1.16 Curvilinear Coordinates - Faculty of Engineering

Section 1.16

1.16 Curvilinear Coordinates

Up until now, a rectangular Cartesian coordinate system has been used, and a set of
orthogonal unit base vectors ei has been employed as the basis for representation of
vectors and tensors. This basis is independent of position and provides a simple
formulation. Two exceptions were in §1.6.10 and §1.14.4, where cylindrical and
spherical coordinate systems were used. These differ from the Cartesian system in that
the cylindrical and spherical base vectors do depend on position. However, although the
directions of these base vectors may change with position, they are always orthogonal to
each other. In this section, arbitrary bases, with base vectors not necessarily orthogonal
nor of unit length, are considered. It will be seen how these systems reduce to the special
cases of orthogonal (e.g. cylindrical and spherical systems) and Cartesian systems.

1.16.1 Curvilinear Coordinates

A Cartesian coordinate system is defined by the fixed base vectors e1,e2 ,e3 and the
coordinates (x1, x2 , x3 ) , and any point p in space is then determined by the position
vector x = xiei (see Fig. 1.16.11). This can be expressed in terms of curvilinear
coordinates (Θ1, Θ2 , Θ3 ) by the transformation (and inverse transformation)

( )Θi = Θi x1, x2 , x3 (1.16.1)

xi = xi (Θ1, Θ2 , Θ3 )

In order to be able to solve for the Θi given the xi , and to solve for the xi given the Θi ,

it is necessary and sufficient that the following determinants are non-zero – see Appendix
1.A.2 (the first here is termed the Jacobian J of the transformation):

J ≡ det ⎡ ∂x i ⎤ = ∂x i , det ⎡ ∂Θ i ⎤ = ∂Θ i = 1, (1.16.2)
⎢ ∂Θ j ⎥ ∂Θ j ⎢ ∂x j ⎥ ∂x j J
⎣ ⎦ ⎣ ⎦

the last equality following from (1.15.2, 1.10.18d). If Θ1 is varied while holding Θ2 and
Θ3 constant, a space curve is generated called a Θ1 coordinate curve. Similarly, Θ2
and Θ3 coordinate curves may be generated. Three coordinate surfaces intersect in
pairs along the coordinate curves. On each surface, one of the curvilinear coordinates is
constant.

Note:
• This Jacobian is the same as that used in changing the variable of integration in a volume

integral, §1.7; from Cartesian coordinates to curvilinear coordinates, one has

∫ ∫dx1dx2dx3 → JdΘ1dΘ2dΘ3

VV

1 superscripts are used here and in much of what follows for notational consistency (see later)

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Section 1.16

Θ2 − curve

x3 g2
Θ3 = const
g3
Θ3 − curve p g1
e3 x g1
Θ1 − curve

e1 e2 x2

x1

Figure 1.16.1: curvilinear coordinate system and coordinate curves

1.16.2 Base Vectors in the Moving Frame

Covariant Base Vectors

( )From §1.6.2, writing x = x Θi , tangent vectors to the coordinate curves at x are given

by2

gi = ∂x = ∂x m em Covariant Base Vectors (1.16.3)
∂Θ i ∂Θi

( )with inverse ei = ∂Θm / ∂xi g m . The gi emanate from the point p and are directed

towards the site of increasing coordinate Θi . They are called covariant base vectors.

Increments in the two coordinate systems are related through

dx = dx dΘ i = g i dΘi
∂Θ i

Note that the triple scalar product g1 ⋅ (g 2 × g3 ), Eqns. 1.2.15-16, is equivalent to the

determinant in 1.16.2,

2 in the Cartesian system, with the coordinate curves parallel to the coordinate axes, these equations reduce

( )trivially to ei = ∂x m / ∂xi e m = δ mie m

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Section 1.16

(g1 )1 (g1 )2 (g1 )3 = J = det ⎡ ∂x i ⎤ (1.16.4)
g1 ⋅ (g 2 × g3 ) = (g 2 )1 (g 2 )2 (g 2 )3 ⎢ ∂Θ j ⎥
(g3 )2 (g3 )3 ⎣ ⎦
(g3 )1

so that the condition that the determinant does not vanish is equivalent to the condition
that the vectors gi are linearly independent, and so the gi can form a basis.

