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olc.tex 1

Orthogonal Curvilinear Coordinates

Phyllis R. Nelson

Electrical and Computer Engineering

California State Polytechnic University, Pomona

Spring 2014

1 Introduction

This article reviews the basic mathematical and visualization skills you willneed in ECE 302. You have already seen this material in your requiredmath courses. The purpose of this review is to present these concepts ina form that improves your ability to use them to describe electromagneticphenomena. My goal is to present this mathematics as a useful languagefor describing, visualizing, and understanding concepts such as forces, fields,currents, and waves that are central to electromagnetics.

In introductory electromagnetics, students become familiar with the ba-sic concepts, build intuition, and develop computational skills. Thus, manyexamples and problems are symmetrical. In the appropriate coordinate sys-tem, symmetry reduces the dimensionality of the equations from three toone or two, eliminating much of the computational complexity. Many ofthe standard electromagnetics examples are symmetric in spherical or cylin-drical coordinates, rather than in the Cartesian (rectangular) coordinateswith which you are probably most comfortable. All of the problems andexamples in this course can be solved using Cartesian coordinates. How-ever, the resulting expressions are unnecessarily complicated. Working outsuch expressions exercises a student’s ability to carry out mathematical ma-nipulations, but does little to build an understanding of the fundamentalconcepts. Using mathematics to model the world should clarify rather thancomplicate. One of my challenges is to teach you how to use vector calculusin that way.

In electromagnetics, math is used as a language. As with any language,the more familiar it becomes, the better a tool it becomes for describing and

P. R. Nelson Spring 2014

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understanding the phenomena we study. Because symmetry is such a usefulapproach for simplifying expressions and exposing the underlying electro-magnetics concepts, it is important to be comfortable with the basic vectorand vector calculus operations in the three most common coordinate systems(rectangular, cylindrical, and spherical). Orthogonal curvilinear coordinates(OLC) is a formalism that emphasizes the similarities of these three coordi-nate systems rather than their differences. This approach helps in choosingthe appropriate coordinate system for a particular situation, expressing theresulting equations in the simplest possible form, and therefore facilitatingunderstanding and insight. The purpose of this article is to present orthog-onal curvilinear coordinates as they are used in electromagnetics.

1.1 Writing Coulomb’s law in various coordinate systems

The following examples illustrate the advantages and disadvantages of dif-ferent choices of coordinate systems for writing Coulomb’s law, the formulafor the force on a point charge q1 caused by another point charge q2 locateda distance d away. The magnitude of this force (in SI units) is

|~F12| =q1q24πǫd2

(1)

and the direction (assuming that both q1 and q2 are positive) is repulsive(the vector points in the direction from q2 toward q1). The charges and theforce are shown in Figure 1

~F12

bq2

bq1

Figure 1: Coulomb’s law for two point charges

1.1.1 Coulomb’s law in rectangular coordinates

In this example, I’ll implement equation (1), Coulomb’s law, to calculate themagnitude and direction of the force on a charge q1 at (x1, y1, z1) caused bya charge q2 at (x2, y2, z2). The distance between the charges is

d =√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 (2)


olc.tex 3

Substituting (2) into(1) gives

|~F12| =q1q2

4πǫ [(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2]

(3)

Next I need to find the unit vector that points from q2 to q1.

u =x(x1 − x2) + y(y1 − y2) + z(z1 − z2)√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2

(4)

Finally, I can combine the magnitude expression in (3) with the unitvector from (4) to give

~F12 =q1q2[x(x1 − x2) + y(y1 − y2) + z(z1 − z2)]

4πǫ [(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2]

3/2(5)

The result in equation (5) is completely general. That is, substitutingvalues for q1, q2, x1, y1, z1, x2, y2, and z2 into the equation will give anumerical result for each of the components of the vector ~F12. For example,if q2 is at the origin then

~F12 =q1q2[x(x1) + y(y1) + z(z1)]

4πǫ [(x1)2 + (y1)2 + (z1)2]

3/2(6)

You should check this result for practice!

