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Vector Analysis for quantum mechanics Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: February 03, 2014)
Here we show Mathematica programs for the calculations of Laplacian, gradient,
divergence, and rotation in the spherical co-ordinates and the cylindrical co-ordinates, as well as the Cartesian co-ordinates. These programs are very useful for the discussions on a variety of topics, such as the central field problem on the electron motion near a hydrogen atom. ________________________________________________________________________ 1. Gradient
(; scalar) Nabra, gradient, del The gradient is defined as
),,(ˆˆˆzyxz
zy
yx
x
; gradient of the scalar
((Example))
f = f(r)
with 222 zyxr .
dr
df
dr
df
r
zzyyxxdr
df
r
z
r
dr
dfz
y
r
dr
dfy
x
r
dr
dfx
z
fz
y
fy
x
fxrf
rr
ˆ
)ˆˆˆ(1
ˆˆˆ
ˆˆˆ)(
where
r
x
x
r
, r
y
y
r
, r
z
z
r
.
2. Divergence Now we define the divergence of the vector as
2
z
F
y
F
x
F zyx
F .
(A) F is a real scalar. (B) Definition of solenoid
0 B . B is said to be soloenoid. _______________________________________________________________________ 3. F
We define the rotation of the vector as
zyx FFFzyx
zyx
ˆˆˆ
FW .
________________________________________________________________________ 4. Successive application of (i) )( This is defined by a Laplacian,
2
2
2
2
2
22
zyx
.
The equation 02 is called as the Laplace equation. (ii) is irrotational.
0 , since
0
ˆˆˆ
)(
zyx
zyx
zyx
.
3
Thus is irrotational. (iii) )( F is solenoid.
0)(
zyx FFFzyx
xxx
F .
(iv) Formula
FFF 2)()( . ((Proof))
We use the formula given by
CBACABCBA )()()( . with A , B , and C = F. Then we find
FFF 2)()( . ((Example)) Calculations
If )2,2,( 2 yzxzyx A , find A , )( A , and ))(( A . We use the Mathematica.
4
Clear"Gobal`";
Needs"VectorAnalysis`"SetCoordinatesCartesianx, y, z;
A1 x2 y, 2 x z , 2 y z;
CurlA12 x 2 z, 0, x2 2 z
CurlCurlA10, 2 2 x, 0
CurlCurlCurlA10, 0, 2
__________________________________________________________________ 5. Line and surface integral
We consider about the line integral
PQ
dI rA ,
where dsd r and the tangential component is assumed to be AS'
PQ
sdsAI .
If the contour is closed, we can write down as
5
rA d .
In general the line integral depends on the choice of path. If F ( ; scalar)
P
Q
)()( PQddIPQPQ
rrF .
This value does not depend on the path of integral. 6. Surface Integral n normal vector to the surface
dad na . (da; area element) Then the surface integral is defined by
SS
dad nFaF .
6
Fig. Right-hand rule for the positive normal. If F corresponds to the magnetic field; F = B,
S
daB is a magnetic flux through the area element S.
B
a
7. Gauss's theorem Here we define the volume integral as
7
V
d ,
where is a scalar. (i) Gauss's theorem
SV
dd aFF .
First we consider the physical interpretation of F . Suppose that F = J (current density). The current coming out through ABCD is
dydzdxx
JJdydzJ x
xxdxxx )|(| 0
.
The current coming in through EFGH is equal to
dydzJ xx 0| .
Thus the net current through EFGH is equal to
dxdydzx
J x
.
8
Thus the net current along the x direction through this small region is
dxdydzx
J x
.
Similarly for the y and z components, we have the net current along the y-direction and z-direction through the small region as
dxdydzy
J y
, dxdydzz
J z
,
respectively. Therefore the net current coming out through the volume element
dxdydzd can be expressed by
dd
surfaceSix
)( JaJ .
Summing over all parallel-pipes, we find that aJ d terms cancel out for all interior faces. Only the contributions of the exterior surface survive.
volume
surfaceexterior
dd )( JaJ ,
or
VA
dd JaJ . (Gauss’ theorem)
9
(ii) Gauss' theorem Let F be a continuous and differentiable vector throughtout a region V of the space. Then
VSS
ddad FnFaF ,
where the surface integral is taken over the entire surface that encloses V. ((Example-1)) In the maxwell's equation, we have
10
0
E ,
where is the charge density. From the Gauss's law, we have
SVV
ddd aEE
0
.
