Hyperbolicity of a model for polyatomic gases in relativistic extended thermodynamics

Abstract

The hyperbolicity of a model for relativistic extended thermodynamics of polyatomic gases is here proved for every time direction $\xi _\alpha$; moreover, an expression for the production terms of the field equation is proposed which encloses as particular cases the variant of the Anderson–Witting model already known in the literature, but also a new variant of the Marle model. With these expressions, it is shown that the field equations satisfy automatically the requirement of a non-negative entropy production.

Appendices

Some useful integrals

In the principal text of this article, we need some integrals. They are the following ones:

$$\begin{aligned} \begin{aligned}&V = m \, c \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, f \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} , \\&T^\theta = \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, f \, \left( 1 \, + \, \frac{\mathcal {I}}{m \, c^2} \right) \, p^\theta \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} , \\&A_E^{\beta \theta }= \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, f_E \, \left( 1 \, + \, \frac{\mathcal {I}}{m \, c^2} \right) ^2 \, p^\beta p^\theta \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} , \\&A_{22}^{\mu \nu \beta \gamma }= \frac{c}{m^3} \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, f_E \, \left( 1 \, + \, \frac{2 \, \mathcal {I}}{m \, c^2} \right) ^2 \, p^\mu p^\nu p^\beta p^\gamma \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} . \end{aligned} \end{aligned}$$

We can integrate these expressions by using the representation theorems and with the same methodology used in [4]. So we obtain, for example:

$$\begin{aligned} V_E = 4 \, \pi \, m^3 \, c^3 \, e^{- 1 - \, \frac{m}{k_B} \, \lambda _E } \int _{0}^{+ \infty } \, J_{2,0} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} , \end{aligned}$$

where $\gamma ^*= \gamma \, \left( 1 \, + \, \frac{\mathcal {I}}{m \, c^2} \right)$. But from eq. (26) of [4], we have

$$\begin{aligned} n = 4 \, \pi \, m^3 \, c^3 \, e^{- 1 - \, \frac{m}{k_B} \, \lambda _E } \int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} , \end{aligned}$$

so that the above expression becomes

$$\begin{aligned} V_E = n \, \frac{\int _{0}^{+ \infty } \, J_{2,0} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } . \end{aligned}$$

(31)

