# On the derivation of new non-classical hydrodynamic equations for
Hamiltonian particle systems.

Raffaele Esposito
M&MOCS - Universita’ dell’Aquila<EMAIL_ADDRESS>and  Rossana Marra
Dipartimento di Fisica and Unità INFN, Università di Roma Tor Vergata
<EMAIL_ADDRESS>
###### Abstract.

We consider a Hamiltonian system of particles, interacting through of a smooth
pair potential. We look at the system on a space scale of order
$\varepsilon^{-1}$, times of order $\varepsilon^{-2}$, and mean velocities of
order $\varepsilon$, with $\varepsilon$ a scale parameter, under initial
conditions where the system is in a local Gibbs state with parameters
corresponding to density and temperature with gradients of order $1$. Assuming
that the phase space density of the particles is given by a suitable series in
$\varepsilon$ the behavior of the system under this rescaling is described, to
the lowest order in $\varepsilon$, by new non-classical hydrodynamic equations
that cannot be derived from the compressible Navier-Stokes equations in the
small Mac number limit. The analogous equations in kinetic theory are called
ghost effect equations.

R. Marra is supported by INFN

##  1\.  Introduction

The problem of deriving the hydrodynamical equations from the Hamiltonian
equations of motion of atoms, in the limit when a scale parameter
$\varepsilon$ is small, is one of the main open problems of non-equilibrium
Statistical Mechanics. The compressible Navier-Stokes system (CNSE) is a
phenomenological description of the dissipative hydrodynamics and the
incompressible Navies-Stokes-Fourier system (INSF) can be derived from the
compressible one in the low Mach number limit. Unfortunately the CNSE has no
scaling space-time invariance and hence cannot be obtained from a microscopic
description while such an obstruction is not present for the INSF system and,
assuming the temperature constant at the $\varepsilon^{0}$ order, a formal
derivation from particles was given in [13], while a rigorous proof at the
level of the Boltzmann equation was obtained, among the others, in [7], [9].

The situation is very different when the assumption of constant temperature at
$\varepsilon^{0}$ order is removed. In this case, a set of new physically
relevant hydrodynamical equations has been derived formally starting from the
Boltzmann equation [27][19][20][7][3]. They are characterized by a correction
to the Navier-Stokes stress tensor that depends on derivatives of the
temperature. This effect was already known to Maxwell [23]. The relevance of
this system is that it is non-classical in the sense that it cannot be derived
from the CNSE, and hence in some sense it indicates a failure of the CNSE in
describing the real world. Since these equations are derived from the
Boltzmann equation, the state equation is the one of a perfect gas. Sone has
given to these equations the name of ghost effect system. The name is
suggested by the fact that a vanishingly small velocity field produces finite
size modifications of the usual heat equation. There are situations
(particular geometries, stationary cases etc.) in which the classical heat-
conduction equation fails to correctly describe the temperature field of the
gas. These modifications are confirmed by many numerical experiments. There
has been a big theoretical, numerical and experimental work on this and for
the details we refer to [27] and references therein.

It is a natural question to ask if such kind of equations can be derived also
from a Hamiltonian particle system. In this paper we answer affirmatively to
that, but only at a formal level.

A system of many interacting particles, moving according to the Newton
equations of motion, can be described on a space scale much larger than the
typical microscopic scale (say the range of the interaction) in terms of
density, velocity and temperature fields, satisfying hydrodynamic equations,
like Euler or Navier-Stokes equations. The scale separation and the local
conservation laws are responsible of this reduced description.

In fact, on the macroscopic scale the quantities which are locally conserved
(slow modes) play a major role in the motion of the fluid. The derivation of
the Euler equations is based on the assumption of local equilibrium. On times
of order $\varepsilon^{-1}$, the system is expected to be described
approximately by a local Gibbs measure, with parameters varying on regions of
order $\varepsilon^{-1}$, $\varepsilon$ being a scale parameter. The local
equilibrium assumption implies that the parameters of the local Gibbs measures
satisfy the Euler equations [24], [8], [6]. The microscopic structure (the
potential) appears only in the state equation which links pressure and
internal energy to the other macroscopic parameters. The microscopic locally
conserved quantities converge, as $\varepsilon\rightarrow 0$, by a law of
large numbers, to macroscopic fields. To make this correct, the many particles
Hamiltonian system must have good dynamical mixing properties to approach and
stay in a state close to the local equilibrium. At the moment it is not
understood how to provide such properties. Therefore the only rigorous results
are obtained by adding some noise to the Hamiltonian evolution [25] (see [28]
for a review on the rigorous results for stochastic systems).

The derivation of the Navier-Stokes equations presents many more difficulties.
These equations, which describe the behavior of a fluid in the presence of
dissipative effects, do not have an immediate interpretation in terms of scale
separation. This is not surprising because the NS equations do not have a
natural space-time scale invariance like the Euler equations. In fact to see
the effect of the viscosity and the thermal conduction one has to look at
times such that neighboring regions in local equilibrium exchange a sensible
amount of momentum and energy. Simple considerations show that the right scale
of time is $\varepsilon^{-2}$. On the other hand we cannot hope to find the
compressible Navier-Stokes-Fourier behavior from the particle system under the
parabolic rescaling $x\rightarrow\varepsilon^{-1}x$ and
$t\rightarrow\varepsilon^{-2}t$ since the NS equations are not invariant under
this scaling, due to the presence of the transport terms. A way out is to
consider the incompressible limit simultaneously, because the incompressible
Navier-Stokes-Fourier equations (INSF) have the required scaling invariance.
Along this path, in [13] we gave a formal derivation of the INSF from a
Hamiltonian particle system under the parabolic rescaling, in the low Mach
number regime.

In the paper [13] the main ingredient is the assumption that the non-
equilibrium density solution of the rescaled Liouville equation can be
expressed as a truncated series in the parameter $\varepsilon$. We followed a
procedure inspired to the Hilbert-Chapmann-Enskog expansion [5], used to
construct the solution of the rescaled Boltzmann equation. From the physical
point of view we think of the system as being in local equilibrium with
parameters which are themselves given by a series in $\varepsilon$. However
there is a non-hydrodynamic correction to the local equilibrium which depends
from the non conserved quantities in the system (fast modes) and we assume
that this correction does not affect the first order in the expansion, that is
at the first order the system is still described by a local Gibbs measure with
parameters which differ from constants by terms of order $\varepsilon$. The
parameters conjugate to density and temperature are constant plus terms of
order $\varepsilon$ and the one conjugate to the velocity field is of order
$\varepsilon$. This last is strictly related to the incompressibility
assumption and would be false in the case of finite Mach number. This
assumption is the translation of the Hilbert expansion for the Boltzmann
equation to the particle system case. On the other hand, the non
hydrodynamical corrections in the second order are important on the scale
$\varepsilon^{-2}$ and give rise to the N.S. terms. It is worth to mention
here that very strong and rather uncontrollable assumptions are necessary even
to give sense to the formal calculations below:

  * •

the space of the invariant observables for the microscopic dynamics reduces to
the locally conserved quantities, mass, momentum and energy and functions of
them;

  * •

some equilibrium time correlation functions decay sufficiently fast.

Such assumptions are far from being sufficient for a mathematical proof.

Under the same scaling, in the context of the Boltzmann equations, for initial
conditions such that the density and temperature have gradients of order $1$
the formal limiting equations are different and some new terms depending on
gradients of temperature appear in the momentum equations and the velocity
field is not anymore divergence-less, see e.g. [27][19][7]. The rigorous
derivation for the stationary Boltzmann equation has been obtained recently in
[10].

Here, we try to formally derive the analogous equations as limiting equations
from a system of interacting particles. Again, the main ingredient is the
assumption that the non-equilibrium density can be expressed as a truncated
series in the parameter $\varepsilon$. The difference, to be consistent with
the initial conditions, is that the first term of the expansion is the Gibbs
measure with parameters conjugate to density and temperature depending on time
and position.

This a particular local Gibbs measure and we denote it by $G_{0}$. We stress
that it is not anymore an equilibrium.

If the initial condition is the local Gibbs measure $G_{0}$, plus correction
of order $\varepsilon$, the empirical fields evolve on the macroscopic space-
time scales close to the solution of the non-classical new equations, while if
the initial condition is the global Gibbs measure plus correction of order
$\varepsilon$, the empirical fields evolves on the macroscopic space-time
scales close to the solution of the INSF, since in this case the perturbation
of order $\varepsilon$ is too small to generate the new terms in the limiting
equations.

In Section 2 we introduce the empirical fields and the microscopic evolution
equations for them in terms of the currents. The currents cannot be expressed
back in terms of the hydrodynamical fields, and one has to solve the closure
problem in a suitable way to get the hydrodynamical equations. We consider the
averages versus $F^{\varepsilon}$, solution of the Liouville equation, and the
lowest order in $\varepsilon$ gives the following hydrodynamic equations,
derived in Section 3, for the velocity field $u$, the density $\rho$ and the
internal energy $e$

$\displaystyle\left\\{\begin{array}[]{rcl}\nabla P&=&0,\\\
\rule{0.0pt}{11.99998pt}{\partial_{t}}\rho+\nabla_{x}\cdot(\rho u)&=&0,\\\
\rule{0.0pt}{11.99998pt}\rho\partial_{t}u+u\cdot\nabla_{x}u+\nabla_{x}\mathfrak{p}&=&\nabla_{x}\cdot\left(\tau^{(1)}-\tau^{(2)}\right),\\\
\rule{0.0pt}{18.00005pt}\rho[\partial_{t}e+u\cdot\nabla
e]+P\big{(}\nabla_{x}\cdot
u\big{)}&=&\nabla_{x}\cdot\left(\kappa\dfrac{\nabla_{x}T}{2T^{2}}\right),\end{array}\right.$
(1.5)

where, for $\alpha,\beta=1,\dots d$, with $d$ the space dimension,

$\tau^{(1)}_{\alpha\beta}:=\eta\left(\partial_{\alpha}u^{\beta}+\partial_{\beta}u^{\alpha}-{2\over
d}\delta_{\alpha\beta}\partial_{\alpha}u^{\alpha}\right)+\zeta\delta_{\alpha\beta}\partial_{\alpha}u^{\alpha},$
$\tau^{(2)}_{\alpha\beta}:=[K_{1}({\partial_{\alpha}}T\partial_{\beta}T-{1\over
d}\sum_{\alpha}({\partial_{\alpha}}T)^{2})+{\omega_{1}}\sum_{\alpha}({\partial_{\alpha}}T)^{2}+K_{2}(\partial^{2}_{\alpha\beta}T-{1\over
d}\sum_{\alpha}\partial^{2}_{\alpha\alpha}T)+{\omega_{2}}\sum_{\alpha}\partial^{2}_{\alpha\alpha}T)]$

Here, $P$ and $e$ are the thermodynamical functional pressure and internal
energy which are functions of $\rho,T$ as determined by the Gibbs measure.

These equations are a new set of hydrodynamic equations, which cannot be
derived from the compressible Navier-Stokes equations. The divergence of the
velocity field is not zero as in INSF, even if we sent the Mach number to
zero. There are new ”thermal stress” terms in the equation for the velocity
field. We observe that, even if in the thermal stress tensor $\tau^{(2)}$
there are third order derivatives in $T$, these terms (the last two) appear in
the equation as a gradient and hence can be absorbed in the unknonw pressure
$\mathfrak{p}$ [22][27].

We also show that the entropy associated to the solutions of these equations,
correctly, is increasing in time.

Our derivation gives the viscosity coefficient $\eta$, bulk viscosity $\zeta$
and the conductivity $\kappa$ in terms of the local equilibrium time
correlation functions.

The fluctuation-dissipation theory relates the transport coefficients to time-
integrated correlation functions (see [17], [21]). The expression we find
agrees with the Green-Kubo formula for the shear and bulk viscosity as well
for the conductivity. Moreover, in these new equations, in particular in the
equation for the momentum, there are more transport coefficients and we find
an expression for them in terms of the potential and of a double time integral
in the Appendix. These expression reminds the ones in the second order
dissipative hydrodynamics introduced in [30] ([18]). We remark that the linear
response theory cannot give these new transport coefficients.

In the last section we compare this equations with the ones derived from the
Boltzmann equation.

##  2\. Conservation laws

We consider a system of many, $N$, identical particles of unit mass in a torus
$\Lambda_{\varepsilon}$ of size $\varepsilon^{-1}$ in $\mathbb{R}^{d}$,
interacting via a pair central potential $V$ of finite range. The Newton
equations are

$\begin{split}{d\xi_{i}\over d\tau}(\tau)&=v_{i}(\tau),\\\ {dv_{i}\over
dt}(\tau)&=-\sum_{i\neq j}\nabla V(\xi_{i}-\xi_{j}|),\end{split}$

where $\xi_{i},v_{i},\tau$, denote the microscopic coordinates, velocities and
time, and $i=1,\dots,N$, After rescaling space as $\varepsilon^{-1}$ and time
as $\varepsilon^{-2}$, they become,

$\begin{split}{dx_{i}\over dt}(t)&=\varepsilon^{-1}v_{i}(t),\\\ \\\
{dv_{i}\over dt}(t)&=-\varepsilon^{-2}\sum_{i\neq j}\nabla
V(\varepsilon^{-1}|x_{i}-x_{j}|),\end{split}$ (2.1)

with $x_{i}=\varepsilon\xi_{i}$ the macroscopic coordinates and
$t=\varepsilon^{2}\tau$ the macroscopic time. The number of particles $N$ is
assumed to be of order $\varepsilon^{-d}$ to keep the density finite.

The rescaled Newton equations, with initial data randomly distributed, are
equivalent to the Liouville equation for the evolution of a distribution
$\mu^{\varepsilon}_{0}(x_{1},\cdots,x_{N},v_{1},\cdots,v_{N}):=\mu_{0}(\varepsilon^{-1}x_{1},\cdots,\varepsilon^{-1}x_{N}v_{1},\cdots,v_{N})$
at time $0$ on the $N$ particle phase space

${\partial\over\partial
t}\mu^{\varepsilon}_{t}={\varepsilon^{-2}}\texttrademark{\mathscr{L}}^{*}\mu^{\varepsilon}_{t},$
(2.2)

where ${\mathscr{L}}^{*}$ is the Liouville operator

$\displaystyle{\mathscr{L}}^{*}\phi(x_{1},\cdots,x_{N},v_{1},\cdots,v_{N})=-\sum_{i=1}^{N}\sum_{k=1}^{d}\big{\\{}\varepsilon
v_{i}^{k}{\partial\phi\over\partial x_{i}^{k}}-\sum_{i=\hskip-5.12128pt/\hskip
1.9919ptj}{\partial_{k}}V\Big{(}\varepsilon^{-1}|x_{i}-x_{j}|\Big{)}{\partial\phi\over\partial
v_{i}^{k}}\big{\\}},$ (2.3)

with ${\mathscr{L}}^{*}=-{\mathscr{L}}$ the adjoint of ${\mathscr{L}}$ with
respect to the scalar product induced by the a priori measure
$d\mathbb{Z}={1\over n!}d^{d}x_{1}d^{d}v_{1}$ $\dots d^{d}x_{N}d^{d}v_{N}$. To
be more explicit, we denote the average versus the a priori measure
$d\mathbb{Z}$ by $\Big{\langle}\phi\Big{\rangle}$ and define the scalar
product $\langle\phi,\psi\rangle=\Big{\langle}\phi\psi\Big{\rangle}$ so that

$\langle\phi,{\mathscr{L}}^{*}\psi\rangle=-\langle{\mathscr{L}}\phi,\psi\rangle.$

The total number of particles, the $d$ components of the total momentum and
the total energy are the conserved quantities. We define the corresponding
empirical fields:

empirical density

$z^{0}(x)=\varepsilon^{d}\sum_{i=1}^{N}\delta(x_{i}-x),$ (2.4)

empirical velocity field density

$z^{\alpha}(x)=\varepsilon^{d}\sum_{i=1}^{N}v^{\alpha}_{i}\delta(x_{i}-x),\phantom{...}\alpha=1,\dots,d,$
(2.5)

empirical energy density

$z^{d+1}(x)=\varepsilon^{d}\sum_{i=1}^{N}{1\over
2}\big{[}v^{2}_{i}+\sum_{j\neq
i=1}^{N}V(\varepsilon^{-1}|x_{i}-x_{j}|)\big{]}\delta(x_{i}-x)].$ (2.6)

Their meaning is as follows: The average of the integral of $z^{\alpha}$ over
a small region is equal to the average number of particles, momentum, energy
associated to the region. We will write use the short notation

$z^{\mu}(x)=\varepsilon^{d}\sum_{i=1}^{N}\delta(x_{i}-x)z_{i}^{\mu},$ (2.7)

with

$z_{i}^{0}=1;\phantom{..}z_{i}^{\alpha}=v_{i}^{\alpha},\alpha=1,\dots,d;\phantom{..}z_{i}^{d+1}={1\over
2}[v^{2}_{i}+\sum_{i\neq j=1}^{N}V(\varepsilon^{-1}|x_{i}-x_{j}|)].$

Note that the correct way of writing these quantities would be to add an
$\varepsilon$ index: $z_{\varepsilon}^{\mu}(x)=z^{\mu}(\xi)$. We will omit
this index in the following, being clear from the context whether we are
referring to microscopic or macroscopic variables.