Contravariant Base Vectors

Unlike in Cartesian coordinates, where ei ⋅ e j = δ ij , the covariant base vectors do not
necessarily form an orthonormal basis, and gi ⋅ g j ≠ δ ij . In order to deal with this
complication, a second set of base vectors are introduced, which are defined as follows:
introduce three contravariant base vectors gi such that each vector is normal to one of
the three coordinate surfaces through the point p. From §1.6.4, the normal to the
coordinate surface Θ1 = const is given by the gradient vector grad Θ1 , with Cartesian
representation

grad Θ1 = ∂Θ1 em
∂x m

and, in general, one may define the contravariant base vectors through

gi = ∂Θi em Contravariant Base Vectors (1.16.5)
∂x m

The contravariant base vector g1 is shown in Fig. 1.16.1.

( )As with the covariant base vectors, the triple scalar product g1 ⋅ g2 × g3 is equivalent to

the determinant in 1.16.2,

( ) ( ) ( )g1 1 g1 2 g1 3 = 1 = det ⎡ ∂Θ j ⎤ (1.16.6)
( ) ( ) ( ) ( )g1 ⋅ g2 × g3 = g2 1 g2 2 g2 3 J ⎢ ∂x i ⎥
g3 2 g3 3 ⎣ ⎦
( ) ( ) ( )g3 1

and again the condition that the determinant does not vanish is equivalent to the condition
that the vectors gi are linearly independent, and so the contravariant vectors also form a
basis.

1.16.3 Metric Coefficients

It follows from the definitions of the covariant and contravariant vectors that {▲Problem
1}

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Section 1.16

gi ⋅g j = δ i (1.16.7)
j

which is the defining relationship between reciprocal pairs of general bases. Of course

the gi were chosen precisely because they satisfy this relation. Here, δ ij is again the
Kronecker delta3, with a value of 1 when i = j and zero otherwise.

One needs to be careful to distinguish between subscripts and superscripts when dealing
with arbitrary bases, but the rules to follow are straightforward. For example, each free
index which is not summed over, such as i or j in 1.16.7, must be either a subscript or
superscript on both sides of an equation. Hence the new notation for the Kronecker delta
symbol.

The relation gi ⋅g j = δ i implies that each base vector gi is orthogonal to two of the
j

reciprocal base vectors gi . For example, g1 is orthogonal to both g 2 and g3 .

Unlike the orthogonal base vectors, the dot product of a covariant/contravariant base
vector with another base vector is not necessarily one or zero. Because of their
importance in curvilinear coordinate systems, the dot products are given a special symbol:
define the metric coefficients to be

gij = g i ⋅ g j Metric Coefficients (1.16.8)
g ij = g i ⋅ g j

The following important and useful relations may be derived by manipulating the
equations already introduced: {▲Problem 2}

g i = gij g j (1.16.9)
g i = g ij g j

and {▲Problem 3}

g ij g kj = δ i ≡ g i (1.16.10)
k k

Note here another rule about indices in equations involving general bases: summation can
only take place over a dummy index if one is a subscript and the other is a superscript –
they are paired off as with the j’s in these equations.

The metric coefficients can be written explicitly in terms of the curvilinear components:

g ij = gi ⋅g j = ∂x k ∂x k , g ij = gi ⋅g j = ∂Θ i ∂Θ j (1.16.11)
∂Θ i ∂Θ j ∂x k ∂x k

Note here also a rule regarding derivatives with general bases: the index i on the right
hand side of 1.16.11a is a superscript of Θ but it is in the denominator of a quotient and

3 although in this context it is called the mixed Kronecker delta

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Section 1.16

so is regarded as a subscript to the entire symbol, matching the subscript i on the g on the
left hand side4.

