Coordinate Systems and Transformations & Vector Calculus By: Hanish Garg 12105017 ECE Branch PEC University of Technology
Jan 27, 2015
Coordinate Systems and Transformations
& Vector Calculus
By:Hanish Garg
12105017ECE Branch
PEC University of Technology
Coordinate Systems
• Cartesian or Rectangular Coordinate System
• Cylindrical Coordinate System
• Spherical Coordinate System
Choice of the system is based on the symmetry of the problem.
Cartesian Or Rectangular Coordinates
P (x, y, z)
x
y
z
P(x,y,z)
z
y
x
A vector A in Cartesian coordinates can be written as
),,( zyx AAA or zzyyxx aAaAaA
where ax,ay and az are unit vectors along x, y and z-directions.
Cylindrical CoordinatesP (ρ, Φ, z)
x= ρ cos Φ, y=ρ sin Φ, z=z
z
Φ
z
ρx
y
P(ρ, Φ, z)
z
20
0
A vector A in Cylindrical coordinates can be written as
),,( zAAA orzzaAaAaA
where aρ,aΦ and az are unit vectors along ρ, Φ and z-directions.
zzx
yyx ,tan, 122
The relationships between (ax,ay, az) and (aρ,aΦ, az)are
zz
y
x
aa
aaa
aaa
cossin
sincos
zz
yx
yx
aa
aaa
aaa
cossin
sincosor
zzyxyx aAaAAaAAA )cossin()sincos(
Then the relationships between (Ax,Ay, Az) and (Aρ, AΦ, Az)are
zz
yx
yx
AA
AAA
AAA
cossin
sincos
z
y
x
z A
A
A
A
A
A
100
0cossin
0sincos
In matrix form we can write
Spherical CoordinatesP (r, θ, Φ)
x=r sin θ cos Φ, y=r sin θ sin Φ, Z=r cos θ
20
0
0
r
A vector A in Spherical coordinates can be written as
),,( AAAr or aAaAaA rr
where ar, aθ, and aΦ are unit vectors along r, θ, and Φ-directions.
θ
Φ
r
z
yx
P(r, θ, Φ)
x
y
z
yxzyxr 1
221222 tan,tan,
The relationships between (ax,ay, az) and (ar,aθ,aΦ)are
aaa
aaaa
aaaa
rz
ry
rx
sincos
cossincossinsin
sincoscoscossin
yx
zyx
zyxr
aaa
aaaa
aaaa
cossin
sinsincoscoscos
cossinsincossin
or
Then the relationships between (Ax,Ay, Az) and (Ar, Aθ,and AΦ)are
aAA
aAAA
aAAAA
yx
zyx
rzyx
)cossin(
)sinsincoscoscos(
)cossinsincossin(
z
y
xr
A
A
A
A
A
A
0cossin
sinsincoscoscos
cossinsincossin
In matrix form we can write
cossin
sinsincoscoscos
cossinsincossin
yx
zyx
zyxr
AAA
AAAA
AAAA
Cartesian CoordinatesP(x, y, z)
Spherical CoordinatesP(r, θ, Φ)
Cylindrical CoordinatesP(ρ, Φ, z)
x
y
zP(x,y,z)
Φ
z
rx y
z
P(ρ, Φ, z)
θ
Φ
r
z
yx
P(r, θ, Φ)
Differential Length, Area and Volume
Differential displacement
zyx dzadyadxadl
Differential area
zyx dxdyadxdzadydzadS
Differential VolumedxdydzdV
Cartesian Coordinates
Cylindrical Coordinates
ρρ
ρ
ρ
ρρ
ρ
ρ
ρρ
ρ
Differential Length, Area and Volume
Differential displacement
zdzaadaddl
Differential area
zadddzaddzaddS
Differential Volume
dzdddV
Cylindrical Coordinates
Spherical Coordinates
Differential Length, Area and Volume
Differential displacement
adrarddradl r sin
Differential area
ardrdadrdraddrdS r sinsin2
Differential Volume ddrdrdV sin2
Spherical Coordinates
Line, Surface and Volume Integrals
Line Integral L
dlA.
Surface Integral
Volume Integral
S
dSA.
dvpV
v
• Gradient of a scalar function is a vector quantity.
• Divergence of a vector is a scalar quantity.
• Curl of a vector is a vector quantity.• The Laplacian of a scalar A
f Vector
A.
A
A2
The Del Operator
Del Operator
Cartesian Coordinateszyx a
zay
ax
Cylindrical Coordinates
Spherical Coordinates
zazaa
1
a
ra
rar r
sin
11
VECTOR CALCULUS
GRADIENT OF A SCALAR
DIVERGENCE OF A VECTOR
DIVERGENCE THEOREM
CURL OF A VECTOR
STOKES’S THEOREM
GRADIENT OF A SCALAR Suppose is the temperature at ,
and is the temperature at
as shown.
zyxT ,,1 zyxP ,,1
2P dzzdyydxxT ,,2
The differential distances are the components of the differential distance vector :
dzdydx ,,
zyx dzdydxd aaaL Ld
However, from differential calculus, the differential temperature:
dzz
Tdy
y
Tdx
x
TTTdT
12
GRADIENT OF A SCALAR (Cont’d)
But,
z
y
x
ddz
ddy
ddx
aL
aL
aL
So, previous equation can be rewritten as:
Laaa
LaLaLa
dz
T
y
T
x
T
dz
Td
y
Td
x
TdT
zyx
zyx
GRADIENT OF A SCALAR (Cont’d)
The vector inside square brackets defines the change of temperature corresponding to a vector change in position .
