2. Strain fields in Cartan media

A spatial configuration is a map $q:M\to \mathbb{X}$. In the following we will not consider spatio-temporal configurations, as the inclusion of time follows trivially from the treatment presented here.

A vector field is a map $V:q(M)\to Tq(M)$, and we write $V\in \mathfrak{X}(q(M))$. Given such a vector field, we can always consider a corresponding $T\mathbb{X}$-valued field on $M$, given by $V\circ q:M\to Tq(M)$. The field $V\circ q$ can be considered a section on the vertical bundle of $M\times \mathbb{X}$.

The generalised strain fields

We now introduce the fact that the configuration space $\mathbb{X}$ has a group $G$ that acts on it transitively. This makes $\mathbb{X}$ a homogeneous space. Furthermore, as we will show, this will allow us to trivialise the tangent bundle $T\mathbb{X}$. That is, instead of working with the strain fields ${\chi }_{\alpha }:M\to T\mathbb{X}$, which map to a non-linear space, we will be able to work with vector space-valued fields.

:::info Within the context of fiber bundles, the maps ${\chi }_{\alpha }:M\to T\mathbb{X}$ can be considered a section on the vertical bundle of the fiber bundle $M\times \mathbb{X}$. :::

Let the structure field $\Phi :M\to G$ be defined such that

$$q=\Phi \cdot q_r$$

$d\Phi$ is a $TG$-valued $1$-form on $M$, left translating this to the identity yields the left Maurer-Cartan form

$${\xi }^L={\Phi }^{-1}d\Phi$$

and correspondingly we have the right Maurer-Cartan form ${\xi }^R=d\Phi {\Phi }^{-1}$. These are $\mathfrak{g}$-valued $1$-forms on $M$.

We then have that $dq={\eta }_{\Phi }(d\Phi ){|}_{q_r}=\zeta ({\xi }^R){|}_q$, where we simply followed the same derivation from Sec. 1.1.4.2.

:::info Here we have reused the definitions of $\eta$ and $\zeta$ from Sec. 1, and applying them on the $1$-forms $d\Phi$ and ${\xi }^R$. Since they are vector-valued forms, the definitions extend naturally. For example, for any $\mathfrak{g}$-valued $1$-form $\omega ={\omega }_{\alpha }d{u}^{\alpha }$ on $M$, where ${\omega }_{\alpha }:M\to \mathfrak{g}$, we define

$$\zeta (\omega )=\zeta ({\omega }_{\alpha })d{u}^{\alpha }$$

:::

We may call ${\xi }^L$ (or ${\xi }^R$), the generalised strain. We relate them to strain as

$$\begin{array}{rl}dq& =\zeta ({\xi }^R){|}_q\\ & =\zeta ({\text{Ad}}_{\Phi }{\xi }^L){|}_q\end{array}$$

:::info Recall that $\zeta :\mathfrak{g}\to \mathfrak{X}(\mathbb{X})$ is a representation of $\mathfrak{g}$. For $Y\in \mathfrak{g}$, $\zeta (Y)$ is a vector field on $\mathbb{X}$. We denote $\zeta (Y){|}_x$ as the vector field evaluated at $x\in \mathbb{X}$, so $\zeta (Y){|}_x\in T_x\mathbb{X}$.

If $\omega$ is a $\mathfrak{g}$-valued $1$-form on $M$, then $\zeta (\omega )$ is a $T\mathbb{X}$-valued $1$-form on $M$. For any point $x\in \mathbb{X}$, then $\zeta (\omega ){|}_x$ is a $T_x\mathbb{X}$-valued $1$-form on $M$.

