Orthogonal Curvilinear Coordinates · Mathematical Methods in the Physics Sciences, Chapter 10, Sections 6–9. These approaches tend to be more mathematical and general than the

Vector Calculus: Orthogonal Curvilinear Coordinates 1

Orthogonal Curvilinear Coordinates

Contents

1 Introduction 1

2 Cylindrical Polar Coordinates 12.1 Position Vector . . . . . . . . . . . . . 12.2 Line Element . . . . . . . . . . . . . . 22.3 Surface Elements . . . . . . . . . . . 22.4 Volume Elements . . . . . . . . . . . 22.5 Gradient Operator . . . . . . . . . . . 32.6 Divergence Operator . . . . . . . . . 32.7 Curl Operator . . . . . . . . . . . . . 4

2.7.1 ρρρ Component . . . . . . . . . 42.7.2 φφφ Component . . . . . . . . . 52.7.3 k Component . . . . . . . . . 52.7.4 ∇ × B . . . . . . . . . . . . . . 6

3 Spherical Polar Coordinates 63.1 Position Vector . . . . . . . . . . . . . 63.2 Line Element . . . . . . . . . . . . . . 63.3 Surface Elements . . . . . . . . . . . 63.4 Volume Element . . . . . . . . . . . . 73.5 Gradient Operator . . . . . . . . . . . 73.6 Divergence Operator . . . . . . . . . 73.7 Curl Operator . . . . . . . . . . . . . 7

3.7.1 r component . . . . . . . . . . 83.7.2 θθθ component . . . . . . . . . 83.7.3 φφφ component . . . . . . . . . . 83.7.4 ∇ × B . . . . . . . . . . . . . . 9

4 General Orthogonal Coordinate Systems 9

5 Non-orthogonal Coodinate Systems 9

1 Introduction

Many problems in physics have a central point oraxis. For these, cartesian (x, y, z) coordinates canbe tedious, and it is natural to introduce a coordi-nate system that reflects the shapes and symme-tries of the problem. Examples include cylindricaland spherical polar coordinates, which we shall ex-plore further here, parabolic or hyperbolic coordi-nates, and others. Many are constructed so thatthe corresponding unit vectors (i, j, k), (ρρρ, φφφ, k), etc.,are orthogonal (i.e., perpendicular to one another).Since in these systems lines of constant compo-nents (e.g., constant r) are curved, we refer to such

coordinate systems as “orthogonal curvilinear coor-dinates.” Below is a summary of the main aspectsof two of the most important systems, cylindrical andspherical polar coordinates. Many of the steps pre-sented take subtle advantage of the orthogonal na-ture of these systems.

You can find complementary material in both Ri-ley et al., Mathematical Methods for Physics and En-gineering, Sections 10.9 and 10.10, and in Boas,Mathematical Methods in the Physics Sciences,Chapter 10, Sections 6–9. These approaches tendto be more mathematical and general than the onegiven here.

You will not necessarily be expected to repro-duce the calculations given below. However, muchof it is quite instructive in terms of the way cylindricaland spherical polar coordinates work. Additionally,seeing how the forms of grad, div, and curl, togetherwith line, surface and volume elements, are derivedin different systems helps provide some insight intotheir interpretation.

2 Cylindrical Polar Coordinates

z

y

x

ρz

dρdz

ρ dΦ

ρ

zΦ

dΦΦ

r

Figure 1: Cylindrical polar coordinates and theirvolume element.

To begin, let us recall some basics about cylin-drical polar coordinates (see Figure 1). From this

Vector Calculus: Orthogonal Curvilinear Coordinates 2

figure it is apparent that

x = ρ cos φ (1)

y = ρ sin φ (2)

z = z (3)

It is quite common to use the symbol r instead ofρ, and you will often encounter the notation (r , φ, z)for cylindrical polar coordinates. Here, to avoid allpossible confusion with r in spherical polars, we willuse ρ for the distance from the z-axis.

2.1 Position Vector

The point P indicated by the black dot in Figure 1has position vector

r = ρ ρρρ + z k (4)

At first sight this might seem strange; where isthe φφφ component? But if you look at Figure 1 youcan see that you do need to know three things tolocate P: ρ, z and ρρρ = ρρρ(φ). So the informationabout φ is contained in ρρρ; different φ values giveyou ρρρ unit vectors that point in different directions.A general vector field B will have three componentsB(ρ, φ, z) = Bρ ρρρ + Bφ φφφ + Bz k; the position vector isspecial in this regard.

