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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.

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  • If $w$ is skew-symmetric, then $w'=-w$ – Svetoslav Jul 28 '15 at 16:17
  • Also if $w$ and $S$ commute you are done. See http://math.stackexchange.com/questions/1141231/when-does-a-2-times2-symmetric-matrix-commute-with-a-skew-symmetric-matrix – Svetoslav Jul 28 '15 at 16:21
  • I understand that $w' = -w$ but that does not explain the above as the term which changed sign did not involve a transpose? – 1QuickQuestion Jul 29 '15 at 08:56
  • I guess the term $-w.S$ is equal to $w'.S$ which if they commute equals to $S.w'$. And the term $S.w=w.S$, again if they commute. – Svetoslav Jul 29 '15 at 10:10
  • So $S$ is symmetric and $w$ is skew symmetric. But playing around in excel, I can't get any of your statements above to hold true (using the standard mmult commands with 3x3 matrices). – 1QuickQuestion Jul 29 '15 at 10:36
  • Did you see http://math.stackexchange.com/questions/1141231/when-does-a-2-times2-symmetric-matrix-commute-with-a-skew-symmetric-matrix – Svetoslav Jul 29 '15 at 10:46
  • Yes but I think I am missing something. Using the standard dot product operator I find that $A.B = -(B.A)'$. Overall I find that $S.w - w.S = -(S.w' + w.S)$. – 1QuickQuestion Jul 29 '15 at 11:39
  • I thought that this symbol "." between the matrices is just matrix multiplication, not a dot product/inner product ? – Svetoslav Jul 29 '15 at 11:46
  • Maybe there is a mistake in what you are reading. Because if both expressions are equal, this would mean that it must be equal to 0. I cannot help you. – Svetoslav Jul 29 '15 at 11:52
  • I think your are correct in that it is just matrix multiplication but my previous comment still holds for this too. – 1QuickQuestion Jul 29 '15 at 13:52

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