Explain Dot product with Partial derivatives in Polar-coordinates Related to page 819 prob 4 in this book. I am incorrectly calculating the left-hand-side (def. LHS), some newbie error with commutativity probably. Ideas?
Errors?


*

*I propose $(\hat{e}_r\partial_r)\cdot (\hat{e}_r\partial_r)=\partial_r^2$, wrong?

*$(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta})\cdot (\frac{1}{r}\hat{e}_{\theta}\partial_{\theta})=\frac{1}{r^2}\partial_\theta^2$, wrong?
LHS=?

$$(\hat{e}_{r}\partial_{r}   +  
 \frac{1}{r}\hat{e}_{\theta}\partial_{\theta})   \cdot  
 (\hat{e}_{r}\partial_{r}   +  
 \frac{1}{r}\hat{e}_{\theta}\partial_{\theta})     
 =\partial_{r}\partial_{r}+\frac{1}{r^2}\partial_{\theta}\partial_{\theta}$$
The error is probably in the premise about commutativity, could someone
  indicate what is going totally wrong here?

Right-hand-side=A+B

$$\partial_{r}(r\partial_{r})=1\partial_{r}+r\partial_{r}^2$$
so
$$\begin{cases}
A &=\frac{1}{r}\partial_{r}(r\partial_{r})=\frac{1}{r}\partial_{r}+\partial_{r}^2 \\
B &=\frac{1}{r^2}\partial_\theta^2.\end{cases}$$

 A: The basis itself is a function of $\theta$. 
Equations 1 and 2 are fine, but there is a cross term in the dot product you're interested in.
To work this out you should first convince yourself that 
$$\begin{eqnarray}
\partial_r \hat e_r &=& 0, \\
\partial_r \hat e_\theta &=& 0, \\
\partial_\theta \hat e_r &=& \hat e_\theta \\
\partial_\theta \hat e_\theta &=& -\hat e_r. 
\end{eqnarray}$$
The cross term comes from 
$(\frac{1}{r} \hat e_\theta)\cdot [(\partial_\theta \hat e_r)\partial_r] = \frac{1}{r}\partial_r$. 
Let's look at this in more detail. 
First notice that $\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}$ is the del operator in polar coordinates. 
Thus, you are trying to find the Laplacian in polar coordinates.
FOIL out the expression
$$\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)
\cdot
\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right).$$
You will find four terms, 
$$\begin{eqnarray}
\left(\hat{e}_{r}\partial_{r}\right) 
    \cdot \left(\hat{e}_{r}\partial_{r}\right) 
        &=& \partial_r^2 \\
\left(\hat{e}_{r}\partial_{r}\right) 
    \cdot \left( \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) 
        &=& 0 \\
\left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) 
    \cdot \left(\hat{e}_{r}\partial_{r}\right) 
        &=& \frac{1}{r} \partial_r \\
\left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) 
    \cdot \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) 
        &=& \frac{1}{r^2}\partial_\theta^2.
\end{eqnarray}$$
The first and fourth terms are the ones you have claimed. 
But there is a subtlety. 
For example, 
$$\begin{eqnarray}
\left(\frac{1}{r}\hat{e}_{\theta} \partial_{\theta}\right)
    \cdot \left(\frac{1}{r}\hat{e}_{\theta} \partial_{\theta}\right) &=&
\left(\frac{1}{r}\hat{e}_{\theta}\right)\cdot 
\left( \frac{1}{r} \hat{e}_{\theta} \partial_{\theta}^2 
    + \frac{1}{r} (\partial_\theta \hat e_\theta) \partial_\theta \right)     \\
&=& \left(\frac{1}{r}\hat{e}_{\theta}\right)\cdot
\left( \frac{1}{r} \hat{e}_{\theta} \partial_{\theta}^2
    - \frac{1}{r} \hat e_r \partial_\theta \right) \\ 
&=& \frac{1}{r^2} \partial_\theta^2,
\end{eqnarray}$$
since $\hat e_\theta\cdot \hat e_r = 0$. 
Notice that 
$$\partial_\theta (\hat{e}_{\theta} \partial_{\theta})
= \hat e_\theta \partial_\theta^2 + (\partial_\theta \hat e_\theta)\partial_\theta$$ 
by product rule of differentiation. 
Being similarly careful with the other terms you will find the claimed result, 
$$\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)^2
= \partial_r^2 + \frac{1}{r} \partial_r + \frac{1}{r^2} \partial_\theta^2.$$
