3
$\begingroup$

The stress tensor relates the traction $\vec{t}$ (force per area) on a surface with surface normal $\vec{n}$ usually written as (when disregarding co- and contravariance) $$ t_j = \sigma_{ij} n_i.$$

Usually the stress tensor is described as a contravariant tensor. But since both force and surface normal are contravariant tensors (vectors) shouldn't the stress tensor be a mixed tensor of type (1,1) such that when considering covariance and contravariance it should read $$ t^j = \sigma_{\,i}^j n^i.$$

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Yes, but $g_{ij}= \delta_{ij}$... Indeed, it is by definition a mixed $(1,1)$ tensor, but working in Cartesian orthonormal coordinates you cannot see a difference in the use of different representations. $\endgroup$
    – Valter Moretti
    Commented Feb 11, 2022 at 13:04
  • $\begingroup$ So then saying that "It can be shown that the stress tensor is a contravariant second order tensor" like wikipedia does is meaningless/wrong because if you are in an orthonormal coordinate system there is no distinction, but if you are not, employing transformation rules for a (2,0) tensor would give the wrong results. $\endgroup$
    – DiggingDeep
    Commented Feb 11, 2022 at 15:20
  • 1
    $\begingroup$ A $(1,1)$ tensor is, by definition, a linear map $V \to V$. If the vector space $V$ is equipped with a metric $g$, that metric can be used to construct a vector space isomorphism $g: V \to V^*$. With this isomorphism and the notion of tensor product you can construct analogous isomorphisms between spaces of tensors with equal number of factors, independently of the type. In particular there is a canonical isomorphism $V\otimes V^* \to V \otimes V$. This isomorphism is the one used to pass from ${\sigma^i}_j$ to $\sigma_{ij}$ $\endgroup$
    – Valter Moretti
    Commented Feb 11, 2022 at 15:38
  • $\begingroup$ More precisely $\sigma_{jk} = g_{ki}{\sigma^i}_j$. When working in orthonormal frames with respect to $g$, in components, the action of the isomorphism is the identity. However there are two (actually four) different, but isomophically equivalent, representations of the stress tensor. The physical definition ${\sigma^i}_j$ and $\sigma^{ij}$, ${\sigma_i}^j$, $\sigma_{ij}$. You can appreciate the difference in components only if dealing with non-orthonormal frames. $\endgroup$
    – Valter Moretti
    Commented Feb 11, 2022 at 15:43
  • 1
    $\begingroup$ covariant components transform with the transposed inverse. If referring to bases, $e'_i = {A^j}_i e_j$ and $e'^{*k} = {B_h}^k e^{*h}$ the simultaneous requirements $\langle e_i, e^{*k}\rangle = \delta_i^k$ and $\langle e'_j, e'^{*h}\rangle = \delta_j^h$ imply ${A^j}_i {B_h}^i = \delta^j_h$. In terms of matrices $AB^t = I$, that is $B= (A^{-1})^t$ as I said. $\endgroup$
    – Valter Moretti
    Commented Feb 11, 2022 at 22:05

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.