# Link between Lie Group Statistical Mechanics and Thermodynamics of Continua

## Abstract

**:**

## 1. Introduction

## 2. Affine Tensors

- that assigns a set of components to each affine frame f of an affine space $A\mathcal{T}$ of finite dimension n,
- with a transformation law, when changing of frames, which is an affine or a linear representation of $\mathbb{A}ff\left(n\right)$.

- that assigns a set of components to each G-frame f,
- with a transformation law, when changing of frames, which is an affine or a linear representation of G.

## 3. Momentum as Affine Tensor

**Remark**

**1.**

## 4. Symplectic Action and Momentum Map

**Remark**

**2.**

**Remark**

**3.**

## 5. Lie Group Statistical Mechanics

## 6. Relativistic Thermodynamics of Continua

- the 4-flux of mass $\overrightarrow{\mathit{N}}=\rho \phantom{\rule{4pt}{0ex}}\overrightarrow{\mathit{U}}$ where ρ is the density,
- the 4-flux of entropy $\overrightarrow{\mathit{S}}=\rho \phantom{\rule{0.166667em}{0ex}}s\phantom{\rule{4pt}{0ex}}\overrightarrow{\mathit{U}}=s\phantom{\rule{4pt}{0ex}}\overrightarrow{\mathit{N}}$ where s is the specific entropy,
- Planck’s temperature vector $\overrightarrow{\mathit{W}}=\beta \phantom{\rule{4pt}{0ex}}\overrightarrow{\mathit{U}}$,
- its gradient $\mathbf{f}=\nabla \phantom{\rule{0.166667em}{0ex}}\overrightarrow{\mathit{W}}$ called friction tensor,
- the momentum tensor of a continuum
**T**, a linear map from ${T}_{\mathit{X}}\mathcal{M}$ into itself.

