Since stress is represented by a tensor, it should be invariant under a change in coordinate system. For something like a position vector $\mathbf{r}_{P/O}$ which is also invariant under a change in coordinate system, it's intuitive that it should be invariant since the arrow $\mathbf{r}_{P/O}$ should point from $O$ to $P$ regardless of what coordinate system we use. But for a second order tensor like stress, I find it less intuitive. When we do a stress transformation by rotating the coordinate system, what exactly is staying invariant?

Moreover, when we analyze the stress transformation of a material, we often come up with a diagram like the one below:

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What I notice is that when we change the coordinate system, we are working with a different cube of material. The cube on the left isn't exactly the same cube of material on the right since for the cube on the right, the material was sliced at an angle. So why should anything stay invariant when we change the coordinate system?


2 Answers 2


Nature doesn't care which way we draw our arrows and planes, so some aspects of the stress tensor must be invariant.

For example, so-called hydrostatic or equitriaxial stress (when the stress is equal in all directions; the negative version is called pressure) isn't tied to any coordinate system and therefore must be related to some type of invariance.

Whether a material breaks is independent of what we label as $x$, $y$, and $z$; there must be some characteristic stress threshold that's being exceeded, and this must also be related to some invariant quantity calculated from the 9-element stress tensor.

So far, though, the answer seems to be tautological: the invariants must be related to physical phenomena that are indifferent to the choice of coordinate system orientation. Well, we knew this already.

In fact, for the stress tensor $\boldsymbol{\sigma}$, there are three invariants (variations and combinations of these are also invariant):

  • the trace $\mathrm{tr}(\boldsymbol{\sigma})$,

  • $\frac{1}{2}\left[\mathrm{tr}(\boldsymbol{\sigma})^2-\mathrm{tr}(\boldsymbol{\sigma}^2)\right]$, and

  • the determinant $\mathrm{det}(\boldsymbol{\sigma})$.

These invariants are obtained from the characteristic equation of the eigenvalue problem for the stress matrix.

If the stress tensor is rotated to eliminate shear components and leave diagonal normal stresses $\sigma_1$, $\sigma_2$, and $\sigma_3$ only (called the principal stresses, themselves the eigenvalues of the stress matrix), then the three invariants above can be expressed as

  • $\sigma_1+\sigma_2+\sigma_3$,
  • $\sigma_1\sigma_2+\sigma_1\sigma_3+\sigma_2\sigma_3$, and
  • $\sigma_1\sigma_2\sigma_3$.

We can link these with the invariant characteristics discussed above:

  • The trace divided by -3 is the hydrostatic pressure (for the typical convention that tensile stresses are positive).

  • One widely used failure theory for ductile crystals (the von Mises failure criterion) predicts failure when a certain deviatoric stress causes slip. The critical value is $\sqrt{3}$ times the second invariant of the stress tensor with the hydrostatic stress removed (called the deviatoric stress tensor). The interpretation here is that hydrostatic stress can't really damage solids—it just temporarily squeezes the atoms closer together, whereas deviatoric or nonhydrostatic stress can cause permanent rearrangement.

  • Continuing with material plasticity, the so-called Lode angle, for instance, incorporates the second and third invariants of the deviatoric stress tensor.

  • The third invariant of the original stress tensor provides an indication of how three-dimensional the principle stress state is (i.e., it vanishes for 1D or 2D loading).

You asked about the rotating cubes: don't they contain different regions of the material? No; they are infinitesimal and contain nothing. They represent the stress state of the material at a single point. Put another way, they are to be drawn sufficiently small that it doesn't matter what they encompass.

  • $\begingroup$ Thanks for the insightful answer. I have one last (related) question if you don't mind. Is the role of the stress tensor is so that we can compute t=𝝈n to tell us the force per unit area on any surface around a point? $\endgroup$ Jan 21, 2022 at 4:12
  • $\begingroup$ That's one of the useful roles, sure. $\endgroup$ Jan 21, 2022 at 6:15

It's a tensor. It's not invariant, it's covariant. So, the representation of the stress tensor will change under a change of basis (the obvious example would be to just permute $x, y, z$), but the thing that IS invariaint is some product of the stress tensor with something else like ${\vec V}\cdot {\bf \sigma}\cdot {\vec W}$, where, obviously, the vectors will end up rotating/rescaling in a way that exactly cancels the changes in the components of the stress tensor.


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