One can also write 1.16.11 in the matrix form

⎡ ∂x k ⎤T ⎡ ∂x k ⎤⎥, [ ]g ij ⎡ ∂Θi ⎤⎡ ∂Θ j ⎤ T
⎥ ⎦ ⎥⎢ ∂x k ⎥
[ ]gij = ⎢ ∂Θ i ⎦ ⎢ ∂Θ j = ⎢ ∂x k ⎦⎣ ⎦
⎣ ⎣ ⎣

and, from 1.9.13a,b,

⎜⎜⎝⎛ ⎡ ∂x i ⎤ ⎞⎟⎟⎠ 2 ⎛⎜⎜⎝ ⎡ ∂Θ i ⎤ ⎟⎠⎞⎟ 2 1
⎢ ∂Θ j ⎥ ⎢ ∂x j ⎥ J2
[ ]det gij = det ⎣ ⎦ = J 2, [ ]det g ij = det ⎣ ⎦ = (1.16.12)
(1.16.13)
These determinants play an important role, and are denoted by g:

1
det g ij
[ ] [ ]g = det gij =

Note:

[ ] [ ]• The matrix ∂xk / ∂Θi is called the Jacobian matrix J, so J T J = gij

Scale Factors

The covariant and contravariant base vectors are not unit vectors: in particular, consider
the covariant base vectors and introduce the unit triad gˆ i , with gi = gi ⋅ gi :

gˆ i = gi = gi (no sum) (1.16.14)
gi g ii

The lengths of the covariant base vectors are denoted by h and are called the scale
factors:

hi = gi = gii (no sum) (1.16.15)

1.16.4 The Covariant and Contravariant Components of a
Vector

A vector can now be represented in terms of either basis:

( ) ( )u = ui Θ1, Θ2 , Θ3 gi = u i Θ1, Θ2 , Θ3 gi (1.16.16)

4 the rule for pairing off indices has been broken in (1.12.11) for clarity; more precisely, these equations

( )( ) ( )( )should be written as gij = ∂xm / ∂Θi ∂xn / ∂Θ j δ mn and g ij = ∂Θi / ∂x m ∂Θ j / ∂x n δ mn

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Section 1.16

The ui are the covariant components of u and ui are the contravariant components of

u. Thus the covariant components are the coefficients of the contravariant base vectors
and vice versa – subscripts denote covariance while superscripts denote contravariance.

When u is written with covariant components, u = uigi , it is called a covariant vector.

When u is written with contravariant components, u = uigi , it is called a contravariant
vector.

Analogous to the orthonormal case, where u ⋅ ei = ui {▲Problem 4}:

u ⋅ gi = ui , u ⋅ gi = ui (1.16.17)

Note the following useful formula involving the metric coefficients, for raising or
lowering the index on a vector component, relating the covariant and contravariant
components, {▲Problem 5}

u i = g iju j , ui = giju j (1.16.18)

Physical Components of a Vector

The contravariant and covariant components of a vector do not have the same physical
significance in a curvilinear coordinate system as they do in a rectangular Cartesian
system; in fact, they often have different dimensions. For example, the differential dx of
the position vector has in cylindrical coordinates the contravariant components
(dr, dθ , dz) , that is, dx = dΘ1g1 + dΘ2g 2 + dΘ3g3 with Θ1 = r , Θ2 = θ , Θ3 = z (this
will be discussed in detail below). Here, dθ does not have the same dimensions as the
others. The physical components in this example are (dr, rdθ , dz) .

