chapter 6 vector analysis ( 벡터 해석 )
DESCRIPTION
Mathematical methods in the physical sciences 3rd edition Mary L. Boas. Chapter 6 Vector analysis ( 벡터 해석 ). Lecture 18 Basic vector analysis. 1. Introduction. Vector function, Vector calculus, ex. Gauss’s law. 2. Application of vector multiplication ( 벡터곱의 응용 ). 1) Dot product. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 6 Vector analysis ( 벡터 해석 )
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 18 Basic vector analysis
1. Introduction
Vector function, Vector calculus, ex. Gauss’s law
2. Application of vector multiplication ( 벡터곱의 응용 )
zzyyxx BABABAABBA cos
sin , ABBA
BBB
AAA
kji
BA
zyx
zyx
BA
BA
a) Work
rdFdW
dFFdW
�
cos
b) Torque Fr
rv
v
sinrr
c) Angular velocity
1) Dot product
2) Cross product
- Example
1) Triple scalar product ( 삼중 스칼라곱 )
3. Triple products ( 삼중곱 )
zyx
zyx
zyx
CCC
BBB
AAA
CBA )(
cf. volume of unit cell for reciprocal vectors
321
323
321
132
321
321 2 ,2 ,2
aaa
aab
aaa
aab
aaa
aab
“Volume of the parallelepipe”
cosAheight
BCA
BAC
CBACBA
)(
)(
)()(
zyx
zyx
zyx
CCC
BBB
AAA
CBA )(
)()( CBACBA
So, it does not matter where the dot and cross are.
- An interchange of rows changes just the sign of a determinant.
)( CBA
CbBaCBA�
)(
CBABCA
)()(
some vector in the plane of B and C
2) Triple vector product ( 삼중 벡터곱 )
Prove this!
(Vector equation is true independently of the coordinate system.)
kji
ji
i
zyx
yx
x
AAAA
CCC
BB
3) Application of the triple scalar product
“Torque”
Fr
This question is in one special case, namely when r and F are in a plane perpendicular to the axis.
nFrn
4) Application of the triple vector product
v
sinrr
)( rrm
vrm
prL
Angular momentum
Centripetal acceleration )( ra
4. Differentiation of vectors ( 벡터의 미분 )
)( zyx AAAA
kjidt
dA
dt
dA
dt
dA
dt
Ad zyx
Example 1.
2
2
2
2
2
2
2
2
,
,
dt
zd
dt
yd
dt
xd
dt
rd
dt
vda
dt
dz
dt
dy
dt
dx
dt
rdv
zyxr
kji
kji
kji
1) Differentiation of a vector
)order! of (careful )(
,)(
,)(
Bdt
Ad
dt
BdABA
dt
d
Bdt
Ad
dt
BdABA
dt
d
dt
AdaA
dt
daAa
dt
d
2) Differentiation of product
Example 2. Motion of a particle in a circle at constant speed
.
.,2
2
constvvv
constrrr
Differentiating the above equations,
0or 02
,0or 02
avdt
vdv
vrdt
rdr
“two vectors are perpendicular”
r
va
avvrvar
var
vvar
vr
2
2
2
,0 & 0
0 this,atingDifferenti
,0
3) Other coordinates (e.g., polar)
coord.polar ),(
coord.r rectangula),(
θr ee
ji
:),( ji
:),( θr ee
constant in magnitude and direction
constant in magnitude, but directions changes
cossin
sincos
jie
jie
θ
r
.sincos
,cossin
dt
d
dt
d
dt
d
dt
ddt
d
dt
d
dt
d
dt
d
rθ
θr
ejie
ejie
Example 3. ? , dt
Ad AAA r
θr ee
.dt
dA
dt
dA
dt
dA
dt
dA
dt
Ad
dt
dA
dt
dA
dt
dA
dt
dA
dt
Ad
rr
rr
rθθr
θθ
rr
eeee
ee
ee
Chapter 6 Vector analysis
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 19 Directional derivative; Gradient
5. Fields ( 장 )
Field: region + the value of physical quantity in the region ex) electric field, gravitational field, magnetic field
6. Directional derivative: gradient ( 방향 도함수 ; 기울기벡터 )
),,( zyxT
sT for The change of temperature with distance depends on the direction. directional derivative
functionscalar : ),,( zyx
u
kji
ds
d
zyx,grad
(directional derivative for u: directional unit vector)
ds
dT
1) definition of directional derivative
Example 1. Find the directional derivative
)1,2,2(direction
(1,2,-1)at 2
A
xzyx
)1,2,2(3
1
A
A
u
3
5
)1,1,3()1,2,1(
,)2( 2
u
kjikji
xy
xzxyzyx
2) Meaning of gradient : along it the change (slope) is fastest (steepest).
