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


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


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


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


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


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 + Â 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

________________________________________________________________________


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, ,


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, ,


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))



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

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


SetCoordinatesCartesianx, y, zCartesianx, y, z

RRt_ : t Cost, t Sint, ztDRRt, t FullSimplify

Cost t Sint t t,Sint t Cost t t, zt


Cost t t2 t Sint 2 t t t t,

2 Cost t t Sint t t2 t Cost t t, zt


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


t


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


t t2 t


2 t t t t

Az DRRt, t, 2.uz Simplify

zt

ü Some application


3 t t2 3 t t t 3t


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.


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

Chapter 1 Vector Analysis Masatsugu Sei Suzuki Department...

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Transcript of Chapter 1 Vector Analysis Masatsugu Sei Suzuki Department...