# On two possible ways to recover Ordinary Thermodynamics from Extended
Thermodynamics of Polyatomic gases

F. Demontis, S. Pennisi
Department of di Mathematics and Computer Science, University of Cagliari,
09124 Cagliari, Italy<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We consider two possible ways, i.e., the Maxwellian Iteration and the Chapman-
Enskog Method, to recover Relativistic Ordinary Thermodynamics from
Relativistic Extended Thermodynamics of Polyatomic gases with $N$ moments.
Both of these methods give the Eckart equations which are the relativistic
version of the Navier-Stokes and Fourier laws as a first iteration. However,
these methods do not lead to the same expressions of the heat conductivity
$\chi$, the shear viscosity $\mu$, and the bulk viscosity $\nu$ which appear
as coefficients in the Eckart equations. In particular, we prove that the
expressions of $\chi$, $\mu$, and $\nu$ obtained via the Chapman-Enskog method
do not depend on $N$, while those obtained through the Maxwellian Iteration
depend on $N$. Moreover, we also prove that these two methods lead to the same
results in the nonrelativistic limit

.

##  1 Introduction

Rational Extended Thermodynamics (RET) is an elegant theory appreciated by
mathematicians and physicist. This theory was developed in a systematic way by
Liu and Müller in [1] for the classical case, while the relativistic case was
considered by Liu, Müller and Ruggeri in [2]. Both articles [1] and [2] are
based on few natural assumptions, in fact only universal principles111The
universal principles used to develop the RET are: 1) the entropy principle, 2)
the causality principle and 3) the galileian principle in the classical case
or the relativity principle in the relativistic case. are imposed and, as a
consequence of these principles, the hyperbolicity of the field equations is
established. This is an important achievement because in this way in the
relativistic case the paradox of infinity velocity of the propagating waves is
automatically eliminated. These and other related results can also be found in
the book [3] but they concern only the case of a monoatomic gas. The extension
to the case of polyatomic gases was done in [4] for the classical case and in
[5] for the relativistic case. More details on these generalizations to the
polyatomic case can also be found in the book [6]. However, a physical
observation by Pennisi recently published in [7] has caused a revision of the
previous models both for the classical and relativistic cases leading to new
results that can be found in [8] and [9]. In particular, in the article [9] a
satisfactory model for the Relativistic Extended Thermodynamics of Polyatomic
gases with $N$ moments ($ET^{N}$) has been proposed.

However, so far, Ordinary Thermodynamics (OT) has been succesfully used in
practical applications, so a good test for establishing the validity of RET
consists of finding procedures of approximation which allow us to get the
equations of OT as a first step. In the literature two procedures have been
proposed which realize this objective: the Maxwellian Iteration (MI) and the
Chapman-Enskog Method (CEM).
We will prove that the application of both in the relativistic case of these
procedures leads, as a first iteration, to the Eckart equations [10], which in
[2] are called the relativistic version of the Navier-Stokes and Fourier laws
and are two fundamental laws of Relativistic Ordinary Thermodynamics (ROT).
From now on we refer to the Eckart equations as the Navier-Stokes and Fourier
laws. It is important to remark that in the Navier-Stokes and Fourier
equations the following important quantities appear as coefficients: the heat
conductivity $\chi$, the shear viscosity $\mu$, and the bulk viscosity $\nu$.
The aim of this paper is to show that the expressions of $\chi$, $\mu$, and
$\nu$ obtained via the Chapman-Enskog method do not depend on $N$, whereas
these expressions obtained through the Maxwellian Iteration depend on $N$. In
order to get this result, let us recall the basic facts on the field equation
of Relativistic Extended Thermodynamics of Polyatomic gases and Relativistic
Ordinary Thermodynamics.
Let us start by considering the balance equations of ETN [9]. They are
obtained starting from the Boltzmann equation

$\displaystyle\begin{split}&p^{\alpha}\,\partial_{\alpha}\,f=Q\,,\mbox{with}\quad
Q=\frac{U^{\mu}p_{\mu}}{c^{2}\tau}\left[f_{E}-f-f_{E}\,p^{\gamma}q_{\gamma}\frac{3}{m\,c^{4}\rho\,\theta_{1,2}}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\right]\,,\\\
&f=e^{-1\,-\,\frac{\chi}{k_{B}}}\,,\quad\chi=\sum_{n=0}^{N}\frac{1}{m^{n-1}}\lambda_{\alpha_{1}\cdots\alpha_{n}}p^{\alpha_{1}}\cdots
p^{\alpha_{n}}\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)^{n}\,,\\\
&f_{E}=e^{-1\,-\,\frac{m\,\lambda_{E}\,+\,\frac{U_{\mu}}{T}\,p^{\mu}\,\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)}{k_{B}}}\,,\quad\theta_{1,2}=\frac{3\,p}{\rho^{2}\,c^{4}}\,\left(e\,+\,p\right)\,.\end{split}$
(1)

Here, $f$ is the distribution function, $k_{B}$ the Boltzmann constant, $m$
the relativistic particle mass, $\lambda_{\alpha_{1}\cdots\alpha_{n}}$ are
Lagrange multipliers, $c$ the light speed, $p^{\mu}$ the 4-momentum of the
particle (from now on the greek indexes take the values $0,1,2,3$) such that
$p_{\alpha}p^{\alpha}=m^{2}c^{2}$, $\mathcal{I}$ is the internal energy of the
particle due to rotational and vibrational modes, $\tau$ a relaxation time,
$\lambda_{E}$ the first Lagrange multiplier calculated at equilibrium,
$U^{\mu}$ the 4-velocity such that $U_{\alpha}U^{\alpha}=c^{2}$, $Q$ the
production term in the Boltzmann equation, $T$ the absolute temperature,
$\rho$ the mass density, $p$ the pressure, $e$ the energy, and $q^{\alpha}$
the heat flux such that $U_{\alpha}q^{\alpha}=0$. The name “Lagrange
multipliers” has been used in literature [1, 2, 3] and this terminology is due
to the fact that the distribution function $f$ can be obtained through a
variational principle called the Maximum Entropy Principle (MEP) [5] with
constrained variables. In the case with six moments, the heat flux is zero and
$q^{\alpha}$ replaces the scalar anonimous quantities which are present in
[11] and $q^{\alpha}$ is an unknown function to be determined by imposing that
the production of mass and energy-momentum are zero.
After that, the function $\varphi(\mathcal{I})$ is introduced which measures
”how much” the gas is polyatomic.
Finally, by multiplying (1)1 by $\frac{c}{m^{n-1}}\,p^{\alpha_{1}}\cdots
p^{\alpha_{n}}\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)^{n}\,\varphi(\mathcal{I})$
and integrating the result with respect to $d\,\mathcal{I}\,d\,\vec{P}$ one
obtains the balance equations

$\displaystyle\partial_{\alpha}A^{\alpha}=0\,,\quad\partial_{\alpha}A^{\alpha\alpha_{1}}=0\,,$
(2a)
$\displaystyle\partial_{\alpha}A^{\alpha\alpha_{1}\cdots\alpha_{n}}=I^{\alpha_{1}\cdots\alpha_{n}}\,,\quad\mbox{for}\quad
n=2,\,\cdots\,N\,,$ where $\displaystyle
A^{\alpha_{1}\cdots\alpha_{n+1}}=\frac{c}{m^{n-1}}\int_{\Re^{3}}\int_{0}^{+\infty}f\,p^{\alpha_{1}}\cdots
p^{\alpha_{n+1}}\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)^{n}\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,d\,\vec{P}\,,$
(2b) $\displaystyle
I^{\alpha_{1}\cdots\alpha_{n}}=\frac{c}{m^{n-1}}\int_{\Re^{3}}\int_{0}^{+\infty}Q\,p^{\alpha_{1}}\cdots
p^{\alpha_{n}}\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)^{n}\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,d\,\vec{P}\,.$

Obviously, eqs. (2a)1,2 are particular cases of (2a)3 with $n=0,1$ but it is
better to write them separately because they are the mass and energy-momentum
conservation laws, respectively; their productions $I$ and $I^{\alpha_{1}}$
are zero as a consequence of the definition of $Q$ (see (1)2). Moreover, we
sometimes denote $A^{\alpha}$ with $V^{\alpha}$ and $A^{\alpha\alpha_{1}}$
with $T^{\alpha\alpha_{1}}$.
Then, the Lagrange multipliers $\lambda_{\alpha_{1}\cdots\alpha_{n}}$ are
obtained in terms of the physical variables but in a linear departure from
equilibrium (here denoted with the suffix $E$) which is defined as the status
where $\lambda_{\alpha_{1}\cdots\alpha_{n}}=0$ for $n=2,\,\cdots\,N$ and
$\lambda_{\alpha}^{E}=\frac{U_{\alpha}}{T}$. The calculations for the case
$N=2$ can be found in [9]. These calculations are based on the expression of
the following tensor

$\displaystyle
A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}=\frac{c}{m^{n-1}}\int_{\Re^{3}}\int_{0}^{+\,\infty}f_{E}\,p^{\alpha_{1}}\,\cdots\,p^{\alpha_{n+1}}\,\left(1+\,\frac{\mathcal{I}}{m\,c^{2}}\right)^{n}\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,d\,\vec{P}\,.$
(3)

This tensor is only determined in terms of the energy $e$ which is given by

$\displaystyle\frac{e}{\rho\,c^{2}}=\frac{\int_{0}^{+\infty}J^{*}_{2,2}\,\left(1+\,\frac{\mathcal{I}}{m\,c^{2}}\right)\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J^{*}_{2,1}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}\,,$
(4)

where
$J_{m,n}(\gamma)=\int_{0}^{\infty}e^{-\gamma\cosh{s}}\cosh^{n}{s}\sinh^{m}{s}\,ds,\,\,\gamma=\frac{mc^{2}}{k_{B}T},\,\,J^{*}_{m,n}=J_{m,n}\left[\gamma\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\right]\,.$
Here we report a new short proof of this result because we need to use
$A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}$. In fact, from (3) it follows that

$\displaystyle
d\,A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}=-\,\frac{m}{k_{B}}\,\left(A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}d\,\lambda^{E}\,+\,A^{\alpha_{1}\cdots\alpha_{n+2}}_{E}d\,\lambda^{E}_{\alpha_{n+2}}\right)\,.$
(5)

This equation, written for $n=0$, is

$\displaystyle
d\,\left(\rho\,U^{\alpha_{1}}\right)=-\,\frac{m}{k_{B}}\,\left[\rho\,U^{\alpha_{1}}\,d\,\lambda^{E}\,+\,\left(e\,\frac{U^{\alpha_{1}}U^{\alpha_{2}}}{c^{2}}\,+\,p\,h^{\alpha_{1}\alpha_{2}}\right)d\,\lambda^{E}_{\alpha_{2}}\right]\,,$

whose contraction with $U^{\alpha_{1}}$ allows us to determine

$\displaystyle
d\,\lambda^{E}=-\,\frac{k_{B}}{m\,\rho}\,d\,\rho\,-\,\frac{e}{\rho\,c^{2}}\,U^{\alpha_{2}}\,d\,\lambda^{E}_{\alpha_{2}}\,.$

By substituting this in eq. (5), we find

$\displaystyle
d\,A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}=A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}\,\left(\frac{1}{\rho}\,d\,\rho\,+\,\frac{e\,m}{\rho\,c^{2}k_{B}}\,U^{\gamma}\,d\,\lambda^{E}_{\gamma}\right)\,-\,\frac{m}{k_{B}}\,A^{\alpha_{1}\cdots\alpha_{n+2}}_{E}d\,\lambda^{E}_{\alpha_{n+2}}\,.$

If we take $\rho$ and $\lambda_{\gamma}^{E}$ as independent variables, the
coefficient of $d\,\rho$ shows that $A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}$ is
linear and homogeneous in the variable $\rho$, while the coefficient of
$d\,\lambda^{E}_{\gamma}$ allows us to determine

$\displaystyle
A^{\alpha_{1}\cdots\alpha_{n+2}}_{E}=-\,\frac{k_{B}}{m}\,\frac{\partial\,A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}}{\partial\,\lambda^{E}_{\alpha_{n+2}}}\,+\,\frac{e}{\rho\,c^{2}}\,A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}\,U^{\alpha_{n+2}}\,.$
(6)

Taking into account this result, all the tensors
$A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}$ are determined in terms of the previous
ones. Obviously, we must be careful and express everything in terms of $\rho$
and $\lambda_{\gamma}^{E}$. Regarding $\lambda_{\gamma}^{E}$, we note that

$\displaystyle\lambda_{\gamma}^{E}=\frac{U_{\gamma}}{T}\quad\rightarrow\quad
T=\frac{c}{\sqrt{\lambda_{\delta}^{E}\lambda^{E\delta}}}\,;\quad
U_{\gamma}=\frac{c}{\sqrt{\lambda_{\delta}^{E}\lambda^{E\delta}}}\,\lambda_{\gamma}^{E}\,;\quad
A^{\gamma}_{E}=\rho\,U^{\gamma}=\frac{\rho\,c}{\sqrt{\lambda_{\delta}^{E}\lambda^{E\delta}}}\,\lambda_{\gamma}^{E}\,.$

As a test, let us consider eq. (6) for $n=0$, i.e.,

$\displaystyle T^{\alpha_{1}\alpha_{2}}_{E}$
$\displaystyle=-\,\frac{k_{B}}{m}\,\frac{\partial\,A_{E}^{\alpha_{1}}}{\partial\,\lambda^{E}_{\alpha_{2}}}\,+\,\frac{e}{\rho\,c^{2}}\,A^{\alpha_{1}}_{E}\,U^{\alpha_{2}}$
$\displaystyle=-\,\frac{k_{B}}{m}\left(\frac{\rho\,c}{\sqrt{\lambda_{\delta}^{E}\lambda^{E\delta}}}g^{\alpha_{1}\alpha_{2}}\,-\,\frac{\rho\,c}{\left(\lambda_{\delta}^{E}\lambda^{E\delta}\right)^{3/2}}\lambda^{\alpha_{1}}_{E}\,\lambda_{E}^{\alpha_{2}}\right)\,+\,\frac{e}{\lambda_{\delta}^{E}\lambda^{E\delta}}\lambda^{\alpha_{1}}_{E}\,\lambda_{E}^{\alpha_{2}}$
$\displaystyle=\frac{k_{B}}{m}\,\rho\,T\,h^{\alpha_{1}\alpha_{2}}\,+\,\frac{e}{c^{2}}\,U^{\alpha_{1}}U^{\alpha_{2}}\,.$

So we have obtained the expression for the coefficient of
$U^{\alpha_{1}}U^{\alpha_{2}}$, while the other term gives

$\displaystyle p=n\,k_{B}\,T\,.$ (7)

Note: In the above calculations we have used the property

$\displaystyle\lambda^{E\gamma}\lambda^{E}_{\gamma}=g^{\mu\gamma}\,\lambda^{E}_{\mu}\lambda^{E}_{\gamma}\quad\rightarrow\quad\frac{\partial}{\partial\,\lambda^{E}_{\delta}}\left(\lambda^{E\gamma}\lambda^{E}_{\gamma}\right)=2\,g^{\mu\delta}\,\lambda^{E}_{\mu}=2\,\lambda^{E\delta}\,.$

