Chapter 1 Vector Analysis Masatsugu Sei Suzuki Department...
Transcript of Chapter 1 Vector Analysis Masatsugu Sei Suzuki Department...
Chapter 1 Vector Analysis
Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton
(Date: August 31, 2010) ((Note)) You may find original Mathematica programs in my web site http://bingweb.binghamton.edu/~suzuki/
Johann Carl Friedrich Gauss (30 April 1777 – 23 February 1855) was a German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.
http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss Sir George Gabriel Stokes, 1st Baronet FRS (13 August 1819–1 February 1903), was a mathematician and physicist, who at Cambridge made important contributions to fluid dynamics (including the Navier–Stokes equations), optics, and mathematical physics (including Stokes' theorem). He was secretary, then president, of the Royal Society.
http://en.wikipedia.org/wiki/Sir_George_Stokes,_1st_Baronet 1.1 Fundamentals 1.1.1 Definition of vectors
Vectors are usually indicated by boldface letters, such as A, and we will follow this most common convention. Alternative notation is a small arrow over the letters such as
A
. The magnitude of a vector is also often expressed by AA . The displacement
vector serves as a prototype for all other vectors. Any quantity that has magnitude and direction and that behaves mathematically like he displacement vector is a vector. ((Example))
velocity, acceleration, force, linear momentum, angular momentum, torque electric field, magnetic field, current density, magnetization, polarization electric dipole moment, magnetic moment
By contrast, any quantity that has a magnitude but no direction is called a scalar. ((Example))
length, time, mass, area, volume, density, temperature, energy A unit vector is a vector of unit length; a unit vector in the direction of A is written with
a caret as A , which we read as “A hat.”
AA AAA ˆˆ
(A) A vector r
Fig. The vector r represents the position of a point P relative to another point O as origin.
(B) Negative vector: - r
The negative of a given vector r is a vector of the same magnitude, but opposite direction.
Fig. The vector –r is equal in magnitude but opposite in
direction to r. (C) The multiplication of the vector by a scalar
Fig. The vector kr is in the direction of r and is of magnitude kr, where k = 0.6. (D) A unit vector
Fig. The vector r is the unit vector in the direction of r. Note that rr ˆr . 1.1.2. Vector addition
C = A + B = B + A (commutative)
The sum of two vectors is defined by the geometrical construction shown below. This construction is often called the parallelogram of addition of vectors.
1.1.3. Vector subtraction
C = A – B The subtraction of two vectors is also defined by the geometrical construction shown below.
1.1.4. Sum of three vectors
1.1.5 Sum of many vectors
1.1.6. Imporrtant theorem for the geometry (A) Theorem
When the point P is between the point Q and P on the line connecting the two points
P and Q, the vector OP is expressed in terms of the vectors A and B by
BA OP where + = 1 (>0 and >0).
Fig. BA OP where + = 1. is changed between =
0.1 and 0.9 with = 0.1. We now consider the following case.
b
a
b
a
qOB
pOA
OB
OA
1
1
where p and q are between 0 and 1. From the above theorem, the vector OP is expressed by
)()()()( baba qpp
qOP
where
1
1
qp
From these Eqs. we have
pq
p
pq
pp
1
1
1
)1(
Then OP is expressed by
bapq
pq
pq
ppOP
1
)1(
1
)1(
(B) Bisecting vector
In a triangle of this figure, the angle POR is equal to the angle QOR. The point R is
on the line PQ. What is the expression of OR in terms of the vectors A and B?. Since R
is on the line AB, OR is described by
BA OR (1) where + = 1 (>0 and >0).
The vector OR is also described by
)()ˆˆ(BA
kkORBA
BA (2)
where A and B are the magnitudes of A and B., A and B are the unit vectors for A and B. From Eqs.(1) and (2), we have
B
kA
k
or
B
A (3)
Then we get
BA
ABA
B
1.1.7. Cartesian components of vectors (A) 2D system
Let I and j, and k denote a set of mutually perpendicular unit vectors. Let i and j drawn from a common origin O, give the positive directions along the system of rectangular axes Oxy.
We consider a vector A lying in the xy plane and making an angle with the positive x axis. The vector A can be expressed by
)sin(cos),( jijiA AAAAA yxyx
where
22yx AAA A and
x
y
A
Atan
When the vector B is expressed by
jiB yxzyx BBBB ),(
the sum of A and B is
jiBA )()( yyxx BABA
(B) 3D system
Let i, j, and k denote a set of mutually perpendicular unit vectors. Let i, j, and k drawn from a common origin O, give the positive directions along the system of rectangular axes Oxyz.
An arbitrary vector A can be expressed by
kjiA zyxzyx AAAAAA ),,(
where Ax, Ay, and Az are called the Cartesian components of the vector A. When the vector B is expressed by
kjiB zyxzyx BBBBBB ),,(
the sum of A and B is
kjiBA )()()( zzyyxx BABABA
1.1.8. Scalar product of vectors (A) Definition
The scalar product (or dot product) of the vectors A and B is defined as
coscos AB BABA
where is the angle between A and B and is between 0 and . The scalar product is a scalar and is commutative,
ABBA
_______________________________________________________________________ (B) Magnitude:
When B = A, we have
22A AAA
since = 0. _______________________________________________________________________ (C) Orthogonal (A B):
If
0BA (A ≠0 and B ≠0), we say that A is orthogonal to B or perpendicular to B. ______________________________________________________________________ (C) Projection:
The magnitude of the projection of A on B is A cos. So BA is the product of the projection of A on B with the magnitude of A. We also consider that the magnitude of BA is the product of the projection of on B on A with the magnitude of B.
