Application

I consider the cylindrical coordinates

\[ \begin{align}\begin{aligned}& \xi^1 \equiv \sqrt{x^1 x^1 + x^2 x^2},\\& \xi^2 \equiv \tan^{-1} \left( \frac{x^2}{x^1} \right),\\& \xi^3 \equiv x^3,\end{aligned}\end{align} \]

which are further stretched in a rectilinear way:

\[ \begin{align}\begin{aligned}& X^1 \equiv f_1 \left( \xi^1 \right),\\& X^2 \equiv f_2 \left( \xi^2 \right),\\& X^3 \equiv f_3 \left( \xi^3 \right).\end{aligned}\end{align} \]

Note that \(f_1\) is an arbitrary function and can be non-linear, while the other two functions \(f_2, f_3\) are linear.

Adopting the chain rule:

\[\pder{x^i}{X^j} = \sum_k \pder{x^i}{\xi^k} \pder{\xi^k}{X^j}\]

yields the transformation matrix:

\[\begin{split}\begin{pmatrix} \pder{x^1}{X^1} & \pder{x^1}{X^2} & \pder{x^1}{X^3} \\ \pder{x^2}{X^1} & \pder{x^2}{X^2} & \pder{x^2}{X^3} \\ \pder{x^3}{X^1} & \pder{x^3}{X^2} & \pder{x^3}{X^3} \\ \end{pmatrix} & = \begin{pmatrix} \pder{x^1}{\xi^1} & \pder{x^1}{\xi^2} & \pder{x^1}{\xi^3} \\ \pder{x^2}{\xi^1} & \pder{x^2}{\xi^2} & \pder{x^2}{\xi^3} \\ \pder{x^3}{\xi^1} & \pder{x^3}{\xi^2} & \pder{x^3}{\xi^3} \\ \end{pmatrix} \begin{pmatrix} \pder{\xi^1}{X^1} & \pder{\xi^1}{X^2} & \pder{\xi^1}{X^3} \\ \pder{\xi^2}{X^1} & \pder{\xi^2}{X^2} & \pder{\xi^2}{X^3} \\ \pder{\xi^3}{X^1} & \pder{\xi^3}{X^2} & \pder{\xi^3}{X^3} \\ \end{pmatrix} \\ & = \begin{pmatrix} \cos \xi^2 & - \xi^1 \sin \xi^2 & 0 \\ \sin \xi^2 & \xi^1 \cos \xi^2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} \pder{\xi^1}{X^1} & 0 & 0 \\ 0 & \pder{\xi^2}{X^2} & 0 \\ 0 & 0 & \pder{\xi^3}{X^3} \\ \end{pmatrix} \\ & = \begin{pmatrix} \pder{\xi^1}{X^1} \cos \xi^2 & - \pder{\xi^2}{X^2} \xi^1 \sin \xi^2 & 0 \\ \pder{\xi^1}{X^1} \sin \xi^2 & \pder{\xi^2}{X^2} \xi^1 \cos \xi^2 & 0 \\ 0 & 0 & \pder{\xi^3}{X^3} \\ \end{pmatrix}.\end{split}\]

Note that, because of the linearity of \(f_2\) and \(f_3\), \(\pder{\xi^2}{X^2}\) and \(\pder{\xi^3}{X^3}\) (specifically \(\Delta \theta\) and \(\Delta z\), respectively) are both constant in space. \(\pder{\xi^1}{X^1}\) (\(\Delta r\)), on the other hand, can vary in \(X^1\).

The relation of the basis vectors are

\[ \begin{align}\begin{aligned}& \vec{E}_1 = \vec{e}_1 \pder{\xi^1}{X^1} \cos \xi^2 + \vec{e}_2 \pder{\xi^1}{X^1} \sin \xi^2,\\& \vec{E}_2 = - \vec{e}_1 \pder{\xi^2}{X^2} \xi^1 \sin \xi^2 + \vec{e}_2 \pder{\xi^2}{X^2} \xi^1 \cos \xi^2,\\& \vec{E}_3 = \vec{e}_3 \pder{\xi^3}{X^3}.\end{aligned}\end{align} \]

The scale factors are

\[ \begin{align}\begin{aligned}& H_1 = \sqrt{ \vec{E}_1 \cdot \vec{E}_1 } = \pder{\xi^1}{X^1} \left( = \Delta r \right),\\& H_2 = \sqrt{ \vec{E}_2 \cdot \vec{E}_2 } = \pder{\xi^2}{X^2} \xi^1 \left( = r \Delta \theta \right),\\& H_3 = \sqrt{ \vec{E}_3 \cdot \vec{E}_3 } = \pder{\xi^3}{X^3} \left( = \Delta z \right),\end{aligned}\end{align} \]

and thus the Jacobian determinant leads to

\[J = \xi^1 \pder{\xi^1}{X^1} \pder{\xi^2}{X^2} \pder{\xi^3}{X^3} \left( = r \Delta r \Delta \theta \Delta z \right).\]

The spatial derivatives of \(H_i\) are

\[\begin{split}\begin{pmatrix} \pder{H_1}{X^1} & \pder{H_2}{X^1} & \pder{H_3}{X^1} \\ \pder{H_1}{X^2} & \pder{H_2}{X^2} & \pder{H_3}{X^2} \\ \pder{H_1}{X^3} & \pder{H_2}{X^3} & \pder{H_3}{X^3} \\ \end{pmatrix} = \begin{pmatrix} \pder{H_1}{X^1} & \pder{H_2}{X^1} & 0 \\ 0 & \pder{H_2}{X^2} & 0 \\ 0 & 0 & \pder{H_3}{X^3} \\ \end{pmatrix},\end{split}\]

which can be further simplified as

\[\begin{split}\begin{pmatrix} \pder{H_1}{X^1} & \pder{H_2}{X^1} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix},\end{split}\]

by utilising the linearity of the mapping functions \(f_2, f_3\).