We note also that (6) does not allow us to obtain the expression for the
energy $e$; so to find it we must go back to the definitions (3) for $n=0$ and
$n=1$ and contract them by $U_{\alpha_{1}}$ and
$U_{\alpha_{1}}U_{\alpha_{2}}$, respectively. The results are

$\displaystyle\begin{split}&\rho\,c^{2}=\frac{4\,\pi\,m^{4}c^{5}}{\sqrt{-\,g}}\,e^{-1-\,\frac{m\,\lambda^{E}}{k_{B}}}\int_{0}^{+\infty}J^{*}_{2,1}\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,,\\\
&e\,c^{2}=\frac{4\,\pi\,m^{4}c^{7}}{\sqrt{-\,g}}\,e^{-1-\,\frac{m\,\lambda^{E}}{k_{B}}}\int_{0}^{+\infty}J^{*}_{2,2}\,\left(1+\,\frac{\mathcal{I}}{m\,c^{2}}\right)\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,,\end{split}$

the second one of these expressions, divided by the first one gives the
formula (4) reported above.
From these considerations it is possible to derive the expression for
$A^{\alpha\alpha_{1}\cdots\alpha_{j}}_{E}$ reported in eqs. (29) and (30) of
[7] and rewritten in eqs.(14)-(16) of [9]. For the convenience of the reader
we write this expression below:

$\displaystyle
A^{\alpha_{1}\cdots\alpha_{j+1}}_{E}=\sum_{k=0}^{\left[\frac{j+1}{2}\right]}\rho
c^{2k}\theta_{k,j}\,h^{(\alpha_{1}\alpha_{2}}\cdots
h^{\alpha_{2k-1}\alpha_{2k}}U^{\alpha_{2k+1}}\cdots U^{\alpha_{j+1})}\,,$ (8)

where the round brackets appearing in $h^{(\alpha_{1}\alpha_{2}}\cdots
h^{\alpha_{2k-1}\alpha_{2k}}U^{\alpha_{2k+1}}\cdots U^{\alpha_{j+1})}$ denote
the symmetric part of this tensor, while the scalar coefficients
$\theta_{k,j}$ are defined as follows:

$\displaystyle\theta_{k,j}=\frac{1}{2k+1}\begin{pmatrix}j+1\\\
2k\end{pmatrix}\frac{\int_{0}^{+\infty}J_{2k+2,j+1-2k}^{*}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)^{j}\,\phi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}\,.$
(9)

Moreover, $\theta_{k,j}$ can be determined by the recurrence relations which
use the quantity $\gamma=\frac{m\,c^{2}}{k_{B}T}$:

$\displaystyle\begin{split}&\theta_{0,0}=1\,,\\\
&\theta_{0,j+1}=\frac{e}{\rho\,c^{2}}\,\theta_{0,j}\,-\,\frac{\partial\,\theta_{0,j}}{\partial\,\gamma}\,,\\\
&\\\
&\theta_{h,j+1}=\frac{j+2}{\gamma}\left(\theta_{h,j}+\frac{j+3-2h}{2h}\theta_{h-1,j}\right)\,,\hskip
54.06006pt\mbox{for }h=1,\,\cdots,\,\left[\frac{j+1}{2}\right]\,,\\\ &\\\
&\theta_{\frac{j+2}{2},j+1}=\frac{1}{\gamma}\theta_{\frac{j}{2},\,j}\,,\hskip
179.25244pt\mbox{for $j$ even}\,.\end{split}$ (10)

Regarding the production terms in the balance equations (2b)2 we see that, by
means of (3), it becomes

$\displaystyle\begin{split}&I^{\alpha_{1}\cdots\alpha_{i}}=-\,\frac{U_{\alpha}}{c^{2}\tau}\,\left(A^{\alpha\alpha_{1}\cdots\alpha_{i}}\,-\,A^{\alpha\alpha_{1}\cdots\alpha_{i}}_{E}\right)\,-\,\frac{3}{\rho\,c^{6}\,\tau\,\theta_{1,2}}\,U_{\alpha}\,q_{\beta}\,A^{\alpha\beta\alpha_{1}\cdots\alpha_{i}}_{E}\,,\\\
&\mbox{where}\quad
q^{\beta}=U_{\gamma}\,\left(T^{\gamma\beta}\,-\,T^{\gamma\beta}_{E}\right)\,.\end{split}$
(11)

So far we have described the results of Extended Thermodynamics of polyatomic
gases, as obtained in [9] following the new ideas of [7].
On the other hand, Relativistic Ordinary Thermodynamics (ROT) uses only the
equations (2a) with the following definitions

$\displaystyle\begin{split}&A^{\alpha\alpha_{1}}=T^{\alpha\alpha_{1}}=\frac{e}{c^{2}}\,U^{\alpha}U^{\alpha_{1}}\,+\,(p+\pi)h^{\alpha\alpha_{1}}\,+\,\frac{2}{c^{2}}\,U^{(\alpha}q^{\alpha_{1})}\,+\,t^{<\alpha\alpha_{1}>}\quad\mbox{where}\\\
&\pi=-\nu\,\partial_{\alpha}\,U^{\alpha}\,,\,q^{\beta}=-\,\chi\,h^{\alpha\beta}\left(\partial_{\alpha}\,T\,-\,\frac{T}{c^{2}}\,U^{\mu}\,\partial_{\mu}\,U_{\alpha}\right)\,,\,t_{<\beta\gamma>}=2\,\mu\,h^{\alpha}_{\beta}\,h^{\mu}_{\gamma}\,\partial_{<\alpha}\,U_{\mu>}\,,\end{split}$
(12)

where the angular brackets appearing in $t_{<\beta\gamma>}$ denote the
traceless symmetric part of the tensor.
Here (12)2-4 are the Eckart equations [10]; in particular, following [2]
equation (12)3 corresponds to relativistic version of the Fourier law while
equations (12)2 and (12)4 correspond to the relativistic version of the
Navier-Stokes law. The coefficients $\nu$, $\chi$, $\mu$ are called the bulk
viscosity, the heat conductivity and the shear viscosity, respectively.
These equations have the drawback that they are not hyperbolic but parabolic.
As we have already said, this was the reason for the birth of Extended
Thermodynamics whose equations are hyperbolic.

The paper is organized as follows: In section 2 we briefly recall how the MI
procedures works and show how it is possible to reconstruct the laws of
Relativistic Ordinary Thermodynamics by using this procedure. Moreover, in
section 2 we derive the expressions of the heat conductivity $\chi$, the shear
viscosity $\mu$, and the bulk viscosity $\nu$ in the particular cases $N=3$
and $N=2$ putting in evidence that, if one uses the MI procedure, these
expressions depend on the number of moments $N$. In section 3 we explain how
the CEM procedure works and derive the laws of the ROT by using this
procedure. In section 4, generalizing a well-known result of [12] for
monoatomic case, we prove that in the non-relativistic case the MI and the CEM
procedures lead to the same results. The convergence of $\nu$, $\chi$, $\mu$
in the non relativistic limit is proved in section 4.1. Then, the results
obtained are summarized in section 5. Finally, an appendix is devoted to some
particular integrals used to develop the computations of this paper.

##  2 The Maxwellian Iteration

The Maxwellian Iteration Method was applied to recover OT from Extended
Thermodynamics of monoatomic gases in [1] for the non relativistic case and in
[2] for the relativistic framework. The relativistic case for polyatomic gases
with $N=2$, and for its subsystems with fourteen and six moments, has been
treated in [9]. In the next section MI will be implemented in the case of an
arbitrary number $N$. It works in the following way:

  * •

The eqs. (2a) are considered, but with their left hand sides calculated at
equilibrium and their right hand sides at first order with respect to
equilibrium, i.e.,

$\displaystyle\begin{split}&\partial_{\alpha}A^{\alpha}_{E}=0\,,\quad\partial_{\alpha}A^{\alpha\alpha_{1}}_{E}=0\,,\\\
&\partial_{\alpha}A^{\alpha\alpha_{1}\cdots\alpha_{n}}_{E}=I^{\alpha_{1}\cdots\alpha_{n}}_{MI}\,,\quad\mbox{for}\quad
n=2,\,\cdots\,N\,,\end{split}$ (13)

where the meaning of the subscrpt MI will be introduced in the next item.

  * •

The deviations of the independent variables from equilibrium are calculated in
terms of $\partial_{\alpha}\lambda^{E}$ and
$\partial_{\alpha}\lambda_{\mu}^{E}$ from (13)3; they are called ”first
iterates” and we will denote them with a suffix $MI$. After that, they are
substituted in $T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}$ with
$T^{\alpha\beta}$ given by (12)1.

  * •

The quantities $\partial_{\alpha}\lambda^{E}$ and
$U^{\alpha}U^{\mu}\partial_{\alpha}\lambda_{\mu}^{E}$ are calculated from
(13)1,2 and substituted in the expression of
$T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}$ obtained in the previous step.

In this way one obtains (12) of ROT, with particular expressions
$\nu_{MI}^{N}$, $\chi_{MI}^{N}$, $\mu_{MI}^{N}$ of the bulk viscosity $\nu$,
the heat conductivity $\chi$ and the shear viscosity $\mu$.
However, these expressions depend on the number $N$ of the extended model from
which they come from. For example, in [9] it was found that, for the subsystem
with 14 moments, the values of $\mu$ and $\chi$ remain the same as for the
model with 15 moments (i.e., $N=2$) while $\nu$ changes. This expression for
$\nu$ changes again for its further subsystem with six moments (In this
subsystem with six moments $\mu$ and $\chi$ do not play a role). This is not
fully satisfactory because there is only one Ordinary Thermodynamics and it is
strange that its equations depend on the number $N$ of the extended model from
which they are derived. In the next subsection this Maxwellian iteration will
be described in more detail for an arbitrary number $N$; furthermore, we will
see what is the difference of the expressions of $\mu$, $\chi$ and $\nu$ in
the cases $N=3$ and $N=2$ in the subsections 2.2 and 2.3, respectively.

###  2.1 ROT recovered with the Maxwellian Iteration

It is easy to prove that eqs. (13)1,2 can be written as (see [5] for more
details)

$\displaystyle\begin{split}&V^{\alpha}_{E}\,\partial_{\alpha}\,\lambda^{E}\,+\,T^{\alpha\mu}_{E}\,\partial_{\alpha}\,\lambda^{E}_{\mu}=0\,,\quad
T^{\alpha\beta}_{E}\,\partial_{\alpha}\,\lambda^{E}\,+\,A^{\alpha\beta\mu}_{E}\,\partial_{\alpha}\,\lambda^{E}_{\mu}=0\,.\end{split}$
(14)

The first one of these equations and the second one contracted with
$U_{\beta}$ give a system whose solution is:

$\displaystyle\begin{split}&U^{\alpha}\,\partial_{\alpha}\lambda^{E}=-\,\left|\begin{matrix}\rho&&\frac{e}{c^{2}}\\\
&&\\\
\frac{e}{c^{2}}&&\rho\,\theta_{0,2}\end{matrix}\right|^{-1}\,\left|\begin{matrix}p&&\frac{e}{c^{2}}\\\
&&\\\
\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}&&\rho\,\theta_{0,2}\end{matrix}\right|\,h^{\alpha\delta}\,\partial_{\alpha}\lambda^{E}_{\delta}\,,\\\
&U^{\alpha}U^{\beta}\,\partial_{\alpha}\lambda^{E}_{\beta}=-\,\left|\begin{matrix}\rho&&\frac{e}{c^{2}}\\\
&&\\\
\frac{e}{c^{2}}&&\rho\,\theta_{0,2}\end{matrix}\right|^{-1}\,\left|\begin{matrix}\rho&&p\\\
&&\\\
\frac{e}{c^{2}}&&\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}\end{matrix}\right|\,h^{\alpha\delta}\,\partial_{\alpha}\lambda^{E}_{\delta}\,.\end{split}$
(15)

It is interesting to note that
$h^{\alpha\mu}\,\partial_{\alpha}\,\lambda^{E}_{\mu}=\frac{1}{T}\,h^{\alpha\mu}\,\partial_{\alpha}\,U_{\mu}=-\,\frac{1}{T}\,\partial_{\alpha}\,U^{\alpha}$.
The equation (14)2, contracted with $h^{\beta}_{\delta}$, allows us to
determine

$\displaystyle
h^{\alpha\theta}\,\partial_{\alpha}\lambda^{E}=-\,\frac{2}{3}\,\frac{\rho}{p}\,c^{2}\,\theta_{1,2}\,h^{\theta(\alpha}U^{\delta)}\,\partial_{\alpha}\lambda^{E}_{\delta}\,.$
(16)

Now we consider (13)3, with use of (2b), (1)2 and taking into account that
$q_{\gamma}=U^{\beta}\left(T_{\gamma\beta}\,-\,T_{E\,\gamma\beta}\right)$;
jointly with $V^{\alpha}-V^{\alpha}_{E}=0$,
$U_{\alpha}U_{\beta}\,\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)=0$ we
obtain the system

$\displaystyle\begin{split}&\sum_{m=0}^{N}U_{\alpha}\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}\beta_{1}\cdots\beta_{m}}_{E}\,+\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,g_{\gamma\delta\,U_{\beta}\,A^{\alpha\gamma\alpha_{1}\cdots\alpha_{n}}_{E}A^{\delta\beta\beta_{1}\cdots\beta_{m}}_{E}}\right)\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)=\\\
&=-\,c^{2}\tau\,\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}}_{E}\,\partial_{\alpha}\,\lambda^{E}\,+\,A^{\alpha\alpha_{1}\cdots\alpha_{n}\mu}_{E}\,\partial_{\alpha}\,\lambda^{E}_{\mu}\right)^{MI}\\\
&=-\,c^{2}\tau\,\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}}_{E}\,\partial_{\alpha}\,\lambda^{E}\,+\,A^{\alpha\alpha_{1}\cdots\alpha_{n}\mu}_{E}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,,\,\mbox{for}\quad
n=2\,,\,\cdots\,,N\,,\\\
&\sum_{m=0}^{N}A^{\alpha\beta_{1}\cdots\beta_{m}}_{E}\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)^{MI}=0\,,\\\
&\sum_{m=0}^{N}U_{\alpha}U_{\beta}\,A^{\alpha\beta\beta_{1}\cdots\beta_{m}}_{E}\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)^{MI}=0\,.\end{split}$
(17)

From this system we can obtain
$\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)^{MI}$
as a linear and homogeneous combination of $\partial_{\alpha}\,\lambda^{E}$,
$\partial_{(\alpha}\,\lambda^{E}_{\mu)}$. By using (15) and (16),
$\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)^{MI}$
will be a linear and homogeneous combination of
$h^{\alpha\mu}\,\partial_{\alpha}\,\lambda^{E}_{\mu}$,
$h^{\delta\alpha}\,U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}$ and
$h^{\alpha<\delta}\,h^{\beta>_{3}\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}=h^{\alpha\delta}\,h^{\beta\mu}\,\partial_{<\alpha}\,\lambda^{E}_{\mu>_{3}}$.
By substituting these
$\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)^{MI}$
in

