The problem in the book asks what the curl of $\operatorname{curl}\vec F(\vec r)= \frac {\vec r}{\|\vec r\|}$. Can someone give me a good explanation on why the curl will be zero? I would really appreciate it.
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Did you try plotting the vector field? – mathematics2x2life Dec 11 '13 at 19:47
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I wasn't given a vector field, the problem is copied directly from the book. – Student Dec 11 '13 at 19:48
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The function $\frac{\vec{r}}{\| \vec{r}\|}$ is the gradient of the scalar field $f(\vec{r})=\|\vec{r}\|$. Proceed using the identity $$\text{curl grad } f=0. $$
user1337
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@ user1337: simpler though less general than my approach; in my penchant for abstraction, I lost sight of the fact that $(\vec r / \Vert \vec r \Vert)$ is a gradient! +1! – Robert Lewis Dec 11 '13 at 20:12
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Basically, for any function $f(x, y, z)$ and vector field $\mathbf V(x, y, z)$, we have
$\nabla \times (f \mathbf V) = \nabla f \times \mathbf V + f \nabla \times \mathbf V; \tag{1}$
in the present case, taking $\mathbf V = \vec r$, it is easy to see by direct calculation that
$\nabla \times \mathbf V = \nabla \times \vec r = 0, \tag{2}$
and since
$\nabla (1 / \Vert \vec r \Vert)$ is collinear with $\vec r$, we also have
$\nabla (1 / \Vert \vec r \Vert) \times \vec r = 0; \tag{3}$
plugging everything into (1) gives the desired result.
Though user1337's answer is of course correct, the above shows how this problem fits into a somewhat more general pattern; see my answers to this question.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Robert Lewis
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