The virtual work principle for Cartan media

Classical continuum mechanics

We consider a virtual displacement field $\delta \mathbf{u}$. The virtual work is

$${\int }_{\mathcal{Q}}\mathbf{b}\cdot \delta \mathbf{u}dv+{\int }_{\partial \mathcal{Q}}\mathbf{t}\cdot \delta \mathbf{u}ds={\int }_{\mathcal{Q}}\sigma :\nabla \delta \mathbf{u}dv+{\int }_{\mathcal{Q}}\rho \ddot{\mathbf{u}}\cdot \delta \mathbf{u}dv$$

where $\mathbf{b}$ is the body force, $\mathbf{t}$ is the surface traction, $\sigma$ is the Cauchy stress tensor and $\rho$ is the mass density and $\mathcal{Q}$ is the current configuration.

I think the stress term ends up with an opposite sign to the body force term because it is a reactive force. This can probably be motivated by reading the section in Landau-Lifschitz where they define the stress tensor.

We do integration-by-parts to put it in the form

$${\int }_{\mathcal{Q}}\mathbf{b}\cdot \delta \mathbf{u}dv+{\int }_{\partial \mathcal{Q}}\mathbf{t}\cdot \delta \mathbf{u}ds={\int }_{\mathcal{Q}}\sigma :\nabla \delta \mathbf{u}dv-{\int }_{\mathcal{Q}}\rho \dot{\mathbf{u}}\cdot \delta \dot{\mathbf{u}}dv$$

Reference

In classical continuum mechanics, the strain tensor is found by symmetrising $\nabla \delta \mathbf{u}$. That is $ϵ=(\nabla \delta \mathbf{u}+\nabla \delta {\mathbf{u}}^T)/2$. The $+$ there is basically the group action of the translation group. So is there a way to generalise this to a general group $G$?

Generalised continuum mechanics

We consider a Cartan media configuration $q:M\to \mathbb{X}$. To generalise the virtual work principle, we consider the “response” of a virtual displacement field $\delta q:M\to T\mathbb{X}$ and its differential $d\delta q:TM\to TT\mathbb{X}$ and time-derivative $\delta \dot{q}:TW\to TT\mathbb{X}$, similarly to how $\delta \mathbf{u}$, $\nabla \delta \mathbf{u}$ and $\delta \dot{\mathbf{u}}$ appeared in (1).

Let $\left(,\right)$ denote the contraction of a covector and a vector. So if we have $v:M\to T\mathbb{X}$ and $\alpha :M\to T^{\ast }\mathbb{X}$, then $\left(\alpha ,v\right)$ is a scalar on $M$ (that is, it is a map $M\to \mathbb{R}$). This goes for vectors and covectors in the double tangent and codouble tangent bundles as well. So if $w:M\to TT\mathbb{X}$ and $\beta :M\to T^{\ast }T\mathbb{X}$, then $\left(\beta ,w\right)$ is a scalar.

Let $\mathcal{T}$ be a $T^{\ast }\mathbb{X}$-valued volume form on $M$, we extend the notation so that for any $v:M\to T\mathbb{X}$ we have that $\left(\mathcal{T},v\right)$ is an $\mathbb{R}$ volume form on $M$. If we introduce coordinates $u^{\alpha }:M\to \mathbb{R}$ on $M$, then we may write $\mathcal{T}$ as

$$\mathcal{T}={\mathcal{T}}^{0}dV$$

where ${\mathcal{T}}^0:M\to T^{\ast }\mathbb{X}$, which transforms as a scalar density of weight $1$ and $dV=d{u}^1\wedge \cdots \wedge d{u}^d$, which transforms as a tensor density of weight $-1$. We then have that

$$\left(\mathcal{T},v\right)=\left({\mathcal{T}}^0,v\right)dV$$

Now, let $\mathcal{Q}$ be a $T^{\ast }T\mathbb{X}$-valued $(d-1)$-form on $M$, and let

$$d{\bar{u}}^{\alpha }=d{u}^1\wedge \cdots \wedge d{u}^{\alpha -1}\wedge \cdots \wedge d{u}^{\alpha +1}\wedge \cdots \wedge d{u}^d$$

then $d{\bar{u}}^{\alpha }\wedge d{u}^{\alpha }$ is equal to $dV$ up to sign. So absorb that sign into $d{\bar{u}}^{\alpha }$ such that $d{\bar{u}}^{\alpha }\wedge d{u}^{\alpha }=dV$.

