Chapter I Vector Analysis
.
1.1 Vector Algebra
kAjAiAA zyxˆˆˆ
It is well-known that any vector can be written as
Vectors obey the following algebraic rules:
scalaraisˆˆˆ)( kAjAiAAiv zyx
zzyyxx BABABAABBAv cos)(
eABBAii Commutativ)(
ssociativeCBACBAiii A)(
2)( AAAvi
kBAjBAiBABAi zzyyxxˆ)(ˆ)(ˆ)()(
(vii) The law of cosines
BACLet
B
A C
BBBAAABABACC
2
cos2222 ABBAC
ecommutativnotABBA(ix)
Triple Product
ACBBACCBAi
)(
B
A
C
Acos
volume of parallelepiped
rulecabbac)( BACCABCBAii
zyx
zyx
BBB
AAABA(viii)kji
Position, Displacement, & Separation Vectors
x
y
z
P(x,y,z)
The position vector of a point in 3-D is expressed in
Cartesian coordinates as
kzjyixr ˆˆˆ
The infinitesimal displacement vector is
kdzjdyidxld ˆˆˆ
A source point r is the point where an e. charge is located.
A field point r is the point at which you are calculating electric or magnetic filed.
The separation vector from the source point
to the field point is defined as
rrd
rr
r
r
Source point
field point
Differential Calculus
Ordinary Derivatives It is known that dxdx
dfdf
that is if x is changed by dx, the function changes by df with is the
proportionality factor. dx
df
In other words we say that the derivative is small if the function varies slowly
with x and large if the function varies rapidly with x.
Geometrically we say that is the slope of the graph f(x) vs. x. dx
df
Gradient (Directional Derivative)
Let T be a scalar function of 3-variables, i.e., ),,( zyxTT
)1(dzz
Tdy
y
Tdx
x
TdT
This tells us how T varies as we go a small distance (dx, dy, dz) away from the point (x, y, z). Let us rewrite Eq. (1) as
)2(ˆˆˆˆˆˆ kdzjdyidxkz
Tj
y
Ti
x
TdT
)3(ldTdTor
)4(ˆˆˆ kz
Tj
y
Ti
x
TTwith
is called the gradient of the function T.
The symbol is called the del operator. kz
jy
ix
ˆˆˆ
Like any vector the gradient has a magnitude and a direction. Now Eq. (3) can
be written as
)5(cosldTdT
For fixed , dT is maximum when , that is dT is maximum when we move in the same direction as .
ld
0T
The gradient points in the direction of maximum increase of the function T.
The magnitude gives the slope along this maximal direction. T
is directed normal to the level surface of T through the point being considered, i.e., is perpendicular to the surface T=constant.
T
T
Example Find the gradient of the position vector.
Soution In Cartesian coordinates the magnitude of the position vector is
kz
rj
y
ri
x
rr ˆˆˆ
r
r
rk
r
zj
r
yi
r
xˆˆˆˆ
This means that the distance from the origin increases most rapidly in the
radial direction.
222 zyxr
It is also defined as the net rate of flow per unit volume, i.e.,
densitysourcetheA
Example Find the divergence of the vectors
Soution: Since
kzkr ˆand,ˆ,
kzjyixr ˆˆˆ
Divergence
The divergence of a vector is defined as A
z
A
y
A
x
AkAjAiAk
zj
yi
xA z
yxzyx
ˆˆˆˆˆˆ
The divergence is a measure of how much the vector spreads out (diverge)
from the point in question.
3
z
z
y
y
x
xr
0100ˆ
zyxk
100ˆ
z
z
yxkz
The Curl
The curl of a vector is defined as A
zyx AAA
zyx
kji
A
ˆˆˆ
The curl of a vector is a measure of how much the vector curl around the point
in question.
Example Find the curl of the vectors
Soution:
jxBjxiyA ˆandˆˆ
kky
y
x
xj
xz
yi
z
x
y
xy
zyx
kji
A ˆ2ˆˆ0ˆ0
0
ˆˆˆ
kkyx
xj
xzi
z
x
y
x
zyx
kji
B ˆˆ0ˆ00ˆ0
00
ˆˆˆ
The Del Operator Operations
2
2
2
2
2
22)(
z
T
y
T
x
TTTi
0)( Tii
0)( Aiii
AAAiv 2)(
Integral Calculus
Line Integral The line integral is expressed as
b
aldA
where is a vector function and is an infinitesimal displacement vector
along a path from point a to point b. A
ld
If the path forms a closed loop, a circle is put on the integral sign, i.e.,
ldA
If the line integral is independent on the path followed, the vector is called
conservative A
Example Let find
Soution:
jyxiyA ˆ12ˆ2
b
aldA
from point a=(1,1,0) to point b=(2,2,0) along the solid path (path 1) and along
the dashed path (path 2).
a
x
y b
c
kdzjdyidxld ˆˆˆ
we have for the first path
b
a zyxb
adzAdyAdxAldA
b
adyyxdxy 122
dyyxdxyb
c
c
a 122
Now along the path ac y=1, and along the path cb x=2. This leads to
11101142
1
2
1 dyydxldA
b
a
For the second path we have x=y and this gives
10212 21
22
122
1
2
12 dyyydxxdyyxdxyldA
b
a
102 21
22
12 dyyydxx
Note that the two results are different, i.e., the vector is not conservative.
