polar coordinates

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

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

$x(r,\theta )$$=$$r\cos \theta ,$

$y(r,\theta )$$=$$r\sin \theta ,$

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

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

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

$\theta (x,y)$$=$$\arctan (x,y),$

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

Polar basis.Polar coordinates are equipped with an orthonormal base $\{{𝐞}_{𝐫},{𝐞}_{\theta }\}$, which can be defined in terms of the standard cartesian base $\{𝐢,𝐣\}$ in ${ℝ}^2$ as follows.

$\left[\begin{array}{c}\hfill {𝐞}_{𝐫}\hfill \\ \hfill {𝐞}_{\theta }\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \cos \theta 𝐢+\sin \theta 𝐣\hfill \\ \hfill -\sin \theta 𝐢+\cos \theta 𝐣\hfill \end{array}\right],$

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

$\left[\begin{array}{c}\hfill \frac{d{𝐞}_{𝐫}}{d\theta }\hfill \\ \hfill \frac{d{𝐞}_{\theta }}{d\theta }\hfill \end{array}\right]=\left[\begin{array}{c}\hfill {𝐞}_{\theta }\hfill \\ \hfill -{𝐞}_{𝐫}\hfill \end{array}\right].$

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 $\pi /2$.

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

$𝐫=r{𝐞}_{𝐫}.$

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 Euler’s formula (http://planetmath.org/EulerRelation)

$$\cos \theta +i\sin \theta =e^{i\theta }.$$

Thus, we have the following identification:

$$(r,\theta )\leftrightarrow (x,y)\leftrightarrow x+yi=r\cos \theta +(r\sin \theta )i=r{e}^{i\theta }.$$

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 checkeddirectly using the angle sum identities for $\cos$ and $\sin$.)

Multiplications of polar coordinates have some simple geometricinterpretations. For example, if $R=(1,\alpha )$ and $Q=(r,\beta )$,then $Q\to RQ$ given by$(1,\alpha )(r,\beta )=(r,\alpha +\beta )$ is the rotation of $Q$ byangle $\alpha$. If $S=(t,0)$, then $(t,0)(r,\beta )=(tr,\beta )$ canbe viewed as the scaling of $Q$ along the ray $\overrightarrow{OQ}$by $t$. Note also that multiplication by $(t,0)$ has the same effectas 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 derivativesin polar coordinates. The Jacobian for transforming from rectangular topolar cordinates is

$$\frac{\partial (x,y)}{\partial (r,\theta )}=r$$

so we may compute the integral of a scalar field $f$ as

$$\int f(r,\theta )r𝑑r𝑑\theta .$$

Partial derivative operators transform as follows:

${\displaystyle \frac{\partial }{\partial x}}$$=\cos \theta {\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}$

${\displaystyle \frac{\partial }{\partial y}}$$=\sin \theta {\displaystyle \frac{\partial }{\partial r}}+{\displaystyle \frac{1}{r}}\cos \theta {\displaystyle \frac{\partial }{\partial \theta }}$

${\displaystyle \frac{\partial }{\partial r}}$$=\cos \theta {\displaystyle \frac{\partial }{\partial x}}+\sin \theta {\displaystyle \frac{\partial }{\partial y}}$

${\displaystyle \frac{\partial }{\partial \theta }}$$=-r\sin \theta {\displaystyle \frac{\partial }{\partial x}}+r\cos \theta {\displaystyle \frac{\partial }{\partial y}}$