1.1.2 Coulomb’s law in spherical coordinates

Now consider calculating the force ~F12 in spherical coordinates. The goalis to make the results as simple as possible, so let’s use what we learnedfrom doing the problem in rectangular coordinates: put q2 at the origin. Wecould, of course, convert equation (6) to spherical coordinates, but I hope toconvince you that it is much easier to start by using spherical coordinatesfrom the beginning rather than doing the conversion.

If q2 is at the origin, then the distance d between q1 and q2 is r, theradial coordinate in a spherical system. The unit vector pointing from q2 toq1 is r, the radial unit vector in spherical coordinates, and

~F12 = rq1q24πǫr2

(7)


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1.2 Comparisons

Equations (5), (6), and (7) all give rules for computing the vector ~F12. Let’scompare them to see when each one might be useful. The rectangular formin (5) gives a means of finding both the magnitude and direction of ~F12 forarbitrary locations of the two charges relative to the origin of coordinatesand to each other. The other two do not.

The spherical result in (7) closely resembles the original statement ofCoulomb’s law in equation (1). This form makes it easy to see that all pointsat the same distance r from the charge q2 have the same magnitude of ~F12.Also, there is only one unit vector when using spherical coordinates, so thedirection of the force is easy to visualize. The equivalent rectangular resultin equation (6) has all three unit vectors and three distance components.The magnitude and direction aren’t obvious.

However, spherical coordinates pose a problem if I want to add the forceson q1 from two or more other charges. I can find the magnitudes of the forces~F12 and ~F13 easily from equation (7). Unfortunately, since the unit vectorr points in different directions for the two forces, performing the vectoraddition is not simple in spherical coordinates.

The lesson to be learned from these examples is that no one coordinatesystem is best for all applications. With that statement as motivation, let’slook at how you have been taught to visualize coordinate systems in thepast, what properties those coordinate systems have in common, and howto express the vector calculus operations used in electromagnetics in waysthat aid visualization and understanding.

1.3 How coordinate systems are taught

I’ve learned from teaching this course that students want to use rectangularcoordinates. The reasons for this preference are clear. In elementary school,the number line is introduced to visualize relationships between numbers.Somewhat later, graphing pairs of numbers (and, later, functions of onevariable) is taught using two perpendicular number lines that cross at theirzero points. One of the lines points to the right and the other points up.(Did you ever wonder why it’s done that way? More to the point, didyour teacher ever give a reason for the convention?) There is an underlyingtheme: the coordinate axes define a coordinate system. You have been taughtto rely on a property of rectangular coordinates systems that is not sharedby spherical or cylindrical coordinates. In rectangular coordinates, the

unit vectors don’t change direction from point to point.


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When the cylindrical and spherical systems are introduced, the first exer-cises usually involve working out the transformations between these systemsand rectangular coordinates. The unintentional message is that these newsystems are secondary tools. It’s no wonder that a rectangular system seemsto be better than the others. After all, it’s the one that has been used asthe reference for all others since you first were taught to visualize numericaland mathematical concepts.

The Coulomb’s law example above shows that spherical coordinates havesignificant advantages in helping to understand and visualize a particularresult. You probably had no trouble visualizing the direction of the forcevector even though it was described using spherical coordinates. The keyhere is that the physical system being modeled has spherical symmetry.The charge q2 looks the same from all directions. Matching the physicalsymmetry to the coordinate system symmetry by putting q2 at the origin andusing spherical coordinates made Coulomb’s law simple to write and simpleto understand. The concept of choosing a coordinate system to simplifythe form of the expressions that describe a particular physical situation hasnot been stressed in your previous education, but it is very important forlearning electromagnetics.

Matching the symmetry of a given physical configuration with the appro-priate coordinate system helps build mathematical models that clarify howstuff works. The barrier to treating rectangular, cylindrical, and sphericalcoordinate systems on an equal footing is that the unit vectors are not thebest way to visualize all three systems. Generalized orthogonol curvilinearcoordinates provides the insight that links these systems in a useful way.

2 Generalized orthogonal curvilinear coordinates

Before introducing the general concepts, let’s look at an example that demon-strates the difficulty with using unit vectors to visualize coordinate systems.