The physical components u i of a vector u are defined to be the components along the
covariant base vectors (and hence are obtained from the contravariant components),
referred to unit vectors. Thus,

u = uigi (1.16.19)

3

∑= uihigˆ i ≡ u i gˆ i
i =1

and

u i = u i gii (no sum) Physical Components of a Vector (1.16.20)
The Dot Product
The dot product of two vectors can be written in one of two ways: {▲Problem 6}

u ⋅ v = uivi = uivi Dot Product of Two Vectors (1.16.21)

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Section 1.16

1.16.5 The Vector Cross Product

The triple scalar product is an important quantity in analysis with general bases,
particularly when evaluating cross products. From Eqns. 1.16.4, 1.16.6 and 1.16.12-13,

[ ][ ]g = g1 ⋅ g 2 × g3 2 = det gij (1.16.22)

11

[ ] [ ]= g1 ⋅ g2 × g3 2 = det g ij

Introducing permutation symbols eijk , eijk , one can in general write5

eijk ≡ gi ⋅ g j × g k = ε ijk g , eijk ≡ g i ⋅ g j × g k = ε ijk 1
g

where ε ijk = ε ijk is the Cartesian permutation symbol (Eqn. 1.3.8). The cross product of

the base vectors can now be written in terms of the reciprocal base vectors as (note the
similarity to the Cartesian relation 1.3.11) {▲Problem 7}

gi × g j = eijk g k Cross Products of Base Vectors (1.16.23)
gi × g j = eijk g k

Further, from 1.3.17,

eijk e pqr = ε ijk ε pqr, eijk e pqk = δ i δ j − δ pjδ i (1.16.24)
p q q

The Cross Product
The cross product of vectors can be written as {▲Problem 8}

g1 g2 g3 Cross Product of Two Vectors (1.16.25)

u × v = eijk u i v j g k = g u1 u 2 u 3
v1 v2 v3

= eijk ui v j g k = 1 g1 g2 g3
u1 u2 u3
v2 v3
g v1

5 assuming the base vectors form a right handed set, otherwise a negative sign needs to be included

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Section 1.16

1.16.6 The Covariant, Contravariant and Mixed Components of
a Tensor

Tensors can be represented in any of four ways, depending on which combination of base
vectors is being utilised:

A = Aij g i ⊗ g j = Aij g i ⊗ g j = A⋅ij g i ⊗ g j = Ai⋅ j g i ⊗ g j (1.16.26)

Here, Aij are the contravariant components, Aij are the covariant components, A⋅ij
and Ai⋅ j are the mixed components of the tensor A. On the mixed components, the
subscript is a covariant index, whereas the superscript is called a contravariant index.
Note that the “first” index always refers to the first base vector in the tensor product.

An “index switching” rule for tensors is

Aij δ j = Aik , Aijδ k = Aik (1.16.27)
k j

and the rule for obtaining the components of a tensor A is (compare with 1.9.4),
{▲Problem 9}

(A)ij ≡ Aij = gi ⋅ Ag j

(A)ij ≡ Aij = gi ⋅ Ag j (1.16.28)

( )A i ≡ A⋅ij = gi ⋅ Ag j
⋅j

( )A ⋅j ≡ Ai⋅ j = gi ⋅ Ag j
i

As with the vectors, the metric coefficients can be used to lower and raise the indices on
tensors:

T ij = g ik g jlTkl (1.16.29)
Ti⋅ j = gikT kj

In matrix form, these expressions can be conveniently used to evaluate tensor
components, e.g. (note that the matrix of metric coefficients is symmetric)

[ ] [ ] [ ][ ]T ij = g ik Tkl g lj .

An example of a higher order tensor is the permutation tensor E, whose components are
the permutation symbols introduced earlier:

E = eijk g i ⊗ g j ⊗ g k = eijk g i ⊗ g j ⊗ g k .

