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# 7.1: Polar Coordinates

[ "article:topic", "Polar Coordinates", "authorname:nwalet", "license:ccbyncsa" ]

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# Polar and spherical coordinate systems

$$\newcommand{\pdr}[1]{\dfrac{\partial}{\partial r}}$$ $$\newcommand{\pdtheta}[1]{\dfrac{\partial}{\partial \theta}}$$ $$\newcommand{\pdrho}[1]{\dfrac{\partial}{\partial \rho}}$$ $$\newcommand{\pdphi}[1]{\dfrac{\partial}{\partial \phi}}$$ $$\newcommand{\pdw}[1]{\dfrac{\partial}{\partial w}}$$ $$\newcommand{\pdx}[1]{\dfrac{\partial}{\partial x}}$$ $$\newcommand{\pdy}[1]{\dfrac{\partial}{\partial y}}$$ $$\newcommand{\pdz}[1]{\dfrac{\partial}{\partial z}}$$ $$\newcommand{\pdt}[1]{\dfrac{\partial}{\partial t}}$$ $$\newcommand{\pdtt}[1]{\dfrac{\partial^2}{\partial t^2}}$$ $$\newcommand{\pdyy}[1]{\dfrac{\partial^2}{\partial y^2}}$$ $$\newcommand{\pdww}[1]{\dfrac{\partial^2}{\partial w^2}}$$ $$\newcommand{\pdwz}[1]{\dfrac{\partial^2}{\partial w \partial z}}$$ $$\newcommand{\half}[1]{\dfrac{1}{2}}$$ $$\newcommand{\pdxx}[1]{\dfrac{\partial^2}{\partial x^2}}$$

## Polar coordinates

Polar coordinates in two dimensions are defined by \begin{aligned} x = \rho\cos\phi, y= \rho\sin\phi,\\ \rho = \sqrt{x^2+y^2}, \phi = \arctan(y/x),\end{aligned} as indicated schematically in Fig. [fig:polar].

Using the chain rule we find

\begin{aligned} \pdx{~} &=& \pdx{\rho}\pdrho{~}+\pdx{\phi}\pdphi{~}\nonumber\\ &=& \frac{x}{\rho}\pdrho{~}-\frac{y}{\rho^2}\pdphi{~}\nonumber\\ &=& \cos\phi\pdrho{~}-\frac{\sin\phi}{\rho}\pdphi{~},\\ \pdy{~} &=& \pdy{\rho}\pdrho{~}+\pdy{\phi}\pdphi{~}\nonumber\\ &=& \frac{y}{\rho}\pdrho{~}+\frac{x}{\rho^2}\pdphi{~}\nonumber\\ &=& \sin\phi\pdrho{~}+\frac{\cos\phi}{\rho}\pdphi{~},\end{aligned} We can write \begin{aligned} {\vect{\nabla}} &=& {\unitvec{e}}_\rho \pdrho{~}+\unitvec{e}_\phi \frac{1}{\rho} \pdphi{~}\end{aligned}

where the unit vectors

\begin{aligned} \unitvec{e}_\rho &=& (\cos\phi,\sin\phi), \nonumber\\ \unitvec{e}_\phi &=& (-\sin\phi,\cos\phi), \end{aligned} are an orthonormal set. We say that circular coordinates are orthogonal.

We can now use this to evaluate $$\nabla^2$$,

\begin{aligned} \nabla^2 &= & \cos^2\phi\pdrhor{~}+\frac{\sin\phi\cos\phi}{\rho^2}\pdphi{~} +\frac{\sin^2\phi}{\rho}\pdrho{~}+\frac{\sin^2\phi}{\rho^2}\pdphip{~}+ \frac{\sin\phi\cos\phi}{\rho^2} \pdphi{~}\nonumber\\ &&+\sin^2\phi\pdrhor{~}-\frac{\sin\phi\cos\phi}{\rho^2}\pdphi{~} +\frac{\cos^2\phi}{\rho}\pdrho{~}+\frac{\cos^2\phi}{\rho^2}\pdphip{~}- \frac{\sin\phi\cos\phi}{\rho^2} \pdphi{~}\\ &=&\pdrhor{~}+\frac{1}{\rho}\pdrho{~}+\frac{1}{\rho^2}\pdphip{~}\nonumber\\ &=&\frac{1}{\rho}\pdrho{~} \left(\rho\pdrho{~}\right)+\frac{1}{\rho^2}\pdphip{~}.\nonumber\\\end{aligned}

A final useful relation is the integration over these coordinates.