Example 1 At the point (x, y, z) = (1, 2, 3) the radial unit vec-tor in cylindrical coordinates ρ = (x+ 2y) /

√5, but ρ = (x+ y) /

√2

when (x, y, z) = (2, 2, 3) . Clearly, the direction of ρ changesfrom point to point, as is shown in Figure 2.

2.1 Surfaces of constant coordinate

By now you might see where I’m heading in this discussion. The difficultiesyou have with using spherical and cylindrical coordinates are usually because


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x

y

z

bb

Figure 2: The unit vector ρ takes different directions at different points.

you have trouble visualizing the unit vectors, which change direction frompoint to point. Hopefully by now you are ready to consider a differentapproach, one that applies equally to rectangular, spherical, and cylindricalsystems. That unifying concept is surfaces of constant coordinate.

Example 2 The plane z = 3, the surface of a cylinder of radius2, and the surface of a cone at an angle of 30◦ to the positivez axis can all be constant-coordinate surfaces in the appropriatecoordinate systems. (Can you visualize these surfaces?) Each ofthese surfaces can be described by the value of only one coordinate(z, ρ, or θ) by choosing the appropriate system. If we were touse only the rectangular system, then two of these simple surfaceswould be represented by more complex formulas involving two orthree coordinates: x2 + y2 = 4 and z/

√

x2 + y2 + z2 = cos 30◦.

Example 2 demonstrates that an equation of the form

coordinate = value

determines a surface. Three different types of surfaces are needed to uniquelyspecify a point in three-dimensional space. The rules for associating num-bers (the coordinates) with these surfaces can be used to define a coordinatesystem. That is, the three coordinates of a point can be found by determin-ing the values associated with the three constant-coordinate surfaces (oneof each type) which intersect there.

Example 3 In rectangular coordinates, all three types of sur-faces are planes. These planes can be pictured as stacked sheets


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of paper, one stack perpendicular to each of the coordinate axes.However, in other systems the surfaces are not necessarily pla-nar. The family of constant-r surfaces in spherical coordinatesare concentric spheres, and constant-ρ surfaces are coaxial cylin-ders in cylindrical coordinates.

Now we need to express this new approach to defining a point using math-ematical expressions instead of words and visualizations. In developing thisformalism, I will continue to rely on your familiarity with rectangular coor-dinates to develop the ideas, but my goal is to make cylindrical and sphericalcoordinates as easy to use as rectangular.

Our coordinate systems will be defined by three functions, f1, f2, andf3, each of which describes a family of surfaces. Setting fi = qi definesthe surface by the coordinate (value) qi. Given the numbers qi, each of thefollowing three equations describes a specific surface in three-dimensionalspace, and the three equations taken together define a point, the point wherethey intersect.

f1(x, y, z) = q1

f2(x, y, z) = q2 (8)

f3(x, y, z) = q3

For example, in spherical coordinates f1 describes all the spheres centeredat the origin. (In terms of rectangular coordinates f1 =

√

x2 + y2 + z2.) Inboth rectangular and cylindrical coordinates, the f3 surfaces are the samefamily of parallel planes f3 = z.

Of course, I can’t choose three completely arbitrary functions, and somechoices will be much more useful than others. First of all, useful sets of threefunctions fi will describe surfaces such that, for any choice of numbers qi, thethree surfaces generated by equations (8) will intersect at one and only onepoint, ensuring that the set of three numbers (coordinates) (q1, q2, q3) alwayslocates exactly one point (not zero or more than one). This generalizedcoordinate system is called “curvilinear.”

The variables used to name the functions fi in the general system (curvi-linear) and in the three specific systems are all listed in Table 1. The func-tions for the cylindrical and spherical systems are also described there usingthe familiar rectangular (x, y, z) description of the location of a point.

The curvilinear coordinates I have described so far are too general for ourapplications. They lack an important property shared by the rectangular,


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Curvilinear Rectangular Cylindrical Spherical

f1 x ρ =√

x2 + y2 r =√

x2 + y2 + z2

f2 y φ = tan−1 (y/x) θ = cos−1 (z/r)

f3 z z φ = tan−1 (y/x)

Table 1: The coordinate functions fi

cylindrical, and spherical systems: the three constant-coordinate surfacesare always orthogonal at the point where they intersect. This property is souseful that it is built into all of the allowable functions fi for the coordinatesystems we will be using. Constructing sets of functions with this propertyis not simple, but luckily we don’t have to do that because the rectangular,cylindrical, and spherical systems are orthogonal.