$\displaystyle\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}=-\,\frac{m}{k_{B}}\sum_{m=0}^{N}A^{\alpha\beta\beta_{1}\cdots\beta_{m}}_{E}\left(\lambda_{\beta_{1}\cdots\beta_{m}}\,-\,\lambda_{\beta_{1}\cdots\beta_{m}}^{E}\right)^{MI}\,,$
(18)

we obtain that
$h_{\alpha\beta}\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}$ is a
scalar function which is linear and homogeneous in the independent variables
$h^{\alpha\mu}\,\partial_{\alpha}\,\lambda^{E}_{\mu}$ (a scalar variable),
$h^{\delta\alpha}\,U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}$ (a
3-dimensional vector variable) and
$h^{\alpha\delta}\,h^{\beta\mu}\,\partial_{<\alpha}\,\lambda^{E}_{\mu>_{3}}$
(a variable which is a 3-dimensional second order tensor). For the
representation theorems (see [13] for more details on the representation
theorems) the quantity
$h_{\alpha\beta}\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}$ must be
proportional to
$h^{\alpha\mu}\,\partial_{\alpha}\,\lambda^{E}_{\mu}=\frac{-1}{T}\,\partial_{\alpha}\,U^{\alpha}$.
In this way, (12)1 is obtained with $\nu_{MI}^{N}$ instead of $\nu$.
Similarly,
$h_{\alpha\delta}\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}$ is a
3-dimensional vector function which is linear and homogeneous in the same
independent variables. So, by the representation theorems,
$h_{\alpha\delta}\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}$ must
be proportional to
$2\,h^{\delta\alpha}\,U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}=\frac{-c^{2}}{T^{2}}\,h^{\delta\alpha}\,\left(\partial_{\alpha}\,T\,-\,\frac{T}{c^{2}}\,U^{\mu}\,\partial_{\mu}\,U_{\alpha}\right)$.
In this way, (12)2 is obtained with $\chi_{MI}^{N}$ instead of $\chi$.
Finally,
$h_{\mu}^{<\beta}h^{\gamma>}_{\nu}\,\left(T^{\mu\nu}-T^{\mu\nu}_{E}\right)^{MI}$
is a 3-dimensional traceless second order tensorial function which is linear
and homogeneous in the same independent variables. By the representation
theorems
$h_{\mu}^{<\beta}h^{\gamma>}_{\nu}\,\left(T^{\mu\nu}-T^{\mu\nu}_{E}\right)^{MI}$
must be proportional to
$h^{\alpha}_{\beta}\,h^{\mu}_{\gamma}\,\partial_{<\alpha}\,U_{\mu>}$. In this
way, (12)3 is obtained with $\mu_{MI}^{N}$ instead of $\mu$.
As examples of this procedure, we will consider the particular cases with
$N=3$ and $N=2$ in the next subsections.

###  2.2 The Maxwellian iteration in the case $N=3$.

We begin with the case $N=3$ because the calculations developed in this case
allow us also to treat the case with $N=2$ which will be considered in the
next subsection. By comparing the results obtained in the case $N=3$ with
those obtained in the case $N=2$, it is easy to see that the expressions of
the coefficients of bulk viscosity, heat conductivity and shear stress are
different. This argument allows us to prove that the expressions of $\nu,\chi$
and $\mu$ obtained with the Maxwellian iteration depend on $N$. So far in
literature, this fact has been proved only comparing the case $N=2$ with two
of its subsystems.
The equations (17) and (18) for $N=3$ are:

$\displaystyle\begin{split}&U_{\alpha}\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}}_{E}\,+\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,g_{\gamma\delta\,U_{\beta}\,A^{\alpha\gamma\alpha_{1}\cdots\alpha_{n}}_{E}A^{\delta\beta}_{E}}\right)\left(\lambda\,-\,\lambda^{E}\right)^{MI}+\\\
&+\,U_{\alpha}\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}\beta_{1}}_{E}\,+\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,g_{\gamma\delta\,U_{\beta}\,A^{\alpha\gamma\alpha_{1}\cdots\alpha_{n}}_{E}A^{\delta\beta\beta_{1}}_{E}}\right)\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}+\\\
&+\,\sum_{m=0}^{N}U_{\alpha}\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}\beta_{1}\beta_{2}}_{E}\,+\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,g_{\gamma\delta\,U_{\beta}\,A^{\alpha\gamma\alpha_{1}\cdots\alpha_{n}}_{E}A^{\delta\beta\beta_{1}\beta_{2}}_{E}}\right)\,\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}\,+\\\
&+\,U_{\alpha}\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}\beta_{1}\beta_{2}\beta_{3}}_{E}\,+\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,g_{\gamma\delta\,U_{\beta}\,A^{\alpha\gamma\alpha_{1}\cdots\alpha_{n}}_{E}A^{\delta\beta\beta_{1}\beta_{2}\beta_{3}}_{E}}\right)\,\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}=\\\
&=-\,c^{2}\tau\,\left(A^{\alpha\alpha_{1}\cdots\alpha_{n}}_{E}\,\partial_{\alpha}\,\lambda^{E}\,+\,A^{\alpha\alpha_{1}\cdots\alpha_{n}\mu}_{E}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,,\,\mbox{for}\quad
n=2\,,\,\cdots\,,3\,,\end{split}$ (19) $\displaystyle
A^{\alpha}_{E}\left(\lambda\,-\,\lambda^{E}\right)^{MI}\,+\,A^{\alpha\beta_{1}}_{E}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}\,+\,A^{\alpha\beta_{1}\beta_{2}}_{E}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}\,+$
$\displaystyle+\,A^{\alpha\beta_{1}\beta_{2}\beta_{3}}_{E}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}=0\,,$
$\displaystyle
U_{\alpha}U_{\beta}\,\left[A^{\alpha\beta}_{E}\left(\lambda\,-\,\lambda^{E}\right)^{MI}\,+\,A^{\alpha\beta\beta_{1}}_{E}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}\,+\,A^{\alpha\beta\beta_{1}\beta_{2}}_{E}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}\,+\right.$
$\displaystyle+\left.\,A^{\alpha\beta\beta_{1}\beta_{2}\beta_{3}}_{E}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}\right]=0\,.$
$\displaystyle\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}=-\,\frac{m}{k_{B}}\left[A^{\alpha\beta}_{E}\left(\lambda\,-\,\lambda^{E}\right)^{MI}\,+\,A^{\alpha\beta\beta_{1}}_{E}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}\,+\right.$
$\displaystyle+\left.\,A^{\alpha\beta\beta_{1}\beta_{2}}_{E}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}\,+\,A^{\alpha\beta\beta_{1}\beta_{2}\beta_{3}}_{E}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}\right]\,.$

We note that in the case $N=2$ we have to consider (19)1 only for $n=2$ and to
put $\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}=0$; so the
calculations in this section allow us to consider also the case $N=2$ which is
the object of the next section.

Determination of the bulk viscosity $\nu$. Here we consider eqs. (19)1 with
$n=2$ contracted by $\frac{U_{\alpha_{1}}U_{\alpha_{2}}}{\rho\,c^{6}}$, (19)1
with $n=3$ contracted by
$\frac{U_{\alpha_{1}}U_{\alpha_{2}}U_{\alpha_{3}}}{\rho\,c^{8}}$, (19)1 with
$n=2$ contracted by $\frac{h_{\alpha_{1}\alpha_{2}}}{\rho\,c^{4}}$, (19)1 with
$n=3$ contracted by
$\frac{h_{\alpha_{1}\alpha_{2}}U_{\alpha_{3}}}{\rho\,c^{6}}$, (19)2 contracted
by $\frac{U_{\alpha}}{\rho\,c^{2}}$, (19)3 divided by $\rho\,c^{6}$ and (19)4
contracted by $-\,k_{B}\,\frac{h_{\alpha\beta}}{m\,\rho\,c^{2}}$.
So we obtain a system $\sum_{j=1}^{6}a_{ij}X^{j}=b_{i}$ constituted by 7
equations in the 6 unknowns $X^{1}=\left(\lambda\,-\,\lambda^{E}\right)^{MI}$,
$X^{2}=U^{\beta_{1}}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}$,
$X^{3}=U^{\beta_{1}}U^{\beta_{2}}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$,
$X^{4}=U^{\beta_{1}}U^{\beta_{2}}U^{\beta_{3}}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}$,
$X^{5}=c^{2}h^{\beta_{1}\beta_{2}}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$,
$X^{6}=c^{2}h^{\beta_{1}\beta_{2}}U^{\beta_{3}}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}$
where

$\displaystyle\begin{split}&a_{1k}=\theta_{0,k+1}\,+\,3\,\frac{\theta_{0,3}}{\theta_{1,2}}\,\theta_{0,k}\,,\quad\mbox{for}\quad
k=1,2,3,4.\,;\quad
a_{15}=\frac{1}{10}\,\theta_{1,4}\,+\,\frac{1}{2}\,\frac{\theta_{0,3}\,\theta_{1,3}}{\theta_{1,2}}\,;\\\
&a_{16}=\frac{1}{5}\,\theta_{1,5}\,+\,\frac{9}{10}\,\frac{\theta_{0,3}\,\theta_{1,4}}{\theta_{1,2}}\,;\\\
&b_{1}=-\,\tau\,\left[\theta_{0,2}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\theta_{0,3}U^{\alpha}U^{\mu}\,+\,\frac{1}{6}\,c^{2}\,\theta_{1,3}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]\,;\end{split}$
(20)
$\displaystyle\begin{split}&a_{2k}=\theta_{0,k+2}\,+\,3\,\frac{\theta_{0,4}}{\theta_{1,2}}\,\theta_{0,k}\,,\quad\mbox{for}\quad
k=1,2,3,4.\,;\quad
a_{25}=\frac{1}{15}\,\theta_{1,5}\,+\,\frac{1}{2}\,\frac{\theta_{0,4}\,\theta_{1,3}}{\theta_{1,2}}\,;\\\
&a_{26}=\frac{1}{7}\,\theta_{1,6}\,+\,\frac{9}{10}\,\frac{\theta_{0,4}\,\theta_{1,4}}{\theta_{1,2}}\,;\\\
&b_{2}=-\,\tau\,\left[\theta_{0,3}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\theta_{0,4}U^{\alpha}U^{\mu}\,+\,\frac{3}{10}\,c^{2}\,\theta_{1,4}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]\,;\end{split}$
$\displaystyle\begin{split}&a_{31}=\theta_{1,1}\,+\,\frac{3}{2}\,\frac{\theta_{0,1}\,\theta_{1,3}}{\theta_{1,2}}\,;\,a_{32}=\frac{1}{3}\,\theta_{1,3}\,+\,\frac{3}{2}\,\frac{\theta_{0,2}\,\theta_{1,3}}{\theta_{1,2}}\,;\,a_{33}=\frac{3}{10}\,\theta_{1,4}\,+\,\frac{3}{2}\,\frac{\theta_{0,3}\,\theta_{1,3}}{\theta_{1,2}}\,;\\\
&a_{34}=\frac{4}{15}\,\theta_{1,5}\,+\,\frac{3}{2}\,\frac{\theta_{0,4}\,\theta_{1,3}}{\theta_{1,2}}\,;\,a_{35}=\frac{1}{3}\,\theta_{2,4}\,+\,\frac{1}{4}\,\frac{\left(\theta_{1,3}\right)^{2}}{\theta_{1,2}}\,;\\\
&a_{36}=\frac{1}{3}\,\theta_{2,5}\,+\,\frac{9}{20}\,\frac{\theta_{1,4}\,\theta_{1,3}}{\theta_{1,2}}\,;\\\
&b_{3}=-\,\tau\,\left[\theta_{1,2}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\frac{1}{2}\,\theta_{1,3}U^{\alpha}U^{\mu}\,+\,\frac{5}{3}\,c^{2}\,\theta_{2,3}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]\,;\end{split}$
$\displaystyle\begin{split}&a_{41}=\frac{1}{2}\,\theta_{1,3}\,+\,\frac{9}{10}\,\frac{\theta_{0,1}\,\theta_{0,4}}{\theta_{1,2}}\,;\,a_{42}=\frac{3}{10}\,\theta_{1,4}\,+\,\frac{9}{10}\,\frac{\theta_{0,2}\,\theta_{0,4}}{\theta_{1,2}}\,;\,a_{43}=\frac{1}{5}\,\theta_{1,5}\,+\,\frac{9}{10}\,\frac{\theta_{0,3}\,\theta_{1,4}}{\theta_{1,2}}\,;\\\
&a_{44}=\frac{1}{7}\,\theta_{1,6}\,+\,\frac{9}{10}\,\frac{\theta_{0,4}\,\theta_{1,4}}{\theta_{1,2}}\,;\,a_{45}=\frac{1}{9}\,\theta_{2,5}\,+\,\frac{3}{20}\,\frac{\theta_{1,3}\,\theta_{1,4}}{\theta_{1,2}}\,;\\\
&a_{46}=\frac{1}{7}\,\theta_{2,6}\,+\,\frac{27}{100}\,\frac{\left(\theta_{1,4}\right)^{2}}{\theta_{1,2}}\,;\\\
&b_{4}=-\,\tau\,\left[\frac{1}{2}\,\theta_{1,3}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\frac{3}{10}\,\theta_{1,4}U^{\alpha}U^{\mu}\,+\,\frac{1}{3}\,c^{2}\,\theta_{2,4}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]\,;\end{split}$
$\displaystyle
a_{51}=\theta_{0,0}\,;\,a_{52}=\theta_{0,1}\,;\,a_{53}=\theta_{0,2}\,;\,a_{54}=\theta_{0,3}\,;\,a_{55}=\frac{1}{3}\,\theta_{1,2}\,;\,a_{56}=\frac{1}{2}\,\theta_{1,3}\,;\,b_{5}=0\,;$
$\displaystyle
a_{61}=\theta_{0,1}\,;\,a_{62}=\theta_{0,2}\,;\,a_{63}=\theta_{0,3}\,;\,a_{64}=\theta_{0,4}\,;\,a_{65}=\frac{1}{6}\,\theta_{1,3}\,;\,a_{66}=\frac{3}{10}\,\theta_{1,4}\,;\,b_{6}=0\,;$
$\displaystyle\begin{split}&a_{71}=3\,\theta_{1,1}\,;\,a_{72}=\theta_{1,2}\,;\,a_{73}=\frac{1}{2}\,\theta_{1,3}\,;\,a_{74}=\frac{3}{10}\,\theta_{1,4}\,;\,a_{75}=\frac{5}{3}\,\theta_{2,5}\,;\,a_{76}=\theta_{2,4}\,;\\\
&b_{7}=-\,\frac{k_{B}}{m\,\rho\,c^{2}}\,\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}\,h_{\alpha\beta}\,;\end{split}$

where the expression of $\theta_{k,j}$ has been introduced in (9). By applying
the Rouché-Capelli Theorem, we see that the determinant of the augmented
matrix must be zero; so, from this condition, by calling $D_{j}$ the algebraic
complement on the line $i$, column 7 of this matrix, we find

