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Orthogonal
CurvilinearCoordinates 28.3Introduction
The derivatives div, grad and curl from Section 28.2 can be carried out using coordinate systems otherthan the rectangular Cartesian coordinates. This Section shows how to calculate these derivatives inother coordinate systems. Two coordinate systems - cylindrical polar coordinates and spherical polarcoordinates - will be illustrated.
Prerequisites
Before starting this Section you should . . .
be able to nd the gradient, divergence andcurl of a eld in Cartesian coordinates.
be familiar with polar coordinates
Learning OutcomesOn completion you should be able to . . .
nd the divergence, gradient or curl of a
vector or scalar eld expressed in terms of orthogonal curvilinear coordinates
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1. Orthogonal curvilinear coordinatesThe results shown in Section 28.2 have been given in terms of the familiar Cartesian (x,y,z ) co-ordinate system. However, other coordinate systems can be used to better describe some physicalsituations. A set of coordinates u = u(x,y,z ), v = v(x,y,z ) and w = w(x,y,z ) where the direc-
tions at any point indicated by u , v and w are orthogonal (perpendicular) to each other is referred toas a set of orthogonal curvilinear coordinates . With each coordinate is associated a scale factor
hu , hv or hw respectively where hu = xu 2 + yu 2 + zu 2 (with similar expressions for hv andhw ). The scale factor gives a measure of how a change in the coordinate changes the position of apoint.
Two commonly-used sets of orthogonal curvilinear coordinates are cylindrical polar coordinatesand spherical polar coordinates . These are similar to the plane polar coordinates introduced in
17.2 but represent extensions to three dimensions.
Cylindrical polar coordinatesThis corresponds to plane polar (, ) coordinates with an added z -coordinate directed out of thexy plane. Normally the variables and are used instead of r and to give the three coordinates, and z . A cylinder has equation = constant.The relationship between the coordinate systems is given by
x = cos y = sin z = z
(i.e. the same z is used by the two coordinate systems). See Figure 20.
(x,y,z )
z
x
y
Figure 20
The scale factors h , h and hz are given as follows
h = x2
+y
2
+z
2
= (cos )2 + (sin )2 + 0 = 1h = x
2
+y
2
+z
2
= ( sin )2 + ( cos )2 + 0 = h z =
x
z
2
+y
z
2
+z
z
2
=
(02 + 0 2 + 1 2 ) = 1
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Spherical polar coordinatesIn this system a point is referred to by its distance from the origin r and two angles and . Theangle is the angle between the positive z -axis and the line from the origin to the point. The angle is the angle from the x-axis to the projection of the point in the xy plane.
A useful analogy is of latitude, longitude and height on Earth. The variable r plays the role of height (but height measured above the centre of Earth rather
than from the surface).
The variable plays the role of latitude but is modied so that = 0 represents the NorthPole, = 90 =
2
represents the equator and = 180 = represents the South Pole.
The variable plays the role of longitude.
A sphere has equation r = constant.
The relationship between the coordinate systems is given byx = r sin cos y = r sin sin z = r cos . See Figure 21.
(x,y,z )
r
z
y
x,
Figure 21
The scale factors hr , h and h are given by
h r = xr2
+ yr
2
+ z r
2
= (sin cos )2 + (sin sin )2 + (cos )2 = 1
h = x2
+y
2
+z
2
= (r cos cos )2 + ( r cos sin )2 + ( r sin )2 = rh = x
2
+y
2
+z
2
= ( r sin sin )2 + ( r sin sin )2 + 0 = r sin
HELM (2005):Section 28.3: Orthogonal Curvilinear Coordinates
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2. Vector derivatives in orthogonal coordinatesGiven an orthogonal coordinate system u ,v,w with unit vectors u , v and w and scale factors, hu ,hv and hw , it is possible to nd the derivatives f , F and F .
It is found that
grad f = f = 1h u
f u
u + 1hv
f v
v + 1hw
f w
w
If F = F u u + F v v + F w w then
div F = F = 1
hu hvhw u
(F u hvh w ) + v
(F vhu hw ) + w
(F w h u h v)
Also if F = F u u + F v v + F w w then
curl F = F = 1hu hv hw
hu u h v v hw w
u
v
w
hu F u hv F v hw F w
Key Point 6
In orthogonal curvilinear coordinates, the vector derivatives f
, F
and F
include the scalefactors hu , hv and hw .
