Coordinate Sysytem

Coordinate systems other that Cartesian coordinate system sometimes can be more useful in some cases.

Plane Polar Coordinate System

The unit vectors for plane polar coordinate system are $$\begin{align} \hat{r} &= \cos\theta\hat{i} + \sin\theta\hat{j}\\ \hat{\theta} &= -\sin\theta\hat{i} + \cos\theta\hat{j} \\ \end{align}$$ As they depend on $\theta$, which may not be constant, they may have non-zero derivative, unlike the $\hat{i}$ and $\hat{j}$ in Cartesian coordinate system $$\begin{align} \dot{\hat{r}} &= -\dot{\theta}\sin\theta\hat{i} + \dot{\theta}\cos\theta\hat{j} = \dot{\theta}\hat{\theta} \\ \dot{\hat{\theta}} &= -\dot{\theta}\cos\theta\hat{i} - \dot{\theta}\sin\theta\hat{j} = -\dot{\theta}\hat{r}\\ \end{align}$$ The position vector $\textbf{r}(t)$ is given by $$\textbf{r}(t) = r(t)\hat{r}$$ The velocity is $$\textbf{v}(t) = r(t)\dot{\hat{r}} + \dot{r}(t)\hat{r} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$$ The acceleration is $$\begin{align} \textbf{a}(t) &= \ddot{r}\hat{r} + \dot{r}\dot{\theta}\hat{\theta} + (r\ddot{\theta} + \dot{r}\dot{\theta})\hat{\theta} + r\dot{\theta}(-\dot{\theta}\hat{r})\\ &= (\ddot{r} - r\dot{\theta}^2)\hat{r} + (2\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\theta} \end{align}$$

Example

A particle travels at constant speed with a plane polar coordinates defined by $$r(t)=k(1+\cos\theta(t)).$$ The particle is at $\theta=0$ at $t=0$. Find $\theta$ as a function of $t$.

Solution: $$\begin{align} \dot{r} &= k\dot{\theta}\sin\theta\\ v_0 &= \sqrt{\dot{r}^2+(r\dot{\theta})^2}\\ &= \sqrt{(k\dot{\theta}\sin\theta)^2+(k\dot{\theta}(1+\cos\theta))^2}\\ &= k\dot{\theta}\sqrt{2+2\cos\theta}\\ &= 2k\dot{\theta}\cos\frac{\theta}{2}\\ \int v_0dt &= \int 2k\cos\frac{\theta}{2}d\theta\\ v_0t&=4k\sin\frac{\theta}{2}+C, \end{align}$$ where $C$ is a constant. At $t=0$, $\theta=0 \implies C=0$. So, $$\theta(t)=2\sin^-1\left(\frac{v_0t}{4k}\right)$$

Cylindrical Polar Coordinate System

The unit vectors for plane polar coordinate system are $$\begin{align} &\hat{s} = \cos\phi\hat{i} + \sin\phi\hat{j}\\ &\hat{\phi} = -\sin\phi\hat{i} + \cos\phi\hat{j} \\ &\hat{k} \\ \end{align}$$ As $\hat{s}$ and $\hat{\phi}$ depend on $\phi$, which may not be constant, they may have non-zero derivative, unlike the $\hat{i}$ and $\hat{j}$ in Cartesian coordinate system $$\begin{align} \dot{\hat{s}} &= -\dot{\phi}\sin\phi\hat{i} + \dot{\phi}\cos\phi\hat{j} = \dot{\phi}\hat{\phi} \\ \dot{\hat{\phi}} &= -\dot{\phi}\cos\phi\hat{i} - \dot{\phi}\sin\phi\hat{j} = -\dot{\phi}\hat{s}\\ \end{align}$$ The position vector $\textbf{r}(t)$ is given by $$\textbf{r}(t) = s(t)\hat{s} + z(t)\hat{k}$$ The velocity is $$\textbf{v}(t) = s(t)\dot{\hat{s}} + \dot{s}(t)\hat{s} + \dot{z}\hat{k} = \dot{s}\hat{s} + s\dot{\phi}\hat{\phi} + \dot{z}\hat{k}$$ The acceleration is $$\begin{align} \textbf{a}(t) &= \ddot{s}\hat{s} + \dot{s}\dot{\phi}\hat{\phi} + (s\ddot{\phi} + \dot{s}\dot{\phi})\hat{\phi} + s\dot{\phi}(-\dot{\phi}\hat{s}) + \ddot{z}\hat{k}\\ &= (\ddot{s} - s\dot{\phi}^2)\hat{s} + (2\dot{s}\dot{\phi} + s\ddot{\phi})\hat{\phi} + \ddot{z}\hat{k} \end{align}$$

Spherical Coordinate System

The unit vectors for spherical coordinate system are $$\begin{align} \hat{r} &= \sin\theta\cos\phi\hat{i} + \sin\theta\sin\phi\hat{j} + \cos\theta\hat{k}\\ \hat{\theta} &= -\cos\theta\cos\phi\hat{i} + \cos\theta\sin\phi\hat{j} -\sin\theta\hat{k} \\ \hat{\phi} &= \sin\phi\hat{i} + \cos\phi\hat{j} \\ \end{align}$$ As they depend on $\phi$ and $\theta$, which may not be constant, they may have non-zero derivative, unlike the $\hat{i}$, $\hat{j}$ and $\hat{k}$ in Cartesian coordinate system $$\begin{align} \dot{\hat{r}} &= (\dot{\theta}\cos\theta\cos\phi-\dot{\phi}\sin\theta\sin\phi)\hat{i} + (\dot{\theta}\cos\theta\sin\phi + \dot{\phi}\sin\theta\cos\phi)\hat{j} -\dot{\theta}\sin\theta\dot{k} = \dot{\theta}\hat{\theta} + \dot{\phi}\sin\theta\hat{\phi} \\ \dot{\hat{\theta}} &= -\dot{\theta}\hat{r} + \dot{\phi}\cos\theta\hat{\phi}\\ \dot{\hat{\phi}} &= -\dot{\phi}\sin\theta\hat{r} - \dot{\phi}\cos\theta\hat{\phi} \end{align}$$ The position vector $\textbf{r}(t)$ is given by $$\textbf{r}(t) = r(t)\hat{r}$$ The velocity is $$\textbf{v}(t) = r(t)\dot{\hat{r}} + \dot{r}(t)\hat{r} = \dot{r}\hat{r} + r\dot{\phi}\sin\theta\hat{\phi} + r\dot{\theta}\hat{\theta}$$ The acceleration is $$\begin{align} \textbf{a}(t) = &(\ddot{r} - r\dot{\phi}^2\sin^2\theta - r\dot{\theta}^2)\hat{r}\\ &+(r\ddot{\theta} + 2\dot{r}\dot{\theta} - r\dot{\phi}^2\sin\theta\cos\theta)\hat{\theta}\\ &+(r\ddot{\phi}\sin\theta + 2\dot{r}\dot{\phi}\sin\theta + 2r\dot{\theta}\dot{\phi}\cos\theta) \hat{\phi} \\ \end{align}$$