# 物理代写|固体物理代写Solid-state physics代考|PHYSICS7544

## 物理代写|固体物理代写Solid-state physics代考|Elastic moduli

We have so far introduced four different elastic parameters, namely the Young modulus, the Poisson ratio and the two Lamé coefficients. We previously introduced also the bulk modulus $B$ when we investigated thermal expansion in section 4.2.1. Since $B$ was defined as the inverse of the isothermal compressibility (see appendix $C$ ), that is it deals with volume variations, it is expected to be related to some elastic parameter. In order to elucidate this issue, let us consider a hydrostatic stress $T_{i j}=P \delta_{i j}$ (where $P$ is the macroscopic hydrostatic pressure) and insert it into the constitutive equation (5.39) so as to get
$$\left.\mathbb{S}=\frac{1}{3} \frac{1}{\lambda+\frac{2}{3} \mu} P\right]$$
The connection with equation (C.11) is established by defining
$$B=\lambda+\frac{2}{3} \mu,$$

so that
$$\left.\mathbb{S}=\frac{1}{3 B} P\right] \quad \rightarrow \quad \operatorname{Tr}(\mathbb{S})=\sum_i \epsilon_{i t}=\frac{\Delta V}{V}=\frac{P}{B}$$
which leads to the following definition
$$\frac{1}{B}=\frac{1}{V} \frac{\Delta V}{P}$$
representing the finite difference counterpart of equation (C.11). This result reconciles the thermodynamical and elastic treatment of deformations affecting the system volume and it allows us to recast the stress-strain relation of a homogeneous and isotropic linear elastic medium in the form
$\mathbb{T}=2 \mu \mathbb{S}+\left(B-\frac{2}{3} \mu\right) \operatorname{Tr}(\mathbb{S}) \rrbracket$
$\left.=3 B\left[\frac{1}{3} \operatorname{Tr}(\mathbb{S}) \rrbracket\right]+2 \mu\left[\mathbb{S}-\frac{1}{3} \operatorname{Tr}(\mathbb{S})\right]\right]$,
where the first and second term on the right-hand side are, respectively, named spherical part and deviatoric part of the stress tensor: they describe the hydrostatic volume variation and the change in shape of the solid body subject to $\mathbb{I}$.

## 物理代写|固体物理代写Solid-state physics代考|Thermoelasticity

We have so far implicitly assumed that the deformations are imposed to the system at zero temperature. While this assumption was useful to define a clean purely-elastic problem, we must duly generalise our theory to include stress actions applied at $T>0 \mathrm{~K}$ as well [6].

The starting point is of course the energy balance stated by the first law of thermodynamics (see equation (C.5)) which for a system with volume $V$ in equilibrium at temperature $T$ under some elastic action is written as
$$d \mathcal{U}=V \sum_{i j} T_{i j} d \epsilon_{i j}+T d S,$$
where the mechanical work contributing to the internal energy $\mathcal{U}$ has been written in terms of the stress tensor since we know that this latter describes any possible kind of volume and shape variation of the system. It is easy to reconcile equation (5.47) with the standard thermodynamical formulation by simply considering the case of a hydrostatic stress $T_{i j}=-P \delta_{i j}$, where $P$ is the applied pressure whose negative sign indicates that the mechanical action is compressive. We assume it to operate quasistatically, like anywhere else in the remaining of this chapter. By inserting the hydrostatic stress into equation (5.47) we get
\begin{aligned} d \mathcal{U} &=V \sum_{i j}\left(-P \delta_{i j}\right) d \epsilon_{i j}+T d S \ &=-P V \sum_i d \epsilon_{i i}+T d S \ &=-P V \frac{d V}{V}+T d S \ &=-P d V+T d S \end{aligned}

## 物理代写|固体物理代写Solid-state physics代考|Elastic moduli

$$\left.\mathbb{S}=\frac{1}{3} \frac{1}{\lambda+\frac{2}{3} \mu} P\right]$$

$$B=\lambda+\frac{2}{3} \mu,$$

$$\left.\mathbb{S}=\frac{1}{3 B} P\right] \quad \rightarrow \quad \operatorname{Tr}(\mathbb{S})=\sum_i \epsilon_{i t}=\frac{\Delta V}{V}=\frac{P}{B}$$

$$\frac{1}{B}=\frac{1}{V} \frac{\Delta V}{P}$$

## 物理代写|固体物理代写Solid-state physics代考|Thermoelasticity

$$d \mathcal{U}=V \sum_{i j} T_{i j} d \epsilon_{i j}+T d S,$$

$$d \mathcal{U}=V \sum_{i j}\left(-P \delta_{i j}\right) d \epsilon_{i j}+T d S \quad=-P V \sum_i d \epsilon_{i i}+T d S=-P V \frac{d V}{V}+T d S \quad=-P d V+T d S$$

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