##### Cylindrical Coordinates
c.m.cs.cyl

\((r, \varphi, z)\). Relation to Cartesian coordinates:

\begin{equation} x = r \cos \varphi, y = r \sin \varphi, z = z \label{Gr(1.74)} \end{equation}

The unit vectors are

\begin{equation} \hat{\bf r} = \cos \varphi ~\hat{\bf x} + \sin \varphi~\hat{\bf y}, \hat{\boldsymbol \varphi} = -\sin \theta ~\hat{\bf x} + \cos \varphi~\hat{\bf y}, \hat{\bf z} = \hat{\bf z}. \label{Gr(1.75)} \end{equation}

Infinitesimal displacement:

\begin{equation} d{\bf l} = dr ~\hat{\bf r} + r d\varphi~\hat{\boldsymbol \varphi} + dz ~\hat{\bf z}. \label{Gr(1.77)} \end{equation}

Volume element:

\begin{equation} d\tau = r dr d\varphi dz \label{Gr(1.78)} \end{equation}

Range of parameters: \(r \in [0, \infty[\), \(\varphi \in [0, 2\pi[\) and \(z \in ]-\infty, \infty[\).

###### Gradient

- Gr4(1.79)

\begin{equation} {\boldsymbol \nabla} T = \frac{\partial T}{\partial r}~\hat{\bf r} + \frac{1}{r} \frac{\partial T}{\partial \varphi}~\hat{\boldsymbol \varphi} + \frac{\partial T}{\partial z} ~\hat{\bf z} \tag{cylgrad}\label{cylgrad} \end{equation}

###### Divergence

- Gr4(2.21)

\begin{equation} {\boldsymbol \nabla} \cdot {\bf v} = \frac{1}{r} \frac{\partial}{\partial r} (r v_r) + \frac{1}{r} \frac{\partial v_{\varphi}}{\partial \varphi} + \frac{\partial v_z}{\partial z}. \tag{cyl_div} \label{cyl_div} \end{equation}

###### Curl

- Gr4(2.21)

\begin{align} {\boldsymbol \nabla} \times {\bf v} = \left( \frac{1}{r} \frac{\partial v_z}{\partial \varphi} - \frac{\partial v_{\varphi}}{\partial z}\right) ~\hat{\bf r} + \left( \frac{\partial v_r}{\partial z} - \frac{\partial v_z}{\partial r} \right) ~\hat{\boldsymbol \varphi} \nonumber \\ + \frac{1}{r} \left( \frac{\partial}{\partial r} (r v_{\varphi}) - \frac{\partial v_r}{\partial \varphi} \right) ~\hat{\bf z} \tag{cyl_curl} \label{cyl_curl} \end{align}

###### Laplacian

\begin{equation} {\boldsymbol \nabla}^2 T = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 T}{\partial \varphi^2} + \frac{\partial^2 T}{\partial z^2} \label{Gr(1.82)} \end{equation}

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42