# Plane Polar Coordinates From the [[Unit Circle Definitions of Trigonometric Functions]], we know that if we have a point $(x,y)$ on the two-dimensional unit circle, then if $\theta$ is the angle formed by a line from the origin to the point with the positive direction of the $x$-axis, then $ \begin{align*} x &= \cos \theta\\ y &= \sin \theta \end{align*} $ If we suppose $(x,y)$ is _not_ on the unit circle, but is instead on some other circle centred at the origin and with radius $r$, then it's not hard to see that this simply scales $x$ and $y$ by $r$, so $ \begin{align*} x &= r\cos \theta\\ y &= r\sin \theta \end{align*} $ Thus any point we can describe in terms of cartesian coordinates $(x,y)$ we can describe in terms of $r$, known in this context as the _radial distance_, and the angle $\theta$. These are known as the "plane polar coordinates" of the point. ## Polar Basis Vectors Just as the Cartesian coordinates $(x,y)$ of a point $p$ scale the Cartesian basis vectors $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ such that $p = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}$ so too are there analogous basis vectors for polar coordinates, known by convention as $\hat{\mathbf{r}}$ and $\hat{\mathbf{\theta}}$ (or $\mathbf{e}_r$ and $\mathbf{e}_\theta$), and we can span the plane by scaling the polar basis vectors so if our point $\mathbf{p}$ has polar coordinates $(r,\theta)$ then $\mathbf{p} = r\hat{\mathbf{r}} + \theta\hat{\mathbf{\theta}}$. Notice we are not just renaming variables here- $r$ and $\theta$ are fundamentally different from $x$ and $y$. Whereas $x$ and $y$ specify distance on a uniform grid in the horizontal and vertical distance, $r$ could be seen to "choose" from a set of circles concentric about the origin and $\theta$ says how far anticlockwise we should travel from the horizontal on that circle. This difference becomes particularly important when doing [[Multiple Integrals]] with polar coordinates as instead of rectangles/rectangular prisms, the area or volume elements we sum up in the integration will have a curve to them so extra factors will need to be added to the integral to compensate for this. Just as $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are perpendicular, so too are $\hat{\mathbf{r}}$ and $\hat{\mathbf{\theta}}$, which gives us some insight into how we can convert cartesian into polar basis vectors. $ \begin{align*} \hat{\mathbf{r}} &= \cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}\\ \hat{\mathbf{\theta}} &= \cos \left(\frac{\pi}{2} + \theta\right) \hat{\mathbf{i}}+\sin \left(\frac{\pi}{2} + \theta\right) \hat{\mathbf{j}}\\ &= -\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}} \end{align*} $ In matrix form we can write this as: $ \begin{align*} \begin{bmatrix}\widehat{\mathbf{r}}\\\widehat{\mathbf{\theta}}\end{bmatrix} &= \begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta &\cos \theta\end{bmatrix} \begin{bmatrix}\widehat{\mathbf{i}}\\\widehat{\mathbf{j}}\end{bmatrix} \end{align*} $ That matrix is actually orthogonal meaning $\mathbf{A}^T=\mathbf{A}^{-1}$. So it is very easy to go the other way and say $ \begin{align*} \begin{bmatrix}\widehat{\mathbf{i}}\\\widehat{\mathbf{j}}\end{bmatrix} &= \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta &\cos \theta\end{bmatrix} \begin{bmatrix}\widehat{\mathbf{r}}\\\widehat{\mathbf{\theta}}\end{bmatrix} \end{align*} $ ### Nomenclature $\mathbf{\widehat{r}}$ can also be called $\mathbf{e}_r$ and $\mathbf{\widehat{\theta}}$ is likewise sometimes called $\mathbf{e}_\theta$. ## Defining a polar coordinate system It may seem obvious, but in order to define a polar coordinate system, we need to define the origin ($O$), a horizontal line which is taken as $\theta=0$ and a direction of increasing $\theta$. We will often also define a direction normal to the