Surface Integral The surface integral is expressed as
S SdA
where is a vector function and is an infinitesimal element of area. A
Sd
Again if the surface is closed we put a circle on the integral sign, that is
SdA
The direction of is perpendicular to the surface an directed outward for
closed surfaces and arbitrary for open surfaces. Sd
Example Let
Soution:
Find over the 5-sides of a cube of side 2,
as shown in the figure (excluding the bottom).
kzyjxixzA
3ˆ2ˆ2 2
S SdA
dxdz dxdz
dxdy
y
z
x
For the top side kdydxSd topˆ
dxdyzySdAtop 32
But on the top side z=2 42
0
2
0 ydydxSdAtop
For the right side jdxdzSd right ˆ
dxdzxSdAright 2
122 20
2
0 dzdxx For the left side
jdxdzSd left ˆ
dxdzxSdAleft 2
122 20
2
0 dzdxx
For the front side idzdySd ˆ
xzdydzSdAfront 2
But on the top side x=2 1642
0
2
0 zdzdySdAfront
For the back side idzdySd ˆ
xzdydzSdAback 2
But on the top side x=0 0back SdA
Volume Integral The volume integral is expressed as
V dT
where T is a vector function and d is an infinitesimal element of volume.
The Fundamental Theorem of Calculus
)()( afbfdxdx
dfba
The Fundamental Theorem of Gradient
where T =T(x,y,z) be a scalar function of three variables, then
)()( aTbTldTba
Since the right side of the last equation depends only on the end points and not
on the path followed we conclude that
Corollary 1: is independent on the path followed from a to b. ldTba
Corollary 2 0ldT
Example:
Soution:
Check the fundamental theorem of gradient by taking two paths from point a (0,0,0) to point b (2,1,0).
2xyTLet
The first path is 2-steps: step (i) along the x-axis and then up step (ii).
(i) x
y
b
(ii)
a
(iii)
kdzjdyidxldNow ˆˆˆ
jxyiyTand ˆ2ˆ2
)ii()i( ˆˆ jdyTidxTldTb
a
1
0
2
02 2xydydxy
Bur for step (i) y=0, and for step (ii) x=2 220 ldTb
a
For the 1st path
For the 2nd path dxdyxy21
21
22 20
2432
212
02
41
)iii(2 dxxdxxdxxxydydxyldT
b
a
The Fundamental Theorem of Divergence
It states that the integral of a divergence over a volume is equal to the value of
the function at the boundary.
In another world, the divergence theorem states that the outward flux of a
vector field through a surface is equal to the triple integral of the divergence on
the region inside the surface.
Sv SdAdA
Example: Check the divergence theorem using the vector
kyzjzxyiyA ˆ2ˆ2ˆ 22 Over the unit cube situated at the origin.
Solution:
yx
z
A
y
A
x
AA z
yx 22
1
0
1
0
1
0
222 dxdydzyxdAv
http://en.wikipedia.org/wiki/Fluxhttp://en.wikipedia.org/wiki/Divergence
dxdz dxdz
dxdy
y
z
x dxdy
S SdA
To find we have to calculate the integral over all
the faces:
yzdxdySdAtop 2
But for the top side z=1, so we have
121
0
1
0 ydydxSdAtop
yzdxdySdAbottom 2
But for the top side z=0, so we have 0bottom SdA
dxdzzxySdAright22
But for the right side y=1, so we have 341
021
02 dzzdxxSdAright
dxdzzxySdAleft22
But for the left side y=0, so we have 311
02 dzzSdAleft
dydzySdAfront2
311
0
1
02 dzdyy
dydzySdAback2
311
0
1
02 dzdyy
20131
31
31
34
S
SdA
Stokes' Theorem
ldASdAS
Since the boundary line for any closed surface shrink down to a point, then
0 SdAS
Example: Check the Stokes' theorem using the vector
kyzjyxzA ˆ4ˆ32 22
Over the square surface shown.