From Figure 1 or from (1)–(3) we see that

r = ρ cos φ i + ρ sin φ j + z k (5)

= ρ(cos φ i + sin φ j

)+ z k (6)

= ρ ρρρ + z k (7)

from which we deduce that

ρρρ = cos φ i + sin φ j (8)

From Figure 1 we can find the third unit vector

φφφ = − sin φ i + cos φ j (9)

You can see this either by constructing the neces-sary trigonometry or simply note that φφφ must lie inthe x − y plane and be perpendicular to ρρρ. Indeed,(ρ, φ) are just the plane polar coordinates (r , θ) indisguise.

Finally, from (8) we see that

dρρρdφ

= φφφ (10)

2.2 Line Element

Now that we have established the representation ofa position r we can proceed to consider a displace-ment dr from that position. Graphically with refer-ence to Figure 1, if we increment ρ by an amount dρthen the vector displacement would be dρ ρρρ. Incre-menting φ by an amount dφ would be a vector dis-placement ρ dφ φφφ. And incrementing z by dz wouldbe a vector displacement dz k. An arbitrary dis-placement would be the vector sum of these, i.e.,

dr = dρ ρρρ + ρ dφ φφφ + dz k (11)

Interestingly, and perhaps reassuringly, you can getto (11) by taking the differential of r directly from (7):

dr =∂r∂ρ

dρ +∂r∂φ

dφ +∂r∂z

dz (12)

= dρ ρρρ + ρdρρρdφ

dφ + dz k (13)

= dρ ρρρ + ρ dφ φφφ + dz k (14)

where we have made use of (10).

2.3 Surface Elements

To find a surface element dS we would need to usethe formal machinery we derived in lecture, namelyif we have a surface S parametrised by two variables(u, v) so that a point on the surface is r = r(u, v),then

dS =∂r∂u×∂r∂v

du dv (15)

Since ρρρ, φφφ, k are mutually orthogonal, you could per-form this cross product in cylindrical polars if thatwas convenient, i.e.,

∂r∂u×∂r∂v

=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

ρρρ φφφ k(∂r∂u

)ρ

(∂r∂u

)φ

(∂r∂u

)z(

∂r∂v

)ρ

(∂r∂v

)φ

(∂r∂v

)z

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣(16)

Let’s look at the surfaces of the elemental vol-ume shown in Figure 1. If we start with the bottomsurface, defined by z = constant = zo say, then wecan parametrise this surface by ρ and φ, and a pointon this surface is described simply by (7) with con-stant z, i.e., r(ρ, φ) = ρ ρρρ(φ) + zo k. So we have

∂r∂ρ

= ρρρ + 0 φφφ + 0 k;∂r∂φ

= 0 ρρρ + ρ φφφ + 0 k

Vector Calculus: Orthogonal Curvilinear Coordinates 3

again making use of (10). Thus

dS =

∣∣∣∣∣∣∣∣∣ρρρ φφφ k1 0 00 ρ 0

∣∣∣∣∣∣∣∣∣ dρ dφ = ρ dρ dφ k (17)

Actually, you should be able to look at Figure 1, seethat this face is roughly rectangular and has an areaρ dφ × dρ and that its normal is k and write downimmediately (17).

In a similar way it is now easy to see (I hope)that the surface element for the flat-sided side facenearest the viewer is

dS = dρ dz φφφ (18)

while that of the inner curved face is

dS = ρ dφ dz ρρρ (19)

Note that for all of these pieces of surface, there isan ambiguity of a ± sign depending on the appli-cation at hand. For example, if we were interestedin the surface elements with normals pointing out ofthe volume element, we would need to take the neg-ative of all the above dS expressions for some faces(see Figure 2 below).

2.4 Volume Elements

Again taking advantage of the orthogonality of theunit vectors and looking at Figure 1 we can imme-diately write down the volume of the differential vol-ume depicted there. This volume is simply the prod-uct of the three sides of the (pseudo-)rectangularvolume, i.e.,

dV = ρ dφ dρ dz (20)

You can also deduce this by calculating the Jacobian∂(x, y, z)∂(ρ, φ, z)

as we did in lecture.