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Section 1.16

Physical Components of a Tensor

Physical components of tensors can also be defined. For example, if two vectors a and b
have physical components as defined earlier, then the physical components of a tensor T
are obtained through6

a i = T ij b j . (1.16.30)

As mentioned, physical components are defined with respect to the covariant base
vectors, and so the mixed components of a tensor are used, since

( )Tb i bkgk i j ≡ aigi
= T⋅ j gi ⊗gj = T⋅ j b g
i

as required. It follows from 1.16.22 that

T⋅ i bj = ai (no sum on the g)
j g jj g ii

and so from 1.16.30,

T ij = g ii T⋅ i (no sum) Physical Components of a Tensor (1.16.31)
g jj j

The Identity Tensor

The components of the identity tensor I in a general basis can be obtained as follows:

u = uigi
= g ij u j g i
= g ij (u ⋅ g j )gi
= g ij (gi ⊗ g j )u
≡ Iu

Thus the contravariant components of the identity tensor are the metric coefficients g ij

and, similarly, the covariant components are gij . For this reason the identity tensor is

also called the metric tensor. On the other hand, the mixed components are the

Kronecker delta, δ i (also denoted by g i ). In summary7,
j j

6 these are called right physical components; left physical components are defined through a = bT

7 there is no distinction between δ j , δ i ; they are often written as g j , g i and there is no need to specify
i j i j

which index comes first, for example by g j
⋅i

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Section 1.16

(I)ij = gij ( )I = gij gi ⊗ g j
(I)ij = g ij ( )I = g ij gi ⊗ g j
( )I (1.16.32)
( )I i = δ i = δ i gi ⊗gj = gi ⊗ gi
⋅j j ( )I = δ ij gi ⊗ g j = gi ⊗ gij
( )I
⋅j = δij
i

Symmetric Tensors

A tensor S is symmetric if ST = S , i.e. if uSv = vSu . If S is symmetric, then

S ij = S ji , Sij = S ji , S ⋅ij = S ⋅i = g g im S k
j ⋅m
jk

In terms of matrices,

[ ] [ ] [ ] [ ] [ ] [ ]S ij = S ij T , Si T
Sij = Sij T , S ⋅ij ≠ ⋅j

1.16.7 Generalising Cartesian Relations to the Case of General
Bases

The tensor relations and definitions already derived for Cartesian vectors and tensors in
previous sections, for example in §1.10, are valid also in curvilinear coordinates, for

example AA−1 = I , trA = I : A and so on. Formulae involving the index notation may
be generalised to arbitrary components by:

(1) raising or lowering the indices appropriately
(2) replacing the (ordinary) Kronecker delta δ ij with the metric coefficients gij

(3) replacing the Cartesian permutation symbol εijk with eijk in vector cross products

Some examples of this are given in Table 1.16.1 below.

Note that there is only one way of representing a scalar, there are two ways of
representing a vector (in terms of its covariant or contravariant components), and there
are four ways of representing a (second-order) tensor (in terms of its covariant,
contravariant and both types of mixed components).

Cartesian General Bases

a ⋅ b aibi aibi = a ibi

aB ai Bij (aB) j = ai B⋅ij = a i Bij

(aB) j = ai Bij = a i Bi⋅ j

Ab Aij b j (Ab)i = Aij b j = Ai⋅ jb j

(Ab)i = Aijb j = A⋅ij b j

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Section 1.16

( )AB ij = Aik B⋅kj = Ai⋅k Bkj

( )AB ij = Aik Bk⋅ j = A⋅ik B kj

( )AB ⋅j
Aik Bkj AB i = Ai⋅k Bk⋅ j = Aik B kj

( )AB i = Aik Bkj = A⋅ik B⋅kj
⋅j

a×b ε ijk aib j ( )a × b k = eijk aib j

( )a × b k = eijk aib j
(a ⊗ b)ij = aib j

a⊗b ai b j (a ⊗ b)ij = aib j

(a ⊗ b )⋅ j = aib j
i

(a ⊗ b )i j = aibj
⋅

A:B Aij Bij Aij B ij = Aij Bij = A⋅ij Bi⋅ j = Ai⋅ j B⋅ij
trA ≡ I : A
Aii Ai⋅i = A⋅ii
det A ε ijk Ai1 A j2 Ak 3
AT ε ijk A⋅1i A⋅2j A⋅k3
( )AT ij = Aji
( ) ( )AT ij = Aji , AT ij = A ji