**Theorem**

**1.**

## 7. Planck’s Potential of a Continuum

**T**from

**M**and ζ from z? We work in seven steps:

- Step 1: defining the orbit. To begin with, we consider the momentum as an Galilean tensor, i.e., its components ar modified only by the action of Galilean transformations. In order to calculate the integral (10), the orbit is parameterized thanks to a momentum map. Calculating the infinitesimal generators $Z=(dC,dP)$ by differentiation of (12):$$dC=\left(\begin{array}{c}d{\tau}_{0}\hfill \\ dk\hfill \end{array}\right),\phantom{\rule{2.em}{0ex}}dP=\left(\begin{array}{cc}0\hfill & 0\hfill \\ du\hfill & j\left(d\varpi \right)\hfill \end{array}\right)\phantom{\rule{4pt}{0ex}},$$$$\mu \phantom{\rule{0.166667em}{0ex}}Z=l\xb7d\varpi -q\xb7du+p\xb7dk-e\phantom{\rule{0.166667em}{0ex}}d{\tau}_{0}\phantom{\rule{4pt}{0ex}}.$$$$p=R\phantom{\rule{0.166667em}{0ex}}{p}^{\prime}+m\phantom{\rule{0.166667em}{0ex}}u,\phantom{\rule{2.em}{0ex}}q=R\phantom{\rule{0.166667em}{0ex}}({q}^{\prime}-{\tau}_{0}\phantom{\rule{0.166667em}{0ex}}{p}^{\prime})+m\phantom{\rule{0.166667em}{0ex}}(k-{\tau}_{0}\phantom{\rule{0.166667em}{0ex}}u)\phantom{\rule{4pt}{0ex}},$$$$l=R\phantom{\rule{0.166667em}{0ex}}{l}^{\prime}-u\times \left(R\phantom{\rule{0.166667em}{0ex}}{q}^{\prime}\right)+k\times \left(R\phantom{\rule{0.166667em}{0ex}}{p}^{\prime}\right)+m\phantom{\rule{0.166667em}{0ex}}k\times u\phantom{\rule{4pt}{0ex}},$$$$e={e}^{\prime}+u\xb7\left(R\phantom{\rule{0.166667em}{0ex}}{p}^{\prime}\right)+\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}m{\parallel u\parallel}^{2}\phantom{\rule{4pt}{0ex}}.$$Taking into account (3), the transformation law (6) of the Galilean momentum tensor $\mathit{\mu}$ reads:$$K={K}^{\prime}\phantom{\rule{0.166667em}{0ex}}{P}^{-1}+{K}_{m}(C,P),\phantom{\rule{2.em}{0ex}}L=(P\phantom{\rule{0.166667em}{0ex}}{L}^{\prime}+C\phantom{\rule{0.166667em}{0ex}}{K}^{\prime})\phantom{\rule{0.166667em}{0ex}}{P}^{-1}+{L}_{m}(C,P)\phantom{\rule{4pt}{0ex}},$$$${K}_{m}(C,P)=m\phantom{\rule{0.166667em}{0ex}}\left(-\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}{\parallel u\parallel}^{2},{u}^{T}\right)\phantom{\rule{4pt}{0ex}}.$$
- Step 2: representing the orbit by equations. To obtain them, we have to determine a functional basis. The first step is to calculate their number. We start determining the isotropy group of μ. The analysis will be restricted to massive particles: $m\ne 0$. The components $p,q,l,e$ being given, we have to solve the following system:$$p=R\phantom{\rule{0.166667em}{0ex}}p+m\phantom{\rule{0.166667em}{0ex}}u\phantom{\rule{4pt}{0ex}},$$$$q=R\phantom{\rule{0.166667em}{0ex}}q-{\tau}_{0}(R\phantom{\rule{0.166667em}{0ex}}p+m\phantom{\rule{0.166667em}{0ex}}u)+m\phantom{\rule{0.166667em}{0ex}}k\phantom{\rule{4pt}{0ex}},$$$$l=R\phantom{\rule{0.166667em}{0ex}}l-u\times \left(R\phantom{\rule{0.166667em}{0ex}}q\right)+k\times \left(R\phantom{\rule{0.166667em}{0ex}}p\right)+m\phantom{\rule{0.166667em}{0ex}}k\times u\phantom{\rule{4pt}{0ex}},$$$$u\xb7\left(R\phantom{\rule{0.166667em}{0ex}}p\right)+\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}m{\parallel u\parallel}^{2}=0\phantom{\rule{4pt}{0ex}},$$$$u=\frac{1}{m}\phantom{\rule{0.166667em}{0ex}}(p-R\phantom{\rule{0.166667em}{0ex}}p)\phantom{\rule{4pt}{0ex}},$$$$q=R\phantom{\rule{0.166667em}{0ex}}q-{\tau}_{0}p+m\phantom{\rule{0.166667em}{0ex}}k\phantom{\rule{4pt}{0ex}},$$$$k=\frac{1}{m}\phantom{\rule{0.166667em}{0ex}}(q-R\phantom{\rule{0.166667em}{0ex}}q+{\tau}_{0}p).$$Finally, because of (32), Equation (34) is simplified as follows:$$l=R\phantom{\rule{0.166667em}{0ex}}l-u\times \left(R\phantom{\rule{0.166667em}{0ex}}q\right)+k\times p\phantom{\rule{4pt}{0ex}}.$$Substituting (37) into the last relation gives:$$l=R\phantom{\rule{0.166667em}{0ex}}l-u\times \left(R\phantom{\rule{0.166667em}{0ex}}q\right)+\frac{1}{m}\phantom{\rule{0.166667em}{0ex}}q\times p-\frac{1}{m}\phantom{\rule{0.166667em}{0ex}}\left(R\phantom{\rule{0.166667em}{0ex}}q\right)\times p\phantom{\rule{4pt}{0ex}}.$$Owing to (32) and the definition of the spin angular momentum ${l}_{0}$$${l}_{0}=l-q\times p\phantom{\rule{4pt}{0ex}}/\phantom{\rule{4pt}{0ex}}m\phantom{\rule{4pt}{0ex}},$$$${l}_{0}=R\phantom{\rule{0.166667em}{0ex}}{l}_{0}\phantom{\rule{4pt}{0ex}}.$$These quantity being given, we have to determine the rotations satisfying the previous relation. It turns out that two cases must be considered.
- -
- Generic orbits : massive particle with spin or rigid body. If ${l}_{0}$ does not vanish, the solutions of (38) are the rotations of an arbitrary angle ϑ about the axis ${l}_{0}$. We know by (36) and (37) that u and k are determined in a unique manner with respect to R and ${\tau}_{0}$. The isotropy group of μ can be parameterised by ϑ and ${\tau}_{0}$. It is a Lie group of dimension 2. The dimension of the orbit of μ is $10-2=8$. The maximum number of independent invariant functions is $10-8=2$. A possible functional basis is composed of:$${s}_{0}=\parallel {l}_{0}\parallel \phantom{\rule{4pt}{0ex}},$$$${e}_{0}=e-\frac{1}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{\parallel p\parallel}^{2}\phantom{\rule{4pt}{0ex}},$$
- -
- Singular orbits : spinless massive particle. In the particular case ${l}_{0}=0$, all the rotations of $\mathbb{SO}\left(3\right)$ satisfy (38), then the isotropy group is of dimension 4. By similar reasoning to the case of non vanishing ${l}_{0}$, we conclude that dimension of the orbit is 6 and the number of invariant functions is 4. A possible functional basis is composed of ${e}_{0}$ and the three null components of ${l}_{0}$.