By proceeding in the same way for the other quantities, we obtain

$$\begin{aligned} \begin{aligned}&V^{(1)} = - \, \frac{m}{k_B} \, V_E \, (\lambda - \lambda _E) \, - \, \frac{m \, c}{k_B} \, T^\theta _E \, (\lambda _\theta - \lambda _{E \theta }) \, + \, R^{\beta \gamma } \, \Sigma _{\beta \gamma } , \\&T^\theta _E = T_0 \, U^\theta , \, R_{\beta \gamma } = - \, \frac{c}{k_B} \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, f_E \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) \, p^\beta p^\gamma \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} , \\&T_0 = \frac{n}{c} \, \frac{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \left( 1 \, + \, \frac{ \mathcal {I}}{m \, c^2} \right) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \, R^{\beta \gamma } = R_0 \frac{U^\beta U^\gamma }{c^2} + R_1 \, h^{\beta \gamma } , \\&R_0= - \, \frac{m \, n \, c^2}{k_B} \, \frac{\int _{0}^{+ \infty } \, J_{2,2} (\gamma ^*) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&R_1= - \, \frac{m \, n \, c^2}{3 \, k_B} \, \frac{\int _{0}^{+ \infty } \, J_{4,0} (\gamma ^*) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&T^{(1)\theta } = - \, \frac{m}{k_B} \, T_E^\theta \, (\lambda - \lambda _E) \, - \, \frac{1}{k_B} \, A_E^{\theta \beta } \, (\lambda _\beta - \lambda _{E \beta }) \, + \, T^{\theta \beta \gamma } \, \Sigma _{\beta \gamma } , \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned}&T^{\beta \gamma \theta } = \frac{- 1}{m \, k_B} \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, f_E \, \left( 1 \, + \, \frac{\mathcal {I}}{m \, c^2} \right) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) \, p^\beta p^\gamma p^\theta \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} , \\&A_E^{\beta \gamma } = T_3 \frac{U^\beta U^\gamma }{c^2} + T_4 \, h^{\beta \gamma }, \quad T^{\beta \gamma \theta } = T_1 \, U^\beta U^\gamma U^\theta + 3 \, T_2 \, h^{(\beta \gamma } U^{\theta )} , \\&T_3 = m \, n \, c \, \frac{\int _{0}^{+ \infty } \, J_{2,2} (\gamma ^*) \, \left( 1 \, + \, \frac{ \mathcal {I}}{m \, c^2} \right) ^2 \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&T_4 = \frac{1}{3} \, m \, n \, c \, \frac{\int _{0}^{+ \infty } \, J_{4,0} (\gamma ^*) \, \left( 1 \, + \, \frac{ \mathcal {I}}{m \, c^2} \right) ^2 \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&T_1 = - \, \frac{m \, n}{k_B c} \, \frac{\int _{0}^{+ \infty } \, J_{2,3} (\gamma ^*) \, \left( 1 \, + \, \frac{ \mathcal {I}}{m \, c^2} \right) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&T_2 = - \, \frac{m \, n \, c}{3 \, k_B} \, \frac{\int _{0}^{+ \infty } \, J_{4,1} (\gamma ^*) \, \left( 1 \, + \, \frac{ \mathcal {I}}{m \, c^2} \right) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \end{aligned} \end{aligned}$$

(32)

$$\begin{aligned} \begin{aligned}&A_{22}^{\mu \nu \beta \gamma } = \frac{1}{5} \, C_1 \, h^{( \mu \nu } h^{\beta \gamma )} \, +2 \, C_2 \, h^{( \mu \nu } U^\beta U^{ \gamma )} \, + \, C_3 \, U^\mu U^\nu U^\beta U^{ \gamma } , \\&C_1 = n \, c^4 \, \frac{\int _{0}^{+ \infty } \, J_{6,0} (\gamma ^*) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) ^2 \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&C_2 = n \, c^2 \, \frac{\int _{0}^{+ \infty } \, J_{4,2} (\gamma ^*) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) ^2 \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } , \\&C_3 = n \, \frac{\int _{0}^{+ \infty } \, J_{2,4} (\gamma ^*) \, \left( 1 \, + \, \frac{2 \mathcal {I}}{m \, c^2} \right) ^2 \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I}}{\int _{0}^{+ \infty } \, J_{2,1} (\gamma ^*) \, \phi (\mathcal {I}) \, \mathrm{d} \, \mathcal {I} } . \end{aligned} \end{aligned}$$

In the next subsection, we will find some useful consequences of this expressions.

1.1 Some useful properties of the above integrals

Let us define the function

$$\begin{aligned}&h'= - \, k_B \, c \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, e^{- \, 1 \, - \, \frac{\chi }{k_B}} \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} , \end{aligned}$$

(33)

with $\chi$ given by (3)$_2$, and the quadratic form in the variables $\delta \, \lambda _A$:

$$\begin{aligned} K= \frac{\partial ^2 \, h' }{\partial \, \lambda _B \, \partial \, \lambda _A} \, \delta \, \lambda _A \, \delta \, \lambda _B . \end{aligned}$$

Although they have no physical meaning, they will be useful to find properties which will be used below. Now, introducing the multi-index notation $\lambda _A$ where $A=0,1,2$ indicates the number of index: $\lambda _0=\lambda , \lambda _1= \lambda _{\beta } , \lambda _2= \Sigma _{\beta \gamma }$, we have:

$$\begin{aligned} \frac{\partial \, h'}{\partial \, \lambda _A} = c \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, e^{- \, 1 \, - \, \frac{\chi }{k_B}} \, \frac{\partial \, \chi }{\partial \, \lambda _A} \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} . \end{aligned}$$

