Classical Electrodynamics - National Chiao Tung...
Transcript of Classical Electrodynamics - National Chiao Tung...
Classical Electrodynamics
Chapter 3Boundary-Value Problems in
Electrostatics: II
A First Look at Quantum Physics
2011 Classical Electrodynamics Prof. Y. F. Chen
A First Look at Quantum Physics
2011 Classical ElectrodynamicsProf. Y. F. Chen
Contents§3.1 ▽2 in Different Coordinates§3.2 Laplace Equation in Spherical Coordinates§3.3 Legendre Function§3.4 Ex. Two Hemispheres at Equal and Opposite Potentials, Find the
Potential Outside the Sphere§3.5 Associated Legendre Function §3.6 Spherical Harmonics Function§3.7 Hypergeometric Function§3.8 ▽2Φ in cylindrical coordinates§3.9 Solution of Bessel equation§3.10 General Solution of ▽2Φ=0 in Cylindrical System§3.12. Fourier Analysis of Periodic Function§3.13 Fourier-Bessel series
§3.15 Green Function in Spherical System §3.14 Boundary value problems in cylindrical coordinates
§3. 1 ▽2 in Different Coordinates
Cartesian coordinates
General coordinates
2 2 22
2 2 2x y z
, ,x y z
1 2 3, ,q q q
(1) Gradient
1 1 2 2 3 3
1 1 2 2 3 3
1 1 1dl h dq h dq h dq
h q h q h q
1 2 31 2 3
d dq dq dq dlq q q
d dl
1
1q 2q 3q
1q 2q 3q
2011 Classical Electrodynamics Prof. Y. F. Chen
2 2 3 1 3 1 2
1 2 3 1 1 1 2 2 2 3 3 3
1 h h h h h hh h h q h q q h q q h q
(2) Gauss’ theorem → flux of field
1 2 3 2 3
2 1 3 1 3
3 1 2 1 2
ds h h dq dqds h h dq dqds h h dq dq
1 1 1 1 1 2 3 2 3 2 3, ,E ds E q dq q q h h dq dq
flux a
1 1 1 1 2 3 2 3 2 3, ,E ds E q q q h h dq dq
flux b
2 3 11 2 3 1 2 3
1 2 3 1
1
d
h h E h h h dq dq dqh h h q
2 3 1 1 3 1 1 2 11
1 2 3 1 2 3
1 h h E h h E h h Etotal flux d E dh h h q q q
2(3) in general coordinates
2
From eq. (1) and (2): 1E
1E
1 1h dq2 2h dq
3 3h dqE1
flux aflux b
1 2 3, ,q q q
2011 Classical Electrodynamics Prof. Y. F. Chen
2
2 22 2
1 1sin sinsin sin
rr r r
2
2
22
2 2 2 2
1
1 1 1 1 0sinsin sin
rr r
rr r r r
, ,let r R r P Q
2 2
22 2 2 2 2
1 1 1 0sinsin sin
P QrR PQ RQ RPr r r r
2 22
2 2 2 2 2
2 2
2 2 2 2
1 1 1 1 1 1
1 1 1 1 1 1 1
0
sinsin sin
sinsin sin
P QrRR r r P r Q r
P QrRR r r r P Q
divide by R r P Q
§3. 2 Laplace Equation in Spherical Coordinates
2011 Classical Electrodynamics Prof. Y. F. Chen
22
2
2
2 2
1 0 1 2
1 1 1 1 1 0 1 2
; , , ,
sin ; , , ,sin sin
Q m mQ
letP Q l l l
P Q
(1) 2
22
imd Q m Q Q ed
(2) 2
2
1 1sinsin sin
d d m P l l Pd d
cos , sinlet x dx d
22
21 1 01
:ml
d dP mx l l Pdx dx x
where P x P x associated Legendre functions
0if m
21 1 0
:l
d dPx l l Pdx dxwhere P x P x Legendre functions
3
2011 Classical Electrodynamics Prof. Y. F. Chen
(1) Generating function for Legendre function: from point charge
˙
˙
˙
q
Px nr
x nr
2 2
1 1 1n n 2
,? cosr r r r rr
x xx - x -
2
2
1
1 2
1
1 2
:cos
:cos
if r rr rr r r
if r rr rr r r
2 2
1 1 11 2
1 2
,
cos
tt txr rr r r
Generating function: 2
0
11 2
, ll
lG x t P x t
t tx
§3. 3 Legendre Function
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) Recursion relation(a) Differentiate about ,G x t
0
12 2
0
11 2 1 2
,
ll
l
ll
l
P x t
dG x t x tlP x t
dt t tx t tx
2 1
0 0
1 1 1
1 2
1 2 1
l ll l
l l
l l l l l