This vector is called Gradient of Scalar T.
LddT
For Cartesian coordinate:
zyx z
V
y
V
x
VV aaa
GRADIENT OF A SCALAR (Cont’d)
For Circular cylindrical coordinate:
zz
VVVV aaa
1
For Spherical coordinate:
aaa
V
r
V
rr
VV r sin
11
GRADIENT OF A SCALAR (Cont’d)
EXAMPLE
Find gradient of these scalars:
yxeV z cosh2sin
2cos2zU
cossin10 2rW
(a)
(b)
(c)
SOLUTION TO EXAMPLE
(a) Use gradient for Cartesian coordinate:
zz
yz
xz
zyx
yxe
yxeyxe
z
V
y
V
x
VV
a
aa
aaa
cosh2sin
sinh2sincosh2cos2
SOLUTION TO EXAMPLE (Cont’d)
(b) Use gradient for Circular cylindrical
coordinate:
z
z
zz
z
UUUU
a
aa
aaa
2cos
2sin22cos2
1
2
(c) Use gradient for Spherical coordinate:
a
aa
aaa
sinsin10
cos2sin10cossin10
sin
11
2
r
rW
r
W
rr
WW
SOLUTION TO EXAMPLE (Cont’d)
Illustration of the divergence of a vector field at point P:
Positive Divergence
Negative Divergence
Zero Divergence
DIVERGENCE OF A VECTOR
DIVERGENCE OF A VECTOR (Cont’d)
The divergence of A at a given point P is the outward flux per unit volume:
v
dS
div s
v
A
A A lim0
What is ?? s
dSA Vector field A at closed surface S
DIVERGENCE OF A VECTOR (Cont’d)
Where,
dSdSbottomtoprightleftbackfronts
AA
And, v is volume enclosed by surface S
DIVERGENCE OF A VECTOR (Cont’d)
For Cartesian coordinate:
z
A
y
A
x
A zyx
A
For Circular cylindrical coordinate:
z
AAA z
11A
DIVERGENCE OF A VECTOR (Cont’d)
For Spherical coordinate:
A
r
A
rAr
rrr sin
1sin
sin
11 22
A
DIVERGENCE OF A VECTOR (Cont’d)
Find divergence of these vectors:
zx xzyzxP aa 2
zzzQ aaa cossin 2
aaa coscossincos12
rr
W r
(a)
(b)
(c)
EXAMPLE
39
(a) Use divergence for Cartesian coordinate:
xxyz
xzzy
yzxx
z
P
y
P
x
P zyx
2
02
P
SOLUTION TO EXAMPLE
(b) Use divergence for Circular cylindrical
coordinate:
cossin2
cos1
sin1
11
22
Q
zz
z
z
QQQ z
SOLUTION TO EXAMPLE (Cont’d)
(c) Use divergence for Spherical coordinate:
coscos2
cossin
1
cossinsin
1cos
1
sin
1sin
sin
11
22
22
W
r
rrrr
W
r
W
rWr
rrr
SOLUTION TO EXAMPLE (Cont’d)
It states that the total outward flux of a vector field A at the closed surface S is the same as volume integral of divergence of A.
VV
dVdS AA
DIVERGENCE THEOREM
A vector field exists in the region
between two concentric cylindrical surfaces
defined by ρ = 1 and ρ = 2, with both
cylinders extending between z = 0 and z = 5.
Verify the divergence theorem by evaluating:
aD 3
S
dsD
V
DdV
(a)
(b)
EXAMPLE
(a) For two concentric cylinder, the left side:
topbottomouterinnerS
d DDDDSD
Where,
10)(
)(
2
0
5
01
4
2
0
5
01
3
z
zinner
dzd
dzdD
aa
aa
SOLUTION TO EXAMPLE
160)(
)(
2
0
5
02
4
2
0
5
02
3
z
zouter
dzd
dzdD
aa
aa
2
1
2
05
3
2
1
2
00
3
0)(
0)(
zztop
zzbottom
ddD
ddD
aa
aa
SOLUTION TO EXAMPLE (Cont’d)
Therefore
150
0016010
SDS
d
SOLUTION TO EXAMPLE (Cont’d)
(b) For the right side of Divergence Theorem,
evaluate divergence of D
23 41
D
So,
150
4
5
0
2
0
2
1
4
5
0
2
0
2
1
2
zr
z
dzdddVD
SOLUTION TO EXAMPLE (Cont’d)
CURL OF A VECTOR
The curl of vector A is an axial (rotational) vector whose magnitude is the maximum circulation of A per unit area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum.
maxlim0
a
A
A A ns
s s
dl
Curl
Where,
CURL OF A VECTOR (Cont’d)
dldldacdbcabs
AA
The curl of the vector field is concerned with rotation of the vector field. Rotation can be used to measure the uniformity of the field, the more non uniform the field, the larger value of curl.
