How to interpret stress tensor as a contravariant second order tensor

Question

In a previous question of mine I was lead to believe that the stress tensor was a contravariant second order tensor in the sense of the isomorphism

$$\hom(V^*,V)\to V\otimes V$$

$V\otimes V$ are second order contravariant tensors. The stress tensor is an example of tensor $\hom(V^*,V)$ as it takes a "vector" (the argument $V^*$) and returns vector (the force vector that the plane feels). Now I'm not so sure that viewing the stress tensor as an example of $\hom(V^*,V)$ is tenable.

To explain, consider a linear transformation $f:V\to V$. In the standard basis $B=\{e_1,\dots,e_n\}$ this can be expressed as $[f(v)]_B =M[v]_{B}$. That is it takes coordinates of $v$ with respect to $B$ and returns the results in coordinates with respect to $B$. If $B'=\{e'_1,\dots,e'_n\}$ is another basis related to $B$ by

$$\begin{pmatrix} e'_1 & \cdots & e'_n \end{pmatrix}= \begin{pmatrix} e_1 & \cdots &e_n \end{pmatrix}L$$

then $$L^{-1}[f(v)]_B = L^{-1} M L L^{-1} [v]_{B}$$ $$[f(v)]_{B'} = L^{-1} M L [v]_{B'}$$ The matrix M changes to $L^{-1}ML$ and we say the linear transformation is a mixed second order tensor because there is a $L$ and a $L^{-1}$. One coordinate is transforming contravariantly and the other covariantly.

However with the stress tensor interpreted as a $V^{*} \to V$, a second order contravariant tensor, I would expect to see something like $[f(\omega)]_B =M[\omega]_{B}$ being transformed to $[f(\omega)]_{B'} = L^{-1} M L^{-1} [\omega]_{B'}$.

But what I read instead for the stress $S$ across a surface perpendicular to $\mathbf{n}$ is, using einstein summation notation, $$S=\sigma^{km}(\mathbf{n}\cdot e_k)e_m=\sigma^{km}(\mathbf{n}\cdot \Lambda_k^i\tilde{e}_i) (\Lambda_m^j \tilde{e}_j)= \sigma^{km} \Lambda_k^i \Lambda_m^j (\mathbf{n}\cdot \tilde{e}_i) ( \tilde{e}_j)$$ where $\Lambda = L^{-1}$. This shows $\tilde{\sigma}^{ij}=\sigma^{km} \Lambda_k^i \Lambda_m^j $.

The input $\mathbf{n}$ that I called $\omega$ above makes no appearance as $[\mathbf{n}]_B$ like in the discussion of the linear transformation. How do I reconcile all of this?

Answer 1

If you consider any second order contravariant tensors, as you have written, you can either see it as a multilinear map $:V^{*} \times V^{*} \rightarrow \mathbb{R}$ or as an element of $\textsf{Hom}(V^{*},V)$. To see the equivalence of the transformations let $\lbrace e_{\alpha}\rbrace$ a base for $V$, also denote $\lbrace e^{\beta}\rbrace$ the dual base of $V^{*}$ and $T$ our tensor. In the first case you have that $$ T(\omega,\eta)=T^{\mu\nu}e_{\mu}e_{\nu}(\omega_{\alpha}e^{\alpha},\eta_{\beta}e^{\beta} )=T^{\mu\nu}\omega_{\mu}\eta_{\nu} $$ Where $T^{\mu\nu}=T(e^{\mu},e^{\nu})$Now if $\lbrace e'_{\alpha}\rbrace$ is another base there will be some $\Lambda \in GL(V)$ s.t. $e=\Lambda e'$ or in coordinates $e_{\alpha}= \Lambda^{\beta}_{\alpha}e'_{\beta}$ and denoting by $\overline{\Lambda}=\Lambda^{-1}$, $e^{\alpha}= \overline{\Lambda}_{\beta}^{\alpha}e'^{\beta}$ we get $$ T=T^{\mu\nu} \Lambda^{\sigma}_{\mu} e'_{\sigma} \Lambda^{\rho}_{\nu} e'_{\rho} $$ Hence $T'^{\sigma \rho}=T^{\mu\nu} \Lambda^{\sigma}_{\mu} \Lambda^{\rho}_{\nu}$ or $T'=\Lambda^t T \Lambda$.

On the other hand suppose one consider $T \in \textsf{Hom}(V^{*},V)$. Denote again by T (it is a slight abuse of notation justified however by the canonical isomorphism) its matrix acting on some covector $\omega$, in coordinates, we have $$ T(\omega)=T^{\mu \nu}\omega_{\mu}e_{\nu} $$ Since the coordinates of $\omega$ transform with $\omega_{\alpha}= \Lambda^{\beta}_{\alpha} \omega'_{\beta}$ one has $$ T(\omega)=T^{\mu \nu}\Lambda^{\sigma}_{\mu}\omega'_{\sigma}\Lambda^{\rho}_{\nu}e'_{\rho} $$ resulting in the same transformation.

If $\mathfrak{n}$ is your "normal" vector and given a positive bilineal non-degenerate form $g$ on $V$ (for the product of vectors), your expression is not accurate, the reason is $\mathfrak{n} \cdot e_{\mu}$ is a scalar, not a covector, so $T(\mathfrak{n} \cdot e_{\mu}, \cdot)$ is not well defined. What you can do is use the bilinear form $g$ to construct what is called the musical isomorphism $\flat:V \rightarrow V^{*}$ by $\mathfrak{n}^{\flat}=g(\mathfrak{n},\cdot) \in \textsf{Hom}(V,\mathbb{R}) \simeq V^{*}$. Then you can calculate $T(\mathfrak{n}^{\flat},\cdot)$.

Answer 2

After some thought over my own question I've come up with the following. The bilinear form can be expressed as follows. Let the matrix of the dual basis and basis be $$\tilde{B} = \begin{pmatrix} e^1 \\ \vdots \\ e^n \end{pmatrix}$$ $$B = \begin{pmatrix} e_1 & \cdots & e_n \end{pmatrix}$$ Then the bilinear form applied to vectors $v$ and $w$ can be expressed as $$w \tilde{B}^T A \tilde{B} v =[w]_B^T B^T \tilde{B}^T A \tilde{B} B[v]_B $$ in a new basis we have $B' = BL$, $\tilde{B}' = L^{-1}\tilde{B}$ and this bilinear form becomes

$$[w]_B^T B^T \tilde{B}^T A \tilde{B} B[v]_B = [w]_B^T {L^{-1}}^TL^T B^T \tilde{B}^T {L^{-1}}^TL^T A LL^{-1}\tilde{B} BLL^{-1}[v]_B=[w]_{B'}^T {B'}^T {\tilde{B}'}^T L^TAL \tilde{B'} B'[v]_{B'}$$ and so $A'=L^TAL$ as usual. But notice in this way of writing it the $LL^{-1}$ pairs are added in four times not twice.

With the stress tensor we write analogously that $([n]_{\tilde{B}}\tilde{B}B)(SB^T) = ([n]_{\tilde{B}}LL^{-1}\tilde{B}BLL^{-1})(S{L^{-1}}^TL^TB^T) = ([n]_{\tilde{B}'}\tilde{B}'B'L^{-1})(S{L^{-1}}^T {B'}^T) = ([n]_{\tilde{B}'}\tilde{B}'B')(L^{-1}S{L^{-1}}^T {B'}^T)$ where to keep the similarity with the bilinear form case its necessary to expand the input covector in the dual basis.