The empirical fields satisfy the following local conservation laws, which are
obtained by integrating (2.7) against a smooth $\mathbb{R}^{d}$ test function
$f(x)$, $t$-differentiating and using the Newton equations (LABEL:(2.1)):

${d\over
dt}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})=\varepsilon^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{\alpha=1}^{d}{\partial_{\alpha}f}(x_{i})v_{i}^{\alpha},$
(2.8) ${d\over
dt}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})v_{i}^{\beta}=\varepsilon^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{\alpha=1}^{d}\Big{\\{}{\partial_{\alpha}f}(x_{i})v_{i}^{\alpha}v_{i}^{\beta}-\varepsilon^{-1}\sum_{j\neq
i=1}^{N}{\partial_{\beta}}V(\varepsilon^{-1}|x_{i}-x_{j}|)f(x_{i})\Big{\\}},$
(2.9) ${d\over
dt}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})z_{i}^{d+1}=\varepsilon^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{\alpha=1}^{d}\Big{\\{}{\partial_{\alpha}f}(x_{i})v_{i}^{\alpha}z_{i}^{d+1}-{1\over
2}\varepsilon^{-1}\sum_{i\neq
j=1}^{N}{\partial_{\alpha}}V(\varepsilon^{-1}|x_{i}-x_{j}|)v_{i}^{\alpha}f(x_{i})\Big{\\}}.$
(2.10)

Here $\partial_{\beta}V(\xi)=\partial V(\xi)/\partial\xi_{\beta}$. Because of
the symmetry properties of the potential we can write, as usual, the second
term in the r.h.s. of (2.9) as

$-{1\over 2}\varepsilon^{d-2}\sum_{i\neq
j=1}^{N}{\partial_{\beta}}V(\varepsilon^{-1}|x_{i}-x_{j}|)[f(x_{i})-f(x_{j})].$
(2.11)

Since $f$ is slowly varying on the microscopic scale we can write

$f(\varepsilon\xi_{i})-f(\varepsilon\xi_{j})=\sum_{\gamma=1}^{d}{\partial_{\gamma}f}(x_{i})\varepsilon[\xi_{i}^{\gamma}-\xi_{j}^{\gamma}]+\varepsilon^{2}D_{0}+\varepsilon^{3}D+O(\varepsilon^{4}),$
(2.12)

where

$D_{0}={1\over
2}\sum_{\gamma,\nu=1}^{d}{\partial_{\alpha\nu}^{2}}f(x_{i})[\xi_{i}^{\gamma}-\xi_{j}^{\gamma}][\xi_{i}^{\nu}-\xi_{j}^{\nu}],$
(2.13) $D={1\over
6}\sum_{\gamma,\nu,\alpha=1}^{d}{\partial_{\gamma\alpha\nu}^{3}f}(x_{i})[\xi_{i}^{\gamma}-\xi_{j}^{\gamma}][\xi_{i}^{\nu}-\xi_{j}^{\nu}][\xi_{i}^{\alpha}-\xi_{j}^{\alpha}].$
(2.14)

The last term of (2.9) becomes:

$\begin{split}&{1\over
2}\varepsilon^{-1}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma=1}^{d}{\partial_{\gamma}f}(x_{i})\Psi^{\beta\gamma}(\varepsilon^{-1}|x_{i}-x_{j}|)\\\
+&{1\over
4}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}f}(x_{i})\Phi_{0}^{\beta\nu\gamma}(\varepsilon^{-1}|x_{i}-x_{j}|)\\\
+&{1\over
12}\varepsilon\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu,\alpha=1}^{d}{\partial_{\gamma\nu\alpha}^{3}f}(x_{i})\Phi^{\beta\alpha\gamma\nu}(\varepsilon^{-1}|x_{i}-x_{j}|)+O(\varepsilon^{2}).\end{split}$
(2.15)

with

$\Psi^{\beta\gamma}(\xi)=-\partial_{\beta}V(\xi)\xi^{\gamma},\quad\Phi_{0}^{\beta\nu\gamma}=-\partial_{\beta}V(\xi)\xi^{\gamma}\xi^{\nu},\quad\Phi^{\beta\alpha\gamma\nu}(\xi)=-\partial_{\beta}V(\xi)\xi^{\gamma}\xi^{\alpha}\xi^{\nu}.$
(2.16)

We also set

$\begin{split}&\bar{\Phi}_{0}^{\beta\gamma\nu}(x)={1\over
2}\varepsilon^{d}\sum_{ij=1}^{N}\delta(x_{i}-x)\Phi_{0}^{\beta\gamma\nu}(\varepsilon^{-1}|x_{i}-x_{j}|),\\\
&\bar{\Phi}^{\alpha\beta\gamma\nu}(x)={1\over
2}\varepsilon^{d}\sum_{ij=1}^{N}\delta(x_{i}-x)\Phi^{\alpha\beta\gamma\nu}(\varepsilon^{-1}|x_{i}-x_{j}|),\end{split}$
(2.17)

An analogous computation, which we omit for sake of shortness, can be done for
the energy equation.

The general form of the rescaled local conservation laws is (up to higher
order in $\varepsilon$)

${\partial\over\partial t}\int dxf(x)z^{\beta}(x)=\varepsilon^{-1}\int
dx\sum_{k=1}^{d}{\partial_{k}f}(x)w^{\beta k}(x),$ (2.18)

where $w^{\beta k}$, $\beta=0,\dots,d+1;\ k=1,\dots,d$ are the currents
corresponding to the fields $z^{\beta}$ and are explicitly given by

$w^{0k}(x)=\varepsilon^{d}\sum_{i=1}^{N}\delta(x_{i}-x)v_{i}^{k},$ (2.19)
$\displaystyle\begin{split}w^{\beta
k}(x)&=\varepsilon^{d}\sum_{i=1}^{N}\delta(x_{i}-x)\big{\\{}v_{i}^{\beta}v_{i}^{k}+{1\over
2}\sum_{j=1}^{N}\Psi^{\beta k}(\varepsilon^{-1}|x_{i}-x_{j}|)\big{\\}}\\\
&+{1\over
4}\varepsilon\sum_{\gamma=1}^{d}{\partial_{\gamma}}\Phi_{0}^{k\beta\gamma}(x)+{1\over
12}\varepsilon^{2}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}}\Phi^{k\beta\gamma\nu}(x)+O(\varepsilon^{3}),\quad\beta=1,\dots,d,\\\
w^{d+1\,k}(x)&=\varepsilon^{d}\sum_{i=1}^{N}\big{\\{}v_{i}^{k}z_{i}^{d+1}+{1\over
2}\sum_{j=1}^{N}\sum_{\gamma=1}^{d}\Psi^{\gamma
k}(\varepsilon^{-1}|x_{i}-x_{j}|){1\over
2}[v^{\gamma}_{i}+v^{\gamma}_{j}]\big{\\}}+O(\varepsilon),\end{split}$ (2.20)

We introduce the notation $w^{\beta k}_{i}$ such that

$w^{\beta k}(x)=\varepsilon^{d}\sum_{i=1}^{N}\delta(x_{i}-x)w^{\beta k}_{i}.$

Hence $w^{0k}_{i}=v_{i}^{k}$, and so on.

We also denote the r.h.s. of the first line of (2.20) by

$w_{*}^{\beta
k}(x):=\varepsilon^{d}\sum_{i=1}^{N}\delta(x_{i}-x)\big{\\{}v_{i}^{\beta}v_{i}^{k}+{1\over
2}\sum_{j=1}^{N}\Psi^{\beta k}(\varepsilon^{-1}|x_{i}-x_{j}|)\big{\\}}.$
(2.21)

We stress that we keep terms of order $\varepsilon$ and $\varepsilon^{2}$ in
(2.20) while we do not do the same in (2.21) and that such new terms were not
taken into account in the derivation of the incompressible Navier-Stokes
equation [13] because do not give contribution at the lowest order. In this
new setting instead we will see that the $D$ term gives a contribution at the
lowest order for the momentum equation.

A key point in the argument is the remark that the empirical fields
$z^{\alpha}(\xi)$ are approximate integrals of the motion in the following
sense:

For any smooth function $f(x)$ on the torus $\mathbb{T}_{1}^{d}$, from
previous calculation it follows that

$\mathscr{L}\big{[}\varepsilon^{d}\sum_{i=1}^{N}f(\varepsilon\xi_{i})z_{i}^{\alpha}\big{]}=O(\varepsilon).$
(2.22)

On the other hand, the evolution of a generic observable
$\Phi(x_{1},\cdots,x_{N},v_{1},\cdots,v_{N})$ according to the rescaled Newton
equations is given by

$\partial_{t}\Phi=\varepsilon^{-2}{\mathscr{L}}\Phi.$ (2.23)

Therefore, while a generic observable has a time derivative of order
$\varepsilon^{-2}$, there are special observables as the empirical fields
associated to the mass, momentum and energy, that have a time derivative only
of order $\varepsilon^{-1}$. So they have a comparatively slow evolution and
this justifies the denomination of approximate integrals of motion we used
above.

We stress that we need to assume that the only observables with this property
are the empirical fields (and of course any function of them as well).

Now we turn to the definition of a special class of states, the local Gibbs
states. A (global) Gibbs state in a macroscopic volume $\Lambda$ (hence a
volume $\varepsilon^{-1}\Lambda$ in microscopic variables) is defined as a
probability distribution on the
$R\\!\\!\\!\\!\\!\kern-1.0ptI\,\,_{\Lambda}$-space $\bigcup_{N\geq
0}(\Lambda\times\mathbb{R}^{d})^{N}$, whose density with respect to the a
priori measure $d\mathbb{Z}$ is given by (in the grandcanonical setting)

$\displaystyle
G_{\Lambda}(x_{1},v_{1},\dots,x_{N},v_{N})=Z_{\Lambda}^{-1}\prod_{j=1}^{N}{\tilde{z}}\exp\Big{[}-{(v_{j}-u)^{2}\over
2T}+{1\over T}\sum_{k=\hskip-5.12128pt/\hskip
1.9919ptj}V\Big{(}{|x_{j}-x_{k}|\over\varepsilon}\Big{)}\Big{]}.$ (2.24)

Here $\tilde{z}$ is the activity and $Z_{\Lambda}$ is a normalization factor.
We can write the previous expression in terms of
$\underline{\lambda}=\\{\lambda^{\alpha}\\}$, $\alpha=0,\dots,d+1$, $d+2$ real
numbers called chemical potentials associated to the mass, momentum and energy
empirical fields and parametrizing the family of the Gibbs states

$G(\underline{\lambda})=Z_{\Lambda}^{-1}\exp\\{\sum_{i=1}^{N}\sum_{\mu=0}^{d+1}\lambda^{\mu}z_{i}^{\mu}\\}.$
(2.25)

Their explicit expression is

$\lambda^{0}=\log\tilde{z}-{1\over 2}\beta|u|^{2},\quad\lambda^{\nu}={1\over
T}u^{\nu},\nu=1,\cdots,d\quad\lambda^{d+1}=-{1\over T}.$

A more conventional parametrization is given by $\rho>0$, the mass density,
$u\in R\\!\\!\\!\\!\\!\kern-1.0ptI\,\,^{d}$, the velocity field and $T>0$, the
temperature. Note that for a global Gibbs state, due to the Galilean
invariance, the velocity field is usually assumed to vanish. In the definition
of local Gibbs states, on the contrary we keep a non vanishing velocity field.

The notion of local Gibbs state is related to the one of local equilibrium and
is obtained formally by replacing the constant chemical potentials
$\\{\lambda^{\alpha}\\}$ by some smooth functions
$\\{\lambda^{\alpha}(x,t)\\}$ which are functions of
$x_{i}=\varepsilon\xi_{i}$:

hence the local equilibrium state is defined, for a fixed $\varepsilon$ as

$\tilde{G}(\underline{\lambda})=\tilde{Z}_{\Lambda}^{-1}\exp\\{\sum_{i=1}^{N}\sum_{\mu=0}^{d+1}\lambda^{\mu}(x_{i},t)z_{i}^{\mu}\\},$
(2.26)

or in the equivalent form

$\tilde{G}(\underline{\lambda})={\tilde{Z}_{\Lambda}^{-1}}\exp\left[\int_{\varepsilon^{-1}\mathbb{T}^{d}}d\xi\sum_{\alpha=0}^{d+1}\lambda^{\alpha}(x,t)z^{\alpha}(\xi)\right]={\tilde{Z}_{\Lambda}^{-1}}\exp\varepsilon^{-d}\left[\int_{\mathbb{T}^{d}}dx\sum_{\alpha=0}^{d+1}\lambda^{\alpha}(x,t)z^{\alpha}(\varepsilon^{-1}x)\right],$
(2.27)

with ${\tilde{Z}_{\Lambda}}$ a suitable normalization factor and
${\mathbb{T}^{d}}$ the macroscopic torus.

Note that the local Gibbs states corresponding to any choice of smooth
chemical potentials, being given by the exponential of linear combinations of
the approximate integrals of motion, satisfy the condition (2.22), so that it
results

${\mathscr{L}}^{*}G({\underline{\lambda}(\,\cdot\,,t)})=O(\varepsilon).$
(2.28)

So, in a local Gibbs state, in sufficiently small region and for a
sufficiently small time the state looks like being in equilibrium, hence the
name local equilibrium. This situation is particularly convenient to describe
a regime close to the hydrodynamics and, as discussed in [24], [8], gives
formally the Euler equations. However, this is not sufficient to describe the
Navier-Stokes regime and the ghost regime we are going to discuss here.

To show how the conservation laws give the hydrodynamic equations we follow a
procedure similar to the Hilbert expansion proposed to approximate the
solutions of the Boltzmann equation. Let us start with the phase space
distribution function

$F_{\varepsilon}(x_{1},\dots,x_{N},v_{1},\dots,v_{N},t)=F(\varepsilon^{-1}x_{1},\dots,\varepsilon^{-1}x_{N},v_{1},\dots,v_{N},\varepsilon^{-2}t)$

for the rescaled system, which satisfies the rescaled Liouville equation

${\partial F_{\varepsilon}\over\partial
t}=\varepsilon^{-2}\mathscr{L}^{*}F_{\varepsilon}.$ (2.29)

We prepare the system at time zero as described by a local equilibrium
distribution $\bar{G}_{0}$ defined as

$\displaystyle\bar{G}_{0}$
$\displaystyle=Z_{0}^{-1}\exp\\{\\{\sum_{i=1}^{N}\sum_{\mu=0}^{d+1}\lambda_{0}^{\mu}(x_{i},0)z_{i}^{\mu}\\},$
(2.30) $\displaystyle\lambda_{0}^{\mu}(x,0)$
$\displaystyle=0,\mu=1,\dots,d;\quad\lambda_{0}^{0}(x,0)\neq
0,\lambda_{0}^{d+1}(x,0)\neq 0,$ (2.31)

$\lambda_{0}^{0}$ and $\lambda_{0}^{d+1}$ functions of $(x,t)$ with
$x$-gradient of order $1$. Note that $\bar{G}_{0}$ is not a stationary state
for (2.29).

Writing $F_{\varepsilon}$ as a series in $\varepsilon$,
$F_{\varepsilon}=\sum_{n}\varepsilon^{n}F^{n}$, and substituting it in (2.29)
we get the diverging terms $\varepsilon^{-2}\mathscr{L}^{*}F_{0}$ and
$\varepsilon^{-1}\mathscr{L}^{*}F_{1}$ which we have to take care of.

