Curvilinear Coordinates and Deformation

Curvilinear Coordinates and Deformation

Review of the basics of curvilinear coordinates

I will write this review in Eulerian coordinates. Everything I say applies to Lagrangian
coordinates. I will connect the two in the second part of this note. Most of this is found in
Simmonds’ book.

I define curvilinear coordinates xi by writing the Cartesian coordinates as functions of the
curvilinear coordinates: yi = yi(xk). The differential relation is the important relation, and that
may be written

dy i = ∂y i dx k (1)
∂x k

The partial derivative doesn’t have a universally accepted name, but if we write the vector
equivalent of (1)

dr = dy ie i = ∂yi e idx k = gk dxk (2)
∂x k

we see that ∂y i can be viewed as the set of contravariant components of the base vector gk in the
∂x k

Cartesian system. We can refer to these as the tangent vector set and write them as t i . (Note
k

that this is mathematically equivalent to the deformation gradient tensor.) The differentials dxk

denote the contravariant components of dr in the basis defined by the vectors gk. This idea can
be extended to any vector

v = vˆ iei = vk gk = vk tkiei ⇒ vˆ i = tkivk (3)

There exists a set of inverses that we can call the normal vector set, of which the following is
true

nmi t m = i ⇔ nmi ∂y m = i ⇒ nmi = ∂x i (4)
j j ∂x i j ∂y m

and these allow the relation (3) to be inverted to give vi = nki vˆ k

We know from earlier work that we can define a set of reciprocal basis vectors gk such that

gi ⋅g j = dˆ i , and write vectors in covariant form
j

v = vˆiei = vk gk = v k ∂x k ei ⇒ vˆi = nik vk (5)
∂y i

from which we see, first, that the normal vector set nki denote the covariant components of the
reciprocal basis vector gi in the Cartesian system, and that we have found the rule for
transforming covariant components in general. Finally, the reciprocal relation between the
tangent and normal vector sets allows completion of the transformation relations, which can be
summarized as

vˆi = tki vk , v k = nik vˆ i ; vk = tki vˆi , vˆi = nik vk (6)

All of this was covered in an earlier post. (It is important to note that the basis vectors for the
curvilinear systems may not be dimensionless, which in turn means that the dimensions of the
contravariant and covariant components of a vector written in curvilinear coordinates may not be
those of the physical vector. We will discuss the physical equivalents of the curvilinear basis
vectors at another time.)

In our discussion of deformation we found that ds2 was important. It is important here, too.
I write

ds 2 = dr ⋅dr = gˆijdyi dyj = gˆijtmi t j dx mdx n = gmndx m dx n (7)
n

introducing the general metric tensor gmn, which allows one to express ds2 in terms of the

curvilinear differentials. There is an associated metric tensor that is the inverse of the metric

tensor in the sense that g gim = i = gjm g mi . The metric and associated metric tensors may be
mj j

used to raise and lower indices for vectors expressed in curvilinear coordinates as

v i = gij v j , vi = gijv j (8)

Indeed, this works for tensors of any order, and we will be making use of this from time to time.
(We did so in lecture 7 in Cartesian coordinates when we discussed the connection between the
rotation associated with polar decomposition and the linear rotation that is complementary to
linear strain.) For example

C = g g C⋅⋅ kl
ij⋅⋅ mn
kp lq (9)
ijklmn

On deformation

Suppose there to be a set of Eulerian curvilinear coordinates xI and a set of Lagrangian
curvilinear coordinates XK. There will be metric and associated metric tensors associated with
each of these: gij, gij, GKL, GKL. We are used to describing a deformation by specifying the
position of a material point in an underlying space. The material point is identified by its
Lagrangian position (or label), YK, and the position in space is given in terms of its Eulerian
position coordinates yi. As long as we are using Cartesian coordinates this in unambiguous. It
can also be fairly tedious, as some of the homework problems have inadvertently demonstrated.
I assigned problems involving deformations that could have been more easily described in
curvilinear coordinates.