$\displaystyle\begin{split}&h_{\alpha\beta}\,\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}=-\,\frac{c^{2}m\rho\,\tau}{k_{B}}\cdot\\\
&\cdot\left\\{\frac{D_{1}}{D_{7}}\,\left[\theta_{0,2}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\theta_{0,3}U^{\alpha}U^{\mu}\,+\,\frac{1}{6}\,c^{2}\,\theta_{1,3}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]\right.+\\\
&+\,\frac{D_{2}}{D_{7}}\,\left[\theta_{0,3}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\theta_{0,4}U^{\alpha}U^{\mu}\,+\,\frac{3}{10}\,c^{2}\,\theta_{1,4}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]+\\\
&+\frac{D_{3}}{D_{7}}\,\left[\theta_{1,2}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\frac{1}{2}\,\theta_{1,3}U^{\alpha}U^{\mu}\,+\,\frac{5}{3}\,c^{2}\,\theta_{2,3}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]+\\\
&\left.\,+\,\frac{D_{4}}{D_{7}}\,\left[\frac{1}{2}\,\theta_{1,3}\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}\,+\left(\frac{3}{10}\,\theta_{1,4}U^{\alpha}U^{\mu}\,+\,\frac{1}{3}\,c^{2}\,\theta_{2,4}\,h^{\alpha\mu}\right)\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right]\right\\}\,,\end{split}$

i.e., by using (15),

$\displaystyle\begin{split}&h_{\alpha\beta}\,\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}=-\,3\,\nu^{MI}\,\partial_{\alpha}\,U^{\alpha}\,,\quad\mbox{with}\\\
&\nu^{MI}=\frac{c^{2}m\rho\,\tau}{3\,k_{B}T}\,\left[\left(\frac{D_{1}}{D_{7}}\,\theta_{0,2}+\,\frac{D_{2}}{D_{7}}\,\theta_{0,3}+\frac{D_{3}}{D_{7}}\,\theta_{1,2}\,+\,\frac{1}{2}\,\theta_{1,3}\right)\,\frac{\left|\begin{matrix}\frac{p}{\rho}&&\frac{e}{\rho\,c^{2}}\\\
&&\\\
c^{2}\theta_{1,2}&&\theta_{0,2}\end{matrix}\right|}{\left|\begin{matrix}1&&\frac{e}{\rho\,c^{2}}\\\
&&\\\ \frac{e}{\rho\,c^{2}}&&\theta_{0,2}\end{matrix}\right|}\,\right.+\\\
&+\,\left(\frac{D_{1}}{D_{7}}\,\frac{1}{6}\,c^{2}\,\theta_{1,3}\,+\,\frac{D_{2}}{D_{7}}\,\frac{3}{10}\,c^{2}\,\theta_{1,4}\,+\,\frac{D_{3}}{D_{7}}\,\frac{5}{3}\,c^{2}\,\theta_{2,3}\,+\,\frac{D_{4}}{D_{7}}\,\frac{1}{3}\,c^{2}\,\theta_{2,4}\right)+\\\
&+\left.\,\left(\frac{D_{1}}{D_{7}}\,\theta_{0,3}\,+\,\frac{D_{2}}{D_{7}}\,\theta_{0,4}\,+\,\frac{D_{4}}{D_{7}}\,\frac{3}{10}\,\theta_{1,4}\right)\,\frac{\left|\begin{matrix}1&&\frac{p}{\rho}\\\
&&\\\
\frac{e}{\rho\,c^{2}}&&c^{2}\,\theta_{1,2}\end{matrix}\right|}{\left|\begin{matrix}1&&\frac{e}{\rho\,c^{2}}\\\
&&\\\
\frac{e}{\rho\,c^{2}}&&\theta_{0,2}\end{matrix}\right|}\right]\,.\end{split}$
(21)

Determination of the heat conductivity $\chi$. We consider now eqs. (19)1 with
$n=2$ contracted by
$\frac{h^{\theta}_{\alpha_{1}}U_{\alpha_{2}}}{\rho\,c^{6}}$, (19)1 with $n=3$
contracted by
$\frac{h^{\theta}_{\alpha_{1}}U_{\alpha_{2}}U_{\alpha_{3}}}{\rho\,c^{8}}$,
(19)1 with $n=3$ contracted by
$\frac{h_{\alpha_{1}\alpha_{2}}h^{\theta}_{\alpha_{3}}}{\rho\,c^{6}}$, (19)2
contracted by $\frac{h^{\theta}_{\alpha}}{-\,\rho\,c^{2}}$ and (19)4
contracted by $k_{B}\,\frac{h^{\theta}_{\alpha}U\beta}{\rho\,c^{4}}$.
So we obtain a system $\sum_{j=1}^{4}b_{ij}X^{j\theta}=b_{i}^{\theta}$
constituted by 5 equations in the 4 unknowns
$X^{1\theta}=h^{\theta\beta_{1}}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}$,
$X^{2\theta}=h^{\theta\beta_{1}}U^{\beta_{2}}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$,
$X^{3\theta}=h^{\theta\beta_{1}}U^{\beta_{2}}U^{\beta_{3}}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}$,
$X^{4\theta}=c^{2}h^{\theta\beta_{1}}h^{\beta_{2}\beta_{3}}\,\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}$
with coefficients

$\displaystyle\begin{split}&b_{11}=-\,\frac{1}{3}\,\theta_{1,3}\,;\quad
b_{12}=-\,\frac{1}{5}\,\theta_{1,4}\,+\,\frac{1}{6}\,\frac{\left(\theta_{1,3}\right)^{2}}{\theta_{1,2}}\,;\\\
&b_{13}=-\,\frac{1}{5}\,\theta_{1,5}\,+\,\frac{3}{20}\,\frac{\theta_{1,3}\,\theta_{1,4}}{\theta_{1,2}}\,;\quad
b_{14}=-\,\frac{1}{15}\,\theta_{2,5}\,+\,\frac{1}{10}\,\frac{\theta_{1,3}\,\theta_{2,4}}{\theta_{1,2}}\,;\\\
&b_{1}^{\theta}=\tau\,\left(\frac{1}{3}\,\theta_{1,2}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{1}{3}\,\theta_{1,3}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,;\end{split}$
(22) $\displaystyle\begin{split}&b_{21}=-\,\frac{1}{5}\,\theta_{1,4}\,;\quad
b_{22}=-\,\frac{2}{15}\,\theta_{1,5}\,+\,\frac{1}{10}\,\frac{\theta_{1,4}\,\theta_{1,3}}{\theta_{1,2}}\,;\\\
&b_{23}=-\,\frac{1}{7}\,\theta_{1,6}\,+\,\frac{9}{100}\,\frac{\left(\theta_{1,4}\right)^{2}}{\theta_{1,2}}\,;\quad
b_{24}=-\,\frac{1}{35}\,\theta_{2,6}\,+\,\frac{3}{50}\,\frac{\theta_{1,4}\,\theta_{2,4}}{\theta_{1,2}}\,;\\\
&b_{2}^{\theta}=\tau\,\left(\frac{1}{6}\,\theta_{1,3}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{1}{5}\,\theta_{1,4}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,;\end{split}$
$\displaystyle\begin{split}&b_{31}=-\,\frac{2}{3}\,\theta_{2,4}\,;\quad
b_{32}=-\,\frac{4}{15}\,\theta_{2,5}\,+\,\frac{1}{3}\,\frac{\theta_{2,4}\,\theta_{1,3}}{\theta_{1,2}}\,;\\\
&b_{33}=-\,\frac{1}{7}\,\theta_{2,6}\,+\,\frac{3}{10}\,\frac{\theta_{1,4}\,\theta_{2,4}}{\theta_{1,2}}\,;\quad
b_{34}=-\,\frac{1}{5}\,\theta_{3,6}\,+\,\frac{1}{5}\,\frac{\left(\theta_{2,4}\right)^{2}}{\theta_{1,2}}\,;\\\
&b_{3}^{\theta}=\,\tau\,\left(\theta_{2,3}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{2}{3}\,\theta_{2,4}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,;\end{split}$
$\displaystyle b_{41}=\theta_{1,1}\,;\quad
b_{42}=\frac{2}{3}\,\theta_{1,2}\,;\quad
b_{43}=\frac{1}{2}\,\theta_{1,3}\,;\quad b_{44}=\theta_{2,3}\,;\quad
b_{4}^{\theta}=0\,;$
$\displaystyle\begin{split}&b_{51}=\frac{1}{3}\,\theta_{1,2}\,;\quad
b_{52}=\frac{1}{3}\,\theta_{1,3}\,;\quad
b_{53}=\frac{3}{10}\,\theta_{1,4}\,;\quad
b_{54}=\frac{1}{5}\,\theta_{2,4}\,;\\\
&b_{5}^{\theta}=\frac{k_{B}}{m\,\rho\,c^{4}}\,h^{\theta}_{\alpha}U_{\beta}\,\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)^{MI}\,;\end{split}$

By applying the Rouché-Capelli Theorem, we see that the determinant of the
augmented matrix must be zero; so, from this condition by calling $M_{j}$ the
algebraic complement on the line $i$, column 5 of this matrix, we find

$\displaystyle\begin{split}&h^{\theta}_{\alpha}U_{\beta}\,\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)^{MI}=-\,\frac{c^{4}m\rho\,\tau}{k_{B}}\cdot\\\
&\cdot\left[\frac{M_{1}}{M_{5}}\,\left(\frac{1}{3}\,\theta_{1,2}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{1}{3}\,\theta_{1,3}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\right.+\\\
&+\,\frac{M_{2}}{M_{5}}\,\left(\frac{1}{6}\,\theta_{1,3}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{1}{5}\,\theta_{1,4}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)+\\\
&+\frac{M_{3}}{M_{5}}\,\left.\left(\theta_{2,3}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{2}{3}\,\theta_{2,4}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\right]\,,\end{split}$

i.e., by using (15),

$\displaystyle\begin{split}q^{\theta}&=-\,\chi^{MI}\,h^{\alpha\theta}\left(\partial_{\alpha}\,T\,-\,\frac{T}{c^{2}}\,U^{\mu}\,\partial_{\mu}\,U_{\alpha}\right)\,,\mbox{with}\\\
\chi^{MI}&=\frac{c^{6}m\rho\,\tau}{2\,k_{B}T^{2}}\left[\frac{M_{1}}{M_{5}}\,\left(-\,\frac{2}{3}\,\frac{\rho\,c^{2}}{p}\,\left(\theta_{1,2}\right)^{2}\,+\,\frac{1}{3}\,\theta_{1,3}\right)\right.+\\\
&+\,\frac{M_{2}}{M_{5}}\,\left(-\,\frac{1}{3}\,\frac{\rho\,c^{2}}{p}\,\theta_{1,2}\,\theta_{1,3}\,+\,\frac{1}{5}\,\theta_{1,4}\right)+\,\frac{M_{3}}{M_{5}}\,\left.\left(-\,2\,\frac{\rho\,c^{2}}{p}\,\theta_{1,2}\,\theta_{2,3}\,+\,\frac{2}{3}\,\theta_{2,4}\right)\right]\,,\end{split}$
(23)

Determination of the shear viscosity $\mu$. Let us consider now eqs. (19)1
with $n=2$ contracted by
$\frac{h_{\alpha_{1}<\theta}h_{\psi>\alpha_{2}}}{\rho\,c^{6}}$, (19)1 with
$n=3$ contracted by
$\frac{h_{\alpha_{1}<\theta}h_{\psi>\alpha_{2}}U_{\alpha_{3}}}{\rho\,c^{8}}$
and (19)4 contracted by $\frac{h_{\alpha<\theta}h_{\psi>\beta}}{\rho\,c^{4}}$.
So we obtain a system
$\sum_{j=1}^{2}c_{ij}X^{j}_{<\theta\psi>}=b_{i<\theta\psi>}$ constituted by 3
equations in the 2 unknowns
$X^{1}_{<\theta\psi>}=h^{\beta_{1}}_{<\theta}h^{\beta_{2}}_{\psi>}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$,
$X^{2}_{<\theta\psi>}=h^{\beta_{1}}_{<\theta}h^{\beta_{2}}_{\psi>}U^{\beta_{3}}\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}$,
with coefficients

$\displaystyle c_{11}=\frac{2}{15}\,\theta_{2,4}\,;\quad
c_{12}=\frac{2}{15}\,\theta_{2,5}\,;\quad
b_{1<\theta\psi>}=-\,\frac{2}{3}\,\tau\,\theta_{2,3}\,\partial_{<\theta}\,\lambda^{E}_{\psi>_{3}}\,;$
(24) $\displaystyle c_{21}=\frac{2}{45}\,\theta_{2,5}\,;\quad
c_{22}=\frac{1}{35}\,\theta_{2,6}\,;\quad
b_{2<\theta\psi>}=-\,\frac{2}{15}\,\tau\,\theta_{2,4}\,\partial_{<\theta}\,\lambda^{E}_{\psi>_{3}}\,;$
$\displaystyle c_{31}=\frac{2}{3}\,\theta_{2,3}\,;\quad
c_{32}=\frac{2}{5}\,\theta_{2,4}\,;\quad
b_{3<\theta\psi>}=-\,\frac{k_{B}}{m\,c^{4}\rho}\,h_{\alpha<\theta}h_{\psi>\beta}\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)^{MI}\,.$

From this system we obtain

$\displaystyle\begin{split}&t_{<\theta\psi>}=2\,\mu^{MI}\,\partial_{<\theta}\,U_{\psi>}\,,\mbox{with}\\\
&\mu^{MI}=-\,\frac{c^{4}m\,\rho\,\tau}{2\,k_{B}T}\,\frac{\left|\begin{matrix}\frac{2}{15}\,\theta_{2,4}&&\frac{2}{15}\,\theta_{2,5}&&-\,\frac{2}{3}\,\theta_{2,3}\\\
&&&&\\\
\frac{2}{45}\,\theta_{2,5}&&\frac{1}{35}\,\theta_{2,6}&&-\,\frac{2}{15}\,\theta_{2,4}\\\
&&&&\\\
\frac{2}{3}\,\theta_{2,3}&&\frac{2}{5}\,\theta_{2,4}&&0\end{matrix}\right|}{\left|\begin{matrix}\frac{2}{15}\,\theta_{2,4}&&\frac{2}{15}\,\theta_{2,5}\\\
&&\\\
\frac{2}{45}\,\theta_{2,5}&&\frac{1}{35}\,\theta_{2,6}\end{matrix}\right|}\,.\end{split}$
(25)

Finally, to complete our analysis, we have also to consider eq. (19)1 with
$n=3$ contracted by
$h_{\alpha_{1}}^{<\theta}h^{\psi}_{\alpha_{2}}h_{\alpha_{3}}^{\theta>}$. But
this step will give only $\lambda_{<\beta_{1}\beta_{2}\beta_{3}>_{3}}$ which
does not appear in
$\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)^{MI}$.

###  2.3 The Maxwellian iteration in the case $N=2$.

For this case we have to consider (19)1 only for $n=2$ and put
$\left(\lambda_{\beta_{1}\beta_{2}\beta_{3}}\right)^{MI}=0$ in all the
equations. For example, let us see what happens for the determination of the
bulk viscosity.

