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When presenting the stress tensor (say in a non-relativistic context), it is shown to be a tensor in the sense that it is a linear vector transformation: it operates on a vector $n$ (the normal to a surface), and returns a vector $t_n$ which is the traction vector. It is then shown that conservation of angular momentum leads to symmetry of the matrix.

However, tensors are more more naturally presented a multilinear functions. I wonder:

  • What type of tensor is the Cauchy stress tensor? Is $n$ a vector or a co-vector? What about $t_n$?
  • Is there a way to understand the symmetry when thinking of the stress tensor as a function of two vectors (or two co-vectors), under which it will seems intuitive why $\sigma(A,B) = \sigma(B,A)$?

Edit: To clarify, let's look, for example, at the 1st coordinate of the traction vector $t_n$ of an arbitrary normal $n$: This is $\left<e_1, \sigma(n)\right>$. From symmetry, this is equivalent to $\left<n, \sigma(e_1)\right>$ - the inner product of $n$ with the traction vector of a surface orthogonal to $e_1$. Mathematically, I understand why this is correct. But is there any intuitive meaning as to why these two quantities are the same?

Qmechanic
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R S
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2 Answers2

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Speaking about Cauchy stress tensor in classical mechanics, the answer to your first question is that it does not matter, as you have metric in arbitrary coordinates induced by dot product of underlying Euclidian space.

You can exploit symmetry of Cauchy stress tensor from balance of angular momentum assuming no couple-stresses, i.e. sources of angular momentum. Proof is often proceed by testing traction $\sigma_{ij} n_j$ by to-some-extent-arbitrary field $\phi_i$ and using balances of linear and angular momentum. For example you can examine $\epsilon_{imn}x_n \sigma_{ij} n_j$ as done in the Euler-Cauchy Stress Principle article in Wikipedia. So you can apply $\sigma(A, B)$ to $n$ and whatever quantity so it has meaning multiplying it by force. I don't think that some intuitive statement can be brought into form $\sigma(A,B)=\sigma(B,A)$.

On the other hand in Gurtin, M. E.: An Introduction to Continuum Mechanics (proof of Cauchy's theorem, chapter V, section 14) traction is tested by infinitesimal rigid displacement and balance of angular and linear momentum (in form of virtual work theorem with aforementioned displacement) is used. Then after some manipulation you arrive at $\sigma_{ij}W_{ij}=0$ for all $W$ skew where $W$ is deformation gradient of this displacement.

To make it more intuitive just consider that $\sigma_{ij}\mathbf{v}_{i,j}$ is work done by internal surface forces. No such a work can be done with rigid motions when $\mathbf{v}_{i,j}$ is skew. Hence $\sigma$ must be symmetric.

Emilio Pisanty
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Jan Blechta
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The answer to your first question is that all possible interpretations are OK. In classical continuum mechanics the euclidean metric lets you unambiguously translate between vectors and covectors, so both $n$ and $t_n$ can be either. The choice of type for the "input" and "output" then determines the type of tensor the stress-energy will be.

In more general settings this also holds: talking about tensor symmetry requires specification of a metric, in which case vectors and covectors can be unambiguously identified. If there is no metric present, the only way to speak of tensor symmetry is to view tensors as multilinear forms.

Normally you would choose both $n$ and $t$ (let me drop the ${}_n$) to be vectors, in which case the tensor identity $t=Sn$ reads componentwise as $t^i=S^i_{\phantom{i}j}n^j$, and $S$ should be seen as a linear operator from $E^3$ to itself, and is symmetric in the sense that $\langle u,Sv\rangle=\langle Su,v\rangle$ for all $u,v\in E^3$, where $\langle\cdot,\cdot\rangle$ is the euclidean metric in $E^3$.

This type of symmetry does tend to make some people nervous, though. In this case it can help to take $n$ to be a vector and $t=Sn$ to be a covector, in which case $t_i=S_{ij}n^j$, and the symmetry of $S$ is componentwise $S_{ij}=S_{ji}$. Here $S$ should be seen as a symmetric bilinear form $S(v,n)=S(n,v)$, given by $S(v,n)=(Sn)(v)=t(v)=t_iv^i=v^iS_{ij}n^j$.

Thus the bottom line is that the symmetry is of any type you want it to be, as long as you choose the input and output types of (co)vector appropriately.

Emilio Pisanty
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