I have unit vector fields defined by
$\hat{r} = cos(\theta)\hat{x} + sin(\theta)\hat{y}$
$\hat{\theta} = -sin(\theta)\hat{x} + cos(\theta)\hat{y}$
where $\hat{x} = \frac{\partial}{\partial x}$ and $\hat{y} = \frac{\partial}{\partial y}$
Now to show this, I must show that the Lie Bracket does not commute between $\hat{r}$ and $\hat{\theta} \hspace{5pt} ([\hat{r},\hat{\theta}] \neq 0)$
I can show this, but the solution I am given has the answer given in the following terms
$[\hat{r},\hat{\theta}] = \frac{-\hat{\theta}}{r} \neq 0$
I want to know how they get this form for the answer.
WORK:
- Wrote out commutator
$[\hat{r},\hat{\theta}] = \hat{r}\hat{\theta}-\hat{\theta}\hat{r} $
- Plugged in values for $\hat{r}$ and $\hat{\theta}$ and distributed terms. I also used the product rule as the partial derivative was distributed over a trigonometric angle and another partial derivative (for example $(\frac{\partial}{\partial x}(\cos\theta\frac{\partial}{\partial x})$) to get
$ = -\cos\theta(\frac{\partial}{\partial x}\sin\theta)\frac{\partial}{\partial x}$ - $\cos\theta \sin\theta\frac{\partial^{2}}{\partial x^{2}}$ + $\cos\theta(\frac{\partial}{\partial x}\cos\theta)\frac{\partial}{\partial y}$ + $\cos^{2}\theta\frac{\partial}{\partial x}\frac{\partial}{\partial y}$+... (###)
The first two terms in the distribution are displayed above.
Writing everything out, the terms with $\cos^{2}(\theta)$, $\cos^{2}(\theta)$ and angles next to each other ... $\sin{\theta}\cos\theta$ all cancel because signs were opposite and the trailing partial derivatives commute.
In equation (###) above, notice the first term has the derivative $\frac{\partial}{\partial x}\sin\theta$. I assumed here that $\theta$ depends on $x$ and $y$ respectively, which seems reasonable considering that $x$ and $y$ do change as one moves around the unit circle. Also, if I didn't assume this, all remaining terms would disappear as all remaining terms have derivatives of $\sin\theta$ or $\cos\theta$ with respect to either $x$ or $y$.
The remaining terms don't cancel, but they are not nearly as nice as the simple form given as the answer, and I would like to know how to get that form. Is there a simpler way of computing the Lie Bracket here?