Cross Product of Curl Identity

[EricTheWizard]

Hi, I've been trying to derive the electromagnetic stress tensor on my own, and I've run into a bit of a problem. I have a cross product of a curl $$(\vec{E}\times(\nabla\times\vec{E}))$$ that I need to expand, and the typical $$\vec{A}\times(\vec{B}\times\vec{C})=\vec{B}(\vec{A}\cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B})$$ isn't cutting it, as the book says this special case is $$\vec{E}\times(\nabla\times\vec{E})=\frac{1}{2}\nabla(E^2)-(\vec{E}\cdot\nabla)\vec{E}$$. I've been trying to work this out myself on paper, but to no avail. Can anyone point me to a proof for this or show me how? Much appreciated.

 

[arildno]

Note that we have:
$$\vec{E}\times(\nabla\times\vec{E})=\vec{E}\times((E_{3,2}-E_{2,3})\vec{i}_{1}+(E_{1,3}-E_{3,1})\vec{i}_{2}+(E_{2,1}-E_{1,2})\vec{i}_{3})$$
where E_{i,j} means the i'th component differentiated with respect to the jth variable.

Furthermore, we get, as our i_1-component as we cross-multiply:
$$=(E_{2}(E_{2,1}-E_{1,2})-E_{3}(E_{1,3}-E_{3,1}))\vec{i}_{1}=(E_{2}E_{2,1}+E_{3}E_{3,1}+E_{1}E_{1,1})\vec{i}_{1}-(E_{1}E_{1,1}+E_{2}E_{1,2}+E_{3}E_{1,3})\vec{i}_{1}$$
where I added&subtracted $E_{1}E_{1,1}$.

You should be able to do the other two components now, along with showing the identity you were given

 

[EricTheWizard]

ahh, it was adding and subtracting that $$E_i E_{i,i}$$ that I was missing and was screwing me up. But I managed to work it out now. Thanks for your help!