Since $\frac{\partial \, \chi }{\partial \, \lambda _A}$ does not depend on $\lambda _B$, it follows

$$\begin{aligned} \frac{\partial ^2 \, h'}{\partial \, \lambda _B \, \partial \, \lambda _A} = - \, \frac{c}{k_B} \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, e^{- \, 1 \, - \, \frac{\chi }{k_B}} \, \frac{\partial \, \chi }{\partial \, \lambda _B} \, \frac{\partial \, \chi }{\partial \, \lambda _A} \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} . \end{aligned}$$

and we have

$$\begin{aligned} K= - \, \frac{c}{k_B} \, \int _{\mathfrak {R}^3} \int _{0}^{+ \infty } \, e^{- \, 1 \, - \, \frac{\chi }{k_B}} \, (\delta \, \chi \, )^2 \, \phi (\mathcal {I}) \, \mathrm{d} \overrightarrow{P} \, \mathrm{d} \, \mathcal {I} \, < \, 0 . \end{aligned}$$

(34)

We want now to see its consequences near the equilibrium state. For this aim, we rewrite K as

$$\begin{aligned} \begin{aligned}&K= \frac{\partial ^2 \, h'}{\partial \, \lambda ^2} \, \left( \delta \, \lambda \right) ^2 \, + 2 \, \frac{\partial ^2 \, h' }{\partial \, \lambda \, \partial \, \lambda _\mu } \, \delta \, \lambda \, \delta \, \lambda _\mu + 2 \, \frac{\partial ^2 \, h' }{\partial \, \lambda \, \partial \, \Sigma _{\mu \nu }} \, \delta \, \lambda \, \delta \, \Sigma _{\mu \nu } \\&\qquad + \frac{\partial ^2 h' }{\partial \lambda _\beta \, \partial \lambda _\mu } \, \delta \lambda _\beta \, \delta \lambda _\mu + 2 \, \frac{\partial ^2 \, h' }{\partial \, \lambda _\beta \, \partial \, \Sigma _{\mu \nu }} \, \delta \lambda _\beta \, \delta \Sigma _{\mu \nu } + \frac{\partial ^2 \, h'}{\partial \Sigma _{\beta \gamma } \, \partial \Sigma _{\mu \nu }} \, \delta \Sigma _{\beta \gamma } \, \delta \Sigma _{\mu \nu } . \end{aligned} \end{aligned}$$

By calculating the coefficients of the differentials at equilibrium, it becomes

$$\begin{aligned} \begin{aligned}&K_E= - \, \frac{m}{k_B} \, \left[ V_E \, \left( \delta \, \lambda \right) ^2 \, + 2 \, c \, T^{\mu }_E \, \delta \, \lambda \, \delta \, \lambda _\mu - 2 \, \frac{k_B}{m} \, R^{\mu \nu } \, \delta \, \lambda \, \delta \, \Sigma _{\mu \nu } \right. \\&\quad \left. \, + \, \frac{c}{m} \, A_{E}^{\beta \mu } \, \delta \lambda _\beta \, \delta \lambda _\mu - 2 \, \frac{k_B \, c}{m} \, T^{\beta \mu \nu } \, \delta \, \lambda _\beta \, \delta \, \Sigma _{\mu \nu } + \, A_{22}^{\beta \gamma \mu \nu } \, \delta \, \Sigma _{\beta \gamma } \, \delta \, \Sigma _{\mu \nu } \right] , \end{aligned} \end{aligned}$$

