# Introduction to Mathematical Physics/Energy in continuous media/Generalized elasticity

## Introduction

In this section, the concept of elastic energy is presented. \index{elasticity} The notion of elastic energy allows to deduce easily "strains--deformations" relations.\index{strain--deformation relation} So, in modelization of matter by virtual powers method \index{virtual powers} a power $P$  that is a functional of displacement is introduced. Consider in particular case of a mass $m$  attached to a spring of constant $k$ .Deformation of the system is referenced by the elongation $x$  of the spring with respect to equilibrium. The virtual work \index{virtual work} associated to a displacement $dx$  is

deltWfdx

$\delta W=f.dx$

Quantity $f$  represents the constraint , here a force, and $x$  is the deformation. If force $f$  is conservative, then it is known that the elementary work (provided by the exterior) is the total differential of a potential energy function or internal energy $U$  :

eqdeltaWdU

$\delta W=-dU$

In general, force $f$  depends on the deformation. Relation $f=f(x)$  is thus a constraint--deformation relation .

The most natural way to find the strain-deformation relation is the following. One looks for the expression of $U$  as a function of the deformations using the physics of the problem and symmetries. In the particular case of an oscillator, the internal energy has to depend only on the distance $x$  to equilibrium position. If $U$  admits an expansion at $x=0$ , in the neighbourhood of the equilibrium position $U$  can be approximated by:

$U(x)=a_{0}+a_{1}x^{1}+a_{2}x^{2}+O(x^{2})$

As $x=0$  is an equilibrium position, we have $dU=0$  at $x=0$ . That implies that $a_{1}$  is zero. Curve $U(x)$  at the neighbourhood of equilibrium has thus a parabolic shape (see figure figparabe

figparabe

In the neighbourhood of a stable equilibrium position $x_{0}$ , the intern energy function $U$ , as a function of the difference to equilibrium presents a parabolic profile.

As

$dU=-fdx$

the strain--deformation relation becomes:

$f={\frac {dU}{dx}}$

## Oscillators chains

Consider a unidimensional chain of $N$  oscillators coupled by springs of constant $k_{ij}$ . this system is represented at figure figchaineosc. Each oscillator is referenced by its difference position $x_{i}$  with respect to equilibrium position. A calculation using the Newton's law of motion implies:

$U=\sum {\frac {1}{2}}k_{ij}(x_{i}-x_{j-1})^{2}$

figchaineosc

A coupled oscillator chain is a toy example for studying elasticity.

A calculation using virtual powers principle would have consisted in affirming: The total elastic potential energy is in general a function $U(x_{1},\dots ,x_{N})$ofthedifferences$ x_i$ to the equilibrium positions. This differential is total since force is conservative\footnote{ This assumption is the most difficult to prove in the theories on elasticity as it will be shown at next section} . So, at equilibrium: \index{equilibrium} :

$dU=0$

If $U$  admits a Taylor expansion:

eqdevliUch

$U(x_{1}=0,\dots x_{N}=0)=a+a_{i}x_{i}+a_{ij}x_{i}x_{j}+O(x^{2})$

In this last equation, repeated index summing convention as been used. Defining the differential of the intern energy as:

$dU=f_{i}dx_{i}$

one obtains

$f_{i}={\frac {\partial U}{\partial x_{i}}}.$

Using expression of $U$  provided by equation eqdevliUch yields to:

$f_{i}=a_{ij}x_{j}.$

But here, as the interaction occurs only between nearest neighbours, variables $x_{i}$  are not the right thermodynamical variables. let us choose as thermodynamical variables the variables $\epsilon _{i}$  defined by:

$\epsilon _{i}=x_{i}-x_{i-1}.$

Differential of $U$  becomes:

$dU=F_{i}d\epsilon _{i}$

Assuming that $U$  admits a Taylor expansion around the equilibrium position:

$U=b+b_{i}\epsilon _{i}+b_{ij}\epsilon _{i}\epsilon _{j}+O(\epsilon ^{2})$

and that $dU=0$  at equilibrium, yields to:

$F_{i}=b_{ij}\epsilon _{j}$

As the interaction occurs only between nearest neighbours:

