Show that $\mathrm{rot}\left\{ (\vec e\times \vec r)\times \vec e \right\}=0$, where $\vec e$ be the unit vector.

Question

Question: Show that $\mathrm{rot}\left\{ (\vec e\times \vec r)\times \vec e \right\}=0$, where $\vec e$ be the unit vector.

We have to show $\mathrm{rot}\left\{ (\vec e\times \vec r)\times \vec e \right\}=0$, i.e., $\nabla \times \left\{ (\vec e\times \vec r)\times \vec e \right\}=0$.

Now, $(\vec e\times \vec r)\times \vec e=\vec r-(\vec e\cdot r)\vec e$. What can I do next?

Answer 1

Continuing your line of thought: $$ \underbrace{\nabla\times r}_{\to 0}+\nabla\times((e\cdot r)e) = (\nabla(e\cdot r))\times e + (e\cdot r)\underbrace{\nabla\times e}_{\to 0} = -e\times \nabla(e\cdot r),\\ \nabla(e\cdot r) = (e\nabla)r+(r\nabla)e+e\times(\nabla\times r)+r\times(\nabla\times e)=(e\nabla)r = e,\\ -e\times \nabla(e\cdot r) = -e\times e = 0 $$

You can also do it with tensor notation: $$ \varepsilon_{ijk}\partial_j\varepsilon_{klm}\varepsilon_{lpq}r_pe_qe_m= e_qe_m\varepsilon_{ijk}(\delta_{kq}\delta_{mp}-\delta_{kp}\delta_{mq})\delta_{jp} = e_qe_m\varepsilon_{ipk}(\delta_{kq}\delta_{mp}-\delta_{kp}\delta_{mq})=\\ e_qe_p\varepsilon_{ipq}-e_qe_q\varepsilon_{ipp}=0-0=0 $$

Answer 2

Let us treat $e$ as an arbitrary constant unit vector. I'm going to use some known identities, some of which are listed here. You want to calculate $\nabla \times[ (e\times r)\times e]$, so we can set $A = e\times r$ and $B = e$. Using identity 13 (of the link) we have: $$\nabla \times (A\times B) = A(\nabla \cdot B)-(\nabla \cdot A)B + (B\cdot \nabla)A - (A\cdot \nabla) B$$ Because $e$ is a constant vector, it does not depend on variables $x,y$ and $z$, so that we readily see that $\nabla \cdot B = (A\cdot \nabla) B = 0$. Besides, using identity 9: $$ \nabla \cdot A = \nabla \cdot (e\times r) = r\cdot (\nabla \times e) - e\cdot (\nabla \times r)$$ Again, $\nabla \times e = 0$ and $\nabla \times r = 0$ (it is easy to conclude the latter by explicit calculations, once $r = (x,y,z)$). Finally, we have to evaluate $(B\cdot \nabla) A$. If $e$ is taken to be a vector of the canonical basis for $\mathbb{R}^{3}$, the conclusion that $(B\cdot\nabla) A = 0$ is straightford. For an arbitrary unit vector $e=(e_{x},e_{y},e_{y})$, you can check that: $$e\times r = (z e_{y}-e_{z}y)\hat{x} + (e_{z}x-e_{x}z)\hat{y}+(e_{x}y-e_{y}x)\hat{z}$$ Thus, $(B\cdot \nabla) A$ is given by: $$\bigg{[}e_{x}\frac{\partial}{\partial x}+e_{y}\frac{\partial}{\partial y}+e_{z}\frac{\partial}{\partial z}\bigg{]} (e\times r) = 0$$ as you can easily check.