The right hand side of the first line of (7) should be understood as follows. $\zeta ({\xi }^R){|}_q$ is a map $M\to T\mathbb{X}$, which is given by $p\mapsto \zeta ({\xi }^R){|}_{q(p)}$. :::

Now let

$$\begin{array}{rl}{\xi }^R& =X_{\alpha }^{R}d{u}^{\alpha }\\ {\xi }^L& =X_{\alpha }^{L}d{u}^{\alpha }\end{array}$$

where we call $X_{\alpha }^R,X_{\alpha }^L:M\to \mathfrak{g}$ the right and left generalised strain fields. Then we can write

$$\begin{array}{rl}dq& =\zeta (X_{\alpha }^R){|}_{q}d{u}^{\alpha }\\ & =\zeta ({\text{Ad}}_{\Phi }{X}_{\alpha }^L){|}_{q}d{u}^{\alpha }\end{array}$$

Comparing (2) and (9) we find

$${\chi }_{\alpha }=\zeta (X_{\alpha }^R){|}_q=\zeta ({\text{Ad}}_{\Phi }{X}_{\alpha }^L){|}_q$$

---

We see that we can now describe the strains of Cartan media in terms of the $\mathfrak{g}$-valued $1$-form ${\xi }^L$ (or ${\xi }^R$), instead of the $T\mathbb{X}$-valued $dq$. The benefits of this are clear.

Structure equations

Without providing proof here, it is worth mentioning of course that the generalised strain satisfies

$$d{\xi }^L+[{\xi }^L\wedge {\xi }^L]=0$$

Linear actions

From Sec. 1.2, we see that if the group action is linear, we can rewrite (7) as

$$\begin{array}{rl}dq& =\rho ({\xi }^R)(q)\\ & =\rho ({\text{Ad}}_{\Phi }{\xi }^L)(q)\end{array}$$

where, as before, we have extended the definition of $\rho$ as follows: For any $\mathfrak{g}$-valued $1$-form $\omega ={\omega }_{\alpha }d{u}^{\alpha }$ on $M$, where ${\omega }_{\alpha }:M\to \mathfrak{g}$, let

$$\rho (\omega )=\rho ({\omega }_{\alpha })d{u}^{\alpha }$$

such that $\rho ({\omega }_{\alpha }):\mathbb{X}\to T\mathbb{X}$ as before.

Now let’s say we have a basis $D:{\mathbb{R}}^n\to \mathbb{X}$ for $\mathbb{X}$. Let $q=D\mathbf{q}$ where $\mathbf{q}:M\to {\mathbb{R}}^n$. We then have that $dq=Dd\mathbf{q}$, where $d\mathbf{q}:TM\to T{\mathbb{R}}^n$. We further have that

$$\begin{array}{rl}d\mathbf{q}& =\tilde{\rho }({\xi }^R)\mathbf{q}\\ & =\tilde{\rho }({\text{Ad}}_{\Phi }{\xi }^R)\mathbf{q}\end{array}$$

The body and spatial frames

We saw earlier that the strain fields can be related to the right generalised strain fields as

$${\chi }_{\alpha }=\zeta (X_{\alpha }^R){|}_q$$

and if we have a linear action we can further write this as

$$\begin{array}{rl}{\chi }_{\alpha }& =\rho (X_{\alpha }^R)q\\ & =D\tilde{\rho }(X_{\alpha }^R)\mathbf{q}\end{array}$$

where last line results from choosing a basis.

Now we see that in (14) we evaluate the vector field $\zeta (X_{\alpha }^R)$ at $q$. What happens if we evaluate it at the reference point $q_r$?

$$\zeta (X_{\alpha }^R){|}_{q_r}$$ $$\zeta (X_{\alpha }^R){|}_q=\zeta ({\text{Ad}}_{\Phi }{X}_{\alpha }^L){|}_q$$ $$\zeta (X_{\alpha }^R){|}_{q_r}=\zeta (X_{\alpha }^L){|}_q$$ $$\zeta (X_{\alpha }^R){|}_{q_r}=\zeta (X_{\alpha }^L){|}_q?$$

note to self: see if you can relate N^L directly to a body frame strain field.

$${\chi }_{\alpha }=\zeta (X_{\alpha }^R){|}_q=\zeta ({\text{Ad}}_{\Phi }{X}_{\alpha }^L){|}_q$$