2.5 Gradient Operator

Let’s now turn our attention to the vector calculusoperators, beginning with the gradient. We will dothis by constructing an exact differential of a scalarfunction f by writing

df =∂f∂ρ

dρ +∂f∂φ

dφ +∂f∂z

dz (21)

= ∇f · dr (22)

Δρ

Δz

ρ ΔΦ

dS

1

(ρ+Δρ) ΔΦ

dS

2

dS

3

dS

4

dS

5

dS

6

(ρ, Φ, z)

Figure 2: Volume element in cylindrical polars withsurface elements marked for application of the Di-vergence Theorem

If we now express ∇f in components, i.e., ∇f =(∇f )ρ ρρρ + (∇f )φ φφφ + (∇f )z k, and make use of (7) fordr this gives

df =((∇f )ρ ρρρ + (∇f )φ φφφ + (∇f )z k

)·

·(dρ ρρρ + ρ dφ φφφ + dz k

)= (∇f )ρ dρ + (∇f )φρ dφ + (∇f )z dz (23)

Now dρ, dφ, and dz are all independent, so theircoefficients in (21) and (23) must be equal. Thus

(∇f )ρ =∂f∂ρ

, (∇f )φρ =∂f∂φ

, and (∇f )z =∂f∂z

so we

reach

∇f =∂f∂ρ

ρρρ +1ρ

∂f∂φ

φφφ +∂f∂z

k (24)

2.6 Divergence Operator

We can derive an expression for the divergence op-erator of a vector field B(ρ, φ, z) by applying the Di-vergence Theorem to the elemental volume shownin Figure 1 and expanded in Figure 2. The stepsbelow are essentially the reverse of those used toderive the Divergence Theorem in the first place incartesian coordinates. The underlying concepts areidentical, although the algebra here is a bit moreinvolved, essentially because the lengths of someof the sides of the volume element depend on thevalue of ρ.

Vector Calculus: Orthogonal Curvilinear Coordinates 4

The Divergence Theorem states that$V∇ · B dV =

S

B · dS (25)

When applied to an elemental volume, we can re-move the integration signs to reveal that

∇ · B =6∑

i=1

B · dSi/dV (26)

If we consider dS1 and dS2 to begin with, we cansee from Figure 2 that

dS1 = −ρ∆φ∆z ρρρ

dS2 = +(ρ + ∆ρ)∆φ∆z ρρρ

so that

B · dS1 = −Bρ(ρ, φ∗, z∗)ρ∆φ∆z (27)

B · dS2 =

+Bρ(ρ + ∆ρ, φ∗∗, z∗∗)(ρ + ∆ρ)∆φ∆z (28)

where φ∗ and z∗ denote the values of φ and z forwhich Bρ takes on its average value over dS1, andsimilarly for dS2. By virtue of the mean value theo-rem, φ ≤ φ∗ ≤ (φ + ∆φ), etc. Combining these twoexpressions then yields

B · dS1 + B · dS2 =[(ρ + ∆ρ)Bρ(ρ + ∆ρ, , ) − ρBρ(ρ, , )

]∆φ∆z (29)

where we have suppressed φ∗, etc., quantities forthe sake of brevity. Now as we let the volume shrinkto differential proportions,

∆φ → dφ

∆z → dz

φ∗ → φ

φ∗∗ → φ

z∗ → z

z∗∗ → z

It remains only to let ∆ρ shrink to its limiting differ-ential form:

∆ρ → dρ

(ρ + ∆ρ)Bρ(ρ + ∆ρ) − ρBρ(ρ) →∂(ρBρ)∂ρ

dρ

so that (29) becomes

B · dS1 + B · dS2 =(∂(ρBρ)∂ρ

dρ)

dφ dz (30)

=1ρ

∂(ρBρ)∂ρ

dV (31)

where we have multiplied and divided by ρ in the firstline to form the volume element ρ dρ dφ dz ≡ dV asin (20).

