chapter 8 向量分析 ( vector analysis)

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung

Chapter 8 向量分析 (Vector Analysis)

物理量與符號物理量 :

1. 純量 (scalar quantity): 有大小 , 無方向

例如 : 質量 (mass), 溫度 (temperature), 壓力 (pressure), 能量 (energy)

2. 向量 (vector quantity): 有大小以及方向

例如 : 速度 (velocity), 動量 (momentum), 力矩 (torque)

向量符號 :

1. 一般向量 : 長度

2. 單位向量 : 長度 1

A

x

A



向量的基本運算1. 向量之相等 : 包括大小以及方向的相等 ,

2. 向量之反向 : 大小相等但方向相反 ,

3. 向量之合成 :

4. 向量之倍數 :

5. 向量之純量積 (scalar product): 功 (work) 的計算

6. 向量之向量積 (vector product): 力矩 (torque) 的計算

BA

A

BAC

An

SFW

Fr

向量之合成 :Commutative :

Associative :

ABBAC

)CB(AC)BA(



向量的座標表示法

xA

A

yA

zA

)z,y,x(AAAA

終點表示法 :

分量表示法 : zzyyxxAAAA

單位向量表示法 :

Au

AuAA

2

A

2

A

2

AzyxA

2

A

2

A

2

A

AAA

A zyx

zzyyxx

A

Au

x

y

z



座標軸的轉動 (Rotation of the Coordinate Axes)

X

Y

r

x

y

φ

X’

Y’

x’

y’φ

sinycosxx '

cosysinxy '

2222'2' ryxyx




sinycosxx ' cosysinxy '

1xx 2

xy

Let cosa11 sina

12

sina21

cosa22

212111

'

1xaxax 222121

'

2xaxax

The coefficient aij is the cosine of the angle between xi’ and xj

2

1jjij

'

ixax 2,1i

N dimensions

N

1jjij

'

iVaV N,...,2,1i




The coefficient aij is the cosine of the angle between xi’ and xj

2

1jjij

'

ixax 2,1i

j

'

i

ij x

xa

Using the inverse rotation :

yields

2

1i

'

iijjxax or

ij'

i

j ax

x

N

1jj'

i

jN

1jj

j

'

i'

iV

x

xV

x

xV

N

1jjij

'

iVaV



The orthogonality condition for the direction cosines aij:

jkikiji

aa or jkkijii

aa

k

j

k

'

i

'

i

j

i'

i

k

'

i

j

i x

x

x

x

x

x

x

x

x

x

The Kronecker delta is defined byjk

1jk for j = k

0jk for j k



Vector and vector space

Vector : )x,x,x(x321

)y,y,y(y321

and

1. Vector equality : means xi = yi , i = 1,2,3.yx

zyx 2. Vector addition : means xi + yi = zi , i = 1,2,3.

3. Scalar multiplication : (with a real). )ax,ax,ax(xa321

4. Negative of a vector : )x,x,x(x)1(x321

5. Null vector : there exists a null vector )0,0,0(0



Scalar (Dot) Product

The projection of a vector onto a coordinate axis is a special case of the scalar product of and the coordinate unit vectors :

A

A

xAcosAAx

yAcosAAy

zAcosAAz

zAByABxAB)zByBxB(ABAzyxzyx

ABBAABABABABiiiiiizzyyxx

The scalar product is commutative :

z

x

y

A

B

θ

Definition :

cosABABBABAAB



Distributive Law in the Scalar (Dot) Product

AAA)CB(AACABCABA)CB(A

A

C

B

CB

AC

AB

)CB(AA



Normal vector

n

)yyxx(r

n

r

is a nonzero vector in the x-y plane 0rn

x

y



Invariance of the scalar product under rotations

'

k

'

kk

'

z

'

z

'

y

'

y

'

x

'

xBABABABA

jzjj

izii

jyjj

iyii

jxjj

ixii

BaAaBaAaBaAa

jljilijil

'

k

'

kk

BaAaBA

(using the indices k and l to sum over x,y, and z)

iii

jiijji

jiljlijil

BABA

BA)aa(

iii

'

k

'

kk

BABA

take BAC

)BA()BA(CC

BA2BBAA

)BAC(2

1BA 222

invariant

invariant2AAA

Scalar quantity



Vector (Cross) Product

Definition :

z)BABA(y)BABA(x)BABA(

BBB

AAA

zyx

BACxyyxzxxzyzzy

zyx

zyx

zCyCxCzyx

jkkjiBABAC i,j,k all different and with cyclic permutation of the indices i,j, and k

Magnitude of :C

2222 )BA(BA)BA()BA(CCC 22222222 sinBAcosBABA

Prove it!

sinABC



Parallelogram representation of the vector product

sinABC BAC

x

y

θ

A

B

Bsinθ

ABBA

anticommutation



the vector product under rotations

'

j

'

k

'

k

'

j

'

iBABAC i,j, and k in cyclic order

mjmm

lkll

mkmm

ljll

BaAaBaAa mljmklkmjlm,l

BA)aaaa(

jkikiji

aa If i = 3, then j = 1, k =23312212211

aaaaa

3211232113aaaaa

3113222312aaaaa

l m

233131321233323113322133

'

3BAaBAaBAaBAaBAaBAaC

nn3n

333232131CaCaCaCa

N

1jjij

'

iVaV

C

is indeed a vector !



