Transformation matrix in polar coordinates

I'm trying to write a software widget that allows the user to resize the component, so I can write a transformation matrix $\mathbf T_\text{xy}$ that will map $(x,y)$ to a transformed $(x',y')$, that is

$$\left(\begin{array}{cc} x'\\ y' \end{array}\right)= \mathbf T_\text{xy}\left(\begin{array}{cc} x\\ y \end{array}\right) \tag{1}$$

But my in application it makes it easier to use Polar instead of Cartesian, I can convert $(r,\theta)$ to $(x,y)$ and then use the above transformation to get $(x',y')$ and convert it $(r', \theta ')$, but I was wondering if I there is a way to find a transformation matrix $\mathbf T_{r\theta}$ that will transform directly from $(r,\theta)$ to $(r', \theta ')$ and that is equivalent to the transformation equation above. I failed in deriving such a relation since there is no linear relation between $(x,y)$ and $(r,\theta)$.

In short my question is to find $\mathbf T_{\text{r}\theta}$ in

$$\left(\begin{array}{cc} r'\\ \theta' \end{array}\right)= \mathbf T_{\text{r}\theta}\left(\begin{array}{cc} r\\ \theta \end{array}\right)$$

such that it transforms my points in the same way as eq(1) was doing

Any help is appreciated.

• As you have already noticed $(x,y)$ does not depend linearly on $(r,\theta)$. So I would highly suspect that no such matrix $\mathbf{T}_{r\theta}$ exists. – Jendrik Stelzner Jan 12 '16 at 16:14

There are two different ways you can describe points in the Euclidean plane. Both ways look very similar as they allow you to describe any point by simply a pair of numbers. I'll try to describe the two methods to show you that the representation of a point by $(r,\theta)$ is different from the representation of a point by $(x,y)$. And thus you can't expect them to behave the same way. In particular you can't expect the points $(r,\theta)$ to transform in the same linear fashion as $(x,y)$.

Consider the Euclidean plane $\Bbb E_2$. Choose any two distinct directions and then consider a set $\mathcal V = \{\mathbf {v_1}, \mathbf {v_2}\}$ where each of those vectors is parallel to one of the two directions.

This set is called a basis for the plane because any vector $\mathbf u\in \Bbb E_2$ can be decomposed uniquely into a component parallel to $\mathbf {v_1}$ and a component parallel to $\mathbf {v_2}$: $$\mathbf u = u_1\mathbf {v_1} + u_2\mathbf {v_2}$$

Then the set of numbers $(u_1,u_2)$ can be used to uniquely specify any given point in the plane with respect to the basis $\mathcal V$.

The representation of a point by the pair $(x,y)$ is this type of object. $x$ and $y$ are just the usual names we give to the coordinates of a point in $\Bbb E_2$ when the basis we chose happened to be an orthonormal basis. Denote the orthonormal basis associated with $(x,y)$ as $\mathcal E= \{\mathbf {\hat e_1}, \mathbf {\hat e_2}\}$.

(In this image, the blue vector is $\mathbf {\hat e_1}$ and the orange vector is $\mathbf {\hat e_2}$.)

Then any vector $\mathbf u\in \Bbb E_2$ can be decomposed as $$\mathbf u = x\mathbf {\hat e_1} + y\mathbf {\hat e_2}$$ or it can be more simply represented by the ordered pair $(x,y)$.

Now let's think of another way of representing a point in $\Bbb E_2$. Consider two families of curves that vary smoothly across the plane.

As you can see from the image, we can still describe points in the plane by a pair of numbers. You just need to describe each curve by a particular number -- then for any point of interest you just find the two curves that cross at that point and read off the numbers associated with each.

This is the type of object that $(r,\theta)$ is.

As an exercise, try to locate the unique point on the plot described by $(r,\theta) = (2,105°)$. Do you see that you could also uniquely locate a point that doesn't happen to be the intersection of two of the curves shown in the plot (because it's really hard to draw all infinity curves) such as $(r,\theta) = (1.5, 62°)$?

The pair $(r,\theta)$ isn't the coordinates of a vector with respect to some basis vectors, it's the pair of numbers describing which particular pair of curves (of each pair in the infinite family of curves) intersect at the point $P$.

So $(x,y)$ and $(r,\theta)$ are really are two specific instances of two completely different ways of representing a given point in the plane. As such hopefully it won't be a big surprise to you when I say that while $(x,y)$ can be transformed linearly (by matrices) to some other $(x',y')$, $(r,\theta)$ cannot.

• Thank you, I think I get it now. – Oswald Jan 12 '16 at 18:04