So far we have a method for locating points that is independent of theunit vectors. Now we need a way to find the three unit vectors at any point.

3 Unit vectors

Points in three-dimensional space are defined with ordered triplets of num-bers (q1, q2, q3) by using the functions fi of equation (8) as specified inTable 1. The three functional equations define three surfaces of constantcoordinate which intersect at the point. In order to use this formalism ef-fectively, it is important that you practice until you can easily visualize theshapes of each of the three coordinate surfaces in the rectangular, cylin-drical, and spherical systems. It may be helpful to build yourself physicalmodels using objects like pencils, soup cans, ping pong balls, and cardboard.The better your understanding of these shapes, the more successful you willbe in choosing the best coordinate system for modeling a given real-worldproblem.

If you feel as though you aren’t ready to use orthogonal curvilinear coor-dinates, that’s not surprising. So far there has been no mention of the axesor unit vectors that you have been trained for so long to use to describe andvisualize points and vectors. Although this new approach may feel uncom-fortable for a while, there is an important advantage: locating a point nolonger relies on knowing which way the unit vectors point, so now all threeof our coordinate systems have the same set of rules.


olc.tex 9

In this new formalism, the unit vectors are determined locally usingthe surfaces of constant coordinate at the point of intersection. Let’s startwith a specific example: the surfaces of constant f3 = z in rectangular orcylindrical coordinates. Each value of z defines a plane normal to the z axis.The unit vector z is orthogonal (perpendicular) to all of these planes, andit points in the direction of planes with increasing values of z as shown inFigure 3. The length of the vector z is one unit of z. Since all of the surfacesof constant coordinate are planes in the rectangular system, the unit vectorsx and y can be visualized in the similar way.

x y

z

z = 1.0z = 1.5z = 2.0z = 2.5

z

Figure 3: The unit vector x

Generalizing, the unit vector associated with a particular fi

• is orthogonal to the surface fi = qi at a specific point (q1, q2, q3),

• points in the direction of surfaces with increasing values of qi, and

• is one unit in length.

Some of the coordinates we will be using are defined by angles, not distances.You might think about the meaning of “one unit of length” in such a case.

Example 4 The function f3 = z has the identical meaning inboth cylindrical and rectangular coordinates, so the unit vectoris the same: z. In cylindrical coordinates, f1 represents a fam-ily of infinitely-long coaxial cylinders, each characterized by itsradius f1 = ρ. At any point on one of these cylinders, the unitvector ρ points radially outward, orthogonal to the surface of thecylinder.


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Example 5 The functions f2 = φ in cylindrical coordinates andf3 = φ in spherical coordinates describe a family of half-planeswith the edge along the ρ = 0 surface (the z axis of the rectan-gular system with the same origin). The associated unit vectoris orthogonal to the half-plane, and also tangent to the cylindri-cal or spherical constant-coordinate surfaces, and it points in thedirection of increasing φ.

The names I will be using for the unit vectors in our four coordinatesystems are given in Table 2. Note that these may not be the same as thoseused in your text, and may not be what you used in your math courses.Unfortunately, there is no one universally-accepted convention for namingeither the coordinates or unit vectors. For example, you may see “i, j, k,”“ax, ay, az,” or “ux, uy, uz” for the unit vectors in a rectangular system.Similarly, you may find “r, φ, z” used for the cylindrical system or “R, θ,φ” for the spherical system. I prefer to use the same symbols in typeset textthat I use when working by hand, and will use the symbols in this documentand throughout the course. Since there is no generally-accepted convention,it is useful to get accustomed to seeing different notations. Ultimately,it is important to make sure that you understand each author’s choice ofnotation.


u x ρ r

v y φ θ

w z z φ

Table 2: Unit vectors.

