Background about terms in this question: Hookes law and objective stress rates
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.
Edit - Clarifying where the deviatoric strain rate term came from.
For an isotropic, elastic solid the stress tensor is given by:
$${\sigma}^{ij} = 2\mu{\epsilon}^{ij} + \lambda \delta^{ij}({\epsilon}^{kk})$$
Then the deviatoric stress can be written as:
$${S}^{ij} = 2\mu{\epsilon}'^{ij} + \lambda \delta^{ij}({\epsilon'}^{kk})$$
Given that the deviatoric strain is traceless, the deviatoric stress rate can be written as:
$${\dot{S}}^{ij} = 2\mu\dot{{\epsilon'}}^{ij}$$
The deviatoric strain rate can be rewritten in terms of the strain rate giving:
$${\dot{S}}^{ij} = 2\mu\left[{\dot{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}{\dot{\epsilon}}^{ij}\right]$$