In a similar way, dS3 and dS4 are in the ∓φφφ di-rection and in this case both have area ∆ρ∆z. dS3

is at φ while dS4 is at φ + ∆φ. So the analog to (29)is

B · dS3 + B · dS4 =[Bφ(, φ + ∆φ, ) − Bφ(, φ, )

]∆ρ∆z (32)

which leads to

B · dS3 + B · dS4 =(∂Bφ

∂φdφ

)dρ dz (33)

=1ρ

∂Bφ

∂φdV (34)

Finally, dS5,6 = ∓∆ρ ρ∆φ k so that

B · dS5 + B · dS6 =

[Bz(, , z + ∆z) − Bz(, , z)] ∆ρ ρ∆φ (35)

which leads to

B · dS5 + B · dS6 =(∂Bz

∂zdz

)ρ dρ dφ (36)

=∂Bz

∂zdV (37)

Putting (31), (34), and (37) into the summation in(26) yields the desired result, namely an expressionfor ∇ · B in cylindrical polar coordinates:

∇ · B =1ρ

∂(ρBρ)∂ρ

+1ρ

∂Bφ

∂φ+∂Bz

∂z(38)

2.7 Curl Operator

A similar approach to that in Section 2.6 can be ap-plied to derive an expression for the curl of a vectorfield in cylindrical polar coordinates, this time start-ing from Stoke’s Theorem:�

CB · dr =

"S∇ × B · dS (39)

which for a differential surface element reduces to

nedges∑i=1

B · dri = ∇ × B · dS (40)

We will pick our surface elements from the faces ofthe volume element at ρ, φ, and z shown in Figure 2.

Vector Calculus: Orthogonal Curvilinear Coordinates 5

Δz

ρ ΔΦ

dS

ρ

(ρ, Φ, z)

z+Δz

1

2

3

4

Ф+ ΔΦ

Figure 3: Line integration to determine the ρρρ com-ponent of ∇ × B.

2.7.1 ρρρ Component

The face with normal ρρρ is shown in Figure 3. Thefour edges have dr1 = +ρ∆φ φφφ, dr2 = +∆z k, dr3 =−ρ∆φ φφφ, and dr4 = −∆z k while dSρ = ρ∆φ∆z ρρρ.Thus (40) becomes

B · dr1 + B · dr2 + B · dr3 + B · dr4 =

Bφ(ρ, φ∗, z) ρ∆φ + Bz(ρ, φ + ∆φ, z∗)∆z −

−Bφ(ρ, φ∗∗, z + ∆z) ρ∆φ − Bz(ρ, φ, z∗∗)∆z

= [Bz(ρ, φ + ∆φ, z∗) − Bz(ρ, φ, z∗∗)] ∆z −

−[Bφ(ρ, φ∗∗, z + ∆z) − Bφ(ρ, φ∗, z)

]ρ∆φ

= (∇ × B)ρ ρ∆φ∆z (41)

with φ∗, z∗ again denoting the value within the in-terval for which the relevant component of B takeson its mean value. As we let this surface shrink todifferential proportions, this reduces to:

(∇ × B)ρ ρ dφ dz =(∂Bz

∂φdφ

)dz −

(∂Bφ

∂zdz

)ρdφ

=(1ρ

∂Bz

∂φ−∂Bφ

∂z

)ρ dφ dz (42)

from which we deduce that the ρ component of ∇×Bis

(∇ × B)ρ =1ρ

∂Bz

∂φ−∂Bφ

∂z(43)

Δz

ρ+Δρ

dS

Ф

(ρ, Φ, z)

z+Δz

1

2

3

4

Δρ

Figure 4: Line integration to determine the φφφ com-ponent of ∇ × B.

2.7.2 φφφ Component

The other components of ∇×B follow similarly. Fig-ure 4 shows the φφφ face. Be careful to ensure thatyou go around the edges in a right-handed sensewith respect to dS, which we have chosen here tobe in the +φφφ direction, so that dS = ∆ρ∆z φφφ. Evalu-ating (40) for this face yields

B · dr1 + B · dr2 + B · dr3 + B · dr4 =

= Bz(ρ, , )∆z + Bρ(, , z + ∆z)∆ρ −

−Bz(ρ + ∆ρ, , )∆z − Bρ(, , z)∆ρ

=[Bρ(, , z + ∆z) − Bρ(, , z)

]∆ρ −

− [Bz(ρ + ∆ρ, , ) − Bz(ρ, , )] ∆z

= (∇ × B)φ ∆ρ∆z (44)

Here for brevity and clarity I have omitted the depen-dencies which are either constant along a particularedge or evaluated at some ∗’ed value along themwhile retaining the dependency that characteriseswhich edge we are following.