Determination of the bulk viscosity $\nu$. We consider here eqs. (19)1 with
$n=2$ contracted by $\frac{U_{\alpha_{1}}U_{\alpha_{2}}}{\rho\,c^{6}}$, (19)1
with $n=2$ contracted by $\frac{h_{\alpha_{1}\alpha_{2}}}{\rho\,c^{4}}$, (19)2
contracted by $\frac{U_{\alpha}}{\rho\,c^{2}}$, (19)3 divided by $\rho\,c^{6}$
and (19)4 contracted by $-\,k_{B}\,\frac{h_{\alpha\beta}}{m\,\rho\,c^{2}}$.
So we obtain a system composed by 5 equations in the 6 unknowns
$X^{1}=\left(\lambda\,-\,\lambda^{E}\right)^{MI}$,
$X^{2}=U^{\beta_{1}}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}$,
$X^{3}=U^{\beta_{1}}U^{\beta_{2}}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$,
$X^{4}=c^{2}h^{\beta_{1}\beta_{2}}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$.
The augmented matrix can be obtained by cutting the rows 2 and 4 and columns 4
and 6 from the augmented matrix introduced in subsection 4.1 and its
determinant is given by:

$\displaystyle\left|\begin{matrix}a_{11}\quad&a_{12}\quad&a_{13}\quad&a_{15}\quad&b_{1}\\\
a_{31}\quad&a_{32}\quad&a_{33}\quad&a_{35}\quad&b_{3}\\\
a_{51}\quad&a_{52}\quad&a_{53}\quad&a_{55}\quad&b_{5}\\\
a_{61}\quad&a_{62}\quad&a_{63}\quad&a_{65}\quad&b_{6}\\\
a_{71}\quad&a_{72}\quad&a_{73}\quad&a_{75}\quad&b_{7}\end{matrix}\right|=0\,,$

where the expressions of $a_{ij}$ and $b_{i}$ are the same of the case $N=3$.
By calling $D_{j}$ the algebraic complements of the row $i$, column 5 of the
preceding matrix, we find

$\displaystyle\begin{split}&\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}\,h_{\alpha\beta}=-\,\frac{m\,\rho\,c^{2}\,\tau}{k_{B}}\,\left(\frac{\theta_{0,2}\,D_{1}\,+\,\theta_{1,2}\,D_{2}}{D_{5}}\,\,U^{\alpha}\partial_{\alpha}\,\lambda^{E}+\right.\\\
&+\,\frac{\theta_{0,3}\,D_{1}\,+\,\frac{1}{2}\,\theta_{1,3}\,D_{2}}{D_{5}}\,U^{\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\,+\,c^{2}\,\left.\frac{\frac{1}{6}\,\theta_{1,3}\,D_{1}\,+\,\frac{5}{3}\,\theta_{2,3}\,D_{2}}{D_{5}}\,\,h^{\alpha\mu}\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,,\end{split}$

i.e., by using (15),

$\displaystyle\begin{split}&\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)^{MI}\,h_{\alpha\beta}=-\,3\,\nu^{2MI}\,\partial_{\alpha}\,U^{\alpha}\,,\quad\mbox{where}\\\
&\nu^{2MI}=\frac{c^{2}m\,\rho\,\tau}{3\,k_{B}T}\,\left(\frac{\theta_{0,2}\,D_{1}\,+\,\theta_{1,2}\,D_{2}}{D_{5}}\,\frac{\left|\begin{matrix}\frac{p}{\rho}&&\frac{e}{\rho\,c^{2}}\\\
&&\\\
c^{2}\theta_{1,2}&&\theta_{0,2}\end{matrix}\right|}{\left|\begin{matrix}1&&\frac{e}{\rho\,c^{2}}\\\
&&\\\ \frac{e}{\rho\,c^{2}}&&\theta_{0,2}\end{matrix}\right|}\,+\right.\\\
&+\,c^{2}\,\left.\frac{\frac{1}{6}\,\theta_{1,3}\,D_{1}\,+\,\frac{5}{3}\,\theta_{2,3}\,D_{2}}{D_{5}}+\,\frac{\theta_{0,3}\,D_{1}\,+\,\frac{1}{2}\,\theta_{1,3}\,D_{2}}{D_{5}}\,\frac{\left|\begin{matrix}1&&\frac{p}{\rho}\\\
&&\\\
\frac{e}{\rho\,c^{2}}&&c^{2}\,\theta_{1,2}\end{matrix}\right|}{\left|\begin{matrix}1&&\frac{e}{\rho\,c^{2}}\\\
&&\\\
\frac{e}{\rho\,c^{2}}&&\theta_{0,2}\end{matrix}\right|}\,\,\,\right)\,.\end{split}$
(26)

It is evident that this value of $\nu$ is different from that found in (21)
when $N=3$.

Determination of the heat conductivity $\chi$. We consider now eqs. (19)1 with
$n=2$ contracted by
$\frac{h^{\theta}_{\alpha_{1}}U_{\alpha_{2}}}{\rho\,c^{6}}$, (19)2 contracted
by $\frac{h^{\theta}_{\alpha}}{-\,\rho\,c^{2}}$ and (19)4 contracted by
$k_{B}\,\frac{h^{\theta}_{\alpha}U\beta}{\rho\,c^{4}}$. We obtain a system
$\sum_{j=1}^{2}b_{ij}X^{j\theta}=b_{i}^{\theta}$ constituted by 3 equations in
the 2 unknowns
$X^{1\theta}=h^{\theta\beta_{1}}\left(\lambda_{\beta_{1}}\,-\,\lambda_{\beta_{1}}^{E}\right)^{MI}$
and
$X^{2\theta}=h^{\theta\beta_{1}}U^{\beta_{2}}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$.
The augmented matrix can be obtained by eliminating the rows 2 and 3 and the
columns 3 and 4 from the augmented matrix introduced in subsection 4.2, so its
determinant is given by:

$\displaystyle\left|\begin{matrix}b_{11}\quad&b_{12}\quad&b_{1}^{\theta}\\\
b_{41}\quad&b_{42}\quad&b_{4}^{\theta}\\\
b_{51}\quad&b_{52}\quad&b_{5}^{\theta}\end{matrix}\right|=0\,,$

where $b_{ij}$ and $b_{i}^{\theta}$ are the same of the case $N=3$. By calling
$D_{j}$ the algebraic complements of the line $i$, column 3 of the preceding
matrix, we find

$\displaystyle\frac{k_{B}}{m\,\rho\,c^{4}}\,h^{\theta}_{\alpha}U_{\beta}\,\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)^{MI}=-\,\frac{\tau\,\left(\frac{1}{3}\,\theta_{1,2}\,h^{\theta\alpha}\partial_{\alpha}\,\lambda^{E}\,+\,\frac{1}{3}\,\theta_{1,3}\,h^{\theta\alpha}U^{\mu}\,\partial_{(\alpha}\,\lambda^{E}_{\mu)}\right)\,D_{1}}{D_{3}}\,,$

i.e., by using (15),

$\displaystyle\begin{split}&q^{\theta}=-\,\chi^{2MI}\,h^{\alpha\theta}\left(\partial_{\alpha}\,T\,-\,\frac{T}{c^{2}}\,U^{\mu}\,\partial_{\mu}\,U_{\alpha}\right)\,,\mbox{where}\\\
&\chi^{2MI}=\frac{m\,\rho\,c^{6}\tau}{2\,k_{B}T^{2}}\,\frac{D_{1}}{D_{3}}\,\left(-\,\frac{2}{3}\,\rho\,c^{2}\,\frac{\left(\theta_{1,2}\right)^{2}}{p}\,+\,\frac{1}{3}\,\theta_{1,3}\right)\,,\end{split}$
(27)

It is evident that this value of $\chi$ is different from the value obtained
by using the expression of $\chi$ found in (23) in the case $N=3$.

Determination of the shear viscosity $\mu$. Let us consider now eqs. (19)1
with $n=2$ contracted by
$\frac{h_{\alpha_{1}<\theta}h_{\psi>\alpha_{2}}}{\rho\,c^{6}}$ and (19)4
contracted by $\frac{h_{\alpha<\theta}h_{\psi>\beta}}{\rho\,c^{4}}$. We obtain
a system $\sum_{j=1}^{1}c_{ij}X^{j}_{<\theta\psi>}=b_{i<\theta\psi>}$
constituted by 2 equations in the 1 unknown
$X^{1}_{<\theta\psi>}=h^{\beta_{1}}_{<\theta}h^{\beta_{2}}_{\psi>}\left(\lambda_{\beta_{1}\beta_{2}}\right)^{MI}$.
The augmented matrix can be obtained by eliminating row 2 and column 2 from
the augmented matrix introduced in subsection 4.3 and its determinant is given
by:

$\displaystyle\left|\begin{matrix}c_{11}\quad&b_{1<\theta\psi>}\\\
c_{31}\quad&b_{3<\theta\psi>}\end{matrix}\right|=0\,,$

where $c_{ij}$ and $b_{i<\theta\psi>}$ are the same of the case $N=3$. From
this equation we find

$\displaystyle
t_{<\theta\psi>}=2\,\mu^{MI}\,\partial_{<\theta}\,U_{\psi>}\,,\quad\mbox{with}$
(28)
$\displaystyle\mu^{MI}\,=\frac{1}{3}\,\frac{m\,c^{4}\rho\,\tau}{k_{B}T}\,\frac{c_{31}}{c_{11}}\,\theta_{2,3}=\frac{5}{3}\,\frac{m\,c^{4}\rho\,\tau}{k_{B}T}\,\frac{\theta_{2,3}}{\theta_{2,4}}\,\theta_{2,3}\,,$
(29)

It is evident that this value of $\mu$ is different from the value of $\mu$
furnished by (25) when $N=3$.
These transport coefficients are the same ones of [9] if we take into account
that in [9] the authors call $\omega=\frac{e}{\rho\,c^{2}}$ (see also eq.
$(12)_{2}$) and the quantities $B_{q}$, $B_{2}^{\pi}$, $B^{t}$ present in [9]
whose expressions are reported in equation (44) in terms also of $C_{5}$ which
is described in eq. (34)2, the matrices $N^{\pi}$ and $D_{4}$ which are given
in the equation before equation (30) and the matrices $N_{3}$ and $D_{3}$
which are given in the equation after (32)).

##  3 The Chapman-Enskog Method

This method can be found in the articles [14], [15] and has been further
explained in [11]. We describe how this method works by enclosing the full
expression of the production term which was found in [16] and modified in [9].
In particular, the method starts by considering the following equations

$\displaystyle
p^{\alpha}\,\partial_{\alpha}\,f=Q\,,\quad\partial_{\alpha}V^{\alpha}=0\,,\quad\partial_{\alpha}T^{\alpha\beta}=0\,,$
(30)

i.e., the Boltzmann equation and the conservation laws of mass and momentum-
energy.

Then the following steps have to be followed:

  * 1

The eqs. (30) are considered, but with their left hand sides calculated at
equilibrium and their right hand sides at first order with respect to
equilibrium, i.e.

$\displaystyle\begin{split}&p^{\alpha}\,\partial_{\alpha}\,f_{E}=Q^{(OT)}=\frac{U^{\mu}p_{\mu}}{c^{2}\tau}\left[\left(f_{E}-f\right)^{OT}-f_{E}\,p^{\gamma}q_{\gamma}^{OT}\frac{3}{m\,c^{4}\rho\,\theta_{1,2}}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\right]\,.\\\
&\\\
&\partial_{\alpha}V^{\alpha}_{E}=0\,,\quad\partial_{\alpha}T^{\alpha\beta}_{E}=0\,,\end{split}$
(31)

where the superscript $OT$ denotes that these quantities are the first
iterates defined with this approach.

  * 2

The deviation of the distribution function from its value at equilibrium is
calculated in terms of $\partial_{\alpha}\lambda^{E}$ and
$\partial_{\alpha}\lambda_{\mu}^{E}$ from (31)1 and used in eqs. (2b)1,2 with
$n=0$ and $n=1$. Obviously, in this way $I=0$, $I^{\alpha_{1}}=0$ are
obtained, thus respecting the conservation laws of mass and of momentum-
energy.

  * 3

The quantities $\partial_{\alpha}\lambda^{E}$ and
$U^{\alpha}U^{\mu}\partial_{\alpha}\lambda_{\mu}^{E}$ are calculated from
(31)2,3 and substituted in the expression of $A^{\alpha}\,-\,A^{\alpha}_{E}$,
$T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}$ obtained in the previous step.

We note that in the expression of $Q$ in [11] there are 6 unknown scalars
$a_{i}$ with $i=0,\cdots 5$ which have to be determined. From the third line
on page 116 of [11], by imposing $V^{\alpha}-V^{\alpha}_{E}=0$,
$e\,-\,e_{E}=\frac{U_{\alpha}U_{\beta}}{c^{2}}\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)=0$
the authors find $a_{0}$, $a_{1}$, $a_{3}$. By imposing
$V^{\alpha}-V^{\alpha}_{E}=0$,
$e\,-\,e_{E}=\frac{U_{\alpha}U_{\beta}}{c^{2}}\left(T^{\alpha\beta}-T^{\alpha\beta}_{E}\right)=0$
we find simply that $q_{\gamma}$ is constrained by $q_{\gamma}\,U^{\gamma}=0$.
So, $q_{\gamma}$ replaces the remaining part of Cercignani-Kremer’s unknown
scalars $a_{2}$, $a_{4}$, $a_{5}$. It is interesting that in a model with 14
or more moments, $q_{\gamma}$ becomes exactly the heat flux density. In a
model with six moments, there is no heat flux; in this case $q_{\gamma}$
remains a mathematical tool as the scalars $a_{2}$, $a_{4}$, $a_{5}$ of the
Cercignani-Kremer method. But it cannot be eliminated, otherwise the zero
deviation of $V^{\alpha}$ from its value at equilibrium would be lost.

###  3.1 ROT recovered with the Chapman-Enskog Method

In this subsection we apply the Chapman-Enskog method to the equations of
polyatomic gases with an arbitrary number $N$. In this way we will find (12)
of ROT, with particular expressions $\nu^{OT}$, $\chi^{OT}$, $\mu^{OT}$ of the
bulk viscosity $\nu$, the heat conductivity $\chi$ and the shear viscosity
$\mu$ and we will show that all these coefficients do not depend on $N$.

We have to consider the equations

$\displaystyle\begin{split}&f\,-\,f_{E}=\frac{c^{2}\tau}{k_{B}U^{\mu}p_{\mu}}\,f_{E}\,p^{\delta}\left[m\,\partial_{\delta}\,\lambda^{E}\,+\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\,p^{\nu}\,\partial_{\delta}\,\lambda^{E}_{\nu}\right]-\,3\,f_{E}\,p^{\mu}q_{\mu}\frac{1+\frac{\mathcal{I}}{mc^{2}}}{m\,c^{4}\rho\,\theta_{1,2}}\,,\\\
&V^{\alpha}\,-\,V^{\alpha}_{E}=0\,,\,U_{\alpha}U_{\beta}\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)=0\,,\,\partial_{\alpha}V^{\alpha}_{E}=0\,,\,\partial_{\alpha}T^{\alpha\beta}_{E}=0\,.\end{split}$
(32)

The equations (32)4 and (32)5 are exactly the equations (14) of the MI
approach and so the solution of these equations is given by (15) and (16).