)cos()cos(cos BAAB BABA
(D) The expression of the scalar product using Cartesian components of vectors
Inner product of A and B We now consider two vectors given by
kjiB
kjiA
zyxzyx
zyxzyx
BBBBBB
AAAAAA
),,(
),,(
The scalar product of these two vectors A and B can be expressed in terms of the components
)(
)()(
)()(
kkjkik
kjjjijkijiii
kjikjiBA
zzyzxz
zyyyxyzxyxxx
zyxzyx
BABABA
BABABABABABA
BBBAAA
or
zzyyxx BABABA BA
Here we use the above relations for the inner products of the unit vectors. where
0
0
1
ki
ji
ii
0
1
0
kj
jj
ij
1
0
0
kk
jk
ik
In special cases, the components of A are given by
zzyx
yzyx
xzyx
AAAA
AAAA
AAAA
kkkjkiiA
jkjjjijA
ikijiiiA
The unit vector A of the vector A is expressed by
kkA
jjA
iiA
kjiA
AAA
A
A
A
A
A
AAAA
Azyx
zyx
),,(1ˆ
(E) Law of cosine
cos2
)(2)()(22
222
ABBA
BAC
BABABACC
This is the famous trigonometric relation (law of cosine). 1.1.9 Vector product (A) Definition
This product is a vector rather than scalar in character, but it is a vector in a somewhat restricted sense. The vector product of A and B is defined as
nnBABAC ˆsinˆsin AB
where A is the magnitude of A. B is the magnitude of B. is the angle between A and
B. n is a unit vector, perpendicular to both A and B in a sense defined by the right hand thread rule.
We read A x B as “A cross B.” The vector A is rotated by the smallest angle that will bring it into coincidence with the direction of B. The sense of C is that of the direction of motion of a screw with a right-hand thread when the screw is rotated in the same as was the vector A.
((Note))
The vector C is perpendicular to both A and B. Rotate A into B through the lesser of the two possible angles – curl the fingers of the right hand in the direction in which A is rotated, and the thumb will point in the direction of C = A x B. (B)
Because of the sign convention, B x A is a vector opposite sign to A x B. In other words, the vector product is not commutative,
BAAB .
(C)
It follows from the definition of the vector product that
0AA (D)
The vector product obey the distributive law.
CABACBA )( (E) Cartesian components.
The vectors A and B are expressed by
kjiB
kjiA
zyxzyx
zyxzyx
BBBBBB
AAAAAA
),,(
),,(
Then the vector product A x B is expressed in terms of the Cartesian components
zzyzxz
zyyyxy
zxyxxx
zyxzyx
BABABA
BABABA
BABABA
BBBAAA
)()()(
)()()(
)()()(
)()(
kkjkik
kjjjij
kijiii
kjikjiBA
Here we use the relations,
jki
kji
ii
0
ikj
jj
kij
0
0
kk
ijk
jik
zyx
zyx
xyyxzxxzyzzy
yzxzzyxyzxyx
BBB
AAA
BABABABABABA
BABABABABABA
kji
kji
ijikjkBA
)()()(
It is easier for one to remember if the determinant is used. Using the cofactor, BA can be simplified as
yxzx
zx
zy
zy
zyx
zyx BB
AyAx
BB
AA
BB
AA
BBB
AAA kji
kji
BA .
where a 2x2 determinant is given by bcaddc
ba .
Note
zyx
zyx
zyx
BBB
AAA
CCC
)( BAC
Note that
CBABACACBBACACBCBA )()()()()()( , where the order of A, B, and C is cyclic.
((Mathematica))
A A1, A2, A3A1, A2, A3
B B1, B2, B3B1, B2, B3
CC C1, C2, C3C1, C2, C3
DD D1, D2, D3D1, D2, D3
CrossA, BA3 B2 A2 B3, A3 B1 A1 B3, A2 B1 A1 B2
A.B
A1 B1 A2 B2 A3 B3
CrossCC, CrossA, B Simplify
A2 B1 C2 A1 B2 C2 A3 B1 C3 A1 B3 C3,
B2 A1 C1 A3 C3 A2 B1 C1 B3 C3,B3 A1 C1 A2 C2 A3 B1 C1 B2 C2
CrossCrossCC, DD, CrossA, B Simplify
A3 B1 C2 D1 C1 D2 A2 B1 C3 D1 C1 D3 A1B3 C2 D1 B2 C3 D1 B3 C1 D2 B2 C1 D3,
A3 B2 C2 D1 C1 D2 A1 B2 C3 D2 C2 D3 A2B3 C2 D1 B3 C1 D2 B1 C3 D2 B1 C2 D3,
B3 A2 C3 D1 A1 C3 D2 A2 C1 D3 A1 C2 D3 A3 B2 C3 D1 B1 C3 D2 B2 C1 D3 B1 C2 D3
1.9.3 Application of the vector product (A) Area of parallelogram The magnitude of A x B is the area of the parallelogram.
sinABBA
(B) Volume of a parallelepiped
The scalar given by
V CBA )(
is the volume of parallelepiped (C) Law of sine
We consider the triangle defined by C = A + B, and take the vector product
BABAAABAACA )( The magnitude of both sides must be equal so that
)),(sin(),sin(),sin( BAABBAABCAAC or
C
BA
B
CA )],(sin[),sin(
(Law of sine).
where sin(A, B) denotes the sine of the angle between A and B. 1.10 BAC-CAB rule
)()()( BACCABCBA Similarly the following two identities are also very important. (A)
)}({)}({)()( CBADDBACDCBA (B)
))(())(()()( CBDADBCADCBA _______________________________________________________________
1.2. Advanced topics See Chapter 1S for more detail in the rotation. 1.2.1. Directional cosine aij
e1e2
e3
e1'
e2'
e3'
P
The vector field is defined in terms of the behavior of its components under the rotation of the co-ordinate axes. Here we use the following notation.
1ˆ ex , 2ˆ ey , 3ˆ ez
By the rotation of the co-ordinate system, we have the new co-ordinate system, such as
''ˆ 1ex , ''ˆ 2ey , ''ˆ 3ez
The new vectors 'ie is related to the old vectors je through the following relationship.
3
2
1
333231
232221
131211
3
2
1
3
2
1
'
'
'
e
e
e
e
e
e
e
e
e
aaa
aaa
aaa
a
where A is the 3 x 3 matrix, and
ijji ee , ijji '' ee
The matrix element aij is called the directional cosine. The symbol ij is the Kronecker
delta, and is defined by
1ij for i = j, and 0 for i ≠ j.
Then we have
)'( ijija ee .
The vector {ei} is also expressed by using {ej’}
'
'
'
'
'
'
'
'
'
3
2
1
332313
322212
312111
3
2
1
3
2
11
3
2
1
e
e
e
e
e
e
e
e
e
e
e
e
aaa
aaa
aaa
aa T
,
where aT is the transpose of the matrix a. For simplicity, we can write down
jj
iji a ee ' , '')( jj
jijj
ijT
i aa eee .
