Orthogonal curvilinear coordinates - Mathematica

Orthogonal curvilinear coordinates

B. Lautrup
December 17, 2004

1 Curvilinear coordinates

Let xi with i = 1, 2, 3 be Cartesian coordinates of a point and let ξa with a = 1, 2, 3 be the corresponding
curvilinear coordinates. We shall use ordinary Cartesian vector notation x = (x1, x2, x3) for the Cartesian
coordinates, but not for the curvilinear ones. The two sets of coordinates are connected by a bijective

coordinate transformation

x = f (ξ1, ξ2, ξ3) (1)
The most important quantity is the infinitesimal vector element

dx = Jadξa (2)
(3)
a

where the Ja are the columns of the Jacobian

Ja = ∂x
∂ξa

The Cartesian square norm of the infinitesimal element is

dx2 = gabdξadξb (4)

ab

where

gab = Ja · Jb (5)

is the metric.

2 Orthogonality

A large subclass of interesting coordinate systems are orthogonal, which means that

gab = Ja · Jb = 0 (a = b) (6)
In that case it is better to write

∂x = haea (7)
∂ξa

where ha is a scale factor and ea is a unit vector. These vectors form a local basis in each point

ea · eb = δab eaea = ←→1 (8)

a

where the first equation expresses orthogonality of the basis and the second completeness. This permits
normal vector and matrix algebra to be used in the curvilinear coordinates.

1

3 Basis derivatives

The derivatives ∂eb/∂ξa of the basis vectors play an important role. From eb2 = 1 we get

∂eb · eb = 0 (9)
∂ξa

for all a, b. This shows that all derivatives of a unit vector are orthogonal to the unit vector. Similarly
from the symmetry of the second derivatives we get

∂2x = ∂(haea) = ∂(hbeb) (10)
∂ξa∂ξb ∂ξb ∂ξa

Expanding this becomes

∂ha ea + ha ∂ea = ∂hb eb + hb ∂eb (a = b) (11)
∂ξb ∂ξb ∂ξa ∂ξa

Expanding in the basis, and using that the derivative of a unit vector is orthogonal to the unit vector,
this equation can only be fulfilled for

hb ∂eb = ∂ha ea + λabcec (a = b = c) (12)
∂ξa ∂ξb

where

λabc = λbac (a = b = c) (13)
(14)
Dotting with ec and using that eb · ec = 0 we get (15)
(16)
λabc = hb ∂eb · ec = −hb ∂ec · eb = − hb λacb (17)
∂ξa ∂ξa hc (18)

Combining these two rules we get

λabc = − hb λacb = − hb λcab = ha λcba = ha λbca = −λbac = −λabc
hc hc hc hc

Consequently we have λabc = 0 so that

∂eb = 1 ∂ha ea (a = b)
∂ξa hb ∂ξb

Dotting (11) with ea we get

∂eb · ea = 1 ∂ha (a = b)
∂ξa hb ∂ξb

and using ea · eb = 0 this leads to

∂ea · eb = − 1 ∂ha (a = b)
∂ξa hb ∂ξb

Using completeness this becomes (as may be easily verified)

∂eb = 1 ∂ha ea − δab 1 ∂ha ec (19)
∂ξa hb ∂ξb hc ∂ξc
c

This concludes the analysis of derivatives of the basis vectors.

2

4 Vector operators

Derivative operators transform as

∂ = ∂x ∂ = haea · ∇ (20)
∂ξa ∂ξa ∂x (21)
(22)
where ∇i = ∂/∂xi. Using completeness we get (23)

∇= ea ∂ (24)
ha ∂ξa
a

or

∇a = 1 ∂
ha ∂ξa

4.1 Gradient of scalar field

If f is a scalar field, then ∇af is the gradient in the local basis,
(∇f )a = ∇af

4.2 Divergence of vector field

For the divergence of a vector field v with local components va = ea · v we find

∇·v = eb ∂ · vaea = eb · ∂(vaea) = 1 ∂va + va eb · ∂ea
hb ∂ξb hb ∂ξb ha ∂ξa hb ∂ξb
b a ab a ab

= 1 ∂va + va ∂hb − va ∂ha
ha ∂ξa hahb ∂ξa h2a ∂ξa
a ab a

= 1 ∂va + va ∂hb
ha ∂ξa hahb ∂ξa
a a=b

Introducing h = a ha = h1h2h3 this may be written

∇·v = 1 ∂(vah/ha)
h ∂ξa
a

which is the most compact form of the divergence.

4.3 Curl of vector field

In the local system the curl becomes

(∇ × v)a = ea · eb ∂ × vcec
hb ∂ξb
bc

= 1 ∂vc + vc ea · eb × ∂ec
abc hb ∂ξb hb ∂ξb
bc bc

= 1 ∂vc + vc ea · eb × 1 ∂hb eb − δbc 1 ∂hb ed
abc hb ∂ξb hb hc ∂ξc hd ∂ξd
bc bc d

= abc 1 ∂vc − vb ∂hb = abc 1 ∂vc + vc ∂hc
hb ∂ξb hbhc ∂ξc hb ∂ξb hbhc ∂ξb
bc bc

3

This can be combined into

(∇ × v)a = 1 ∂(vchc) (25)
hbhc ∂ξb
bc

4.4 Gradient of vector field

The vector field gradient ∇v is a tensor with the following components in the local basis

(∇v)ab = 1 ∂v · eb = 1 ∂(vcec) · eb
ha ∂ξa ha ∂ξa
c

= 1 ∂vb + 1 vc ∂ec · eb
ha ∂ξa ha ∂ξa
c

= 1 ∂vb + 1 vc 1 ∂ha ea − δac 1 ∂ha ed · eb
ha ∂ξa ha hc ∂ξc hd ∂ξd
c d

(∇v)ab = 1 ∂vb + δab vc ∂ha − va ∂ha (26)
ha ∂ξa ha hc ∂ξc hahb ∂ξb
c

One may immediately verify that its trace equals the divergence.
Specializing to diagonal and non-diagonal elements we get

(∇v)ab = 1 ∂vb − va ∂ha (a = b) (27)
ha ∂ξa hahb ∂ξb (a = b) (28)

(∇v)aa = 1 ∂va + vc ∂ha
ha ∂ξa hahc ∂ξc
c=a

4.5 Divergence of tensor

It is often necessary to calculate the divergence of a tensor ∇ · ←→t in curvilinear coordinates. We find

(∇ · ←→t )a = eb · ∂(tcdeced) · ea (29)
hb ∂ξb
bcd

Expanding the sum we get

(∇ · ←→t )a = eb · ∂(tcaec) + tbd ∂ed · ea
hb ∂ξb hb ∂ξb
bc bd

= ∇ · ta + 1 tab ∂ha − tbb ∂hb
hahb ∂ξb ∂ξa
b

Using the divergence of a vector this becomes

(∇ · ←→t )a = 1 ∂tba + tba ∂hc + tab ∂ha − tbb ∂hb (30)
hb ∂ξb hbhc ∂ξb hahb ∂ξb hahb ∂ξa
b b=c b=a b=a

4