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Recently, my undergraduate thesis advisor gave me a formula for the Cauchy stress tensor in fluid mechanics: $$\boldsymbol{\sigma}=-P\mathbf{I}+\mu\big(\nabla\boldsymbol u+(\nabla\boldsymbol u)^{\intercal}\big)$$ However, I recall that the Cauchy stress tensor is supposed to be second order covariant, having shape $(0,2)$. But, the objects on the right hand side all have shape $(1,1)$. The distinction is irrelevant when working in rectangular coordinates, but in general, one needs to be careful. I am wondering if the correct formula involves raising or lowering an index on one side or the other? Like, would we compute $$\sigma^i_j=-P\delta^i_j+\mu\big((\nabla u)^i_j+(\nabla u)^j_i\big)$$ And then index lower the result, $$\sigma_{ij}=g_{ik}\sigma^k_j$$ Or perhaps instead of this, the correct formula is actually $$\sigma_{ij}=-P g_{ik}\delta^k_j+\mu\big(g_{ik}(\nabla u)^k_j+g_{kj}(\nabla u)^k_i\big)\\ -P g_{ij}+\mu\big(g_{ik}(\nabla u)^k_j+g_{kj}(\nabla u)^k_i\big)$$ I hope somebody can clear this up for me.

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You are on the right track, but it would be easier if you combined the second term into a single tensor (the rate-of-strain tensor) rather than treating it as multiple terms.

A good reference on this subject is Rutherford Aris's book titled "Vectors, Tensors, and the Basic Equations of Fluid Mechanics". It is available in paperback and is my go-to reference for things like this. Page 180 gives the general equation for the contravariant components of the stress tensor of a Newtonian fluid as

$$ \sigma^{ij} = ( - p + \lambda e^m{}_m ) g^{ij} + 2 \mu e^{ij} \,, $$

where the covariant components of the rate-of-strain tensor are defined as

$$ e_{ij} \equiv \tfrac{1}{2} ( u_{j,i} + u_{i,j} ) \,. $$

Here the commas just refer to the covariant derivative. It seems that in your case, the flow is incompressible, so $e^m{}_m = 0$, and the 2nd term drops out leaving

$$ \sigma^{ij} = - p g^{ij} + 2 \mu e^{ij} \,, $$

This basically is your answer but written in terms of the contravariant components and using the covariant derivative.

Unfortunately most books either use vector notation or use Cartesian index notation (like G. K. Batchelor's "Introduction to Fluid Mechanics" or Landau/Lifshitz's book). Aris's book is one of the few that I know that uses general tensor notation, so it is a good reference for this topic.

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  • $\begingroup$ Thank you, Andrew. My professor recommend the same text to me but I haven't had a chance to pick it up from the library yet. $\endgroup$
    – K.defaoite
    Oct 11, 2021 at 17:37
  • $\begingroup$ Coming back to this, we should have $e^{m}{}_{m}=\nabla \cdot \boldsymbol{u}$, right? $\endgroup$
    – K.defaoite
    Oct 13, 2021 at 16:34
  • $\begingroup$ Yes, $e^m{}_m$ is the divergence of the velocity (the dilatation rate). $\endgroup$ Oct 15, 2021 at 0:03
  • $\begingroup$ I would like to just clear up some terminology. We basically define the viscous stress tensor as the part of the stress that isn't from the pressure, right? So $\tau_{ij}=\sigma_{ij}+pg_{ij}=\lambda e^k{}_kg_{ij}+2\mu e_{ij}$ ? Or is the viscous stress tensor simply the non isotropic part, $2\mu e_{ij}$? $\endgroup$
    – K.defaoite
    Nov 25, 2022 at 11:09
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    $\begingroup$ Yes, you are correct that the viscous stress tensor still has diagonal components. Unfortunately I misremembered when writing my previous post. You bring up an interesting point about how "shear stress" is a misnomer here. "Deviatoric stress" is a more physically accurate term. The division in the constitutive equation is between an isotropic part (pressure) and an anisotropic part (viscosity), and as you correctly point out viscosity can act in the normal directions, so the term "deviatoric stress" is more accurate here. $\endgroup$ Nov 27, 2022 at 13:57

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