x t P x t t tx lP x t
xP P l P lxP l P
1 12 1 1l l ll xP l P lP 4
0
2 20
11 2 1 2
,
ll
l
ll
l
P x t
dG x t t P x tdx t tx t tx
2
0 0
1 2 1
0 0 0 0
1 2
2
l ll l
l l
l l l ll l l l
l l l l
t P x t t tx P x t
P x t P x t P x t x P x t
1 2 12l l l lP P P xP 5
t
Differentiate about ,G x t x
2011 Classical Electrodynamics Prof. Y. F. Chen
(b) Recursion formula 1:
Differentiate eq. (4) by : 1 12 1 2 1 1l l l ll P l xP l P lP
multiply eq. (5) by : l 1 12l l l llP lxP lP lP
6
7
16 7 1 l l ll P xP P
(c) Recursion formula 2:
multiply eq. (5) by : 1l 9
16 9 l l llP xP P
8
1 11 2 1 1 1l l l ll P x l P l P l P
10
(d) Recursion formula 3:
1 18 10 2 1 l l ll P P P
x
11
2011 Classical Electrodynamics Prof. Y. F. Chen
(3) Prove that satisfy lP x
1l l in eq. (8) : 1 1l l llP xP P
multiply eq. (10) by : x 21l l lxlP x P xP
12
13
2112 13 1 0l l lx P xlP lP
14
Differentiate eq. (14) about : 211 0l l l lx P lP xlP lP
15
Use eq. (10)
21 1 0l lx P l l P
21 1 0d dPx l l Pdx dx
x
2011 Classical Electrodynamics Prof. Y. F. Chen
(4) Orthogonality condition: 1
1
02
2 1l m
l mP x P x dx
l ml
(a) if l m
multiply eq. (2) by and integrate: mP x
1 12 2
1 11 1 1 1 0m l l l m l mP x P l l P dx x P P l l PP dx
16
l m in eq. (16)
1 2
11 1 0l m l mx P P m m PP dx
17
1
116 17 1 1 0l ml l m m P x P x dx
1
10l mP x P x dx if l m
2011 Classical Electrodynamics Prof. Y. F. Chen
(b) if l m
1 12 1 1l l ll xP l P lP
1 22 1 1l l ll xP lP l P
eq. (4):
1l l in eq. (4): 18
118 2 1 ll P
21 1 1 12 1 2 1 2 1 1 2 1l l l l ll l xPP l l P P l l P
1 1 1 2
1 1 1 11 1 1
0
2 1 2 1 2 1 1 2 1l l l l ll l xPP dx l l P P dx l l P dx
1 1 2
1 11 12 1 2 1 2 1l l ll l xPP dx l l P dx
19
21 22 1 2 1 2 1 2 1 1l l l l ll l xPP l l P l l PP
1 1 2
11 12 1 2 1 2 1l l ll l xPP dx l l P dx
20
1 12 2
11 1
2 119 202 1l llP dx P dxl
1 1 12 2 2
2 01 1 1
2 1 2 3 2 1 2 3 22 1 2 1 2 1 2 1 2 1l ll l l lP dx P dx P dxl l l l l
118 2 1 ll P
2011 Classical Electrodynamics Prof. Y. F. Chen
(5) Evaluate the integral 1
0 lP x dx
1 1
1 10 0
12 1l l lP x dx P x P x dxl
1 1 1 11 1 1 0 0
2 1 l l l lP P P Pl
21
1 1
0
0ll
PP x dx
l
1 1
1 114 0 0.
l
l l
Plfrom eq P P
l
Substitute into eq. (21)
(a)
(b) Evaluate 1 0lP
Generating function at :0x
2
0
10 01
, ll
lG t P t
t
2
21
1 1 3 2 1 112 2 21 !
n
n
n tnt
2 222
1 1
1 2 1 1 21 1
2 2
!! !! !
n nn n
n nn n
n nt t
n n
22 1
2!
!!!n
nn
n
23
24
22
2011 Classical Electrodynamics Prof. Y. F. Chen
Compare eq. (23) and eq. (24)
2
2
0
0 1
2 2
!
!
l
l
l
if l odd
P lif l even
l
(c) From eq. (25), we known that must be odd in eq. (22)
25
l
Let 2 1l n
121 1
20 1
0 1 11
12 2
!
!
l
ll
l
P lP x dx
l l l
22 2
1 2 21 0 1 22 1 2 1
!, , ,
!
n
n
nn
n n
26
2011 Classical Electrodynamics Prof. Y. F. Chen
(6) Legendre series representation:
0
1
1
2 12
l ll
l l
f x A P x
lwhere A f x P x dx
Ex.
0
1
1
11
1 01 0
for xf x
for x
0
l ll
f x A P x
1 0 1
1 1 0
2 1 2 12 2l l l l
l lA f x P x dx P x dx P x dx
1
10
12
21
22 2
2 12 1 0
2 1 1 1
12 2
4 3 1 2 20 1 2
2 1 2 1
!
!
!, , ,
!
l l
l
l
n
n
ll P x dx P
l
l ll l
n nn
n n
2011 Classical Electrodynamics Prof. Y. F. Chen
2 222 2
4 3 1 2 22 1 2 1
!