Solution: kzixzA ˆ2ˆ24 2
idydzSd ˆ
1
0
21
0
24 dydzxzSdA
S
But on the surface x=0, so we have
34
1
0
21
0
4 dydzzSdAS
lefttoprightbottom
kdzAjdyAkdzAjdyAldA ˆˆˆˆ
132ˆ 10
2 dyyxzjdyAbottom
Along the bottom side x=z=0, so we have
Along the top side x=0, z=1 so we have
132ˆ 10
2 dyyxzjdyAtop
Along the right side x=0, y=1 so we have 341
024ˆ dzyzkdzA
right
Along the left side x=0, y=0 so we have 04ˆ 102 dzyzkdzA
left
34
34 011 ldA
Integration by Parts
It is known that dx
dfg
dx
dgffg
dx
d
Integrating both sides we get dxdx
dfgdx
dx
dgfdxfg
dx
d b
a
b
a
b
a
Using the fundamental theorem of calculus we get
dxdx
dfgfgdx
dx
dgf
b
a
ba
b
a
Example: Evaluate the integral
0
dxxe x
Solution: It is known that
00
dxedx
dxdxxe xx
100
xx edxxe
Curvilinear Coordinates
Spherical coordinates (r, ,)
r: is the distance from the origin (from 0 to ∞)
: the polar angle, is the angle between r and the z-axis (from 0 to )
: the azimuthal angle is the angle between the projection of r to the x-y plane and the x-axis (from 0 to 2)
The relation between the Cartesian coordinates and
the spherical coordinates can be written as
cossinrx
sinsinry cossinrz
The unit vectors associated with the spherical coordinates are related to the
corresponding unit vectors in the Cartesian coordinates as
kjir ˆcosˆsinsinˆcossinˆ
kji ˆsinˆsincosˆcoscosˆ
ji ˆcosˆsinˆ
The infinitesimal displacement vector in spherical coordinates is expressed as
ˆsinˆˆ drrdrdrld
The volume element is expressed as
drddrdldldld r sin2
For the surface elements we have
constantisˆˆ1 drdrdldlSd r
constantisˆsinˆ2 drdrdldlSd r
constantˆsinˆ2
3 isrrddrrdldlSd
To find the volume of a sphere of radius R we have
3342
02
00sin RdddrrdV
R
To find the gradient in spherical coordinates let T=T (r, , ) so
)1(sin drTrdTdrTldTdT r
)2(
dT
dT
drr
TdT
Equating the above two equations we get
rr
TT r ˆ
ˆ1
r
T
rT
,
ˆsin
1
r
T
rT
ˆ
sin
1ˆ1ˆorr
T
rr
T
rr
r
TT
ˆsin
1ˆ1ˆrrrr
rr
Similarly, one can find the divergence and the curl in spherical coordinates
A
rA
rAr
rrA r
sin
1sin
sin
11 22
ArrAA
r
rrr
rA
r sin
ˆsinˆˆ
sin
12
2
2
2222
22
sin
1sin
sin
11
T
r
T
rr
Tr
rrT
The Laplacian is defined as
Cylindrical coordinates (, , )
: is the distance from the z-axis (from 0 to ∞)
: the azimuthal angle is the angle between and the x-axis (from 0 to 2)
z: the distance from the x-y plane (from -∞ to ∞)
The relation between the Cartesian coordinates and
the cylindrical coordinates can be written as
cosx siny zz
The unit vectors associated with the cylindrical coordinates are related to the
corresponding unit vectors in the Cartesian coordinates as
ji ˆsinˆcosˆ
ji ˆcosˆsinˆ
kz ˆˆ
The infinitesimal displacement vector in cylindrical coordinates is expressed as
zdzddld ˆˆˆ
The volume element is expressed as
dzdddldldld z
For the surface elements we have
constantiszˆˆ1 zddzdldlSd
constantisˆˆ2 dzddldlSd zr
constantisˆˆ3 dzddldlSd z
zzˆˆ
1ˆ
The Del
The Divergence z
AAAA z
11
The Curl
zAAAz
A
ˆsinˆˆ
1
2
2
2
2
2
2 11
z
TTTT
The Laplacian
The Dirac Delta Function
Consider the function rr
A ˆ12
011
Now2
22
rr
rrA
2
0
22
0
ˆsinˆ1
and rddrrr
SdAS
But from the divergence theorem we know that
4sin2
00
dd
Sv SdAdA
Her we have a contradiction. The problem is the point r=0, where the vector blows up.
Her we have a contradiction. The problem is the point r=0, where the vector blows up. So we write
rr
r 32
4ˆ
Where (x) is called the Dirac delta function with the following properties:
axax
ax
0
Using theses properties we have
1
dxax
aFdxaxxF
zyxr 3
44 3 vv drdA
As expected
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