We assume that the volume V is formed of sphere with radius r. From the symmetry, E is perpendicular to the sphere surfaces,
rrE eE . Thus we have
S
rr
V
dEd ae
0
.
Since V
dQ , we get
204 r
QEr
. (Coulomb's law)
((Example-2))
11
0 E ,
if = 0. From the Gauss's law, we have
SVV
ddd aEE 00
.
E1n and E2n are the normal components of E1 and E2. Then we have
0)( 21 aEE nn .
Therefore we have the boundary condition for E as
nn EE 21 .
8. Green's theorem
SV
dd a)()( 22 .
((Proof)) In the Gauss's theorem, we put
A . Then we have
12
SVV
dddI aA )()(1 .
Noting that
2)( , we have
SV
ddI a)()( 21 .
By replacing , we also have
SV
ddI a)()( 21 .
Thus we find the Green's theorem
SV
ddII a)()( 2221 .
9. Stokes' theorem
x
y
A x0, y0 B
CD
1
dx
2
3
4 dy
13
2
00
1
00
4321
1234
)},(),({)},(),({
)(
dyyxFydxxFdxdyyxFyxF
dyFdxFdyFdxF
ncirculatiod
yyxx
yxyx
lF
Note that
dxx
FyxFydxxF
dyy
FyxFdyyxF
yx
yyy
yx
xxx
00
00
,
00
,
00
),(),(
),(),(
Then we have
dxdydxdyy
F
x
Fncirculatio z
xy )()()( 1234 F
.
We can write down this as
1)()( aFeFlF ddxdyd z
sidesFour
,
where
dxdyd zea 1 . Imagine that paths 1 and 2 are expanded out until they coalesce with path C (or path 3). Since the line integrals of F along the potions that 1 and 2 have in common will cancel each other,
2,1
21
21
)()()( aFaFaF
lFlF
lF
ddd
dd
dC
.
14
1 2
C Now let the surface S be divided up into a large number N of elements.
The above idea is extended to arrive at
SC
dd aFlF )( .
((Stoke's theorem)) Let S be a surface of any shape bounded by a closed curve C. If F is a vector, then
SSC
dadd nFaFlF )()( .
15
________________________________________________________________________ 10 Curvilinear co-ordinates 10.1 General definition
We consider that new co-ordinate (q1, q2, q3) are related to (x, y, z) through
),,(
),,(
),,(
321
321
321
qqqzz
qqqyy
qqqxx
or
),,(
),,(
),,(
33
22
11
zyxqq
zyxqq
zyxqq
Since
j
jj
dqq
dqq
dqq
dqq
drrrr
r 33
22
11
,
we have
ji
jiijji
jiji
dqdqgdqdqqq
ddds,,
2 )(rr
rr ,
where
jijijijiij q
z
q
z
q
y
q
y
q
x
q
x
qqg
rr
(second rank tensor).
We now consider the general coordinate system. The relation between the constants h1, h2, and h3 and the tensor gij will be discussed later.
16
e1
e2
e3
h1dq1e1
h2dq2e2
h3dq3e3
dr
iiiidqh
dqhdqhdqh
dsdsdsd
e
eee
eeer
333222111
332211
ji
ijiji idqdqhhddds
,
2 )( eerr ,
or we have
)(iijiij hhg ee ,
or
2iii hg .
Then
)(ii
jjii
ij
gg
gee .
Now we limit ourselves to orthogonal co-ordinate system.
17
ijg for i ≠ j.
In order to simplify the notation, we use 2iii hg , so that
i
iidqhds 22
iiiidqh
dqhdqhdqh
dsdsdsd
e
eee
eeer
333222111
332211
Where e1, e2, and e3 are unit vectors which are perpendicular to each other.
3333
2222
1111
1
1
1
sqh
sqh
sqh
rre
rre
rre
where
jiiii qq
gh
rr2 ,
or
222
iiiii
iii q
z
q
y
q
x
qqgh
rr. (second rank tensor).
The volume element for an orthogonal curvilinear coordinate system is given by
321321333222111 )}(){( dqdqdqhhhdqhdqhdqhdV eee .
10.2 Spherical coordinate (i) Unit vectors
The position of a point P with Cartesian coordinates x, y, and z may be expressed in terms of r, , and of the spherical coordinates;
x rsin cos , y rsin sin , z rcos ,
18
or
zy rrr eeer x cossinsincossin ,
jjj
jjj
j dqhdsd
3
1
3
1
eer .
x
y
z
q
dq
f df
r
drr cosq
r sinq
r sinq df
rdq
er
ef
eq
19
sin
1
222
222
222
rzyx
h
rzyx
h
r
z
r
y
r
xhr
or
drdrdrdhdhdrhd rrrr eeeeeer sin ,
zyreee
re xr cossinsincossin
,
zyreee
re x
sinsincoscoscos1
,
yree
re x
cossinsin
1
.
