Tensor calculus, gradient of skew tensor

[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 that's 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

 

[lippyka]

Yes, since ##\Lambda## is a constant skew tensor, this implies that the gradient of it is zero, i.e., ##grad \Lambda = 0##. Therefore, we have that $$grad (\Lambda r) = \Lambda.$$