Letting the surface shrink to its differential formyields

(∇ × B)φ dρ dz =

=(∂Bρ

∂zdz

)dρ −

(∂Bz

∂ρdρ

)dz (45)

from which we can see that

(∇ × B)φ =∂Bρ

∂z−∂Bz

∂ρ(46)

Vector Calculus: Orthogonal Curvilinear Coordinates 6

Φ+ΔΦ

Δρ

ρ ΔΦ

(ρ+Δρ) ΔΦ

dS

z

(ρ, Φ, z)

ρ+Δρ

1

2

3

4

Figure 5: Line integration to determine the k com-ponent of ∇ × B.

2.7.3 k Component

There remains only the k-component of ∇ × B tocalculate, by reference to Figure 5. Following thenow-familiar pattern, dSz = ∆ρ ρ∆φ k and

B · dr1 + B · dr2 + B · dr3 + B · dr4 =

= Bρ(, φ, )∆ρ + Bφ(ρ + ∆ρ, , ) (ρ + ∆ρ)∆φ −

−Bρ(, φ + ∆φ, )∆ρ − Bφ(ρ, , ) ρ∆φ

=[(ρ + ∆ρ)Bφ(ρ + ∆ρ, , ) − ρBφ(ρ, , )

]∆φ

−[Bρ(, φ + ∆φ, ) − Bρ(, φ, )

]∆ρ

= (∇ × B)z ρ∆ρ∆φ (47)

Again shrinking to differential size yields

(∇ × B)z ρ dρ dφ =

=(∂(ρBφ)∂ρ

dρ)

dφ −(∂Bρ

∂φdφ

)dρ (48)

from which we deduce

(∇ × B)z =1ρ

∂(ρBφ)∂ρ

−1ρ

∂Bρ

∂φ(49)

2.7.4 ∇ × B

If we now collect together (43), (46), and (49) we seethat in cylindrical polar coordinates the curl takes theform

∇ × B =(1ρ

∂Bz

∂φ−∂Bφ

∂z

)ρρρ +

+(∂Bρ

∂z−∂Bz

∂ρ

)φφφ +

+(1ρ

∂(ρBφ)∂ρ

−1ρ

∂Bρ

∂φ

)k (50)

z

y

x

r

dr

r sin θr

ΦФ

θ

dФ

dθ

r sin θ dФ

r dθ

Φθ

Figure 6: Spherical polar coordinates and their vol-ume element

3 Spherical Polar Coordinates

All the methods we applied in the previous sectionfor cylindrical polar coordinates can be applied in thesame way to spherical polar coordinates.

To begin, let us recall some basics about spheri-cal polar coordinates (see Figure 6). From this figureit is apparent that

x = r sin θ cos φ (51)

y = r sin θ sin φ (52)

z = r cos θ (53)

Note carefully that the two angles, θ and φ, areintrinsically different. θ is a polar angle, that mea-sures inclination with respect to an axis. φ is anazimuthal angle, that measures a rotation about anaxis.

3.1 Position Vector

The point P indicated by the black dot in Figure 6has position vector

r = x i + y j + z k = r r (54)

This might look even stranger than (4), but youshould now expect position vectors in curvilinear co-ordinates to have information contained within the(non-constant) unit vectors. Here r = r (θ, φ) con-tains all the direction information about r. That is, to

Vector Calculus: Orthogonal Curvilinear Coordinates 7

get to a point P, you travel a distance r = |r| in the rdirection.

From (54) we can write down

r = sin θ cos φ i + sin θ sin φ j + cos θ k (55)

Differentiating this with respect to θ and φ will leadus to unit vectors in the direction of increasing θ andincreasing φ respectively:

∂r∂θ

= cos θ cos φ i + cos θ sin φ j − sin θ k

= θθθ (56)

∂r∂φ

= − sin θ sin φ i + sin θ cos φ j + 0 k

= sin θ(− sin φ i + cos φ j

)= sin θ φφφ (57)

You should convince yourself that θθθ and φφφ are in-deed unit vectors and that (56) and (57) give direc-tions that agree with what trigonometry would tellyou from Figure 6.

