11.1 Coordinate Changes

11.1.1 General

Suppose that starting at Cartesian coordinates, we make a switch to new coordinates, like cylindrical or spherical ones? What is involved such a change in coordinates?

To find out, as always take the Cartesian coordinates $(x,y,z)$ to form a vector

$${\skew0\vec r}= x {\hat\imath}+ y {\hat\jmath}+ z {\hat k}$$

in physical space. Let the new coordinates be called $(u_1,u_2,u_3)$. You can take these to form a vector $\vec{u}$ in an artificial parameter space. Note that that vector has no physical meaning; it is just a concise way of writing the three coordinates.

The Jacobian matrix is defined as the matrix of derivatives

$$\fbox{$\displaystyle \frac{\partial{\skew0\vec r}}{\part... ...rtial z}{\mathstrut\partial u_3} \end{array} \right) $}$$

Note that often it is easier to differentiate the new coordinates with respect to the physical ones, rather than vice versa. If so, you can find the Jacobian matrix as the inverse:

$$\fbox{$\displaystyle \frac{\partial{\skew0\vec r}}{\part... ...frac{\partial\vec u}{\partial{\skew0\vec r}}\right)^{-1} $}$$

This is based on the fact that the product of the two matrices above equals $\partial{\skew0\vec r}/\partial{\skew0\vec r}$, which is the unit matrix.

Any small change ${\rm d}\vec{u}$ in the artificial vector $\vec{u}$ corresponds to a small change in physical position ${\rm d}{\skew0\vec r}$ given by:

$$\fbox{$\displaystyle {\rm d}\vec{\skew0\vec r}= \frac{\partial{\skew0\vec r}}{\partial\vec u} {\rm d}\vec u $}$$

So if the Jacobian matrix is well defined, and its determinant nonzero, any small change in physical space ${\rm d}{\skew0\vec r}$ will correspond to a unique displacement ${\rm d}\vec{u}$ in coordinate space, and vice versa. If however the coefficients of the Jacobian matrix are undefined at a point, or the determinant is zero, the mapping is called locally singular at that point.

So the determinant of the Jacobian matrix is very important. This determinant is called the Jacobian $J$:

$$\fbox{$\displaystyle J \equiv \left\vert\frac{\partial... ...l z}{\mathstrut\partial u_3} \end{array} \right\vert $}$$

If you would rather differentiate the new coordinates with respect to the old, the determinant will give the inverse of $J$.

Now suppose you take a little block ${\rm d}{}u_1{\rm d}{}u_2{\rm d}{}u_3$ in the artificial parameter space. In physical space, that little block will correspond to a little parallelepiped with sides

$$\fbox{$\displaystyle {\rm d}{\skew0\vec r}_1 \equiv \fra... ... \frac{\partial {\skew0\vec r}}{\partial u_3} {\rm d}u_3 $}$$

The volume of this small parallelepiped, call it ${\rm d}V$ is given by the scalar triple product of its three sides, which equals the determinant of these three sides:

$${\rm d}V= {\rm d}{\skew0\vec r}_1\cdot({\rm d}{\skew0\vec ... ...{\partial u_3} \right\vert {\rm d}u_1{\rm d}u_2{\rm d}u_3$$

But the determinant above is just the Jacobian $J$! So it follows that if you would rather integrate some function $f$ in new coordinates instead of physical ones, you must simply add a Jacobian:

$$\fbox{$\displaystyle \mathop{\int\kern-7pt\int\kern-7pt\... ...nolimits f \vert J\vert { \rm d}u_1{\rm d}u_2{\rm d}u_3 $}$$

11.1.2 Orthogonal coordinates

Orthogonal coordinates are a special case of the new coordinates discussed in the previous subsection. For orthogonal coordinates, the little parallelepiped in physical space corresponding to a little block ${\rm d}{}u_1{\rm d}{}u_2{\rm d}{}u_3$ in coordinate space has orthogonal sides. So it too is a little block, rather than just a little parallelepiped.

Since the three sides are proportional to the derivatives $\partial{\skew0\vec r}/\partial{}u_1$, $\partial{\skew0\vec r}/\partial{}u_2$, and $\partial{\skew0\vec r}/\partial{}u_3$, these derivatives must be orthogonal for orthogonal coordinates. That means that if you write them as magnitudes $h_1$, $h_2$, and $h_3$ times unit vectors ${\hat\imath}_1$, ${\hat\imath}_2$, and ${\hat\imath}_3$,

$$\fbox{$\displaystyle \frac{\partial {\skew0\vec r}}{\par... ... {\skew0\vec r}}{\partial u_3} \equiv h_3 {\hat\imath}_3 $}$$

then these three unit vectors are orthogonal at each point. So you can use them as a local orthogonal coordinate system. And write any vector $\vec{v}$ at the point in the form

$$\fbox{$\displaystyle \vec v = v_1 {\hat\imath}_1 + v_2 {\hat\imath}_2 + v_3 {\hat\imath}_3 $}$$

You can then convert the gradient, divergence, and curl operators to the new coordinates. These formulas will involve the magnitudes, called metric indices $h_1$, $h_2$, and $h_3$. Since you can find these formulae in any mathematical handbook, they will not be discussed here.

But often you want to find other derivatives in the new coordinates. To do so, you must know how to find the derivatives of the unit vectors ${\hat\imath}_1$, ${\hat\imath}_2$, and ${\hat\imath}_3$. These formulae are not soeasy to find, so they will be given here. For any $i$ equal to 1, 2, or 3, and any $j$ equal to 1, 2, or 3,

$$\fbox{$\displaystyle \frac{\partial{\hat\imath}_i}{\part... ...1}{h_k} \frac{\partial h_i}{\partial u_k} {\hat\imath}_k $}$$

The derivation is in {D.1}.