I'm going through Chorin and Marsden's derivation of the Navier-Stokes equations in A Mathematical Introduction to Fluid Mechanics. There are three assumptions made about the Cauchy stress tensor $\pmb\sigma$ (which I am paraphrasing):

1) $\pmb \sigma$ depends only on the velocity gradient $\nabla \mathbf{u}$. So written as a linear transformation $\pmb \sigma = \pmb \sigma (\nabla \mathbf{u})$

2) $\pmb \sigma$ is an isotropic tensor function (rotationally invariant) so that $$\pmb \sigma (U \ \nabla \mathbf u \ U^T)= U \ \pmb \sigma (\nabla \mathbf u) U^T$$ for any proper orthogonal matrix $U$.

3) $\pmb \sigma$ is symmetric.

After stating these assumptions, they conclude that because of 3), it follows from 1) and 2) that $\pmb \sigma$ can depend only on the symmetric part of $\nabla \mathbf u$. In other words, if you write $$\nabla \mathbf u = D +W$$ where $D=\frac 1 2 (\nabla \mathbf u + \nabla \mathbf u^T) $ (the symmetric part of $\nabla \mathbf u$) and $W = \frac 1 2 (\nabla \mathbf u - \nabla \mathbf u^T) $ (the antisymmetric part of $\nabla \mathbf u$), then $\pmb \sigma (W)= \mathbf 0$, i.e. $\pmb \sigma$ maps antisymmetric second order tensors to the zero second order tensor.

I am not sure how the above argument works and would appreciate some help in seeing it.

Note: I've seen arguments which use the fact that $\pmb \sigma$ must satisfy in component form

$$\sigma_{ij}= -p \delta_{ij} + C_{ijrs} u_{r,s}$$

where $p$ is the pressure, and it ends up that the fourth order tensor $\mathbf C$ must take a special form (because it is isotropic). From this special form, you can end up deducing that $\pmb \sigma$ maps antisymmetric second order tensors to the zero second order tensor.

I'm interested in seeing an argument that doesn't rely on the component analysis and also doesn't jump immediately to the special representation of isotropic second order tensor functions in terms of its principal invariants.


1 Answer 1


Since no one has answered this:

Assumption (1) is natural: The velocity gradient contains all relevant information pertaining to deformation in an Eulerian description of the continuum.

Assumption (3) reflects the fact that for "standard" continua, the symmetry of the Cauchy stress tensor is a proxy for the balance of angular momentum.

The fact that $\mathbf{\sigma}$ depends only on $sym(\mathbf{\nabla u})$ has nothing to do with material isotropy; I'm quite surprised to hear you say that a text by Marsden makes that claim.

Every 2nd order tensor like $L=\nabla u$ can be additively decomposed into symmetric and skew-symmetric parts. For $\nabla u$, these are the [symmetric] rate of deformation tensor $D$ and the [skew-symmetric] spin tensor $W$.

All information related to rates of change of size / shape are contained in $\mathbf{D}$. In fact one can show that the rate of change of squared length of a deforming vector element $\mathbf{dx}$ depends entirely on $\mathbf{D}$ $$ \frac{D}{Dt}(ds^2) = 2 \mathbf{dx \cdot {D} \, dx} $$ where $\frac{D}{Dt}$ is the usual material derviative.

If $\mathbf{D}$ is zero, the deformation is instantaneously a rigid-body deformation, and $\mathbf{W}$ is related to the spin rate.

Another way of seeing this is from is from the power balance or 1st law of thermodynamics: The stress power density is $\mathbf{\sigma : L}$; however, the spin tensor does not contribute to the stress power density at all, i.e. $\mathbf{\sigma : W} = 0$, so one can replace $\mathbf{\sigma : L}$ with $\mathbf{\sigma : D}$.

For further details, please see any text on continuum mechanics (rather than a fluids text), specifically the chapters dealing with theories of constitutive modeling and rates of deformation.


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