In coordinates, let

$$\mathcal{Q}={\mathcal{Q}}^{\alpha }d{\bar{u}}^{\alpha }$$

Now, let $w$ be a $TT\mathbb{X}$-valued $1$-form on $M$. In coordinates we write it as $w=w_{\alpha }d{u}^{\alpha }$. We then define the wedge contraction as

$$\left(\mathcal{Q}\wedge w\right)=\left({\mathcal{Q}}^{\alpha },w_{\alpha }\right)d{\bar{u}}^{\alpha }\wedge d{u}^{\alpha }$$

We can then finally write down a generalisd virtual work principle

$${\int }_M\left(\mathcal{T},\delta q\right)={\int }_M\left(\mathcal{Q}\wedge d(\delta q)\right)-{\int }_M\left(\mathcal{S},\delta \dot{q}\right)$$

where we have ommitted the generalised surface traction for now.

The lift

Let $\eta :M\to \mathfrak{g}$ such that $\delta q=\eta q$. So we have that $\left({\mathcal{T}}^0,\delta q\right)=\left({\mathcal{T}}^0,\eta q\right)$. We thus have a map $\mathfrak{g}\to \mathbb{R}$, defined by $\eta \mapsto \left({\mathcal{T}}^0,\eta q\right)$. This is a dual Lie algebra field $T:M\to {\mathfrak{g}}^{\ast }$. We can thus write the contraction of $T$ and $\eta$ as

$$⟨T,\eta ⟩=\left({\mathcal{T}}^0,\eta q\right)$$

We define the corresponding ${\mathfrak{g}}^{\ast }$-valued volume-form as $\bar{T}=TdV$.

In a previous note we saw that $dq={\xi }^{R}q$. We thus have that $d(\delta q)=\delta {\xi }^{R}q$. So

$$(\mathcal{Q}\wedge d(\delta q))=⟨{\bar{Q}}^R\wedge \delta {\xi }^R⟩$$

Now, let $\delta {\xi }^R={\text{Ad}}_{\Phi }\delta {\xi }^L$ and ${\bar{Q}}^R={\text{Ad}}_{\Phi }^{\ast }{\bar{Q}}^L$. Since $⟨A,Y⟩=⟨{\text{Ad}}_g^{\ast }A,{\text{Ad}}_{g}Y⟩$ we can write

$$(\mathcal{Q}\wedge d(\delta q))=⟨{\bar{Q}}^L\wedge \delta {\xi }^L⟩$$

Similarly, as $\delta \dot{q}=\delta N^{R}q$ we have that

$$(\mathcal{S},\delta \dot{q})=⟨{\bar{S}}^R,\delta N^R⟩=⟨{\bar{S}}^L,\delta N^L⟩$$

We thus have

$${\int }_M⟨\bar{T},\eta ⟩={\int }_M⟨{\bar{Q}}^L\wedge \delta {\xi }^L⟩-{\int }_M⟨{\bar{S}}^L,\delta N^L⟩$$

Dropping the L’s we have

$${\int }_M⟨\bar{T},\eta ⟩={\int }_M⟨\bar{Q}\wedge \delta \xi ⟩-{\int }_M⟨\bar{S},\delta N⟩$$

:::warning Note that $\xi$ here is not defined over the kinematic base space $W=[0,T]\times M$, but rather just $M$. :::

In components, we can write

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

We then have

$${\int }_M⟨T,\eta ⟩dV={\int }_M⟨Q^{\alpha },\delta X_{\alpha }⟩dV-{\int }_M⟨S,\delta N⟩dV$$

Compare this to the expression we have in the paper

Cosserat systems

We have $\mathbf{r}:M\to {\mathbb{E}}^3$ and ${\mathbf{e}}_i:M\to T{\mathbb{E}}^3$. We also write $q=(\mathbf{r},E):M\to OF({\mathbb{E}}^3)$, where $OF({\mathbb{E}}^3)$ is the orthonormal frame bundle.

