# Homework Help: Tensor algebra, divergence of cross product

1. Jul 30, 2014

### Telemachus

Hi there. I wanted to demonstrate this identity which I found in a book of continuum mechanics:

$curl \left ( \vec u \times \vec v \right )=div \left ( \vec u \otimes \vec v - \vec v \otimes \vec u \right )$

I've tried by writting both sides on components, but I don't get the same, I'm probably making some mistake.

$\displaystyle curl \left ( \vec u \times \vec v \right )= \epsilon_{ijk} \frac{\partial}{\partial x_j}(\epsilon_{kji}u_j v_i)= \epsilon_{ijk} \left [ \epsilon_{kji} \frac{\partial u_j}{\partial x_j}v_i +u_j \epsilon_{kji} \frac{\partial v_i}{\partial x_j} \right ]$

I've used that $(\vec u \times \vec v )_k=\hat e_k \cdot \epsilon_{ijk} u_j v_k \hat e_i=\delta_{ki} \epsilon_{ijk} u_j v_k=\epsilon_{kji}u_j v_i$ I'm not sure if this is right.

Then, using the epsilon delta identity I've got $rot \left ( \vec u \times \vec v \right )=6 \left [ \frac{\partial u_j}{\partial x_j}v_i +u_j \frac{\partial v_i}{\partial x_j} \right ]=6 \left [ (div \vec u) \vec v + (grad \vec v) \vec u \right ]$

Expanding the right hand side of the identity in components:

$div \left ( \vec u \otimes \vec v - \vec v \otimes \vec u \right ) = \frac{\partial}{\partial x_j}(u_i v_j -v_i u_j)=v_j \frac{\partial u_i}{\partial x_j}+ u_i \frac{\partial v_j}{\partial x_j} -u_j \frac{\partial v_i}{\partial x_j}- v_i \frac{\partial u_j}{\partial x_j}=(grad \vec u) \vec v+\vec u div \vec v - (grad \vec v) \vec u- \vec v div \vec u$

PD: The title should be curl of cross product instead of divergence. Sorry.

2. Jul 30, 2014

### vela

Staff Emeritus
You shouldn't have indices that appear more than twice. $i$ appears 3 times; $j$, 4 times.

3. Jul 30, 2014

### Telemachus

ups, I see. Thank you vela.

4. Aug 1, 2014

### Fredrik

Staff Emeritus
It is, but I wouldn't do those first two steps. I think of $(\vec u\times\vec v)_k=\varepsilon_{kij}u_jv_k$ as the definition of $\vec u\times\vec v$, so it would have been my starting point. It makes sense to take this as the definition, since it implies that $\vec u\times\vec v=(\vec u\times\vec v)_k e_k =\varepsilon_{kij}u_jv_k e_k$.