i( ) ( )AT AT ⋅j
⋅j = A⋅ji ≠ A⋅ij , i = A⋅ij ≠ A⋅ji

Table 1.16.1: Tensor relations in Cartesian and general curvilinear coordinates

1.16.8 Line, Surface and Volume Elements

In order to carry out integration along curves, over regions, or throughout volumes, it is
necessary to have expressions for the length of a line element Δs , the area of a surface
element ΔS and the volume of a volume element ΔV , in terms of the increments in the
curvilinear coordinates ΔΘ1, ΔΘ2 , ΔΘ3 .

The Metric
Consider a differential line element, Fig. 1.16.2,

dx = dxiei = dΘigi (1.16.33)

The square of the length of this line element, denoted by (Δs)2 and called the metric of

the space, is then

( ) ( )(Δs)2 = dx ⋅ dx = dΘigi ⋅ dΘ jg j = gij dΘidΘ j (1.16.34)

This relation (Δs)2 = gij dΘi dΘ j is called the fundamental differential quadratic form.

The gij ' s can be regarded as a set of scale factors for converting increments in Θi to
changes in length.

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Section 1.16

Θ3 − curve dx dΘ2g2
x3
dΘ1g1 Θ2 − curve
dΘ3g3

x

Θ1 − curve
x1 x2

Figure 1.16.2: a line element in space

Surface Area and Volume Elements

The surface area ΔS1 of a face of the elemental parallelepiped on which Θ1 is constant
(to which g1 is normal) is, using 1.7.6,

ΔS1 = (ΔΘ2g 2 ) × (ΔΘ3g3 ) (1.16.35)
= g 2 × g3 ΔΘ2ΔΘ3

= (g 2 × g3 )⋅ (g 2 × g3 )ΔΘ2ΔΘ3
= (g 2 ⋅ g 2 )(g3 ⋅ g3 ) − (g 2 ⋅ g3 )(g 2 ⋅ g3 )ΔΘ2ΔΘ3

= g 22 g33 − (g 23 ) 2 ΔΘ2ΔΘ3
= g g11 ΔΘ2ΔΘ3

and similarly for the other surfaces. (1.16.36)
The volume ΔV of the parallelepiped is

ΔV = g1 ⋅ g 2 × g3 ΔΘ1ΔΘ2ΔΘ3 = g ΔΘ1ΔΘ2ΔΘ3

1.16.9 Orthogonal Curvilinear Coordinates

Orthogonal curvilinear coordinates are considered in this section). In this case,

[ ]gij = ⎢⎡⎢h012 0 0⎤
h22 ⎥
gij = g i ⋅ g j = δ ij g i g j = δ ij hi h j , 0 ⎥ (1.16.37)

⎢⎣ 0 0 h32 ⎦⎥

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Section 1.16

The contravariant base vectors are collinear with the covariant, but the vectors are of
different magnitudes:

gi = hi gˆ i , gi = 1 gˆ i (1.16.38)
hi

It follows that

( )Δs 2 = h12 dΘ12 + h22 dΘ 2 + h32 dΘ 2
2 3

ΔS1 = h2h3ΔΘ2ΔΘ3 (1.16.39)
ΔS2 = h3h1ΔΘ3ΔΘ1
ΔS3 = h1h2ΔΘ1ΔΘ2

ΔV = h1h2h3ΔΘ1ΔΘ2ΔΘ3
Examples
1. Cylindrical Coordinates

( )Consider the cylindrical coordinates, (r,θ , z) = Θ1, Θ2 , Θ3 , cf. §1.6.10, Fig. 1.16.3:

x1 = Θ1 cos Θ2 ( ) ( )Θ1 = x1 2 + x2 2
x 2 = Θ1 sin Θ2 , ( )Θ2 = tan −1 x2 / x1
x3 = Θ3
Θ3 = x3

with Θ1 ≥ 0, 0 ≤ Θ2 < 2π , − ∞ < Θ3 < ∞
which give

J = det ⎡ ∂x i ⎤ = Θ1
⎢ ∂Θ j ⎥
⎣ ⎦

so that there is a one-to-one correspondence between the Cartesian and cylindrical

coordinates at all point except for Θ1 = 0 (which corresponds to the axis of the cylinder).
These points are called singular points of the transformation.

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Section 1.16

x3 Θ3 − curve e3 g3 g2
e1 • g1
• Θ2 − curve
Θ3 Θ1 − curve e2

x2

x1 Θ2 Θ1

Figure 1.16.3: Cylindrical Coordinates

The unit vectors and scale factors are {▲Problem 11}

h1 = g1 = 1 (= 1) gˆ 1 = g1 (= er )
h2 = g 2 = Θ1 (= r)
h3 = g3 = 1 (= 1) gˆ 2 = g2 (= eθ )
Θ1
gˆ 3 = g3 (= e z )

The physical components of a vector v are v1, Θ1v 2 , v3 and one has

Metric: ( ) ( ) ( ) ( )( )Δs 2 = dΘ1 2 + Θ2dΘ2 2 + dΘ3 2 = dr 2 + (rdθ )2 + dz 2
Surface Element:
Volume Element: ΔS1 = Θ1ΔΘ2ΔΘ3

ΔS2 = ΔΘ3ΔΘ1

ΔS3 = Θ1ΔΘ1ΔΘ2

ΔV = Θ1ΔΘ1ΔΘ2ΔΘ3 (= rΔrΔθΔz)

2. Spherical Coordinates

( )Consider the spherical coordinates, (r,θ ,φ ) = Θ1, Θ2 , Θ3 , cf. §1.6.10, Fig. 1.16.4:

x1 = Θ1 sin Θ2 cos Θ3 ( ) ( ) ( )Θ1 = x1 2 + x2 2 + x3 2
x 2 = Θ1 sin Θ2 sin Θ3 , ( ) ( ) ( )Θ2 = tan −1⎛⎝⎜ x1 2 + x2 2 / x3 2 ⎟⎞⎠
x3 = Θ1 cos Θ2
( )( ) ( )Θ3 = tan −1 x2 2 / x1 2

with

Θ1 ≥ 0, 0 ≤ Θ2 ≤ π , 0 ≤ Θ3 < 2π

which give

( )J=det ⎡ ∂x i ⎤ = Θ1 2 sin Θ2
⎢ ∂Θ j ⎥
⎣ ⎦

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Section 1.16

so that there is a one-to-one correspondence between the Cartesian and spherical
coordinates at all point except for the singular points along the x3 axis.

x3 Θ1 − curve g1

Θ2 • Θ3 − curve • g3
Θ1 Θ2 − curve e3 g2
x2
e2
Θ3 e1

x1

Figure 1.16.4: Spherical Coordinates
The unit vectors and scale factors are {▲Problem 11}

h1 = g1 = 1 (= 1) gˆ 1 = g1 (= er )
h2 = g 2 = Θ1 (= r) (= eθ )
gˆ 2 = g2
Θ1 (= eφ )

h3 = g3 = Θ1 sin Θ2 (= r sinθ ) gˆ 3 = g3
Θ1 sin Θ2

The physical components of a vector v are Θ1, Θ1v2 , Θ1 sin Θ2v3 , and one has

Metric: ( ) ( ) ( )( )Δs 2 = dΘ1 2 + Θ1dΘ2 2 + Θ1 sin Θ2dΘ3 2
Surface Element:
Volume Element: ( )= dr 2 + (rdθ )2 + (r sinθdφ )2