3) Relation between scalar function and gradient
“The vector grad. is perpendicular to the surface =const.”
u
0s
lim
0s
,0
0s
Example 3. surface x^3y^2z=12. find the tangent plane and normal line at (1,-2,3)
kjikji 4123623 23322
23
yxyzxzyxw
zyxw
1
3
3
2
9
1
,0)3()2(3)1(9
zyx
xyx
4) other coordinates (e.g., polar)
rryx
1θr eeji
cf. Cylindrical & Spherical coord.
φθr ˆsin
1ˆ1ˆ
T
r
T
rr
TT
zφr ˆˆ1
ˆz
TT
rs
TT
cylindrical
spherical
7. Some other expressions involving grad. ( 을 포함하는 다른 표현들 )
zyxzyx
zyx
kjikji
kji
)(
1) vector operator
2) divergence of V
z
V
y
V
x
V
VVVzyx
VV
zyx
zyx
),,(),,(div
3) curl of V
zyx
zyx
VVVzyx
VVVzyx
VV
kji
),,(),,(curl
4) Laplacian
2
2
2
2
2
2
2
),,(),,(
grad
zyx
zyxzyx
div
conductionheat ofequation or equation diffusion theis 1
equation. wave theis 1
equation.Laplace' is 0
22
2
2
22
2
ta
ta
5) and etc.
VV
VVV
2)(
)()()(
)()( VVV
Chapter 6 Vector analysis
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 20 Line integral & Green’s theorem
8. Line integrals ( 선적분 )
integrating along a given curve. only one independent variable!
1) definition
rF d
Example 1. F=(xy)i-(y2)j, find the work from (0,0) to (2,1)
)( 2
2
dyyxydxW
dyyxydxrdF
dydxrd
rdFdW
ji
path 1 (straight line)
dxdyxy2
1 ,
2
1 1
2
1
2
1
2
12
0
22
0
21
dxxxdxxdyyxydxW
path 2 (parabola) xdxdyxy2
1 ,
4
1 2
3
2
2
1
4
1
4
12
0
222
2
0
22
xdxxdxxxdyyxydxW
path 3 (broken line)
0dx
0dy
1
0
2
3
1)00(
ydyyy
3
52
3
1 ,2)011( 3
2
0
Wdxxy
path 4 (parameter) x=2t^2, y=t^2
x: (0,2) t: (0,1)
6
72242
1
0
222224 tdtttdtttdyyxydxW
1)
2)
1)
2)
Example 2. Find the value of
22 yx
ydxxdyI
path 1 (polar coordinate ) r=1 (constant) so, only d may be considered.
d
dd
yx
ydxxdy
1
)sin(sin)(coscos22
1 ,cos ,sin
sin ,cos22
yxddyy
ddxx
0
1 dI
path 2
1) 2)
(0,1)
(-1,0) (1,0)
1xyxy 1
2)12arctan(
)1(
)1( 0
1
0
12222
x
xx
dxxxdx
yx
ydxxdy
2)12arctan(
)1(
)1( 1
0
1
02222
xxx
dxxxdx
yx
ydxxdy
2I
## Question: Would you compare between example 1 and 2?
2) Conservative fields (F or V) ( 보존장 )
- Example 1 : depends on the path. nonconservative field- Example 2 : does not depend on the path. conservative field
field veconservatifor condition sufficient andnecessary ,0curl F
0 this,From
,,similarly and ,
, Using
,,
2
22
F
x
F
z
F
y
F
z
F
x
F
xy
W
y
F
xy
W
yx
W
z
WF
y
WF
x
WF
z
W
y
W
x
WWF
zxzyyx
zyx
kji
.for which findcan we,0 if ,Conversely
0, If
WFWF
FWF
3) Potential () ( 퍼텐셜 )
potentialscalar :)(
field veconservati : ,
W
FF
B
A
sdF
for A: a proper reference point
cf. Electric field, gravitational field conservative
Example 3. Show that F is conservative, and find a scalar potential.
kji )13()2( 223 xzxzxyF
0
132 223
xzxzxyzyx
F
kji
1) F is conservative.