Due to the initial conditions $\lambda_{0}^{\beta}(x,0)=0$, for
$\beta=1,\dots,d$, in absence of external forces we have
$\lambda_{0}^{\beta}(x,t)=0$ at the lowest order, and we are forced to choose
$F_{0}$ as a local Gibbs state $G_{0}$ with $\lambda_{0}^{0}$ and
$\lambda_{0}^{4}$ functions of $(x,t)$ to be determined while
$\lambda_{0}^{\beta}=0$, for $\beta=1,\dots,d$.

To single out the non-hydrodynamic contribution to $F_{\varepsilon}$ let us
decompose $F_{\varepsilon}$ in a part which is Gibbsian with parameters slowly
depending on the microscopic variables and depending on $\varepsilon$ by means
of a series in $\varepsilon$, and a remainder. More explicitly, we put

$F_{\varepsilon}=G_{\varepsilon}+\varepsilon G_{0}R_{\varepsilon},$ (2.32)

with

$G_{\varepsilon}=Z_{\varepsilon}^{-1}\exp\\{\sum_{i=1}^{N}\sum_{\mu=0}^{d+1}\lambda_{\varepsilon}^{\mu}(x_{i},t)z_{i}^{\mu}\\},$
$\lambda_{\varepsilon}^{\mu}(x,t)=\sum_{n=0}^{\infty}\varepsilon^{n}\lambda^{\mu}_{n}(x,t);\phantom{...}\lambda^{\mu}_{0}=0,\mu=1,\dots,d;\quad\lambda_{0}^{0}(x,t)\neq
0,\lambda_{0}^{d+1}(x,t)\neq 0,$ (2.33)

where $\lambda_{n}^{\mu}$ are functions of $(x,t)$ to be determined. Note
that, if $\lambda_{0}^{0}$ and $\lambda_{0}^{d+1}$ do not depend on $(x,t)$,
we are back to to the case of the incompressible Navier-Stokes-Fourier system
discussed in [13].

In our context $G_{0}$, the zero order term in the expansion, is a particular
time dependent local equilibrium, but not all the relevant terms are included
in $G_{0}$. We include all the hydrodynamic terms in $G_{\varepsilon}$ and we
can assume that in $R_{\varepsilon}$ there are no terms which are combinations
of the invariant quantities $z^{\alpha}$ with coefficients depending on the
macroscopic variables, since these terms are already present in
$G_{\varepsilon}$. $R_{\varepsilon}$ represents the non-equilibrium part of
the distribution $F_{\varepsilon}$, which takes into account the fast modes in
the system, namely the non-conserved quantities. They appear at the
hydrodynamic level only through dissipative effects and determine the
expression of transport coefficients.

To make this concept more precise we refer to [29], [28] where it is
introduced the Hilbert space of the local observables equipped with the scalar
product

$(\phi,\psi)=\int
dx[\langle\phi\tau_{x}\psi\rangle_{G_{0}}-\langle\phi\rangle_{G_{0}}\langle\psi\rangle_{G_{0}}].$
(2.34)

Here $\langle\cdot\rangle_{G_{0}}$ is the average on the local Gibbs measure
$G_{0}$. In terms of this scalar product we define the projector on the
invariant space as

${\mathscr{P}\phi}=\sum_{\mu,\nu=0}^{d+1}(\phi,z^{\mu})(z,z)^{-1}_{\mu\nu}z^{\nu},$
(2.35)

where $(z,z)^{-1}$ denotes the inverse of the matrix with elements $\langle
z_{\mu}z_{\nu}\rangle_{G_{0}}$.

We assume that $R_{\varepsilon}$ has no component on the invariant space, in
other words we ask

${\mathscr{P}}[R_{\varepsilon}]=0.$ (2.36)

We also assume that

$G_{0}R_{\varepsilon}(t)=G_{0}R_{1}+\varepsilon
G_{0}R_{2}(t)+O(\varepsilon^{2}).$ (2.37)

We need explicit expressions for $R_{i},i=1,2$, in terms of the empirical
fields, so that, inserting (2.32) in the conservation laws averaged with
respect to $F_{\varepsilon}$, we can get close equations for the empirical
fields up to order $\varepsilon$. To find such expressions, we insert the
expansion (2.32) for $F_{\varepsilon}$ in the Liouville equation (2.29)

$\displaystyle\begin{split}\partial_{t}G_{\varepsilon}+\varepsilon\partial_{t}(G_{0}R_{\varepsilon})=\varepsilon^{-2}\mathscr{L}^{*}G_{\varepsilon}+\varepsilon^{-1}\mathscr{L}^{*}G_{0}R_{\varepsilon}\\\
=\varepsilon^{-2}\mathscr{L}^{*}G_{0}+\varepsilon^{-1}\mathscr{L}^{*}G_{1}+\varepsilon^{-1}\mathscr{L}^{*}G_{0}R_{1}+\mathscr{L}^{*}G_{0}R_{2}+O(\varepsilon).\end{split}$
(2.38)

By (2.28), $\varepsilon^{-1}\mathscr{L}^{*}G_{0}=O(1)$. In fact, the
expression of $\mathscr{L}^{*}G_{0}$ can be computed as

$\mathscr{L}^{*}G_{0}=\varepsilon
G_{0}\sum_{i=1}^{N}\sum_{\mu=0,d+1}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{i})w^{\mu\gamma}_{i}=\varepsilon
G_{0}\sum_{i=1}^{N}[{\partial_{\gamma}\lambda_{0}^{0}}(x_{i})w^{0\gamma}_{i}+{\partial_{\gamma}\lambda_{0}^{d+1}}(x_{i})w^{{d+1}\gamma}_{i}].$
(2.39)

Next we compute $\mathscr{L}^{*}G_{1}$. We write

$G_{1}=G_{0}g_{1},$ (2.40)

where

$g_{1}=\sum_{j=1}^{N}\sum_{\mu=0}^{d+1}\lambda_{1}^{\mu}(x_{j},t)[z_{j}^{\mu}-<z_{j}^{\mu}>_{G_{0}}].$
(2.41)

Since $g_{1}$ is a linear combination of the invariant quantities $z$ with
coefficients depending on the macroscopic variables, the action of
$\mathscr{L}^{*}$ on it gives a linear combination of the currents $w$ with a
factor $\varepsilon$:

$-\varepsilon^{-1}\mathscr{L}^{*}G_{1}=\sum_{i=1}^{N}\sum_{\mu=0}^{d+1}\sum_{\gamma=1}^{d}G_{0}{\partial_{\gamma}\lambda_{1}^{\mu}\
}(x_{i},t)w^{\mu\gamma}_{i}+\varepsilon^{-1}g_{1}\mathscr{L}^{*}G_{0}=O(1).$
(2.42)

We conclude that $\varepsilon^{-1}\mathscr{L}^{*}G_{1}$,
$\partial_{t}G_{\varepsilon}$ and $\mathscr{L}^{*}G_{0}R_{2}$ are of order at
least $1$.

If we multiply by $\varepsilon$ (2.38), in the limit $\varepsilon\rightarrow
0$, we have

$\displaystyle\varepsilon^{-1}\mathscr{L}^{*}G_{0}+\mathscr{L}^{*}G_{0}R_{1}=0.$
(2.43)

Moreover, it is easy to check that ${\mathscr{L}}^{*}G_{0}$ is odd by the
exchange $v\rightarrow-v$ because $w^{\mu\gamma}_{i}$, for
$\mu=0,d+1\quad\gamma=1\cdots d$ is odd in $v$. Hence, the condition (2.43)
can determine only the odd part of $R_{1}$ that we call $R_{1}^{a}$.

To find $R_{1}^{s}$, the even part of $R_{1}$, we apply again
${\mathscr{L}}^{*}$ to (2.43) to get, in the limit $\varepsilon\rightarrow 0$,

$\mathscr{L}^{*}\mathscr{L}^{*}G_{0}R_{1}^{s}=-\varepsilon^{-1}\mathscr{L}^{*}\mathscr{L}^{*}G_{0}.$
(2.44)

The term $\varepsilon^{-1}\mathscr{L}^{*}\mathscr{L}^{*}G_{0}$ is even (see
Appendix A.1) so that this determines $R_{1}^{s}$. It has a part of order
$\varepsilon$ which enter in the form of the new transport coefficients in the
momentum equation and a part of order $1$ which does not give contribution in
the equation. The expression of
$\varepsilon^{-2}\mathscr{L}^{*}\mathscr{L}^{*}G_{0}$ is very complicate and
involves two space derivatives of the chemical potentials. To summarize,
$R_{1}=R_{1}^{a}+R_{1}^{s}$ with $R_{1}^{a}$ odd of order $1$ and $R_{1}^{s}$
even solutions of (2.43) and (2.44).

Now we go back to (2.38) and get, in the limit $\varepsilon\rightarrow 0$,

${\mathscr{L}}^{*}G_{0}R_{2}=-\varepsilon^{-1}{\mathscr{L}}^{*}G_{1}-\partial_{t}G_{0}.$
(2.45)

This condition determines $R_{2}$ in the limit $\varepsilon\rightarrow 0$ as

${\mathscr{L}^{*}G_{0}R_{2}}=G_{0}\sum_{i=1}^{N}\sum_{\mu=0}^{d+1}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{1}^{\mu}}(x_{i},s)w^{\mu\gamma}_{i}-\varepsilon^{-1}g_{1}\mathscr{L}^{*}G_{0}-\partial_{t}G_{0}.$
(2.46)

In order the previous equations to be satisfied the r.h.s. cannot have
components on the null space. This condition will be satisfied a-posteriori
using the fact that $(\rho,e,u)$ is solution of the hydrodynamical equations
(in Appendix A.2).

We assume that there exists a unique solution $R_{1}^{a}(t)$ to (2.43) and
$R_{2}$ to (2.46) such that ${\mathscr{P}}R_{1}^{a}(t)=0$,
${\mathscr{P}}R_{2}(t)=0$, which are expressed formally in terms of
${\mathscr{L}^{*}}^{-1}$. We assume also that there exists a unique solution
to (2.44) with ${\mathscr{P}}R_{1}^{s}(t)=0$ which is expressed formally in
terms of ${\mathscr{L}^{*}}^{-1}{\mathscr{L}^{*}}^{-1}$. This is the
assumption we really need on the inverse of $\mathscr{L}^{*}$ to get the
result.

For sake of simplicity, we consider from now on a particular form for $g_{1}$,
namely we put to zero $\lambda_{1}^{0}$ and $\lambda_{1}^{d+1}$:

$\lambda_{1}^{0}=0,\quad\lambda_{1}^{d+1}=0.$ (2.47)

This is sufficient to obtain the limiting equation for $(\rho,e,u)$. Assuming
a general form of $g_{1}$ we would also get equations for the first
corrections $\rho_{1}$ and $e_{1}$, which are not needed to obtain the ghost
equations (1.5).

##  3\. Hydrodynamic equations

The incompressible limit corresponds to the assumption that the velocity field
is small compared with the sound speed. In other words we assume that
$U^{\mu}(x,t)\equiv\langle z^{\mu}(x)\rangle_{F_{\varepsilon}(t)}$,
$\mu=1,\dots,d$, starts with a term of order $\varepsilon$. Under the
assumptions on $F_{\varepsilon}$, this corresponds to choose
$\lambda_{0}^{\mu}=0$ for $\mu=1,\dots,d$. Moreover, it results
$U^{\mu}(x,t)=\varepsilon\rho T\lambda_{1}^{\mu}(x,t)+O(\varepsilon^{2})$ with
$T$, the temperature of the Gibbs state $G_{0}$, given by
$(\lambda_{0}^{d+1})^{-1}$ and $\rho$ the density of the Gibbs state $G_{0}$
corresponding to the chemical potential $\lambda_{0}^{0}$. We denote by
$u^{\mu}(x,t)$ the rescaled velocity field given by
$u^{\mu}(x,t)=T\lambda_{1}^{\mu}(x,t)$. For sake of simplicity, we will not
write the explicit dependence on time in the chemical potentials from now on.

We will use the notation
$\Big{\langle}f\Big{\rangle}_{G_{0}}=\Big{\langle}G_{0}f\Big{\rangle}$ where
$\Big{\langle}\cdot\Big{\rangle}$ is the already introduced integration w.r.t.
the a priori measure $d\mathbb{Z}={1\over n!}d^{d}x_{1}d^{d}v_{1}$ $\dots
d^{d}x_{N}d^{d}v_{N}$.

###  3.1. Continuity equation

Let’s derive first the continuity equation. To obtain it, we start from the
conservation law for the empirical density (2.8) and we take the expectation
with respect to the non-equilibrium measure $F_{\varepsilon}(t)$

$\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})\Big{\rangle}_{F_{\varepsilon}(t)}-\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})\Big{\rangle}_{F_{\varepsilon}(0)}=\varepsilon^{-1}\int_{0}^{t}ds\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})v_{i}^{k}\Big{\rangle}_{F_{\varepsilon}(s)}.$
(3.1)

Using (2.37) and the fact that
$\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})\Big{\rangle}_{G_{0}}=\int
dxf(x)\rho(x,t)$ we see that the l.h.s. of (3.1) gives

$\int dxf(x)\rho(x,t)-\int dxf(x)\rho(x,0)+O(\varepsilon).$

The term

$\varepsilon^{-1}\int_{0}^{t}ds\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})v_{i}^{k}\Big{\rangle}_{G_{0}},$

in r.h.s. of (3.1) does not appear since $G_{0}$ is Gaussian in the velocities
with zero mean and the rhs of (3.1) becomes

$\int_{0}^{t}ds\Big{\langle}g_{1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})v_{i}^{k}\Big{\rangle}_{G_{0}}+\int_{0}^{t}ds\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})v_{i}^{k}R_{1}\Big{\rangle}_{G_{0}}+O(\varepsilon).$
(3.2)

The second term in (3.2) is $0$ by using that ${\mathscr{P}}R_{1}=0$. Now we
discuss the first term. By (2.41),

$\int_{0}^{t}ds\Big{\langle}g_{1}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})v_{i}^{k}\Big{\rangle}_{G_{0}}=\int_{0}^{t}ds\Big{\langle}\sum_{j=1}^{N}\sum_{\mu=1}^{d}\lambda^{\mu}_{1}(x_{j})z^{\mu}_{j}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})v_{i}^{k}\Big{\rangle}_{G_{0}}.$
(3.3)

We use the choice we made for $\lambda_{1}$ so that the l.h.s. of (3.3)
becomes

$\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}\sum_{\mu,k=1}^{d}\lambda^{\mu}_{1}(x_{i}){\partial_{\mu}f}(x_{i})v_{i}^{\mu}v_{i}^{k}\Big{\rangle}_{G_{0}}.$
(3.4)

Since the average on $G_{0}$ of $v_{i}^{\mu}v_{i}^{k}$ contributes only for
$k=\mu$, in the limit $\varepsilon\rightarrow 0$ we have, using
$u^{\mu}(x,t)=T\lambda_{1}^{\mu}(x,t)$,

$-\int_{0}^{t}ds\int
dx\sum_{\mu=1}^{d}{\partial_{\mu}(\rho\/u^{\mu})}(x,t)f(x),$ (3.5)

for any test function $f$ and for any $t$. Hence

$\rho_{0}(x,t)-\rho(x,0)=-\int_{0}^{t}\text{div}\/(\rho u),$ (3.6)

or in differential form

$\partial_{t}\rho=-\text{div}\/(\rho u).$

###  3.2. Pressure

We examine now the second conservation law (2.9). By averaging as before, for
$\beta=1,\dots,d$, we get, in the limit $\varepsilon\rightarrow 0$,

$\displaystyle\begin{split}&\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})v_{i}^{\beta}\Big{\rangle}_{F_{\varepsilon}(t)}-\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})v_{i}^{\beta}\Big{\rangle}_{F_{\varepsilon}(0)}\\\
&=\varepsilon^{-1}\int_{0}^{t}ds\Big{\langle}\varepsilon^{d}\int
dx\sum_{k=1}^{d}{\partial_{k}f}(x)w^{\beta
k}(x)\Big{\rangle}_{F_{\varepsilon}}+O(\varepsilon).\end{split}$ (3.7)

Using the assumptions on $F_{\varepsilon}(t)$ we see that the l.h.s. is of
order $\varepsilon$, since the term of order $1$ vanishes because
$\lambda_{0}^{\mu},\mu=1,2,3$ are zero.