!
nn
l n
n nB a V
n n
V
V
a 0
2
2
,V
aV
10
1, cosl lll
r B Pr
General solution:
At :r a 10
1, cosl lll
a B Pa
Multiply in both side of eq. (27) and integrate
27
coslP
22 2
1
1 0 1
0 1 1
1 0 0
264 3 1 2 22 1 2 1
2 1 2 12 2
2 1 2 12
.!
!
, cos sin ,
n
n
ll ll
l l l
from eqn n
Vn n
B l la P d a P x dxa
l V P x dx P x dx l V P x dx
2 2 2
2 1 121 2 20 0
4 3 1 2 21 32 1 22 1
!, cos cos cos
!
n n
l l nl nl l
n n a ar B P V P V Pr n r rn
(1) Method I:
§3.4 Ex. Two Hemispheres at Equal and Opposite Potentials, Find the Potential Outside the Sphere
2011 Classical Electrodynamics Prof. Y. F. Chen
1
0
1
0
1 1 2 14 4
1 2 14
, , , , cos , sin
cos , sin
lD
lS Sl
l
lSl
G ar r da l P a d dr r
al P a d dr
(2) Method II : use Green’s function
2 2
2
1 1 1n n
n n,
??
DaG r r
r r r ra r ar r r rr a r
2 22 2
12
0 0
1 1
1 2 1 2
1 1
: ,
cos cos
cos cos
D
ll
l ll l
for r a G r rr r rr a arr r a rr rr
r aP Pr r a rr
1 21 2
10 0
1 1
1 10 0
1 11n
1
cos cos
cos cos
l llD D
l lll lr a
l l
l ll ll l
G G a al P l Pr r a r a
a al P l Pr r
1
10
2 1 cosl
lll
al Pr
(a)
282011 Classical Electrodynamics Prof. Y. F. Chen
cos cos cos sin sin cos
Set 0cos
cos cos cos
cos cos cos
cosl l
l ll
P P
A P
Set in eq.(28), we get 0 0 1l
l lA
Set in eq.(28), we get0
29
cos cos cosl l lP A P
1cosl l l lP A P A
cos cos cos cos cosl l l lP P P P
31
(b) Evaluate coslP
30
Finally, from eq. (29) and eq. (30)
32
2011 Classical Electrodynamics Prof. Y. F. Chen
(c) Substitute eq. (32) into eq. (28)
1
0
12
0 00
1
0
1 2 14
1 2 14
1 2 12
, , cos , sin
cos cos , sin
cos cos ,
l
lSl
l
l ll
l
l ll
ar l P a d dr
al P P a d dr
al P P ar
0sin d
2011 Classical Electrodynamics Prof. Y. F. Chen
22
21 1 01
:ml
dP xd mx l l P xdx dx x
where P x P x associated Legendre functions
(1) Prove that 2 21m m
ml lm
dP x x P xdx
2 21m
mlP x x y x
12 22 21 12
m mm
ld mP x x y x x y xdx
2 2 12 2 2 22 2 22 2 1 2 1 1
m m mm
ld P x m m x x y x mx x y x x y xdx
34
33
21 34 2 33 1 mlx x l l P x
21 2 1 1 1 0x y m x y m m y l l y 35
(a) Let
§3. 5 Associated Legendre Function
2011 Classical Electrodynamics Prof. Y. F. Chen
(c) Compare eq. (35) and eq. (36)
m
m
ldy x P xdx
2 2
2 2
1
1
mm
l
m m
lm
P x x y x
dx P xdx
(b) Differentiate eq. (35) by m times
0
m k m kmmkm k m k
k
d d duse f g C f gdx dx dx
2 1 1
2 1 121 2 1 2 2 1 0m m m m m m
m m m m m m
l l l l l ld d d d d dx P mx P m m P x P m P l l Pdx dx dx dx dx dx
2 1
2 121 2 1 1 1 0m m m m
m m m m
l l l ld d d dx P m x P m m P l l Pdx dx dx dx
36
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) Other properties
2 2 22 2
1
1
11 1 12
1
22 1
!!!
!!
m mm l m lml lm l l m
mm ml l
m ml l ll
d dP x x P x x xdx l dx
l mP x P x
l m
l mP x P x dx
l l m
2011 Classical Electrodynamics Prof. Y. F. Chen
2 1
4!
