# How the curl of position vector is zero

I'm a new learner of tensor product. I have seen a question that asked to show that curl of a position vector is zero. $\nabla \times r=0$

If we write the equation using epsilon, we get, $$\nabla \times r= \epsilon_{ijk} \partial_{j}r_k$$

How it could be zero?

Is that equation a special case? We get that equal to zero only if any of the indices are equal.

As you've said, if two of the indices are equal, then the equation vanishes. This is because the Levi-Civita symbol vanishes. However, if they are all different, then we have

$$j \neq k \implies \partial_j r_k = 0 \implies \nabla \times r = 0$$

Because the coordinates of the position vector are independent (i.e. have no partial dependence).

You're very close to getting to the answer. Here: \begin{align} [\nabla \times \vec{r}]_i & = \epsilon_{ijk} \partial_j r_k \\ & = \epsilon_{ijk} \delta_{jk}\end{align}. You can do the contraction manually to see that you get zero, or note that $\epsilon_{ijk}$ is antisymmetric under the interchange of any two indices. Any such tensor will be traceless for any pair of indices.

Instead of using the Levi-Civita symbol, $\epsilon_{ijk}$, I prefer to write it more explicitly as $\hat x_i\cdot (\hat x_j \times \hat x_k)$, where $\hat x_i$ is the Cartesian unit vector along the $x_i$ axis. We proceed now to write the curl of the position vector.

The $x_i$ component of the curl of the position vector $\vec r=\sum_{k=1}^3 \hat x_k x_k$ can be written

$$\hat x_i\cdot \nabla \times \vec r =\hat x_i\cdot (\hat x_j\partial_j)\times (\hat x_k x_k) \tag 1$$

where summation over repeated indices in $(1)$ is implied. Continuing, we have

\begin{align} \hat x_i\cdot \nabla \times \vec r &=\hat x_i\cdot (\hat x_j \times \hat x_k)\partial_j(x_k) \tag 2\\\\ &=\hat x_i\cdot (\hat x_j \times \hat x_k)\delta_{ij} \tag 3\\\\ &=\hat x_i\cdot (\hat x_j\times \hat x_j) \tag 4\\\\ &=0 \tag 5 \end{align}

where $\delta_{ij}=1$ for $i=j$ and $0$ otherwise is the Kronecker Delta.

In going from $(2)$ to $(3)$, we noted that the partial $\frac{\partial x_k}{\partial x_j} =\delta_{ij}$.

In going from $(3)$ to $(4)$, we exploited the sifting property of the Kronecker Delta, which sifted our the $k$ index at $k=j$.

In going from $(4)$ to $(5)$, we simply recognized that $\hat x_j \times \hat x_j=0$ for all $j$.

And we are done!