((Note))
1 aaT ((Proof)) From
ijji '' ee
we have
ijk
jkiklk
kljliklk
lkjlikl
ljlk
kikji aaaaaaaa ,,
)())(('' eeeeee
or
ijk
jkikaa ,
where aT is the transpose of the matrix a;
100
010
001
332313
322212
312111
333231
232221
131211
aaa
aaa
aaa
aaa
aaa
aaa
aaT
In other words, we have
Iaaaa TT or
1 aaT
Note that
kjk
kikjk
ikT
ijijT aaaaaa )()()(
or
ijkjk
kiaa .
1.2.2 Two dimensional rotation
e1
e2
O
e1'
e2'
q
2221212
2121111
'
'
eee
eee
aa
aa
with
cos)'.(
sin)'.(
sin)'.(
cos)'.(
2122
2121
1212
1111
ee
ee
ee
ee
a
a
a
a
or
cossin
sincos
2221
1211
aa
aaa
((Note)) Mathematica
R RotationMatrix Simplify
Cos, Sin, Sin, Cos
R MatrixForm
Cos SinSin Cos
1.2.3 Three dimensional rotation
Rotations of the body frame are defined to have a countercloskwise sense, with the rotations carried out in the following order. 1. First, make a rotation by an angle about the initial z axis. ( – –). 2. Then, make a second rotation by an angle about the body (= ’) axis, called
the line of nodes. (' – ' – '). 3. Finally, make a third rotation by an angle about the body ' (= z’) axis. (x' - y' -
z').
100
0cossin
0sincos
R
cossin0
sincos0
001
R
100
0cossin
0sincos
R
The net result in the body frame is
cossincossinsin
cossinsinsincoscoscossincoscossincos
sinsinsincoscoscossinsincossincoscos
333231
232221
131211
aaa
aaa
aaa
RRRR
x
y
z
x
h
z
f
Fig.1 Rotation by an angle about the initial z axis. ( – –).
x
y
z
xh
z
x '
h'
z '
q
Fig.2 A second rotation by an angle about the body (= ’) axis, called the line of
nodes. (' – ' – ').
x
y
z
x'
y'z'
x '
h'z '
y
Fig.3 A third rotation by an angle about the body ' (= z’) axis. (x' - y' - z').
((Mathematica)) R RotationMatrix, 0, 0, 1 Simplify;
R MatrixForm
Cos Sin 0Sin Cos 0
0 0 1
R RotationMatrix, 1, 0, 0 Simplify;
R MatrixForm
1 0 00 Cos Sin0 Sin Cos
R RotationMatrix, 0, 0, 1 Simplify;
R MatrixForm
Cos Sin 0Sin Cos 0
0 0 1 S R.R.R Simplify;
S MatrixForm
Cos Cos Cos Sin Sin Cos Sin Cos Cos Sin Sin SinCos Cos Sin Cos Sin Cos Cos Cos Sin Sin Cos Sin
Sin Sin Cos Sin Cos ______________________________________________________________________ 1.2.4 Definition of vector
Suppose that the vector r can be expressed by
i
iii
ii xx ''eer
for the old and new co-ordinate systems, respectively. Then we have
i
iiji i j
ijijiijjj i
iijjj
jj xxaaxaxx '''')(')'(,
ereeeer
Then we have
j
jiji xax ' ,
or
axx '
We may write (Cartesian co-ordinate)
j
iij x
xa
'
Note that we also have
''1 xaxax T or
j
jjij
jijT
i xaxax '')( .
Also we may write
'j
iji x
xa
, or 'i
jij x
xa
Using the above notations, we get the original definition;
kkk
lkklkl
lk ikilikl
lkiklkili
i kkki
llli
iii xxaaxxaaaxax ''''''''''
,,,,
eeeeee
Now we consider more general case in order to get the definition of vector.
x1
x2
x3
P
Q
A
Suppose that
'' i
iii
ii yyOP ee
'' i
iii
ii zzOQ ee
where
j
jiji yay ' , j
jiji zaz '
Then we have
i
iiii
iii yzyzOPOQPQ ')''()( eeA
Since
j
jjii a 'ee ,
the expression of A can be rewritten as
i
iiiji
jjiii yzayz ')''(')(,
ee .
By the interchange between i and j in the left-hand side,
i
iiii j
iijjjji
jjiii yzayzayz ')''(')(')(,
eee .
Therefore we get
j
jjijii yzayz )('' .
Since the component of A is given by
iii yzA , and ''' iii yzA
in the old and new co-ordinate systems, we can write
j
jiji AaA ' .
In summary, under the rotation of the co-ordinate system,
jj
iji a ee ' , or '' jj
ijijj
ijT
i aa eee
the components of the vector are transformed through
jj
iji AaA '
((Example))
(1) Newton's second law
We consider how the Newton's second law transforms under the rotation of the coordinate by the angle around the z axis. From the definition of the vector for r, we have a relation between the old coordinates and new coordinates,
z
y
x
z
y
x
100
0cossin
0sincos
'
'
'
since r is a real vector. In the old system, the Newton’s second law states that
2
2
2
2
2
2
dt
zddt
yddt
xd
m
F
F
F
z
y
x
In the new system, the Newton's second law should be written as
'
'
'
2
2
'
'
'
z
y
x
dt
dm
F
F
F
z
y
x
Then we have
z
y
x
z
y
x
F
F
F
z
y
x
dt
dm
z
y
x
dt
dm
F
F
F
100
0cossin
0sincos
100
0cossin
0sincos
100
0cossin
0sincos
2
2
2
2
'
'
'
This means that the force is a real vector. In other words, if Newton's second law is correct one set of axes, they are also valid on any other set of axes. (2) Angular momentum
The angular momentum is defined by
zzyyxx
zyx
zyx
LLL
ppp
zyx eee
eee
prL
where
xyz
zxy
yzx
ypxpL
xpzpL
zpypL
Now we consider how the angular momentum transforms under the rotation of the coordinate by the angle around the z axis. The angular momentum in the new coordinate is
''''''
'''
''' '''
'''
zzyyxx
zyx
zyx
LLL
ppp
zyx eee
eee
prL
with
'''
'''
'''
''
''
''
xyz
zxy
yzx
pypxL
pxpzL
pzpyL
Using
z
y
x
z
y
x
100
0cossin
0sincos
'
'
'
z
y
x
z
y
x
p
p
p
p
p
p
100
0cossin
0sincos
'
'
'
We can show that
zxyz
yxzxy
yxyzx
LpypxL
LLpxpzL
LLpzpyL
'''
'''
'''
''
cossin''
sincos''
or
''
'
'
100
0cossin
0sincos
z
y
x
z
y
x
L
L
L
L
L
L
This means that the angular momentum is a vector. ((Mathematica))
R x, y, z; P px, py, pz; L CrossR, P;
A RotationMatrix, 0, 0, 1;
RN A.R
x Cos y Sin, y Cos x Sin, z
PN A.P
px Cos py Sin, py Cos px Sin, pz
LN CrossRN, PN Simplify
pz y py z Cos pz x px z Sin,pz x px z Cos pz y py z Sin,
py x px y
LN A.L Simplify
0, 0, 0 ________________________________________________________________________ 1.2.5 Scalar product
The scalar is invariant under the rotation of the co-ordinate system. We show that the scalar product BA is scalar;
'' BABA We start with the definition of the vectors,
jj
iji AaA ' . kj
iki AaB ' .