CURL OF A VECTOR (Cont’d)
For Cartesian coordinate:
zyx
zyx
AAAzyx
aaa
A
zxy
yxz
xyz
y
A
x
A
z
A
x
A
z
A
y
AaaaA
CURL OF A VECTOR (Cont’d)
z
z
AAAz
aaa
A1
z
zz
AA
z
AA
z
AA
a
aaA
1
1
For Circular cylindrical coordinate:
CURL OF A VECTOR (Cont’d)
For Spherical coordinate:
ArrAA
rrr
r
sinsin
12
aaa
A
a
aaA
r
rr
A
r
rA
r
r
rAA
r
AA
r
)(1
sin
11sin
sin
1
CURL OF A VECTOR (Cont’d)
zx xzyzxP aa 2
zzzQ aaa cossin 2
aaa coscossincos12
rr
W r
(a)
(b)
(c)
Find curl of these vectors:
EXAMPLE
(a) Use curl for Cartesian coordinate:
zy
zyx
zxy
yxz
xyz
zxzyx
zxzyx
y
P
x
P
z
P
x
P
z
P
y
P
aa
aaa
aaaP
22
22 000
SOLUTION TO EXAMPLE
(b) Use curl for Circular cylindrical coordinate
z
z
zzz
zz
z
z
y
Q
x
z
Q
z
aa
a
aa
aaaQ
cos3sin1
cos31
00sin
11
3
2
2
SOLUTION TO EXAMPLE (Cont’d)
(c) Use curl for Spherical coordinate:
a
aa
a
aaW
22
2
cos)cossin(1
coscos
sin
11cossinsincos
sin
1
)(1
sin
11sin
sin
1
rr
r
r
r
rrr
r
r
W
r
rW
r
r
rWW
r
WW
r
r
r
rr
SOLUTION TO EXAMPLE (Cont’d)
a
aa
a
aa
sin1
cos2
cossin
sin
2cos
sincossin2
1
cos01
sinsin2cossin
1
3
2
r
rr
rr
r
rr
r
r
r
SOLUTION TO EXAMPLE (Cont’d)
STOKE’S THEOREM
The circulation of a vector field A
around a closed path L is equal to the
surface integral of the curl of A over
the open surface S bounded by L that A
and curl of A are continuous on S.
SL
dSdl AA
STOKE’S THEOREM (Cont’d)
By using Stoke’s Theorem, evaluate
for dlA
aaA sincos
EXAMPLE
Stoke’s Theorem,
SL
dSdl AA
where, and
zddd aS Evaluate right side to get left side,
zaA
sin11
SOLUTION TO EXAMPLE (Cont’d)
941.4
sin11
0
0
60
30
5
2
aA zS
dddS
SOLUTION TO EXAMPLE (Cont’d)
Verify Stoke’s theorem for the vector field
for given figure by evaluating: aaB sincos
(a) over the semicircular contour. LB d
(b) over the surface of semicircular contour.
SB d
EXAMPLE
(a) To find LB d
321 LLL
dddd LBLBLBLB
Where,
dd
dzddd z
sincos
sincos
aaaaaLB
SOLUTION TO EXAMPLE
So
202
1
sincos
2
0
2
0
0
00,0
2
01
LB
zzL
ddd
4cos20
sincos
0
0,200
2
22
LB
zzL
ddd
SOLUTION TO EXAMPLE (Cont’d)
202
1
sincos
0
2
2
00,0
0
23
r
zzL
ddd
LB
Therefore the closed integral,
8242 LB d
SOLUTION TO EXAMPLE (Cont’d)
(b) To find SB d
z
z
z
zz
a
aaa
a
aa
aaB
11sin
sinsin1
00
cossin1
0cossin01
sincos
SOLUTION TO EXAMPLE (Cont’d)
Therefore
82
1cos
1sin
11sin
0
2
0
2
0
2
0
0
2
0
aaSB
dd
ddd zz
SOLUTION TO EXAMPLE (Cont’d)
LAPLACIAN OF A SCALAR
The Laplacian of a scalar field, V written as: V2
Where, Laplacian V is:
zyxzyx z
V
y
V
x
V
zyx
VV
aaaaaa
2
For Cartesian coordinate:
2
2
2
2
2
22
z
V
y
V
x
VV
For Circular cylindrical coordinate:
2
22
22 11
z
VVVV
LAPLACIAN OF A SCALAR (Cont’d)
LAPLACIAN OF A SCALAR (Cont’d)
For Spherical coordinate:
2
2
22
22
22
sin
1
sinsin
11
V
r
V
rr
Vr
rrV
EXAMPLE
Find Laplacian of these scalars:
yxeV z cosh2sin 2cos2zU
cossin10 2rW
(a)
(b)
(c)
You should try this!!
SOLUTION TO EXAMPLE
yxeV z cosh2sin22
02 U
2cos21
cos102 r
W
(a)
(b)
(c)
Thank You !!!