3.2 Line Element

Looking at Figure 6 to reveal the displacement vec-tors resulting from increments dr , dθ, and dφ, or tak-ing the differential dr of the position vector (54) andmaking use of (56)–(57) leads to the expression forthe line element in spherical polar coordinates:

dr = dr r + r dθ θθθ + r sin θ dφ φφφ (58)

3.3 Surface Elements

With reference to Figure 6 we can write down thebasic surface elements of the three faces that meetat the black dot (r , θ, φ):

dSr = r2 sin θ dθ dφ r (59)

dSθ = r sin θ dφ dr θθθ (60)

dSφ = r dθ dr φφφ (61)

We will see these again in Sections 3.6 and 3.7when we work out the divergence and curl in spher-ical polar coordinates.

3.4 Volume Element

The volume of the element shown in Figure 6 is theproduct of the three orthogonal edges, i.e.,

dV = r2 sin θ dr dθ dφ (62)

Δr

r sin θ ΔФ

r

Δ

θ

dS

1

dS

2

dS

3

dS

4

dS

5

dS

6

(r,θ,Φ)

Figure 7: Volume element in spherical polars withsurface elements marked for application of the Di-vergence Theorem

3.5 Gradient Operator

We can find the gradient through the exact differen-tial as for cylindrical polars:

df =∂f∂r

dr +∂f∂θ

dθ +∂f∂φ

dφ (63)

= ∇f · dr (64)

= (∇f )r dr + (∇f )θ r dθ +

(∇f )φ r sin θ dφ (65)

Comparing (63) and (65) gives

∇f =∂f∂r

r +1r∂f∂θθθθ +

1r sin θ

∂f∂φ

φφφ (66)

3.6 Divergence Operator

We derived the form of the divergence in sphericalpolar coordinates in lecture by a method that mim-ics closely that employed in Section 2.6. Here wewill outline the key steps which follow from the di-vergence theorem applied to the volume depicted inFigure 7.

Recall (26)

∇ · B =6∑

i=1

B · dSi/dV

Retaining only the dependencies related to the facein question we have

B · dS1 = −Br (r , , ) r∆θ r sin θ∆φ (67)

Vector Calculus: Orthogonal Curvilinear Coordinates 8

B · dS2 = +Br (r + ∆r , , ) ×

×(r + ∆r)∆θ (r + ∆r) sin θ∆φ (68)

B · dS3 = −Bθ(, θ, )∆r r sin θ∆φ (69)

B · dS4 = +Bθ(, θ + ∆θ, ) ×

∆r r sin(θ + ∆θ)∆φ (70)

B · dS5 = −Bφ(, , φ)∆r r∆θ (71)

B · dS6 = +Bφ(, , φ + ∆φ)∆r r∆θ (72)

(73)

in which you should recognise the forms of the dif-ferent surface elements given in (59)-(61) with mi-nus signs in some places to ensure that all the dS’spoint out of the volume. Summing these pairwiseand letting the volume shrink to differential propor-tions yields

6∑i=1

B · dS =∂(r2Br )∂r

dr sin θ dθ dφ +

+∂(sin θBθ)

∂θdθ r dr dφ +

+∂Bφ

∂φdφ r dr dθ (74)

Dividing this summation by dV = r2 sin θ dr dθ dφthen produces the desired expression:

∇ · B =1r2

∂(r2Br )∂r

+1

r sin θ∂(sin θBθ)

∂θ+

1r sin θ

∂Bφ

∂φ(75)

3.7 Curl Operator

As in the calculation of ∇×B in Section 2.7, we shallapply Stoke’s Theorem to path integrals aroundfaces of the volume element (Figure 7) to find thecomponents of ∇×B in spherical polar coordinates.The three faces, area elements dS, and correspond-ing right-handed paths are shown in Figure 8.

