# 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 ${\displaystyle P}$ that is a functional of displacement is introduced. Consider in particular case of a mass ${\displaystyle m}$ attached to a spring of constant ${\displaystyle k}$.Deformation of the system is referenced by the elongation ${\displaystyle x}$ of the spring with respect to equilibrium. The virtual work \index{virtual work} associated to a displacement ${\displaystyle dx}$ is

deltWfdx

${\displaystyle \delta W=f.dx}$

Quantity ${\displaystyle f}$ represents the constraint , here a force, and ${\displaystyle x}$ is the deformation. If force ${\displaystyle f}$ 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 ${\displaystyle U}$ :

eqdeltaWdU

${\displaystyle \delta W=-dU}$

In general, force ${\displaystyle f}$ depends on the deformation. Relation ${\displaystyle f=f(x)}$ 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 ${\displaystyle U}$ as a function of the deformations using the physics of the the problem and symmetries. In the particular case of an oscillator, the internal energy has to depend only on the distance ${\displaystyle x}$ to equilibrium position. If ${\displaystyle U}$ admits an expansion at ${\displaystyle x=0}$, in the neighbourhood of the equilibrium position ${\displaystyle U}$ can be approximated by:

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

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

figparabe

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

As

${\displaystyle dU=-fdx}$

the strain--deformation relation becomes:

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

## Oscillators chains

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

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

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 ${\displaystyle U(x_{1},\dots ,x_{N})$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} :

${\displaystyle dU=0}$

If ${\displaystyle U}$ admits a Taylor expansion:

eqdevliUch

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

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

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

one obtains

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

Using expression of ${\displaystyle U}$ provided by equation eqdevliUch yields to:

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

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

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

Differential of ${\displaystyle U}$ becomes:

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

Assuming that ${\displaystyle U}$ admits a Taylor expansion around the equilibrium position:

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

and that ${\displaystyle dU=0}$ at equilibrium, yields to:

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

As the interaction occurs only between nearest neighbours:

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

so:

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

This does correspond to the expression of the force applied to mass ${\displaystyle i}$ :

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

if one sets ${\displaystyle k=-b_{ii}=-b_{ii+1}}$.

secmaterelast

## Tridimensional elastic material

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

${\displaystyle x=a+X}$

Vector ${\displaystyle X}$ represents the deformation.

Remark:

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

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

Definition: The deformation tensor SPH is the symmetric part of the tensor gradient of ${\displaystyle X}$.

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

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

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

with

${\displaystyle dU=-P_{i}dt}$

Tensor ${\displaystyle u_{i,j}^{s}}$ is called rate of deformation tensor. It is the symmetric part of tensor ${\displaystyle u_{i,j}}$. 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:

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

Thus:

dukij

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

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

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

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

eqrhoel

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

We have\footnote{

footdensi

Indeed:

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

and from the properties of the particulaire derivative:

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

Now,

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

From the mass conservation law:

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

}

eqdudt

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

Thus

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

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

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

thus:

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

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

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

it is a generalized Hooke law\index{Hooke law}. The ${\displaystyle b_{ijkl}}$'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 ${\displaystyle u(x,y)}$. } ${\displaystyle n}$. 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 ${\displaystyle n}$. In a continuous model, this yields to a vector field ${\displaystyle n}$. Internal energy of the nematic is a function of the vector field ${\displaystyle n}$ and its partial derivatives.

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

${\displaystyle U=\int ud\tau }$

with

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

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

eqsansder

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

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

• Rotation invariance. Functional ${\displaystyle u_{1}}$ should be rotation invariant.

${\displaystyle u_{1}(\partial _{i}n_{j})=u_{1}(R_{ik}R_{jl}\partial _{k}n_{l})}$

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

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

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

${\displaystyle u_{1}=k_{0}{\mbox{ div }}n}$

• Invariance under the transformation ${\displaystyle n}$ maps to ${\displaystyle -n}$ . The energy of distortion is independent on the sense of ${\displaystyle n}$, that is ${\displaystyle u_{1}(n)=u_{1}(-n)}$. This implies that the constant ${\displaystyle k_{0}}$ 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 ${\displaystyle u_{2}}$. general form for ${\displaystyle u_{2}}$ is:

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

Let us consider how symmetries can simplify this last form.

• Invariance under the transformation ${\displaystyle n}$ maps to ${\displaystyle -n}$ . This invariance condition is well fulfilled by ${\displaystyle u_{2}}$.
• Rotation invariance. The rotation invariance condition implies that:

${\displaystyle L_{ijk}=R_{il}R_{jm}R_{kn}L_{lmn}}$

It is known that there does not exist any third order isotrope tensor in ${\displaystyle R^{3}}$, but there exist a third order isotrope pseudo tensor: the signature pseudo tensor ${\displaystyle e_{jkl}}$ (see appendix secformultens). This yields to the expression:

${\displaystyle u_{2}=k_{1}e_{ijk}n_{k}\partial _{i}n_{j}=k_{1}n{\mbox{ rot }}n.}$

• {\bf Invariance of the energy with respect to the axis transformation ${\displaystyle x\rightarrow -x}$, ${\displaystyle y\rightarrow -y}$, ${\displaystyle z\rightarrow -z}$.} The energy of nematic crystals has this invariance property\footnote{ Cholesteric crystal doesn't verify this condition.} . Since ${\displaystyle e_{ijk}}$ is a pseudo-tensor it changes its signs for such transformation.

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

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

and ${\displaystyle u_{4}}$:

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

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

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