# Divergence of traceless matrix

• I
jouvelot
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?
Pierre

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The trace of $M$ is $M_{aa}$. By symmetry, $$\frac{\partial M_{ab}}{\partial \hat{n}_a} = \frac{\partial M_{aa}}{\partial \hat{n}_b}$$

jouvelot
Thanks. This was pretty simple; I should have gotten that ;)

Bye,

Pierre