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I have a missing $\frac{1}{r}\partial_r$ -term (notice the question mark) but cannot see why, could someone hint where I am doing mistake.

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Your $\varphi$'s look like backwards 3's! – anon Mar 25 '12 at 13:27
up vote 2 down vote accepted

$$\partial_\varphi e_r=e_\varphi\implies\frac{1}{r}e_\varphi\partial_\varphi \cdot e_r \partial_r =\frac{1}{r}(e_\varphi\cdot e_\varphi)\partial_r=\frac{1}{r}\partial_r$$

You evaluated this to zero. If you write this out fully (ignoring the $1/r$ in front), you have

$$\begin{pmatrix}-\sin\varphi \; \frac{\partial}{\partial\varphi} \\ \cos\varphi \; \frac{\partial}{\partial\varphi}\end{pmatrix} \cdot\begin{pmatrix} \cos\varphi \; \frac{\partial}{\partial r} \\ \sin\varphi \frac{\partial}{\partial r}\end{pmatrix} = -\sin\varphi \; \frac{\partial}{\partial \varphi}\left(\cos\varphi \; \frac{\partial}{\partial r}\right)+\cos\varphi \; \frac{\partial}{\partial\varphi}\left(\sin\varphi\frac{\partial}{\partial r}\right)$$

$$=-\sin\varphi \left(-\sin\varphi\frac{\partial}{\partial r}+\cos\varphi\frac{\partial^2}{\partial \varphi \partial r}\right)+\cos\varphi\left(\cos\varphi\frac{\partial}{\partial r}+\sin\varphi \frac{\partial^2}{\partial \varphi \partial r}\right)=\frac{\partial}{\partial r}.$$

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$\frac{1}{r}e_\phi\partial_\phi \cdot e_r \partial_r = \frac{1}{r}\partial_\phi \partial_r (e_\phi \cdot e_r)=\frac{1}{r}\partial_\phi \partial_r (0)=0\not =\frac{1}{r}\partial_r$, what is wrong here? Premise is $$e_r=\begin{pmatrix}\cos(\phi) & \sin(\phi)\end{pmatrix}\implies e_\phi = \begin{pmatrix} -\sin(\phi) & \cos(\phi)\end{pmatrix}$$ because $e_r$ and $e_\phi$ are orthogonal. – hhh Mar 25 '12 at 13:32
@hhh: Pay attention to the order of the partial derivative operators and the vectors. You cannot switch the order like you just did. If you could do that, then $1=\partial_x(1\cdot x)=x\cdot\partial_x1=0$. – anon Mar 25 '12 at 13:37
@hhh: I added a more explicit description of what's going on. – anon Mar 25 '12 at 13:50

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