We write the current $w^{\beta k}$ using the decomposition (2.20) as $w^{\beta
k}_{*}+\varepsilon{\partial_{\gamma}}\bar{\Phi}_{0}^{k\beta\gamma}(x)+O(\varepsilon^{2})$,
with $w_{*}^{\beta k}$ given by (2.21). Using the assumptions (2.32) and
(2.37) we have:

$\varepsilon^{-1}\Big{\langle}w_{*}^{\beta
k}\Big{\rangle}_{F_{\varepsilon}}=\varepsilon^{-1}\Big{\langle}w_{*}^{\beta
k}\Big{\rangle}_{G_{\varepsilon}}+\Big{\langle}w_{*}^{\alpha\beta}R_{\varepsilon}\Big{\rangle},$
(3.8)

so that

$\varepsilon^{-1}\Big{\langle}w^{\beta
k}\Big{\rangle}_{F_{\varepsilon}}=\varepsilon^{-1}\Big{\langle}w^{\beta
k}_{*}\Big{\rangle}_{G_{\varepsilon}}+\Big{\langle}\sum_{\gamma=1}^{d}{\partial_{\gamma}}\bar{\Phi}_{0}^{k\beta\gamma}(x)\Big{\rangle}_{G_{0}}+\Big{\langle}w_{*}^{\alpha\beta}R_{\varepsilon}\Big{\rangle}_{G_{0}}+O(\varepsilon).$
(3.9)

The second term on the r.h.s. is zero since the term
$\bar{\Phi}_{0}^{k\beta\gamma}(x)={1\over
2}\varepsilon^{d}\sum_{i,j=1}^{N}\delta(x_{i}-x)\Phi_{0}^{\beta\gamma\nu}(\varepsilon^{-1}|x_{i}-x_{j}|)$
is antisymmetric under the exchange $i\rightarrow j$ while $G_{0}$ is
symmetric.

We introduce the currents $\tilde{w}^{\beta k}$ as given by $w^{\beta k}_{*}$
with the velocities $v_{i}$ replaced by $\tilde{v}_{i}=v_{i}-\varepsilon
u(x_{i})$. Then

$w_{*}^{\beta k}(x)={\tilde{w}}^{\beta
k}(x)+\varepsilon^{2}u^{k}(x)u^{\beta}(x)+\varepsilon
u^{k}(x)\tilde{v}^{\beta}+\varepsilon u^{\beta}(x)\tilde{v}^{k}.$ (3.10)

For the symmetry of the measure $G_{\varepsilon}$ we have
$\Big{\langle}\tilde{w}^{\beta
k}(x)\Big{\rangle}_{G_{\varepsilon}}=O(\varepsilon^{2})$, if $k\neq\beta$. The
average of $\tilde{w}^{\beta\beta}$, $\beta=1,\dots,d$, with respect the local
Gibbs state $G_{\varepsilon}$ is, by the virial theorem, the thermodynamic
pressure $P_{\varepsilon}$ in the state $G_{\varepsilon}$ [26].

Now we consider the term
$\Big{\langle}w_{*}^{\alpha\beta}R_{\varepsilon}\Big{\rangle}_{G_{0}}$ and use
(2.39), (2.43) and the ‘identity’
${\mathscr{L}^{*}}^{-1}{\mathscr{L}^{*}}R_{1}^{a}=R_{1}^{a}$. For sake of
shortness we omit the time dependence in the chemical potentials. We obtain

$\Big{\langle}w_{*}^{\alpha\beta}R_{\varepsilon}\Big{\rangle}_{G_{0}}=\Big{\langle}\tilde{w}_{*}^{\alpha\beta}R_{1}^{a}\Big{\rangle}_{G_{0}}+O(\varepsilon)=\Big{\langle}\sum_{i=1}^{N}\sum_{\gamma=1}^{d}[{\partial_{\gamma}\lambda_{0}^{0}}(x_{i})w^{0\gamma}_{i}+{\partial_{\gamma}\lambda_{0}^{d+1}}(x_{i})w^{{d+1}\gamma}_{i}]{\mathscr{L}}^{-1}\tilde{w}^{\alpha\beta}\Big{\rangle}_{G_{0}}+O(\varepsilon).$
(3.11)

We have

$\int
dx{\partial_{k}f}(x)\Big{\langle}w_{*}^{\alpha\beta}R_{\varepsilon}\Big{\rangle}_{G_{0}}=\int
dx{\partial_{k}f}(x)\Big{\langle}{\mathscr{L}}^{-1}w_{*}^{\alpha\beta}\sum_{i=1}^{N}\sum_{\gamma=1}^{d}[{\partial_{\gamma}\lambda_{0}^{0}}(x_{i})w^{0\gamma}_{i}+{\partial_{\gamma}\lambda_{0}^{d+1}}(x_{i})w^{{d+1}\gamma}_{i}]\Big{\rangle}_{G_{0}}+O(\varepsilon)$
$=\varepsilon^{-d}\int dx{\partial_{k}f}(x)\int
dy\sum_{\nu=0,d+1}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\nu}\
}(y)\Big{\langle}{\mathscr{L}}^{-1}w_{*}^{\alpha\beta}(x)w^{\nu\gamma}(y)\Big{\rangle}_{G_{0}}+O(\varepsilon).$

The symmetries of the microscopic current-current correlations imply (see
[28]) that the cross correlations between $\mu=0,d+1$ and $\beta=1,2,3$
vanish.

Summarizing, eq. (3.7) implies for
$P^{\varepsilon}=\Big{\langle}w_{*}^{\beta\beta}\Big{\rangle}_{G_{\varepsilon}}$

$\varepsilon^{-1}\int dx\nabla f(x)P^{\varepsilon}(x,t)=O(\varepsilon).$
(3.12)

Since $P^{\varepsilon}$ is a function of the thermodynamic parameters
$\lambda_{\varepsilon}$, we can expand it in series of $\varepsilon$ as
$\sum_{k}\varepsilon^{k}P_{k}$, where $P_{k}={{d^{k}P^{\varepsilon}\over
d\varepsilon^{k}}\big{|}_{\varepsilon=0}}$ We have that $P_{0}$ is a function
of the $\lambda_{0}^{0}$ and $\lambda_{0}^{d+1}$, while
$P_{1}=\sum_{\mu=0}^{d+1}{\partial
P^{\varepsilon}\over\partial\lambda_{\varepsilon}^{\mu}}\big{|}_{\varepsilon=0}\lambda_{1}^{\mu}$.
In order to fulfill (3.12) for any test function $f$,

$\nabla P_{0}=0,\quad\nabla P_{1}=0.$

In particular, the pressure $P_{0}(\rho,T)$ is a function of $\rho,T$
quantities conjugate to $\lambda_{0}^{0}$ and $\lambda_{0}^{d+1}$
respectively, and has to be constant versus $x$. In the following, we will
denote $P_{0}$ simply as $P$.

###  3.3. Momentum equation

To determine the equation for $u^{\mu}(x,t)$, which is of order $\varepsilon$,
we have to rescale the empirical velocity field. This means that we have to
look at the empirical field

$\bar{z}^{\alpha}(x)=\varepsilon^{-1}\varepsilon^{d}\sum_{i}v^{\alpha}_{i}\delta(x_{i}-x),\phantom{...}\alpha=1,\dots,d.$
(3.13)

We proceed as we did before to obtain (3.7), but we have to look at the
explicit form of the terms $O(\varepsilon)$ and $O(\varepsilon^{2})$ because
they have to be divided by $\varepsilon^{2}$. We have:

$\displaystyle\begin{split}&\Big{\langle}\varepsilon^{d-1}\sum_{i=1}^{N}f(x_{i})v_{i}^{\beta}\Big{\rangle}_{F_{\varepsilon}(t)}-\Big{\langle}\varepsilon^{d-1}\sum_{i=1}^{N}f(x_{i})v_{i}^{\beta}\Big{\rangle}_{F_{\varepsilon}(0)}=\\\
&\varepsilon^{-2}\int_{0}^{t}ds\Big{\langle}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})\\{v_{i}^{k}v_{i}^{\beta}+{1\over
2}\sum_{j\neq i}\Psi^{\beta
k}(\varepsilon^{-1}(x_{i}-x_{j}))\\}\Big{\rangle}_{F_{\varepsilon}(s)}+\\\
&+\varepsilon^{-1}\int_{0}^{t}ds\Big{\langle}{1\over
4}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}f}(x_{i})\Phi_{0}^{\nu\beta\gamma}(\varepsilon^{-1}(x_{i}-x_{j}))\Big{\rangle}_{F_{\varepsilon}(s)}\\\
&+\int_{0}^{t}ds\Big{\langle}{1\over
12}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu,\alpha=1}^{d}{\partial_{\gamma\nu\alpha}^{3}f}(x_{i})\Phi^{\alpha\beta\gamma\nu}(\varepsilon^{-1}(x_{i}-x_{j}))\Big{\rangle}_{F_{\varepsilon}(s)}\\\
\end{split}$ (3.14)

The l.h.s. of (LABEL:(3.10)) converges to

$\int dxf(x)\rho[u(x,t)-u(x,0)],$ (3.15)

as $\varepsilon\rightarrow 0$.

To get the equation for the velocity field we have to compute the non-
equilibrium average of the velocity current tensor $w^{\beta k}$ but now there
is a factor $\varepsilon^{-2}$ in front of it. Therefore we see that in this
case also the terms of order $\varepsilon^{2}$ in (2.32) have to be taken into
account.

We start by discussing the term containing $D_{0}$ namely the second line in
the r.h.s. of (LABEL:(3.10)). The lowest order term in $\varepsilon$ is

$\varepsilon^{-1}\int_{0}^{t}ds\Big{\langle}{1\over
2}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}f}(x_{i})\Phi_{0}^{\beta\gamma\nu}(\varepsilon^{-1}(x_{i}-x_{j}))[G_{0}+\varepsilon
G_{0}g_{1}+\varepsilon G_{0}R_{1}]\Big{\rangle}.$

The diverging term

$\varepsilon^{-1}\int_{0}^{t}ds\Big{\langle}{1\over
2}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma,\nu}^{2}f}(x_{i})\Phi_{0}^{\beta\gamma\nu}(\varepsilon^{-1}(x_{i}-x_{j}))\Big{\rangle}_{G_{0}}.$

is zero by symmetry property of the potential. In fact,
$\Phi_{0}^{\beta\gamma\nu}$ is antysimmetric in the exchange
$\xi_{i}\rightarrow\xi_{j}$ while $G_{0}$ is symmetric.

The term of order zero in $\varepsilon$ is

$\int_{0}^{t}ds\Big{\langle}{1\over
2}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}f}(x)\Phi_{0}^{\beta\gamma\nu}(\varepsilon^{-1}(x_{i}-x_{j}))[G_{0}g_{1}+G_{0}R_{1}^{a}]\Big{\rangle}.$

For the same reasons the part involving $g_{1}$ is zero because all the
conserved quantities are symmetric. We are left with

$\int_{0}^{t}ds\Big{\langle}{1\over
4}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}f}(x_{i})\Phi_{0}^{\beta\gamma\nu}(\varepsilon^{-1}(x_{i}-x_{j}))R_{1}^{a}\Big{\rangle}_{G_{0}}.$

We can safely change $\Phi_{0}^{\beta\gamma\nu}$ in
$\tilde{\Phi}_{0}^{\beta\gamma\nu}:=\Phi_{0}^{\beta\gamma\nu}-{\mathscr{P}}\Phi_{0}^{\beta\gamma\nu}$
since ${\mathscr{P}}R_{1}$ is zero.

By (2.36) we can use the ‘identity’
${(\mathscr{L}^{*}})^{-1}{\mathscr{L}^{*}}R_{1}^{a}=R_{1}^{a}$ and the
expression of $R_{1}^{a}$ to get

$\int_{0}^{t}ds\Big{\langle}{1\over
4}\varepsilon^{d}\sum_{i,j=1}^{N}\sum_{\gamma,\nu=1}^{d}{\partial_{\gamma\nu}^{2}f}(x_{i}){\mathscr{L}}^{-1}\tilde{\Phi}_{0}^{\beta\gamma\nu}(\varepsilon^{-1}(x_{i}-x_{j}))\sum_{k=1}^{d}[{\partial_{k}\lambda_{0}^{0}}(x,s)w^{0k}_{i}+{\partial_{k}\lambda_{0}^{d+1}}(x,s)w^{{d+1}k}_{i}]\Big{\rangle}_{G_{0}}.$

This term is zero by oddness in the velocity.

We observe that the term containing $D$, third line in (LABEL:(3.10)) is not
$0$ because the lowest order is given by an average with respect to $G_{0}$
(Gibbs measure with non constant parameters). The other terms are of higher
order, hence do not contribute.

The explicit expression is

$\displaystyle\begin{split}&{1\over 6}\int
dx\sum_{\alpha,\gamma,\nu=1}^{d}{\partial_{\alpha\gamma\nu}^{3}f}(x)\Big{\langle}\bar{\Phi}_{\beta\alpha\gamma\nu}(x)\Big{\rangle}_{F_{\varepsilon}}=\\\
&\Big{\langle}{1\over 12}\varepsilon^{d}\sum_{i\neq
j=1}^{N}\sum_{\alpha,\gamma,\nu=1}^{d}{\partial_{\alpha\gamma\nu}^{3}f}(\varepsilon\xi_{i})\partial_{\beta}V(|\xi_{i}-\xi_{j}|)[\xi_{i}^{\gamma}-\xi_{j}^{\gamma}][\xi_{i}^{\nu}-\xi_{j}^{\nu}][\xi_{i}^{\alpha}-\xi_{j}^{\alpha}]\Big{\rangle}_{G_{0}}+o(\varepsilon).\end{split}$
(3.16)

By symmetry, only the terms of the form
$\Big{\langle}\bar{\Phi}_{\alpha\alpha\beta\beta}\Big{\rangle}_{G_{0}}$ are
different from zero. Define
$\Phi_{\alpha\beta}:=\bar{\Phi}_{\alpha\alpha\beta\beta}$ and
$\hat{\Phi}_{\alpha\beta}:=\Big{\langle}\Phi_{\alpha\beta}\Big{\rangle}_{G_{0}}$.

$\hat{\Phi}_{\alpha\beta}$ is a function of $\rho,T$ and using the condition
on $\nabla P_{0}=0$ it can be seen as a function of $T$ only, so only the
gradient of $T$ will appear in the final term. By two integrations by parts we
move two $x$-derivatives in the main term in (3.16) on
$\hat{\Phi}_{\alpha\beta}$ and (3.16) becomes

${1\over 6}\int
dx\sum_{\alpha=1}^{d}{\partial_{\alpha}f}{\partial_{\alpha\beta}^{2}}\hat{\Phi}_{\alpha\beta}(\rho,T).$

We have

${\partial_{\alpha}}\hat{\Phi}_{\alpha\beta}={\partial\hat{\Phi}_{\alpha\beta}\over\partial
T}{\partial_{\alpha}}T:=\hat{\Phi}^{\prime}_{\alpha\beta}{\partial_{\alpha}}T,$

and

$\partial^{2}_{\alpha\beta}\hat{\Phi}_{\alpha\beta}=\partial_{\beta}(\hat{\Phi}^{\prime}_{\alpha\beta}{\partial_{\alpha}}T)={\partial_{\alpha}}T\partial_{\beta}(\hat{\Phi}^{\prime}_{\alpha\beta})+\hat{\Phi}^{\prime}_{\alpha\beta}\partial^{2}_{\alpha\beta}T$
$={\partial_{\alpha}}T{\partial\hat{\Phi}^{\prime}_{\alpha\beta}\over\partial
T}\partial_{\beta}T+\hat{\Phi}^{\prime}_{\alpha\beta}\partial^{2}_{\alpha\beta}T=\hat{\Phi}^{\prime\prime}_{\alpha\beta}{\partial_{\alpha}}T\partial_{\beta}T+\hat{\Phi}^{\prime}_{\alpha\beta}\partial^{2}_{\alpha\beta}T.$