For the orbits with $m=0$, the reader is referred to [6] (pp. 440, 441).To physically interpret the components of the momentum, let consider a coordinate system ${X}^{\prime}$ in which a particle is at rest and characterized by the components ${p}^{\prime}=0$, ${q}^{\prime}=0$, ${l}^{\prime}={l}_{0}$ and ${e}^{\prime}={e}_{0}$ of the momentum tensor. Let us consider another coordinate system $X=P{X}^{\prime}+C$ with a Galilean boost v and a translation of the origin at $k={x}_{0}$ (hence ${\tau}_{0}=0$ and $R={1}_{{\mathbb{R}}^{3}}$), providing the trajectory equation:$$x={x}_{0}+v\phantom{\rule{0.166667em}{0ex}}t\phantom{\rule{4pt}{0ex}},$$$$p=m\phantom{\rule{0.166667em}{0ex}}v,\phantom{\rule{1.em}{0ex}}q=m\phantom{\rule{0.166667em}{0ex}}{x}_{0},\phantom{\rule{1.em}{0ex}}l={l}_{0}+q\times v,\phantom{\rule{1.em}{0ex}}e={e}_{0}+\frac{m}{2}{\parallel v\parallel}^{2}\phantom{\rule{4pt}{0ex}},$$- -
- The quantity p, proportional to the mass and to the velocity, is the linear momentum.
- -
- The quantity q, proportional to the mass and to the initial position, provides the trajectory equation. It is called passage because indicating the particle is passing through ${x}_{0}$ at time $t=0$.
- -
- The quantity l splits into two terms. The second one, $q\times v\phantom{\rule{4pt}{0ex}}=x\times m\phantom{\rule{0.166667em}{0ex}}v\phantom{\rule{4pt}{0ex}}=x\times p$, is the orbital angular momentum. The first one, ${l}_{0}=l-\phantom{\rule{4pt}{0ex}}q\times p\phantom{\rule{4pt}{0ex}}/\phantom{\rule{4pt}{0ex}}m$, is the spin angular momentum. Their sum, l, is the angular momentum.

