Column vectors in polar coordinates?

Question

So, we can represent the Cartesian vector $r= x\hat{x}+ y\hat{y}$ as the column vector where the entries are:

$$\begin{bmatrix} x\\ y\\ \end{bmatrix} $$

How would the polar coordinates position vector :$r= r\hat{r}$ be represented as the entries of a column vector? Like this? $$\begin{bmatrix} r\\ 0\\ \end{bmatrix} $$

I want to know this because I wish to discuss linear maps in different coordinates. For instance a rotation in Cartesian coordinates around the z axis is $$r' = R\begin{bmatrix} x\\ y\\ \end{bmatrix} $$

But the same operation in polar coordinates doesn't make much sense in this representation, since only the angle changes.

Answer 1

So polar coordinate $(r,\theta)$ is a parametrization of $\mathbb R^2$ i.e. $$g:\mathbb R_+\times[0,2\pi)\to\mathbb R^2,(r,\theta)\mapsto(r\cos\theta,r\sin\theta)$$ Definitely, you can represent the polar coordinate as a column vector. However, this is not a linear basis so you cannot represent a point by linear summation of the basis vectors like this $$ \hat r =x \hat x+y \hat y $$

As you said, a linear transform $T$ viewed in a nonlinear paramatrization is not linear, so it's not surprising that it cannot be represented as a matrix. $$ [x',y']^T=T[x,y]^T\\ [r',\theta']^T=g(Tg^{-1}([r,\theta]^T)) $$

Nevertheless, if you study differential geometry, even if the whole map is nonlinear, you could represent local infinitesimal changes (tangent vectors) by a linear transform (aka Jacobian, pushforward, differential map) $$ d\mathbb r'=dg\circ T\circ dg^{-1}d\mathbb r $$

Answer 2

Let's take an example familiar to physicists: planar orbits. Sure, the position $\vec{r}=r\hat{r}$ only has one component, but the velocity and acceleration are the less boring - and more to the point, less confusing - two-component results $\vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$ (with $\dot{}$ denoting time derivatives) and $\vec{a}=(\ddot{r}-r\dot{\theta}^2)\vec{r}+\frac1r\frac{d}{dt}(r^2\dot{\theta})\hat{\theta}$. I won't rehearse the full proof of these, but what's going here is that$$\hat{r}=\hat{x}\cos\theta+\hat{y}\sin\theta,\,\hat{\theta}=-\hat{x}\sin\theta+\hat{y}\cos\theta$$are orthogonal unit vectors in terms of which we can describe a lot more than position, and they satisfy$$\dot{\hat{r}}=\dot{\theta}\hat{\theta},\,\dot{\hat{\theta}}=-\dot{\theta}\hat{r}$$by the chain rule. So now any arbitrary vector-valued function in the plane can be spanned by whichever basis you prefer, be it $\{\hat{x},\,\hat{y}\}$ or $\{\hat{r},\,\hat{\theta}\}$. But each position in the plane has its own proposal for the latter basis, always so that $\vec{r}\parallel\hat{r}$.