Orthogonal Curvilinear Coordinates

Let $x_1$ , $x_2$ , $x_3$ be a set of standard right-handed Cartesian coordinates. Furthermore, let $u_1(x_1, x_2, x_3)$ , $u_2(x_1, x_2, x_3)$ , $u_3(x_1, x_2, x_3)$ be three independent functions of these coordinates which are such that each unique triplet of $x_1$ , $x_2$ , $x_3$ values is associated with a unique triplet of $u_1$ , $u_2$ , $u_3$ values. It follows that $u_1$ , $u_2$ , $u_3$ can be used as an alternative set of coordinates to distinguish different points in space. Because the surfaces of constant $u_1$ , $u_2$ , and $u_3$ are not generally parallel planes, but rather curved surfaces, this type of coordinate system is termed curvilinear.

Let $h_1=\vert\nabla u_1\vert^{-1}$ , $h_2=\vert\nabla u_2\vert^{-1}$ , and $h_3=\vert\nabla u_3\vert^{-1}$ . It follows that ${\bf e}_1=h_1\,\nabla u_1$ , ${\bf e}_2=h_2\,\nabla u_2$ , and ${\bf e}_3=h_3\,\nabla u_3$ are a set of unit basis vectors that are normal to surfaces of constant $u_1$ , $u_2$ , and $u_3$ , respectively, at all points in space. Note, however, that the direction of these basis vectors is generally a function of position. Suppose that the ${\bf e}_i$ , where $i$ runs from 1 to 3, are mutually orthogonal at all points in space: that is,

$${\bf e}_i\cdot {\bf e}_j = \delta_{ij}.$$ (C.1)

In this case, $u_1$ , $u_2$ , $u_3$ are said to constitute an orthogonal coordinate system. Suppose, further, that

$${\bf e}_1\cdot {\bf e}_2\times {\bf e}_3 = 1$$ (C.2)

at all points in space, so that $u_1$ , $u_2$ , $u_3$ also constitute a right-handed coordinate system. It follows that

$${\bf e}_i\cdot{\bf e}_j\times {\bf e}_k = \epsilon_{ijk}.$$ (C.3)

Finally, a general vector ${\bf A}$ , associated with a particular point in space, can be written

$${\bf A} = \sum_{i=1,3} A_i\,{\bf e}_i,$$ (C.4)

where the ${\bf e}_i$ are the local basis vectors of the $u_1$ , $u_2$ , $u_3$ system, and $A_i={\bf e}_i\cdot {\bf A}$ is termed the $i$ th component of ${\bf A}$ in this system.

Consider two neighboring points in space whose coordinates in the $u_1$ , $u_2$ , $u_3$ system are $u_1$ , $u_2$ , $u_3$ and $u_1+d u_1$ , $u_2+d u_2$ , $u_3+d u_3$ . It is easily shown that the vector directed from the first to the second of these points takes the form

$$d {\bf x} = \frac{du_1}{\vert\nabla u_1\vert}\,{\bf e}_1 + \frac{... ...\frac{du_3}{\vert\nabla u_3\vert}\,{\bf e}_3=\sum_{i=1,3} h_i\,du_i\,{\bf e}_i.$$ (C.5)

Hence, from (C.1), an element of length (squared) in the $u_1$ , $u_2$ , $u_3$ coordinate system is written

$$d{\bf x}\cdot d{\bf x}= \sum_{i=1,3} h_i^{\,2}\,du_i^{\,2}.$$ (C.6)

Here, the $h_i$ , which are generally functions of position, are known as the scale factors of the system. Elements of area that are normal to ${\bf e}_1$ , ${\bf e}_2$ , and ${\bf e}_3$ , at a given point in space, take the form $dS_1 = h_2\,h_3\,du_2\,du_3$ , $dS_2 = h_1\,h_3\,du_1\,du_3$ , and $dS_3 = h_1\,h_2\,du_1\,du_2$ , respectively. Finally, an element of volume, at a given point in space, is written $dV = h\,du_1\,du_2\,du_3$ , where

$$h = h_1\,h_2\,h_3.$$ (C.7)

It can be seen that [see Equation (A.176)]

$$\nabla\times \nabla u_i = 0,$$ (C.8)

and

$$\nabla\cdot \left(\frac{h_i^{\,2}}{h}\,\nabla u_i\right) = 0.$$ (C.9)