By using (31) and (32), our expression becomes

$$\begin{aligned} - \, \frac{k_B}{c} \, K_E= \sum _{a,b=1}^{3} \, P^{ab} X_a \, X_b + \sum _{a,b=1}^{2} \, Q^{ab} X_{a \mu }\, X^{b \mu } + \, \frac{2}{15} \, m \, C_1 \, X_{\mu \nu } X^{\mu \nu } , \end{aligned}$$

where $X_1= \delta \, \lambda$, $X_2= U^\mu \delta \, \lambda _\mu$, $X_3= U^\mu U^\nu \delta \, \Sigma _{\mu \nu }$, $X_{1 \mu }= h_{\mu }^\nu \delta \, \lambda _\nu$, $X_{2 \mu }= h_{\mu }^\nu U^\delta \, \delta \, \Sigma _{\nu \delta }$, $X_{\mu \nu } = \delta \, \Sigma _{< \mu \nu >_3}$, and moreover, the matrices $P^{ab}$ and $Q^{ab}$ are

$$\begin{aligned} \begin{aligned}&P = \left( \begin{array}{ccc} \frac{m}{c} \, V_E &{} m \, T_0 &{} - \, \frac{k_B}{c^3} \, (R_0 +R_1) \\ m \, T_0 &{} \frac{T_3}{c^2} &{} - \, k_B \, \left( T_1 + \, \frac{1}{c^2} \, T_2 \right) \\ - \, \frac{k_B}{c^3} \, (R_0 +R_1) &{} \, - \, k_B \, \left( T_1 + \, \frac{1}{c^2} \, T_2 \right) &{} \, \, \frac{m}{c} \, \left( \frac{1}{9} \, \frac{C_1}{c^4} \, + \, \frac{2}{3} \, \frac{C_2}{c^2} \, + C_3 \right) \end{array}\right) , \\&Q = \left( \begin{array}{cc} T_4 &{} - 2 \, k_B \, T_2 \\ &{} \\ - 2 \, k_B \, T_2 &{} \, \frac{4}{3} \, \frac{m}{c} \, C_2 \end{array}\right) . \end{aligned} \end{aligned}$$

Since we have $K < 0$, we have also that P and Q are definite positive. But $V_E>0$, $T_4>0$ are immediate consequences of (31) and following equations; so these properties are equivalent to the following ones

$$\begin{aligned} {\tilde{M}}_1&= \left| \begin{matrix} V_E &{} c \, T_0 \\ &{} \\ c \, T_0 &{} \frac{T_3}{m \, c} \\ \end{matrix}\right| >0 , \end{aligned}$$

(35)

$$\begin{aligned} {\tilde{M}}_3&= \left| \begin{matrix} V_E &{} c \, T_0 &{} \frac{k_B}{m \, c^2} \, (R_0 +R_1) \\ &{}&{} \\ c \, T_0 &{} \frac{T_3}{m \, c} &{} \frac{k_B \, c}{m} \, \left( T_1 + \, \frac{1}{c^2} \, T_2 \right) \\ &{}&{} \\ \frac{k_B}{m \, c^2} \, (R_0 +R_1) &{} \, \frac{k_B \, c}{m} \, \left( T_1 + \, \frac{1}{c^2} \, T_2 \right) &{} \, \, \quad \frac{1}{9} \, \frac{C_1}{c^4} \, + \, \frac{2}{3} \, \frac{C_2}{c^2} \, + C_3 \end{matrix}\right|> 0 ,\\ {\tilde{M}}_2&= \left| \begin{matrix} T_4 &{} 2 \, k_B \, T_2 \\ &{} \\ 2 \, k_B \, T_2 &{} \, \frac{4}{3} \, \frac{m}{c} \, C_2 \end{matrix}\right| >0 . \end{aligned}$$

The first one of these equations has been used above, after (20).

A theorem useful in the above considerations

In the main part of this article, we had to exploit the condition

$$\begin{aligned} \sum _{h=0}^{n} a_h \left( \xi _0 \right) ^{2h} \, \left( \xi _1 \right) ^{2n-2h}> 0 , \, \forall \, \left( \xi _1 \right) ^2 , \end{aligned}$$

(36)

and with $\xi _0 = \sqrt{1+(\xi _1)^2}$. We prove now the following

Theorem: “The above condition is equivalent to $a_n>0$ and $x_i \le 1$ for all real roots of $f(x)= \sum _{h=0}^{n} a_h \, x^h$”.