- Step 3: parameterizing the orbit. If the particle has an internal structure, introducing the moment of inertia matrix $\mathcal{J}$ and the spin ϖ, we have, according to König’s theorem:$${l}_{0}=\mathcal{J}\phantom{\rule{0.166667em}{0ex}}\varpi ,\phantom{\rule{2.em}{0ex}}{e}_{0}=\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}\varpi \xb7\left(\mathcal{J}\phantom{\rule{0.166667em}{0ex}}\varpi \right)\phantom{\rule{4pt}{0ex}}.$$Hence each orbit defines a particle of mass m, spin ${s}_{0}$, inertia $\mathcal{J}$ and can be parameterized by 8 coordinates, the 3 components of q, the 3 components of p and the 2 components of the unit vector n defining the spin direction, thanks to the momentum map ${\mathbb{R}}^{3}\times {\mathbb{R}}^{3}\times {\mathbb{S}}^{2}\to {\mathfrak{g}}^{*}:(q,p,n)\mapsto \mu =\psi (q,p,n)$ such that:$$l=\frac{1}{m}\phantom{\rule{0.166667em}{0ex}}q\times p+{s}_{0}n,\phantom{\rule{2.em}{0ex}}e=\frac{1}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{\parallel p\parallel}^{2}+\frac{{s}_{0}^{2}}{2}\phantom{\rule{0.166667em}{0ex}}n\xb7\left({\mathcal{J}}^{-1}\phantom{\rule{0.166667em}{0ex}}n\right)\phantom{\rule{4pt}{0ex}}.$$The corresponding measure is $d\lambda ={d}^{3}q\phantom{\rule{0.166667em}{0ex}}{d}^{3}p\phantom{\rule{0.166667em}{0ex}}{d}^{2}n$. For simplicity, we consider further only a singular orbit of dimension 6 representing a spinless particle of mass m, which corresponds to the particular case ${l}_{0}=0$ then $n=0$. It can be parameterized by 6 coordinates, the 3 components of q and the 3 components of p thanks to the map:$$\psi :{\mathbb{R}}^{3}\times {\mathbb{R}}^{3}\to {\mathfrak{g}}^{*}:(q,p)\mapsto \mu =\psi (q,p)\phantom{\rule{4pt}{0ex}},$$$$l=\frac{1}{m}\phantom{\rule{0.166667em}{0ex}}q\times p,\phantom{\rule{2.em}{0ex}}e=\frac{1}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{\parallel p\parallel}^{2}\phantom{\rule{4pt}{0ex}}.$$
- Step 4: modelling the deformation. Statistical mechanics is essentially based on a set of discrete particles and, in essence, incompatible with continuum mechanics. Thus, according to usual arguments, the passage from the statistical mechanics to continuum mechanics is obtained by equivalence between the set of N particles (in huge number) and a box of finite volume V occupied by them, large with respect to the particle size but so small with respect to the continuous medium that it can be considered as infinitesimal. Let us consider N identical particles contained in V, large with respect to the particles but representing the volume element of the continuum thermodynamics. The motion of the matter being characterized by (13), let us consider the change of coordinate$$t={t}^{\prime},\phantom{\rule{2.em}{0ex}}x=\phi ({t}^{\prime},{s}^{\prime})\phantom{\rule{4pt}{0ex}}.$$$$\frac{\partial X}{\partial {X}^{\prime}}=P=\left(\begin{array}{cc}1\hfill & 0\hfill \\ v\hfill & F\hfill \end{array}\right)\phantom{\rule{4pt}{0ex}}.$$Besides, we suppose that the box of initial volume ${V}_{0}$ is at rest in the considered coordinate system ($v=0$) and the deformation gradient F is uniform in the box, then:$$dx=F\phantom{\rule{0.166667em}{0ex}}d{s}^{\prime}\phantom{\rule{4pt}{0ex}}.$$$$p={F}^{-T}{p}^{\prime}\phantom{\rule{4pt}{0ex}}.$$$$q=m\phantom{\rule{0.166667em}{0ex}}x\phantom{\rule{4pt}{0ex}}.