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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 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?

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Since you're talking about normal vectors, you're guaranteed to have a metric. When you have a metric, there isn't really an important distinction between vectors and covectors; you can use the metric to convert them back and forth in a natural way. –  Ben Crowell May 19 '13 at 22:25
    
While I can convert them when I have a metric, I still think there is a distinction in the way to think about them. This is particularly true when generalizing to SR or GR, where they don't have the same coordinates. –  R S May 19 '13 at 22:44
    
In some cases, the idea is that even if the stress-energy tensor you get by the definition $\frac{\delta S}{\delta g_{\mu\nu}}$ is not symmetric, you can add another piece to make it symmetric and still satisfy the necessary properties. eg: en.wikipedia.org/wiki/… –  Siva May 20 '13 at 2:57
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@Silva: The stress-tensor when derived from $\frac{\delta S}{\delta g_{ab}}$ is always symmetric (by definition), it's when you derived it as Noether current under translations that it may not be symmetric and need to be added a peculiar divergence (as explained in your link). –  Learning is a mess May 22 '13 at 8:28
    
Ben Crowell gives you the main answer : once you have a metric, You could always change, for instance, a 2_covariant tensor into a 2_contravariant tensor or a 1-covariant, 1_contravariant tensor, thanks to the metrics : $\sigma_{ij} = g_{ik} \sigma^k_j = g_{ik} g_{jl} \sigma^{kl} $ . Of course, the first and last expression of the tensor are best fit for expressing the symmetry of the tensor.Four your second point, saying that $\sigma(A,B) = \sigma(B,A) \, \forall B,A$ is exactly the same thing that saying that $\sigma$ is symmetric. So it does not bring you something new or interesting. –  Trimok May 28 '13 at 13:31

2 Answers 2

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.

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Thanks - the Gurtin explanation sounds promising. Can you write some more details, or alternatively link to the relevant parts of the book? –  R S May 28 '13 at 8:44
    
@RS: I wouldn't rather try to condense here book's content. Gurtin camouflages linear and angular momentum balance into virtual work theorem and then uses it in proof of symmetry. I improved a reference. Can you reach the book? –  Jan Blechta May 28 '13 at 14:54
    
I'll try. Also see the clarification to my question, if you haven't already. –  R S May 28 '13 at 18:17
    
That's only reformulation of symmetry. As I said I see very good physical self-contained reason in that $\sigma_{ij}\mathbf{v}_{i,j}$ is work done by internal surface forces and must be zero for rigid motions. I would say that this is intuitive. –  Jan Blechta May 28 '13 at 18:44

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.

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Thanks for the comment. I'm not sure this answers my question - I edited it for clarification. –  R S May 28 '13 at 8:36
    
-1 as this does not answer the second part of the question. It only explains what is a symmetry of tensor. –  Jan Blechta May 28 '13 at 14:27
    
@JanBlechta Thanks for the downvote. I find it nicely sportsmanlike to dismiss answers that did not anticipate future changes to the original question. –  Emilio Pisanty May 29 '13 at 10:41
    
You're welcome. As you only give intuitive physical reason of stress tensor symmetry I change my vote. –  Jan Blechta May 29 '13 at 11:36
    
@JanBlechta Welcome to physics.stackexchange. Downvoting works differently to what you may expect from other sites. People normally only downvote answers that are plainly not suitable to the question. Partial answers, or answers that have not yet adapted to changes in the question, are not usually downvoted. I commend you for your efforts in answering the OP's original and revised question, and would ask you to be a sportsman and let the community decide the relative merit of both answers. –  Emilio Pisanty May 29 '13 at 12:25

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