Consider virtual displacement fields $\delta \mathbf{r}:M\to T{\mathbb{E}}^3$ and $\delta {\mathbf{e}}_i:M\to TT{\mathbb{E}}^3$. Collectively, we have that $\delta q:M\to TOF({\mathbb{E}}^3)$.

Each of the virtual displacements will give rise to a generalised body force volume form, generalised stress $(d-1)$-form, as well as a generalised momentum volume form. Each of these will act on the virtual displacements to form a scalar volume form on $M$.

Let us first consider the body force. Let $f$ be a $T^{\ast }{\mathbb{E}}^3$-valued volume form on $M$, then $f\cdot \delta \mathbf{r}$, where $\cdot$ is the Euclidean inner product, is a scalar volume form on $M$.

Let $m$ be a $T^{\ast }T{\mathbb{E}}^3$-valued volume-form on $M$. Then $m\cdot \delta {\mathbf{e}}_i$ is a scalar volume form on $M$ for each $i$.

Now we move on to the stress. We need a $(d-1)$-form $\mathcal{F}$ that contracts and wedges with $\delta \mathbf{r}$ to get a scalar volume form on $M$. In coordinates, we may write

$$\mathcal{F}={\mathcal{F}}^{\alpha }d{\bar{u}}^{\alpha }$$

where $d{\bar{u}}^{\alpha }$ is defined such that $d{\bar{u}}^{\alpha }\wedge d{u}^{\alpha }=dV$, and where $dV=d{u}^1\wedge \cdots \wedge d{u}^d$. We also write

$$d(\delta \mathbf{r})=\delta {\mathbf{\theta }}_{\alpha }^{s}d{u}^{\alpha }$$

We can then define the wedge inner product

$$\mathcal{F}\hat{\cdot }d(\delta \mathbf{r})={\mathcal{F}}^{\alpha }\cdot \delta {\mathbf{\theta }}_{\alpha }d{\bar{u}}^{\alpha }\wedge d{u}^{\alpha }={\mathcal{F}}^{\alpha }\cdot \delta {\mathbf{\theta }}_{\alpha }dV$$

Similarly, let $\mathcal{M}$ be a $(d-1)$-form that contracts and wedges with $\delta {\mathbf{e}}_i$ to get a scalar volume form $\mathcal{M}\hat{\cdot }\delta {\mathbf{e}}_i$ on $M$, for each $i$.

Finally, let $\mathcal{P}$ and $\mathcal{L}$ be volume forms on $M$ that contracts with $\delta \dot{\mathbf{r}}$ and $\delta {\dot{\mathbf{e}}}_i$ to form scalar volume forms on $M$.

We can now write down the virtual work principle

$$\begin{array}{r}{\int }_{M}f\cdot \delta \mathbf{r}+\sum _i{\int }_{M}m\cdot \delta {\mathbf{e}}_i={\int }_M\mathcal{F}\hat{\cdot }d\delta \mathbf{r}+\sum _i{\int }_M\mathcal{M}\hat{\cdot }d\delta {\mathbf{e}}_i\\ -{\int }_M\mathcal{P}\cdot \delta \dot{\mathbf{r}}-\sum _i{\int }_M\mathcal{L}\cdot \delta {\dot{\mathbf{e}}}_i\end{array}$$

Note about vector-valued forms

Let $E$ a vector-bundle with base-space $M$ and $\pi :E\to M$. We denote the space of $E$-valued $p$-forms as ${\Omega }^p(M,E)$.

