# Tensor calculus, gradient of skew tensor

1. Aug 19, 2014

### Telemachus

Hi there. I was dealing with the derivation on continuum mechanics for the conservation of angular momentum. The derivation I was studying uses an arbitrary constant skew tensor $\Lambda$. It denotes by $\lambda$ its axial vector, so that $\Lambda=\lambda \times$

Then it defines $w(x)=\lambda \times r=\Lambda r$

So that $grad (\Lambda r)=\Lambda$

And thats the doubt I have.

When I do $grad (\Lambda r)$ I have (I use that $r=x-x_0$):

$\displaystyle grad (\Lambda r)=\frac{\partial}{\partial x_k} ( \Lambda_{ij} r_j ) = \frac{\partial \Lambda_{ij} } {\partial x_k} r_j + \frac {\partial r_j} {\partial x_k} \Lambda_{ij} = \frac{\partial \Lambda_{ij}} {\partial x_k} (x_j-x_{0j})+\frac{\partial (x_j-x_{0j})}{\partial x_k}\Lambda_{ij}=(grad \Lambda ) r+\Lambda$

Now, the fact that $\Lambda$ was constant determines that the gradient is zero? that was the doubt, I recognize that I didn't noticed before the fact that the tensor was constant until I written this post :p

Last edited: Aug 19, 2014