4 Right-handedness

There is another property of the surfaces, coordinates, and unit vectors thatwe haven’t considered yet, and it has to do with the following question. Whyare the functions fi used in a particular order? Specifically, why is the samefunction (φ) used as f2 in a cylindrical system and f3 in a spherical system?This choice is not arbitrary. Orthogonality is not the only property ourthree coordinate systems have in common. They also share the convention


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that the vector product (cross product) of the first and second unit vectorsmust equal the third unit vector (u × v = w). So we have x × y = z andρ × φ = z. In spherical coordinates r × φ = −θ and r × θ = φ. Thusf2 = θ and f3 = φ. This “right-handedness” is an arbitrary convention forour coordinate systems. However, once we have made that choice for onesystem, making the same choice for the rest of our systems is important forensuring that the expressions for the laws of electromagnetics have the sameform in all three coordinate systems.

The choice of the first coordinate (f1) is arbitrary, and is also determinedby convention. Once f1 is chosen, the order of assignment of the constant-coordinate surface functions f2 and f3 must give a right-handed system asdetermined from the definition of the vector cross product. That is, inrectangular coordinates we could choose (x, y, z), (y, z, x), or (z, x, y), whichare all right-handed, but not (x, z, y), which is left-handed.

The condition for right-handedness in the generalized orthogonal curvi-linear coordinate system notation is

u× v = w (9)

5 Scale factors and differential distances

There is another way in which rectangular coordinates are special: all threeof the fi are constant-distance surfaces. In both cylindrical and spherical co-ordinates, at least one of the fi represent constant-angle surfaces. Constant-distance surfaces have the property that the differential change dqi in thecoordinate qi is the same as the differential distance dli between the surfacesfi = qi and fi = qi + dqi. However, constant-angle surfaces have the morecomplex relationship

dli = hidqi (10)

where hi is the appropriate radius.

Example 6 Figure 4 shows two lines rotated by an angle dφ.The distance between the two lines depends on the radial dis-tance from the axis. If dφ is small enough that curvature can beneglected, the distance is ρ dφ.

The scale factors hi for each of the coordinate systems we will be usingare given in Table 3. Note that the rectangular system, the only systemin which all of the fuctions represent constant-distance surfaces, is also theonly one for which all hi = 1.


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b

0 1 cm 3 cm

dφ 3 dφ cmdφ cm

Figure 4: The distance between surfaces of constant angular coordinatedepends on the distance to the axis.


h1 1 1 1

h2 1 ρ r

h3 1 1 r sin θ

Table 3: The scale factors hi.

To solve electromagnetic problems you will need to be able to carry outone-, two-, and three-dimensional line, surface, and volume integrals. Thedifferential distance given in equation (10), together with the scale factors inTable 3, help in writing down the differential distances, areas, and volumesrequired for these operations. The following examples demonstrate the useof these scale factors.

The arbitrary differential displacement vector is constructed by movinga differential distance in each of the unit vector directions.

~dl = u h1dq1 + v h2dq2 + w h3dq3 (11)

In rectangular coordinates this vector is

~dl = x dx+ y dy + z dz (12)

and in spherical coordinates it is

~dl = r dr + y r dθ + φ r sin θdφ (13)

Example 7 The differential unit of area over a constant-ρ(constant-f1) surface in cylindrical coordinates is

dl2 dl3 = (h2dq2)(h3dq3) = (ρ dφ) dz


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as shown in Figure 5. Thus, the surface area of a cylinder ofradius 2 cm and height 5 cm is

A =

∫

cylinder

dA =

∫

5 cm

0

∫

2π

0

(2 cm) dφ dz = 20π cm2

x

y

z

ρ

dz

ρ dφ

Figure 5: A differential constant-ρ surface in cylindrical coordinates.