Let us now consider equation (32)2 contracted with $\frac{U_{\alpha}}{c^{2}}$.
By using (32)1 contracted with
$m\,c\,\varphi(\mathcal{I})\,\frac{U_{\alpha}}{c^{2}}\,p^{\alpha}$ and
integrated in $d\,\mathcal{I}\,d\,\vec{P}$ it becomes

$\displaystyle
0=\frac{U_{\alpha}}{c^{2}}\left(V^{\alpha}\,-\,V^{\alpha}_{E}\right)=\frac{m\,\tau}{k_{B}}\left(V^{\alpha}_{E}\,\partial_{\alpha}\lambda^{E}\,+\,T^{\alpha\delta}_{E}\,\partial_{\alpha}\lambda^{E}_{\delta}\right)\,-\,\frac{3}{\rho\,c^{6}\,\theta_{1,2}}\,U_{\alpha}\,q_{\mu}\,T^{\alpha\mu}_{E}\,,$

which is an identity for eqs. (15) (see also the first equation after (32)).
To impose eq. (32)2 contracted with $h_{\alpha}^{\theta}$, we need the tensors
(49) and their representations (50) of the appendix. By using (32)1 contracted
with $m\,c\,\varphi(\mathcal{I})\,h_{\alpha}^{\theta}\,p^{\alpha}$ and
integrated in $d\,\mathcal{I}\,d\,\vec{P}$ we find

$\displaystyle
0=h_{\alpha}^{\theta}\left(V^{\alpha}\,-\,V^{\alpha}_{E}\right)=\frac{m\,\tau}{k_{B}}\left(h_{\alpha}^{\theta}\,A^{*\alpha\delta}\,\partial_{\delta}\lambda^{E}\,+\,h_{\alpha}^{\theta}\,A^{*\alpha\delta\nu}\,\partial_{\delta}\lambda^{E}_{\nu}\right)\,-\,\frac{3}{\rho\,c^{4}\,\theta_{1,2}}\,h_{\alpha}^{\theta}\,q_{\mu}\,T^{\alpha\mu}_{E}\,,$

from which we desume

$\displaystyle q^{\theta}=-\,\frac{m\,\tau
c^{6}}{3\,k_{B}}\,\frac{\rho^{2}}{p}\,\theta_{1,2}\left(\theta_{1,1}^{*}\,h^{\alpha\theta}\,\partial_{\alpha}\lambda^{E}+\,\frac{2}{3}\,\theta_{1,2}^{*}\,h^{\theta(\delta}U^{\nu)}\,\partial_{\delta}\lambda^{E}_{\nu}\right)=-\,\chi\,h^{\theta\alpha}\,\left(\partial_{\alpha}\,T\,-\,\frac{T}{c^{2}}\,U^{\mu}\,\partial_{\mu}\,U_{\alpha}\right)\,,$
(33) $\displaystyle\mbox{with}\quad\chi=-\,\frac{m\,\tau
c^{8}}{9\,k_{B}T^{2}}\,\frac{\rho^{2}}{p}\,\theta_{1,2}\left(\theta_{1,2}^{*}\,-\,\frac{\rho\,c^{2}}{p}\,\theta_{1,2}\theta_{1,1}^{*}\right)\,,$

where in the last passage we have used (16) and
$\lambda_{\nu}^{E}=\frac{U\nu}{T}$ . We see here that $q^{\theta}$, replaces
the Cercignani-Kremer’s scalars which did not have a clear physical meaning.
They cannot simply be put equal to zero (as in [17]), otherwise the physical
requirement $V^{\alpha}-V^{\alpha}_{E}=0$ would be violated.
We now impose eq. (32)3, by using (32)1 contracted with
$U_{\alpha}U_{\beta}\,c\,p^{\alpha}p^{\beta}\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\,\varphi(\mathcal{I})$
and integrated in $d\,\mathcal{I}\,d\,\vec{P}$; we find

$\displaystyle
0=\frac{m\,c^{2}\tau}{k_{B}}\left(U_{\beta}T^{\beta\delta}_{E}\,\partial_{\delta}\lambda^{E}+U_{\beta}A^{\beta\delta\nu}_{E}\,\partial_{\delta}\lambda^{E}_{\nu}\right)-\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,A^{\mu\alpha\beta}_{E}q_{\mu}\,U_{\alpha}U_{\beta}\,.$

This is an identity for eqs. (15) (see also the second equation after (32)).
We now proceed evaluating the other components of
$T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}$. We use (32)1 contracted with
$h_{\alpha}^{\theta}U_{\beta}\,c\,p^{\alpha}p^{\beta}\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\,\varphi(\mathcal{I})$
and integrate in $d\,\mathcal{I}\,d\,\vec{P}$ to find

$\displaystyle\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)h_{\alpha}^{\theta}U_{\beta}=\frac{m\,c^{2}\tau}{k_{B}}\left(h_{\alpha}^{\theta}T^{\alpha\delta}_{E}\,\partial_{\delta}\lambda^{E}+h_{\alpha}^{\theta}A^{\alpha\delta\nu}_{E}\,\partial_{\delta}\lambda^{E}_{\nu}\right)-\,\frac{3}{c^{4}\rho\,\theta_{1,2}}\,A^{\mu\alpha\beta}_{E}q_{\mu}\,h_{\alpha}^{\theta}U_{\beta}=$
$\displaystyle=-\,\frac{m\,c^{2}\tau}{k_{B}}\left(p\,h^{\theta\delta}_{E}\,\partial_{\delta}\lambda^{E}+\frac{2}{3}\,\rho\,c^{2}\,\theta_{1,2}\,h^{\theta(\delta}U^{\nu)}\,\partial_{\delta}\lambda^{E}_{\nu}\right)-\,q^{\theta}=-\,q^{\theta}\,,$
(34)

where, in the last passage, we have used (16). The result is an identity. We
note that, in the 6 moments model, the left hand side of (3.1) is zero, so
that the right hand side is $-\,q^{\theta}$ must be zero; but we have said,
after eq. (33) that in this case the physical requirement
$V^{\alpha}-V^{\alpha}_{E}=0$ would be violated. This means that this approach
cannot be applied to the case of 6 moments. This is not surprising because it
has been shown in eq. (19) of [18] (see also [19]) that the optimal choices of
moments are $N=0$ (trivial case with only the conseration law of mass), $N=1$
(only the 5 Euler’s Equations where there is no production term), $N=2$ (the
15 moments model), $N=3$ (the 35 moments model) and so on. The 6 moments model
is not present in this hierarchy, but it can be considered a subsystem of the
15 moments model by putting $q^{\theta}=0$ (forgetting the role it played in
building the model and simply eliminating eq. (33)). From this perspective the
article [17] can be considered correct.
Finally, we multiply eq. (32)1 by
$h_{\alpha}^{\theta}\,h_{\beta}^{\psi}\,cp^{\alpha}p^{\beta}\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\,\varphi(\mathcal{I})$
and integrate in $d\,\mathcal{I}\,d\,\vec{P}$; so we obtain

$\displaystyle
h_{\alpha}^{\theta}\,h_{\beta}^{\psi}\left(T^{\alpha\beta}\,-\,T^{\alpha\beta}_{E}\right)=h_{\alpha}^{\theta}\,h_{\beta}^{\psi}\left[\frac{m\,\tau}{k_{B}}\,\left(A^{*\delta\alpha\beta}\partial_{\delta}\,\lambda^{E}+\,A^{*\delta\alpha\beta\nu}\,\partial_{\delta}\,\lambda^{E}_{\nu}\right)\,-\frac{3}{c^{4}\,\rho\,\theta_{1,2}}\,q_{\mu}\,A^{\mu\alpha\beta}_{E}\right]=$
$\displaystyle=\frac{m\,\tau}{k_{B}}\left[\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}^{*}\,h^{\theta\psi}U^{\delta}\partial_{\delta}\,\lambda^{E}+\left(\frac{1}{6}\,\rho\,c^{2}\,\theta_{1,3}^{*}\,h^{\theta\psi}\,U^{\delta}\,U^{\nu}\,+\,\rho\,c^{4}\,\theta_{2,3}^{*}\,h^{(\theta\psi}\,h^{\delta\nu)}\right)\partial_{\delta}\,\lambda^{E}_{\nu}\right]\,.$

This equation, contracted with $h_{\theta\psi}$ gives

$\displaystyle\Pi=\frac{m\,\tau}{k_{B}}\left[\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}^{*}\,U^{\delta}\partial_{\delta}\,\lambda^{E}+\left(\frac{1}{6}\,\rho\,c^{2}\,\theta_{1,3}^{*}\,U^{\delta}\,U^{\nu}\,+\,\frac{5}{9}\,\rho\,c^{4}\,\theta_{2,3}^{*}\,h^{\delta\nu}\right)\partial_{\delta}\,\lambda^{E}_{\nu}\right]=-\,\nu\,\partial_{\alpha}\,U^{\alpha}\,,$
(35) $\displaystyle\mbox{with}\quad\nu=$
$\displaystyle-\,\frac{m\,\tau}{k_{B}}\,\left[\left|\begin{matrix}\rho&&\frac{e}{c^{2}}\\\
&&\\\
\frac{e}{c^{2}}&&\rho\,\theta_{0,2}\end{matrix}\right|^{-1}\,\left(\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}^{*}\,\left|\begin{matrix}p&&\frac{e}{c^{2}}\\\
&&\\\
\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}&&\rho\,\theta_{0,2}\end{matrix}\right|\,+\right.\right.$
$\displaystyle\left.\left.+\frac{1}{6}\,\rho\,c^{2}\,\theta_{1,3}^{*}\,\,\left|\begin{matrix}\rho&&p\\\
&&\\\
\frac{e}{c^{2}}&&\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}\end{matrix}\right|\right)\,-\,\frac{5}{9}\,\rho\,c^{4}\,\theta_{2,3}^{*}\right]\,,$
(36)

where eqs. (15) have been used. Moreover, contracting equation (3.1) with
$h_{\theta}^{<\gamma}\,h_{\psi}^{\phi>}=h_{\theta}^{\gamma}\,h_{\psi}^{\phi}\,-\,\frac{1}{3}\,h_{\theta\psi}h^{\gamma\phi}$
it gives us

$\displaystyle
t_{<\gamma\phi>}=2\,\mu\,h^{\alpha}_{<\beta}\,h^{\mu}_{\gamma>}\,\partial_{\alpha}\,U_{\mu}\,,\quad\mbox{with}\quad\mu=\frac{1}{3}\,\frac{m\,\tau}{k_{B}T}\,\rho\,c^{4}\,\theta_{2,3}^{*}\,.$
(37)

The equations (33), (35) and (37)1 are those of Relativistic Ordinary
Thermodynamics.
In conclusion, with this approach we have obtained the equations of
Relativistic Ordinary Thermodynamics with heat conductivity, bulk viscosity
and shear viscosity given respectively by (33)2, (35)2 and (37)2.
It is evident from these expressions that they do not depend on the number of
moments of the extended model from which they are derivated.

##  4 The non relativistic approach

In this case the balance equation found in eq. (19)-(20) of [19] are

$\displaystyle\partial_{t}\,H_{s}^{i_{1}\cdots
i_{h}}+\partial_{k}\,H_{s}^{ki_{1}\cdots i_{h}}=J_{s}^{i_{1}\cdots
i_{h}}\,\,\mbox{with}\quad s=0,\cdots,N,\mbox{and}\quad h=0,\cdots,N-s.$ (38)

In particular, $H_{0}=\rho$ is the mass density,
$H_{0}^{i_{1}}=\rho\,v^{i_{1}}$ where $v^{i_{1}}$ is the velocity and
$H_{1}=2\,\rho\,\epsilon\,+\,\rho\,v^{2}$ where $\epsilon$ is the energy
density. All the variables are expressed in integral form as

$\displaystyle H_{s}^{i_{1}\cdots
i_{h}}=m\,\int_{\mathbb{R}^{3}}\int_{0}^{+\,\infty}f\,\xi^{i_{1}}\,\cdots\,\xi^{i_{h}}\,\left(\frac{2\,\mathcal{I}}{m}\,+\,\xi^{2}\right)^{s}\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,d\,\vec{\xi}\,.$
(39)

The expression of $H_{s}^{ki_{1}\cdots i_{h}}$ is the same of equation (39)
but with a further factor $\xi^{k}$ inside the integral; the expression of
$J_{s}^{i_{1}\cdots i_{h}}$ is the same of (39) but with the production
density $Q=-\,\frac{f-f_{E}}{\tau}$ instead of the distribution function $f$.
This distribution function has the form

$\displaystyle
f=e^{-1-\,\frac{m}{k_{B}}\,\chi}\,,\quad\chi=\sum_{h=0}^{N}\sum_{s=0}^{N-h}\lambda^{s}_{i_{1}\cdots
i_{h}}\xi^{i_{1}}\,\cdots\,\xi^{i_{h}}\,\left(\frac{2\,\mathcal{I}}{m}\,+\,\xi^{2}\right)^{s}\,.$
(40)

We prove now that the Chapman-Enskog method and the Maxwellian Iteration
method give the same result for polyatomic gases and with whatever number of
moments. This was already proved in [12] but only for monoatomic gases with 14
moments.
Let us start with the Chapman-Enskog Method where the Boltzamnn equation and
the conservation laws of mass, momentum and energy are considered:

$\displaystyle\partial_{t}\,f+\xi^{k}\partial_{k}\,f=-\,\frac{f-f_{E}}{\tau}\,,\,\partial_{t}\,H_{0}+\partial_{k}\,H_{0}^{k}=0\,,\,\partial_{t}\,H_{0}^{i}+\partial_{k}\,H_{0}^{ki}=0\,,\,\partial_{t}\,H_{1}+\partial_{k}\,H_{1}^{k}=0\,.$
(41)

After that, the following steps are followed:

  1. 1.

The left hand sides of (41) are calculated at equilibrium, while the right
hand sides at first order with respect to equilibrium

$\displaystyle\begin{split}&\partial_{t}\,f_{E}+\xi^{k}\partial_{k}\,f_{E}=\frac{f-f_{E}}{\tau}\,,\\\
&\partial_{t}\,H_{0}+\partial_{k}\,H_{0}^{k}=0\,,\,\partial_{t}\,H_{0}^{i}+\partial_{k}\,H_{0E}^{ki}=0\,,\,\partial_{t}\,H_{1}+\partial_{k}\,H_{1E}^{k}=0\,.\end{split}$
(42)

  2. 2.

The derivatives with respect to time of the independent variables $\rho$,
$v^{i}$, $T$ are obtained from (42)2-4 and substituted in (42)1 which, after
that, depends only on the independent variables and on their derivatives with
respect to $x^{k}$.

  3. 3.

The new eq. (42)1 is multiplied by
$m\,\xi^{i_{1}}\,\xi^{i_{2}}\,\varphi(\mathcal{I})$ and integrated with
respect to $d\,\mathcal{I}\,d\,\vec{\xi}$.
$\overline{\partial_{t}\,H_{0E}^{i_{1}i_{2}}}+\partial_{k}\,H_{0E}^{ki_{1}i_{2}}=-\,\left(H_{0}^{i_{1}i_{2}}\,-\,H_{0E}^{i_{1}i_{2}}\right)\,\frac{1}{\tau}$same
variable
As a result of the above figure we get the Navier-Stokes equations with a
precise expression of the bulk viscosity and of the shear viscosity (in the
above figure the overline denotes the fact that, after derivating with respect
to time, the above found time derivatives of the independent variables have
been substituted).
Similarly, multiplication of the new eq. (42)1 by
$m\,\xi^{i_{1}}\,\left(\frac{2\,\mathcal{I}}{m}\,+\,\xi^{2}\right)^{s}\,\varphi(\mathcal{I})$
and integration with respect to $d\,\mathcal{I}\,d\,\vec{\xi}$ gives
$\overline{\partial_{t}\,H_{1E}^{i_{1}}}+\partial_{k}\,H_{1E}^{ki_{1}}=-\,\left(H_{1}^{i_{1}}\,-\,H_{1E}^{i_{1}}\right)\,\frac{1}{\tau}$same
variable
from which the Fourier equation with a precise expression of the heat
conductivity. The expressions found for the bulk viscosity, the shear
viscosity and the heat conductivity do not depend on the number $N$ because
they use equation which are present in every model with $N\geq 1$.