Then we have
j
jjkj
jkkjkj i
ikijkji j k
kikjijii
i BABAaaBABaAaBA,,
))(('''' BA
Then we have
BABA i
iiBA''
Thus BA is a scalar. ________________________________________________________________________ 1.2.6. Vector product
Here there still remains the problem of verifying that
BAC is indeed a vector. ((Proof))
Under the rotation of the co-ordinate system,
'''''
'
'
jkkjii
mmkmkk
lljljj
BABACC
BaBB
AaAA
where i, j, and k are in cyclic order.
313212111
313212111
12213231
22213113
3133
21232332
3332
2322
313123332113332131232333223223
323223332212322131222131223221
,2332
23321
)()()(
)()()(
)()()(
)()()(
)(
'''''
CaCaCa
BAaBAaBAa
BABAaa
aaBABA
aa
aaBABA
aa
aa
BAaaaaBAaaaaBAaaaa
BAaaaaBAaaaaBAaaaa
BAaaaa
BABAC
mlmlmlml
323222121
323222121
12211211
32313113
1113
31332332
1312
3332
311133133113133111332313321233
321233133212123111322111321231
,3113
31132
)()()(
)()()(
)()()(
)()()(
)(
'''''
CaCaCa
BAaBAaBAa
BABAaa
aaBABA
aa
aaBABA
aa
aa
BAaaaaBAaaaaBAaaaa
BAaaaaBAaaaaBAaaaa
BAaaaa
BABAC
mlmlmlml
333232131
333232131
12212221
12113113
2123
11132332
2322
1312
312113231113231121132323122213
322213231212221121122121122211
,1221
12213
)()()(
)()()(
)()()(
)()()(
)(
'''''
CaCaCa
BAaBAaBAa
BABAaa
aaBABA
aa
aaBABA
aa
aa
BAaaaaBAaaaaBAaaaa
BAaaaaBAaaaaBAaaaa
BAaaaa
BABAC
mlmlmlml
Thus BAC is a real vector. We note that
3332
232211 aa
aaa ,
3133
212312 aa
aaa
, 3231
222113 aa
aaa ,
1312
333221 aa
aaa
, 1113
313322 aa
aaa ,
1211
323123 aa
aaa
2322
131231 aa
aaa ,
, 2123
111332 aa
aaa
, 2221
121133 aa
aaa
((Note)) The above relations among {aij} can be derived in the following way.
321
'
eee
ee
j
jjii a
Now we consider about the vectors e1, e2 x e3, respectively.
'''' 33122111111 eeeee aaaaj
jj
')(')(')(
)'''()'''(
)'()'(
313222312233121332123323322
333223113332222112
3232
eee
eeeeee
eeee
aaaaaaaaaaaa
aaaaaa
aam
mml
ll
Thus we find
3332
232211 aa
aaa ,
1312
333221 aa
aaa ,
2322
131231 aa
aaa ,
Similarly we have
3133
212312 aa
aaa ,
1113
313322 aa
aaa ,
2123
111332 aa
aaa ,
3231
222113 aa
aaa ,
1211
323123 aa
aaa
2221
121133 aa
aaa .
______________________________________________________________________ 1.2.7. Tensor
Ohm’s law;
k
kiki EJ
is the tensor of second rank.
333231
232221
131211
scalar: tensor of rank zero vector: tensor of first rank
((Definition of tensor of the second rank))
lk
kljlikij aa,
'
where Only in Cartesian coordinates we have
'
'
i
j
j
iij x
x
x
xa
.
So there is no difference between contravariant and covariant transformation. In other systems, this in general does not apply, so the distinction between contavariant and covariant is real.
lk
klj
l
i
kij C
x
x
x
xC
, ''' Covariant wrt i, j
lk
kl
l
j
k
iij Ax
x
x
xA
,
''' Contravariant wrt i, j
lk
kl
j
l
k
ii Bx
x
x
xB j
, '
'' Contravariant wrt i, covariant wrt j.
Summation convention:
kl
j
l
k
ii Bx
x
x
xB j
'
''
The Kronecker delta ij is really a mixed tensor of second rank i
j .
We have, using the summation convention
'
'
'
'
j
k
k
i
j
l
k
ikl x
x
x
x
x
x
x
x
by definition of the Kronecker delta. Now we have
ij
j
i
j
k
k
i
x
x
x
x
x
x'
'
'
'
'
by direct partial differentiation of the right-hand side (chain rule). Hence,
kl
j
l
k
iij x
x
x
x '
''
________________________________________________________________________ 1.2.8. 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
(A) Geometrical interpretation
Let us give a geometrical interpretation of .
zdzydyxdxd ˆˆˆ r From the definition, we have
dzz
dyy
dxx
dd
r
Fig. The normal vector n which is perpendicular to the line PQ on the surface.
is perpendicular to the surface ( = constant).