Triple Scalar Product

)CBCB(A)CBCB(A)CBCB(A)CB(Axyyxzzxxzyyzzyx

)BA(C)AC(B

)CA(B)AB(C)BC(A

The dot and the cross may be interchanged :

CBA)CB(A

zyx

zyx

zyx

CCC

BBB

AAA

scalar



Parallelepiped representation of triple scalar product

x

y

z

A

C

B

CB

)CB(A

Volume of parallelepiped defined by , , and A

B

C



Construction of a reciprocal crystal lattice

Let , , and (not necessarily mutually perpendicular) represent the vectors that define a crystal lattice. The distance from one lattice point to another may be written as

a

b

c

cnbnanrcba

With these vectors we may form the reciprocal lattices :

cba

cba '

cba

acb '

cba

bac '

We see that is perpendicular to the plane containing and and has a magnitude proportional to .

'a

b

c

1a

1ccbbaa '''

0bcaccbabcaba ''''''

Fourier space



Triple Vector Product

)CB(A

CB

x

y

z

)CB(A

C

A

B

CyBx)CB(A

CAyBAx0)]CB(A[A

CAzx BAzy

)BACCAB(z)CB(A

)BACCAB(

BAC-CAB rule



Proof : z = 1 in )BACCAB(z)CB(A

Let us denote cosCB cosAC cosBA

2222 )]CB(A[)CB(A)]CB(A[ 2222 )BA(BA)BA(

22 )]CB(A[cos1 2222 )CB(CB)CB(

]CBCABA2)BA()CA[(z 222

)coscoscos2cos(cosz 222

)coscoscos2cos(coszcos1)]CB(A[ 22222

The volume is symmetric in αβ,γ z2 = 1 z = ± 1

For the special case y)yx(x 1z



Gradient Suppose that φ(x,y,z) is a scalar point function which is independent of the rotation of the coordinate system.

)x,x,x()x,x,x(321

'

3

'

2

'

1

'

jj

ijj

'

i

j

j

'

i

321

'

i

'

3

'

2

'

1

'

xa

x

x

xx

)x,x,x(

x

)x,x,x(

N

1jj'

i

j'

iV

x

xV

We construct a vector with components : j

x

zz

yy

xx

or

zz

yy

xx

A vector differential operator

)potential(force


is a vector having the direction of the maximum space rate of change of φ.


A Geometrical Interpretation

P Qdr

z

x

y

y

z

x

φ(x,y,z)= C

φ= C2 > C1

φ= C1P

Q

zz

yy

xx

dzzdyydxxrd

0ddzz

dyy

dxx

rd

is perpendicular to rd

rd)(CCCd12

For a given dφ, is a minimum when it is chosen parallel to (cosθ = 1).

rd

For a given , is a maximum when is chosen parallel to .

rd d rd



Exercise : 試求曲面上一點 (2,2,8) 之切面與法線方程式 (88 台大造船 )

)yx(8z 222

Solution : 取 , 而曲面在之法向量為 :k8j2i2r0

0r

N

kz2jy16ix16)zy8x8( 222

kj2i2k16j32i32N

根據直線與平面之點向式 :

切面 :

法線 :

0zy2x20)8z()2y(2)2x(2N)rr(0

1

8z

2

2y

2

2x0N)rr(

0



Example : Calculate the gradient of f(r) = )zyx(f 222

z

)r(fz

y

)r(fy

x

)r(fx)r(f

x

r

dr

)r(df

x

)r(f

r

x

)zyx(

x

x

)zyx(

x

r2/1222

2/1222

r

x

dr

)r(df

x

)r(f

r

y

dr

)r(df

y

)r(f

r

z

dr

)r(df

z

)r(f

dr

dfr

dr

df

r

r

dr

df

r

1)zzyyxx()r(f



Divergence

x

y

)t(r

)tt(r

r

vt

rlim

0t

Differentiating a vector function

vt

)t(r)tt(r

dt

rdlim

0t

: vector

: differential property

z

V

y

V

x

VV zyx

Scalar

Vector

z

VfV

z

f

y

VfV

y

f

x

VfV

x

f)fV(

z)fV(

y)fV(

x)Vf( z

z

y

y

x

xzyx

VfV)f()Vf(



Example : Calculate the divergence of f(r) r

)]r(zf[z

)]r(yf[y

)]r(xf[x

))r(fr(

dr

)r(df

r

z

dr

)r(df

r

y

dr

)r(df

r

x)r(f3

222

dr

)r(dfr)r(f3

if 1nr)r(f

1n2n1n1n r)2n(r)1n(rr3)rr(



A Physical Interpretationz

x

y

G

C

A

E

B

F

H

D

dy

dz

dx

Consider )v( )z,y,x(v

: the velocity of a compressible fluid

)z,y,x( : the density of a compressible fluid

The rate of flow in (EFGH) = dydzv0xx

The rate of flow out (ABCD) = dydzvdxxx

dydz]dx)v(x

v[0xxx

Expand in a Maclaurin seriesNet rate of flow out|x = dxdydz)v(

x x

Net rate of flow out = dxdydz))v((dxdydz)]v(z

)v(y

)v(x

[zyx

)v( : the net flow of the compressible fluid out of the volume element dxdydz per unit

volume per unit time. (divergence)