3. Cylindrical polar coordinatesIn cylindrical polar coordinates (,,z ), the three unit vectors are , and z with scale factors
h = 1, h = , hz = 1.The quantities and are related to x and y by x = cos and y = sin . The unit vectors are = cos i + sin j and = sin i + cos j . In cylindrical polar coordinates,
grad f = f = f
+ 1
f
+ f z
z
The scale factor is necessary in the -component because the derivatives with respect to aredistorted by the distance from the axis = 0.
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If F = F + F + F z z then
div F = F = 1
(F ) +
(F ) + z
(F z )
curl F = F = 1
z
z
F F F z
.
Example 20
Working in cylindrical polar coordinates, nd f for f = 2
+ z 2
Solution
If f = 2 + z 2 then f
= 2,f
= 0 andf z
= 2z so f = 2 + 2z z .
Example 21Working in cylindrical polar coordinates nd
(a) f for f = 3 sin
(b) Show that the result for (a) is consistent with that found working inCartesian coordinates.
Solution
(a) If f = 3 sin then f
= 32 sin , f
= 3 cos and f z
= 0 and hence,
f = 32 sin + 2 cos .
(b) f = 3 sin = 2 sin = ( x 2 + y2 )y = x 2 y + y3 so f = 2xyi + ( x 2 + 3 y2 ) j .Using cylindrical polar coordinates, from (a) we have
f = 32 sin + 3 cos = 3 2 sin (cos i + sin j ) + 2 cos ( sin i + cos j )= 32 sin cos 2 sin cos i + 32 sin2 + 2 cos2 j= 22 sin cos i + 32 sin2 + 2 cos2 j = 2xyi + (3 y2 + x2 ) j
So the results using Cartesian and cylindrical polar coordinates are consistent.
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Example 22Find F for F = F + F + F z z = 3 + z + z sin z . Show that theresults are consistent with those found using Cartesian coordinates.
Solution
Here, F = 3 , F = z and F z = z sin so
F = 1
(F ) +
(F ) + z
(F z )
= 1
(4 ) +
(z ) + z
(2 z sin )
= 1
43
+ 0 + 2
sin = 4 2 + sin
Converting to Cartesian coordinates,
F = F + F + F z z = 3 + z + z sin z
= 3 (cos i + sin j ) + z ( sin i + cos j ) + z sin k= ( 3 cos z sin )i + ( 3 sin + z cos) + zk= 2 ( cos ) sin z i + 2 ( sin ) + cos z j + sin zk
= (x2
+ y2
)x yz i + (x2
+ y2
)y + xz j + yzk= ( x 3 + xy 2 yz )i + ( x 2 y + y3 + xz ) j + yzk
So
F = x
(x 3 + xy 2 yz ) + y
(x 2 y + y3 + xz ) + z
(yz )
= (3 x 2 + y2 ) + ( x 2 + 3 y2 ) + y = 4x 2 + 4 y2 + y= 4( x 2 + y2 ) + y= 4 2 + sin
So F is the same in both coordinate systems.
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Example 23Find F for F = 2 + z sin + 2z cosz .
Solution
F = 1
z
z
F F F z
= 1
z
z
2 z sin 2z cos
= 1
2z cos z
z sin + z
2
2z cos + z
z sin
2
= 1
( 2z sin sin ) + (0) + z (z sin )
= (2z sin )
+
z sin
z
Engineering Example 2
Divergence of a magnetic eld
Introduction
A magnetic eld B must satisfy B = 0. An associated current is given by:
I = 10 ( B )
Problem in words
For the magnetic eld (in cylindrical polar coordinates , , z )
B = B 0
1 + 2 + z
show that the divergence of B is zero and nd the associated current.
Mathematical statement of problem
We must
(a) show that B = 0 (b) nd the current I = 10
( B )
HELM (2005):Section 28.3: Orthogonal Curvilinear Coordinates
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Mathematical analysis
(a) Express B as (B , B , B z ); then
B =
1
(B
) +
B +
B zz
= 1
(0) +
B 0
1 + 2 +
z
( )
= 1
[0 + 0 + 0] = 0 as required.