B
A
B
A
dzxzdyxdxzxyrdFW )13()2( 223
(0,0,0)
(x,y,z)
a. find the point where the field (or potential) is zero.b. do line integral to an arbitrary point along the path with which the
integration is easiest.
i) dxii) dy
iii) dz
(x,0,0)
(x,y,0)
i) only dx
0)(
0 ,0
0
x
x
W
dzdyzy
ii) only dy
y
yxdyx
dzdxzconstx
0
22
0,0.,
iii) only dz
z
zxzdzxz
dydxconstyx
0
32 )13(
0.,,zxzyxW 32
2) Scalar potential of F
Example 4. scalar potential for the electric field of a point charge q at the origin
r
q
r
q
r
rdrq
rdrrdrrrd
r
rdrqrdE
rr
q
r
r
r
q
r
qE
rr
3
r to3
r to
322
22)(For
re
AdAAdAcf
.
9. Green’s theorem in the plane ( 평면에서의 Green 정리 )
- The integral of the derivative of a function is the function.
)()()( afbfdxxfdx
db
a
1) Definition of Green theorem
sderivative partial first continuous withfunction a : ),( ),,( yxQyxP
dyyaQybQdxdyx
yxQdxdy
x
yxQ d
cy
d
c
b
axA
)],(),([),(),(
Area integral:
Line integral:
d
c
c
dC
d
c
dyyaQybQ
dyyaQdyybQdyyxQ
)],(),([
),(),(),(
CA
Qdydxdyx
Q
cf.
dxcxPdxPdxdyy
yxPdxdy
y
yxP d
cy
b
a
b
axA
)],(),([),(),(
Similarly,
d
a
a
bC
a
a
dxdxPcxP
dxdxPdxcxPdxyxP
)],(),([
),(),(),(
CA
Pdxdxdyy
P
theoremsGreen' ,)(
AC
dxdyy
P
x
QQdyPdx
This relation is valid even for an irregular shape!!
“Using Green’s theorem we can evaluate either a line integral around a closed path or a double integral over the area inclosed, whichever is easier to do.”
Example 1. F=xyi-y2j, find the work from (0,0) to (2,1) and back
)( 2
2
dyyxydxW
dyyxydxrdF
dydxrd
rdFdW
ji
For a closed path, 132 WW
path 2 (parabola) xdxdyxy2
1 ,
4
1 2
3
2
2
1
4
1
4
12
0
222
2
0
22
xdxxdxxxdyyxydxW
(previous section)
path 3 (broken line)
0dx
0dy
1
0
2
3
1)00(
ydyyy
3
52
3
1 ,2)011( 3
2
0
Wdxxy
1)
2)1)
2)
Example 1. F=xyi-y2j, find the work from (0,0) to (2,1) and back
)( 2
2
dyyxydxW
dyyxydxrdF
dydxrd
rdFdW
ji
For a closed path, 132 WW
Using Green’s theorem,
1
0
2
0
22
1
)()(
y
y
xA
AA
xdxdyxdxdy
dxdyxyy
yx
dyyxydxW
AC
dxdyy
P
x
QQdyPdxcf )(.
Example 2. dxdyy
F
x
FdyFdxFW x
A
y
A
yx )()(
0 If
y
F
x
Fxy ( z-component of curl F = 0),
then, W from one point to another point is independent of the path. (F : conservative field)
AC
dxdyy
P
x
QQdyPdxcf )(.
- Two useful way to apply Green’s theorem to the integration of vector functions
yxyx VVVPVQ jiV where, ,
0 with ,div
zyx V
y
V
x
V
y
P
x
QV
0
wherenormal), (outward
(tangent)
22
dsd
dydxdsdxdyds
dydxd
nr
jin
jir
dsdxdyVVdyVdxVQdyPdx yxxy nVjiji )()(
AA
dsdxdy nVVdiv
Divergence theorem
ddxdydz nVVdiv
a) Divergence theorem
AC
dxdyy
P
x
QQdyPdxcf )(.
b) Stoke’s theorem
yxxy VVVPVQ jiV where, ,
0 with ,)(curl
zxy V
y
V
x
V
y
P
x
QkV
rVjiji ddydxVVdyVdxVQdyPdx yxyx )()(
AA
ddxdy rVkV)(curl
Stoke’s theorem
rVnV dd)(curl
AC
dxdyy
P
x
QQdyPdxcf )(.
