Proof That the Curl of the Radial Vector is Zero
\[\mathbf{r} = x \mathbf{i} + y \mathbf{j} +z \mathbf{k}\]
is zero because\[\begin{equation} \begin{aligned} & (\frac{\partial}{\partial x} \mathbf{i}+\frac{\partial}{\partial y} \mathbf{j}+ \frac{\partial}{\partial k} \mathbf{k}) \times (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \\ &= (\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}) \mathbf{i} + (\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}) \mathbf{j} + (\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}) \mathbf{k}=0 \end{aligned} \end{equation}\]
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. 
