArBY
r
AB
(6) 25
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27
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29
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where A’s and B’s are chosen to satisfy BCs.
Note: φlm = rlYlm in first term is a homogeneous polynomial in x,y,z with complex
coefficients
2
0
00
11
10
11
11
2
22 22
2,0
22
21
2222 2
22
1
4
33
cos
44
33
sin ( )
44
53cos 1 5
()(2
4216
15 15
sin cos ( )
88
15 15
sin ( )
32 32
i
i
i
rY
rY r z
tY re x iy
rY r z x y
rY re x iy z
rY re x iy
2
)
1
2
3
4
5
6
7
(7)
So rlYlm are complex lth order polynomials in x,y,z. Ylm are complex polynomials is sinθ ,
cosθ , sin
, cos
Note: φlmout is also determined by inversion theorem
121
11 1 1
(,,) (,,) (,) ( (,,))
out in in
lm lm lm lm
ll
rY
rr r r r
(8) 8
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10
11
12
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So the complex homogeneous polynomials play a fundamental role in both solutions.
Note: Different approach to using polynomials relative to Taylor expansion
Taylor expansion coefficients of f(θ ,
) depends on all partial derives at one point.
Fourier expansion coefficients depend on <Ylm|f>. which depends on f everywhere.
and picking orthogonal polynomials over the domain
1cos 1, 0 2
(9) 18
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Example: Solution to LE outside sphere of radius a on which φ (a,θ, φ) is specified.
Solution
1
0
1
(, , ) ( , )
l
lm lm
l
lml
rBY
r
(10) 23
24
25
with B’s found by using orthonormality of Ylm’s
3
21
1*
01
(cos) (,)(,,)
0
l
lm lm
lm
Badd Y a
A
(11) 1
2 Example: Inside solution
0
(, , ) ( , )
l
l
lm lm
lml
rArY
(12) 3
21
*
01
1(cos) (,)(,,)
0
lm lm
l
lm
AddY a
a
B
(13) 4
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11
12
21
0
Generating Function for Ylm
Can help in doing integrals and having other insights.
Consider a lth order polynomial polynomial (a
•x
)
l
that solves the LE. We will derive an
expression for this solutions in terms of the Ylm.
212
() () (1)()
ll l
ax laax ll a ax
(14) 13
14
2
So
2 222
()()
x
yz xyxy
aaaa aiaaia a
z
15
16
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19
20
(15)
Note: ax,ay,az must generally be complex. a
is called a null vector. There is no solution
other than zero for a
real.
Defining two more complex numbers
22
,
x
y
aia aia
xy
21
22
2
a
(16)
gives
222
z
a
(17) 23
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We will always have a solution to LE if we pick
z
a
27
28 So we will have a solution with the complex vector
2222
(, ,
22
ai
)
(18) 29
30
Thus two arbitrary generally complex parameters ,
determine all (a
• x
)
polynomials that satisfy the LE.
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Document Summary
Ap 216 lecture 9 electrostatic problems with spherical symmetry. We now demonstrate the use of the ylm in electrostatics. Now separate variables solves the le if r f r y r. Because of the presence of l to l(l+1) and the second to l(l+1). 2 we set the first term equal. L y l m rf r l l l. The solution of the first depending on which l values is chosen in the second. First equation is solved by rfl = rl+1 and rfl = 1 rl to yield. So a general solution to le is f r l. 30 where a"s and b"s are chosen to satisfy bcs. Note: lm = rlylm in first term is a homogeneous polynomial in x,y,z with complex coefficients. So rlylm are complex lth order polynomials in x,y,z. Ylm are complex polynomials is sin , cos , sin , cos . Note: lm out is also determined by inversion theorem r out lm.
