# [Physics] the curl of $k\hat{r}/r^n$

calculusclassical-mechanicsdifferentiationhomework-and-exercisesVector Fields

I'm trying to find the curl of ${\bf F}(r) = k \hat{r}/r^n$. I think that this converts to:

$$k\left(\frac{\hat{x}}{r} + \frac{\hat{y}}{r} + \frac{\hat{z}}{r}\right)\frac{1}{(x^2 + y^2 + z^2)^{n/2}}$$

I think that the determinant should look like:

$$\hat{x}\left(\frac{\partial F_z}{\partial y} – \frac{\partial F_y}{ \partial z}\right) + \hat{y} \left(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\right) + \hat{z}\left(\frac{\partial F_y}{\partial x} – \frac{\partial F_x}{\partial y}\right)$$

I know that either each of the contents of the brackets equals zero or each individual derivative equals zero.

If I assume that it should be $\hat{x}/r$, the first derivative equals:

$$\frac{\partial F_z}{\partial y} = k\left(-(n+2)y(x^2+ y^2 + z^2)^{-(n+4)/2}\right)$$

Intuitively speaking, the field ${\bf F}(r)$ is radial, so its curl is zero,

$$\nabla \times {\bf F} = 0$$

Below I show a proof, which is based on two facts

1. $$\nabla \times (\phi{\bf A}) = \phi (\nabla \times {\bf A}) + (\nabla \phi)\times {\bf A} \tag{1}$$

2. $$\nabla \times {\bf r} = {\bf 0} \tag{2}$$

this can be shown to be true by using

$$\nabla\times {\bf r} = \epsilon_{ijk}\partial_i x_j\hat{e}_k = \epsilon_{ijk}\delta_{ij}\hat{e}_k = \epsilon_{iik}\hat{e}_k = {\bf 0}$$

Taking this into account

\begin{eqnarray} \nabla \times \left(k\frac{\hat{r}}{r^n} \right) &=& \nabla \times \left(k\frac{{\bf r}}{r^{n + 1}} \right) \\ &\stackrel{(1)}{=}& \frac{k}{r^{n+1}}(\nabla\times {\bf r}) + k\left(\nabla\frac{1}{r^{n+1}} \right)\times {\bf r} \\ &\stackrel{(2)}{=}& -k(n+1)\frac{{\bf r}}{r^{n+2}}\times {\bf r} = 0 \end{eqnarray}