Challenging Cauchy's Stress Tensor: Objectivity and Generalization of Divergence Theorem

I'm investigating the limitations of the Cauchy stress tensor model in classical continuum mechanics, specifically focusing on its compliance with the principle of material frame indifference (MFI) and exploring potential generalizations beyond its inherent assumptions. A key concern lies in simplifying a general control volume to an infinitesimal orthogonal element (i.e., cuboid) when deriving the local form of the momentum balance equations. This simplification may not accurately represent the complex, multi-dimensional nature of the stress field, leading me to believe that the Cauchy stress tensor might be a linearization or approximation of a more intricate reality.

My primary questions revolve around the objectivity and mathematical representation of stress:

• Can we establish the objectivity of the Cauchy stress tensor without simplifying the control volume, thereby demonstrating its independence from the observer's reference frame?
• Do generalizations of the divergence theorem exist that can effectively handle the stress vector, which, from a differential geometry perspective, maps from a five-dimensional space (position and plane orientation) to a three-dimensional space (force components)?

To illustrate the challenge, let's consider the integral form of the mass conservation law:

$$\frac{d}{dt} \int_{\forall} \rho(\vec{r}, t) d\forall = -\oint_{\partial \forall} \rho(\vec{r}, t) \, \vec{v}(\vec{r}, t) \cdot d\vec{A}, \tag{1}$$

where:

• $$\forall$$ represents the control volume, a fixed region in space.
• $$\partial \forall$$ denotes the closed surface boundary of the control volume.
• $$\rho(\vec{r}, t)$$ is the material density at position $$\vec{r}$$ and time $$t$$.
• $$\vec{v}(\vec{r}, t)$$ is the material velocity at position $$\vec{r}$$ and time $$t$$.
• $$d\forall$$ represents a differential volume element.
• $$d\vec{A}$$ represents a differential area vector pointing outward from the control surface

The equation expresses that the mass change rate within the control volume is balanced by the net flux of mass across its boundaries. We can readily convert this integral form into its corresponding partial differential equation (PDE) form using the Gauss-Ostrogradsky divergence theorem, which allows us to transform the surface integral of the mass flux into a volume integral of the divergence of the mass flux density:

$$\frac{\partial \rho(\vec{r}, t)}{\partial t} + \nabla \cdot (\rho(\vec{r}, t) \, \vec{v}(\vec{r}, t)) = 0, \tag{2}$$

This PDE, known as the continuity equation, describes the local mass balance at every point within the continuum.

However, a similar conversion for the integral form of the momentum conservation law proves more challenging:

$$\begin{split} \frac{\partial}{\partial t} \int_{\forall} \rho(\vec{r}, t) \vec{v}(\vec{r}, t) d\forall = \oint_{\partial \forall} \vec{\sigma}(\vec{r}, t, \hat{n}) dA \\ + \int_{\forall} \vec{b}(\vec{r}, t) d\forall - \oint_{\partial \forall} \rho(\vec{r}, t) \vec{v}(\vec{r}, t) (\vec{v}(\vec{r}, t) \cdot d\vec{A}), \end{split} \tag{3}$$

where:

• $$\vec{\sigma}(\vec{r}, t, \hat{n})$$ is the stress vector acting on a plane with unit normal $$\hat{n}$$ at position $$\vec{r}$$ and time $$t$$.
• $$\vec{b}(\vec{r}, t)$$ is the body force per unit volume acting at position $$\vec{r}$$ and time $$t$$.
• $$dA$$ represents a differential area element on the control surface.

The key issue arises from the stress vector's dependence on both position and surface orientation, rendering standard integral transformation theorems like the divergence theorem inadequate. Cauchy's approach circumvents this issue by simplifying the control volume to a cuboid, effectively eliminating the dependence on surface orientation. However, this simplification might not accurately reflect the true nature of stress in materials, motivating the exploration of alternative formulations and generalizations that can address this challenge.

I'm interested in exploring alternative ways to understand and represent stress that accounts for its full dimensionality and potential non-linear dependencies on the surface normal. Any insights or suggestions for relevant mathematical tools and theoretical frameworks would be greatly appreciated.