plane of the polar coordinates, which we may call $\widehat{n}$ or $\mathbf{k}$. The procedure for chosing the direction of positive $\mathbf{k}$ is given [[#Definition of clockwise/anticlockwise|below]], but in general, it is often possible to just assume rotation is anticlockwise such that $\mathbf{k}$ is pointing in a direction that is conveniently upwards. ## Position and polar basis vectors Notice that unlike $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$, which remain fixed, the polar basis vectors $\hat{\mathbf{r}}$ and $\hat{\mathbf{\theta}}$ vary with position, because for any point $p$, $\hat{\mathbf{r}}$ will point in the direction of the line from the origin through that point, and $\hat{\mathbf{\theta}}$ will be perpendicular to this in an anticlockwise direction. Since in most kinematics situations, position is a function of time, this means the basis vectors too are a function of time and thus they are often written $\hat{\mathbf{r}}(t)$ and $\hat{\mathbf{\theta}}(t)$ to emphasise this. The idea of a coordinates system that rotates, with basis vectors which depend on position and/or time may not be immediately intuitive, so here are a couple of video presentations of the basic idea of polar unit vectors that I found particularly clear.   ## Derivatives Given that the polar basis vectors move with time, it's natural to wonder what their derivatives might be, and there is indeed an important, and rather beautiful, relationship between the basis vectors and their time derivatives. $ \begin{align*} \text{If we have}\\ \mathbf{e}_r &= \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\\ \text{and}~ \mathbf{e}_\theta &= -\sin \theta \mathbf{i} + \cos \theta \mathbf{j},\\[8px] \text{then, by the chain rule,}\\ \frac{\mathrm{d}}{\mathrm{d}t} \mathbf{e}_r &= -\sin \theta \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{i} + \cos \theta \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{j}\\ &= \frac{\mathrm{d}\theta}{\mathrm{d}t}\left( -\sin \theta \mathbf{i} + \cos \theta \mathbf{j}\right)\\ &= \mathbf{e}_\theta \frac{\mathrm{d}\theta}{\mathrm{d}t}\\ \text{and similarly,} \frac{\mathrm{d}}{\mathrm{d}t} \mathbf{e}_\theta &= -\cos \theta \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{i} - \sin \theta \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{j}\\ &= - \frac{\mathrm{d}\theta}{\mathrm{d}t} \left( \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\right)\\ &= - \mathbf{e}_r \frac{\mathrm{d}\theta}{\mathrm{d}t}. \end{align*} $ Or to put it in Newton's notation, we have $ \begin{align*} \mathbf{\dot{e}}_r &= \dot{\theta} \mathbf{e}_\theta \\ \text{and}~\mathbf{\dot{e}}_\theta &= - \dot{\theta} \mathbf{e}_r. \end{align*} $ Since these time-derivatives are vectors, their natural interpretation is that $\mathbf{e}_r$ varies by the angular speed in the direction of $\mathbf{e}_\theta$, and $\mathbf{e}_\theta$ varies by the angular speed in the opposite of the radial direction (i.e. towards the centre of the circle). In both cases this makes complete sense but bears thinking about. # Orientation of the plane/Definition of clockwise/anticlockwise Given that $\hat{\mathbf{\theta}}$ is defined as pointing perpendicular to $\hat{\mathbf{r}}$ in the direction of anticlockwise rotation this gives rise to the problem that if an observer was on one side of a plane they might see rotation as perhaps anticlockwise, but an identical observer on the opposite side of the plane looking at the same rotation would consider that rotation as clockwise. Therefore a more precise definition is needed. Both the problem and the solution are excellently explained here.  ## Aside There is a lovely irony in the video above if you understand how the lightboard works. The lightboard used in the presentation is a transparent whiteboard with the instructor standing on the opposite side of the glass to the camera. After the filming is done it is mirrored so we (the observer) see the writing the "right" way round. This has the effect of mirroring certain things so for example the instructor in the angular velocity will actually be writing with her right hand, even though it appears to be her left, when she describes rotations as clockwise or counterclockwise they will actually appear to her to be the other way around, and she demonstrates the right hand rule using her left hand. In short she is suffering the exact paradox that she explains of the two observers on opposite sides of a plane seeing motion as going in opposite directions - justifying the whole "into the board"/"out of the board" $\hat{\mathbf{n}}$ thing. ## Solution Procedure 1. take your right hand and get your thumb, first and second fingers to be perpendicular to one another. [^1] 2. curl your two smallest finders in the direction of the rotation 3. Point your index finder in the direction of $\hat{\mathbf{r}}$. 4. Your middle finger is now pointing in the direction of $\hat{\mathbf{\theta}}$. 5. As a bonus, your thumb is pointing in the direction of $\hat{\mathbf{n}}$. This is the unit vector that uniquely defines the orientation of the plane of rotation. You never know when that might come in handy (eg for taking a cross product of two vectors in the plane of $\hat{\mathbf{r}}$ and $\hat{\mathbf{\theta}}$) and you and an observer on the other side of the page will share a conception of $\hat{\mathbf{n}}$, so you'll have something to talk about. 6. If your thumb/$\hat{\mathbf{n}}$ is pointing "Into the page" (denoted $\otimes$) the motion is clockwise from your perspective and $\dot{\theta}<0$. If your thumb/$\hat{\mathbf{n}}$ is pointing "out of the page" (denoted $\odot$) the motion is anticlockwise from your perspective and $\dot{\theta}>0$. 7. The observer on the other side of the page may disagree with you about whether or not the motion is clockwise or anticlockwise but they will share the same 3 basis vectors you have so their kinematics should work out the same as yours. [^1] If you're like me, make sure it's your actual right hand not your "other right" hand. # $\nabla$ and friends in polar coordinates Suppose we have some function $f(x,y)$ and we want to take $\mathbf{\mathop{grad}} f$.That's obviously very easy, we just take $\mathbf{\nabla} f$, but what if we have $f(r,\theta)$, a function which takes position in polar coordinates? Or what if we have $f_x$ and we want $f_\theta$, or vice versa? Luckily this is pretty simple using the chain rule. $ \begin{align*} f_r &= f_x x_r + f_y y_r~\text{by the chain rule}\\ \text{but}~x &= r \cos \theta\\ \text{and}~y &= r \sin \theta\\ \text{so}~x_r &= \cos \theta\\ \text{and}~y_r &= \sin \theta.\\ \text{Thus,}~f_r &= \cos \theta f_x + \sin \theta f_y\\ \text{Likewise,}~x_\theta &= -r \sin \theta,\\ \text{and}~y_\theta &= r \cos \theta,\\ \text{so}~f_\theta &= -\sin \theta f_x + r\cos \theta f_y\\ \end{align*} $ So, in matrix form, $ \begin{align*} \begin{bmatrix}\frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta}\end{bmatrix} &= \begin{bmatrix}\cos \theta & \sin \theta \\ -r \sin \theta & r \cos \theta\end{bmatrix} \begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y}\end{bmatrix} \end{align*} $ Additionally, this can be inverted to give $ \begin{align*} \begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y}\end{bmatrix} &= \begin{bmatrix}\cos \theta & -\frac{1}{r}\sin \theta \\ \sin \theta & \frac{1}{r} \cos \theta\end{bmatrix} \begin{bmatrix}\frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta}\end{bmatrix} \end{align*} $ Therefore $ \begin{align*} \mathbf{\nabla} f &= \widehat{\mathbf{r}}\frac{\partial}{\partial r} + \frac{1}{r} \widehat{\mathbf{\theta}}\frac{\partial}{\partial \theta} \end{align*} $