This can be described using a matrix A as
z
y
x
z
y
xr
e
e
e
e
e
e
A
e
e
e
0cossin
sinsincoscoscos
cossinsincossin
.
or by using the inverse matrix A-1 as
e
e
e
e
e
e
A
e
e
e
A
e
e
e rrT
r
z
y
x
0sincos
cossincossinsin
sincoscoscossin1 .
or
eeee sincoscoscossin rx ,
eeee cossincossinsin ry ,
eee sincos rz .
20
in the spherical coordinate systems.
(ii)
From the definition of , we have
3
1
3
1 j jjj
j jj qhs
ee ,
or,
sin
11
rrrr eee ,
where is a scalar function of r, , and . (iii) A
When a vector A is defined by
eeeA AAA rr .
The divergence is given by
21
)]()sin()sin([sin
1
)]()()([1
22
rAArArrr
AhhAhhAhhrhhh
r
rrrr
A
or
A
rA
rAr
rr r
sin
1)(sin
sin
1)(
1 22A .
(iv) A
A is given by
ArrAA
r
rr
rAhAhAh
r
hhh
hhhr
r
rr
rr
r
sin
sin
sin
112
eeeeee
A
(v) Laplacian
]sin
1)(sin)([sin
sin
1
)]sin
1()(sin)sin([
sin
1
)]()()([1
2
22
2
22
2
rr
rr
rr
rr
h
hh
h
hh
rh
hh
rhhhrr
rr
or
2
2
2222
2
2
2222
22
sin
1)(sin
sin
1)(
1
sin
1)(sin
sin
1)(
1
rrr
rr
rrrr
rr
We can rewrite the first term of the right hand side as
)(1
)(1
22
2
rrrr
rrr
,
22
which can be useful in shortening calculations. Note that we also use the expression for the operator
}sin
1)(sin
sin
1{
1)(
1
sin
1)(sin
sin
1)(
1
2
2
2222
2
2
2
2222
22
rrr
rr
rrrr
rr
11. Mathematica-1
We derive the above formula using the Mathematica. We use the Spherical co-rdinate. We need a Vector Analysis Package. We also need SetCordinatinates.In this system the vector is expressed in terms of (Ar, Aq, Af)
Clear"Gobal`"; Gra : Grad, r, , , "Spherical" &;
Diva : Div, r, , , "Spherical" &;
Curla : Curl, r, , , "Spherical" &;
Lap : Laplacian, r, , , "Spherical" &;
eq1 Lapr, , Simplify
1
r2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,
eq2 Grar, , 1,0,0r, , ,
0,1,0r, , r
,Csc 0,0,1r, ,
r
23
A Arr, , , Ar, , , Ar, , ;
eq3 CurlaA Simplify
1rAr, , Cot
Csc A0,0,1r, , A0,1,0r, , ,
1rAr, , Csc Ar0,0,1r, ,
r A1,0,0r, , ,
1rAr, , Ar0,1,0r, , r A1,0,0r, ,
eq3 DivaA Simplify
1r2 Arr, ,
Ar, , Cot Csc A0,0,1r, , A0,1,0r, , r Ar1,0,0r, ,
_______________________________________________________________________ 12. Quantum mechanical orbital angular momentum The orbital angular momentum in the quantum mechanics is defined by
)( riprL , using the expression
sin
11
rrrr eee ,
in the spherical coordinate. Then we have
)sin
1(
)sin
11()(
ee
eeeerL
i
rrrrii rr
.
The angular momentum Lx, Ly, and Lz (Cartesian components) can be described by
24
]sin
1)sinsincoscos(cos)cossin([
zyyi eeeeeL xx.
or
)coscot(sin
iLx ,
)sincotcos(
iLy ,
iLz .
We define L+ and L- as
)cot(
ieiiLLL iyx ,
and
)cot(
ieiiLLL i
yx .
We note that the operator can be expressed using the operator L as
2r
i
rr
Lre
.
The proof of this equation is given as follows.
)sin
1()
sin
1(
)(
eeeeeLr
rri r
,
or
rrrri rr
eeeLr
sin
11)(2
,
or
25
2
)(
r
i
rr
Lre
.