( )ΔS1 = Θ1 2 sin Θ2ΔΘ2ΔΘ3

ΔS2 = Θ1 sin Θ2ΔΘ3ΔΘ1

ΔS3 = Θ1ΔΘ1ΔΘ 2

( ) ( )ΔV = Θ1 2 sin Θ2ΔΘ1ΔΘ2ΔΘ3 = r 2 sinθΔrΔθΔφ

1.16.10 Rectangular Cartesian (Orthonormal) Coordinate System

In an orthonormal Cartesian coordinate system, gi = gi = ei , gij = δ ij , g = 1 , hi = 1 and
eijk = ε ijk (= ε ijk ) .

1.16.11 Problems

1. Derive the fundamental relation gi ⋅g j = δ i .
j

2. Show that gi = gij g j [Hint: assume that one can write gi = aik g k and then dot both

sides with g j .]

Solid Mechanics Part III 149 Kelly

Section 1.16

3. Use the relations 1.16.9 to show that g ij g kj = δ i . Write these equations in matrix
k

form.

4. Show that u ⋅ gi = ui .

5. Show that ui = giju j .

6. Show that u ⋅ v = uivi = u ivi

7. Use the relation eijk ≡ gi ⋅ g j × g k = ε ijk g to derive the cross product relation

( )gi × g j = eijk g k . [Hint: show that gi × g j = gi × g j ⋅ gk gk .]

8. Derive equation 1.16.25 for the cross product of vectors

9. Show that (A)ij = gi ⋅ Ag j .

10. Given g1 = e1, g 2 = e2 , g3 = e1 + e3 , v = e1 + e2 + e3 . Find gi , gij , eijk , vi , v j (write

the metric coefficients in matrix form).

11. Derive the scale factors for the (a) cylindrical and (b) spherical coordinate systems.

12. Parabolic Cylindrical (orthogonal) coordinates are given by
(( ) ( ) )x1
= 1 Θ1 2 − Θ2 2 , x2 = Θ1Θ2 , x3 = Θ3
2

with

− ∞ < Θ1 < ∞, Θ2 ≥ 0, − ∞ < Θ3 < ∞

Evaluate:

(i) the scale factors

(ii) the Jacobian – are there any singular points?

(iii) the metric, surface elements, and volume element

Verify that the base vectors gi are mutually orthogonal.

[These are intersecting parabolas in the x1 − x2 plane, all with the same axis]
13. Repeat Problem 7 for the Elliptical Cylindrical (orthogonal) coordinates:

x1 = a cosh Θ1 cos Θ2 , x 2 = a sinh Θ1 sin Θ2 , x3 = Θ3

with

Θ1 ≥ 0, 0 ≤ Θ2 < 2π , − ∞ < Θ3 < ∞

[These are intersecting ellipses and hyperbolas in the x1 − x2 plane with foci at

x1 = ±a .]
14. Consider the non-orthogonal curvilinear system illustrated in Fig. 1.16.5, with

transformation equations

Θ1 = x1 − 1 x2
3

Θ2 = 2 x2
3

Θ3 = x3
Derive the inverse transformation equations, i.e. xi = xi (Θ1, Θ2 , Θ3 ) ,

the Jacobian matrices

J = ⎡ ∂x i ⎤⎥, J −1 = ⎡ ∂Θi ⎤ ,
⎢ ∂Θ j ⎦ ⎢ ⎥
⎣ ⎣ ∂x j ⎦

the covariant and contravariant base vectors, the matrix representation of the metric

[ ] [ ] [ ] [ ]coefficients gij , g ij from 1.16.8, verify that J TJ = gij , J −1J −T = g ij and evaluate

g.

Solid Mechanics Part III 150 Kelly