, cos!
m imlm l
l mlY P el m
(1)
2 1
0 1
1, , ,
, , cos
ml m lm
lm l m ll mm
Y Y
Y Y d d
(2) Addition theorem
˙
˙
˙P
P
x
y
z
42 1
cos , ,l
l lm lmm l
P Y Yl
if z-axis is along : n , ,l
lm m lmm l
Y B Y
Let 0 0: and m
0 0 02 104
, ,lm llY B Y B
0 0, ,lm lB Y Y d 37
38
《Prove》
§3. 6 Spherical Harmonics Function
2011 Classical Electrodynamics Prof. Y. F. Chen
Similarly, as z-axis is along , can be expressed asn coslP
04
2 1cos , ,
l
l l lm lmm l
P Y b Yl
04
2 1, cos , ,lm lm l lm lb Y P d Y Y d
l 39
From eq. (37), eq. (38), and eq. (39)
04 4
2 1 2 14
2 1
,
cos , , ,
lm lm
l l
l lm lm lm lmm l m l
b B Yl l
P b Y Y Yl
2011 Classical Electrodynamics Prof. Y. F. Chen
§3. 7 Hypergeometric Function
2
21 1 0d dz z u z z u z u zdz dz
General solution:
0
01 2 1 1
, , ;!
,
, , ; :
kk k
k k
k
u z F z zk
where k
F z Hypergeometric Function
Hypergeometric equation:
(1) Solve hypergeometric equation
Let 0
s kk
ku z z C z
2 1 0z u zu u zu u z
1
0 01 1 1 0s k s k
k kk k
s k s k s k C z s k s k s k C z
1
0 01 0s k s k
k kk k
s k s k C z s k s k s k C z
10 1
0 11 1 1 0s k s s k
k kk k
s k s k C z s s s C z s k s k s k C z
01 0
1s or
s s ss
40
2011 Classical Electrodynamics Prof. Y. F. Chen
(a) As 0s
11 1k kk k C k k k C
11 1 1k kk k C k k C 1k k
1
1 11k k
k kC C
k k
Choose 0 1C
0 1 2
1 2 1
, , ,!
k kk
k
k
C kk
where k
10
, , ;!
, , ; :
kk k
k k
u z z F zk
F z Hypergeometric Function
41
2011 Classical Electrodynamics Prof. Y. F. Chen
(b) As 1s
1
1 1 1 11 2k k
k kC C
k k
Choose
1 10 1 2
2
1 2 1
, , ,!
k kk
k
k
C kk
where k
2
0
1 11 1 2
2, , ;
!kk k
k k
u z z F zk
From eq. (41)
11 1 1 1 1 2k kk k C k k C
0 1C
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) Relation between hypergeometric equation and
Legendre equation
2
221 2 1 0d dx P x x P x v v P x
dx dx Legendre equation:
2
111 2
2 114
xzxz
xz z
2
21 1 2 1 0d dz z P z z P z v v P zdz dz
43
Compare eq. (42) and eq. (43)
11
vv
2
21 1 0d dz z u z z u z u zdz dz
Hypergeometric equation: 42
2
0
11 1 1112 2 2
, , ;!
kk k
vk
v vx x xP z F v vk
Let in Legendre equation
2011 Classical Electrodynamics Prof. Y. F. Chen
2(1 ) " 2 ' ( 1) 0: (cos ) 0
x P xP Pwith boundary condition P
Legendre's differential equation :
1 20
( ) ( 1)1 1( ) ( , 1,1; )2 ! ! 2
kk k
k
x xP x Fk k
0
( ) ( 1) 1 cos(cos ) cos 0! ! 2
kk k
k
Pk k
P
z
(3) The fields in a Conical hole or near a sharp point:
→ Azimuthal symmetry Problems
General solution:
The power series form has the problem of divergence
2011 Classical Electrodynamics Prof. Y. F. Chen
1222
0
20
221 0
0 2
2210
0 2
2
2 sin , 0,1,!