• I'm not sure if I grasp your concern. The fact that the stress tensor $\sigma_{ij}$ is defined as the $i$-th component over the surface with $j$-th normal vector doesn't requieres any assumpion. You are integrating any shape you like. It's like saying that $\mathrm{d}x\mathrm{d}y$ requieres some assumpion on the volume of integration since it's a paralelogram. $\iiint \partial_{j} \sigma_{ij} \, \mathrm{d}V = \iint \sigma_{ij} n_{j} \, \mathrm{d}A$ holds always. Commented May 3 at 2:31
• @Gilgamesh, I'm hesitant to employ Einstein notation because it doesn't help with my confusion between tensors and matrices. My concerns essentially come down to a few fundamental questions. For instance, we applied the divergence theorem to transform Equation 1 from its integral form to Equation 2, the PDE form. I wonder if there are any generalizations of the divergence theorem that could similarly transform Equation 3?
Commented May 3 at 8:04
• @Gilgamesh Secondly, I remain unconvinced that Cauchy's stress tensor model is the most comprehensive representation of a continuum. Simplifying the general control volume to an infinitesimal cuboid results in a loss of information. To verify this, I would like to test whether Cauchy's model adheres to the principle of material frame indifference (MFI). If it does, then Cauchy's tensor should be an objectivity field. Consequently, starting from two different frames of reference, such as two rotated Cartesian coordinate systems, should yield identical results.
Commented May 3 at 8:12
• 1. The PDE of (3) is known as Cauchys's momentum equation which can be derived with the divergence theorem (you don't need a generalization) and Reynold's transport theorem (which is essentially Leibniz rule for diff under int). I think your LHS needs a time derivative. Commented May 3 at 15:54
• 2. It's the same than considering $\mathrm{d}A$ as a paralelogram (even when you integrate over a sphere, for example). The fact that is infinitesimal and the jacobian allows you to integrate any shape, not just square macroscopic objects. 3. What does is mean "yield identical results"? The stress-tensor will transform as a tensor, but the EOM will be covariant (same functional form). Commented May 3 at 16:04

There are a lot of questions here. Firstly, a good framework for multi-dimensional integration is differential forms. You have asked about generalization of Gauss theorem: Generalized Stokes Theorem

One of the things you will find is that integrals of kind: $$\int_V d^3 r \,\rho\mathbf{v}$$ Are problematic since you are trying to integrate a vector field over space, yet the basis of that vector field may itself be space-dependent, so it is not clear whether this expression always makes sense. In flat space it can make sense if you use Cartesian basis, in general probably not. I am saying this because expressions such as this, do not fit nicely into differential forms formalism, precisely for the reason I outlined.

One can make it awfully complicated by trying to project vectors along tangents of congruences of curves, lets make it simpler. We are in flat space, so let us use the Cartesian basis which is coordinate independent (unlike spherical basis, for example).

Next, lets move to differential forms, since a lot of things there become explicit. We can talk about density three-form $$d^3\rho$$ and density as a function that allows one to connect density three-form to volume three-form:

$$d^3\rho=\rho\,dx\wedge dy\wedge dz$$

You will need something similar for momentum density along every basis vector, lets just do it for $$x$$:

$$d^3\pi^{x}=\pi^x\,dx\wedge dy\wedge dz=\rho v^{x}\,dx\wedge dy\wedge dz$$

and the body force density: $$d^3 b^{x}=b^{x}\,dx\wedge dy\wedge dz$$

Now you can introduce the two-form of stress along x-axis: $$d^2\sigma^x=\sigma^{x}_{ij}\,dx^{i}\wedge dx^{j}=\sigma^{x}_{xy}\,dx\wedge dy+\sigma^{x}_{xz}\,dx\wedge dz+\dots$$

Now we can talk about integrals over two-dimensional surface $$\Sigma$$ in 3d space:

$$\int_\Sigma \sigma^{x}_{ij}\,dx^{i}\wedge dx^{j}$$

The whole point of differential forms is to define how to do it. Briefly, 2d surface is broken into small enough triangles and then integral of each of those is defined (you pull-back your form into space where each triangle is in a canonical form and you can do simple integration).

Then there is the final term, which is about the surface integral that characterizes influx. The conservation for the three-form $$\rho \,dx\wedge dy\wedge dz$$ can be stated as:

$$\partial_t \left(\rho\,dx\wedge dy\wedge dz\right) +\mathbf{d}\left(\rho\star\mathbf{v}\right)=0$$

Where $$\mathbf{v}=v^x dx+v^y dy+v^z dz$$ is the result of applying the musical isomorphism to the velocity vector (making it 'down-stairs'), $$\star\mathbf{v}=v^x\,dy\wedge dz+\dots$$ is the Hodge star and $$\mathbf{d}$$ is the exterior derivative for which Generalized Stokes theorem is defined.

So then your last term for the momentum density becomes:

$$\int_V \mathbf{d}\left(\pi^{x}\star\mathbf{v}\right)=\oint_{\partial V} \pi^{x}\star\mathbf{v}$$

Pulling it all together:

$$\frac{d}{dt}\int_V \pi^{x}\,dx\wedge dy\wedge dz=\oint_{\partial V} \sigma^x_{ij}dx^i\wedge dx^j+\int_V b^x dx\wedge dy\wedge dz-\oint_{\partial V} \pi^{x}\star \mathbf{v}$$

NB! Note that when you write using differential forms then things like $$\rho$$ can be scalars and things like $$\sigma$$ can be tensors. The density transformations are handled by $$dx\wedge\dots$$ parts, if you write it the way you did, you need to keep track of appropriate Jacobians all the time

• Thanks a lot for being kind and posting this answer. In the meantime I'm trying to learn what differential forms are. They are not easy to swallow.