Example 8 The differential unit of volume in spherical coordi-nates, which is illustrated in Figure 6 is

dl1 dl2 dl3 = (h1dq1)(h2dq2)(h3dq3) = (dr)(rdθ)(r sin θ dφ)

Thus, the volume of a sphere of radius R0 is

V =

∫

sphere

dτ =

∫

2π

0

∫ π

0

∫ R0

0

r2 sin θ dr dθ dφ

=

∫

2π

0

dφ

∫ π

0

sin θ dθ

∫ R0

0

r2 dr

= (2π)(2)

(

R3

0

3

)

=4π

3R3

0 (14)

6 Unit vector conversions

Equation 11 provides a method for expressing one set of unit vectors in termsof another. This method requires that you can transfrom the coordinates.


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x

y

z

rdθdr

r sin θ dφ

Figure 6: A differential volume in spherical coordinates

Example 9 Suppose we need to express the unit vectors of cylin-drical coordinates in terms of the rectangular unit vectors. Theconversion can be done by writing a differential length in bothsystems.

~dl = ρ dρ+ φ ρdφ+ z dz = x dx+ y dy + z dz

= x

(

∂x

∂ρdρ+

∂x

∂φdφ

)

+ y

(

∂y

∂ρdρ+

∂y

∂φdφ

)

+ z dz

=

(

x∂x

∂ρ+ y

∂y

∂ρ

)

dρ+

(

x∂x

∂φ+ y

∂y

∂φ

)

dφ+ z dz (15)

Using the coordinate transformations x = ρ cosφ and y = ρ sinφ,

ρ = x cosφ+ y sinφ (16)

φ ρ = (−x sinφ+ y cosφ) ρ (17)

Test your understanding by finding x and y in the cylindrical sys-tem (the inverse of the transformation done above). The resultis

x = ρ cosφ− φ sinφ (18)

y = ρ sinφ+ φ cosφ (19)

7 Vector derivatives

Now we come to a much tougher test of my claim that orthogonal curvilinearcoordinates will simplify the vector calculus used in electromagnetics: thegradient, divergence, curl, and Laplacian. These operators give the vector


olc.tex 15

and scalar derivatives of scalar and vector fields (scalar and vector functionsof position).

LetS = S(q1, q2, q3) (20)

be an arbitrary scalar field like temperature or voltage and

~V = uV1(q1, q2, q3) + vV2(q1, q2, q3) + wV3(q1, q2, q3) (21)

be an arbitrary vector function like force or the electric field.

7.1 Gradient

The gradient is one of the vector derivative operations. It is used, for ex-ample, to find the force given the potential energy. Since the electric field isforce per unit charge and the voltage (electrostatic potential) is electrostaticpotential energy per unit charge, the gradient is also used to find the electricfield given the voltage.

To understand the gradient, consider how a change dS in the scalarfunction S depends on changes dx, dy, and dz in the coordinates.

dS =

(

∂S

∂x

)

dx+

(

∂S

∂y

)

dy +

(

∂S

∂z

)

dz (22)

This expression can be written as the dot product of the differential lengthvector

~dl = x dx+ y dy + z dz (23)

and a second vector called the gradient ~∇S of S, where

~∇S = x∂S

∂x+ y

∂S

∂y+ z

∂S

∂z(24)

That is, a differential change in the function S where the change is taken inthe direction of ~dl is

dS = ~∇S · ~dl =∣

∣

∣

~∇S∣

∣

∣

∣

∣

∣

~dl∣

∣

∣cosα ⇒ dS

∣

∣

∣

~dl∣

∣

∣

=∣

∣

∣

~∇S∣

∣

∣cosα (25)

where α is the angle between the vectors ~∇S and ~dl.Suppose I choose a constant length for ~dl and rotate its direction about a

point so that only the angle α between ~∇S and ~dl changes. Then Equation 25tells us that the change dS in the function S at this point in the distance


olc.tex 16

|~dl| is maximum when α = 0 (~dl has the same direction as ~∇S). Thatis, ~∇S is a vector whose magnitude is the maximum value of dS/dl andwhose direction is the direction in which dS/dl is maximum. Translatingthis concept into the generalized OLC formalism using the expression forthe differential distance in the direction of a unit vector gives

~∇S = u1

h1

∂S

∂q1+ v

1

h2

∂S

∂q2+ w

1

h3

∂S

∂q3(26)

Example 10 The surface defined by h = x2y is plotted in Fig-ure 7. The gradient of h is ~∇h = x 2xy + y x2. A vector plot ofthe gradient is shown in Figure 8.