( ) ( 1) 1 cos(cos )( !) 2
( ) ( 1)2 1 cos sin! 2
( ) ( 1)2 1 cos sin! 2
2 1 1 cos, 1, ; sin2 2
k k
kk k
k
kkk k
k kk
k k
k k
d kk
Pk
dk
dk
F
20
2 1 2 11 ( 1 ) ( 1 ), 1, ; ( )2 2 1
1( ) (1 ) ( 2 ) ( 1) 02
d
w w w wF w f ww
f w satisfy w w f w f f
(a) Another useful forms of Legendre:
2011 Classical Electrodynamics Prof. Y. F. Chen
2 2 2
2
2 1 2 1 12 2 2 2 2
2 2
1 12 2
2
sin sin sin ,2 2
1 cos sin sin sin sin2 2 2
1 1 sin cos2 2
(cos sin ) (cos sin ) cos[( ) ]1, 1, ;2 2cos cos
cos[( ) ] co2(cos )cos
let
w i
w
i iF w
P
22
02
12
12 22
02 2
122
2 202 2
122
0
ssin cos
cos[( ) ] sin2 12sin sin
cos[( ) ]22 sin sin
cos[( ) ]22cos 2cos
d
d
d
d
Integral form of Legendre
2011 Classical Electrodynamics
(b) Asymptotic form of Legendre:
12 1(cos ) sin (1 ( )) 0sin 2 4
1 1 3 1, 1,2 . . 1,2 4 4 2
P O
n n e g n
(c) Electric potential and electric field :
1
1
1
( , ) (cos )
0, ( , ) (cos )
(cos )
1 sin (cos )
k
kkk
r
r A r P
at r r Ar P
E Ar Pr
E Ar Pr
2011 Classical Electrodynamics Prof. Y. F. Chen
(cos ) versusP
Difference between the three forms of Legendre
2011 Classical Electrodynamics Prof. Y. F. Chen
§3. 8 ▽2Φ in cylindrical coordinates
2 22
2 2 2
1 1 0z
, ,let z R Q Z z
2 22
2 2 2
1 1 0R QZ Q RZ Z RQz
2 22
2 2 2
1 1 1 1 1 0R Q ZR Q Z z
Divide by R Q Z z
22
2
22
2
1 0 1 2
1
; , , ,Q m mQ
letZ k
Z z
2011 Classical Electrodynamics Prof. Y. F. Chen
(1) 2
22
imd Q m Q Q ed
(3)2
22
1 1 0mR kR
2 22
2 2
1 0d R dR mk Rd d
let x k
2 2 2
2 2 22 2 2
1 1 0 0d R dR m d R dRR x x x m R Bessel equationdx x dx x dx dx
2
22
kzd Z k Z Z z edz
(2)
2011 Classical Electrodynamics Prof. Y. F. Chen
§3. 9 Solution of Bessel equation
2
2 2 22 0d y dyx x x v y
dx dx
Let 0
s kk
ky x x C x
2 2
0 01 0s k s k
k kk k
s k s k s k v C z C z
2 2 10 1
22
2 2
1 1 1
1 0
s s
s k s kk k
k k
s s s v C z s s s v C z
s k s k s k v C z C z
21 0:sz term s s s v s v
(1) Series solution
2011 Classical Electrodynamics Prof. Y. F. Chen
(a) As s v
2k K
Choose
01 1 12 1
1 1 1
:
, !v
k
x Gamma function
C where n n n nv
n k n n
2 2
112 1!
KK K vC
K v k
2
10
11 2!
:
K v K
vK
v
xy x J xK v K
J x Bessel Function
221 k kv k v k v k v C C
21
2k kC Ck k v
2 2 2 2 22
02 2 2
02
1 12 2 2 2
1 1 12 2 1 1 2 1
112 1 2 1!
K K K
KK
C C CK K v K K v
CK K v K K v v
CK K v K v v v
Prof. Y. F. Chen
(b) As s v
2
20
11 2!
K v K
vK
xy x J xK v K
Similarly
(2) Properties of solution
if v m integer
1 1
1
2 2
0 0 1 1
1 11
2 2! ! ! !
K Km K m Km
mK K
x xJ xK K m K K m
1 mmJ x
1let K m K
m mJ x and J x are linear dependent
mWe must find another solution which is linear independent to J x
Define Neumann function:
cossin
m mm
m J x J xN x
m
2
10
2
20
11 2
11 2
!
!
K v K
vK
K v K
vK
xy x J xK v K
xy x J xK v K
Solution of Bessel equaiton
(a)
2011 Classical Electrodynamics Prof. Y. F. Chen
(b) Asymptotic form of Bessel and Neumann function
22 4
22 4
cos
sin
m
m
nJ x xx
As xnN x x
x
2011 Classical Electrodynamics Prof. Y. F. Chen
Two linear independent solution
(3) Summary in solution of Bessel equation
2
10
2
20
11 2
11 2
!
!
K v K
vK
K v K
vK
xy x J xK v K
xy x J xK v K
(a) if v integer
Two linear independent solution
2
10
2
12! !
cossin
K m K
mK
m mm
xy x J xK m K
m J x J xy x N x
m
(b) if v integer m
Asymptotic form
22 4
22 4
cos
sin
m
m
nJ x xx
nN x xx
2011 Classical Electrodynamics Prof. Y. F. Chen
(c) Figure of Bessel and Neumann function
2011 Classical Electrodynamics Prof. Y. F. Chen
§3. 10 General Solution of in Cylindrical System2 0
˙
x
y
z
a
L
(1) Boundary condition:
2 22
2 2 2
1 1 0z
2 22
2 2 2
1 1 1 1 1 0R Q ZR Q Z z
, , z R Q Z z
0
02
, ,
, ,
a z
Lz
22
2
22
2
2 22
2 2
1 0
;
m
d Q m Qd
d Z k Zdz
d R dR mk Rd d
R J k
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) Boundary condition:
0
02
, ,
, ,
a z
Lz
22
2
22
2
2 22
2 2
1 0
;
m m
m
d Q m Qd
d Z k Zdz
d R dR mk Rd d
R J ik I k
where I x is modified Bessel function
2 2
0 0
12 2! ! ! !
K m K m Km
mK K
ix i xJ ixK m K K m K
Modified Bessel function
2
0
12! !
m Km
m mK
xI x i J ixK m K
mI x
2011 Classical Electrodynamics Prof. Y. F. Chen
2
0
11 2!
