Proof for Vector Identity

Question

I am currently studying electrodynamic and came across the following vectoridentity, but I am unable to prove it:

$$ \vec{f} \times ( \nabla \times \vec{f} ) -\vec{f}(\nabla\cdot\vec{f}) = \nabla \cdot (1/2 \cdot f^2 \delta - \vec{f} \otimes \vec{f} ) $$

$$ \vec{f} \times (rot(\vec{f})) -\vec{f}(\nabla\cdot\vec{f})= div (1/2 \cdot f^2 E - \vec{f} \otimes \vec{f} ) $$

The second equation is the same as the first.

Edit: I totally forgot the term $ -\vec{f}(\nabla\cdot\vec{f}). $

$ \delta $ , E is the unit tensor (I first wrote it as Kronecker symbol $ \delta_{ij} $).

Switched + to - on the right side between the tensors.

Answer 1

First equation is inconsistent. What are $i,j$? Why do they only appear on the right-hand side? Are you familiar with notation? Tensor product? I would suggest proving this component-by-component, at first. And then looking at Levi-Civitas ( https://en.wikipedia.org/wiki/Levi-Civita_symbol) and their contractions.

with Levi-Civitas you get (given trivial metric and in Cartesian coordinates):

$\mathbf{f}\times\boldsymbol{\nabla}\times\mathbf{f} = \mathbf{e}_i \epsilon_{ijk}\,f_j\, \epsilon_{kst} \,\partial_s f_t = \epsilon_{kij}\epsilon_{kst}\,\mathbf{e}_i f_j\partial_s f_t = \left(\delta_{is}\delta_{jt}-\delta_{it}\delta_{js}\right)\,\mathbf{e}_i f_j\partial_s f_t=\dots$

where $\mathbf{e}_{i=1,2,3}=\mathbf{\hat{x}},\,\mathbf{\hat{y}},\,\mathbf{\hat{z}}$

Answer 2

You notation in the first expression does not make sense.

If we use standard index notation, we have, from the triple product identity $$ {\bf a}\times({\bf b}\times {\bf c}) ={\bf b}({\bf c}\cdot {\bf a}) - {\bf a}({\bf b}\cdot {\bf c}) $$ and bearing in mind that ${\bf b}\equiv \nabla$ acts only on "${\bf c}$" that $$ ({\bf f}\times(\nabla\times {\bf f}))_i= f_i (\partial_j f_j)-f_i(\partial _j f_j) = \frac 12 \partial_i (f_jf_j)-f_i(\partial _j f_j)\\ $$ $$ = \frac 12 \nabla |f|^2-({\bf f}\cdot \nabla){\bf f} $$

Answer 3

After your edit, the identity is now correct. A fast check with Mathematica gives indeed

Notice that here we must understand the divergence of the two-tensor as acting over each row vector. In index notation, this is:

\begin{equation} \partial_i \left( \frac{1}{2} \, f_k \, f_k \, \delta_{ji} \, - \, f_j \, f_i \right) \, = \, \epsilon_{jkl} \, f_k \, \epsilon_{lmn} \, \partial_m \,f_n \, - \, f_j \, \partial_k \, f_k \end{equation}

Use the contraction properties of the Levi-Civita symbol and you will be able to work it out.