Cauchy's Stress Theorem

Here is a simple explanation of the proof for Cauchy's Stress Theorem. It is easiest to begin in 2D. Consider a triangle surface element within a motionless fluid in 2D. It looks like this:

Note that in the above, we have accidentally reversed the indices of $\sigma_{ij}$.

We know that the in-fluid stresses in the normal directions are given by the stress tensor, with $\sigma_{ij}$ being the force per unit area in the ith direction on a surface whose normal points in the jth direction. Within our motionless fluid, all the forces are balanced, so for example, we can write in the $x_1$ direction: \[ F_1 \delta s = \sigma_{11} \delta x_2 + \sigma_{12} \delta x_1. \]

The left hand side gives you the force in the $x_1$th direction on the slanted face with length $\delta s$ (the reason why you multiply by $\delta s$ is that $F_1$ is force per unit area, and where we assume the length in the coordinate direction out of the page is unity). The right hand side gives you the force in the $x_1$th direction on the other two faces.

We can write this as \begin{align*} F_1 \delta s &= \left(\sigma_{11} \frac{\delta x_2}{\delta s} + \sigma_{12} \frac{\delta x_1}{\delta s}\right) \delta s \\ &= \left(\sigma_{11} \cos\theta + \sigma_{12} \sin\theta \right) \delta s \\ &= \sigma_{1j}n_j \delta s. \end{align*}

The $x_2$ direction is done similarly. In other words, assuming no other forces or fluid movement, then the stress tensor is related to the forces (per unit area) by \[ F_i = \sigma_{ij} n_j. \]

Note that in your lecture notes, you use $t_i = F_i$ for the force. Now bringing in fluid motion doesn't really change much. We assume that our little triangle is moving at a velocity $\mb{u}$, and we may have body forces $\mb{g}$. Also, we assume that we may not necessarily have $F_i = \sigma_{ij} n_j$, but the forces may be unbalanced. Newton's law for the ith direction implies that \[ \Bigl[ \rho \frac{D \mb{u}_i}{Dt} \Bigr] \delta V = \Bigl[ \rho \mb{g}_i\Bigr] \delta V + \Bigl[ F_i - \sigma_{ij} n_j\Bigr] \delta S. \]

However, if $L$ is a representative length of the triangle, then the `volume' (in this case, area) $\delta V = O(L^2)$ and $\delta S = O(L)$. In other words, as $L \to 0$, any contributions to the forces due to acceleration and body forces are minor, and the leading order balance requires that \[ F_i = \sigma_{ij} n_j. \]

The proof for the three-dimensional case is almost exactly the same. Instead of considering a triangle, you now consider a tetrahedron. The only tricky bit is to show that if the slanted face has normal $\mb{n}$ and area $\delta S$, then the other faces along the coordinate axes have area $\mb{n}_j \delta S$. This is not quite trivial to prove (you can draw some pictures to test), but to prove it, you can apply the divergence theorem to the tetrahedron and the vector field $\mb{G} = \mb{e}_j$.