K m K
mK
xJ xK m K
2
0
12! !
K m K
K
xK K m
(1) Recursion relation
44
22
0
22
0
1 12
1 12
! !
! !
K m Km Km
mK
K m Km K
mK
x J x xK K m
x J x xK K m
2 2 12 1
0 0
1
2 2 12 1
0 0
1 2 112 1 2
1 2 112 1 2
! ! ! !
! ! ! !
K Km K m Kmm Km
mK K
mm
K Km K m Kmm K
mK K
m K xd xx J x xdx K K m K K m
x J x
K xd xx J x xdx K K m K K m
1 2 1
10
11 2! !
K m Kmm
mK
x x x J xK K m
45
46
§3. 11 Bessel Function
2011 Classical Electrodynamics Prof. Y. F. Chen
(a) Recursion formula 1:
From eq. (45):
11
m m m mm m m m
d x J x mx J x x J x x J xdx
1m m mm J x J x J xx
(b) Recursion formula 2: From eq. (46):
11
m m m mm m m m
d x J x mx J x x J x x J xdx
1m m mm J x J x J xx 48
47
(c) Recursion formula 3:
1 1247 48 m m mm J x J x J xx
(d) Recursion formula 4: 1 147 48 2 m m mJ x J x J x
49
50
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) Orthogonality condition:
220
1
0
2
a
m mn m mnm mn
n nJ k J k d a J k a n n
2
22
1 0m mn mn m mnd d mJ k k J k
d d
Bessel equation:
multiply Bessel equation by and integrate:
n n in eq. (51)
52
m mnJ k
2
220
0a
m mn m mn mn m mn m mnd d mJ k J k k J k J k d
d d
00
22
20
0
22
20
0
0
aa
m mn m mn mn m mn m mn
a
mn m mn m mn
a
m mn mn m mn mn mn m mn m mn
a
mn m mn m mn
d dJ k J k k J k J k dd d
mk J k J k d
J k a ak J k a k k J k J k d
mk J k J k d
0
22
200
a
m mn mn m mn mn mn m mn m mn
a
mn m mn m mn
J k a ak J k a k k J k J k d
mk J k J k d
51
integrate by part
Prof. Y. F. Chen
2 2
00
a
mn m mn m mn mn m mn m mn mn mn m mn m mnk aJ k a J k a k aJ k a J k a k k J k J k d
(a) if n n
2 2
00
0a
mn m mn m mn mn m mn m mn mn mn m mn m mnk aJ k a J k a k aJ k a J k a k k J k J k d
0
0a
m mn m mnJ k J k d (b) if n n
2 20
a mn m mn m mn mn m mn m mnm mn m mn
mn mn
k aJ k a J k a k aJ k a J k aJ k J k d
k k
0
2 2
2 2
2
2
2
lim
lim
mn mn
mn mn
mn mn
a
m mn m mnk k
mn m mn m mn mn m mn m mn
k kmn mn
mn m mn m mn m mn m mn mn m mn m mn k k
mn
m mn m mn
J k J k d
k aJ k a J k a k aJ k a J k ak k
k a J k a J k a aJ k a J k a k a J k a J k a
k
a J k a J k a
0
53
51 52
0
2011 Classical Electrodynamics Prof. Y. F. Chen
1 12 0m mn m mn m mnmn
m J k a J k a J k ak a
1 1 12 2m mn m mn m mn m mnJ k a J k ax J k a J k a
eq. (49):
eq. (50): 54
Substitute eq. (54) into eq. (53)
2
210 2
a
m mn m mn m mnaJ k J k d J k a
2011 Classical Electrodynamics Prof. Y. F. Chen
(3) Generating function of Bessel function:
2
0
12
,! !
K m Km m
mm m K
xG x t J x t tK K m
Let 0:m K K K
0 0 0 0
0 0
22
1 12 2 2 2
1 12 2
,! ! ! !
! !
