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We are stuck with this question here because I cannot understand the following results. I find it hard to visualize this, let alone deduce from that. How to do it?

Objective to Attack The closely Related Problems with Orthogonal Basis and Dot Products In Polar-coordinates

  1. $\left(\hat{e}_{r}\partial_{r}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)= 0$

  2. $\left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) \cdot \left(\hat{e}_{r}\partial_{r}\right) = \frac{1}{r} \partial_r$

  3. $\partial_\theta \hat e_r = \hat e_\theta$

  4. $\partial_\theta \hat e_\theta = -\hat e_r$

Trials

  1. I have some errors there, related to 3-4 apparently.

$$ \left(\hat{e}_{r}\partial_{r}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) = \left(\hat{e}_{r}\partial_{r}\right) \cdot \frac{1}{r}+ \left(\hat{e}_{r}\partial_{r}\right) \cdot \left( \hat{e}_{\theta}\partial_{\theta} \right) \not = \frac{-\hat{e}_r}{r^2}+ \left(\hat{e}_{r}\cdot\hat{e}_\theta \right) \partial_{r} \partial_{\theta} $$

Perhaps Related

  1. Explain Dot product with Partial derivatives in Polar-coordinates

  2. Show geometrically $\hat{e}_R=\frac{(x,y,z)}{r}$

  3. Visual Ways to Remember Cross-product in $\mathbb F^3$?

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1 Answer 1

up vote 2 down vote accepted

Some hints: (not a complete solution)

  • I guess once you think about it you find it clear that $$\hat{e}_{r}\cdot\hat{e}_\theta=0$$ or in words the unit vector along the radial direction is orthogonal to the unit vector along the angular direction.

  • I guess the confusion you have originates from the following fact: the unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ themselves depend on the coordinate $(r,\theta)$. This dependence is usually not made explicit but you should always keep that in mind. If you think about it then I guess it is clear. For $\theta=0$ the radial unit vector points along the $x$-axis whereas for $\theta=\pi/2$ the unit vector points along the $y$-axis.

  • The last two points hold for an arbitrary rectangular coordinate system. What is special for the polar coordinate system is that even though $\hat{e}_{r,\theta}$ depend on $\theta$ they do not depend on $r$, i.e., $$\hat{e}_{r,\theta} = \hat{e}_{r,\theta} (\theta).$$

  • The important relations $\partial_\theta \hat e_r(\theta) = \hat e_\theta (\theta)$ and $\partial_\theta \hat e_\theta(\theta) = -\hat e_r(\theta)$, you can check by taking the explicit expressions $$\begin{array}\ \hat{e}_r(\theta)&= \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} & \hat{e}_\theta(\theta)&= \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} \end{array}$$

With these remarks, it is easy to show that (here, I make the dependence of the unit vectors on the coordinates explicit) $$\left(\hat{e}_{r}(\theta)\partial_{r}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta}(\theta) \partial_{\theta}\right) = \underbrace{\hat{e}_{r}(\theta) \cdot \hat{e}_{\theta}(\theta)}_{=0} \,\left(\partial_{r} \frac{1}{r} \partial_{\theta} \right) =0 .$$ The other results follow similarly (but I will leave the proof up to you).

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Why is there no $r$ in the $e_r(\theta)$? –  hhh Mar 25 '12 at 11:38
1  
@hhh: There is a typo. The equation you wrote is correct. Also notice that $$\hat e_r = \frac{{\bf r}}{r} = \hat e_x \frac{x}{r} + \hat e_y \frac{y}{r},$$ but $x/r = \cos\theta$ and $y/r = \sin\theta$, so there is no $r$. –  user26872 Mar 25 '12 at 15:54

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