Now, mathematically speaking, nothing changes if we swap the order of the
steps 2 and 3; but in this way we obtain the steps of the Maxwellian
Iteration. So we can say that, in the non relativistic case, the Chapman-
Enskog Method and the Maxwellian Iteration give the same result.
So it is natural to ask why in the relativistic case the two methods give
different results. To understand it, let us repeat the same steps in the
relativistic case.
With the Chapman-Enskog Method the Boltzamnn equation and the conservation
laws of mass, momentum-energy, with the left hand sides calculated at
equilibrium, are considered:

$\displaystyle\begin{split}&p^{\alpha}\,\partial_{\alpha}\,f_{E}=\frac{U_{\mu}}{c^{2}\tau}\left[\left(f_{E}-f\right)\,p^{\mu}-f_{E}\,q_{\gamma}\,\frac{3}{m\,c^{4}\rho\,\theta_{1,2}}\,p^{\mu}p^{\gamma}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\right]\,.\\\
&\partial_{\alpha}V^{\alpha}=0\,,\quad\partial_{\alpha}T^{\alpha\beta}=0\,,\end{split}$
(43)

After that, the following steps are followed:

  1. 1.

The deviation of the distribution function from its value at equilibrium is
calculated in terms of $\partial_{\alpha}\lambda_{E}$ and
$\partial_{\alpha}\lambda_{E}^{\mu}$ from (43)1 and used in the definition of
$T^{\alpha\beta}$ which now becomes

$\displaystyle
T^{\alpha\beta}-T^{\alpha\beta}_{E}=-c^{3}\tau\,\int_{\mathbb{R}^{3}}\int_{0}^{+\infty}\frac{p^{\alpha}\partial_{\alpha}f_{E}}{p^{\mu}U_{\mu}}p^{\alpha}p^{\beta}\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,d\,\vec{p}-3\frac{q_{\gamma}}{c^{4}\theta_{1,2}}A^{\gamma\alpha\beta}_{E}\,,$
(44)

  2. 2.

The quantities $\partial_{\alpha}\lambda_{E}$ and
$U^{\alpha}U^{\mu}\partial_{\alpha}\lambda_{E}^{\mu}$ are calculated from
(43)2,3 and substituted in (44). From the resulting expression, the bulk
viscosity, the shear viscosity and the heat conductivity can be obtained and
they do not depend on the number $N$ because they use equation which are
present in every model with $N\geq 1$.

There is also the oppotunity to modify a little the procedure, by taking the
last term in (43)1 to the left hand side before calculating the left hand
sides at equilibrium; in this case it will disappear and, consequently, also
the last term in (44) will be no more present.
Instead of this, with the Maxwellian Iteration

  1. 1.

The conservation laws of mass, momentum-energy, and the balance equation for
the triple tensor with the left hand sides calculated at equilibrium, are
considered:

$\displaystyle\partial_{\alpha}V^{\alpha}_{E}=0\,,\quad\partial_{\alpha}T^{\alpha\beta}_{E}=0\,,$
(45)
$\partial_{\alpha}\,A_{E}^{\alpha\beta\gamma}=-\,\left(A^{\mu\beta\gamma}\,-\,A_{E}^{\mu\beta\gamma}\right)\,\frac{U_{\mu}}{c^{2}\tau}\,-\,U_{\mu}q_{\delta}A_{E}^{\mu\delta\beta\gamma}\,\frac{3}{\tau\,c^{6}\rho\,\theta_{1,2}}\,.$Same
variable but different from $T^{\beta\gamma}$.Variable different from
$T_{E}^{\beta\gamma}$ and from $A_{E}^{\mu\beta\gamma}$.

  2. 2.

Some derivatives of the independent variables are obtained from (45)1,2 and
substituted in (45)3.

  3. 3.

The new eq. (45)3 is used to obtain $\Pi$, $q^{\alpha}$, $t^{<\beta\gamma>}$
and, consequently, the bulk viscosity, the shear viscosity and the heat
conductivity. This fact could give rise to some doubts because these
coefficients should be obtained from
$T^{\beta\gamma}\,-\,T_{E}^{\beta\gamma}$, not from
$U_{\mu}\left(A^{\mu\beta\gamma}\,-\,A_{E}^{\mu\beta\gamma}\right)$. Moreover,
$U_{\mu}\left(A^{\mu\beta\gamma}\,-\,A_{E}^{\mu\beta\gamma}\right)$ depends
not only on $\Pi$, $q^{\alpha}$, $t^{<\beta\gamma>}$ but also on other
variables whose number increases for increasing values of $N$. So $\Pi$,
$q^{\alpha}$, $t^{<\beta\gamma>}$ must be isolated from the other variables
and this means solving some algebraic linear systems depending on $N$. It is
therefore not surprising that the solution also depends on $N$. Obviously,
this is consequence of the form of the production term in the right hand side
of (43). It remains open the problem to find another expression which respect
the requirements of zero production of mass and of momentum-energy, and whose
consequent Maxwellian Iteration does not depend on $N$. We can say that
another possible expression is

$\displaystyle
Q=-\,\frac{U^{\alpha}_{L}\,p^{\alpha}}{c^{2}\tau}\,\left(f-f_{E}\right)\,,$
(46)

where $U^{\alpha}_{L}$ is the 4-velocity in the Landau-Liftschiz frame as
reported in [20], [21]. But in [22] it was proved that, up to first order with
respect to equilibrium, the expression (46) is equivalent to the right hand
side of the present eq. (43). So nothing changes by adopting the production
term (46).

In any case, the two procedures have to give the same result at the non
relativistic limit. In fact, from $U^{\alpha}U_{\alpha}=c^{2}$,
$p^{\alpha}p_{\alpha}=m^{2}c^{2}$ we have the following decompositions:

$\displaystyle U^{\alpha}=\Gamma(v)\left(c\,,v^{i}\right)\,,\quad
p^{\alpha}=m\,\Gamma\left(\frac{p}{m}\right)\left(c\,,\frac{p^{i}}{m}\right)\,,\quad\mbox{with}\quad\Gamma(v)=\left(1-\,\frac{v^{2}}{c^{2}}\right)^{-1/2}\,.$
(47)

Consequently, the limit for $c\,\rightarrow\,+\infty$ of (43) is

$\displaystyle
m\,\left(\partial_{t}\,f+\xi^{k}\partial_{k}\,f\right)=-\frac{m}{\tau}\,\left(f-f_{E}\right)\,,$
(48)

as in eq. (41)1. It follows that both the results of the Chapman-Enskog Method
and the Maxwellian Iteration have the same non relativistic limit. In the next
section we compute the non relativistic limit.

###  4.1 The non relativistic limit of $\chi$, $\mu$ and $\nu$

In this section we prove the convergence in the non relativistic limit of the
heat conductivity $\chi$, the shear viscosity $\mu$, and the bulk viscosity
$\nu$.

In the previous sections we have introduced the new variables
$\theta_{1,1}^{*}$, $\theta_{1,2}^{*}$, $\theta_{1,3}^{*}$, $\theta_{2,3}^{*}$
which have not studied so far in literature. In order to compute the non
relativistic limit of $\chi$, $\mu$ and $\nu$ it is necessary to analyze the
non relativistic limit of these new quantitities.
Taking into account (52), we have

$\displaystyle\theta_{1,2}^{*}=3\,\theta_{1,1}\,,\,\theta_{1,3}^{*}=2\,\theta_{1,2}\,.$

So we need only the non relativistic limit of $\theta_{1,1}^{*}$,
$\theta_{2,3}^{*}$ given by (54). To evaluate them, let us consider the
expression of $J_{4,-1}$, i.e.,

$\displaystyle\begin{split}&J_{4,-1}=\int_{0}^{+\infty}e^{-\gamma\,\cosh\,s}\frac{\sinh^{4}\,s}{\cosh\,s}\,d\,s=\int_{0}^{+\infty}e^{-x}\,e^{-\gamma}\frac{\left[\left(\frac{x}{\gamma}+1\right)^{2}-\,1\right]^{\frac{3}{2}}}{\frac{x}{\gamma}+1}\,\frac{d\,x}{\gamma}=\\\
&=\frac{e^{-\gamma}}{\gamma}\int_{0}^{+\infty}e^{-x}\,\frac{\left(\frac{x}{\gamma}+2\right)^{\frac{3}{2}}}{\frac{x}{\gamma}+1}\,\left(\frac{x}{\gamma}\right)^{\frac{3}{2}}\,d\,x\,,\end{split}$

where in the first passage we have changed the integration variable according
to the law
$\cosh\,s=\frac{x}{\gamma}+1\quad\rightarrow\quad\sinh\,s\,d\,s=\frac{d\,x}{\gamma}$.
Now the Mac-Laurin expansion of the function
$g(y)=\frac{\left(y+2\right)^{\frac{3}{2}}}{y+1}$ around $y=0$ is

$\displaystyle g(y)$
$\displaystyle=2\,\sqrt{2}\left(1\,-\frac{1}{4}\,y\right)\,+\,y^{2}(\cdots)\,\,\rightarrow\,\,\gamma^{\frac{5}{2}}e^{\gamma}\,J_{4,-1}=$
$\displaystyle=2\,\sqrt{2}\,\int_{0}^{+\infty}e^{-x}\,\left(1\,-\frac{1}{4}\,\frac{x}{\gamma}\,+\,\frac{1}{\gamma^{2}}(\cdots)\right)\,x^{\frac{3}{2}}\,d\,x=$
$\displaystyle=2\,\sqrt{2}\,\left[\Gamma\left(\frac{5}{2}\right)\,-\,\frac{1}{4}\Gamma\left(\frac{7}{2}\right)\frac{1}{\gamma}\,+\,\frac{1}{\gamma^{2}}(\cdots)\right]=$
$\displaystyle=2\,\sqrt{2\,\pi}\,\left[\frac{3}{4}\,-\,\frac{15}{32}\,\frac{1}{\gamma}\,+\,\frac{1}{\gamma^{2}}(\cdots)\right]\,.$

where in the last passage we have used the Gamma function

$\displaystyle\Gamma(s)=\int_{0}^{+\infty}e^{-x}\,x^{s-1}\,d\,x$

defined for $s>0$ and satisfying the relations $\Gamma(s+1)=s\,\Gamma(s)$,
$\,\,\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$. In a similar way we can
obtain the expansion of $J_{2,1}$ or we can read it on page 21 of [5] and it
is

$\displaystyle
J_{2,1}=2\,\sqrt{2\,\pi}\,e^{-\gamma}\,\gamma^{-1/2}\,\left[\frac{1}{4\,\gamma}\,+\,\frac{15}{32}\,\frac{1}{\gamma^{2}}\,+\,\frac{105}{512}\,\frac{1}{\gamma^{3}}\,-\,\frac{315}{32\,\cdot\,128}\,\frac{1}{\gamma^{4}}\,+\,\cdots\right]\,.$

It follows that

$\displaystyle\frac{\int_{0}^{+\infty}J_{4,-1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}$
$\displaystyle=\frac{\int_{0}^{+\infty}\gamma^{\frac{5}{2}}e^{\gamma}\,J_{4,-1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\gamma\,\int_{0}^{+\infty}\gamma^{\frac{3}{2}}e^{\gamma}\,J_{2,1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}=\frac{\int_{0}^{+\infty}\left(\frac{\gamma}{\gamma^{*}}\right)^{\frac{5}{2}}e^{\gamma-\gamma^{*}}\,e^{\gamma^{*}}\,\gamma^{*\,\frac{5}{2}}\,J_{4,-1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\gamma\,\int_{0}^{+\infty}\left(\frac{\gamma}{\gamma^{*}}\right)^{\frac{3}{2}}e^{\gamma-\gamma^{*}}\,e^{\gamma^{*}}\,\gamma^{*\,\frac{3}{2}}\,J_{2,1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}=$
$\displaystyle=\frac{\int_{0}^{+\infty}\left(\frac{\gamma}{\gamma^{*}}\right)^{\frac{5}{2}}e^{\gamma-\gamma^{*}}\,2\,\sqrt{2\,\pi}\,\left[\frac{3}{4}\,-\,\frac{15}{32}\,\frac{1}{\gamma^{*}}\,+\,\frac{1}{\gamma^{*2}}(\cdots)\right]\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\gamma\,\int_{0}^{+\infty}\left(\frac{\gamma}{\gamma^{*}}\right)^{\frac{3}{2}}e^{\gamma-\gamma^{*}}\,\,2\,\sqrt{2\,\pi}\,\left[\frac{1}{4}\,+\,\frac{15}{32}\,\frac{1}{\gamma^{*}}\,+\,\frac{1}{\gamma^{*2}}(\cdots)\right]\,\varphi(\mathcal{I})\,d\,\mathcal{I}}\,,$

and, consequently,

$\displaystyle\gamma\,\left(\gamma\,\frac{\int_{0}^{+\infty}J_{4,-1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}\,-\,3\right)=$
$\displaystyle=\frac{\int_{0}^{+\infty}\left(\frac{\gamma}{\gamma^{*}}\right)^{\frac{3}{2}}e^{\gamma-\gamma^{*}}\,\left[\frac{3}{4}\gamma\,\left(\frac{\gamma}{\gamma^{*}}\,-\,1\right)\,-\,\frac{15}{32}\,\left(\frac{\gamma}{\gamma^{*}}\right)^{2}\,-\,\frac{45}{32}\,\frac{\gamma}{\gamma^{*}}\,+\,\frac{1}{\gamma^{*}}(\cdots)\right]\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}\left(\frac{\gamma}{\gamma^{*}}\right)^{\frac{3}{2}}e^{\gamma-\gamma^{*}}\,\left[\frac{1}{4}\,+\,\frac{1}{\gamma^{*}}(\cdots)\right]\,\varphi(\mathcal{I})\,d\,\mathcal{I}}\,.$

Since we have

$\displaystyle
e^{\gamma-\gamma^{*}}=e^{-\frac{\mathcal{I}}{k_{B}T}}\,,\,\gamma\,\left(\frac{\gamma}{\gamma^{*}}\,-\,1\right)=\frac{\gamma}{\gamma^{*}}\left(\gamma-\gamma^{*}\right)=-\,\frac{\gamma}{\gamma^{*}}\,\frac{\mathcal{I}}{k_{B}T}\,,$

we can write

$\displaystyle\lim_{\gamma\,\rightarrow\,+\infty}\,\gamma\,\left(\gamma\,\frac{\int_{0}^{+\infty}J_{4,-1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}\,-\,3\right)=-\,3\,\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}-\,\frac{15}{2}\,.$

Consequently, we get

$\displaystyle\frac{\int_{0}^{+\infty}J_{4,-1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}=3\,\frac{1}{\gamma}\,+\,\left(-\,3\,\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}-\,\frac{15}{2}\right)\frac{1}{\gamma^{2}}\,+\,\frac{1}{\gamma^{3}}(\cdots)\,,$
$\displaystyle\theta_{1,1}^{*}=\frac{1}{\gamma}\,+\,\left(-\,\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}-\,\frac{5}{2}\right)\frac{1}{\gamma^{2}}\,+\,\frac{1}{\gamma^{3}}(\cdots)=$
$\displaystyle=\frac{p}{\rho\,c^{2}}\,+\,\left(-\,\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}-\,\frac{5}{2}\right)\frac{1}{\gamma^{2}}\,+\,\frac{1}{\gamma^{3}}(\cdots)\,.$

