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Next: Cylindrical Coordinates Up: Non-Cartesian Coordinates Previous: Introduction

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. Since 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 which 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: i.e.,

\begin{displaymath}
{\bf e}_i\cdot {\bf e}_j = \delta_{ij}.
\end{displaymath} (1595)

In this case, $u_1$, $u_2$, $u_3$ are said to constitute an orthogonal coordinate system. Suppose, further, that
\begin{displaymath}
{\bf e}_1\cdot {\bf e}_2\times {\bf e}_3 = 1
\end{displaymath} (1596)

at all points in space, so that $u_1$, $u_2$, $u_3$ also constitute a right-handed coordinate system. It follows that
\begin{displaymath}
{\bf e}_i\cdot{\bf e}_j\times {\bf e}_k = \epsilon_{ijk}.
\end{displaymath} (1597)

Finally, a general vector ${\bf A}$, associated with a particular point in space, can be written
\begin{displaymath}
{\bf A} = \sum_i A_i\,{\bf e}_i,
\end{displaymath} (1598)

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

\begin{displaymath}
d {\bf x} = \frac{du_1}{\vert\nabla u_1\vert}\,{\bf e}_1 + \...
...{\vert\nabla u_3\vert}\,{\bf e}_3=\sum_i h_i\,du_i\,{\bf e}_i.
\end{displaymath} (1599)

Hence, from (1595), an element of length (squared) in the $u_1$, $u_2$, $u_3$ coordinate system is written
\begin{displaymath}
d{\bf x}\cdot d{\bf x}= \sum_i h_i^{\,2}\,du_i^{\,2}.
\end{displaymath} (1600)

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
\begin{displaymath}
h = h_1\,h_2\,h_3.
\end{displaymath} (1601)

Note that [see Equation (1481)]

\begin{displaymath}
\nabla\times \nabla u_i = 0,
\end{displaymath} (1602)

and
\begin{displaymath}
\nabla\cdot \left(\frac{h_i^{\,2}}{h}\,\nabla u_i\right) = 0.
\end{displaymath} (1603)

The latter result follows from Equations (1480) and (1481) because $(h_1^{\,2}/h)\,\nabla u_1= \nabla u_2\times \nabla u_3$, etc. Finally, it is easily demonstrated from (1595) and (1597) that
$\displaystyle \nabla u_i\cdot \nabla u_j$ $\textstyle =$ $\displaystyle h_i^{\,-2}\,\delta_{ij},$ (1604)
$\displaystyle \nabla u_i\cdot\nabla u_j\times \nabla u_k$ $\textstyle =$ $\displaystyle h^{-1}\,\epsilon_{ijk}.$ (1605)

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

\begin{displaymath}
\nabla \phi = \sum_i \frac{\partial \phi}{\partial u_i}\,\na...
...i\frac{1}{h_i}\,\frac{\partial \phi}{\partial u_i}\,{\bf e}_i.
\end{displaymath} (1606)

Hence, the components of $\nabla\phi$ in the $u_1$, $u_2$, $u_3$ coordinate system are
\begin{displaymath}
(\nabla\phi)_i = \frac{1}{h_i}\,\frac{\partial \phi}{\partial u_i}.
\end{displaymath} (1607)

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

$\displaystyle \nabla\cdot{\bf A}$ $\textstyle =$ $\displaystyle \sum_i\nabla\cdot(A_i\,{\bf e}_i) =\sum_i\nabla\cdot (h_i\,A_i\,\...
...\nabla\!\cdot\!\left(\frac{h}{h_i}\,A_i\,\frac{h_i^{\,2}}{h}\,\nabla u_i\right)$  
  $\textstyle =$ $\displaystyle \sum_i\frac{h_i^{\,2}}{h}\nabla u_i\cdot\nabla\left(\frac{h}{h_i}...
..._i
\frac{1}{h}\,\frac{\partial}{\partial u_i}\!\left(\frac{h}{h_i}\,A_i\right),$ (1608)

where use has been made of Equations (1479), (1603), and (1604). Thus, the divergence of ${\bf A}$ in the $u_1$, $u_2$, $u_3$ coordinate system takes the form
\begin{displaymath}
\nabla\cdot{\bf A} =\sum_i
\frac{1}{h}\,\frac{\partial}{\partial u_i}\!\left(\frac{h}{h_i}\,A_i\right).
\end{displaymath} (1609)

We can write

$\displaystyle \nabla\times {\bf A}$ $\textstyle =$ $\displaystyle \sum_k\nabla\times (A_k\,{\bf e}_k)=\sum_k \nabla\times (h_k\,A_k\,\nabla u_k)
=\sum_k\nabla (h_k\,u_k)\,\times \nabla u_k$  
  $\textstyle =$ $\displaystyle \sum_{j,k}\frac{\partial (h_k\,A_k)}{\partial u_j}\,\nabla u_j\times\nabla u_k,$ (1610)

where use has been made of Equations (1483), (1602), and (1606). It follows from (1605) that
\begin{displaymath}
(\nabla\times {\bf A})_i = {\bf e}_i\cdot\nabla\times {\bf A...
...ijk}\,\frac{h_i}{h}\,\frac{\partial (h_k\,A_k)}{\partial u_j}.
\end{displaymath} (1611)