Chapter 6 Vector analysis
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 21 Divergence and Divergence theorem
10. Divergence and divergence theorem ( 발산과 발산정리 )
z
V
y
V
x
V
VVVzyx
VV
zyx
zyx
),,(),,(div
flow of a gas, heat, electricity, or particles
vV
nV
coscos
cos
))()((
Vv
AvtAvt
Avt
: flow of water
amount of water crossing A’ for t
1) Physical meaning of divergence
),,( zyx VVVV
- Rate at which water flows across surface 1 dydzV x )1(
- Rate at which water flows across surface 2 dydzV x )2(
- Net outflow along x-axis dydzdxx
VdydzVV x
xx
)]1()2([
axis-z along ,
axis-y along ,
dxdydzz
V
dzdxdyy
V
z
y
In this way,
dxdydzdxdydzdxdydzz
V
y
V
x
V zyx VV
div
“Divergence is the net rate of outflow per unit volume at a point.”
(a) positive divergence for positive charge (or negative divergence for a negative charge)(b) zero divergence(c) positive divergence along the z-axis
cf. (from ‘Griffiths’)
Example 1.4 in Figure 1.18
s?divergenceir then the..............
ˆ
,ˆ
,ˆˆˆ
zv
zv
zyxrv
z
zyx
c
b
a
.1
0
3
c
b
a
v
v
v
cf. (from ‘Griffiths’)
= (source density) minus (sink density)
= net mass of fluid being created (or added via something like a minute
sprinkler system) per unit time per unit volume
= density of fluid = mass per unit volume
/t = time rate of increase of mass per unit volume
Rate of increase of mass in dxdydz = (rate of creation) minus (rate of outward flow)
V
V
t
dxdydzdxdydzdxdydzt
1) If there is no source or sinks, 0
continuity ofEquation ,0
t
V
V,0 If
t
cf. 0div ,div BD
2) Example of the divergence 1
2)
Consider any closed surface.
ddrd sin2
Mass of fluid flowing out through d is Vn d.
Total outflow: dnV
For volume element d=dxdydz, the outflow from d is dV
d
od
d
dd
dd
of surface
of surface
1lim nVV
nVV
another definition of divergence
3) Example of the divergence 2
inclosingsurface volume
dd nVV
4) Divergence theorem
5) Example of the divergence theorem
? , dzyx nVkjiV
haddd
z
z
y
y
x
x
dd
3
cylinder of volume
cylinder of surface
33
,3
VnV
V
VnV
If we directly evaluate dnV
rji
n
k
a
yx :surface curved
:top
6) Gauss’s law
electric field at r due to a point chage q at (0,0)
unit) mksin in vacuum 1091/4 constant, dielectric : (
law sCoulomb' 4
9
2
reE
r
q
reDED24
,r
qε
qddrr
qDdAdDd
dAr
ddd
4
1
4cos
1sin
22
2
nD
) inside ( 444
angle solidtotal
σ surface closed
qqπ
qd
π
qdσ
nD
i
ii
i
qdσdσ
σ surface closed
σ surface closed
nDnDFor multi-sources,
surface closed theinside charge total
σ surface closed
dσnD
by boundedvolume
σ surface closed
ddσ
nD Gauss’s Law
by boundedvolume
by boundedvolume
σ surface closed
ddσ
DnDUsing the divergence theorem,
ext D
7) Example of gauss’s law. E=?
a) For electrostatic problem, E=0 insideb) For symmetry, E should be vertical.
area) surface( , DnDDnD d
total charge inside is C(surface area) for C surface charge density.
C
C
D
D area) (surfacearea) surface(
Chapter 6 Vector analysis
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 22 Curl and Stoke’s theorem
11. Curl and Stoke’s theorem ( 회전이론과 Stoke’s 정리 )
zyx
zyx
VVVzyx
VVVzyx
VV
kji
),,(),,(curl
v
sinrr
0 since
)()(
3)(
)()()(
x
z
x
y
zyxzyx
z
z
y
y
x
x
zyxzyx ωkjikjikjirω
ωωrω
rωωrrωv
rωv
ωrωv 2)( “Curl v gives the angular velocity.”