From 2222zyx LLL L , we have
)](sinsin
1
sin
1[ 2
2
222
L ,
where the proof is given by Mathematica. Using
)( 2222
2
rr
rr
L,
we can also prove that
L
i
rrr )1(2 .
((Note))
)(11
)(11
)(11
2
22
22
22
222
22
222
2
rrrr
rr
rrr
rr
rrr
L
L
L
13. Mathematica ((Arfken))
_______________________________________________________________________ ((Arfken))
26
________________________________________________________________________ ((Arfken))
________________________________________________________________________ ((Arfken))
________________________________________________________________________ ((Arfken))
((Arfken))
________________________________________________________________________ ((Mathematica))
27
Vector analysisAngular momentum in the spherical coordinatesHere we use the angular momentum operatior in the unit of Ñ=1
Clear"Global`"; ux Sin Cos, Cos Cos, Sin;
uy Sin Sin, Cos Sin, Cos;
uz Cos, Sin, 0; ur 1, 0, 0;
Lap : Laplacian, r, , , "Spherical" &;
Gra : Grad, r, , , "Spherical" &;
Diva : Div, r, , , "Spherical" &;
Curla : Curl, r, , , "Spherical" &;
L : Crossur r, Gra & Simplify;
Lx : ux.L & Simplify; Ly : uy.L & Simplify;
Lz : uz.L & Simplify;
Arfken 2.5 .16 a
Lr, , Simplify
0, Csc 0,0,1r, , , 0,1,0r, ,
28
Arfken 2.5.14, 2.5.16(b)
Lz
Lxr, , FullSimplify
Cos Cot 0,0,1r, , Sin 0,1,0r, , Lxr, , Lyr, , Simplify
Cos Sin Cot 0,0,1r, , 0,1,0r, , Lxr, , Lyr, , TrigToExp FullSimplify
Cos Sin Cot 0,0,1r, , 0,1,0r, , Lzr, , Simplify
0,0,1r, , Arfkem 2.5 .15
29
LxLyr, , LyLxr, , Lzr, , Simplify
0
Arfken 2.5.16(c)
eq1 LxLxr, , LyLyr, , LzLzr, , Expand FullSimplify
Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , eq2 r2 Lapr, , Dr2 Dr, , , r, r Simplify
Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , eq1 eq2 Simplify
0
Arfken 2.5.17(a)
30
eq3 ur Dr, , , r Cross1
rur, Lr, , Simplify
1,0,0r, , ,0,1,0r, ,
r,
Csc 0,0,1r, , r
eq4 Grar, ,
1,0,0r, , ,0,1,0r, ,
r,
Csc 0,0,1r, , r
eq3 eq4 Simplify
0, 0, 0
Arfken 2.5.17 (b)
eq5 ur r Lapr, , Simplify
1rCsc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, , , 0, 0
eq6 Grar, , r Dr, , , r Simplify
31
2 1,0,0r, , r 2,0,0r, , ,
0,1,0r, , r
1,1,0r, , ,
Csc 0,0,1r, , r 1,0,1r, , r
eq7 CurlaLr, , Simplify
1r
Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , ,
0,1,0r, , r 1,1,0r, ,
r,
Csc 0,0,1r, , r 1,0,1r, ,
r
seq1 eq5 eq6 eq7 Simplify
0, 0, 0The form of “2y(r) can be expressed by
Arfken 2.5.18
eq8 Lapr FullSimplify
2 rr
r
eq9 1
r2Dr2 Dr, r, r FullSimplify
2 rr
r
eq10 1
rDr r, r, 2 FullSimplify
2 rr
r
14. Radial momentum operator in the quantum mechanics (a) In classical mechanics, the radial momentum of the radius r is defined by
32
)(1
pr r
prc .
(b) In quantum mechanics, this definition becomes ambiguous since the component
of p and r do not commute. Since pr should be Hermitian operator, we need to define as the radial momentum of the radius r is defined by
)(2
1
rrprq
rpp
r .
This symmetric expression is indeed the canonical conjugate of r.
irprp rqrq
.
Note that
rrr
irr
iprq
1
)()1
)(( .
((Mathematica))
33
Clear"Global`"; ur 1, 0, 0;
Gra : Grad, r, , , "Spherical" &;
Diva : Div, r, , , "Spherical" &;
prc :—
ur.Gra & ;
prcr, , — 1,0,0r, ,
prq : —
2ur .Gra —
2Diva ur &;
prqr, , Simplify
— r, , r 1,0,0r, ,
r
Commutation relation
prqr r, , r prqr, , Simplify
— r, , 15. Expression of the Laplacian in the spherical coordinate Here we redefine prq simplily as
)(2
1
rrpr
rpp
r .