We can apply this result in (33)2, jointly with
$\theta_{1,2}^{*}=3\,\theta_{1,1}$ , and have that the heat conductivity has
the form

$\displaystyle\chi=-\frac{m\,\tau
c^{8}}{9\,k_{B}T^{2}}\,\frac{\rho^{2}}{p}\,\theta_{1,2}$
$\displaystyle\left[3\,\theta_{1,1}\,-\,\theta_{1,2}+\,\frac{\rho\,c^{2}}{p}\,\theta_{1,2}\left(\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}+\,\frac{5}{2}\right)\frac{1}{\gamma^{2}}-\frac{\rho\,c^{2}}{p}\,\theta_{1,2}\frac{1}{\gamma^{3}}(\cdots)\right]\,.$

Moreover, from eq. (11) of [23], we have that the exact expressions

$\displaystyle\theta_{1,1}=\frac{p}{\rho}\,\frac{1}{c^{2}}\,,\,\theta_{1,2}=3\,\frac{p}{\rho}\,\frac{1}{c^{2}}\,+\,3\,\frac{p}{\rho}\,g_{1}\,\frac{1}{c^{4}}\quad\mbox{with}\quad
g_{1}=\frac{e-\rho c^{2}\,+\,p}{\rho}\,,$

so that we find

$\displaystyle\chi=-\,\frac{m\,\tau
c^{2}}{k_{B}T^{2}}\,\rho\,\left[-\,\frac{p}{\rho}\,g_{1}\,+\,\left(\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}+\,\frac{5}{2}\right)\left(\frac{p}{\rho}\right)^{2}\,-\,\frac{p^{3}}{\rho^{3}c^{2}}(\cdots)\right]\,.$

By performing similar calculations, we find that

$\displaystyle
g_{1}=\left(\frac{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\frac{\mathcal{I}}{k_{B}T}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}e^{-\,\frac{\mathcal{I}}{k_{B}T}}\,\varphi(\mathcal{I})\,d\,\mathcal{I}}+\,\frac{5}{2}\right)\,\frac{p}{\rho}+\,\frac{1}{\gamma}\,(\cdots)\,,$

so that the above expression of the heat conductivity $\chi$ has a finite non
relativistic limit.
To evaluate the non relativistic limit of the shear viscosity $\mu$ in (37)2,
we need the expression of $\theta_{2,3}^{*}$ and of $J_{6,-1}$. With similar
computations we obtain that $c^{4}\,\theta_{2,3}^{*}$ and $\mu$ have a finite
limit

$\displaystyle\lim_{c\,\rightarrow\,+\infty}c^{4}\,\theta_{2,3}^{*}=3\,\left(\frac{p}{\rho}\right)^{2}\quad\mbox{and}\quad\lim_{c\,\rightarrow\,+\infty}\mu=\,\tau\,p\,.$

Moreover, with similar computations we obtain that $\nu$ is convergent in the
non relativistic limit.

##  5 Summary

In this article we have described how it is possible to reconstruct, as a
first iteration, the laws of the Relativistic Ordinary Thermodynamics starting
from the laws of the Relativistic Extended Thermodynamics of polyatomic gases
by using two different iteration methods. In literature, two procedures are
used which are the so-called Maxwellian Iteration and the Chapman-Enskog
Method. Both of these methods lead to the relativistic version of the Navier-
Stokes and Fourier laws, i.e, the so-called Eckart equations as a first
iteration. It is well known that the relativistic version of the Navier-Stokes
and Fourier laws are two fundamental laws of Relativistic Ordinary
Thermodynamics and in these equations the following remarkable physical
quantities appear as coefficients: the heat conductivity $\chi$, the shear
viscosity $\mu$, and the bulk viscosity $\nu$. We have proved that the
expressions of $\chi$, $\mu$, and $\nu$ obtained via the Chapman-Enskog method
do not depend on $N$, whereas these expressions obtained through the
Maxwellian Iteration depend on $N$. In order to make clear this difference we
describe our main results giving more details.
First of all, we observe that we have found the following expressions for the
shear viscosity $\mu$ by using the Maxwellian Iteration method in the case
$N=3$ (see eq. (25)) and in the case $N=2$ (see eq. (28))

$\displaystyle\mu$
$\displaystyle=\frac{5}{3}\,\frac{m\,c^{4}\rho\,\tau}{k_{B}T}\,\frac{\theta_{2,3}}{\theta_{2,4}}\,\theta_{2,3}\,,\,\quad\text{case}\,N=2\,\,$
$\displaystyle\mu$
$\displaystyle=-\,\frac{c^{4}m\,\rho\,\tau}{2\,k_{B}T}\,\frac{\left|\begin{matrix}\frac{2}{15}\,\theta_{2,4}&&\frac{2}{15}\,\theta_{2,5}&&-\,\frac{2}{3}\,\theta_{2,3}\\\
&&&&\\\
\frac{2}{45}\,\theta_{2,5}&&\frac{1}{35}\,\theta_{2,6}&&-\,\frac{2}{15}\,\theta_{2,4}\\\
&&&&\\\
\frac{2}{3}\,\theta_{2,3}&&\frac{2}{5}\,\theta_{2,4}&&0\end{matrix}\right|}{\left|\begin{matrix}\frac{2}{15}\,\theta_{2,4}&&\frac{2}{15}\,\theta_{2,5}\\\
&&\\\
\frac{2}{45}\,\theta_{2,5}&&\frac{1}{35}\,\theta_{2,6}\end{matrix}\right|}\,,\,\quad\text{case}\,N=3$

It is immediate to realize that these two expressions of $\mu$ are, in
general, different each other and then we can conclude that the Maxwellian
Iteration give us in the relativistic case a result depending, in general, on
the number of moments $N$. Analogous conclusions can be reached by observing
the different expressions of $\chi$ (compare eqs. (23) and (27)) and $\nu$
(compare eqs. (21) and (26)) when one uses the Maxwellian Iteration method in
the relativistic cases for polyatomic gases in the cases $N=3$ and $N=2$,
rispectively.

Let us now look at the expressions obtained for $\chi$, $\nu$ and $\mu$ by
using the Chapman-Enskog Method in the relativistic case for a polyatomic gas
with an arbitrary value of $N$ (which are written below for the convenience of
the reader)

$\displaystyle\nu$
$\displaystyle=-\frac{m\,\tau}{k_{B}}\,\left[\left|\begin{matrix}\rho&&\frac{e}{c^{2}}\\\
&&\\\
\frac{e}{c^{2}}&&\rho\,\theta_{0,2}\end{matrix}\right|^{-1}\,\left(\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}^{*}\,\left|\begin{matrix}p&&\frac{e}{c^{2}}\\\
&&\\\
\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}&&\rho\,\theta_{0,2}\end{matrix}\right|\,+\right.\right.$
$\displaystyle+\left.\,\frac{1}{6}\,\rho\,c^{2}\,\theta_{1,3}^{*}\,\,\left|\begin{matrix}\rho&&p\\\
&&\\\
\frac{e}{c^{2}}&&\frac{1}{3}\,\rho\,c^{2}\,\theta_{1,2}\end{matrix}\right|\right)\,\left.-\,\frac{5}{9}\,\rho\,c^{4}\,\theta_{2,3}^{*}\right]\,,$
$\displaystyle\chi$ $\displaystyle=-\,\frac{m\,\tau
c^{8}}{9\,k_{B}T^{2}}\,\frac{\rho^{2}}{p}\,\theta_{1,2}\left(\theta_{1,2}^{*}\,-\,\frac{\rho\,c^{2}}{p}\,\theta_{1,2}\theta_{1,1}^{*}\right)\,,$
$\displaystyle\mu$
$\displaystyle=\frac{1}{3}\,\frac{m\,\tau}{k_{B}T}\,\rho\,c^{4}\,\theta_{2,3}^{*}\,,$

where $\theta_{k,j}$ and $\theta^{*}_{k,j}$ are introduced in eqs. (9) and
(51), rispectively. Since these expressions of $\chi$, $\nu$ and $\mu$ do not
depend on $N$ we can conclude that the Chapman-Enskog Method furnishes results
which do not depend on the number of the moments $N$ in the relativistic case.
The convergence of $\nu,\mu,\chi$ in the non relativistic limit have been
proved in section 4.1.

Moreover, in section 4 it has been proved that the Maxwellian Iteration and
the Chapman-Enskog Method lead at the same results in nonrelativistic case.

Finally, we want to conclude this section with an important observation. Of
course, if one uses the Maxwellian Iteration Method for polyatomic gas in the
relativistic case the results depend on the choice of the production term $Q$
defined in equation (1). So a natural problem (still open) is the
determination of a specific function $Q$ such that the requirements of zero
production of mass and of momentum-energy are satisfied and whose consequent
Maxwellian Iteration does not depend on the number of the moments $N$. This
remains an open problem.

Another open problem is the limit in the case of the Maxwellian Iteration for
$N$ going to infinity. Obviously, if this would be possible, the result will
not depend on $N$, but we do not know if this will be the same of Ordinary
Thermodynamics obtained with the Chapman-Enskog method being based on the non
proved paradigma that“the truncated moment approach is to consider an
approximation of the Boltzmann equation that corresponds to infinite moments”.
In past times it was thought that this was the solution of the problem of the
loss of hyperbolicity of the field equations (See [24], [25]. But this was not
true; in fact, Struchtrup ([26], [27]) proved that with an increasing number
of moments the hyperbolicity region does not increase. In [24] a second order
approach with respect to equilibrium was tested and the result showed that the
hyperbolicity region increases. Finally, in [7], in section 2, the following
facts have been proved:

  1. 1.

For every truncated system there exists full hyperbolicity if only the
Lagrange multipliers are taken as independent variables;

  2. 2.

The hyperbolicity region appears only as a consequence of the approximations
(usually at first order with respect to equilibrium) involved in the
tranformation from the Lagrange multipliers to physical variables;

  3. 3.

The full hyperbolicity is recovered if the above transformation is performed
up to whatever order with respect to equilibrium.

So the full hyperbolicity is not achieved by increasing the number of moments
(or taking its limit for $N$ going to infinity), but eliminating the
approximations in the passage from the Lagrange multipliers to physical
variables.
Based on this fact, we are not able to anticipate how the model with infinite
moments can influence the present problem of the transition to Ordinary
Thermodynamics. It needs to be tested.

## Acknowledgments

The authors would like to thank the two anonymous refeees and C. van der Mee
whose suggestions, comments and remarks helped us to improve the quality of
the paper. The authors have been partially supported by INdAM-GNFM. Moreover,
one of the authors (FD) has been partially supported by Ministero
dell’Università e della Ricerca of Italy (MIUR) under the PRIN project
“2022TEB52W - The charm of integrability: from nonlinear waves to random
matrices”.

##  Appendix A Some integrals necessary for recovering OT with the Chapman-
Enskog Method

We define

$\displaystyle
A^{*\alpha_{1}\cdots\alpha_{n+1}}=\frac{c^{3}}{m^{n-2}}\int_{\Re^{3}}\int_{0}^{+\infty}\frac{f_{E}}{U_{\mu}p^{\mu}}\,p^{\alpha_{1}}\cdots
p^{\alpha_{n+1}}\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)^{n-1}\,\varphi(\mathcal{I})\,d\,\mathcal{I}\,d\,\vec{P}\,.$
(49)

We see that $A^{*\alpha_{1}\cdots\alpha_{n+1}}_{E}$ is like
$A^{\alpha_{1}\cdots\alpha_{n+1}}_{E}$ but with the function to be integrated
which

is now divided by $\cosh\,s\left(1\,+\,\frac{\mathcal{I}}{m\,c^{2}}\right)$.
Consequently, we find the expressions corresponding to (8), (9), i.e.,

$\displaystyle
A^{*\alpha_{1}\cdots\alpha_{n+1}}_{E}=\sum_{k=0}^{\left[\frac{n+1}{2}\right]}\rho
c^{2k}\theta_{k,n}^{*}\,h^{(\alpha_{1}\alpha_{2}}\cdots
k^{\alpha_{2k-1}\alpha_{2k}}U^{\alpha_{2k+1}}\cdots U^{\alpha_{n+1})}\,.$ (50)

where the scalar coefficients $\theta_{k,n}^{*}$ are

$\displaystyle\theta_{k,n}^{*}=\frac{1}{2k+1}\begin{pmatrix}n+1\\\
2k\end{pmatrix}\frac{\int_{0}^{+\infty}J_{2k+2,n-2k}^{*}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)^{n-1}\,\phi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}\,,$
(51)

where
$J_{m,n}(\gamma)=\int_{0}^{\infty}e^{-\gamma\cosh{s}}\cosh^{n}{s}\sinh^{m}{s}\,ds,\,\,\gamma=\frac{mc^{2}}{k_{B}T},\,\,J^{*}_{m,n}=J_{m,n}\left[\gamma\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\right]$.
By comparing this last equation with (9), we find that

$\displaystyle\theta_{k,n}^{*}=\frac{n+1}{n+1-2k}\,\theta_{k,n-1}\,,\quad\mbox{for
every $k$ such that}\quad n+1>2k\,.$ (52)

From this last equation it follows that only the expressions for $n+1=2k$ are
present (which means that only the case $n$ odd has to be considered), and for
these cases, eq. (51) gives

$\displaystyle\theta_{k,2k-1}^{*}=\frac{1}{2k+1}\,\frac{\int_{0}^{+\infty}J_{2k+2,-1}^{*}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)^{2k-2}\,\phi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}\,.$
(53)

The expressions are necessary with $k=1$ and with $k=2$, i.e.,

$\displaystyle\theta_{1,1}^{*}=\frac{1}{3}\,\frac{\int_{0}^{+\infty}J_{4,-1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}\,,\quad\theta_{2,3}^{*}=\frac{1}{5}\,\frac{\int_{0}^{+\infty}J_{6,-1}^{*}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)^{2}\,\phi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}\,.$
(54)

The derivative of (54)1 with respect to $\gamma$ gives

$\displaystyle\frac{\partial\,\theta_{1,1}^{*}}{\partial\,\gamma}=\frac{-1}{3}\,\frac{\int_{0}^{+\infty}J_{4,0}^{*}\,\left(1+\frac{\mathcal{I}}{mc^{2}}\right)\,\phi(\mathcal{I})\,d\,\mathcal{I}}{\int_{0}^{+\infty}J_{2,1}^{*}\,\phi(\mathcal{I})\,d\,\mathcal{I}}\,+\,\theta_{1,1}^{*}\,\frac{e}{\rho\,c^{2}}=-\,\frac{1}{\gamma}\,+\,\theta_{1,1}^{*}\,\frac{e}{\rho\,c^{2}}\,.$

From this result it follows

$\displaystyle\frac{e}{\rho\,c^{2}}=\frac{1}{\gamma\,\theta_{1,1}^{*}}\,+\,\frac{\partial\,}{\partial\,\gamma}\,\ln\theta_{1,1}^{*}\,.$
(55)

Since in literature everything has been expressed in terms of
$\frac{e}{\rho\,c^{2}}$ and its derivatives, we see that now everything is
expressed in terms of $\theta_{1,1}^{*}$, its derivative and of
$\theta_{2,3}^{*}$.

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