$$$$d\lambda ={m}^{3}{d}^{3}x\phantom{\rule{0.166667em}{0ex}}{d}^{3}p\phantom{\rule{0.166667em}{0ex}}{d}^{2}n={m}^{3}{d}^{3}{s}^{\prime}\phantom{\rule{0.166667em}{0ex}}{d}^{3}{p}^{\prime}\phantom{\rule{0.166667em}{0ex}}{d}^{2}n\phantom{\rule{4pt}{0ex}}.$$$$Z=(-W,0)\phantom{\rule{4pt}{0ex}}.$$$$W=\beta \phantom{\rule{0.166667em}{0ex}}U=\left(\begin{array}{c}\beta \hfill \\ 0\hfill \end{array}\right)\phantom{\rule{4pt}{0ex}}.$$$$\mu \phantom{\rule{0.166667em}{0ex}}Z=\beta \phantom{\rule{0.166667em}{0ex}}e\phantom{\rule{4pt}{0ex}},$$$$\mu \phantom{\rule{0.166667em}{0ex}}Z=\frac{\beta}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{\parallel p\parallel}^{2}=\frac{\beta}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{\parallel {F}^{-T}{p}^{\prime}\parallel}^{2}=\frac{\beta}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{p}^{\prime T}{\mathcal{C}}^{-1}{p}^{\prime}\phantom{\rule{4pt}{0ex}}.$$$$z=ln\left({m}^{3}{I}_{0}{I}_{1}{I}_{2}\right)\phantom{\rule{4pt}{0ex}},$$$${I}_{0}={\int}_{{V}_{0}}{d}^{3}{s}^{\prime}={V}_{0}\phantom{\rule{4pt}{0ex}},$$$${I}_{1}={\int}_{{\mathbb{R}}^{3}}{e}^{-\frac{\beta}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{p}^{\prime T}{\mathcal{C}}^{-1}{p}^{\prime}}{d}^{3}{p}^{\prime}\phantom{\rule{4pt}{0ex}},$$$${I}_{2}={\int}_{{\mathbb{S}}^{2}}{d}^{2}n=4\phantom{\rule{0.166667em}{0ex}}\pi \phantom{\rule{4pt}{0ex}}.$$$$z=\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}ln(det\left(\mathcal{C}\right))-\frac{3}{2}\phantom{\rule{0.166667em}{0ex}}ln\beta +{C}^{te}\phantom{\rule{4pt}{0ex}},$$As pointed out by Barbaresco [17], there is a puzzling analogy between the integral occuring in (10) and Koszul–Vinberg characteristic function [18,19]:$${\psi}_{\mathsf{\Omega}}\left(Z\right)={\int}_{{\mathsf{\Omega}}^{*}}{e}^{-\mu \phantom{\rule{0.166667em}{0ex}}Z}\phantom{\rule{0.166667em}{0ex}}d\lambda \phantom{\rule{4pt}{0ex}},$$
- Step 5: identification. It is based on the following result.
**Theorem 2.**The transformation law of the temperature vector $\widehat{\overrightarrow{\mathit{W}}}$ is the same as the one of affine maps $\mathbf{\Theta}$ on the affine space of momentum tensors through the identification:$$Z=(-W,0),\phantom{\rule{2.em}{0ex}}z=m\phantom{\rule{0.166667em}{0ex}}\zeta \phantom{\rule{4pt}{0ex}},$$**Proof.**First of all, let us verify that the form $Z=(-W,0)$ does not depend on the choice of the affine frame. Indeed, starting from ${Z}^{\prime}=(-{W}^{\prime},0)$ and applying the adjoint representation (5) with $d{C}^{\prime}=-{W}^{\prime}$ and $d{P}^{\prime}=0$, we find that $dC=-W$ and $dP=0$ with:$$W=P\phantom{\rule{0.166667em}{0ex}}{W}^{\prime}\phantom{\rule{4pt}{0ex}}.$$$$z={z}^{\prime}-\theta \left(a\right)\phantom{\rule{0.166667em}{0ex}}Ad\left(a\right)\phantom{\rule{0.166667em}{0ex}}{Z}^{\prime}={z}^{\prime}+{K}_{m}P\phantom{\rule{0.166667em}{0ex}}{W}^{\prime}\phantom{\rule{4pt}{0ex}}.$$$$\widehat{W}=\left(\begin{array}{c}W\\ \zeta \end{array}\right)=\left(\begin{array}{c}\beta \\ w\\ \zeta \end{array}\right)\phantom{\rule{4pt}{0ex}}.$$$$\left(\begin{array}{c}W\\ \zeta \end{array}\right)=\left(\begin{array}{cc}P& 0\\ {F}_{1}P& 1\end{array}\right)\phantom{\rule{4pt}{0ex}}\left(\begin{array}{c}{W}^{\prime}\\ {\zeta}^{\prime}\end{array}\right)\phantom{\rule{4pt}{0ex}},$$ - Step 6: boost method. For the box at rest in the coordinate system X, the temperature 4-vector is given by (46):$$W=\left(\begin{array}{c}\beta \\ 0\end{array}\right)\phantom{\rule{4pt}{0ex}}.