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:

$${\dot{S}}^{ij} = 2\mu\left[{\dot{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}{\dot{\epsilon}}^{ij}\right]$$

  • $\begingroup$ Can you link to the reference where you saw the indices written out? Also, where did the isotropic part of your stress go? You have the time derivative of the deviatoric strain there, but nothing about $\dot{\epsilon}_{kk}$ $\endgroup$ Commented Jul 28, 2015 at 17:11
  • $\begingroup$ Reference to where the indices are written out. I can provide others if needed. Added a section to the question which explains the origins of the deviatoric stress rate. $\endgroup$ Commented Jul 29, 2015 at 9:55
  • $\begingroup$ Sorry for the delay. Thesis deadlines call... Regardless, the answer is below. $\endgroup$ Commented Aug 6, 2015 at 0:49

2 Answers 2


It seems that you confused the Jaumann derivative $\overset{o}{{S}}$ (in your notation $\overset{\bigtriangleup}{{S}}$) with the time derivative ${\dot{S}}$

$$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} -{S} \cdot {w} +{w} \cdot {S}$$

See how it is derived in "http://www.continuummechanics.org/cm/corotationalderivative.html". Using the argument that $w^T = -w$ we get

$$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} +{S} \cdot {w^T} +{w} \cdot {S}$$

Moreover, when you "clarified" where does the deviatoric strain come from it is not quite clear to me all the steps you followed. Because actually is not the time derivative that equals the term $\left({\dot{{\epsilon}}}^{ij} - \frac{1}{3}{\delta}^{ij}{\dot{{\epsilon}}}^{ij} \right)$ but the Jaumann derivative. For this you can take a look at Hookes law and objective stress rates

More concretely, the derivation of the formula for the Jaumann stress

$$\overset{o}{{S}^{ij}} = 2\mu\left[{\overset{o}{\epsilon}^{ij}} - \frac{1}{3}\delta^{ij}{\overset{o}{\epsilon}^{ij}}\right]$$

is derived in Hooke's law wikipedia and in "An Introduction to Continuous Mechanics, Klaus Hackl, Mehdi Goodarzi". However, here I show a summary of the derivation of the Jaumann stress.

The definition of deviatoric stress is just

$$\sigma = \begin{pmatrix} \frac{\sigma_{11}}{3} & 0 & 0\\ 0 & \frac{\sigma_{22}}{3} & 0\\ 0 & 0 & \frac{\sigma_{33}}{3}\\ \end{pmatrix} + \begin{pmatrix} \sigma_{11} - \frac{\sigma_{11}}{3} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22} - \frac{\sigma_{22}}{3} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}- \frac{\sigma_{33}}{3}\\ \end{pmatrix}$$

the second term of the previous equation. If we apply Hooke's law to it (which relates stress $\sigma$ and strain $\varepsilon$ linearly)

$${{S}} = 2\mu\left({{\epsilon} - \frac{1}{3}\mathrm{tr}(\varepsilon)}\right)$$

With the Einstein notation

$${{S}}^{ij} = 2\mu\left[{{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}\epsilon^k{}_k\right]$$

where $\epsilon$ is the so called strain tensor.

Finally, we can compute the Jaumann derivative to get what we wanted.

  • $\begingroup$ yes you're right! $\endgroup$
    – kamipeer
    Commented Apr 22, 2016 at 11:34
  • $\begingroup$ You are correct that I initially confused the Jaumann derivative with the time derivative. Could you clarify why: $$\overset{\circ}{{S^{ij}}} = 2\mu\left[{\dot{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}{\dot{\epsilon}}^{ij}\right]$$ This is the only piece that I am missing. I do not know why this is the objective Jaumann deviatoric stress rate (as opposed to simply the time derivative of the deviatoric stress). $\endgroup$ Commented Apr 22, 2016 at 16:07

TL;DR The expression in the paper is correct. It's just written weirdly.

Take a look at where the "=" is in your index equation. You'll notice that your $\dot{\mathbf{S}}$ and your spin terms are on opposite sides, so you should expect it to look a little goofy. Here's the algebra (in direct notation since I hate indices):

Let $\mathbf{\epsilon}' = \mathbf{\epsilon} - \frac{1}{3}\mathrm{tr}(\epsilon) \mathbf{1}$

$$ \dot{\mathbf{S}} = 2\mu\,\dot\epsilon' + \mathbf{SW}^T + \mathbf{WS} $$

$$ \dot{\mathbf{S}} - \mathbf{SW}^T - \mathbf{WS} = 2\mu\,\dot\epsilon' $$

Recall that $\mathbf{W} = -\mathbf{W}^T$. If you didn't know this, you should read an introductory textbook on continuum mechanics. It should clarify why this is the case.

Thus, the second line can be rewritten as

$$ \dot{\mathbf{S}} + \mathbf{SW} - \mathbf{WS} = 2\mu\,\dot\epsilon' $$

and you get your Jaumann deviatoric stress rate back.

  • $\begingroup$ Thanks for the answer, please point out if I am being slow but shouldn't the above be: [ $\overset{\bigtriangleup}{{S}} = {\dot{S}} +{S} \cdot {w'} +{w} \cdot {S}$ ] [ $\overset{\bigtriangleup}{{S}} - {S} \cdot {w'} - {w} \cdot {S} = {\dot{S}}$ ] As $W = -W^T$: [ $\overset{\bigtriangleup}{{S}} + {S} \cdot {w} - {w} \cdot {S} = {\dot{S}}$ ] Which isn't the same as: [ ${\dot{S}} + {S} \cdot {w} - {w} \cdot {S} = \overset{\bigtriangleup}{{S}}$] $\endgroup$ Commented Aug 6, 2015 at 11:02
  • $\begingroup$ No. If you look at the index equation from the paper, you'll notice that dS/dt is on the left hand side alone. By moving the spin terms back to the left-hand side, you get back your full jaumann rate on the LHS. It's important to realize that $dS/dt = \dot{S}$. $\endgroup$ Commented Aug 6, 2015 at 11:12
  • $\begingroup$ In continuum mechanics, when you write out a constitutive equation as a rate, you must make sure that the constitutive law transforms properly under a rigid change in frame. I won't go into details here, since it's a little beyond the power of the comments section, but this results in the requirement that you write constitutive equations as $objectiveRate = isotropicFunc(variables$. This requirement is called "material frame indifference," if you'd like to look it up. $\endgroup$ Commented Aug 6, 2015 at 11:23
  • $\begingroup$ Thank you for the comments. I need to look into this more as I am clearly missing something. I thought we were looking for the objective rate of stress ($\overset{\bigtriangleup}{{S}}$) and we found that by a combination of the non objective stress rate (${\dot{S}} = 2\mu\dot{{\epsilon'}}$) and a product of the rotation tensor and the stress. I'll upvote when I have the rep. $\endgroup$ Commented Aug 6, 2015 at 11:33
  • $\begingroup$ O sorry for the confusion. $\dot{S}$ is always $dS/dt$. It is just that constitutive laws must be written in terms of an objective rate, so that is where the extra terms come in. These terms don't show up in small-deformation elasticity, so you may not have seen them before. I highly recommend "The Mechanics and Thermodynamics of Continua" if you're looking for a good textbook on the subject. It has a very thorough treatment of constitutive theory, and has a bit of a solid mechanics tilt to it since it was written by Prof Anand at MIT. $\endgroup$ Commented Aug 6, 2015 at 11:39

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