In fact, in the particular case $\xi _1 =0$, the above condition becomes $a_n>0$ and there remains to exploit it for $\xi _1 \ne 0$. For this case, we define

$$\begin{aligned} x = \frac{\left( \xi _0 \right) ^{2}}{\left( \xi _1 \right) ^{2}} = 1 + \, \frac{1}{\left( \xi _1 \right) ^{2}} , \end{aligned}$$

and see that x takes all the values belonging to the interval $] \, 1 , \, + \infty [$ because it is a decreasing function of $\left( \xi _1 \right) ^{2}$ and its limits for $\left( \xi _1 \right) ^{2}$ going to zero or to $+ \infty$ are $+ \infty$ and 1, respectively.

So, for $\xi _1 \ne 0$, the condition (36) can be written as

$$\begin{aligned} f(x) = \sum _{h=0}^{n} a_h \, x^ h> 0 , \, \forall \, x > 1 . \end{aligned}$$

(37)

We see now that our condition $x_i \le 1$ is necessary because, if there is a real root ${\bar{x}}$ of f(x) with ${\bar{x}}>1$, then (37) is violated in $x={\bar{x}}$.

Our condition $x_i \le 1$ is also sufficient because in this case in $] \, 1 , \, + \infty [$ the function f(x) has always the same sign, and moreover,

$\lim _{x \rightarrow + \infty } f(x)= + \infty >0$ (thanks to $a_n>0$). This fact confirms that $f(x) >0$ in $] \, 1 , \, + \infty [$.

• The particular case $n=1$.

The condition $a_n>0$ becomes $a_1>0$; moreover, f(x) has only the root ${\bar{x}}= - \, \frac{a_0}{a_1}$ so that the second condition $x_i \le 1$ becomes $a_1 + a_0 \ge 0$. We can conclude that, in the case $n=1$, the condition (36) becomes

$$\begin{aligned} a_1>0 , \quad a_1 + a_0 \ge 0 . \end{aligned}$$

• The particular case $n=2$.

The condition $a_n>0$ becomes $a_2>0$; moreover, if $(a_1)^2 - 4 \, a_0 \, a_2 <0$, then f(x) has no real roots and also our second condition is satisfied.

If $(a_1)^2 - 4 \, a_0 \, a_2 \ge 0$, then f(x) has two real roots (or only one double root) $x_1 \le x_2$. Since $x_1 \le 1$ and $x_2 \le 1$, we have $\frac{x_1 + x_2}{2} \, \le 1$, i.e. $2 \, a_2 + a_1 \ge 0$. Moreover, we have $f(1) \ge 0$ (because, if $f(1) <0$, we have also $f(x) <0$ in a right neighbourhood of 1) and this condition can be expressed as $a_2 + a_1 + a_0 \ge 0$. These conditions are also sufficient because $f(1) \ge 0$ implies that $1 \, \in \, ] \, - \infty , \, x_1 \, [$ or $1 \, \in \, ] \, x_2 , \, + \infty \, [$ and the first one of these eventualities cannot occur because $\frac{x_1 + x_2}{2} \, \le 1$. We can conclude that, in the case $n=2$, the condition (36) becomes

$$\begin{aligned} \left\{ \begin{matrix} a_2>0 , \, a_0 + a_1 + a_2 \ge 0 , \, a_1 + 2\, a_2 \ge 0 &{} \quad \text{ if } \quad (a_1)^2 - 4 a_0 a_2 \ge 0 , \\ &{} \\ a_2 >0 , &{} \quad \text{ if } \quad (a_1)^2 - 4 a_0 a_2 < 0 . &{} \end{matrix} \right. \end{aligned}$$