>From this symmetric gradient tensor, do we have one easy way to calculate
> the following "the second invariant of rate-of-strain tensor"?
There is SymmetricTensor::second_invariant  which is defined as I2 =
1/2[ (trace sigma)^2 - trace (sigma^2) ]
and SymmetricTensor::first_invariant  which just the trace I_1=trace
According to Wikipedia , your \Pi_\gamma can then be expressed as
\Pi_\gamma^2 = 1/2(I1^2 -2 I2).
> Jaekwang Kim
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