We show that
2
2222
rpr
L ,
by using the Mathematica. The total Hamiltonian H can be expressed by
)(22
1)(
2)(
2
12
222
22 rV
mrp
mrV
mrV
mH r
Lp
.
The effective potential Veff(r) is defined as
34
)1(2
)()(2
2
llmr
rVrVeff
.
((Mathematica)) Spherical coordinate Clear"Global`"; Clearrux Sin Cos, Cos Cos, Sin;
uy Sin Sin, Cos Sin, Cos;
uz Cos, Sin, 0; ur 1, 0, 0;
L :
— Crossur r, Grad, r, , ,
"Spherical" & Simplify;
Lx : ux.L &; Ly : uy.L &;
Lz : uz.L &;
Lap : Laplacian, r, , , "Spherical" &;
Gra : Grad, r, , , "Spherical" &;
Diva : Div, r, , , "Spherical" &;
Cur : Curl, r, , , "Spherical" &;
prq : —
21, 0, 0.Gra
—
2Diva 1, 0, 0 & ;
35
prqr, , Simplify
— r, , r 1,0,0r, ,
r
eq1 Nestprq, r, , , 2 Simplify;
eq2 1
r2LxLxr, , LyLyr, ,
LzLzr, , FullSimplify;
eq12 eq1 eq2 Simplify
1
r2 —2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,
eq3 —2 Lapr, , Simplify
1
r2 —2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,
eq12 eq3 FullSimplify
0 ________________________________________________________________________ 16 Cylindrical coordinates
The position of a point in space P having Cartesian coordinates x, y, and z may be expressed in terms of cylindrical co-ordinates
cosx , siny , z = z. The position vector r is written as
zyx zeeer sincos ,
36
dzdddqhd zj
jjj eeeer
3
1
,
where
1
1
3
2
1
zhh
hh
hh
The unit vectors are written as
zz
z
yx
yx
zzh
h
h
err
e
eerr
e
eerr
e
1
cossin11
sincos1
The above expression can be described using a matrix A as
z
y
x
z
y
x
z e
e
e
e
e
e
A
e
e
e
100
0cossin
0sincos
.
or by using the inverse matrix A-1 as
zz
T
zz
y
x
e
e
e
e
e
e
A
e
e
e
A
e
e
e
100
0cossin
0sincos1 .
37
17. Differential operations in the cylindrical coordinate
The differential operations involving are as follows.
zz
eee
1,
38
zVz
VV
1
)(1
V ,
z
z
VVVz
eee
V1
,
2
2
2
2
22 1
)(1
z
.
where V is a vector and is a scalar. 18 Mathematica for the cylindrical coordinate Vector analysisAngular momentum in the cylindrical coordinatesHere we use the angular momentum operatior in the unit of Ñ=1
Clear"Global`"; r1 z2 2 ; ux Cos, Sin, 0;
uy Sin, Cos, 0; uz 0, 0, 1; r , 0, z;
ur 1
r1, 0, z; Gra : Grad, , , z, "Cylindrical" &;
Lap : Laplacian, , , z, "Cylindrical" &;
Curla : Curl, , , z, "Cylindrical" &;
Diva : Div, , , z, "Cylindrical" &;
Vector analysis in the cylindrical coordinate
eq1 Lap, , z Simplify
0,0,2, , z 0,2,0, , z2
1,0,0, , z
2,0,0, , z
eq2 Gra, , z Simplify
1,0,0, , z,0,1,0, , z
, 0,0,1, , z
39
B B, , z, B, , z, Bz, , z;
eq3 CurlaB Simplify
B0,0,1, , z Bz0,1,0, , z
,
B0,0,1, , z Bz1,0,0, , z,
B, , z B0,1,0, , z B1,0,0, , z
eq4 DivaB Simplify
Bz0,0,1, , z B, , z B0,1,0, , z
B1,0,0, , z
Angular momentum in the cylindrical coordinate
L : Crossr, Gra &; Lx : ux.L & Simplify;
Ly : uy.L & Simplify;
Lz : uz.L & Simplify;
L, , z Simplify
z 0,1,0, , z
,
0,0,1, , z z 1,0,0, , z, 0,1,0, , z
40
Lx, , z FullSimplify
Sin 0,0,1, , z z Cos 0,1,0, , z
z Sin 1,0,0, , z
Ly, , z FullSimplify
Cos 0,0,1, , z z Sin 0,1,0, , z
z Cos 1,0,0, , z
Lx, , z Ly, , z FullSimplify
2 0,0,1, , z z 0,1,0, , z z 1,0,0, , z
Lx, , z Ly, , z FullSimplify
2 0,0,1, , z z 0,1,0, , z z 1,0,0, , z
Lz, , z Simplify
0,1,0, , z
41
The commutation of the angular momentum in the cylindrical coordinate
eq5 LxLy, , z LyLx, , z Lz, , z Simplify
0
L2 in the cylindrical coordinate
eq6 LxLx, , z LyLy, , z Lz Lz, , z FullSimplify
2 z 0,0,1, , z 1
2 4 0,0,2, , z z2 2 0,2,0, , z z2 2 1,0,0, , z
z 2 1,0,1, , z z 2,0,0, , z
Schrodionger equation with a speific wavefunction in the cylindrical coorrdinate
H1 Exp kz z Exp m ;
1
2 LapH1 V H1 Simplify
kz zm m2 kz2 2 2 2 V 2 2
19. Expression of the Laplacian in the cylindrical coordinate Here we redefine prq simply as
)(2
1
rrpr
rpp
r ,
with the commutation relation
ipr r ],[ , where
22 zr , zzeer .
