Let $\mathbf{x}$ be the position vector, $\mathbf{a}$ be a constant vector. I need to show that:


The problem is, I keep getting $3\,\mathbf{a}$ instead, and I cannot figure out why. My work:

$$\begin{aligned}(\text{curl}(\mathbf{a}\times\mathbf{x}))_i &=\epsilon_{ijk}\frac{\partial}{\partial x_j}\epsilon_{kpq}a_px_q\\ &=(\delta_{ip}\delta_{jq}-\delta_{iq}\delta_{jp})\frac{\partial}{\partial x_j}a_px_q\\ &=\frac{\partial}{\partial x_j}a_ix_j-\frac{\partial}{\partial x_j}a_jx_i\\ &=3\,a_i\end{aligned}$$ I am probably making a silly mistake but I cannot see it.


1 Answer 1


Up to your last line you are fine. But $$ \frac{\partial}{\partial x_j}x_j = 3 $$ and $$ \frac{\partial}{\partial x_j}x_i = \delta_{ij} $$ So you next to last line simplifies to $$ 3a_i - a_i = 2a_i$$


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