$b_{ij}=0{\mbox{ si }}i\neq j\pm 1$

so:

$F_{i}=b_{ii}\epsilon _{i}+b_{ii+1}\epsilon _{i+1}$

This does correspond to the expression of the force applied to mass $i$  :

$F_{i}=-k(x_{i}-x_{i-1})-k(x_{i+1}-x_{i})$

if one sets $k=-b_{ii}=-b_{ii+1}$ .

secmaterelast

## Tridimensional elastic material

Consider a system $S$  in a state $S_{X}$  which is a deformation from the state $S_{0}$ . Each particle position is referenced by a vector $a$  in the state $S_{0}$  and by the vector $x$  in the state $S_{X}$ :

$x=a+X$

Vector $X$  represents the deformation.

Remark:

Such a model allows to describe for instance fluids and solids.

Consider the case where $X$  is always "small". Such an hypothesis is called small perturbations hypothesis (SPH). The intern energy is looked as a function $U(X)$ .

Definition: The deformation tensor SPH is the symmetric part of the tensor gradient of $X$ .

$\epsilon _{ij}={\frac {1}{2}}(X_{i,j}-X_{j,i})$

At section secpuisvirtu it has been seen that the power of the admissible intern strains for the problem considered here is:

$P_{i}=\int K_{ij}^{s}u_{i,j}^{s}d\tau$

with

$dU=-P_{i}dt$

Tensor $u_{i,j}^{s}$  is called rate of deformation tensor. It is the symmetric part of tensor $u_{i,j}$ . It can be shown [ph:fluid:Germain80] that in the frame of SPH hypothesis, the rate of deformation tensor is simply the time derivative of SPH deformation tensor:

$u_{i,j}^{s}={\frac {d\epsilon _{ij}}{dt}}$

Thus:

dukij

$dU=-\int K_{ij}^{s}d\epsilon _{ij}d\tau .$

Function $U$  can thus be considered as a function $U(\epsilon _{ij})$ . More precisely, one looks for $U$  that can be written:

$U=\int \rho e_{l}d\tau$

where $e_{l}$  is an internal energy density with\footnote{ Function $U$  depends only on $\epsilon _{ij}$ .} whose Taylor expansion around the equilibrium position is:

eqrhoel

$\rho e_{l}=a+a_{ij}\epsilon _{ij}+a_{ijkl}\epsilon _{ij}\epsilon _{kl}$

We have\footnote{

footdensi

Indeed:

${\frac {d}{dt}}U={\frac {d}{dt}}\int \rho e_{l}d\tau$

and from the properties of the particulaire derivative:

${\frac {d}{dt}}\int \rho e_{l}d\tau =\int {\frac {d}{dt}}(\rho e_{l}d\tau )$

Now,

${\frac {d}{dt}}(\rho e_{l}d\tau )=e_{l}{\frac {d}{dt}}(\rho d\tau )+\rho d\tau {\frac {d}{dt}}e_{l}$

From the mass conservation law:

${\frac {d}{dt}}\rho =0$

}

eqdudt

${\frac {dU}{dt}}=\int \rho ({\frac {d}{dt}}e_{l})d\tau$

Thus

$dU=\int \rho de_{l}d\tau$

Using expression eqrhoel of $e_{l}$  and assuming that $dU$  is zero at equilibrium, we have:

$dU=\int \rho [a_{ijkl}\epsilon _{ij}d\epsilon _{kl}+a_{ijkl}d\epsilon _{ij}\epsilon _{kl}]d\tau$

thus:

$dU=\int \rho b_{ijkl}\epsilon _{kl}d\epsilon _{ij}d\tau$

with $b_{ijkl}=a_{ijkl}+a_{klij}$ . Identification with equation dukij, yields to the following strain--deformation relation:

$K_{ij}^{s}=b_{ijkl}\epsilon _{kl}$

it is a generalized Hooke law\index{Hooke law}. The $b_{ijkl}$ 's are the elasticity coefficients.

