# polar coordinates

Let $x,y$ be Cartesian coordinates for $\mathbbmss{R}^{2}$.

Then $r\geq 0$, $\theta\in[0,2\pi)$ related to $(x,y)$ by

 $\displaystyle x(r,\theta)$ $\displaystyle=$ $\displaystyle r\cos\theta,$ $\displaystyle y(r,\theta)$ $\displaystyle=$ $\displaystyle r\sin\theta,$

are the polar coordinates for $(x,y)$. It is simply written $(r,\theta)$.

The polar coordinates of Cartesian coordinates $(x,y)\in\mathbbmss{R}^{2}\setminus\{0\}$ are

 $\displaystyle r(x,y)$ $\displaystyle=$ $\displaystyle\sqrt{x^{2}+y^{2}},$ $\displaystyle\theta(x,y)$ $\displaystyle=$ $\displaystyle\arctan(x,y),$

where $\arctan$ is defined here (http://planetmath.org/OperatornamearcTanWithTwoArguments).

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.

 $\displaystyle\begin{bmatrix}\mathbf{e_{r}}\\ \mathbf{e_{\theta}}\end{bmatrix}=\begin{bmatrix}\hphantom{-}\cos\theta\mathbf{% i}+\sin\theta\mathbf{j}\\ -\sin\theta\mathbf{i}+\cos\theta\mathbf{j}\end{bmatrix},$

where $\mathbf{e_{r},e_{\theta}}$ are so-called radial and traverse or angular vector, respectively. Since these vectors are variable in direction, they are differentiable. In fact,

 $\displaystyle\begin{bmatrix}\frac{d\mathbf{e_{r}}}{d\theta}\\ \frac{d\mathbf{e_{\theta}}}{d\theta}\end{bmatrix}=\begin{bmatrix}\hphantom{-}% \mathbf{e_{\theta}}\\ -\mathbf{e_{r}}\end{bmatrix}.$

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 $\pi/2$.

For an arbitrary point of polar coordinates $(r,\theta)$, its position vector comes given by the single equation

 $\displaystyle\mathbf{r}=r\mathbf{e_{r}}.$

When the Euclidean plane $\mathbbmss{R}^{2}$ is identified with $\mathbbmss{C}$ by

 $(x,y)\leftrightarrow x+yi,$

it is possible to define multiplications on $\mathbbmss{R}^{2}$.  Via polar coordinates, the formula for this multiplication becomes very simple, thanks to Euler’s formula (http://planetmath.org/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 $(r_{1}r_{2},\theta_{1}+\theta_{2})$. (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 $t$.

For more on polar coordinates, including their construction and extensions on domain of polar coordinates $r$ and $\theta$, see here (http://planetmath.org/ConstructionOfPolarCoordinates).

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:

 $\displaystyle{\partial\over\partial x}$ $\displaystyle=\cos\theta{\partial\over\partial r}-{1\over r}\sin\theta{% \partial\over\partial\theta}$ $\displaystyle{\partial\over\partial y}$ $\displaystyle=\sin\theta{\partial\over\partial r}+{1\over r}\cos\theta{% \partial\over\partial\theta}$ $\displaystyle{\partial\over\partial r}$ $\displaystyle=\cos\theta{\partial\over\partial x}+\sin\theta{\partial\over% \partial y}$ $\displaystyle{\partial\over\partial\theta}$ $\displaystyle=-r\sin\theta{\partial\over\partial x}+r\cos\theta{\partial\over% \partial y}$
Title polar coordinates PolarCoordinates 2013-03-22 15:12:16 2013-03-22 15:12:16 CWoo (3771) CWoo (3771) 14 CWoo (3771) Definition msc 51-01 DerivationOfRotationMatrixUsingPolarCoordinates CylindricalCoordinates ArgumentOfProductAndQuotient