# 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?

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

2. $(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta})\cdot (\frac{1}{r}\hat{e}_{\theta}\partial_{\theta})=\frac{1}{r^2}\partial_\theta^2$, 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

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

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Have you tried applying the formulas for $\hat{e}_R$, $\hat{e}_\theta$, and $\hat{e}_\phi$ from my answer to your other question? – robjohn Mar 25 '12 at 2:54
@robjohn: gathered issues here, to clarify things -- I have to solve the sub-problems before going into the major probs. – hhh Mar 25 '12 at 10:56

The basis itself is a function of $\theta$.
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, $$\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)^2 = \partial_r^2 + \frac{1}{r} \partial_r + \frac{1}{r^2} \partial_\theta^2.$$