$$$$\overline{W}=\left(\begin{array}{c}\beta \\ \beta \phantom{\rule{0.166667em}{0ex}}v\end{array}\right)\phantom{\rule{4pt}{0ex}},$$$$\overline{z}=z+\frac{m\phantom{\rule{0.166667em}{0ex}}\beta}{2}\phantom{\rule{0.166667em}{0ex}}{\parallel v\parallel}^{2}=z+\frac{m}{2\phantom{\rule{0.166667em}{0ex}}\beta}\phantom{\rule{0.166667em}{0ex}}{\parallel w\parallel}^{2}\phantom{\rule{4pt}{0ex}}.$$$$z=\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}ln(det\left(\mathcal{C}\right))-\frac{3}{2}\phantom{\rule{0.166667em}{0ex}}ln\beta +\frac{m}{2\phantom{\rule{0.166667em}{0ex}}\beta}\phantom{\rule{0.166667em}{0ex}}{\parallel w\parallel}^{2}+{C}^{te}\phantom{\rule{4pt}{0ex}}.$$$${e}_{int}=e-\frac{1}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}{\parallel p\parallel}^{2}\phantom{\rule{4pt}{0ex}},$$$$s=\frac{3}{2}\phantom{\rule{0.166667em}{0ex}}ln{e}_{int}+\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}ln(det\left(\mathcal{C}\right))+{C}^{te}\phantom{\rule{4pt}{0ex}},$$$$\beta ={\displaystyle \frac{\partial s}{\partial e}}={\displaystyle \frac{3}{2\phantom{\rule{0.166667em}{0ex}}{e}_{int}}},\phantom{\rule{2.em}{0ex}}w=-gra{d}_{\phantom{\rule{0.166667em}{0ex}}p}\phantom{\rule{0.166667em}{0ex}}s={\displaystyle \frac{3}{2\phantom{\rule{0.166667em}{0ex}}{e}_{int}}}\phantom{\rule{0.166667em}{0ex}}{\displaystyle \frac{p}{m}}\phantom{\rule{4pt}{0ex}}.$$
- Step 7: link between z and ζ. As z is an extensive quantity, its value for N identical particles is ${z}_{N}$ = $N\phantom{\rule{0.166667em}{0ex}}z$. Planck’s potential ζ being a specific quantity, we claim that:$$\zeta =\frac{{z}_{N}}{N\phantom{\rule{0.166667em}{0ex}}m}=\frac{z}{m}=\frac{1}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}ln(det\left(\mathcal{C}\right))-\frac{3}{2\phantom{\rule{0.166667em}{0ex}}m}\phantom{\rule{0.166667em}{0ex}}ln\beta +\frac{1}{2\phantom{\rule{0.166667em}{0ex}}\beta}\phantom{\rule{0.166667em}{0ex}}{\parallel w\parallel}^{2}+{C}^{te}\phantom{\rule{4pt}{0ex}}.$$$$\mathcal{H}=\rho \phantom{\rule{0.166667em}{0ex}}\left(\frac{3}{2}\phantom{\rule{0.166667em}{0ex}}\frac{{k}_{B}T}{m}+\frac{1}{2}{\parallel v\parallel}^{2}\right),\phantom{\rule{2.em}{0ex}}p=\rho v,\phantom{\rule{2.em}{0ex}}\sigma =-q\phantom{\rule{0.166667em}{0ex}}{1}_{{\mathbb{R}}^{3}}\phantom{\rule{4pt}{0ex}},$$$$q=\frac{\rho}{m}\phantom{\rule{0.166667em}{0ex}}{k}_{B}T=\frac{N}{V}\phantom{\rule{0.166667em}{0ex}}{k}_{B}T\phantom{\rule{4pt}{0ex}}.$$$$\frac{\partial \mathcal{H}}{\partial t}}+div\phantom{\rule{4pt}{0ex}}\left(\mathcal{H}v-\sigma v\right)=0,\phantom{\rule{2.em}{0ex}}\rho \phantom{\rule{0.166667em}{0ex}}{\displaystyle \frac{dv}{dt}}=-grad\phantom{\rule{0.166667em}{0ex}}q,\phantom{\rule{2.em}{0ex}}{\displaystyle \frac{\partial \rho}{\partial t}}+div\phantom{\rule{0.166667em}{0ex}}\left(\rho \phantom{\rule{0.166667em}{0ex}}v\right)=0\phantom{\rule{4pt}{0ex}}.$$

**Remark**

**4.**

## 8. Conclusions

## Conflicts of Interest

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De Saxcé, G.
Link between Lie Group Statistical Mechanics and Thermodynamics of Continua. *Entropy* **2016**, *18*, 254.
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De Saxcé G.
Link between Lie Group Statistical Mechanics and Thermodynamics of Continua. *Entropy*. 2016; 18(7):254.
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De Saxcé, Géry.
2016. "Link between Lie Group Statistical Mechanics and Thermodynamics of Continua" *Entropy* 18, no. 7: 254.
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