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Next: 3-D Cylindrical Polar Coordinates Up: Non-Cartesian Coordinates Previous: Summary of 2D Cartesian

2D Cylindrical Polars $(r, \theta )$

To study a new coordinate system we need to obtain the relevent scale factors. To this we, as above, analyse how the point $(r, \theta )$ moves when

\begin{displaymath}
r \qquad \rightarrow \qquad r + \delta r,
\end{displaymath}

and

\begin{displaymath}
\theta \qquad \rightarrow \qquad \theta + \delta \theta.
\end{displaymath}

Thus, from Figure 1.15, we see that ${\bf r}$ moves to ${\bf r} + {\bf\delta r}$, where
\begin{displaymath}
{\bf\delta r} = \delta r {\bf e}_{r} + r\delta \theta {\bf e}_{\theta}.
\end{displaymath} (1.30)

Figure 1.15: Increasing $r$ and $\theta $ by the respetive amounts $\delta r$ and $\delta \theta $ gives the change in position, ${\bf
\delta r}$. Note that the distance round the circle is $r \delta \theta $.

This time the scale factors are (the coefficients of $\delta r$ and $\delta \theta $) $1$ and $r$. The infinitesimal change in area is
\begin{displaymath}
dS = r dr d\theta.
\end{displaymath} (1.31)

The gradient operator is now
\begin{displaymath}
\nabla f = {\partial f\over \partial r}{\bf e}_{r} + {1\over
r}{\partial f \over \partial \theta}{\bf e}_{\theta},
\end{displaymath} (1.32)

while
\begin{displaymath}
\nabla \cdot {\bf A} = {1\over r}{\partial \over \partial r...
...ht ) + {1\over r}{\partial A_{\theta}\over \partial
\theta}.
\end{displaymath} (1.33)

Note that the dimensions must be correct. $\theta $ has no dimensions. Finally, for the 2D case considered here,
\begin{displaymath}
\nabla \times {\bf A} = \left ({1\over r}{\partial \over \p...
...r}{\partial A_{r}\over
\partial \theta}\right ) {\bf e}_{z}.
\end{displaymath} (1.34)


next up previous
Next: 3-D Cylindrical Polar Coordinates Up: Non-Cartesian Coordinates Previous: Summary of 2D Cartesian
Prof. Alan Hood
2000-11-06