We show that
42
2
2222
rpr
L ,
by using the Mathematica. ((Mathematica))
Clear"Global`"; r1 z2 2 ; ux Cos, Sin, 0;
uy Sin, Cos, 0; uz 0, 0, 1; r , 0, z; ur 1
r1, 0, z;
Gra : Grad, , , z, "Cylindrical" &;
Lap : Laplacian, , , z, "Cylindrical" &;
Curla : Curl, , , z, "Cylindrical" &;
Diva : Div, , , z, "Cylindrical" &;
Angular momentum in the cylindrical coordinate
L : — Crossr, Gra &; Lx : ux.L & Simplify;
Ly : uy.L & Simplify;
Lz : uz.L & Simplify;
The linear momentum in the cylindrical coordinate
prq : —
2ur.Gra —
2Diva ur & ;
prq, , z Simplify
— , , z z 0,0,1, , z 1,0,0, , z
z2 2
Commutation relation in the cylindrical coordinate
r1 prq, , z prqr1 , , z Simplify
— , , zProof of
-Ñ2“2= pr 2 + L2
r2
eq1 Nestprq, , , z, 2 Simplify
1
z2 2 —2 2 z 0,0,1, , z z2 0,0,2, , z 2 1,0,0, , z 2 z 1,0,1, , z 2,0,0, , z
eq20 LxLx, , z LyLy, , z LzLz, , z FullSimplify;
eq2 eq20
r12 Simplify
43
1
2 z2 2 —2 2 z 2 0,0,1, , z z2 0,2,0, , z
3 0,0,2, , z z2 1,0,0, , z 0,2,0, , z 1,0,0, , z 2 z 1,0,1, , z z2 2,0,0, , z
eq12 eq1 eq2 FullSimplify
1
2 —2 0,2,0, , z 1,0,0, , z 0,0,2, , z 2,0,0, , z
L2 in the cylindrical coordinate
eq3 —2 Lap, , z Expand FullSimplify
1
2 —2 0,2,0, , z 1,0,0, , z 0,0,2, , z 2,0,0, , z
eq12 eq3 Simplify
0 19 Jacobian
321321321321 ),,(
),,(dqdqdqhhhdqdqdq
qqq
zyxdxdydzdV
.
Jacobian determinant is defined as;
321
321
321
321 ),,(
),,(
q
z
q
z
q
zq
y
q
y
q
yq
x
q
x
q
x
qqq
zyx
.
(a) Spherical coordinate
ddrdrddrdhhhdqdqdqhhh r sin2321321 .
(b) Cylindrical co-ordinate
dzdddzddhhhdqdqdqhhh z 321321 .
((Mathematica))
44
This is the program to determine the Jacobian determinant.
JacobianDeterminant[pt, coordsys]:
to give the determinant of the Jacobian matrix of the transformation from the coordinate system coordinate system to the Cartesian coordinate system at the point pt.
Clear"Gobal`"Jacobian determinant for transformation from cylindrical to Cartesian coordinates:
jdet JacobianDeterminant, , z, Cylindrical
Jacobian determinant for transformation from cylindrical to Spherical coordinates:
jdet JacobianDeterminantr, , , Sphericalr2 Sin
________________________________________________________________________ APPENDIX Formula
45