Therefore

$\displaystyle\begin{split}&{1\over 6}\int
dx\sum_{\alpha=1}^{d}{\partial_{\alpha}f}{\partial_{\alpha\beta}^{2}}\hat{\Phi}_{\alpha\beta}(\rho,T)=\\\
&\sum_{\alpha=1}^{d}\Big{[}\int
dx\partial_{\beta}f[Y_{1}({\partial_{\alpha}}T\partial_{\beta}T-\delta_{\alpha\beta}{1\over
d}\sum_{\gamma=1}^{d}({\partial_{\gamma}}T)^{2})+{\bar{\omega}_{1}}\delta_{\alpha\beta}\sum_{\gamma=1}^{d}({\partial_{\gamma}}T)^{2}\\\
&+Y_{2}(\partial^{2}_{\alpha\beta}T-\delta_{\alpha\beta}{1\over
d}\sum_{\gamma=1}^{d}\partial^{2}_{\gamma\gamma}T)+{\bar{\omega}_{2}}\delta_{\alpha\beta}\sum_{\gamma=1}^{d}\partial^{2}_{\alpha\gamma}T)]\Big{]},\end{split}$
(3.17)

where

$Y_{1}={1\over
6}\hat{\Phi}_{\alpha\beta}^{\prime\prime},\quad\alpha\neq\beta;\quad
Y_{2}={1\over
6}\hat{\Phi}^{\prime}_{\alpha\beta}\quad\alpha\neq\beta;\quad\bar{\omega}_{1}={1\over
6d}\sum_{\gamma=1}^{d}\hat{\Phi}_{\gamma\gamma}^{\prime\prime},\quad\bar{\omega}_{2}={1\over
6d}\sum_{\gamma=1}^{d}\hat{\Phi}^{\prime}_{\gamma\gamma}.$ (3.18)

Finally, we turn to first term in the rhs of (LABEL:(3.10)) and compute the
contributions for the currents $w_{*}$ and use the definition (3.10) of
$\tilde{w}$:

$\displaystyle\begin{split}&\varepsilon^{-2}\Big{\langle}w_{*}^{\beta
k}(x)\Big{\rangle}_{G_{\varepsilon}}+\Big{\langle}w_{*}^{\beta
k}(x)R_{\varepsilon}\Big{\rangle}_{G_{0}}=\varepsilon^{-2}\Big{\langle}\tilde{w}^{\beta
k}\Big{\rangle}_{G_{\varepsilon}}+\rho
u^{\beta}u^{k}+\varepsilon^{-1}\Big{\langle}w_{*}^{\beta
k}R_{\varepsilon}\Big{\rangle}_{G_{0}}+O(\varepsilon)=\\\
&\varepsilon^{-2}P_{0}+\varepsilon^{-1}P_{1}+P_{2}+\rho
u^{\beta}u^{k}+\varepsilon^{-1}\Big{\langle}{G_{0}R_{1}}w_{*}^{\beta
k}(x)\Big{\rangle}+\Big{\langle}{G_{0}R_{2}}w_{*}^{\beta
k}(x)\Big{\rangle}+O(\varepsilon).\end{split}$ (3.19)

The first two terms in the second line of (3.19) do not contribute because
$P_{0}$ and $P_{1}$ are constant. The fourth term in the rhs of (3.19) gives
the non linear transport term, while $P_{2}$ represents the second order
correction to the thermodynamic pressure $P_{\varepsilon}$ and gives rise to
the unknown pressure $\mathfrak{p}$ appearing in equation (1.5).

Now we pass to examine the terms involving $R_{i}$. To compute these terms,
let us first introduce $\bar{w}^{\beta k}=w_{*}^{\beta
k}-\mathscr{P}w_{*}^{\beta k}$ and notice that (2.36) implies

$\Big{\langle}{R_{1}}\mathscr{P}w^{\alpha\beta}(x)\Big{\rangle}_{G_{0}}=0,\quad\Big{\langle}{R_{2}}\mathscr{P}w^{\alpha\beta}(x)\Big{\rangle}_{G_{0}}=0.$
(3.20)

By (2.36) we can use the ‘identity’
${(\mathscr{L}^{*}})^{-1}{\mathscr{L}^{*}}G_{0}R_{i}=G_{0}R_{i},\quad i=1,2$
and (2.46). First of all, let’s consider the diverging term due to $R_{1}$ and
start from the contribution due to $R_{1}^{a}$:

$\varepsilon^{-1}\Big{\langle}\sum_{j=1}^{N}[{\partial_{\gamma}\lambda_{0}^{0}}(x)w^{0\gamma}_{j}+{\partial_{\gamma}\lambda_{0}^{4}}(x,s)w^{4\gamma}_{j}]\varepsilon^{d}\sum_{i=1}^{N}{\partial_{k}f}(x_{i}){\mathscr{L}}^{-1}\bar{w}^{\beta
k}_{i}\Big{\rangle}_{G_{0}}.$

The symmetries of the microscopic current-current correlations imply (see
[28]) that the cross correlations between $\mu=0,d+1$ and $\beta=1,2,3$
vanish. Hence this term is zero.

On the other hand the term involving $\varepsilon^{-1}R_{1}^{s}$,

$\varepsilon^{-1}\Big{\langle}R_{1}^{s}\varepsilon^{d}\sum_{i=1}^{N}\bar{w}^{\beta
k}_{i}{\partial_{k}f(x_{i})}\Big{\rangle}_{G_{0}},$ (3.21)

by our assumption is of order $1$ and has to be computed.

This term gives in the hydrodynamic equations a term similar to the one due to
$D$ and will add new terms to the transport coefficients $Y_{1}$ and $Y_{2}$.
The explicit expression is computed in the Appendix A.1. The total transport
coefficients in the equation will be denoted by $K_{1}$ and $K_{2}$.

We discuss now the last term in the r.h.s of (3.19) involving $R_{2}$,

$\displaystyle\begin{split}&\int
dx\sum_{k=1}^{d}{\partial_{k}f}\Big{\langle}\bar{w}^{\beta
k}(x)(\mathscr{L}^{*})^{-1}{\mathscr{L}^{*}}G_{0}R_{2}\Big{\rangle}=\Big{\langle}{\mathscr{L}^{*}}G_{0}R_{2}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}\\\
=&-\varepsilon^{-1}\Big{\langle}\Big{[}G_{0}{\mathscr{L}^{*}}g_{1}+g_{1}{\mathscr{L}^{*}}G_{0}-\varepsilon\partial_{t}G_{0}\Big{]}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}.\end{split}$ (3.22)

The term $\partial_{t}G_{0}{\mathscr{L}}^{-1}\dots$ in the r.h.s. does not
contribute because $\mathscr{L}^{-1}$ is orthogonal to $\partial_{t}G_{0}$.

The second term is

$\displaystyle\begin{split}\Big{\langle}g_{1}&\sum_{j=1}^{N}\sum_{\gamma=1}^{d}[{\partial_{\gamma}\lambda_{0}^{0}}(x)w^{0\gamma}_{j}+{\partial_{\gamma}\lambda_{0}^{d+1}}(x)w^{{d+1}\gamma}_{j}]{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}\\\
=&\sum_{\ell=1}^{N}\sum_{\mu,\gamma=1}^{d}\sum_{j=1}^{N}{\partial_{\gamma}\lambda_{0}^{0}}(x)\Big{\langle}\lambda_{1}^{\mu}(x_{\ell})z_{\ell}^{\mu}w^{0\gamma}_{j}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}\\\
&+\sum_{\ell=1}^{N}\sum_{\mu,\gamma=1}^{d}\sum_{j=1}^{N}\lambda_{1}^{\mu}(x_{\ell}){\partial_{\gamma}\lambda_{0}^{d+1}}(x)\Big{\langle}z_{\ell}^{\mu}w^{{d+1}\gamma}_{j}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}.\end{split}$ (3.23)

In (3.26) will appear a term of a similar form. All the terms together
((3.23)+second term in (3.26)) give in the hydrodynamic equation a term of the
form

$\sum_{j,k,l=1}^{d}\partial_{j}[\alpha_{\beta jkl}(T)u_{k}\partial_{\ell}T],$

which is Galileian invariant for any temperature only if $\alpha_{\beta
jkl}=0$ (see [3] page 1071). So, $\alpha_{\beta jkl}$ has to be zero based on
this physical consideration. In the Appendix A.3 we will prove that this is
indeed the case.

We are left with the first term

$\displaystyle\begin{split}\Big{\langle}\sum_{j=1}^{N}\sum_{\mu=1}^{d}\sum_{l,k=1}^{d}{\partial_{l}\lambda_{1}^{\mu}}(x_{j})\bar{w}^{\mu
l}_{j}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}.\\\ \end{split}$ (3.24)

We remark that ${\mathscr{L}}^{-1}$ is “well defined” on $\bar{w}$ by the
assumptions discussed in section 2. The substitution of the current $w$ with
$\bar{w}$ is correct because the range of $\mathscr{L}^{-1}$ is orthogonal to
$\mathscr{P}w$.

Since $\lambda_{1}^{\mu}=\displaystyle{{u^{\mu}\over T}}$ and $T$ are not
costant, we have two different contribution to (3.24)

$\displaystyle\begin{split}\sum_{j=1}^{N}\sum_{\mu=1}^{d}\sum_{l,k=1}^{d}\Big{\langle}{\partial_{l}\lambda_{1}^{\mu}}(x_{j})\bar{w}^{\mu
l}_{j}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}\\\ \end{split}$ (3.25)
$=\sum_{j=1}^{N}\sum_{\mu=1}^{d}\sum_{l,k=1}^{d}\Big{\langle}[{1\over
T}{\partial_{l}u^{\mu}}(x_{j})-u^{\mu}{1\over
T^{2}}{\partial_{l}T}(x_{j})]\bar{w}^{\mu
l}_{j}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}.$ (3.26)

The second term will combine with the terms in (3.23) to eliminate the term in
the equation which is not Galileian invariant (see Appendix A.3).

We discuss here the first term in (3.26). To find the expression of the
transport coefficients we consider

$\displaystyle\begin{split}\Big{\langle}\sum_{j=1}^{2}\sum_{\mu,l=1}^{d}&{1\over
T}{\partial_{l}u^{\mu}}(x_{j})w^{\mu
l}_{j}{\mathscr{L}}^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}_{G_{0}}=\\\
&\sum_{\mu,k,l=1}^{d}\int dy{1\over
T}{\partial_{l}u^{\mu}}(y)\int\varepsilon^{-d}dz{\partial_{k}f}(z)\Big{\langle}\bar{w}^{\mu
l}(y){\mathscr{L}}^{-1}\bar{w}^{\beta k}(z)\Big{\rangle}_{G_{0}}.\end{split}$

Since the Gibbsian state $G_{0}$ is invariant under translations on
$\mathbb{R}^{d}$ we have

$\text{r.h.s. of \eqref{(3.16)}}=\sum_{\mu,k,l=1}^{d}\int dy{1\over
T}{\partial_{l}u^{\mu}}(y)\int
d\xi{\partial_{k}f}(y+\varepsilon\xi)\Big{\langle}\bar{w}^{\mu
l}(0){\mathscr{L}}^{-1}\bar{w}^{\beta k}(\xi)\Big{\rangle}_{G_{0}},$ (3.27)

where we have changed the variable $z$ in $\xi=\varepsilon^{-1}(z-y)$
absorbing the factor $\varepsilon^{-d}$. Hence, provided that
$\langle\bar{w}^{\mu l}(0){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(\xi)\rangle_{G_{0}}$ decay fast enough for large $\xi$, up to
$O(\varepsilon)$ we have

$\text{r.h.s. of \eqref{(3.16)}})=\sum_{\mu,k,l=1}^{d}\int dy{1\over
T}{\partial_{l}u^{\mu}}(y){\partial_{k}f}(y)\int d\xi\Big{\langle}\bar{w}^{\mu
l}(0){\mathscr{L}}^{-1}\bar{w}^{\beta k}(\xi)\Big{\rangle}_{G_{0}}.$ (3.28)

To conclude the argument we transform ${\mathscr{L}^{-1}}$ in a time-integral
of $\exp[t\mathscr{L}]$:

$-\Big{\langle}\bar{w}^{\mu l}(0){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(\xi)\Big{\rangle}_{G_{0}}=\int_{0}^{\infty}d\tau\Big{\langle}\bar{w}^{\mu
l}(\xi,\tau)\bar{w}^{\beta k}(0,0)\Big{\rangle}_{G_{0}}.$ (3.29)

In this formula with an abuse of notation we are using the symbol
$\bar{w}^{\mu}(\xi)$ for the microscopic current. We notice that before the
change of variable to microscopic variables $\xi$ the meaning of the current
$\bar{w}^{\mu}$ was indeed
$\bar{w}^{\mu}_{\varepsilon}(x)=\bar{w}^{\mu}(\varepsilon^{-1}x)$.

We have that (see [28])

$\int d\xi\Big{\langle}\bar{w}^{\mu l}(\xi,\tau)\bar{w}^{\beta
k}(0,0)\Big{\rangle}_{G_{0}}=c(\tau)[\delta_{kl}\delta_{\beta
k}+\delta_{k\mu}\delta_{\beta l}]+c^{\prime}(\tau)\delta_{\beta
k}\delta_{l\mu}.$ (3.30)

Therefore the time integral of (3.30) has only two independent coefficients

$\int_{0}^{\infty}d\tau c(\tau)=2\eta T;\phantom{...}\int_{0}^{\infty}d\tau
c^{\prime}(\tau)=2T(\zeta-{2\over d}\eta),$ (3.31)

where $\eta$ and $\zeta$ are the shear viscosity and the bulk viscosity
respectively.

They are finite if the correlations decay sufficiently fast to make the time
integrals in (3.31) convergent. Notice that the subtraction of
$\mathscr{P}\tilde{w}^{\alpha\beta}$ has been crucial, because the self-
correlation of the slow part of the current does not decay in time.

Since $\text{div}\ u\neq 0$, the term proportional to the bulk viscosity does
appear in the limiting equation. The computation gives also the Green-Kubo
formula for the bulk viscosity $\zeta$

$\zeta={1\over 2d^{2}T}\int_{0}^{\infty}d\tau\int
d\xi\Big{[}\Big{\langle}\sum_{\alpha}\bar{w}^{\alpha\alpha}(\xi,\tau)\sum_{\gamma}\bar{w}^{\gamma\gamma}(0,0)\Big{\rangle}_{G_{0}}-\Big{\langle}\sum_{\alpha}\bar{w}^{\alpha\alpha}\Big{\rangle}_{G_{0}}\Big{\langle}\sum_{\gamma}\bar{w}^{\gamma\gamma}\Big{\rangle}_{G_{0}}\Big{]}.$
(3.32)

The usual expression given in [28] is recovered using the explicit form of the
projector $\mathscr{P}$. The viscosity is given by

$\eta={1\over 2T}\int_{0}^{\infty}d\tau\int
d\xi\Big{\langle}\bar{w}^{12}(\xi,\tau)\bar{w}^{12}(0,0).\Big{\rangle}$ (3.33)

Putting all the terms together we have the following equation to the lowest
order in $\varepsilon$:

$\displaystyle\begin{split}\int dxf(x)\rho&[u^{\beta}(x,t)-u^{\beta}(x,0)]=\\\
&\int_{0}^{t}ds\Big{[}\int dy\sum_{k=1}^{d}{\partial_{k}f}(y)\\{\rho
u^{\beta}(y,s)u^{k}(y,s)-\eta[{\partial_{k}u^{\beta}}]^{Tr}(y,s)-p-\zeta\sum_{\gamma=1}^{d}{\partial_{\gamma}u^{\gamma}}(y,s)\\}\\\
+&\int
dx\sum_{\alpha=1}^{d}{\partial_{\alpha\beta\alpha}^{3}f}(x)\hat{\Phi}_{\alpha\beta}(x,s)+\int
dx\sum_{k=1}^{d}{\partial_{k}f}(x)\varepsilon^{-1}\Big{\langle}R_{1}^{s}\bar{w}^{\beta
k}\Big{\rangle}_{G_{0}}\Big{]},\end{split}$ (3.34)

for any test function $f$, where $[A]^{Tr}$ means zero trace of $A$. The
differential form is then

$\partial_{t}(\rho u)+\nabla\cdot(u\otimes u)+\nabla
p=\nabla\cdot(\tau^{(1)}-\tau^{(2)})$
$\tau^{(1)}_{\alpha\beta}:=\eta\left(\partial_{\alpha}u^{\beta}+\partial_{\beta}u^{\alpha}-{2\over
d}\delta_{\alpha\beta}\partial_{\alpha}u^{\alpha}\right)+\zeta\delta_{\alpha\beta}\partial_{\alpha}u^{\alpha},$

$\tau^{(2)}_{\alpha\beta}:=[K_{1}({\partial_{\alpha}}T\partial_{\beta}T-{1\over
d}\sum_{\alpha=1}^{d}({\partial_{\alpha}}T)^{2})+{\omega_{1}}\sum_{\alpha=1}^{d}({\partial_{\alpha}}T)^{2}+K_{2}(\partial^{2}_{\alpha\beta}T-{1\over
d}\sum_{\alpha=1}^{d}\partial^{2}_{\alpha\alpha}T)+{\omega_{2}}\sum_{\alpha=1}^{d}\partial^{2}_{\alpha\alpha}T)],$
(3.35)

with

$K_{i}=Y_{i}+Z_{i},\quad\omega_{i}=\bar{\omega}_{i}+{\mathscr{\phi}}_{i},\quad
i=1,2,$

computed in the Appendix A.1.

