Why are the x-components of the stress tensor ($\sigma_{xx}, \sigma_{yx}, \text{and } \sigma_{zx}$) different?

Question

I understand the stress tensor at a specific point $P$ as follows. You draw a cube with its center at $P$ and look at its faces. The face with its normal vector in the positive $x$ direction has a certain force acting on it. $\sigma_{xx}$ is the limit of the $x$-component of that force over the area of the face as the cube shrinks to size $0$. Similarly, $\sigma_{xy}$ is the limit of the $y$-component and so on.

Specifically, the force on a face is essentially the sum of the forces on all the particles on the face. If the force inside the material is roughly continuous (is that true for the insides of materials?) then as you take the limit as the cube shrinks to size $0$, shouldn't the force on all the particles on each face of the cube be essentially the same, thus leading to the forces on each face being the same and thus making, for example, $\sigma_{xx}=\sigma_{yx}=\sigma_{zx}$?

It seems to me somewhat obvious that all stress tensor components whose second element is $x$ should be the same (along with $y$ and $z$) but given how that is not stated in physics books that is definitely not the case.

Where does my logic go wrong? Also, could people give me some examples of where the forces on the different faces of the cube are clearly different even as the cube size gets extremely small? I would like to gain a better understanding of stresses.

Answer 1

I think that there is a little confusion here at several places. First, $\sigma_{xx}$ is the normal (not shear) force on the x-face of the tiny cube you mentioned. The corresponding normal force on the y-face of the cube is not $\sigma_{xy}$. $\sigma_{xy}$ is the shearing force on the x-plane in the y-direction (or, equivalently, the shearing force on the y-plane in the x-direction). The corresponding normal force on the y-face is $\sigma_{yy}$ and the corresponding normal force on the z-face is $\sigma_{zz}$. I believe that there are a total of six independent stress components in the stress tensor, 3 normal stress components and 3 shear stress components: $\sigma_{xx}$, $\sigma_{yy}$, $\sigma_{zz}$, $\sigma_{xy}$, $\sigma_{xz}$, and $\sigma_{yz}$.

Now as for your question of why the stress components aren't simply all the same as you shrink the imaginary cube to zero size, it's because these stress components have directionality and solids have strength. Suppose you have a large steel brick and place it into a huge mechanical press that puts an enormous compressive force on the opposite faces of the brick in the x-direction. Now imagine a small, tiny cubical volume element at the center of the steel cube. No matter how small of a tiny cubical volume element you consider, the compressive stress across the x-faces of that tiny imaginary cube is always going to be larger than the compressive stresses on the y- or z-faces of that same cube. (i.e., $\sigma_{xx}$>$\sigma_{yy}$ and $\sigma_{xx}$>$\sigma_{zz}$ no matter how small of a tiny, imaginary cube you are considering for stress analysis). Indeed, even in the limit in which the imaginary cube vanishes to zero size and you are considering the values of $\sigma_{xx}$ and $\sigma_{yy}$ and $\sigma_{zz}$ at the exact same point, $\sigma_{xx}$ will be greater than either $\sigma_{yy}$ or $\sigma_{zz}$. Why? Because they refer to different directions.

Answer 2

The stress tensor is a continuum mechanics concept, and you should expect it to break down (or become far more complicated) at the particle scale anyway.

However, a problem with your logic is that stress isn't really concerned with the net force on particles. Assume that your material is stationary: the net force on every single particle is zero, regardless of whether there is stress present or not. (I'm also not sure you're aware that $\sigma_{xy}=\sigma_{yx}$, nor that you appreciate $\vec\sigma[-\hat n]=-\vec\sigma[\hat n]$.)

Instead, stress is all about the relationship between adjacent particles (and how they respond to this deviating from their relaxed or "preferred" configuration). Think of a (stationary) material body that is subjected to large (static) external forces, and imagine surgically removing a cube (or other test volume) from the interior of this body. When you remove this cube, what forces would you have to apply to the cube's sides to keep it a cube (that is, to maintain it in exactly the same configuration as before you removed it from the bulk, and prevent it relaxing or "springing" back to some other shape)? Alternatively, what forces would you have to apply to the internal surfaces of the void (the hole left in the middle of the original body) to prevent it from deforming in response to removal of that piece of material?

A simple example (described in Landau & Lifshitz, and probably any other elasticity textbook) is a block of material in a vice, such that the stress tensor is homogeneous (uniform and constant, and the same for any sized cube) and $\sigma_{xx}$ is its only non-zero component (i.e. $\sigma_{xx}\neq\sigma_{yx}$).

Answer 3

I think an analogy is helpful here.

If you had some mass you could discuss the rate at which mass leaves a volume. It could have a mass density $\rho$ and a volume current density $\vec J$ and then the total mass in a region $R$ is $$M=\iiint_R\rho dV.$$

And if the region is fixed you get:

$$\frac{dM}{dt}=\frac{d}{dt}\int_R\rho dV=\int_R\frac{\partial \rho}{\partial t} dV.$$

Alternatively you can use current density $\vec J$

$$\frac{dM}{dt}=-\oint_{\partial R}\vec J\cdot \hat n dS =-\int_R\vec \nabla \cdot \vec J dV$$

And the important thing is that there are three different components to $\vec J$ specifying the direction of the flow of mass as well as the overall rate.

The same happens with charge, you can have a density of charge $\rho$ and a charge current density $\vec J$ and the current has three components specifying the direction of the flow of mass as well as the overall rate.

Now you can think of the stress tensor as three vectors. The first vector tells you the current of x-momentum. The second vector tells you the current of y-momentum. And the third vector tells you the current of z-momentum.

Of you take the divergence of each of those vectors you get a scalar field that tells you the rate at which the momentum density is changing at that point. There are three components of momentum. So three scalar fields telling you how each component changes in time. And each component has a current which has its own three components telling the direction as well as the overall rate of flow.

So basically the $x$ component of momentum $\vec p\cdot \hat x$ can flow in the $\hat x,$ $\hat y,$ or $\hat z$ directions and so we need three components of the stress to tell us about that flow. And we need three more to tell us how $\vec p\cdot \hat y$ can flow in the $\hat x,$ $\hat y,$ or $\hat z$ directions. And three more to tell us how $\vec p\cdot \hat z$ flows in the $\hat x,$ $\hat y,$ or $\hat z$ directions. So nine components are telling us that because we have a 3d flow of three things.

So really it's about how 3d vectors flow in 3d. And sure, momentum changes because of forces, so forces are the reason the momentum is flowing. But stepping back and nothing that a flow of momentum requires three vector fields tells you what is going on.