If we choose two points P and Q on the surface const)(r , where rdPQ in the limit of 0rd . Since
0 rdd , we find that is perpendicular to the surface ( = constant). It is called the normal vector. ((Example-1)) Find a unit normal to the surface 422 xzyx at the point P (2, -2, 3).
)2,,22()2( 22 xxzxyxzyx A . A = (-2, 4, 4) at the point P. Then a unit normal to the surface is (-1/3, 2/3, 2/3). Another unit normal is (1/3, -2/3, -2/3).
((Example-2)) Find an equation for the tangent plane to the surface
07432 2 xxyxz at the point P r0 = (1, -1, 2).
)4,3,432()432( 22 xzxyzxxyxz A . Then a normal to the surface at the point P is A = (7, -3, 8) at the point P. The equation of a plane passing through a point P, which is perpendicular to A is (r – r0).A = 0; 0)2(8)1(3)1(7 zyx
(B) Vector
Next we will verify that is a vector. is scalar, which means the invariant under the rotation of the coordinate system.
)()'(' ii xx .
j jij
i
j
j jii xa
x
x
xxx
'''
',
since
'i
jij x
xa
Thus is a real vector (contravariant vector) (C) Plotting of equi-potential lines and vector fields Now we consider a rather simple 2D function,
xy The gradient operating on this function generate the vector field
),( xyF Using the Matematica, we make a plot of the equipotential lines of in the x-y plane (ContourPlot) and a plot of the field lines of F in the x-y plane (StreamPlot). The field lines are perpendicular to the equipotential lines of constant .
x
y
O
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3
1.2.9 Divergence Now we define the divergence of the vector as
z
F
y
F
x
F zyx
F
(A) F is a real scalar. Under the rotation of the co-ordinate system
j
jiji FaF '
Then we have
lj l
jklij
lj l
j
k
lij
jjij
kk
i
x
Faa
x
F
x
xaFa
xx
F
,, ')(
''
'
where
j
iij x
xa
'.
Then we have
j j
j
lj l
jjl
lj l
j
iilij
lji l
jilij
i i
i
x
F
x
F
x
Faa
x
Faa
x
F
,,,,
)('
'
Thus F is a scalar. (B) Definition of solenoid
0 B B is said to be soloenoid. _______________________________________________________________________ 1.2.10 F
Now we define the rotation of the vector as
zyx FFFzyx
zyx
ˆˆˆ
FW
We show that F is a real vector. ((Proof))
Now we put
'
'
'
''
3
2
2
31 x
F
x
FW
(For simplicity we use x1 = x, x2 = y, x3 = z)
lj l
jlj
j l
jlj
j
jj x
Faa
x
F
x
xa
x
Fa
x
F
,23
23
23
2
3
'''
'
Similarly, we have
lj l
jlj
j l
jlj
j
jj x
Faa
x
F
x
xa
x
Fa
x
F
,32
32
32
3
2
'''
'
Then we have
lj l
jljlj x
Faaaa
x
F
x
FW
,3223
3
2
2
31 )(
'
'
'
''
or
2
332232233
1
331232133
3
233222332
1
231222132
3
133212331
2
1322122311
)()(
)()(
)()('
x
Faaaa
x
Faaaa
x
Faaaa
x
Faaaa
x
Faaaa
x
FaaaaW
or
313212111
2
1
1
213
1
3
3
112
3
2
2
311
2
311
1
312
3
211
1
213
3
112
2
1131
)()()(
'
WaWaWa
x
F
x
Fa
x
F
x
Fa
x
F
x
Fa
x
Fa
x
Fa
x
Fa
x
Fa
x
Fa
x
FaW
Therefore,
j
jijWaWi'
which means that F is a real vector. ________________________________________________________________________ 1.2.11 Successive application of (A) )( This is defined by a Laplacian,
2
2
2
2
2
22
zyx
The equation 02 is called as the Laplace equation. (B) is irrotational.
0 since
0
ˆˆˆ
)(
zyx
zyx
zyx
Thus is irrotational. (C) )( F is solenoid.
0)(
zyx FFFzyx
xxx
F
(D) Formula
FFF 2)()( ((Proof))
We use the formula given by
CBACABCBA )()()( with A , B , and C = F. Then we find
FFF 2)()( 1.2.12 Examples (A) Electromagnetic wave equation
Derivation of electromagnetic wave equation from Maxwell’s equation. James Clerk Maxwell (13 June 1831 – 5 November 1879) was a Scottish theoretical physicist and mathematician. His most important achievement was classical electromagnetic theory, synthesizing all previously unrelated observations, experiments and equations of electricity, magnetism and even optics into a consistent theory. His set of equations—Maxwell's equations—demonstrated that electricity, magnetism and even light are all manifestations of the same phenomenon: the electromagnetic field. From that moment on, all other classic laws or equations of these disciplines became simplified cases of Maxwell's equations. Maxwell's work in electromagnetism has been called the "second great unification in physics", after the first one carried out by Isaac Newton.
http://en.wikipedia.org/wiki/James_Clerk_Maxwell Maxwell’s equations in vacuum (in SI units);
BE
EJB
E
B
t
t
)(
0
00
0
Suppose that = 0 and J = 0. Then we have
)()(1
2
2
2EBEB
ttct
or
EE2
2
2
1)(
tc
.
where 00
1
c is the velocity of light.
Since
EEE 2)()( 0 E
we have
EE2
2
22 1
tc
.
This equation is called as electromagnetic wave equation. Similarly we have
BB2
2
22 1
tc
.
(B) Calculations
If )2,2,( 2 yzxzyx A , find A , )( A , and ))(( A . We use the Mathematica.
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
__________________________________________________________________ 1.2.13 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
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. 1.2.14 Surface Integral n normal vector to the surface
dad na (da; area element) Then the surface integral is defined by
SS
dad nFaF
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
1.2.15 Gauss's theorem Here we define the volume integral as
V
d
where is a scalar. (A) 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
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)
(B) 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
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))
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
1.2.16 Green's theorem
SV
dd a)()( 22
((Proof)) In the Gauss's theorem, we put
A Then we have
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
1.2.17 Stokes' theorem
x
y
A x0, y0 B
CD
1
dx
2
3
4 dy
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
.
