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In classical continuum mechanics, the stress tensor is said to be of type/valence (1,1) and I do not see why.

If I am correct, its maps a vector $n$ defined in $\mathbb{R}^3$ (which is the normal to a given infinitesimal surface) to another vector $t$ (which is a force) also defined in $\mathbb{R}^3$. We can then consider a second direction $m$ defined in $\mathbb{R}^3$ so that the usual scalar product $m^\top\cdot t$ provides a real number (which is physically the projection of the force $t$ on direction $m$, ie a linear combination of the stress components).

Coming back to the definition of a tensor $T$ of valence (1,1), it takes two vectors $v\in V$ and $w\in V^*$ ($V^*$ being the dual of $V$) such that $T(w,v)\in\mathbb{R}$. Where I am confused in the above example comes from the fact that I do not really see $m$ as the dual of $n$ but it is probably the case through the usual scalar product I guess. Am I correct?

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There's a construction on finite-dimensional, real inner product spaces relating rank 2 tensors and linear operators that I think answers your question. The construction essentially shows how to obtain a rank two tensor from a linear operator. Since $\mathbb R^3$ with the standard inner product is a real inner product space, and since the general construction isn't so hard, we might as well be more general.

Inner product space construction.

Let a finite-dimensional real inner product space $(V,\langle\cdot,\cdot\rangle)$ be given, then we claim that there is a way to associate a tensor of type $(0,2)$ to every linear transformation on $V$. In addition, if we let $T^k_l(V)$ denote the vector space of all tensors of type $(k,l)$ on $V$, then one has the following isomorphisms: \begin{align} T^2_0(V) \cong T^1_1(V) \cong T^0_2(V) \end{align} In fact, in index notation, the isomorphisms written here just correspond to appropriately raising and lower indices. Combining these observations, we see that each linear operator on $V$ gives rise to three tensors of types $(2,0)$, $(1,1)$, and $(0,2)$, and they're all essentially the same object. Here's the nitty gritty of how you associate a tensor of type $(0,2)$ to a linear operator. For each linear operator $L$ on $V$, we can define a mapping $T:V\times V\to \mathbb R$ as follows: \begin{align} T_L(v,w) = \langle v,L(w)\rangle \end{align} I claim that $T$ is a type $(0,2)$ tensor. Recall that a type $(k,l)$ tensor $S$ is a multilinear function $S:(V^*)^k\times V^l\to\mathbb R$. Therefore, it suffices to show that the function $T_L$ defined above is bilinear. Linearity in its first argument follows immediately from linearity in the first argument of the inner product, while linearity in its second argument follows from linearity in the second argument of the inner product combined with linearity of $L$.

To see that this coincides with the index notation we often use as physicists, let $\{e_1,e_2,\dots, e_n\}$ denote an orthonormal basis for $V$, then the components of the tensor $T_L$ are \begin{align} (T_L)_{ij} = T_L(e_i, e_j) = \langle e_i,T(e_j)\rangle = \langle e_i,L_{kj}e_k\rangle = L_{kj}\delta_{ik} = L_{ij} \end{align} We see that the components of the tensor that we defined correspond precisely to those of the linear transformation.

Contact with your notation.

If we denote the linear operator that you refer to as the stress tensor with the letter $L$, and if we define the corresponding bilinear map $T_L$, as above, then using your notation for the vectors, the equation I used to define $T_L$ would be written as follows: \begin{align} T_L(\mathbf m, \mathbf n) = \mathbf m^t L(\mathbf n) \end{align} The right hand side is the same as $\langle \mathbf m, L(\mathbf n)\rangle$ since the standard inner product on $\mathbb R^3$ is simply given by \begin{align} \langle \mathbf v, \mathbf w\rangle = \mathbf v^t\mathbf w \end{align} Incidentally, in light of this fact notice that transposition effectively associates a dual vector to each vector in $\mathbb R^3$.

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  • $\begingroup$ Thanks. Are you essentially saying that a bilinear function can always be seen as the composition of a linear function and an inner product in finite-dimensional spaces? And in your derivations, where are the stress tensor, and the two directions $\bf n$ and $\bf m$? $\endgroup$ – pluton Sep 3 '13 at 19:19
  • $\begingroup$ @pluton See my edits at the end; hopefully that makes it more clear. $\endgroup$ – joshphysics Sep 3 '13 at 20:36
  • $\begingroup$ It seems to me that the transpose operator makes things quite unclear. $\endgroup$ – pluton Sep 3 '13 at 23:31
  • $\begingroup$ @pluton You mean the general use of the transpose in writing the Euclidean inner product? I personally kind of like it. $\endgroup$ – joshphysics Sep 3 '13 at 23:33
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    $\begingroup$ @pluton $\mathbf m$ and $\mathbf n$ are both in $V = \mathbb R^3$. However, if you define the mapping $f_{\mathbf m}:V\to \mathbb R$ by $f_{\mathbf m}(\mathbf n) = \mathbf m^t \mathbf n$, the linear functional $f_{\mathbf m}$ is an element of $V^*$; it is in this sense that the transpose associates a dual vector to each vector. Also, yes $L$ only has one argument; the construction shows that you construct a $(0,2)$ tensor out of $L$, which has two arguments. Then, the equivalence of $(1,1)$ and $(0,2)$ tensors mentioned at the start gives you what you want. $\endgroup$ – joshphysics Sep 4 '13 at 0:26

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