I was wondering how the distance function $(\Delta s)^2 = (\Delta r)^2 + (r \Delta \theta)^2$ can be shown to be invariant under the rotation matrix $ \begin{pmatrix} cos\ \theta & - sin\ \theta \\ sin\ \theta & cos\ \theta \end{pmatrix}$.
I know that $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2$ is invariant under the above rotation matrix because $(\Delta s')^2 = (\Delta x')^2 + (\Delta y')^2 = (\Delta x\ cos\ \theta\ - \Delta y\ sin\ \theta)^2 + (\Delta x\ sin\ \theta\ + \Delta y\ cos\ \theta)^2 = (\Delta x)^2 + (\Delta y)^2.$
How do we do the same for the polar coordinates?
The key thing to realize is that a rotation transformation in polar coordinates is simply vector addition and that there is no matrix operation which represents this transformation. It is the equivalent of a translation in euclidean coordinates. That is to say that the rotation transformation $f$ takes $$f(r,\theta)=(r, \theta) +(0, c)= (r,\theta+c)$$ with $c \in \mathbb{R}$. Therefore none of the terms in $\Delta s$ change when we apply this transformation.