# 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
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 calculation using virtual powers principle would have consisted in affirming: The total elastic potential energy is in general a function $U(x_{1},\dots ,x_{N})<math>ofthedifferences$x_i</math> 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
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:
$u=K_{1}({\mbox{ div }}n)^{2}+K_{2}(n.{\mbox{ rot }}n)^{2}+K_{3}(n\wedge {\mbox{ rot }}n)^{2}$