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 )()( .
________________________________________________________________________
1.3 Curvilinear co-ordinates 1.3.1 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.
e1
e2
e3
h1dq1e1
h2dq2e2
h3dq3e3
dr
i
iiidqhdqhdqhdqhdsdsdsd eeeeeeer 333222111332211
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.
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
1.3.2 Spherical coordinete (A) 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 or
r rsin cosex rsin siney r cosez
dr e jj1
3
ds j e jj1
3
hjdqj
x
y
z
q
dq
f df
r
drr cosq
r sinq
r sinq df
rdq
er
ef
eq
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
A Sin Cos, Sin Sin, Cos,
Cos Cos, Cos Sin, Sin,
Sin, Cos, 0;
A MatrixForm
Cos Sin Sin Sin CosCos Cos Cos Sin Sin
Sin Cos 0
Ainv InverseA Simplify;
Cos Sin, Cos Cos, Sin,
Sin Sin, Cos Sin, Cos,
Cos, Sin, 0
Ainv MatrixForm
Cos Sin Cos Cos SinSin Sin Cos Sin Cos
Cos Sin 0
A.Ainv Simplify
1, 0, 0, 0, 1, 0, 0, 0, 1 The time derivatives er, e, and e are obtained as
eeer sin
eee cos r
)cos(sin eee r
We note that
0
rre
, e
er
,
eer sin
0
re
ree
,
ee
cos
.
(B)
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 . (C) A
When a vector A is defined by
eeeA AAA rr
The divergence is given by
)]()sin()sin([sin
1
)]()()([1
22
rAArArrr
AhhAhhAhhrhhh
r
rrrr
A
or
A
rA
rAr
rr r
sin
1)(sin
sin
1)(
1 22A
(D) A
A is given by
ArrAA
r
rr
rAhAhAh
r
hhh
hhhr
r
rr
rr
r
sin
sin
sin
112
eeeeee
A
(E) Laplacian
)]sin
1()(sin)sin([
sin
1
)]()()([1
22
2
rr
rr
h
hh
h
hh
rh
hh
rhhhrr
rr
or
2
2
2222
22
sin
1)(sin
sin
1)(
1
rrr
rrr
We can rewrite the first term of the right hand side as
)(1
)(1
22
2r
rrrr
rr
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
((Mathematica))
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`";
Needs"VectorAnalysis`";
SetCoordinatesSphericalr, , ;
Vector analysisGrad, Curl, Laplacian which are expressed in terms of the spherical coordin ates
eq1 Laplacianr, , Simplify
1
r2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,
eq2 Gradr, ,
1,0,0r, , ,0,1,0r, ,
r,
Csc 0,0,1r, , r
A Arr, , , Ar, , , Ar, , ;
eq3 CurlA 1
r2 Csc r Ar, , Cos r A0,0,1r, , r Sin A0,1,0r, , ,
1
rCsc Ar, , Sin Ar0,0,1r, ,
r Sin A1,0,0r, , ,
Ar, , Ar0,1,0r, , r A1,0,0r, , r
eq3 DivA1
r2 Cscr Ar, , Cos 2 r Arr, , Sin r A0,0,1r, ,
r Sin A0,1,0r, , r2 Sin Ar1,0,0r, , _______________________________________________________________________ 1.3.3 Velocity and acceleration in the spherical coordinate
The velocity (v) and acceleration (a) in the spherical co-ordinates are given by
sinrv
rv
rvr
cos2sin2sin
cossin2
sin2
222
rrra
rrra
rrrar
((Mathematica))
We drive the above formula using the Mathematica.
Velocity and acceleration in the spherical coordinates
Clear"Gobal`" "VectorAnalysis`"
SetCoordinatesCartesianx, y, zCartesian x, y, z
RRt_ : rt Sint Cost, rt Sint Sint, rt CostDRRt, t FullSimplify
Cost Sint rt Cost rt t rt Sint Sint t,
Sint Sint rt Cost rt t Cost rt Sint t,
Cost rt rt Sint t
DRRt, t, 1 FullSimplify
Cost Sint rt Cost rt t rt Sint Sint t,
Sint Sint rt Cost rt t Cost rt Sint t,
Cost rt rt Sint t
DRRt, t, 2 FullSimplify
Cost 2 t Cost rt rt Sint t Cost rt t Sint Cost rt t2 t2 rt
Sint 2 rt t rt t, SintSint rt t2 t2 rt Cost 2 rt t rt t
Cost 2 Sint rt Cost rt t t rt Sint t,
Cost rt t2 rt Sint 2 rt t rt t
DRRt, t, 3 FullSimplify
Cost 3 Sint t 2 rt t rt t rt t t Cost3 t rt 3 rt t rt t3 3 t t2 3t Sint
Cost 3 rt t2 t2 3 rt t t t t r3t Sint 3 t rt rt t rt 3 t2 t t3 3t,
Sint Sint 3 rt t2 t2 3 rt t t t t r3t
Cost 3 t rt 3 rt t rt t3 3 t t2 3t Cost 3 Cost t 2 rt t rt t rt t t
Sint 3 t rt 3 rt t rt 3 t2 t t3 3t,
Cost 3 t rt t rt t r3t Sint 3 t rt rt t rt t3 3t
Unit vectors along the r, q, and f directions (Cartesian coordinate)
ur rt RRt Simplify
Cost Sint, Sint Sint, Cost
u t RRt rt Simplify
Cost Cost, Cost Sint, Sint
u t RRt rt Sint Simplify
Sint, Cost, 0
ur.u
0
ur.u Simplify
0
ü Velocity and kinetic energy in the spherical coordinates
Vr DRRt, t.ur Simplify
rt
V DRRt, t.u Simplify
rt t
V DRRt, t.u Simplify
rt Sint t
K1 m
2Vr2 V2 V2 Simplify
1
2m rt2 rt2 t2 Sint2 t2
ü Acceleration in the spherical coordinate
Ar DRRt, t, 2.ur Simplify
rt t2 Sint2 t2 rt
A DRRt, t, 2.u Simplify
2 rt t rt Cost Sint t2 t
A DRRt, t, 2.u Simplify
2 Sint rt t rt 2 Cost t t Sint t
ü Some application
Sr DRRt, t, 3.ur Simplify
1
26 rt t2 Sint2 t2
3 rt t Sin2 t t2 2 t 2 Sint2 t t 2 r3t
S DRRt, t, 3.u Simplify
1
26 t rt rt 3 Sin2 t t2 6 t
rt 2 t3 6 Cost2 t t2 3 Sin2 t t t 2 3t
S DRRt, t, 3.u Simplify
3 Sint t rt 3 rt 2 Cost t t Sint t rt 3 Sint t2 t Sint t3
3 Cost t t 3 Cost t t Sint 3t _______________________________________________________________________ 1.3.4 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
]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
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
1.3.5 Mathematica Arfken 2-5-13
Show that
- ‰ Ñ (x ∂∂y
- y ∂∂x
) = -‰ Ñ ∂
∂f
This is the quantum mchanical operator corresponding to the z-componenet of orbital angular momentum. _______________________________________________________________________ Arfken 2-5-14
With the quantum mechanical orbial angular momentum operator defined as L = r × p= r × (-ÂÑ—), show that
(a) Lx +  Ly = - Ñ ‰Âf ( ∑∑q
+ Â cotq ∑∑f
)
(b) Lx + Â Ly = - Ñ ‰-Âf ( ∑∑q
- Â cotq ∑∑f
)
________________________________________________________________________ Arfken 2-5-15
Verify that L× L = ‰ L in spherical polar coordinates. L = -ÂÑ( r × —), the quantum mechanical orbital angular momentm operator ________________________________________________________________________
Arfken 2-5-16
(a) Show that
L = - Â Ñ (r × “) = Â Ñ (eq 1sinq
∑∑f
- ef ∑∑q
)
(b) Resolving eq and ef into Cartesiancomponents, determine Lx , Ly , and Lz in terms of q, f, and their derivatives.