As indicated schematically in Fig. [fig:polar2], the surface related to a change $$\rho \rightarrow \rho + \delta \rho$$, $$\phi \rightarrow \phi+\delta\phi$$ is $$\rho \delta \rho \delta\phi$$. This leads us to the conclusion that an integral over $$x,y$$ can be rewritten as $\int_V f(x,y) dx dy = \int_V f(\rho\cos\phi,\rho\sin\phi) \rho d\rho d\phi$

## spherical coordinates

Spherical coordinates are defined as \begin{aligned} x = r \cos\phi\sin\theta,\; y= r \sin\phi\sin\theta,\; z =r \cos\theta,\\ r = \sqrt{x^2+y^2+z^2},\; \phi = \arctan(y/x), \; \theta=\arctan\left(\frac{\sqrt{x^2+y^2}}{z}\right),\end{aligned} as indicated schematically in Fig. [fig:spherical].

Using the chain rule we find

\begin{aligned} \pdx{~} &=& \pdx{r}\pdr{~}+\pdx{\phi}\pdphi{~}+\pdx{\theta}\pdtheta{~}\nonumber\\ &=& \frac{x}{r} \pdr{~}-\frac{y}{x^2+y^2}\pdphi{~} +\frac{xz}{r^2\sqrt{x^2+y^2}}\pdtheta{~} \nonumber\\ &=& \sin\theta\cos\phi\pdr{~}-\frac{\sin\phi}{r\sin\theta} \pdphi{~} +\frac{\cos\phi\cos\theta}{r} \pdtheta{~},\\ \pdy{~} &=& \pdy{r}\pdr{~}+\pdy{\phi}\pdphi{~}+\pdy{\theta}\pdtheta{~}\nonumber\\ &=& \frac{y}{r} \pdr{~}+\frac{x}{x^2+y^2}\pdphi{~} +\frac{yz}{r^2\sqrt{x^2+y^2}}\pdtheta{~} \nonumber\\ &=& \sin\theta\sin\phi\pdr{~}+\frac{\cos\phi}{r\sin\theta} \pdphi{~} +\frac{\sin\phi\cos\theta}{r} \pdtheta{~},\\ \pdz{~} &=& \pdz{r}\pdr{~}+\pdz{\phi}\pdphi{~}+\pdz{\theta}\pdtheta{~}\nonumber\\ &=& \frac{z}{r} \pdr{~} -\frac{\sqrt{x^2+y^2}}{r^2}\pdtheta{~} \nonumber\\ &=& \sin\theta\sin\phi\pdr{~}-\frac{\sin\theta}{r} \pdtheta{~}.\\\end{aligned}

once again we can write $$\vect{\nabla}$$ in terms of these coordinates.

\begin{aligned} {\vect{\nabla}} &=& \unitvec{e}_r \pdr{~}+\unitvec{e}_\phi \frac{1}{r\sin\theta}\pdphi{~} + \unitvec{e}_\theta \frac{1}{r}\pdtheta{~}\end{aligned} where the unit vectors \begin{aligned} \unitvec{e}_r &=& (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta), \nonumber\\ \unitvec{e}_\phi &=& (-\sin\phi,\cos\phi,0), \nonumber\\ \unitvec{e}_\theta &=& (\cos\phi\cos\theta,\sin\phi\cos\theta,-\sin\theta).\end{aligned}

are an orthonormal set. We say that spherical coordinates are orthogonal.

We can use this to evaluate $$\Delta={\vect{\nabla}}^2$$,

$\Delta = \frac{1}{r^2}\pdr{~}\left(r^2 \pdr{~}\right) +\frac{1}{r^2} \frac{1}{\sin\theta} \pdtheta{~} \left( \sin\theta\pdtheta{~} \right) + \frac{1}{r^2}\pdphip{~}$

Finally, for integration over these variables we need to know the volume of the small cuboid contained between $$r$$ and $$r+\delta r$$, $$\theta$$ and $$\theta + \delta\theta$$ and $$\phi$$ and $$\phi+\delta\phi$$. The length of the sides due to each of these changes is $$\delta r$$, $$r \delta \theta$$ and $$r \sin \theta \delta \theta$$, respectively. We thus conclude that

$\int_V f(x,y,z) dx dy dz = \int_V f(r,\theta,\phi) r^2\sin\theta dr d\theta d\phi.$