###  3.4. Energy equation

Using the arguments developed before it is possible also to find the equation
for the energy.

We consider the conservation law for the energy (2.10).

Therefore we look for the equation for the quantity $z_{i}^{d+1}$. By (2.10)we
have

${d\over
dt}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})z_{i}^{d+1}=\varepsilon^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}{\partial_{k}f}(x_{i})w^{d+1\,k}_{i}+O(\varepsilon).$
(3.36)

We take the average of both sides w.r.t. $F_{\varepsilon}$. We have

$\big{\langle}\varepsilon^{d}\sum_{i=1}^{N}f(x_{i})z_{i}^{{d+1}}\big{\rangle}_{F_{\varepsilon}}=\int
dxf(x)\rho(x)e(x)+O(\varepsilon),$

and, by using the expression of
$F_{\varepsilon}=G_{\varepsilon}+G_{0}R_{1}+O(\varepsilon)$,

$\varepsilon^{-1}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\big{\langle}{\partial_{k}f}(x_{i})w^{d+1\,k}_{i}\big{\rangle}_{F_{\varepsilon}}=\int
dx{\partial_{k}f}(x)[\varepsilon^{-1}\big{\langle}w^{d+1\,k}(x)\big{\rangle}_{G_{\varepsilon}}+\big{\langle}G_{0}R_{1}w^{d+1\,k}(x)\big{\rangle}]+O(\varepsilon).$

We need to evaluate $\big{\langle}\varepsilon
w^{{d+1}\,k}(x)G_{0}R_{1}\big{\rangle}$ and
$\varepsilon^{-1}\big{\langle}w^{{d+1}\,k}(x)\big{\rangle}_{G_{\varepsilon}}$.

The first of them gives the diffusive correction. We introduce
$\tilde{w}^{{d+1}\,k}$ and $\tilde{z}^{d+1}$ defined by (2.6) with $v$
replaced by $\tilde{v}$ and

$w_{i}^{{d+1}\,k}=\tilde{w}_{i}^{{d+1}\,k}+\varepsilon\big{\\{}u^{k}(x_{i})z_{i}^{d+1}+\sum_{\gamma=1}^{d}\big{(}u^{\gamma}\tilde{v}_{i}^{\gamma}\tilde{v}_{i}^{k}+\sum_{j\neq
i}\Psi^{\gamma k}(\varepsilon^{-1}|x_{i}-x_{j}|){1\over
2}[u^{\gamma}(x_{i})+u^{\gamma}(x_{j})]\big{)}\big{\\}}.$ (3.37)

Then,

$\big{\langle}\tilde{w}^{{d+1}\,k}(x){\mathscr{L}^{*}}^{-1}{\mathscr{L}^{*}}G_{0}R_{1}\big{\rangle}=-\varepsilon^{-1}\big{\langle}{\mathscr{L}}^{-1}\bar{w}^{{d+1}\,k}(x)\mathscr{L}^{*}G_{0}\big{\rangle},$

and by (2.43) ( only $R_{1}^{a}$ enters since $w^{d+1}$ is odd),

$\Big{\langle}\bar{w}^{{d+1}\,k}G_{0}R_{1}^{a}\Big{\rangle}=-\Big{\langle}\sum_{j=1}^{N}\sum_{l=1}^{d}\sum_{\mu=0,4}{\partial_{l}\lambda_{0}^{\mu}}(x_{j})\bar{w}^{\mu
l}_{j}{\mathscr{L}}^{-1}\bar{w}^{{d+1}\,k}(x)\Big{\rangle}_{G_{0}}+O(\varepsilon),$
(3.38)

where
$\bar{w}^{{d+1}\,k}=\tilde{w}^{{d+1}\,k}-\mathscr{P}(\tilde{w}^{{d+1}\,k})$.
Because of time-reversal and rotation invariance of the Gibbs state the only
correlations different from zero are (see [28])

$\int
dx\Big{\langle}\bar{w}^{{d+1}\,k}(x,\tau)\bar{w}^{{d+1}\,l}(0,0)\Big{\rangle}_{G_{0}}=\delta_{lk}a(\tau),$
(3.39)

and $\int d\tau a(\tau)=2\kappa T^{2}$. Therefore the conductivity $\kappa$ is
given by

$\kappa={1\over 2dT^{2}}\int d\tau\Big{\\{}\int
d\xi\Big{\langle}\sum_{k,l=1}^{d}w^{{d+1}\,k}(x,\tau)w^{{d+1}\,l}(0,0)\Big{\rangle}-d(T(e+P)^{2}/\rho\Big{\\}}.$
(3.40)

We observe that, since $\lambda_{\varepsilon}=-(T_{\varepsilon})^{-1}$,
$\lambda_{0}^{d+1}$ is given by $-T^{-1}$.

Using the previous arguments one can see that the second term in (3.37) gives
no contribution to $\langle w^{{d+1}\,k}R_{\varepsilon}\rangle$ in the limit
$\varepsilon\rightarrow 0$.

The mean of the energy current on the Gibbs state $G_{\varepsilon}$, i.e.
$\langle w^{{d+1}\,k}\rangle_{G_{\varepsilon}}$, is nothing but
$(\rho_{\varepsilon}e_{\varepsilon}+P_{\varepsilon})u_{\varepsilon}$ (see
[28]), where $e_{\varepsilon}=\langle z^{d+1}\rangle_{G_{\varepsilon}}$.
Hence, $\varepsilon^{-1}\langle w^{{d+1}\,k}\rangle_{G_{\varepsilon}}=(\rho
e+P)u+o(\varepsilon)$ with $e$ the order zero term for the internal energy and
$P$ the order zero term for the pressure.

Summarizing, if we take the average of (3.36) with respect $F_{\varepsilon}$
we obtain

$\displaystyle\begin{split}&\int dxf(x)[\rho e(x,t)-\rho e(x,0)]=\\\
&\int_{0}^{t}ds\int dxf(x)\\{-\text{div}[(\rho e+P)u](x,s)+\nabla(k\nabla
T)(x,s)\\}.\end{split}$ (3.41)

In differential form

$\partial_{t}(\rho e)=-\text{div}[(\rho e+P)u]+\nabla(k\nabla T).$

By using the equation for $\rho$ we get also

$\rho[\partial_{t}e+u\cdot\nabla e]=-P\nabla\cdot u+\nabla(k\nabla T).$

###  3.5. Entropy

Now we show that these equations imply the growth in time of the total
thermodynamic entropy and hence the second law of the thermodynamics. Let
$s(\rho,e)$ denote the entropy as function of the density $\rho$ and the
internal energy $e$ and notice that ${\partial s\over\partial e}={1\over
T};\quad{\partial s\over\partial\rho}=-{P\over T\rho^{2}}$. We have

$\partial_{t}(\rho s)+\nabla\cdot(\rho
us)=\rho\partial_{t}s+s\partial_{t}\rho+\nabla\cdot(\rho us)={\partial
s\over\partial e}\rho\partial_{t}e+{\partial
s\over\partial\rho}\rho\partial_{t}\rho+s\partial_{t}\rho+\nabla\cdot(\rho
us)$ $={1\over T}\rho\partial_{t}e-{P\over
T\rho}\partial_{t}\rho-s\nabla\cdot(\rho u)+\nabla\cdot(\rho us).$

By using the equation for the energy

$\partial_{t}(\rho s)+\nabla\cdot(\rho us)=-{1\over T}\rho u\cdot\nabla
e-{P\over T}\nabla\cdot u+{1\over T}\nabla(k\nabla T)+{P\over
T\rho}\nabla\cdot(\rho u)-u\rho\cdot\nabla s$ $=\nabla({k\over T}\nabla
T)+{1\over T^{2}}|\nabla T|^{2}.$

We have used the identity

$\rho u\cdot[{1\over T}\nabla e+{P\over T\rho^{2}}\nabla\rho]=\rho
u\cdot\nabla s$

Hence, by integrating over $x$ on a torus

$\partial_{t}\int dx\rho s=\int dx{1\over T^{2}}|\nabla T|^{2}\geq 0$

##  4\. Comparison with the Boltzmann case

It is well known that the incompressible Navier-Stokes-Fourier equations for
the perfect gas can be derived from the Boltzmann equation under a suitable
diffusive scaling (scale space as $\varepsilon^{-1}$ and time as
$\varepsilon^{-2}$) and taking the Mach number proportional to $\varepsilon$
in the limit $\varepsilon\rightarrow 0$. Under the same scaling but taking
initial conditions with gradient of density and temperature of order $1$
and/or diffusive boundary condition with gradient of temperature of order $1$
the formal limiting equation are different (usually called ghost effect
equations following Sone)

$\displaystyle\left\\{\begin{array}[]{rcl}\nabla P&=&\rho T,\\\
\rule{0.0pt}{11.99998pt}\partial_{t}{u}+u\cdot\nabla
u+\nabla\mathfrak{p}&=&\nabla\cdot\left(\bar{\tau}^{(1)}-\bar{\tau}^{(2)}\right),\\\
\rule{0.0pt}{11.99998pt}{\partial_{t}}\rho+\nabla\cdot(\rho u)&=&0,\\\
\rule{0.0pt}{18.00005pt}{3\over 2}\partial_{t}P(t)+{5\over
2}P\big{(}\nabla\cdot u\big{)}&=&\nabla\cdot\left(\bar{\kappa}\dfrac{\nabla
T}{2T^{2}}\right),\end{array}\right.$ (4.5)

for $d=3$, where $\bar{\kappa}(T)>0$ is the heat conductivity,

$\tau^{(1)}_{ij}:=\lambda\left(\partial_{i}u_{j}+\partial_{j}u_{i}-{2\over
3}\delta_{ij}\partial_{i}u_{i}\right),$ $\tau^{(2)}_{ij}:={\lambda^{2}\over
P}\Big{(}\bar{K}_{1}[\big{(}\partial_{i}\partial_{j}T\big{)}-{1\over
3}\delta_{ij}\sum_{i}\partial^{2}_{i}T]+{\bar{K}_{2}\over
T}[\big{(}\partial_{i}T\big{)}\big{(}\partial_{j}T\big{)}-{1\over
3}\delta_{ij}\sum_{i}(\partial_{i}T)^{2}]\Big{)}$

for some smooth function $\lambda[T]>0$, the viscosity coefficient, and
positive constants $K_{1}$ and $K_{2}$.

To give the expressions of the transport coefficients define the quantities

$\displaystyle\mu(x,v):={\rho(x)\over\big{(}2\pi{T}(x)\big{)}^{3/2}}\exp\bigg{(}-{\left|v\right|^{2}\over
2{T}(x)}\bigg{)},$ (4.6)
$\displaystyle\bar{\mathscr{A}}:=v\cdot\left(\left|v\right|^{2}-5T\right)\sqrt{\mu}\in
R^{3},\quad{\mathscr{A}}:=\mathscr{L}^{-1}\left[\bar{\mathscr{A}}\right]\in
R^{3},$ (4.7) $\bar{\mathscr{B}}=\bigg{(}v\otimes v-{\left|v\right|^{2}\over
3}\mathbf{1}\bigg{)}\sqrt{\mu}\in\mathbb{R}^{3\times
3},\quad\mathscr{B}=L^{-1}\bar{\mathscr{B}}\in\mathbb{R}^{3\times 3},$

$\displaystyle\bar{k}I:=\int_{R^{3}}\left({\mathscr{A}}\otimes\bar{\mathscr{A}}\right)\mathrm{d}v,\quad\lambda:={1\over
T}\int_{R^{3}}\mathscr{B}_{ij}\bar{\mathscr{B}}_{ij}\ \ \text{for}\ \ i\neq
j.$ (4.9)

where $Lf{1\over\sqrt{\mu}}=\mathscr{L}(\sqrt{\mu}f)$ and $\mathscr{L}$ is the
linearized Boltzmann operator around the local Maxwellian $\mu$ with zero mean
for hard sphere,

${\lambda^{2}\over P}\bar{K}_{1}={1\over T^{2}}\int
dv\mathscr{B}_{ij}v_{i}{\mathscr{A}}_{j},\quad i\neq j,$

${\lambda^{2}\over TP}\bar{K}_{2}={1\over T^{4}}\int
dv\mathscr{B}_{ij}\Big{[}\Gamma({\mathscr{A}}_{i},{\mathscr{A}}_{j})+v_{i}\sqrt{\mu}^{-1}{\partial\over\partial
T}({\sqrt{\mu}\over T^{2}}{\mathscr{A}}_{j})\Big{]}\quad i\neq j$

where $\Gamma$ is the collision operator in the Boltzmann equation.

We want to compare the equations we get for the Hamiltonian particle system
and the ones derived from Boltzmann, taking into account that the state
equation in the second case is the one for the perfect gas $P(\rho,T)=\rho T$,
since the Boltzmann equation describes a rarefied gas. The continuity equation
in kinetic theory is the same as for particle system and also the condition
$\nabla P=0$.

The equation for the energy if we consider the state equation of a perfect
case becomes:

For a perfect gas $e={3\over 2}T$ and

${3\over 2}\partial_{t}P(t)=-{5\over 2}P\nabla\cdot u+\nabla(k\nabla T).$

If the domain is a torus, $\partial_{t}P(t)=0$ and we get

${5\over 2}P\nabla\cdot u=\nabla(k\nabla T).$

If we compare the equation for the momentum we see that also in this case the
structure of the equation is the same included the new thermal stress terms.

Obviously, the transport coefficients are different and we are not able to
compare them, since the Boltzmann equation is modelling a gas of hard sphere
and for the particles we were considering a smooth potential. However, we
notice that if we represent ${\mathscr{L}}^{-1}$ as

${\mathscr{L}}^{-1}f=\int_{0}^{\infty}dse^{-s{\mathscr{L}}}f(s)$

the transport coefficients $\lambda$ and $\bar{\kappa}$ are also in kinetic
theory expressed as time correlations of currents

$\lambda=-{1\over T}\int_{0}^{\infty}ds<(|v_{i}v_{j}-{1\over
3}|v|^{2}I)(s)(|v_{i}v_{j}-{1\over 3}|v|^{2}I)(0)>_{\mu},$ (4.10)
$\bar{\kappa}=-\int_{0}^{\infty}ds<{1\over 2}(|v|^{2}-5T)v_{i}(s){1\over
2}(|v|^{2}-5T)v_{i}(0)>_{\mu}.$ (4.11)

Notice that $\zeta$ in the kinetic theory is zero and we realise that also
$\omega_{1}$ and $\omega_{2}$ are not present in kinetic theory. The
expression of the transport coefficients as given by the Green-Kubo formulas
are similar with the difference that in kinetic theory is present the
linearized Boltzmann operator around the local Maxwellian equilibrium while
for the particles the same role is played by the Liouville operator. Moreover,
the averages are taken versus the local Maxwellian equilibrium with zero
velocity $\mu$ in kinetic theory and versus the local Gibbs equilibrium
$G_{0}$ for particles.

Moreover, in kinetic theory $\lambda$ and $\bar{\kappa}$ are well defined due
to the hypocoericvity property of $L$, while for the particle system we need
ergodic and mixing properties of the Liouville operator ${\mathscr{L}}$.

On the other hand, the new transport coefficients $K_{1}$ and $K_{2}$ have two
different contributions: one involving spatial correlations, which are not
present in kinetic theory, and one involving space and double time
correlations which are similar in this respect to the ones present in kinetic
theory.

To conclude , we remark that also in the Boltzmann case the entropy grows in
time as showed by Bobylev [3].

Finally, we want to stress that the argument for the particles is completely
formal and some rigourous results in the case of the incompressible Navier-
Stokes equations can be obtained for stochastic systems of particles on the
lattice [14],[15]. The model of stochastic termal gas on the lattice in [2]
could be useful for deriving the new equations.

In the case of the kinetic theory the ghost effect equations were obtained in
one-dimensional stationary cases, [1], [4], [12]. A recent advance has been
obtained in the stationary case in a general domain for a rarefied gas in
contact with a thermal reservoir with non-homogeneous temperature: the
rigourous proof of the hydrodynamic limit [11] [10].

## Appendix A

###  A.1.  Transport coefficients

We examine more closely the expression of the transport coefficients
$K_{1},K_{2}$ in (3.35) and first we compute $Y_{1},Y_{2}$ in (3.17) and then
the part $Z_{i},i=1,2$ due to the term (3.21).