The continuity equation : 0)v(t

ρ(x,y,z,t)



Exercise : For a particle moving in a circular orbit tsinrytcosrxr

(a) Evaluate (b) Show that (The radius r and the angular velocity are constant)

rr

0rr 2

tsinrytcosrxr tcosrytsinrxr

rtsinrytcosrxr 222

(a)

)tcosrytsinrx()tsinrytcosrx(rr

22222 rz)tsinrtcosr(z

(b) tsinrytcosrxr

0rr 2 Proof !



Curl Definition :

zyx

xyzxyz

VVVzyx

zyx

)Vy

Vx

(z)Vx

Vz

(y)Vz

Vy

(xV

)Vz

f

z

VfV

y

f

y

Vf()]fV(

z)fV(

y[)Vf(

y

y

z

z

yzx

xxV)f(Vf

V)f(Vf)Vf(



A Physical Interpretation

x

y

x0+dx, y0

x0+dx, y0+dyx0, y0+dy

x0, y0 1

2

3

4

Circulation around a differential loop

4

yy3

xx2

yy1

xx1234d)y,x(Vd)y,x(Vd)y,x(Vd)y,x(Vncirculatio

dxdy)y

V

x

V(

)dy)(y,x(V)dx](dyy

V)y,x(V[

dy]dxx

V)y,x(V[dx)y,x(V

xy

00y

x

00x

y

00y00x

circulation per unit areaz

V

Vorticity ( 渦度向量 ):V

0V

V

is labeled irrotational



A Physical Interpretation

0V

V

is labeled irrotational

(the gravitational and electrostatic forces)

32 r

rC

r

rCV

21mGmC Newton’s law of universal gravitation

0

21

4

qqC

Coulomb’s law of electrostatics

V)f(Vf)Vf(

Calculate: ))r(fr(

r)]r(f[r)r(f))r(fr(

0

zyxzyx

zyx

r

)dr

df(r)r(f

0rr)dr

df(



Example : verify that

)B(A)A(BB)A(A)B()BA(

Gradient of a Dot Product

BACCAB)CB(A BAC-CAB rule

B)A()BA()B(A

A)B()BA()A(B

)B(A)A(BB)A(A)B()BA(



Successive Applications of

The divergence of the gradient : the Laplacian of

)z

zy

yx

x()z

zy

yx

x(

2

2

2

2

2

2

zyx

When φ is the electrostatic potential 0

Laplace’s equation of electrostatics

2

in the European literature



Example : Calculate )r(g

Example :dr

)r(dgr)r(g )

dr

dgr()r(g

Example :dr

)r(dfr)r(f3))r(fr(

replacingdr

)r(dg

r

1)r(f

dr

)dr

)r(dgr1

(dr

dr

)r(dg

r

3)

dr

dgr()r(g

2

2

2

2

2 dr

)r(gd

dr

)r(dg

r

2]

dr

)r(gd

r

1

dr

)r(dg

r

1[r

dr

)r(dg

r

3

If nr)r(g 2n2n2n

2

n2n

n r)1n(nr)1n(nnr2dr

rd

dr

dr

r

2)r(

ttancons)r(g n = 0 0)r(g

r

1)r(g n = -1 0)r(g

A consequence of physics



Successive Applications of : The curl of the gradient

zyx

zyx

zyx

0)xyyx

(z)zxxz

(y)yzzy

(x222222

All gradients are irrotational A mathematical identity !



Successive Applications of : V

The divergence of a curl

zyxVVVzyx

zyx

V

)y

V

x

V(

z)

z

V

x

V(

y)

z

V

y

V(

xxyxzyz

0zy

V

zx

V

zy

V

yx

V

zx

V

yx

Vx

2

y

2

x

2

z

2

y

2

z

2

All curls are solenoidal



Successive Applications of :

VV)V(

BACCAB)CB(A

BAC-CAB rule

Example : Maxwell’s equation (in vacuum)

0B

0E

t

EB

00

t

BE

t

BB

t

)E(t

E2

2

00

EEE

2

2

00 t

EE

The electromagnetic vector wave equation



Successive Applications of :

Exercise : 試證明 P)a(a)P(P)a(a)P()Pa(

(74 台大材料 , 清華材料 )

)Pa()Pa()Pa(Pa

P)a(a)P(P)a(a)P(PPaa

BACCAB)CB(A

P)a(a)P(P)a(a)P(

chapter 8 向量分析 ( vector analysis)

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