-3 -2 -1 0 1 2 3-3-2

-1 0

1 2

3

-30-20-10

0 10 20 30

x2 * y 12 10 8 6 4 2 0 -2 -4 -6 -8 -10 -12

x

y

Figure 7: The surface x2y.

7.2 The “Del” Operator

Equations 24 and 26 can be written using a vector differentiation operator

~∇ = u1

h1

∂

∂q1+ v

1

h2

∂

∂q2+ w

1

h3

∂

∂q3(27)

As we have seen above, applying ~∇ to a scalar function gives a vector func-tion: the gradient. Now I’ll explore what happens when this operator is


olc.tex 17

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3

y

x

Figure 8: The gradient of x2y.

applied to a vector function. There are two possibilities, the scalar (dot)product and the vector (cross) product. Each of them is used in electromag-netics.

7.3 Divergence

The divergence describes the “sources” and “sinks” of a vector field. Oneof the uses of the divergence in electromagnetics is to find the distributionof charge that creates an electric field. Two of Maxwell’s four equations usethe divergence to relate the electric and magnetic fields to sources.

Since you have a lot of everyday experience with water, I’ll use water flowto illustrate the divergence. First consider water flowing in a pipe. Assumethat the velocity ~v of the water is constant across the cross section of thepipe and points along the axis. If the pipe ends with a cut perpendicular tothe axis as shown on the left in Figure 9, then the weight of water collectedin a bucket in a time t is W = η|~v|Atg where η is the density of water, A isthe area of the open end of the pipe, and g is the gravitational acceleration.That is, the rate of flow of water weight into the bucket per unit area of thepipe is W/(At) = ηg|~v|.

Now suppose the end of the pipe is cut at an angle θ to the axis as shownon the right in Figure 9. The rate at which water is supplied to the end


olc.tex 18

of the pipe doesn’t change, so the rate at which water is collected in thebucket should be the same even though the area Aθ of the opening is larger.Some thought should convince you that it is the projection of the area ofthe opening in the direction of the water velocity that matters. That is, |~v|should be replaced with v · n where n is a vector normal to the opening ofthe pipe. Then W/t = (ηg~v) · (nA) = ηg|~v|Aθ cos θ = ηg|~v|A as before.

~v ~vθ

Figure 9: Water flow out of a pipe.

This water flow example shows that, for a vector function ~V , ~V · ~dσ isthe proportion of |~V | that “flows through” the differential surface area dσ.(In the example, ~V = ηg~v and ~dσ = ndA.)

Now consider the total “flow” of ~V “through” the surface of the differen-tial volume with sides of lengths dx, dy, and dz that is shown in Figure 10.On the face whose normal is x, ~dσ = x dy dz. On the face parallel to it,

x

y

z

dydx

dz

Figure 10: A differential valume element.

~dσ = −x dy dz. Thus, for this pair of faces, the total “flow” of ~V out of the


olc.tex 19

volume is

~V · ~dσ∣

∣

∣

x=

(

~V∣

∣

∣

x=dx− ~V

∣

∣

∣

x=0

)

· (x dy dz) =(

∂Vx∂x

dx

)

dy dz (28)

Adding the contributions of the other two pairs of faces gives the total “flow”of ~V out of this volume.

~V · ~dσ =

(

∂Vx∂x

+∂Vy

∂y+

∂Vz∂z

)

dx dy dz =(

~∇ · ~V)

dx dy dz (29)

Thus, ~∇ · ~V is a local measure (at a particular point) of the source of ~V perunit volume. We have, as expected,

~∇ · ~V =∂Vx∂x

+∂Vy

∂y+

∂Vz∂z

(30)

or, for the general OLC system,

~∇ · ~V =1

h1h2h3

[

∂

∂q1(V1h2h3) +

∂

∂q2(V2h3h1) +

∂

∂q3(V3h1h2)

]

(31)

Another image for the divergence is that it measures how much ~V spreadsout from a point. A two-dimensional visualization aid is to imagine leavesfloating on the surface of a pond or stream. If the leaves spread out awayfrom a point, then that point has a positive divergence. If they move towardthe point, then it has a negative divergence.