$\bullet$ Computation of $Y_{i},i=1,2$.

$Y_{2}(x,t)={\partial\over\partial
T}\hat{\Phi}_{\alpha\beta}(x,t),\alpha\neq\beta$ is defined in (3.18). We have

$\hat{\Phi}_{\alpha\beta}:=\Big{\langle}\bar{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}=\Big{\langle}{G_{0}}(x,t)\bar{\Phi}_{\alpha\beta}(x)\Big{\rangle}$

where

$\bar{\Phi}_{\alpha\beta}(x)={1\over 2}\varepsilon^{d}\sum_{i,j=1,i\neq
j}^{N}\delta(x-x_{i})\partial_{\beta}V(\varepsilon^{-1}(x_{i}-x_{j}))[\xi_{i}^{\beta}-\xi_{j}^{\beta}][\xi_{i}^{\alpha}-\xi_{j}^{\alpha}]^{2}.$

We put ${\partial\over\partial
T}\lambda_{0}^{\mu}=(\lambda_{0}^{\mu})^{\prime}$ and
${\partial^{2}\over\partial
T^{2}}\lambda_{0}^{\mu}=(\lambda_{0}^{\mu})^{\prime\prime}$ for $\mu=0,d+1$.
Then,

$Y_{2}=\Big{\langle}{\partial G_{0}\over\partial
T}{}\bar{\phi}_{\alpha\beta}\Big{\rangle}=\Big{\langle}\sum_{k=1}^{N}z^{0}_{k}(\lambda_{0}^{0})^{\prime}(x_{k})\bar{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}-\Big{\langle}\sum_{k=1}^{N}z^{0}_{k}(\lambda_{0}^{0})^{\prime}(x_{k})\Big{\rangle}_{G_{0}}\Big{\langle}\bar{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}$
$+\Big{\langle}\sum_{k=1}^{N}z^{d+1}_{k}(\lambda_{0}^{d+1})^{\prime}(x_{k})\bar{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}-\Big{\langle}\sum_{k=1}^{N}z^{d+1}_{k}(\lambda_{0}^{d+1})^{\prime}(x_{k})\Big{\rangle}_{G_{0}}\Big{\langle}\bar{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}.$

Now call $(\tilde{z}^{\mu})=z^{\mu}-\Big{\langle}z^{\mu}\Big{\rangle}_{G_{0}}$
and
$\tilde{\Phi}_{\alpha\beta}=\bar{\Phi}_{\alpha\beta}-\Big{\langle}\bar{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}$.
Then,

$Y_{2}=\Big{\langle}\sum_{k=1}^{N}\tilde{z}^{0}_{k}(\lambda_{0}^{0})^{\prime}(x_{k})\tilde{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}+\Big{\langle}\sum_{k=1}^{N}\tilde{z}^{d+1}_{k}(\lambda_{0}^{d+1})^{\prime}(x_{k})\tilde{\Phi}_{\alpha\beta}\Big{\rangle}_{G_{0}}.$

Let $h(x)$ be a test function and compute $\int dxh(x)Y_{2}(x)$.

$\int dxh(x)Y_{2}(x)=\varepsilon^{-d}\sum_{\mu=0,4}\int dxh(x)\int
dy(\lambda_{0}^{\mu})^{\prime}(y)\Big{\langle}\tilde{z}^{\mu}(y)\tilde{\Phi}_{\alpha\beta}(x)\Big{\rangle}_{G_{0}}.$

We change variable $x=y+\varepsilon\zeta$ to absorb $\varepsilon^{-d}$ and get

$\int dxh(x)Y_{2}(x)=\sum_{\mu=0,4}\int
dyh(y)(\lambda_{0}^{\mu})^{\prime}(y)\int
d\zeta\Big{\langle}\tilde{z}^{\mu}(0)\tilde{\Phi}_{\alpha\beta}(\zeta)\Big{\rangle}_{G_{0}}.$

In conclusion

$Y_{2}(y)=\sum_{\mu=0,4}(\lambda_{0}^{\mu})^{\prime}(y)\int
d\zeta\Big{\langle}\tilde{z}^{\mu}(0)\tilde{\Phi}_{\alpha\beta}(\zeta)\Big{\rangle}_{G_{0}}.$

For $\alpha\neq\beta$

$Y_{1}={\partial^{2}\hat{\Phi}_{\alpha\beta}\over\partial T^{2}}={\partial
Y_{2}\over\partial T}=\sum_{\mu=0,4}(\lambda_{0}^{\mu})^{\prime\prime}(y)\int
d\zeta\Big{\langle}\tilde{z}^{\mu}(0)\tilde{\Phi}_{\alpha\beta}(\zeta)\Big{\rangle}_{G_{0}}$
$+\sum_{\mu,\nu=0,4}\int
dy(\lambda_{0}^{\mu})^{\prime}(y){(\lambda_{0}^{\nu})}^{\prime}(y)\int
d\zeta\int
d\zeta^{\prime}\Big{\langle}\tilde{z}^{\mu}(0)\tilde{z}^{\nu}(\zeta^{\prime})\tilde{\Phi}_{\alpha\beta}(\zeta)\Big{\rangle}_{G_{0}}.$

$\bullet$ Computation of $Z_{i},i=1,2$.

Now we compute the contribution to the transport coefficients due to the term

$C:=\varepsilon^{-1}\Big{\langle}R_{1}^{s}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{x}f}(x_{i})\Big{\rangle}_{G_{0}}.$

We use the expression of $R_{1}^{s}$ given by (2.44)

$C=\varepsilon^{-1}\Big{\langle}G_{0}R_{1}^{s}\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})\Big{\rangle}$
$=\varepsilon^{-2}\Big{\langle}{\mathscr{L}^{*}}^{-1}{\mathscr{L}^{*}}^{-1}[{\mathscr{L}^{*}}{\mathscr{L}^{*}}G_{0}]\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}]{\partial_{k}f}(x_{i})\Big{\rangle}$
$=\varepsilon^{-1}\Big{\langle}{\mathscr{L}^{*}}[G_{0}\sum_{\ell=1}^{N}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{\ell})w^{\mu\gamma}_{\ell}]{\mathscr{L}}^{-1}{\mathscr{L}}^{-1}[\varepsilon^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}\bar{w}^{\beta
k}_{i}{\partial_{k}f}(x_{i})]\Big{\rangle}.$

We compute now

$A:=\varepsilon^{-1}[{\mathscr{L}^{*}}G_{0}]\sum_{\ell=1}^{N}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{\ell})w^{\mu\gamma}_{\ell}.$
$\displaystyle\begin{split}&A=[\varepsilon^{-1}{\mathscr{L}^{*}}G_{0}]\sum_{\ell=1}^{N}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{\ell})w^{\mu\gamma}(x_{\ell})+\varepsilon^{-1}G_{0}{\mathscr{L}^{*}}[\sum_{\ell=1}^{N}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{\ell})w^{\mu\gamma}(x_{\ell})]\\\
&=G_{0}[\sum_{\ell=1}^{N}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\partial_{\gamma}\lambda_{0}^{\mu}}_{\ell}w^{\mu\gamma}_{\ell}][\sum_{s=1}^{N}\sum_{\mu=0,4}\sum_{\nu=1}^{d}{\partial_{\nu}\lambda_{0}^{\mu}}(x_{s})w^{\mu\nu}_{s}]\\\
&+\varepsilon^{-1}G_{0}\sum_{\ell=1}^{N}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\mathscr{L}^{*}}[{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{\ell})w^{\mu\nu}_{\ell}].\end{split}$
(A. 1.1)

The first term in the rhs of (A. 1.1) is

$\varepsilon\Big{[}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}\varepsilon^{-d}\int
dx(\lambda_{0}^{\mu})^{\prime}\sum_{\gamma=1}^{d}\partial_{\gamma}T(x)w^{\mu\gamma}(x)\quad\varepsilon^{-d}\int
dy(\lambda_{0}^{\mu})^{\prime}\sum_{\nu=1}^{d}\partial_{\nu}T(y)w^{\mu\nu}(y)$
$+2\varepsilon^{-d}\int
dx(\lambda_{0}^{0})^{\prime}\sum_{\gamma=1}^{d}\partial_{\gamma}T(x)w^{0\gamma}(x)\quad\varepsilon^{-d}\int
dy(\lambda_{0}^{d+1})^{\prime}\sum_{\nu=1}^{d}\partial_{\nu}T(y)w^{d+1\nu}(y)\Big{]}.$

The contribution to $C$ is

$\sum_{\mu=0,4}\sum_{\gamma,\nu=1}^{d}\varepsilon^{-d}\int
dx(\lambda_{0}^{\mu})^{\prime}\partial_{\gamma}T(x)\varepsilon^{-d}\int
dy(\lambda_{0}^{\mu})^{\prime}\partial_{\nu}T(y)$ $\times\int
dz\partial_{k}f(z)\int_{0}^{\infty}ds\int_{0}^{\infty}dt\Big{\langle}w^{\mu\gamma}(x0)w^{\mu\nu}(y0)\bar{w}^{\beta
k}(zs+t)\Big{\rangle}_{G_{0}}$ $+2\varepsilon^{-d}\sum_{\gamma,\nu=1}^{d}\int
dx(\lambda_{0}^{0})^{\prime}\partial_{\gamma}T(x)\varepsilon^{-d}\int
dy(\lambda_{0}^{d+1})^{\prime}\partial_{\nu}T(y)$ $\times\sum_{k=1}^{d}\int
dz\partial_{k}f(z)\int_{0}^{\infty}ds\int_{0}^{\infty}dt\Big{\langle}w^{0\gamma}(x0)w^{d+1\nu}(y0)\bar{w}^{\beta
k}(zs+t)\Big{\rangle}_{G_{0}}.$

By the change of variables $x=z+\varepsilon\xi,\quad
y=z+\varepsilon\xi^{\prime},\quad s+t=\tau$

$=\sum_{\mu=0,d+1}\sum_{\gamma,\nu,k=1}^{d}\int
dz\partial_{k}f(z)(\lambda_{0}^{\mu})^{\prime}\partial_{\gamma}T(z)(\lambda_{0}^{\mu})^{\prime}\partial_{\nu}T(z)$
$\times\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int d\xi\int
d\xi^{\prime}\Big{\langle}w^{\mu\gamma}(\xi\tau)w^{\mu\nu}(\xi^{\prime}\tau)\bar{w}^{\beta
k}(00)\Big{\rangle}_{G_{0}}$ $+2\sum_{\gamma,\nu,k=1}^{d}\int
dz\partial_{k}f(z)(\lambda_{0}^{0})^{\prime}\partial_{\gamma}T(z)(\lambda_{0}^{d+1})^{\prime}\partial_{\nu}T(z)$
$\times\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int d\xi\int
d\xi^{\prime}\Big{\langle}w^{0\gamma}(\xi\tau)w^{d+1\nu}(\xi^{\prime}\tau)\bar{w}^{\beta
k}(00)\Big{\rangle}_{G_{0}}.$

These terms give rise in the equation for $u^{\beta}$ to terms of the form

$-2\sum_{\gamma,\nu,k=1}^{d}\partial_{k}\Big{[}[(\lambda_{0}^{0})^{\prime}\partial_{\gamma}T(z)(\lambda_{0}^{d+1})^{\prime}\partial_{\nu}T(z)]\alpha^{0d+1}_{\gamma\nu\beta
k}\Big{]}$

and for $\mu=0,d+1$

$-\sum_{\gamma,\nu,k=1}^{d}\partial_{k}\Big{[}[(\lambda_{0}^{\mu})^{\prime}\partial_{\gamma}T(z)(\lambda_{0}^{\mu})^{\prime}\partial_{\nu}T(z)]\alpha^{\mu\mu}_{\gamma\nu\beta
k}\Big{]},$

where

$\alpha^{0d+1}_{\gamma\nu\beta
k}:=\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int d\xi\int
d\xi^{\prime}\Big{\langle}w^{0\gamma}(\xi\tau)w^{d+1\nu}(\xi^{\prime}\tau)\bar{w}^{\beta
k}(00)\Big{\rangle}_{G_{0}}$

$=a_{1}(t,z)\delta_{\beta
k}\delta_{\gamma\nu}+a_{2}(tz)\delta_{\beta\nu}\delta_{\gamma
k}+a_{3}(t,z)\delta_{\beta\gamma}\delta_{\nu k},$

$\alpha^{00}_{\gamma\nu\beta k}:=\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int
d\xi\int
d\xi^{\prime}\Big{\langle}w^{0\gamma}(\xi\tau)w^{0\nu}(\xi^{\prime}\tau)\bar{w}^{\beta
k}(00)\Big{\rangle}_{G_{0}}$ $=b_{1}(t,z)\delta_{\beta
k}\delta_{\gamma\nu}+b_{2}(tz)\delta_{\beta\nu}\delta_{\gamma
k}+b_{3}(t,z)\delta_{\beta\gamma}\delta_{\nu k},$
$\alpha^{d+1d+1}_{\gamma\nu\beta
k}:=\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int d\xi\int
d\xi^{\prime}\Big{\langle}w^{d+1\gamma}(\xi\tau)w^{d+1\nu}(\xi^{\prime}\tau)\bar{w}^{\beta
k}(00)\Big{\rangle}_{G_{0}}$ $=c_{1}(t,z)\delta_{\beta
k}\delta_{\gamma\nu}+c_{2}(tz)\delta_{\beta\nu}\delta_{\gamma
k}+c_{3}(t,z)\delta_{\beta\gamma}\delta_{\nu k}$

Since the potential is central we have $a_{2}=a_{3}$, $b_{2}=b_{3}$
$c_{2}=c_{3}$ [28].

Now we compute the second term in (A. 1.1)

$B:=\varepsilon^{-1}G_{0}\sum_{\ell=1}^{d}\sum_{\mu=0,d+1}\sum_{\gamma=1}^{d}{\mathscr{L}^{*}}[{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{\ell})w^{\mu\gamma}_{\ell}].$

We have for

$\mu=0:$

$\quad\quad\varepsilon^{-1}\sum_{\ell=1}^{d}\sum_{\mu=0,4}\sum_{\gamma=1}^{d}{\mathscr{L}^{*}}[{\partial_{\gamma}\lambda_{0}^{0}}(x_{\ell})w^{0\gamma}_{\ell}]=\varepsilon^{-d}\sum_{\gamma,\nu=1}^{d}\int
dx\partial^{2}_{\nu\gamma}\lambda_{0}^{0}(x)w^{\nu\gamma}(x).$

The corresponding term in the equation for $u^{\beta}$ is

$-\sum_{\gamma,\nu,k=1}^{d}\nabla_{k}\Big{[}[(\lambda_{0}^{0})^{\prime}\partial^{2}_{\nu\gamma}T(z)+(\lambda_{0}^{0})^{\prime\prime}\partial_{\gamma}T(z)\partial_{\nu}T(z)]h^{0}_{\gamma\nu\beta
k}\Big{]},$

with

$h^{0}_{\gamma\nu\beta k}=\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int
d\xi\Big{\langle}\bar{w}^{\nu\gamma}(\xi\tau)\bar{w}^{\beta
k}(00)\Big{\rangle}_{G_{0}}=h_{1}(t,z)\delta_{\beta
k}\delta_{\gamma\nu}+h_{2}(tz)\delta_{\beta\nu}\delta_{\gamma
k}+h_{3}(t,z)\delta_{\beta\gamma}\delta_{\nu k}.$

The most difficult term is

$\mu=d+1$:

$\quad\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}{\mathscr{L}^{*}}[{\partial_{\gamma}\lambda_{0}^{d+1}(x_{\ell})}w^{d+1\gamma}_{\ell}]:=\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}{\mathscr{L}^{*}}[g_{\gamma}(x_{\ell})w^{d+1\gamma}_{\ell}],$

where, for semplicity, we put
$g_{\gamma}(x)=\partial_{\gamma}\lambda_{0}^{d+1}(x)$.

