## Proof That the Curl of the Radial Vector is Zero

The curl of the radial vector
$\mathbf{r} = x \mathbf{i} + y \mathbf{j} +z \mathbf{k}$
is zero because
\begin{aligned} & (\frac{\partial}{\partial x} \mathbf{i}+\frac{\partial}{\partial y} \mathbf{j}+ \frac{\partial}{\partial k} \mathbf{k}) \times (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \\ &= (\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}) \mathbf{i} + (\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}) \mathbf{j} + (\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}) \mathbf{k}=0 \end{aligned}

To show this using tensor notation write

$\mathbf{\nabla} \times \mathbf{r}==e_{ijk} \partial_i x_k$

$\partial_i x_k =0$
when
$i \neq k$
and
$\partial_i x_k =1$
when
$i=k$
. We can write nbsp;
$\partial_i x_k =\delta_{ik}$

$\mathbf{\nabla} \times \mathbf{r}=e_{ijk} \delta_{ik} =e_{ijk} \delta_{ii} =0$

since
$e_{ijk}=0$
when any
$i,j,k$
are equal.