(b) To nd the current evaluate
I = 10
( B ) = 10
1
z
z
B B Bz
=
z
z
0 B02
1 + 2
= 10
0 + 0 + B 0
2
1 + 2 z
= 10 B 0
1 + 43
(1 + 2 )2 z
Interpretation
The magnetic eld is in the form of a helix with the current pointing along its axis.
xy
z
B
I
Figure 22
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Example 24A magnetic eld B is given by B = 2 + kz . Find B and B .
Solution
B = 1
0 +
2 + z
k = 1
[0 + 0 + 0] = 0
B = 1
z
z
B
B
Bz
= 1
z
z
20
k
= 0
All magnetic elds satisfy B = 0 i.e. an absence of magnetic monopoles. There is a class of magnetic elds known as potential elds that satisfy B = 0
Task skUsing cylindrical polar coordinates, nd f for f = 2 z sin
Your solution
Answer
[2
z sin ] + 1
[
2
z sin ] +
z [
2
z sin ]z = 2z sin + z cos +
2
sin z
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Task skUsing cylindrical polar coordinates, nd f for f = z sin 2
Your solution
Answer
[2sin2] +
1
[2sin2] + z
[2sin2]z = 2
z cos2 + sin2z
Task skFind F for F = cos sin + z z
i.e. F = cos , F = sin , F z = z
(a) First nd the derivatives
[F ],
[F ], z
[F z ]:
Your solution
Answer2 cos , cos , 2
(b) Now combine these to nd
F :Your solution
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Answer
F = 1
(F ) +
(F ) + z
(F z )
= 1
(
2 cos ) + ( sin ) +
z (
2 z )
= 1
2 cos cos + 2
= cos + 2
Task skFind F for F = F + F + F z z = 3 + z + z sin z . Show that theresults are consistent with those found using Cartesian coordinates.
(a) Find the curl F :Your solution
Answer
1
z
z
3 2 z z sin
= ( z cos ) z sin + 2z z
(b) Find F in Cartesian coordinates:Your solution
AnswerUse = cos i +sin j, = sin i +cos j to get F = ( x 3 + xy 2 yz )i +( x 2 y + y3 + xz ) j + yzk
(c) Hence nd F in Cartesian coordinates:Your solution
Answer(z x)i yj + 2zk
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(d) Using = cos i + sin j and = sin i + cos j , show that the solution to part (a) is equalto the solution for part (c):
Your solution
Exercises
1. For F = + ( sin + z ) + z z , nd F and F .
2. For f = 2 z 2 cos2, nd ( f ).
Answers
1. 1 + cos + , z cos + (2 sin + z )z
2. 0
4. Spherical polar coordinatesIn spherical polar coordinates (r,, ), the 3 unit vectors are r , and with scale factors hr = 1,h = r , h = r sin . The quantities r , and are related to x, y and z by x = r sin cos ,y = r sin sin and z = r cos . In spherical polar coordinates,
grad f = f = f r
r + 1r
f
+ 1
r sin f
If F = F r r + F + F
then
div F = F = 1
r2
sin
r
(r 2 sin F r ) +
(r sin F ) +
(rF )
curl F = F = 1
r 2 sin
r r r sin
r
F r rF r sin F
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Example 25In spherical polar coordinates, nd f for
(a) f = r (b) f = 1
r (c) f = r 2 sin( + )
Solution
(a) f = f
r r +
1r
f
+ 1r sin
f
= (r )
r r +
1r
(r )
+ 1r sin
(r )
= 1 r = r
(b) f = f
r r +
1r
f
+ 1r sin
f
= ( 1r )
r r +
1r
( 1r )
+ 1r sin
( 1r )
= 1r 2
r
(c) f = f r
r + 1r
f
+ 1r sin
f
= (r sin( + )
r r +
1r
(r sin( + )
+ 1r sin
(r 2 sin( + )
= 2 r sin( + ) r + 1r
r 2 cos( + ) + 1
r sin r 2 cos( + )
= 2 r sin( + ) r + r cos( + ) + r cos( + )
sin
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Engineering Example 3
Electric potential
Introduction
There is a scalar quantity V , called the electric potential, which satises
V = E
Problem in words
Given the electric potential, nd the electric eld.
