Conversion of polar coordinate differential 1-forms to xy-plane

Question

I am new to differential geometry (and StackExchange!) and am having some trouble with the conversion of the polar differential one-forms: $dr$ and $d \theta $. How do I express these in terms of regular $x, y$ coordinates?

Answer 1

Starting from $$x = r \cos \theta \\ y = r \sin \theta$$ just differentiate $x$ and $y$ in the $(r,\theta)$ coordinate system: $$\begin{align} dx &= \frac{\partial x}{\partial r} dr + \frac{\partial x}{\partial \theta} d \theta &= \cos (\theta) dr - r \sin (\theta) d\theta &= \frac xr dr - y\ d\theta\\ dy &=\cdots&= \sin(\theta)dr + r \cos (\theta)d\theta &= \frac yr dr + x\ d\theta. \end{align}$$

At each point $(x,y,r=\sqrt{x^2 + y^2})$ this is a $2 \times 2$ linear system relating $(dx,dy)$ and $(dr, d\theta)$. Solving it for $(dr,d\theta)$ gives

$$\begin{align} dr &= \frac x r dx + \frac y r dy \\ d\theta &= \frac1{r^2} (x\ dy -y\ dx). \end{align}$$