K K K K K
K K K K
K K
K K
xxtt
xt x xt xG x tK K t K K t
xt xK K t
e e
1
2,x t
mtm
mG x t e J x t
55
2011 Classical Electrodynamics Prof. Y. F. Chen
(4) Integral representation of Bessel function
1
2,x t
mtm
mG x t e J x t
Let i
x kt e
2 sini ik e e ik im
mm
e e J k e
2 2 2
0 0 0
2
sin
mm
ik im im im im imm m
m me e d J k e e d J k e e d
Multiply in both side of eq. (56) and integrateime
56
2
0
12
sinik immJ k e e d
2011 Classical Electrodynamics Prof. Y. F. Chen
(5) Expansion of a plane wave in terms of Bessel function
˙
,x yk k k
,r x y
x
y
cosx yk r k x k y k
plane wave: cosx yi k x k y ikik re e e
Let and substitute it into i
x k
t ie
2 cos,
i ik ie ie mik imm
mG x t e e J k i e
,G x t
cos mik imik rm
me e J k i e
2011 Classical Electrodynamics Prof. Y. F. Chen
(6) Completeness relation of cylindrical wave:
21
2x y x yi k x k y i k x k y
x yx x y y e e dk dk
2
0
12
m imm m
m
x x y y
J k J k i e kdk
22 cos cos
0 0
22
0 0
22
0 0
2
12
12
12
mm
ik ik
m mim imm m
m m
m m i m m i m mm m
m m
e e kdkd
J k i e J k i e kdkd
J k J k i e kdk e d
2
0
12
m imm m
mJ k J k i e kdk
0 0
12
im
m
m m m m
e
k kJ k J k kdk J k J k d
k
Prof. Y. F. Chen
§3. 12 Fourier Analysis of Periodic Function(1) periodic function:
2 ni xL
nn
f x f x L f x F e
57
22
2 2
2 2
n nnL Li x i xL L
n nL Ln
f x e dx F e dx F L
2
22
2
1
ni xL
nn
nL i xL
n L
f x F e
F f x e dxL
(2) Helmholtz equation
with periodic B. C.
2
22 0d k x
dx
x x L
General solution: 21 ni x
L
nx e
L
(3) 21 ni x xL
nx x e
L
12
in
ne
2;x L
2011 Classical Electrodynamics Prof. Y. F. Chen
(4) Fourier integral
2 2 2 2 2
2 2
2 2
1 2 12
n n n n nL Li x i x i x i x i xL L L L L
n L Ln n n
f x F e f x e dx e f x e dx eL L
As 2
2n
dkLL
n kL
12
n n
n
ik x ik x ikx
n
F k
f x dk f x e dx e F k e dk
12
ikx
ikx
f x F k e dk
F k f x e dx
2011 Classical Electrodynamics Prof. Y. F. Chen
(5) Discrete Fourier Transform (DFT)
2
1 0 1 2 1, , , ,i nN NZ Z e n N
2 11 1 1 0N NZ Z Z Z Z
0
22 1
1
1 0 1 2 1, , ,i sN N
s
Z e
Z Z Z root is Z e s N
58
Let from eq. (58) s n n
21
0
1 0,N i m n n
Nnn
me n n N
N
01
2
3
45
67
8
9
1N
2 2 21 1 1 1 1
0 0 0 0 0
1 1
m
N N N N Ni m n n i mn i mnN N N
n n nn n nn n m m n
F
f f f e f e eN N
21
0
21
0
1
N i mnN
n mm
N i mnN
m nn
f F e
F f eN
59
unit circle
2011 Classical Electrodynamics Prof. Y. F. Chen
21
0
21
0
1
N i mnN
n mm
N i mnN
m nn
f F e
F f eN
Discrete Fourier Transform (DFT) Fourier Series
2
22
2
1
ni xL
nn
nL i xL
n L
f x F e
F f x e dxL
n
n n
LdxN
x nL Nf x f
2011 Classical Electrodynamics Prof. Y. F. Chen
22
2
1 0
0:
mn m mn
m mn
d d mk J kd d
with boundary condition J k a
2
21
0 0 0
2 m mn nn
a a a
m mn n m mn m mn n m mn m mnn n
a J k a
f J k d F J k J k d F J k J k d
2 2 01
2
n m mnn
a
n m mnm mn
f F J k
F f J k da J k a
Arbitrary function: n m mnn
f F J k
(1) Expansion of by a series of Bessel function f
§3. 13 Fourier-Bessel series
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) 2-D Fourier-Bessel series
0 1
, sin cosmn mn m mnm n
V A m B m J k
2
0 0
2
0 00 1
2 2
0 00 0
0
, sin
sin sin sin cos
sin sin sin cos
mn mm
a
m mn
a
mn mn m mn m mnm n
mn mnm m
A
V m J k d d
A m m B m m J k J k d d
A m m d B m m d
01
01
22
12
a
m mn m mnn
a
mn m mn m mnn
mn m mn
J k J k d
A J k J k d
aA J k a
2
2 2 0 01
2 , sina
mn m mnm mn
A V m J k d da J k a
(a) coefficient mnA
2011 Classical Electrodynamics Prof. Y. F. Chen
0 1
2
2 2 0 01
2
2 2 0 01
2
2
, sin cos
, sin
, cos
mn mn m mnm n
a
mn m mnm mn
a
mn m mnm mn
