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?
Best Answer
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 $T(\mathfrak{n}^{\flat},\cdot)$.