Mathematical statement of problem
For a point charge, the potential V is given by
V = Q4 0 r
Verify, using spherical polar coordinates, that E = V is indeed E = Q4 0 r 2
r
Mathematical analysis
In spherical polar coordinates:
V = V
r r + 1r
V
+
1r sin
V
= V
r r as the other partial derivatives are zero
= r
Q4 0 r
r
= Q
4 0 r 2r
Interpretation
So E = Q
4 0 r 2r as required.
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Example 26Using spherical polar coordinates, nd F for the following vector functions.
(a) F = r r (b) F = r 2 sin r (c) F = r sin r + r 2 sin + r cos
Solution
(a)
F = 1r 2 sin
r
(r 2 sin F r ) +
(r sin F ) +
(rF )
= 1
r2
sin
r(r 2 sin r ) +
(r sin 0) +
(r 0)
= 1r 2 sin
r
(r 3 sin ) +
(0) +
(0) = 1
r 2 sin 3r 2 sin + 0 + 0 = 3
Note :- in Cartesian coordinates, the corresponding vector is F = xi + yj + zk with F = 1 + 1 + 1 = 3 (hence consistency).
(b)
F = 1r 2 sin
r
(r 2 sin F r ) +
(r sin F ) +
(rF )
= 1r 2 sin
r
(r 2 sin r 2 sin ) +
(r sin 0) +
(r 0)
= 1r 2 sin
r
(r 4 sin2 ) +
(0) +
(0)
= 1r 2 sin
4r 3 sin2 + 0 + 0 = 4r sin
(c)
F = 1
r2
sin
r
(r 2 sin F r ) +
(r sin F ) +
(rF )
= 1r 2 sin
r
(r 2 sin r sin ) +
(r sin r 2 sin ) +
(r r cos )
= 1r 2 sin
r
(r 3 sin2 ) +
(r 3 sin sin ) +
(r 2 cos )
= 1r 2 sin
3r 2 sin2 + r 3 cos sin + 0 = 3 sin + r cot sin
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Example 27Find F for the following vector elds F .
(a) F = r k r , where k is a constant (b) F = r 2 cos r + sin + sin 2
Solution
(a)
F = 1r 2 sin
r r r 2 sin
r
F r rF r2 sin F
= 1r 2 sin
r r r 2 sin
r
r k r 0 r 2 sin 0
= 1r 2 sin
(0)
(0) r +
(r k ) r
(0) r
+ r
(0)
(r k ) r 2 sin
= 0 r + 0 + 0 = 0
(b)
F = 1r 2 sin
r r r sin
r
F r rF r sin F
= 1
r 2 sin
r r r sin
r
r 2 cos r sin r sin sin2
= 1r 2 sin
(r sin3 )
(r sin ) r +
(r 2 cos ) r
(r sin3 ) r
+ r
(r sin )
(r 2 cos ) r sin
= 1r 2 sin
3r sin2 cos + 0 r + 0 sin3 r + (sin + r sin ) r sin
= 3sin cos
r
r sin2
r
+ (1 + r 2 )
r
sin
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Task skUsing spherical polar coordinates, nd f for
(a) f = r 4
(b) f = rr 2 + 1
(c) f = r 2 sin2 cos
Your solution
Answer(a) 4r 3 r ,
(b) 1 r 2
(1 + r 2 )2r ,
(c) r
(r 2 sin2 cos ) r + 1r
(r 2 sin2 cos ) + 1
r 2 sin
(r 2 sin2 cos )
= 2 r sin2 cos r + 2 r cos2 cos 2r cos sin
Exercises
1. For F = r sin r + r cos + r sin , nd F and F .
2. For F = r 4 cos r + r 4 sin , nd F and F .
3. For F = r 2 cos r + cos nd ( F ).
Answers
1. cos(cot + cosec) + 3 sin , cot 2
sin r 2sin + (2cos cos )
2. 0, 2r 5 sin
3. 0
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