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(As I understand it ... qualifies every sentence in what follows).. a stress tensor is a rank 2 tensor that maps a unit vector normal to a surface to the stress (or traction) vector corresponding to that surface. A rank 2 tensor can be represented by a 3x3 matrix, and that matrix maps the components of the unit vector to the components of a stress (or traction) vector.

A rank 2 tensor can be written as a dyad, that is, the vector dyadic product of two vectors. Is there a geometric interpretation of the two vectors making up the dyad corresponding to the stress tensor?

Related question - the product of a dyad $UV$ and a vector $D$, say $UV$ dot $D$, corresponds to the matrix product of the matrix representing the dyad and the vector $D$, and is always a vector that equals $sU$ where $s$ is a scalar, so the stress (or traction) for any surface at a point always points in the same direction. T or F?

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  • $\begingroup$ Note that not all rank 2 tensors can be written as dyads, in the same way that only matrices of matrix rank 1 can be decomposed as a column vector times a row vector. Any rank 2 tensor in $\mathbb R^3$ can be decomposed as the sum of three dyads, but this is not generally unique and tends not to have physical meaning. $\endgroup$ – Emilio Pisanty Feb 23 '15 at 22:21
  • $\begingroup$ Thanks. Looks like I gotta back up. Can I define a tensor as an element of a vector product space? Note: there is a series of vids on YouTube 'What is a tensor' that seems to do just that. However, the answer must be no, as an element of a vector product space would be the the tensor product of two vectors. So ... what the heck is the definition of a 2nd order tensor? $\endgroup$ – DrWill Feb 23 '15 at 23:16
  • $\begingroup$ Can't edit previous comment so I'll add Definition 1 Let V be a vector space over the field F, and let V∗ be its dual space. A tensor on V is a multilinear map T:V∗×V∗×⋯×V∗×V×V×...×V→F. $\endgroup$ – DrWill Feb 23 '15 at 23:47
  • $\begingroup$ and the multilinear map is a product of linear maps on V and V* (?) and each linear map corresponds to a vector (?), so a rank 2 tensor does seem to me to correspond to product of vectors. $\endgroup$ – DrWill Feb 23 '15 at 23:56
  • $\begingroup$ Each linear map on $V^\ast$ corresponds to a vector on $V$. A rank 2 tensor is indeed (naturally isomorphic to) a linear map $T:V^\ast\times V\to F$. Dyads are also of this form: if $\phi\in V^\ast$ and $v\in V$, then $T=v\otimes \phi$ acts as $T(\psi,w)=\psi(v)\phi(w)$. However, not all $T$s are of this form, because this one is degenerate: there are a bunch of $w\in\mathrm{ker}(\phi)$ and a bunch of $\phi\in\mathrm{null}(v)$ (the nullifier of $v$) for which $T(\psi,w)=0$. A bunch of other tensors (i.e. the identity) don't have this property, so they can't be written this way. $\endgroup$ – Emilio Pisanty Feb 24 '15 at 10:58
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A rank-2 tensor is a linear combination of dyadic products, simply because the space of all such tensors is spanned by the dyadic products of the basis vectors of the underlying vector space. Each dyadic product is also known as a rank-1 operator, where rank here refers to the matrix rank rather than the order of the tensor. On inner product vector spaces they are usually denoted as $$\theta_{x,y}(z):=(y,z)x$$ but when the product is between a vector and a covector one can replace the inner product with the natural pairing between the vector space and its dual.

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