Cylindrical coordinates

The cylindrical coordinates are defined as

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

or equivalently

\[ \begin{align}\begin{aligned}& x^1 = X^1 \cos X^2,\\& x^2 = X^1 \sin X^2,\\& x^3 = X^3.\end{aligned}\end{align} \]

The transformation matrix is

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

or

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

As a consequence, the basis vectors are related by

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

or

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

The scale factors are

\[ \begin{align}\begin{aligned}& H_1 = \sqrt{ \vec{E}_1 \cdot \vec{E}_1 } = 1,\\& H_2 = \sqrt{ \vec{E}_2 \cdot \vec{E}_2 } = X^1,\\& H_3 = \sqrt{ \vec{E}_3 \cdot \vec{E}_3 } = 1,\end{aligned}\end{align} \]

and their spatial derivatives are

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

The Jacobian determinant is

\[J = H_1 H_2 H_3 = X^1.\]

Thus the normalised basis vectors are

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

or inversely

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

Although not limited, I consider the velocity vector

\[\vec{u} \equiv \tder{\vec{x}}{t} = \sum_i u^i \vec{e}_i\]

as an example to see how the components are related. Then the contravariant components are

\[ \begin{align}\begin{aligned}& U^1 = u^1 \cos X^2 + u^2 \sin X^2,\\& U^2 = - u^1 \frac{\sin X^2}{X^1} + u^2 \frac{\cos X^2}{X^1},\\& U^3 = u^3,\end{aligned}\end{align} \]

while the normalised components are

\[ \begin{align}\begin{aligned}& \hat{U}^1 = u^1 \cos X^2 + u^2 \sin X^2,\\& \hat{U}^2 = - u^1 \sin X^2 + u^2 \cos X^2,\\& \hat{U}^3 = u^3.\end{aligned}\end{align} \]