Example 11 The vector function ~V = ρ ρ2 is plotted in Fig-ure 11. The divergence ~∇· ~V = 3ρ is plotted in Figure 12. Thesetwo images show that for larger distance away from the originthe increasing magnitude of ~V requires larger and larger local“sources” (divergence).

7.4 Curl

The curl gives the local vorticity (rotation) of a vector field. Two of Maxwell’sfour equations relate the electric and magnetic fields through the curl. Oneuse of these equations is to derive the wave equation that describes electro-magnetic waves.

In two dimensions, the leaves on the surface of a pond or stream canalso demonstrate the curl. If a leaf rotates, then the water flow at thatpoint has a curl. The formula for the curl in OLC is given in Equation 32,


olc.tex 20

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3

Figure 11: The vector function ~V = ρ ρ2.

-3 -2 -1 0 1 2 3-3-2

-1 0

1 2

3

0 2 4 6 8

10 12 14

Divergence 12 8 4

x

y

Figure 12: The divergence ~∇ · ~V = 3ρ of the vector function ~V = ρ ρ2.


olc.tex 21

and Example 12 gives an opportunity to try this formula in two coordinatesystems.

~∇× ~V =1

h1h2h3

∣

∣

∣

∣

∣

∣

∣

∣

∣

h1u h2v h3w

∂∂q1

∂∂q2

∂∂q3

h1V1 h2V2 h3V3

∣

∣

∣

∣

∣

∣

∣

∣

∣

(32)

Example 12 The vector function ~V = x y − y x = φ(−ρ) isplotted in Figure 13. The curl ~∇× ~V = z(−2) everywhere. Takethe point (2, 0) as an example. If ~V represents water flow on thesurface of a pond, then a leaf at that point would rotate cloclwisebecause the flow to the right is faster than the flow to the left.The vector direction of the associated torque is given by the right-hand rule, so it points to −z.

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3

y

x

Figure 13: The vector function ~V = x y − y x = φ(−ρ).


olc.tex 22

8 Second derivatives

There are five possible second derivatives: the divergence and curl of thegradient of a scalar field, the gradient of the divergence of a vector field, andthe divergence and curl of the curl of a vector field. Each of these operationswill be examined in the following sections. Three of these expressions leadto formulas which I have not derived. These derivations are easily found intextbooks on vector calculus or mathematical physics.

8.1 Divergence of the gradient of a scalar

The divergence of the gradient of a scalar field is called the Laplacian. Thisoperation is so widely used that it has its own symbol, ∇2. The Laplacian isused in electromagnetics in the relationship between voltage and charge andin the wave equation. It is also used in the diffusion equation that describesheat conduction.

The Laplacian of a scalar field is

∇2S = ~∇ · ~∇S =

1

h1h2h3

[

∂

∂q1

(

h2h3h1

∂S

∂q1

)

+∂

∂q2

(

h3h1h2

∂S

∂q2

)

+∂

∂q3

(

h1h2h3

∂S

∂q3

)]

(33)

8.2 Curl of the gradient of a scalar

This operation gives a useful identity:

~∇×(

~∇S)

= 0 (34)

8.3 Gradient of the divergence of a vector

The operation ~∇(

~∇ · ~V)

is not common in physics. However, it is important

to note that this is not the Laplacian of a vector.

8.4 Divergence of the curl of a vector

This operation also leads to a useful identity:

~∇ ·(

~∇× ~V)

= 0 (35)


olc.tex 23

8.5 Curl of the curl of a vector

This operation gives a definition of the Laplacian of a vector. Using thedefinition of ~∇, it can be shown that

∇2~V = ~∇(

~∇ · ~V)

− ~∇× ~∇× ~V (36)

This is a tricky concept. The Laplacian of a vector is a vector. In rectangularcoordinates, the derivatives of the unit vectors with respect to position areall zero, so this operation takes the simple form

∇2~v = x∇2Vx + y∇2Vy + z∇2Vz (37)

It is important to remember that this simplification does not apply to othercoordinate systems.


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