Re: Derivative of a tensor

Correcting some errors:

T is symmetric so the second term disappears

where z = [0 0 1]'

I can go no further and suspect I have made an error before this point.

I would really like to get an answer of the form:

but that seems unlikely as I can see no way to conjure up the partial derivative w.r.t. t.

Perhaps there is something special about the Maxwell Stress Tensor that allows me to do this?

This problem is related to my earlier post:

http://mathhelpforum.com/calculus/22...em-fields.html

Re: Derivative of a tensor

Can you be a little more specific about your T and S here so that we can help better? I think the negative sign in the end should not be there. And if S is the poynting vector and T is the stress tensor, then you have where f is the force per unit volume for the charge distribution. If your f is 0, then you get exactly the form you are looking for. Let me know if this helped.