# 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 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:

$U(x)=a_0+a_1x^1+a_2x^2+O(x^2)$

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 $U(x_1,\dots, x_N)$of the differences$x_i$ 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

$U(x_1=0,\dots x_N=0)=a+a_i x_i+a_{ij}x_ix_j+O(x^2)$

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:

$dU=F_id\epsilon_i$

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

$dU=-P_idt$

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,

$\frac{d}{dt} (\rho e_ld\tau)=e_l\frac{d}{dt} (\rho d\tau) + \rho d\tau\frac{d}{dt}e_l$

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:

$U=\int u d\tau$

with

$u=u_1(\partial_in_j)+u_2(n_i\partial_jn_k)+ u_3(\partial_i n_j\partial_kn_l)+ u_4(n_pn_q\partial_i n_j\partial_kn_l)+\dots$

The most general form of $u_1$ for a linear dependence on the derivatives is:

eqsansder

$u_1(\partial_in_j)= K_{ij} \partial_i n_j$

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 know 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:

$u_2=L_{ijk}n_k\partial_in_j$

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:

$u_2=k_1e_{ijk}n_k\partial_in_j=k_1n\mbox{ rot } n.$

• {\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:

$u=K_1 (\mbox{ div } n)^2+K_2 (n.\mbox{ rot } n)^2+K_3(n \wedge \mbox{ rot } n)^2$