Curl and gradient properties for $f ( r)\vec r$

Question

I need to show that the curl of $f( r) \vec{r}$ is $0$.

I think I can use this property:

$$\operatorname{curl}(Av) = \operatorname{grad}(A)\times v+A \operatorname{curl}(v)$$

I have started working through it but I am getting stuck. Does the curl of $\vec r $ equal $0$? And if so, why?

Also, is $\operatorname{grad}(r) = r \vec r$?

Answer 1

It can be easily proved using the Einstein summation notation: $$ \nabla \times \left(r\cdot f(r)\right) = \epsilon_{ijk}\partial_j\left[r_k f(r)\right]\\ = \epsilon_{ijk} \left[\left(\partial_j r_k\right)f(r) + r_k \partial_j f(r)\right]\\ = \epsilon_{ijk}\left[\delta_{jk}f(r) + \frac{df}{dr} r_k \left(\frac{r_j}{r}\right)\right]\\ = 0 + \frac{f'(r)}{r} \epsilon_{ijk}r_j r_k\\ = \frac{f'(r)}{r} \vec{r} \times \vec{r}\\ = 0 $$

Answer 2

The curl of $\vec r$ is indeed equal to zero, and the easiest way to see is probably $$ \operatorname{grad} \left(\frac12 r^2\right) = \vec r$$ (Every conservative field has zero curl). Generalizing the above to $$ \operatorname{grad} (F(r)) = F'(r) \frac{\vec r}{r}$$ suggests a direct way to attack your problem: find $F$ such that $F'(t)=rf(t)$. Actually, you don't need to find it explicitly: the existence of such $F$, guaranteed by the fundamental theorem of calculus, is all that's needed. Since $f(r)\vec r$ has potential function $F(r)$, its curl is zero.