Remark: Calculation of the footnote footdensi show that calculations done at previous section secchampdslamat should deal with volumic energy densities.

secenernema

## Nematic material

A nematic material\index{nematic} is a material [ph:liqcr:DeGennes74] whose state can be defined by vector field\footnote{ State of smectic materials can be defined by a function $u(x,y)$ . } $n$ . This field is related to the orientation of the molecules in the material (see figure figchampnema)

figchampnema

Each molecule orientation in the nematic material can be described by a vector $n$ . In a continuous model, this yields to a vector field $n$ . Internal energy of the nematic is a function of the vector field $n$  and its partial derivatives.

Let us look for an internal energy $U$  that depends on the gradients of the $n$  field:

$U=\int ud\tau$

with

$u=u_{1}(\partial _{i}n_{j})+u_{2}(n_{i}\partial _{j}n_{k})+u_{3}(\partial _{i}n_{j}\partial _{k}n_{l})+u_{4}(n_{p}n_{q}\partial _{i}n_{j}\partial _{k}n_{l})+\dots$

The most general form of $u_{1}$  for a linear dependence on the derivatives is:

eqsansder

$u_{1}(\partial _{i}n_{j})=K_{ij}\partial _{i}n_{j}$

where $K_{ij}$  is a second order tensor depending on $r$ . Let us consider how symmetries can simplify this last form.

• Rotation invariance. Functional $u_{1}$  should be rotation invariant.
$u_{1}(\partial _{i}n_{j})=u_{1}(R_{ik}R_{jl}\partial _{k}n_{l})$

where $R_{mn}$  are orthogonal transformations (rotations). We thus have the condition:

$K_{ij}=R_{ik}R_{jl}K_{kl},$

that is, tensor $K_{ij}$  has to be isotrope. It is known that the only second order isotrope tensor in a three dimensional space is $\delta _{ij}$ , that is the identity. So $u_{1}$  could always be written like:

$u_{1}=k_{0}{\mbox{ div }}n$
• Invariance under the transformation $n$  maps to $-n$  . The energy of distortion is independent on the sense of $n$ , that is $u_{1}(n)=u_{1}(-n)$ . This implies that the constant $k_{0}$  in the previous equation is zero.

Thus, there is no possible energy that has the form given by equation eqsansder. This yields to consider next possible term $u_{2}$ . general form for $u_{2}$  is:

$u_{2}=L_{ijk}n_{k}\partial _{i}n_{j}$

Let us consider how symmetries can simplify this last form.

• Invariance under the transformation $n$  maps to $-n$  . This invariance condition is well fulfilled by $u_{2}$ .
• Rotation invariance. The rotation invariance condition implies that:
$L_{ijk}=R_{il}R_{jm}R_{kn}L_{lmn}$
It is known that there does not exist any third order isotrope tensor in $R^{3}$ , but there exist a third order isotrope pseudo tensor: the signature pseudo tensor $e_{jkl}$  (see appendix secformultens). This yields to the expression:
$u_{2}=k_{1}e_{ijk}n_{k}\partial _{i}n_{j}=k_{1}n{\mbox{ rot }}n.$
• {\bf Invariance of the energy with respect to the axis transformation $x\rightarrow -x$ , $y\rightarrow -y$ , $z\rightarrow -z$ .} The energy of nematic crystals has this invariance property\footnote{ Cholesteric crystal doesn't verify this condition.} . Since $e_{ijk}$  is a pseudo-tensor it changes its signs for such transformation.

There are thus no term $u_{2}$  in the expression of the internal energy for a nematic crystal. Using similar argumentation, it can be shown that $u_{3}$  can always be written:

$u_{3}=K_{1}({\mbox{ div }}n)^{2}$

and $u_{4}$ :

$u_{4}=K_{2}(n.{\mbox{ rot }}n)^{2}+K_{3}(n\wedge {\mbox{ rot }}n)^{2}$

Limiting the development of the density energy $u$  to second order partial derivatives of $n$  yields thus to the expression:

$u=K_{1}({\mbox{ div }}n)^{2}+K_{2}(n.{\mbox{ rot }}n)^{2}+K_{3}(n\wedge {\mbox{ rot }}n)^{2}$