If the curvilinear systems are well-behaved, nonsingular systems, then one can specify a
deformation by mapping the Lagrangian coordinates XK to their Eulerian equivalents xi. In other
words, the material points can be identified by their coordinate values XK instead of YK, the

former being unique if the coordinate system makes sense, and the location in space can be
specified by the coordinate values xi. This being the case, a deformation can be specified in the

form

( )xi = x i XK (10)

I can define a deformation gradient tensor as in the Cartesian case

FKi = ∂ xi (11)
∂X K

and write

ds 2 = gij dx idx j = gij FKi FLj dX K dX L = CKL dX K dX L (12)

where CKL denotes the same Lagrangian deformation tensor, but now written in Lagrangian
curvilinear coordinates. Note also that the Eulerian metric tensor does not have a hat on; it is the
metric tensor for the Eulerian curvilinear system. The physical definition of strain remains the
same, and we have

( )ds2 − dS2 = 2EKLdX KdXL = CKL − GKL dXKdX L (13)

and again, there is no hat on the metric tensor. The strain thus defined is the Lagrangian strain
written in the Lagrangian curvilinear coordinates.

We used CKL to understand deformation in Cartesian coordinates by finding the eigenvalues
and noting that these represented the squares of the semiaxes of a deformed initial unit sphere.
those eigenvalues were dimensionless because the element of the Cartesian deformation tensor
were dimensionless. The latter follows from the fact that dS2 has dimensions of length squared,
and dYK and dYL have dimensions of length, so that CKL must be dimensionless to maintain
dimensional consistency. However, dXK and dXL do not necessarily have the dimension of length
(angle is quite common), so CKL may not be dimensionless, and its eigenvalues may not be
dimensionless. Do note that GKL has the same dimensions element for element as CKL. What
should we do?

To answer this question, consider the dot product of two vectors, supposedly of the same
dimension. This corresponds to GKLdXKdXL, so is relevant to our inquiry. Write

A ⋅ B = GKL AK BL = AL B L = AK BK = GKL AK BL (14)

The product of a covariant and contravariant components of vectors of the same dimension has

the correct dimensions. Of course, AKBL is a covariant tensor, and there is nothing remarkable
about it. If CKL has the “wrong” dimensions, GKMCML will be a mixed tensor with the correct
dimensions. We apply this to the eigenvalue relation as follows

( ) ( )( ) ( )det CKL − GKL = 0 ⇒ det G KM CML − GML = det C⋅KL − K =0 (15)
L

The eigenvalues we want are the eigenvalues of the mixed tensor. (If you solve the eigenvalue

relation in the form given in the first half of equation (15) you will get the same eigenvalues, but

you need to remember that you do not subtract from the diagonal terms. You must subtract
GKK [no sum] from the Kth diagonal element. This will work out automatically if you form the

invariants using the metric tensor following the relations

( )IC 1 1 eRST GKN
= G MN CMN , IIC = 2 G MN G PQ − G MPG NQ CMN CPQ, IIIC = 6 eIJK G ILG JM CRL CSM CTN (16)

which automatically sorts out the dimensions.

Of course one can draw deformation ellipsoids without finding the eigenvalues, but it is
rather more involved for the curvilinear systems. You need to have a mapping from the unit
sphere in the Cartesian Lagrangian coordinates to the deformed sphere in the Cartesian Eulerian
coordinates. This generally requires the inversion of the Lagrangian curvilinear system. If that
is possible (it is always possible in principle, but not always practical), then the following chain
produces what is needed.

( ) ( ( )) ( ( ( )))yi = yi xk = yi xk X K = yi xk X K YL ⇒ dyi = ∂yi ∂x k ∂X K dY L = tki FKk NLK dY L (17)
∂ xk ∂X K ∂Y L

and when dYL describes a sphere dyi will describe the deformation ellipsoid.