V A m B m J k
A V m J k d da J k a
B V m J k d da J k a
60
(b) coefficient mnB
2
0 0
2
0 00 1
, cos
cos sin cos cos
a
m mn
a
mn mn m mn m mnm n
V m J k d d
A m m B m m J k J k d d
similarly
2
2 2 0 01
2 , cosa
mn m mnm mn
B V m J k d da J k a
2011 Classical Electrodynamics Prof. Y. F. Chen
˙
x
y
z
aL
2 22
2 2 2
1 1 0
0
0 0
, ,
: , ,
, , ,
z
a z
with boundary condition z
z L V
(1) General solution: 0 1
, , sin cos sinhmn mn m mn mnm n
z A m B m J k k z
0 1
, , , sin cos sinhmn mn m mn mnm n
z L V A m B m J k k L
mn mnA and B can be obtained from eq. (60)
61
§3. 14 Boundary value problems in cylindrical coordinates
2011 Classical Electrodynamics Prof. Y. F. Chen
Fourier series → Fourier transform
Fourier-Bessel series → Hankel transform
(2) Hankel transform
In eq. (61), let
:sinh
mn
kzmn
ak k
zk z e
0
0, , sin cos kz
m m mm
z A k m B k m J k e dk
0
0, sin cosm m m
mV A k m B k m J k dk
2
0 0
2
0 0 00
2 2
0 00 0
0
, sin
sin sin sin cos
sin sin sin cos
m mm
m
m m m mm
m mm m
A k
V m J k d d
A k m m B k m m J k dk J k d d
A k m m d B k m m d
0 0 m mJ k J k d dk
(a) coefficient mA k
2011 Classical Electrodynamics Prof. Y. F. Chen
2
0 0, sinm m
kA k V m J k d d
(b) coefficient mB k
2
0 0
2
0 0 00
, cos
cos sin cos cos
m
m m m mm
V m J k d d
A k m m B k m m J k dk J k d d
Similarly
2
0 0, cosm m
kB k V m J k d d
0 0 0
mm m m m
k kk
k k A kA k J k J k d dk A k dk
k k
62
00
2
0 0
2
0 0
, sin cos
, sin
, cos
m m mm
m m
m m
V A k m B k m J k dk
kA k V m J k d d
kB k V m J k d d
2011 Classical Electrodynamics Prof. Y. F. Chen
2 4,G x x x - x
2 2
2 2 2 2 2
1 1 1 1 4sin , cos cossin sin
r G r rr r r r
x x
We know that 0
, , cos cosl
lm lml m l
Y Y
63
, , , ,l
l lm lml m l
G g r r Y Y
x x
Substitute into eq. (63) ,G x x
2
2 2 2
11 4,l
l l r rd r g r rr dr r r
§3. 15 Green Function in Spherical System
2011 Classical Electrodynamics Prof. Y. F. Chen
(1) at : r r
2
2 2
1
11 0,
: ,
l
lll l l
l ld r g r rr dr r
B rgeneral solution g r r A r r
r
→ Consider three cases:
(a) Outside the sphere:
a
r
1
1
: ,
: ,
lll l l
ll l
B ra r r g r r A r r
rB r
r r g r rr
2 1
1
1
: ,
: ,
ll
l l l
ll l
aa r r g r r A r rr
B rr r g r r
r
0,lfrom g r a r
2 1
1 1
1: ,l
ll l l l
ageneral form g r r C r rr r
64
2011 Classical Electrodynamics Prof. Y. F. Chen
(b) Inside the sphere:
r
b
1
0 : ,
: ,
ll l
lll l l
r r g r r A r r
B rr r b g r r A r r
r
1 2 1
0
1
: ,
: ,
ll l
l
l l l l
r r g r r A r r
rr r b g r r B rr b
0,lfrom g r b r
1 2 1
1: ,l
ll l l l
rgeneral form g r r C r rr b
(c) Concentric sphere (spherical shell):
a
b
65
From eq. (64) and eq. (65)
2 1
1 1 2 1
1: ,ll
ll l l l l
rageneral form g r r C rr r b
2011 Classical Electrodynamics Prof. Y. F. Chen
(2) at : r r
2
2 2
11 4,l
l ld r g r r r rr dr r
Consider the cases of concentric sphere
2
2
14,
r r
lr r
l l r rd r g r r dr drdr r r
0
2
2
1 4
,
, ,
r
lr
r r
l lr r
d rg r rdr
l ld rg r r dr g r r drdr r r
Upper limit: ,r r
r rr r
2 1 1
1 2 11,
l ll
l l l l lr
d a d rrg r r C rdr r dr r b
66
2 1
1 1 2 1
1 1l l
ll l l l
a rC r l lr r b
67
2011 Classical Electrodynamics Prof. Y. F. Chen
lower limit: ,r r
r rr r
2 1 1
12 1
1,l l
ll l l l l
r
d d a rrg r r C rdr dr r r b
2 1 1
1 2 1
11l l
ll l l l
a rC l r lr r b
68
From eq. (66), eq. (67), and eq. (68)
2 1
4
2 1 1l l
Calb
2 1
1 1 2 12 10
142 1 1
, , , ,
, ,
l
l lm lml m l
llllm lm l
l l lll m l
G g r r Y Y
Y Y rarr r bal
b
x x
Finally, we obtain
2011 Classical Electrodynamics Prof. Y. F. Chen