- #1
jouvelot
- 53
- 2
Assume that ##\partial M_{ab}/\partial \hat{n}_c## is completely symmetric in ##a, b## and ##c##. Then, it is stated in the book I read that the divergence of the traceless part of ##M## is proportional to the gradient of the trace of ##M##. More precisely,
$$ \partial /\partial \hat{n}_a (M_{ab} - \delta_{ab} {\rm Tr} (M)/2) = \partial ({\rm Tr} (M)/2)/\partial \hat{n}_b .$$ Can anyone provide some hints on why this is true, please?
Thanks in advance.
Pierre
$$ \partial /\partial \hat{n}_a (M_{ab} - \delta_{ab} {\rm Tr} (M)/2) = \partial ({\rm Tr} (M)/2)/\partial \hat{n}_b .$$ Can anyone provide some hints on why this is true, please?
Thanks in advance.
Pierre