(c) From Lx2 + Ly
2 + Lz2 , show that
L2
Ñ2 = - 1sinq
∑∑q
(sinq ∑∑q
) - 1sin2 q
∑2
∑q2 = - r2 “2 + ∑
∑rr2 ∑
∑r
or
—2= - L2
Ñ2 r2 + 1r2
∑∑r
r2 ∑∑r
This identity is useful in relating orbital angular momentum. ________________________________________________________________________ Arfken 2-5-17
With L = - Â Ñ (r × “) , verify the operator identities
(a) “ = e r ∑∑r
- Â r μ Lr2
(b) r “2 = - “ (1 + r ∑∑r
) = Â “ × L
________________________________________________________________________
Clear"Gobal`" "VectorAnalysis`"
SetCoordinatesSphericalr, , Spherical r, ,
ClearL — Crossur r, Grad &
— ur rGrad 1 &
Lx : ux. — Crossur r, Grad & Simplify
Ly : uy. — Crossur r, Grad & Simplify
Lz : uz. — Crossur r, Grad & Simplify
Lxr, , Simplify
— Cos Cot 0,0,1r, , Sin 0,1,0r, ,
Lyr, , Simplify
— Cot Sin 0,0,1r, , Cos 0,1,0r, ,
Arfken Problem 2-5-13
Lzr, , Simplify
— 0,0,1r, ,
Arfken Problem 2-5-14
Lx r, , Ly r, , FullSimplify
— Cot 0,0,1r, , 0,1,0r, ,
Lx r, , Ly r, , FullSimplify
— Cos Sin Cot 0,0,1r, , 0,1,0r, ,
Arfken Problem 2-5-15
Lx Lyr, , Ly Lxr, , — Lzr, , Expand FullSimplify
0
Ly Lzr, , Lz Lyr, , — Lxr, , Expand FullSimplify
0
Lz Lxr, , Lx Lzr, , — Lyr, , Expand FullSimplify
0
Arfken Problem 2-5-16 (a)
Lr, , Simplify
0, — Csc 0,0,1r, , , — 0,1,0r, ,
Arfken Problem 2-5-16 (b)
Lxr, , Simplify
— Cos Cot 0,0,1r, , Sin 0,1,0r, ,
Lyr, , Simplify
— Cot Sin 0,0,1r, , Cos 0,1,0r, ,
Lzr, , Simplify
— 0,0,1r, ,
Arfken Problem 2-5-16 (c)
seq1 LxLxr, , FullSimplify
1
4—2 3 Cos2 Csc2 Sin2 0,0,1r, ,
4 Cot Cos2 Cot 0,0,2r, , 0,1,0r, , Sin2 0,1,1r, , 4 Sin2 0,2,0r, ,
seq2 LyLyr, , FullSimplify
1
4—2
3 Cos2 Csc2 Sin2 0,0,1r, , 4 Cot SinSin Cot 0,0,2r, , 0,1,0r, ,
2 Cos 0,1,1r, , 4 Cos2 0,2,0r, ,
seq3 LzLzr, , Simplify
—2 0,0,2r, ,
seq123 seq1 seq2 seq3 Expand FullSimplify
—2
Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, ,
seq4 —2 r2 Laplacianr, , —2 Dr2 Dr, , , r, r Simplify
—2
Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, ,
seq123 seq4 Simplify
0
Arfken Problem 2-5-17(a)
Gradr, ,
—
1
r2 Crossr, 0, 0, Lr, , Simplify
1,0,0r, , , 0, 0
Arfken Problem 2-5-17(b)
—CurlLr, , r, 0, 0 Laplacianr, ,
Gradr, , r r r, , Expand FullSimplify
0, 0, 0 ______________________________________________________________________ 1.3.6 Radial momentum operator in the quantum mechanics (a) In classical mechanics, the radial momentum of the radius r is defined by
)(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))
Clear"Gobal`" "VectorAnalysis`"
SetCoordinatesSphericalr, , Spherical r, ,
Clearprc : — 1, 0, 0.Grad &prcr, , — 1,0,0r, ,
prq : —2
1, 0, 0.Grad —
2Div 1, 0, 0 &
prqr, , Simplify
— r, , r 1,0,0r, ,
r
((Commutation relation))
prqr r, , r prq r, , Simplify
— r, , ________________________________________________________________________
ü
Arfken 2-5-18
Show that the following three forms (spherical coordinates) of —2y(r) are equvalent:
(a) 1
r2
d
dr [ r2 dyr
dr]; (b)
1
r
d2
dr2 [ r y (r)]; (c)
d2 y rdr2
+ 2
r
dy rdr
Clear"Gobal`" "VectorAnalysis`"
SetCoordinatesSphericalr, , Spherical r, ,
Clear
kr : 1
rDr , r &
krkrr Expand
2 rr
r
1
rDr r, r, 2 Simplify
2 rr
r
1
r2Dr2 Dr, r, r Simplify
2 rr
r
_______________________________________________________________________ 1.3.7 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
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
We note that
e
e
,
e
e
,
0e
, 0e
.