3.7.1 r component

The bottom sketch in Figure 8 shows the face of thevolume element at radial distance r and that there-fore has its surface element directed in the r direc-tion. (We take it to be in the +r direction here, whichis opposite to what is sketched as dS1 in Figure 7).Applying Stoke’s Theorem to this face gives

4∑i=1

B · dri = Bθ(, , φ) r∆θ +

Δr

r sin θ ΔФ

r Δθ

dSr

dSθ

dSФ

(r,θ,Φ)

(r,θ,Φ)

(r,θ,Φ)

r Δθ

Ф+ΔФ

Ф+ΔФ

r sin (θ+Δθ) ΔФ

Δr

(r+Δr) sin θ ΔФ

(r+Δr) Δθ

1

1

4

2

2

1

3

3

2

4

4

3

r sin θ ΔФ

Figure 8: Three faces of the volume element inspherical polar coordinates with paths indicated forderivation of the θθθ (top), φφφ (middle), and r (bottom)components of ∇ × B.

+Bφ(, θ + ∆θ, ) r sin(θ + ∆θ)∆φ −

−Bθ(, , φ + ∆φ) r∆θ −

−Bφ(, θ, ) r sin θ∆φ

= (∇ × B)r r2 sin θ∆θ∆φ (76)

Dividing by r2 sin θ∆θ∆φ, re-arranging and letting∆θ → dθ and ∆φ→ dφ gives

(∇ × B)r =1

r sin θ

(∂(sin θBφ)

∂θ−∂Bθ

∂φ

)(77)

3.7.2 θθθ component

The top sketch in Figure 8 shows the face of thevolume element at polar angle θ and that thereforehas its surface element directed in the θθθ direction.(We take it to be in the +θθθ direction here, which isopposite to what is sketched as dS3 in Figure 7).

Vector Calculus: Orthogonal Curvilinear Coordinates 9

Applying Stoke’s Theorem to this face gives

4∑i=1

B · dri = Bφ(r , , ) r sin θ∆f +

+Br (, , φ + ∆φ)∆r −

−Bφ(r + ∆r , , ) (r + ∆r) sin θ∆φ −

−Br (, , φ)∆r

= (∇ × B)θ r sin θ∆φ∆r (78)

Dividing by r sin θ∆φ∆r , re-arranging, and letting∆φ→ dφ and ∆r → dr yields

(∇ × B)θ =(

1r sin θ

∂Br

∂φ−

1r∂(rBφ)∂r

)(79)

3.7.3 φφφ component

The middle sketch in Figure 8 shows the face of thevolume element at azimuthal angle φ and that there-fore has its surface element directed in the φφφ direc-tion. (We take it to be in the +φφφ direction here, whichis opposite to what is sketched for the front face dS5

in Figure 7). Applying Stoke’s Theorem to this facegives

4∑i=1

B · dri = Br (, θ, )∆r +

+Bθ(r + ∆r , , ) (r + ∆r)∆θ −

−Br (, θ + ∆θ, )∆r −

−Bθ(r , , ) r ∆θ

= (∇ × B)φ r ∆θ∆r (80)

Dividing by r∆θ∆r , re-arranging, and letting ∆θ →dθ and ∆r → dr yields

(∇ × B)φ =1r

(∂(rBθ)∂r

−∂Br

∂θ

)(81)

3.7.4 ∇ × B

Collecting the components of ∇ × B from (77), (79),and (81) gives our final result

(∇ × B) =1

r sin θ

(∂(sin θBφ)

∂θ−∂Bθ

∂φ

)r +

+(

1r sin θ

∂Br

∂φ−

1r∂(rBφ)∂r

)θθθ +

+1r

(∂(rBθ)∂r

−∂Br

∂θ

)φφφ (82)

4 General Orthogonal CoordinateSystems

The methodology used in this handout can be ap-plied to other orthogonal curvilinear coordinate sys-tems. You will also find this done in several books bypure mathematical manipulations. All start from thecornerstone of coordinate systems, namely a set ofunit vectors e1, e2, e3, and the corresponding scalefactors h1, h2, and h3 that convert differentials in thecoordinates (u1, u2, u3) into vector differential line el-ements, i.e.,

dr = h1 du1 e1 + h2 du2 e2 + h3 du3 e3 (83)

You might try to replicate the approach here for sucha general system, and then compare your answerwith the results in Riley et al., or Boas.

5 Non-orthogonal Coodinate Sys-tems

It is possible to define coordinate systems in whichthe three base unit vectors are neither straight nororthogonal. We will not study such systems here.It is possible to derive corresponding vector calcu-lus expressions for such systems using the funda-mental machinery we have developed in the course.This must be done carefully; the biggest complica-tion is that dot products between such unit vectorsare not zero.