polar coordinatesPolarCoordinates2005-04-24 10:15:202008-01-20 23:39:02Definition% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{cor}{Corollary}\PMlinkescapeword{relations}
Let $x,y$ be Cartesian coordinates for $\R^2$.
Then $r\ge 0$, $\theta\in [0,2\pi)$ related to $(x,y)$ by
\begin{eqnarray*}
x(r,\theta) &=& r \cos \theta, \\
y(r,\theta) &=& r \sin \theta,
\end{eqnarray*}
are the \emph{polar coordinates} for $(x,y)$. It is simply written $(r,\theta)$.
\begin{center}
\includegraphics{polar_4.eps}
\end{center}
The polar coordinates of Cartesian coordinates
$(x,y) \in \R^2\setminus\{0\}$ are
\begin{eqnarray*}
r(x,y) &=& \sqrt{x^2+ y^2}, \\
\theta(x,y) &=& \arctan (x,y),
\end{eqnarray*}
where $\arctan$ is defined \PMlinkname{here}{OperatornamearcTanWithTwoArguments}.
\textbf{Polar basis.}
Polar coordinates are equipped with an orthonormal base $\{\mathbf{e_r,e_\theta}\}$, which can be defined in terms of the standard cartesian base $\{\mathbf{i,j}\}$ in $\mathbbmss{R}^2$ as follows.
\begin{align*}
\begin{bmatrix}\mathbf{e_r}\\ \mathbf{e_\theta}\end{bmatrix}=
\begin{bmatrix}\hphantom{-}\cos\theta\mathbf{i}+\sin\theta\mathbf{j}\\
-\sin\theta\mathbf{i}+\cos\theta\mathbf{j}\end{bmatrix},
\end{align*}
where $\mathbf{e_r,e_\theta}$ are so-called {\em radial} and {\em traverse} or {\em angular} vector, respectively. Since these vectors are variable in direction, they are differentiable. In fact,
\begin{align*}
\begin{bmatrix}\frac{d\mathbf{e_r}}{d\theta}\\ \frac{d\mathbf{e_\theta}}{d\theta}\end{bmatrix}=
\begin{bmatrix}\hphantom{-}\mathbf{e_\theta}\\ -\mathbf{e_r}\end{bmatrix}.
\end{align*}
The geometrical action of the derivative operator $d/d\theta$ is like a rotation operator that rotates each base vector a counter-clockwise angle equals to $\pi/2$.
\textbf{Position vector.} For an arbitrary point of polar coordinates $(r,\theta)$, its position vector comes given by the single equation
\begin{align*}
\mathbf{r}=r\mathbf{e_r}.
\end{align*}
\textbf{Relations with complex numbers.}
When the Euclidean plane $\R^2$ is identified with $\C$ by
$$(x,y)\leftrightarrow x+yi,$$ it is possible to define multiplications on $\R^2$.\, Via polar coordinates, the formula for this multiplication becomes very simple, thanks to \PMlinkname{Euler's formula}{EulerRelation}
$$\cos{\theta}+i\sin{\theta}=e^{i\theta}.$$
Thus, we have the following identification:
$$(r,\theta)\leftrightarrow(x,y)\leftrightarrow x+yi=
r\cos{\theta}+(r\sin{\theta})i=re^{i\theta}.$$ If $P=(r_1,\theta_1)$
and $Q=(r_2,\theta_2)$, the product of $P$ and $Q$ works out to be
$(r_1r_2,\theta_1+\theta_2)$.
(Even if one is not familiar with the complex exponential, this assertion may be checked
directly using the angle sum identities for $\cos$ and $\sin$.)
Multiplications of polar coordinates have some simple geometric
interpretations. For example, if $R=(1,\alpha)$ and $Q=(r,\beta)$,
then $Q\rightarrow RQ$ given by
$(1,\alpha)(r,\beta)=(r,\alpha+\beta)$ is the rotation of $Q$ by
angle $\alpha$. If $S=(t,0)$, then $(t,0)(r,\beta)=(tr,\beta)$ can
be viewed as the scaling of $Q$ along the ray $\overrightarrow{OQ}$
by $t$. Note also that multiplication by $(t,0)$ has the same effect
as multiplication by the scalar $t$.
\begin{center}
\includegraphics{polar_5.eps}
\end{center}
For more on polar coordinates, including their construction and extensions on domain of polar coordinates $r$ and $\theta$, see \PMlinkname{here}{ConstructionOfPolarCoordinates}.
\textbf{Calculus in polar coordiantes.}
For reference, here are some formulae for computing integrals and derivatives
in polar coordinates. The Jacobian for transforming from rectangular to
polar cordinates is
\[
{\partial (x, y) \over \partial (r, \theta)} = r
\]
so we may compute the integral of a scalar field $f$ as
\[
\int f (r, \theta) r \, dr \, d\theta .
\]
Partial derivative operators transform as follows:
\begin{align*}
{\partial \over \partial x} &= \cos \theta {\partial \over \partial r} -
{1 \over r} \sin \theta {\partial \over \partial \theta} \\
{\partial \over \partial y} &= \sin \theta {\partial \over \partial r} +
{1 \over r} \cos \theta {\partial \over \partial \theta} \\
{\partial \over \partial r} &= \cos \theta {\partial \over \partial x} +
\sin \theta {\partial \over \partial y} \\
{\partial \over \partial \theta} &= - r \sin \theta {\partial \over \partial x} +
r \cos \theta {\partial \over \partial y}
\end{align*}