Though a seemingly obscure object, the vector-valued $p$-form is fairly easy to understand. For each $x\in M$, a regular $1$-form on $M$ yields a linear map $T_{x}M\to \mathbb{R}$. An $E$-valued $1$-form on $M$ correspondingly yields a map $T_{x}M\to F_x$, where $F_x={\pi }^{-1}(x)$ is the fiber at $x\in M$. A vector-valued $p$-form yields an antisymmetric linear map $T_{x}M\times \underset{p\text{ times}}{\underbrace{\dots }}\times T_{x}M\to E$ at each $x\in M$.

If the vector bundle is trivial, as is the case for Cartan media, the situation is simpler still. Cartan media configurations are sections on the fiber bundle $M\times \mathbb{X}$. Velocities and deformation fields on Cartan media are sections on $M\times T\mathbb{X}$. Body forces are sections on $M\times T^{\ast }\mathbb{X}$.

:::info $M\times T\mathbb{X}$ is the vertical bundle of the trivial fiber buncle $M\times \mathbb{X}$, which which we write as $V(M\times \mathbb{X})=M\times T\mathbb{X}$. :::

Now, consider a $T\mathbb{X}$-valued $1$-form $\omega$ on $M$. At each $x\in M$, ${\omega }_x$ is a linear map ${\omega }_x:TM\to T\mathbb{X}$. If $u^{\alpha }$ are coordinates on $M$ then we can expand it as

$$\omega ={\omega }_{\alpha }d{u}^{\alpha }$$

where ${\omega }_{\alpha }:M\to T\mathbb{X}$ are sections on $M\times T\mathbb{X}$. For any vector-field $v=v^{\alpha }\frac{\partial }{\partial u^{\alpha }}$ on $M$, we have

$$\omega (v)={\omega }_{\alpha }{v}^{\beta }d{u}^{\alpha }(\frac{\partial }{\partial u^{\beta }})={\omega }_{\alpha }{v}^{\beta }{\delta }_{\beta }^{\alpha }={\omega }_{\alpha }{v}^{\beta }$$

where ${\omega }_{\alpha }{v}^{\beta }$ is now a section on $M\times T\mathbb{X}$.

The wedge product

Let us go back to the vector bundle $E$. The wedge product of a $E$-valued $p$-form with a $E$-valued $q$-form is a $E\otimes E$-valued $(p+q)$-form.

So $p=q=1$ for example, and we have $\omega ={\omega }_{\alpha }d{u}^{\alpha }$ and $\upsilon ={\upsilon }_{\alpha }d{u}^{\alpha }$, then

$$\omega \wedge \upsilon =({\omega }_{\alpha }\otimes {\upsilon }_{\beta })d{u}^{\alpha }\wedge d{u}^{\beta }$$

The wedge contraction

In the virtual work principle, we need to wedge and then contract covector-valued $(d-1)$-forms with vector-valued $1$-forms.

Consider a $T^{\ast }\mathbb{X}$-valued $1$-form $\omega$ and a $T\mathbb{X}$-valued $1$-form $\upsilon$. Their wedge product is

$$\omega \wedge \upsilon =({\omega }_{\alpha }\otimes {\upsilon }_{\beta })d{u}^{\alpha }\wedge d{u}^{\beta }$$

For each $\alpha ,\beta$, we have that ${\omega }_{\alpha }\otimes {\upsilon }_{\beta }$ is a rank-$2$ tensor. That is, for each $x\in M$ we have a map $({\omega }_{\alpha }\otimes {\upsilon }_{\beta }{)}_x:T\mathbb{X}\times T^{\ast }\mathbb{X}\to \mathbb{R}$. The $(1,1)$-contraction of this tensor, which yields a scalar on $M$, is given by $({\omega }_{\alpha },{\upsilon }_{\beta })\in C^{\infty }(M)$.

So this leads us to define the wedge contraction

$$(\omega \wedge \upsilon )=({\omega }_{\alpha },{\upsilon }_{\beta })d{u}^{\alpha }\wedge d{u}^{\beta }$$

such that $(\omega \wedge \upsilon )$ is now a regular $\mathbb{R}$-valued $2$-form on $M$.