28 3 Orthog Cvlnr Coords

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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

HELM (2005):Section 28.3: Orthogonal Curvilinear Coordinates

37
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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

38 HELM (2005):Workbook 28: Differential Vector Calculus


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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


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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

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+ 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 )


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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

HELM (2005): 53

28 3 Orthog Cvlnr Coords

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