Jaumann deviatoric stress rate.

127 Views Asked by At

Being a bit cheeky as I asked this question over on Physics but didn't get a response. https://physics.stackexchange.com/questions/196393/jaumann-deviatoric-stress-rate

Background about terms in this question: https://physics.stackexchange.com/questions/123797/hookes-law-and-objective-stress-rates?newreg=69e424e670524203809b9f248a5d9ef7

From my understading, the Jaumann rate of deviatoric stress is written as: $dS/dt = \overset{\bigtriangleup}{{S}} = {\dot{S}} +{S} \cdot {w} -{w} \cdot {S}$

Yet when I see it in practice it is written as:

$\mathrm{d}{{S}}^{ij}/ \mathrm{d}t = 2\mu\left({\dot{{\epsilon}}}^{ij} - \frac{1}{3}{\delta}^{ij}{\dot{{\epsilon}}}^{ij} \right)+{{S}}^{ik}{{\Omega}}^{jk}+{{\Omega}}^{ik}{{S}}^{kj}$

To me that says:

$dS/dt = \overset{\bigtriangleup}{{S}} = {\dot{S}} +{S} \cdot {w'} +{w} \cdot {S}$

My question is - where did the spin tensor transpose and plus sign come from?

I thought that this might have something to do with Oldroyd and convective stress rates but that uses the tensor of velocity gradients rather than the spin tensor.