The latter result follows from Equations (A.175) and (A.176) because $(h_1^{\,2}/h)\,\nabla u_1= \nabla u_2\times \nabla u_3$ , et cetera. Finally, it is easily demonstrated from Equations (C.1) and (C.3) that

$$\nabla u_i\cdot \nabla u_j =h_i^{\,-2}\,\delta_{ij},$$ (C.10)

$$\nabla u_i\cdot\nabla u_j\times \nabla u_k = h^{-1}\,\epsilon_{ijk}.$$ (C.11)

Consider a scalar field $\phi(u_1,u_2,u_3)$ . It follows from the chain rule, and the relation ${\bf e}_i = h_i\,\nabla u_i$ , that

$$\nabla \phi = \sum_{i=1,3} \frac{\partial \phi}{\partial u_i}\,\nabla u_i=\sum_{i=1,3}\frac{1}{h_i}\,\frac{\partial \phi}{\partial u_i}\,{\bf e}_i.$$ (C.12)

Hence, the components of $\nabla\phi$ in the $u_1$ , $u_2$ , $u_3$ coordinate system are

$$(\nabla\phi)_i = \frac{1}{h_i}\,\frac{\partial \phi}{\partial u_i}.$$ (C.13)

Consider a vector field ${\bf A}(u_1,u_2,u_3)$ . We can write

$$\nabla\cdot{\bf A} =\sum_{i=1,3}\nabla\cdot(A_i\,{\bf e}_i) =\sum_{i=1,3}\nabla\cdot... ...\nabla\!\cdot\!\left(\frac{h}{h_i}\,A_i\,\frac{h_i^{\,2}}{h}\,\nabla u_i\right)$$

$$=\sum_{i=1,3}\frac{h_i^{\,2}}{h}\nabla u_i\cdot\nabla\left(\frac{... ...3} \frac{1}{h}\,\frac{\partial}{\partial u_i}\!\left(\frac{h}{h_i}\,A_i\right),$$ (C.14)

where use has been made of Equations (A.174), (C.9), and (C.10). Thus, the divergence of ${\bf A}$ in the $u_1$ , $u_2$ , $u_3$ coordinate system takes the form

$$\nabla\cdot{\bf A} =\sum_{i=1,3} \frac{1}{h}\,\frac{\partial}{\partial u_i}\!\left(\frac{h}{h_i}\,A_i\right).$$ (C.15)

We can write

$$\nabla\times {\bf A} = \sum_{k=1,3}\nabla\times (A_k\,{\bf e}_k)=\sum_{k=1,3} \nabla\times (h_k\,A_k\,\nabla u_k) =\sum_{k=1,3}\nabla (h_k\,u_k)\,\times \nabla u_k$$

$$=\sum_{j,k=1,3}\frac{\partial (h_k\,A_k)}{\partial u_j}\,\nabla u_j\times\nabla u_k,$$ (C.16)

where use has been made of Equations (A.178), (C.8), and (C.12). It follows from Equation (C.11) that

$$(\nabla\times {\bf A})_i = {\bf e}_i\cdot\nabla\times {\bf A} = \... ...,k=1,3}\epsilon_{ijk}\,\frac{h_i}{h}\,\frac{\partial (h_k\,A_k)}{\partial u_j}.$$ (C.17)

Hence, the components of $\nabla\times {\bf A}$ in the $u_1$ , $u_2$ , $u_3$ coordinate system are

$$(\nabla\times{\bf A})_i = \sum_{j,k=1,3}\epsilon_{ijk}\,\frac{h_i}{h}\,\frac{\partial (h_k\,A_k)}{\partial u_j}.$$ (C.18)

Now, $\nabla^{\,2}\phi = \nabla\cdot\nabla\phi$ [see Equation (A.172)], so Equations (C.12) and (C.15) yield the following expression for $\nabla^{\,2}\phi$ in the $u_1$ , $u_2$ , $u_3$ coordinate system:

$$\nabla^2\phi =\sum_{i=1,3}\frac{1}{h}\,\frac{\partial}{\partial u_i}\!\left(\frac{h}{h_i^{\,2}}\,\frac{\partial\phi}{\partial u_i}\right).$$ (C.19)

The vector identities (A.171) and (A.179) can be combined to give the following expression for $({\bf A}\cdot\nabla)\,{\bf B}$ that is valid in a general coordinate system:

$$({\bf A}\cdot\nabla) \,{\bf B} = \frac{1}{2}\left[\nabla({\bf A}\cdot{\bf B}) - \nabla\times ({\... ...\bf B}) - (\nabla\cdot{\bf A})\,{\bf B} + (\nabla\cdot{\bf B})\,{\bf A} \right.$$

$$\left.\phantom{=} - {\bf A}\times(\nabla\times {\bf B})-{\bf B}\times (\nabla\times {\bf A})\right].$$ (C.20)