Ex.
rV dn circulatio
curl V 0
curl V 0
curl V 0
curl V = 0
curl V = 0
1) meaning of curl
vs.
cf. (from ‘Griffiths’)
Example 1.5
curls? their then,................
1.19b) (Fig. ˆ
1.19a) (Fig. ˆˆ
yv
yxv
x
xy
b
a
.ˆ
,ˆ2
zv
zv
a
a
cf. (from ‘Griffiths’)
dd
ddd
dd
nV
ddxdyd
around0
around
1lim)(
,)(curl)(curl
rV
nVkVrV
surface
boundingcurve
)( dd nVrVStoke’s theorem
goodbad
Example 1.
.0, hemisphere over the
?)( ,24
2222
zazyx
dzxy nVkjiV
kV 3
(a) integrate the expression at it stands
(b) use Stoke’s theorem and evaluate the line integral around the circle
(c) use Stoke’s theorem to say that the integral is the same over any
surface bounded by the circle, for example, the planar area inside the
circle.
33
kknV
kn
Ampere’s law IdC
rH
H : magnetic fieldC : closed curveI : current
r
I
IrrddC
2
22
0
H
HHrH
JH
nJnH
nHrH
nJJnJrH
dd
dd
dIdId
C
C
)(
)(
cf.densitiycurrent :,
: one of the Maxwell equations
For a specific case,
Conservative fields
‘simply connected’ if a simple closed curve in the region can be shrunk to a point without encountering any points not in the region.
If the components of F have continuous first partial derivatives in a simple connected region, any one implies all the others.
a) curl F = 0b) closed line integral = 0c) F conservatived) F = grad W, W single valued
Vector potential
0 ,definitionBy
0potential vector : for ,0div
0 ,definitionBy
0potentialscalar :for ,0curl
A
BAAVV
EVV
W
WW
Example 2. ),2(2)( 22 zxzyzyzx kjiV ?, AAV
02222
)2()2()( 22
xzzx
zxxz
yzy
yzzx
divV
zyx AAAzyx
kji
AV
There are many A’s to satisfy this equation. For convenience, set one component A_x =0.
kji
kji
AV
x
A
x
A
z
A
y
A
AAzyx
yzyz
zy0
i)
ii)
x
Azxz
x
Ayz yz
2 ,2 2
),(2
),(
2
122
zyfxyzA
zyfzxxzA
z
y
,2
z
A
y
Ayzx yz
z
fx
y
f
z
fxzx
y
fxz
z
A
y
Ayzx yz
1221222 22
There are many ways to select f1 and f2.
)2
12()( then,
,2
1,0 takingIf
222
221
zyxyzzxxz
zyff
kjA
Generalization for A potential vector : for ,0div AAVV
For A_x=0,x
AV
x
AV y
zz
y
, ,AV
),,( ),,( zygdxVAzyfdxVA yzzy
),( ,
,0divFor
),(
zyhdxx
VV
x
V
z
V
y
V
zyhdxz
V
y
V
z
A
y
AV
xx
xzy
zyyzx
V
cf. When we know one A, all others are of the form,
AVA 0u , u
2
2
2
2
22 11
ˆˆˆ1
11
ˆˆ1
ˆ
z
TT
ss
Ts
ssT
Laplacian
vrvvzs
s
r
Curl
z
vv
ssv
ss
Divergence
z
TT
ss
TT
Gradient
zs
zs
zφs
v
v
zφs
cf. Cylindrical coordinate
.sin
1sin
sin
11
sin
ˆsinˆˆ
sin
1
sin
1sin
sin
11
ˆsin
1ˆ1ˆ
2
2
2222
22
2
22
T
r
T
rr
Tr
rrT
Laplacian
vrrvvr
rr
r
Curl
v
rv
rvr
rr
Divergence
T
r
T
rr
TT
Gradient
r
r
φθr
v
v
φθr
cf. Spherical coordinate
Homework
Chapter 6
3-17, 4-5, 6-10, 10-14, 11-16