ee , ee , 0ze (time derivative)
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
1.3.8 Differential operations in the cylindrical coordinate The differential operations involving are as follows.
zz
eee
1
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. 1.3.9 Mathematica
We use the cylindrical co-ordinate. We need a Vector Analysis Package. We also need SetCordinatinates.In this system the vector is expressed in terms
of (Ar, Af, Az)
Clear"Gobal`"Needs"VectorAnalysis`"SetCoordinatesCylindrical, , zCylindrical, , z
Vector analysisGrad, Curl, Laplacian which are expressed in terms of the cylindrical coordinates
eq1 Laplacian, , z Simplify
0,0,2, , z 0,2,0, , z2
1,0,0, , z
2,0,0, , z
eq2 Grad, , z
1,0,0, , z,0,1,0, , z
, 0,0,1, , z
B B, , z, B, , z, Bz, , zB, , z, B, , z, Bz, , z
eq3 CurlB
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
eq3 DivB1
B, , z Bz0,0,1, , z B0,1,0, , z B1,0,0, , z
1.3.10 Velocity and acceleration in the cylindrical coordinates
The velocity (v) and acceleration (a) in the cylindrical co-ordinates are given by
zv
v
v
z
za
a
a
z
2
2
((Mathematica))
We drive the above formula using the Mathematica.
Velocity and acceleration in the cylindrical coordinates
Clear"Gobal`" "VectorAnalysis`"
SetCoordinatesCartesianx, y, zCartesianx, y, z
RRt_ : t Cost, t Sint, ztDRRt, t FullSimplify
Cost t Sint t t,Sint t Cost t t, zt
DRRt, t, 2 FullSimplify
Cost t t2 t Sint 2 t t t t,
2 Cost t t Sint t t2 t Cost t t, zt
DRRt, t, 3 FullSimplify
Cost 3 t t t t t 3t Sint 3 t t t t t t3 3t,
Sint 3 t t t t t 3t Cost 3 t t 3 t t t t3 3t, z3t
Unit vectors along the r, f, and z directions
u t RRt Simplify
Cost, Sint, 0
u t RRt t Simplify
Sint, Cost, 0
uz zt RRt Simplify
0, 0, 1
ü Velocity and kinetic energy in the cylindrical coordinates
V DRRt, t.u Simplify
t
V DRRt, t.u Simplify
t t
Vz DRRt, t.uz Simplify
zt
K1 m
2V2 V2 Vz2 Simplify
1
2m zt2 t2 t2 t2
ü Acceleration in the spherical coordinate
A DRRt, t, 2.u Simplify
t t2 t
A DRRt, t, 2.u Simplify
2 t t t t
Az DRRt, t, 2.uz Simplify
zt
ü Some application
S DRRt, t, 3.u Simplify
3 t t2 3 t t t 3t
S DRRt, t, 3.u Simplify
3 t t t t t t3 3t
Sz DRRt, t, 3.uz Simplify
z3t ______________________________________________________________________
1.3.11 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)) 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`" "VectorAnalysis`"
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
1.3.12 Plane polar coordinate for 2D system
The point P is located at (r, ), where r is the distance from the origin and is the measured counterclockwise from the reference line (the x axis).
x
y
P
r
q
ereq
x
y
ereq
We introduce the unit vectors given by
yx
yxr
ee
e
eee
cossin
)cos,sin()2
sin(),2
(cos(
sincos)sin,(cos
.
These expressions can be rewritten using a matrix A as
y
x
y
xr Ae
e
e
e
e
e
cossin
sincos,
and using A-1 as
e
e
e
e
e
e rr
y
xA
cossin
sincos1 .
Note that
e
e
r re
e
The position vector (displacement vector) is given by
yx rrrr eer sincos)sin,cos(
The velocity and acceleration are
eea
eev
)2()( 2
rrrr
rr
r
r
or
rv
rrvr
ˆ
ˆ
v
v
)(1
2ˆ
ˆ
2
2
rdt
d
rrra
rrrar
a
a
((Note)) Velocity along the e direction
r
dt
dsv
rdds
Velocity along the re direction
rdt
drvr .
x
y
P
r
q
dq
drds
((Note))
ˆˆ)cos,sin()sin,(cos)cossin,sincos(
)sin,cos(
rrrrrrrrr
rr
rv
r
sicoscoscossin,cossinsinsincos( 22 rrrrrrrrrr ra
or
)sincoscos2sin,cossinsin2cos( 22 rrrrrrrr r or
jrrrrirrrr ˆ)sincoscos2sin(ˆ)cossinsin2cos( 22 r
ˆ)2(ˆ)()ˆcosˆsin)(2()ˆsinˆ)(cos( 22 rrrrrjirrjirr r ((Mathematica))
R rt Cost, rt SintCost rt, rt Sint
V DR, t Simplify
Cost rt rt Sint t,
Sint rt Cost rt t
A DR, t, 2 Simplify
2 Sint rt t Cost rt rt Cost t2 Sint t, 2 Cost rt t Sint rt rt Sint t2 Cost t
ru Cost, SintCost, Sint
u Sint, CostSint, Cost
A.ru Simplify
rt t2 rt
A.u Simplify
2 rt t rt t
V.ru Simplify
rt
V.u Simplify
rt t 1.13.13 Angular momentum The angular momentum is defined by
)()()]([)( 2 eeeeeeevrprL rrrrr mrmrvvvrmm
1.13.14 Circular motion (r = constant)
We consider a circular motion with r = constant. since 0r and 0r .
raa
ra
t
r
2
rvv
v
t
r
0
In summary, we have
vrvv t
dt
dvraa
r
vra
t
r
2
2