Coordinate Systems
COORDINATE SYSTEMS• RECTANGULAR or Cartesian
• CYLINDRICAL
• SPHERICAL
Choice is based on symmetry of problem
Examples:Sheets - RECTANGULAR
Wires/Cables - CYLINDRICAL
Spheres - SPHERICAL
To understand the Electromagnetics, we must know basic vector algebra and coordinate systems. So let us start the coordinate systems.
Cylindrical Symmetry Spherical Symmetry
Visualization (Animation)
Orthogonal Coordinate Systems:
3. Spherical Coordinates
2. Cylindrical Coordinates
1. Cartesian Coordinates
P (x, y, z)
P (r, θ, Φ)
P (r, Φ, z)
x
y
zP(x,y,z)
Φ
z
rx y
z
P(r, Φ, z)
θ
Φ
r
z
yx
P(r, θ, Φ)
Rectangular CoordinatesOr
X=r cos Φ,Y=r sin Φ,Z=z
X=r sin θ cos Φ,Y=r sin θ sin Φ,Z=z cos θ
Cartesian CoordinatesP(x, y, z)
Spherical CoordinatesP(r, θ, Φ)
Cylindrical CoordinatesP(r, Φ, z)
x
y
zP(x,y,z)
Φ
z
rx y
z
P(r, Φ, z)
θ
Φ
r
z
yx
P(r, θ, Φ)
Cartesian coordinate system• dx, dy, dz are infinitesimal
displacements along X,Y,Z.• Volume element is given by
dv = dx dy dz• Area element is
da = dx dy or dy dz or dxdz• Line element is
dx or dy or dzEx: Show that volume of a cube
of edge a is a3.
P(x,y,z)
X
Y
Z
3
000adzdydxdvV
aa
v
a
dxdy
dz
Cartesian Coordinates
Differential quantities:
Length:
Area:
Volume:
dzzdyydxxld ˆˆˆ
dxdyzsd
dxdzysddydzxsd
z
y
x
ˆ
ˆˆ
dxdydzdv
AREA INTEGRALS
• integration over 2 “delta” distances
dx
dy
Example:
x
y
2
6
3 7
AREA = 7
3
6
2
dxdy = 16
Note that: z = constant
Cylindrical coordinate system (r,φ,z)
X
Y
Z
rφ
Z
Spherical polar coordinate system
• dr is infinitesimal displacement along r, r dφ is along φ and dz is along z direction.
• Volume element is given by dv = dr r dφ dz
• Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π
Ex: Show that Volume of a Cylinder of radius ‘R’ and height ‘H’ is π R2H .
φ is azimuth angle
Cylindrical coordinate system (r,φ,z)
X
Y
Z
rφ
r dφ
dz
dr
r dφ
dr
dφ
Volume of a Cylinder of radius ‘R’ and Height ‘H’
HR
dzdrdr
dzddrrdvV
R H
v
2
0
2
0 0
Try yourself: 1) Surface Area of Cylinder = 2πRH . 2) Base Area of Cylinder (Disc)=πR2.
Differential quantities:
Length element:
Area element:
Volume element:
dzardadrald zr ˆˆˆ
rdrdasd
drdzasddzrdasd
zz
rr
ˆ
ˆˆ
dzddrrdv
Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π
Cylindrical Coordinates: Visualization of Volume element
Spherically Symmetric problem (r,θ,φ)
X
Y
Z
r
φ
θ
Spherical polar coordinate system (r,θ,φ)
• dr is infinitesimal displacement along r, r dθ is along θ and r sinθ dφ is along φ direction.
• Volume element is given by dv = dr r dθ r sinθ dφ
• Limits of integration of r, θ, φ are
0<r<∞ , 0<θ <π , o<φ <2πEx: Show that Volume of a
sphere of radius R is 4/3 π R3 .
P(r, θ, φ)
X
Y
Z
r
φ
θ
drP
r dθ
r sinθ dφ
θ is zenith angle( starts from +Z reaches up to –Z) , φ is azimuth angle (starts from +X direction and lies in x-y plane only)
r cos θ
r sinθ
Volume of a sphere of radius ‘R’
33
0 0
2
0
2
2
342.2.
3
sin
sin
RR
dddrr
dddrrdvV
R
v
Try Yourself:1)Surface area of the sphere= 4πR2 .
Spherical Coordinates: Volume element in space
Points to rememberSystem Coordinates dl1 dl2 dl3
Cartesian x,y,z dx dy dzCylindrical r, φ,z dr rdφ dzSpherical r,θ, φ dr rdθ r sinθdφ
• Volume element : dv = dl1 dl2 dl3• If Volume charge density ‘ρ’ depends only on ‘r’:
Ex: For Circular plate: NOTEArea element da=r dr dφ in both the coordinate systems (because θ=900)
drrdvQv l 24
Quiz: Determine a) Areas S1, S2 and S3.b) Volume covered by these surfaces.
Radius is r,Height is h,
X
Y
Z
r
dφ
S1S2
S3
21
hrdzrddrVb
rrddrSiii
rhdzdrSii
rhdzrdSia
Solution
h r
r
r h
h
)(2
..)
)(2
.3)
2)
)(1))
:
12
2
0 0
12
2
0
0 0
120
2
1
2
1
2
1
Vector Analysis
• What about A.B=?, AxB=? and AB=?• Scalar and Vector product:
A.B=ABcosθ Scalar or (Axi+Ayj+Azk).(Bxi+Byj+Bzk)=AxBx+AyBy+AzBz
AxB=ABSinθ n Vector(Result of cross product is always perpendicular(normal) to the planeof A and B
A
B
n
Gradient, Divergence and Curl
• 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.
f Vector
xAA
.
The Del Operator
Fundamental theorem for divergence and curl
• Gauss divergence theorem:
• Stokes curl theorem
v s
daVdvV .).(
s l
dlVdaVx .).(
Conversion of volume integral to surface integral and vice verse.
Conversion of surface integral to line integral and vice verse.
Gradient:gradT: points the direction of maximum increase of the function T.
Divergence:
Curl:
Operator in Cartesian Coordinate System
kzTj
yTi
xTT ˆˆˆ
y zxV VV
Vx y z
kyV
xV
jxV
zVi
zV
yVV xyzxyz ˆˆˆ
kVjViVV zyxˆˆˆ
where
as
Operator in Cylindrical Coordinate System
Volume Element:
Gradient:
Divergence:
Curl:
dzrdrddv
zzTˆT
rr
rTT
1
1 1 zr
V VV rV
r r r z
zVrVrr
ˆrV
zVr
zVV
rV rzrz
11
zVVrVV zr ˆˆˆ
Operator In Spherical Coordinate System
Gradient :
Divergence:
Curl:
ˆTsinr
ˆTr
rrTT
11
2
2
sin1 1 1sin sin
rr V VVV
r r r r
ˆVrV
rr
ˆrVr
Vsinr
rV
Vsinsinr
V
r
r
1
111
ˆˆˆ VVrVV r
The divergence theorem states that the total outward flux of a vector field F through the closed surface S is the same as the volume integral of the divergence of F.
Closed surface S, volume V, outward pointing normal
Basic Vector Calculus
2
( )
0, 0
( ) ( )
F G G F F G
F
F F F
Divergence or Gauss’ Theorem
SV
SdFdVF
dSnSd
Oriented boundary L
n
Stokes’ Theorem
S L
ldFSdF
Stokes’s theorem states that the circulation of a vector field F around a closed path L is equal to the surface integral of the curl of F over the open surface S bounded by L