# Why are stresses of continuum systems described via a tensor?

The tittle pretty much says enough. I have always been told so but no one really motivated it.

So, I would like to know why do we use a tensor to describe the stresses in continuum mechanics.

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It is a quite famous theorem due to Cauchy.

Consider an internal portion $S$ of a continuous body $C$. There are two kinds of forces acting on it: Forces proportional to the mass, of the form $$\int_V \mu(x) \vec{f}(x) d^3x\tag{0}$$ where $\vec{f}(x)$ is the density of force acting on $x \in V$. And forces acting through the surface $\partial V$, the boundary of $V$, due to the remaining part of continuous body surrounding $V$. Those forces have the form $$\int_{\partial V} \vec{s}(x, \vec{n}(x)) dS(x)\:,$$ where $x \in \partial V$ and $\vec{n}(x)$ is the outward unit vector normal to $\partial V$ at $x$. The vector $\vec{s}(x,\vec{n})$ is the stress at $x$ in the direction $\vec{n}$. For a fixed $\vec{n}$, it is noting but a surface density of force at $x$. Actually, since $V$ is any portion of the continuous body $C$, $x\in C$ is a generic point and $\vec{n} \in \mathbb S^2$ a generic direction.

The above mentioned theorem due to Cauchy establishes that, under suitable (very mild) physical and mathematical hypotheses on $C$, there exists a symmetric (Cartesian) tensor field $C \ni x \mapsto \sigma_{ij}(x)$ such that, for every $x\in C$ and $\vec{n} \in \mathbb S^2$, $$s_i(x, \vec{n}) = \sum_{j=1}^3\sigma_{ij}(x) n_j\:.\tag{1}$$ $\sigma$ is called the stress tensor (field) of the continuous body.

The said physical hypotheses are just that only the two kinds of forces act on every portion of the continuous body and that the usual Newton's laws of mechanics hold for every portion of the body. Mathematical hypotheses instead simply concern the regularity of the involved functions.

It is important to stress that the result is valid for every type of continuous body, not necessarily elastic or plastic or fluid. In the first couple of cases, however, $\sigma$ is a function of the deformation of the body.

ADDENDUM. Cauchy's theorem proof is quite easy. One writes down "F=ma" for a sequence of portions $V_n \subset C$ of continuous body tending, for $n\to +\infty$, to a fixed common point $x\in \cap_n\partial V_n$ while preserving the direction $\vec{n}$ at $x$. In the considered limit the mass force as in (0) as well as acceleration disappear more quickly than surface forces. Hence the surface forces must have vanishing resultant. As one sees, this last fact is mathematically equivalent to say that the function $$\vec{n} \mapsto \vec{s}(x,\vec{n}) \tag{2}$$ extended by linearity to generic vectors $\vec{n}$ (thus also with $|\vec{n}|\neq 1$) is linear.

A well known theorem proves that any linear map is described by a tensor. Therefore there is a tensor $\sigma(x)$ describing (2) as in (1). Symmetry of $\sigma$ can subsequently (easily) be proved from the third law of Newtonian dynamics.

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Applying a force in the $x$-direction might change the shape of the material in the $y$-direction. The only way to capture such an effect is through a tensor.

If you have a general force acting on your body $$\vec F = (F_x, F_y, F_z)^T$$ and you are interested in the reaction of the body by looking at its deformation $$\vec \epsilon = (\epsilon_x, \epsilon_y, \epsilon_z)^T$$ the deformation in the x-direction $\epsilon_x$ should (in the Hook limit) depend linearly on the forces, same for the $\epsilon_y$ and $\epsilon_z$, i.e. $$\epsilon_x = E_{xx} F_x + E_{xy} F_y + E_{xz} F_z$$ and similarly $$\epsilon_y = E_{yx} F_x + E_{yy} F_y + E_{yz} F_z$$ $$\epsilon_z = E_{zx} F_x + E_{zy} F_y + E_{zz} F_z.$$

These three equations can be captured by a vector equation $$\vec \epsilon = \tilde E \vec F$$ or in components $$\epsilon_i = \sum_{k=1}^3 E_{ik} F_k$$

Of course, there can be the special case that all off-diagonal elements are zero, then you just recover $\epsilon_x = E_{xx} F_{x}$ and similar for $y, z$.

Finally, how do we know that $\tilde E$ is a tensor and not just a set of numbers? For this we need to look at transformation properties. The Force obviousely is a vector, i.e. under rotations (employing Einstein's convention) $$F_i \to R_{ij} F_j \quad \text{or} \quad \vec F \to \mathbf R \vec F$$ also $\vec \epsilon$ should behave as a proper vector $$\epsilon_i \to R_{ij} \epsilon_j \quad \text{or} \quad \vec \epsilon \to \mathbf R \vec \epsilon$$. Plugging this into our vector equation we find after rotation $$\mathbf R \vec \epsilon = \tilde E'\mathbf R \vec F.$$ In order for the right-hand side to be a vector it has to have the form $\mathbf R (\tilde E \vec F)$, which implies that under rotations $$\tilde E \to \tilde E' =\mathbf R \tilde E\mathbf R^{-1}$$ which is exactly the defining property of a tensor.

(note on notation, I used $\tilde{}$ to identify tensors and bold face for rotations matrices).

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Could you please develop a little more this statement? – Anarchist Birds Worship Fungus May 23 '14 at 12:27
@silvrfück Added to post to develop the statement "a little more" :P – Neuneck May 23 '14 at 12:58
thanks, that was good – Anarchist Birds Worship Fungus May 23 '14 at 22:10

Take a flat, rectangular material. Remaining 2D, you can apply forces at the edges in different ways. You can exert:

• a "squeezing" force along vertical edges, that will be the $xx$ component of the tensor,

• a "squeezing" force along horizontal edges, that will be the $yy$ component of the tensor,

• a shear force along the vertical edges, that will be the $xy$ component of the tensor,

• a shear force along the vertical edges, that will be the $yx$ component of the tensor.

In 3D, pairs of directions in ${x,y,z}$ give nine components. Thus you can see why you need $d^2$ components to describe the stress at any point.

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Stress has to be a tensor because it describes momentum flux.

Think about the analog of current $j_i$ and mass given by a density $\rho$. The component $j_i$ gives the component of the mass current in the $i$th direction. To fill out the analogy, we will mention the continuity equation $\frac{d\rho}{dt} = - \partial_i j_i$.

Now let's think about momentum. Like mass, it is conserved but can move from one place to another. There we can define a momentum density which I will just call $p_i$ and a momentum current. Now this momentum current will have two indices: the first specifying which component of momentum (because each component is conserved indiviually) and the other specifying which component of the current. So the second index is analogous to the single index of the current we had in the case of mass. The momentum current is denoted $\sigma_{ij}$. To continue the analogy we would expect a continuity equation $\frac{dp_i}{dt} = -\partial_j \sigma_{ij}$. In fact, because of a sign convention the continuity equation is $\frac{dp_i}{dt} = \partial_j \sigma_{ij}$, but it is the exact same idea.

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