Making use of Equations (C.13), (C.15), and (C.18), as well as the easily demonstrated results

$${\bf A}\cdot{\bf B} =\sum_{i=1,3} A_i\,B_i,$$ (C.21)

$${\bf A}\times {\bf B} =\sum_{j,k=1,3} \epsilon_{ijk}\,A_j\,B_k,$$ (C.22)

and the tensor identity (B.16), Equation (C.20) reduces (after a great deal of tedious algebra) to the following expression for the components of $({\bf A}\cdot\nabla)\,{\bf B}$ in the $u_1$ , $u_2$ , $u_3$ coordinate system:

$$[({\bf A}\cdot\nabla)\,{\bf B}]_i= \sum_{j=1,3}\left(\frac{A_j}{h... ...ial u_i} + \frac{A_i\,B_j}{h_i\,h_j}\,\frac{\partial h_i}{\partial u_j}\right).$$ (C.23)

Note, incidentally, that the commonly quoted result $[({\bf A}\cdot\nabla)\,{\bf B}]_i={\bf A}\cdot\nabla B_i$ is only valid in Cartesian coordinate systems (for which $h_1=h_2=h_3=1$ ).

Let us define the gradient $\nabla{\bf A}$ of a vector field ${\bf A}$ as the tensor whose components in a Cartesian coordinate system take the form

$$(\nabla {\bf A})_{ij} = \frac{\partial A_i}{\partial x_j}.$$ (C.24)

In an orthogonal curvilinear coordinate system, the previous expression generalizes to

$$(\nabla{\bf A})_{ij} = [({\bf e}_j\cdot\nabla)\,{\bf A}]_i.$$ (C.25)

It thus follows from Equation (C.23), and the relation $({\bf e}_i)_j = {\bf e}_i\cdot {\bf e}_j=\delta_{ij}$ , that

$$(\nabla{\bf A})_{ij} = \frac{1}{h_j}\,\frac{\partial A_i}{\partia... ...\delta_{ij}\sum_{k=1,3}\frac{A_k}{h_i\,h_k}\,\frac{\partial h_i}{\partial u_k}.$$ (C.26)

The vector identity (A.177) yields the following expression for $\nabla^{\,2}{\bf A}$ that is valid in a general coordinate system:

$$\nabla^{\,2}{\bf A} = \nabla(\nabla\cdot{\bf A}) - \nabla\times(\nabla\times {\bf A}).$$ (C.27)

Making use of Equations (C.15), (C.18), and (C.19), as well as (C.21) and (C.22), and the tensor identity (B.16), the previous equation reduces (after a great deal of tedious algebra) to the following expression for the components of $\nabla^{\,2}{\bf A}$ in the $u_1$ , $u_2$ , $u_3$ coordinate system:

$$(\nabla^{\,2}{\bf A})_i = \nabla^{\,2} A_i +\sum_{j=1,3}\left\{\frac{2}{h_i\,h_j}\left(\f... ...ac{\partial h_j}{\partial u_i}\,\frac{\partial}{\partial u_j}\right) A_j\right.$$

$$\phantom{=}+\frac{h}{h_i\,h_j^{\,2}}\left[\frac{A_j}{h_i^{\,2}}\,... ... u_j}\,\frac{\partial}{\partial u_j}\!\left( \frac{h_j^{\,2}}{h}\right) \right]$$

$$\phantom{=}+\frac{A_j}{h_i}\,\frac{h}{h_j^{\,3}}\left[\frac{1}{h_... ...ial^{\,2}}{\partial u_i\,\partial u_j}\!\left(\frac{h_j^{\,2}}{h}\right)\right]$$

$$\phantom{=}\left.-\frac{A_i}{h_i\,h_j^{\,2}}\left[\frac{2}{h_i}\l... ...partial u_j}\right)^2-\frac{\partial^2 h_i}{\partial u_j^{\,2}}\right]\right\}.$$ (C.28)

Note, again, that the commonly quoted result $(\nabla^{\,2} {\bf A})_i=\nabla^{\,2} A_i$ is only valid in Cartesian coordinate systems (for which $h_1=h_2=h_3=h=1$ ).