Hence, the components of $\nabla\times {\bf A}$ in the $u_1$, $u_2$, $u_3$ coordinate system are
\begin{displaymath}
(\nabla\times{\bf A})_i = \sum_{j,k}\epsilon_{ijk}\,\frac{h_i}{h}\,\frac{\partial (h_k\,A_k)}{\partial u_j}.
\end{displaymath} (1612)

Now, $\nabla^2\phi = \nabla\cdot\nabla\phi$ [see (1477)], so Equations (1606) and (1609) yield the following expression for $\nabla^2\phi$ in the $u_1$, $u_2$, $u_3$ coordinate system:

\begin{displaymath}
\nabla^2\phi =\sum_i\frac{1}{h}\,\frac{\partial}{\partial u_...
...\frac{h}{h_i^{\,2}}\,\frac{\partial\phi}{\partial u_i}\right).
\end{displaymath} (1613)

The vector identities (1476) and (1484) can be combined to give the following expression for $({\bf A}\cdot\nabla){\bf B}$ that is valid in a general coordinate system:

$\displaystyle ({\bf A}\cdot\nabla) {\bf B}$ $\textstyle =$ $\displaystyle \frac{1}{2}\left[\nabla({\bf A}\cdot{\bf B}) - \nabla\times ({\bf...
...\bf B})
- (\nabla\cdot{\bf A})\,{\bf B} + (\nabla\cdot{\bf B})\,{\bf A} \right.$  
    $\displaystyle \left.
- {\bf A}\times(\nabla\times {\bf B})-{\bf B}\times (\nabla\times {\bf A})\right].$ (1614)

Making use of Equations (1607), (1609), and (1612), as well as the easily demonstrated results
$\displaystyle {\bf A}\cdot{\bf B}$ $\textstyle =$ $\displaystyle \sum_i A_i\,B_i,$ (1615)
$\displaystyle {\bf A}\times {\bf B}$ $\textstyle =$ $\displaystyle \sum_{j,k} \epsilon_{ijk}\,A_j\,B_k,$ (1616)

and the tensor identity (1500), Equation (1614) 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:
\begin{displaymath}[({\bf A}\cdot\nabla)\,{\bf B}]_i= \sum_j\left(\frac{A_j}{h_j...
...A_i\,B_j}{h_i\,h_j}\,\frac{\partial h_i}{\partial u_j}\right).
\end{displaymath} (1617)

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

\begin{displaymath}
(\nabla {\bf A})_{ij} = \frac{\partial A_i}{\partial x_j}.
\end{displaymath} (1618)

In an orthogonal curvilinear coordinate system, the above expression generalizes to
\begin{displaymath}
(\nabla{\bf A})_{ij} = [({\bf e}_j\cdot\nabla)\,{\bf A}]_i.
\end{displaymath} (1619)

It thus follows from (1617), and the relation $({\bf e}_i)_j = {\bf e}_i\cdot {\bf e}_j=\delta_{ij}$, that
\begin{displaymath}
(\nabla{\bf A})_{ij} = \frac{1}{h_j}\,\frac{\partial A_i}{\p...
...\sum_k\frac{A_k}{h_i\,h_k}\,\frac{\partial h_i}{\partial u_k}.
\end{displaymath} (1620)

The vector identity (1482) yields the following expression for $\nabla^2{\bf A}$ that is valid in a general coordinate system:

\begin{displaymath}
\nabla^2{\bf A} = \nabla(\nabla\cdot{\bf A}) - \nabla\times(\nabla\times {\bf A}).
\end{displaymath} (1621)

Making use of Equations (1609), (1612), and (1613), as well as (1615) and (1616), and the tensor identity (1500), the above 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:
$\displaystyle (\nabla^2{\bf A})_i$ $\textstyle =$ $\displaystyle \nabla^2 A_i +\sum_j\left\{\frac{2}{h_i\,h_j}\left(\frac{1}{h_i}\...
...ac{\partial h_j}{\partial u_i}\,\frac{\partial}{\partial u_j}\right) A_j\right.$  
    $\displaystyle +\frac{h}{h_i\,h_j^{\,2}}\left[\frac{A_j}{h_i^{\,2}}\,\frac{\part...
... u_j}\,\frac{\partial}{\partial u_j}\!\left(
\frac{h_j^{\,2}}{h}\right) \right]$  
    $\displaystyle +\frac{A_j}{h_i}\,\frac{h}{h_j^{\,3}}\left[\frac{1}{h_j}\,\frac{\...
...partial^2}{\partial u_i\,\partial u_j}\!\left(\frac{h_j^{\,2}}{h}\right)\right]$  
    $\displaystyle \left.-\frac{A_i}{h_i\,h_j^{\,2}}\left[\frac{2}{h_i}\left(\frac{\...
...partial u_j}\right)^2-\frac{\partial^2 h_i}{\partial
u_j^{\,2}}\right]\right\}.$ (1622)

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 $h_1=h_2=h_3=h=1$).


next up previous
Next: Cylindrical Coordinates Up: Non-Cartesian Coordinates Previous: Introduction
Richard Fitzpatrick 2012-04-27