${\mathscr{H}}:=\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}{\mathscr{L}^{*}}[g_{\gamma}(x_{\ell})w^{d+1\gamma}_{\ell}]$
$=-\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\Big{\\{}\varepsilon\sum_{k=1}^{d}v_{\ell}^{k}\partial_{k}g_{\gamma}(x_{\ell})w^{d+1\gamma}_{\ell}+\sum_{i=1}^{N}\sum_{k=1}^{d}v_{i}^{k}g_{\gamma}(x_{\ell}){\partial\over\partial
x_{i}^{k}}w^{d+1\gamma}_{\ell}\Big{\\}}$ (A. 1.2)
$-\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell})\Big{\\{}\sum_{i=1}^{N}\sum_{k=1}^{d}\sum_{j\neq
i}\partial_{k}V(\varepsilon^{-1}(x_{i}-x_{j})){\partial\over\partial
v_{i}^{k}}w^{d+1\gamma}_{\ell}\Big{\\}}.$ (A. 1.3)

The first term is of order $\varepsilon$ and will appear in $C$. The others
has to be examined and we start with the second term in (A. 1.2)

$-\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{i=1}^{N}\sum_{k=1}^{d}v_{i}^{k}g_{\gamma}(x_{\ell}){\partial\over\partial
x_{i}^{k}}w^{d+1\gamma}_{\ell}$
$=-\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{k=1}^{d}g_{\gamma}(x_{\ell})\Big{[}{1\over
2}\sum_{j}\sum_{\nu}(\partial_{k}\Psi^{\nu\gamma}(\varepsilon^{-1}(x_{\ell}-x_{j}))(v_{j}^{k}-v_{\ell}^{k}){1\over
2}(v_{\ell}^{\nu}+v_{j}^{\nu})+\sum_{i}{1\over
2}v_{\ell}^{\gamma}v_{i}^{k}{\partial\over\partial
x_{i}^{k}}z_{\ell}^{d+1\gamma}\Big{]}.$ (A. 1.4)

The first term will not give contribution to $C$ because of the Gaussian
integration on the velocities. Now we pass to the term in (A. 1.3) and add the
last term in in (A. 1.4)

$\displaystyle\begin{split}&\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell})\sum_{i=1}^{N}\Big{\\{}\sum_{k=1}^{d}\sum_{j\neq
i}\partial_{k}V(\varepsilon^{-1}(x_{i}-x_{j})){\partial\over\partial
v_{i}^{k}}w^{d+1\gamma}_{\ell}+{1\over
2}v_{i}^{k}v_{\ell}^{\gamma}{\partial\over\partial
x_{i}^{k}}z_{\ell}^{d+1\gamma}\Big{\\}}\\\
=&\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell})\sum_{i=1}^{N}\Big{\\{}{1\over
2}\sum_{k=1}^{d}v_{i}^{k}v_{\ell}^{\gamma}{\partial\over\partial
x_{i}^{k}}z_{\ell}^{d+1\gamma}+\sum_{k=1}^{d}\sum_{j\neq
i}\partial_{k}V(\varepsilon^{-1}(x_{i}-x_{j}))\Big{[}z^{d+1}_{\ell}{\partial\over\partial
v_{i}^{k}}v_{\ell}^{\gamma}\\\ &+v_{\ell}^{\gamma}{\partial\over\partial
v_{i}^{k}}{1\over 2}|v_{\ell}|^{2}+{1\over
2}\sum_{s=1}^{N}\sum_{\nu=1}^{d}\Psi^{\nu\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s}){1\over
2}{\partial\over\partial
v_{i}^{k}}[v_{\ell}^{\nu}+v_{s}^{\nu}])\big{]}\Big{\\}}.\end{split}$ (A. 1.5)

We begin to deal with the second term in the rhs of (A. 1.5)

${\mathscr{M}}:=\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell})z^{d+1}_{\ell}\sum_{j\neq\ell}\partial_{\gamma}V(\varepsilon^{-1}(x_{\ell}-x_{j})).$

By using the antisymmetry of the gradient of the potential under the exchange
$\ell\rightarrow j$ we get

${\mathscr{M}}={1\over
2}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{j\neq\ell,1}^{N}[g_{\gamma}(x_{\ell})z_{\ell}^{d+1}-g_{\gamma}(x_{j})z_{j}^{d+1}]{\partial_{\gamma}V}(\varepsilon^{-1}(x_{\ell}-x_{j}))$
$={1\over
2}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{j\neq\ell,1}^{n}\Big{[}[g_{\gamma}(x_{\ell})-g_{\gamma}(x_{j})]z_{\ell}^{d+1}+g_{\gamma}(x_{j})[z_{\ell}^{d+1}-z_{j}^{d+1}]{\partial_{\gamma}V}(\varepsilon^{-1}(x_{\ell}-x_{j}))\Big{]}$
$={1\over
2}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{j\neq\ell,1}^{N}\Big{[}[g_{\gamma}(x_{\ell})-g_{\gamma}(x_{j})]z_{\ell}^{d+1}+{1\over
2}[g_{\gamma}(x_{j})+g_{\gamma}(x_{\ell})[z_{\ell}^{d+1}-z_{j}^{d+1}]{\partial_{\gamma}V}(\varepsilon^{-1}(x_{\ell}-x_{j}))\Big{]}.$

As usual, we replace $[g_{\gamma}(x_{\ell})-g_{\gamma}(x_{j})]$ with
$\varepsilon\partial_{s}g_{\gamma}(x_{\ell})(\xi^{s}_{\ell}-\xi^{s}_{j})$ so
that the first term is of order $\varepsilon$ and will give contribution to
the equation, while the second term is not of order $\varepsilon$ and we have
to show that does not give contribution.

${\mathscr{M}}={1\over
2}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{j\neq\ell,1}^{N}\Big{[}\varepsilon\sum_{s=1}^{d}[z_{\ell}^{d+1}\partial_{s}g_{\gamma}(x_{\ell})\Psi^{\gamma
s}(\varepsilon^{-1}(x_{\ell}-x_{j}))$ $+{1\over
4}[g_{\gamma}(x_{j})+g_{\gamma}(x_{\ell})][z_{\ell}^{d+1}-z_{j}^{d+1}]\partial_{\gamma}V(\varepsilon^{-1}(x_{\ell}-x_{j}))\Big{]}$
(A. 1.6)

In $z_{k}^{d+1}={1\over 2}\big{[}v^{2}_{k}+\sum_{j\neq
i=1}^{N}V(\varepsilon^{-1}|x_{k}-x_{j}|)$ the second term $\sum_{j\neq
i=1}^{N}V(\varepsilon^{-1}|x_{k}-x_{j}|)$ is

$\sum_{\ell,j=1}^{N}-V(\varepsilon^{-1}(x_{j}-x_{t}))]\partial_{\gamma}V(\varepsilon^{-1}(x_{\ell}-x_{j}))[g_{\gamma}(x_{j})+g_{\gamma}(x_{\ell})]$
(A. 1.7)

and will not give contribution to $C$ since the term
$\sum_{t=1}^{N}[V(\varepsilon^{-1}(x_{\ell}-x_{t}))$ under the average on the
position in $C$ will not depend on $\ell$ by isotropy.

Then, the second line in (A. 1.6) becomes

${1\over 2}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\sum_{j\neq\ell,1}^{n}{1\over
4}[g_{\gamma}(x_{j})+g_{\gamma}(x_{\ell})]{1\over
2}[|v_{\ell}|^{2}-|v_{j}|^{2}]{\partial_{\gamma}V}(\varepsilon^{-1}(x_{\ell}-x_{j})).$

This term will not give contribution to $C$ because of the Gaussian
integration on the velocities. Now, we compute the first and third terms term
in (A. 1.5).

$\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell})\sum_{i=1}^{N}\Big{\\{}\sum_{k=1}^{d}\sum_{j\neq
i}\partial_{k}V(\varepsilon^{-1}(x_{i}-x_{j}))v_{\ell}^{\gamma}{\partial\over\partial
v_{i}^{k}}{1\over 2}|v_{\ell}|^{2}+{1\over
2}v_{i}^{k}v_{\ell}^{\gamma}{\partial\over\partial
x_{i}^{k}}z_{\ell}^{d+1\gamma}$ (A. 1.8)

$=\sum_{\ell=1}^{N}\sum_{k=1}^{d}g_{\gamma}(x_{\ell}){1\over
2}\biggl{\\{}-2v_{\ell}^{\gamma}v_{\ell}^{k}\sum_{j:j\neq\ell}^{1,N}{\partial_{k}V}(\varepsilon^{-1}(x_{\ell}-x_{j}))+\sum_{j:j\neq\ell}^{1,N}v_{\ell}^{\gamma}\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j}))(v_{\ell}^{k}-v_{j}^{k})\biggr{\\}}$
$=-\sum_{\ell=1}^{N}\sum_{k=1}^{d}g_{\gamma}(x_{\ell}){1\over
2}\sum_{j:j\neq\ell}^{1,N}v_{\ell}^{\gamma}\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j}))(v_{\ell}^{k}+v_{j}^{k}).$

By using the antisymmetry of $\partial_{k}V$ we get

$-{1\over
4}\sum_{\gamma=1}^{d}\sum_{\ell=1}^{N}[g_{\gamma}(x_{\ell})v_{\ell}^{\gamma}-g_{\gamma}(x_{j})v_{j}^{\gamma}]\\{\sum_{j:j\neq\ell}^{1,N}\sum_{k=1}^{d}\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j}))(v_{\ell}^{k}+v_{j}^{k})\\}$
$=-{1\over
4}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\Big{[}[v_{\ell}^{\gamma}[g_{\gamma}(x_{\ell})-g_{\gamma}(x_{j})]+g_{\gamma}(x_{j})[v_{\ell}^{\gamma}-v_{j}^{\gamma}]\Big{]}\\{\sum_{j:j\neq\ell}^{1,N}\sum_{k=1}^{d}\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j})){1\over
2}(v_{\ell}^{k}+v_{j}^{k})\\}$ $=-{1\over
4}\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}\Big{[}\varepsilon
v_{\ell}^{\gamma}\sum_{\nu=1}^{d}\partial_{\nu}g_{\gamma}(x_{\ell})\sum_{j:j\neq\ell}^{1,N}\sum_{k=1}^{d}\Psi^{\nu
k}(\varepsilon^{-1}(x_{\ell}-x_{j})){1\over 2}(v_{\ell}^{k}+v_{j}^{k})\Big{]}$
(A. 1.9)
$+g_{\gamma}(x_{j})[v_{\ell}^{\gamma}-v_{j}^{\gamma}]\Big{]}\\{\sum_{j:j\neq\ell}^{1,N}\sum_{k=1}^{d}\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j})){1\over
2}(v_{\ell}^{k}+v_{j}^{k})\\}.$

The first term is of order $\varepsilon$ and will appear in $C$ while the
second term is zero after average on the velocities in $C$.

Finally, we discuss the last term in (A. 1.5).

$\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell})\sum_{k=1}^{d}\sum_{j\neq
i}\partial_{k}V(\varepsilon^{-1}(x_{i}-x_{j})){1\over
2}\sum_{s=1}^{N}\sum_{\nu=1}^{d}\Psi^{\nu\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s}){1\over
2}{\partial\over\partial v_{i}^{k}}[v_{\ell}^{\nu}+v_{s}^{\nu}])\big{]}$
$=\sum_{\ell=1}^{N}\sum_{\gamma=1}^{d}g_{\gamma}(x_{\ell}){1\over
4}\sum_{s=1}^{N}\sum_{k=1}^{d}\sum_{j}[\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j}))+\partial_{k}V(\varepsilon^{-1}(x_{s}-x_{j}))]\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s})\big{]}$

The first term is equal to

$={1\over
8}\sum_{\ell,j=1}^{N}\sum_{\gamma,k=1}^{d}\sum_{s=1}^{N}\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j}))\\{[g_{\gamma}(x_{\ell})\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s})-g_{\gamma}(x_{j})\Psi^{k\gamma}(\varepsilon^{-1}(x_{j}-x_{s})]$
$={1\over
8}\sum_{\ell,j=1}^{N}\sum_{\gamma,k=1}^{d}\sum_{s=1}^{N}[\partial_{k}V(\varepsilon^{-1}(x_{\ell}-x_{j}))[g_{\gamma}(x_{\ell})-g_{\gamma}(x_{j})]\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s})$
$+g_{\gamma}(x_{j}[\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s}))-\Psi^{k\gamma}(\varepsilon^{-1}(x_{j}-x_{s}))]\\}$

The first term is of order $\varepsilon$ and the second has the same symmetry
properties as the term dealt with in (A. 1.7) and hence does not give
contribution. The other term is

$\sum_{\ell,j=1}^{N}\sum_{\gamma,k=1}^{d}g_{\gamma}(x_{\ell}){1\over
4}\sum_{s=1}^{N}\partial_{k}V(\varepsilon^{-1}(x_{s}-x_{j}))\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s}))$
$=\sum_{\ell,j=1}^{N}\sum_{\gamma,k=1}^{d}g_{\gamma}(x_{\ell})\sum_{s=1}^{N}\partial_{k}V(\varepsilon^{-1}(x_{s}-x_{j}))[\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{s}))-\Psi^{k\gamma}(\varepsilon^{-1}(x_{\ell}-x_{j}))]$

and does not give contribution to $C$ for the same reasons.

Putting all the terms of order $\varepsilon$ together in (A. 1.2),(A. 1.6) and
(A. 1.9) we have that

$\varepsilon\varepsilon^{-d}\int
dx\sum_{s,\gamma=1}^{d}\partial_{s}g_{\gamma}(x)(v^{\gamma}w^{d+1s})(x)+\varepsilon\varepsilon^{-d}\int
dx\sum_{m,\gamma=1}^{d}\partial_{m}g_{\gamma}(x){\mathscr{A}}^{\gamma m}(x)$
$+\varepsilon\varepsilon^{-d}\sum_{s,\gamma=1}^{d}\int
dx\partial_{s}g_{\gamma}(x){\mathscr{C}}^{\gamma
s}(x)+\varepsilon\varepsilon^{-d}\sum_{s,\gamma=1}^{d}\int
dx\partial_{s}g_{\gamma}(x){\mathscr{N}}^{\gamma s}(x),$

with

${\mathscr{C}}^{\gamma s}(x)=\sum_{\ell,j=1}^{N}\delta(x_{\ell}-x){1\over
4}\sum_{\nu,\gamma=1}^{d}\Psi^{\nu\gamma}(\varepsilon^{-1}(x_{\ell}-x_{j}))v_{\ell}^{s}{1\over
2}[v_{\ell}^{\nu}+v_{j}^{\nu}],$ ${\mathscr{A}}^{\gamma
m}(x)=\sum_{\ell}\delta(x-x_{\ell})z_{\ell}^{d+1}\sum_{s}\Psi^{\gamma
m}(\varepsilon^{-1}(|x_{\ell}-x_{s}|),$ ${\mathscr{N}}^{\gamma
s}(x)=\sum_{\ell}\delta(x-x_{\ell})\sum_{s,j}\partial_{m}V(\varepsilon^{-1}(x_{\ell}-x_{j}))\Psi^{s\gamma
m}(\varepsilon^{-1}(|x_{\ell}-x_{s}|).$

The contribution to $C$ is then

$\sum_{m,k,\gamma=1}^{d}\int
dz\partial_{k}f(z)\Big{[}\partial^{2}_{m\gamma}(\lambda_{0}^{d+1})(z)\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int
d\xi\Big{\langle}(v^{\gamma}w^{d+1m}+{\mathscr{A}}^{\gamma
m}(x))(\xi,\tau)\bar{w}^{\beta k}(0,0)\Big{\rangle}_{G_{0}}\Big{]}$
$=\sum_{m,k,\gamma=1}^{d}\int
dz\partial_{k}f(z)\sum_{m}\Big{[}(\lambda_{0}^{d+1})^{\prime\prime}\partial_{\gamma}T\partial_{m}T+(\lambda_{0}^{d+1})^{\prime}\partial^{2}_{m\gamma}T(z)d_{\gamma\mu\beta
k}\Big{]},$

with

$d_{\gamma\mu\beta k}:=\int_{0}^{\infty}ds\int_{s}^{\infty}d\tau\int
d\xi\Big{\langle}([v^{\gamma}w^{d+1m}+{\mathscr{A}}^{\gamma
m}(x)](\xi,\tau)]\bar{w}^{\beta k}(0,0)\Big{\rangle}_{G_{0}}$
$=d_{1}(t,z)\delta_{\beta
k}\delta_{\gamma\mu}+d_{2}(tz)\delta_{\beta\mu}\delta_{\gamma
k}+d_{3}(t,z)\delta_{\beta\gamma}\delta_{\mu k};\quad d_{2}=d_{3}$

and for the term ${\mathscr{C}}$: $\varepsilon^{-d}\varepsilon\int
dx{\mathscr{C}}^{\ s\gamma}(x)\partial_{s}g_{\gamma}(x)$ in the equation

$\partial_{k}[((\lambda_{0}^{d+1})^{\prime}\partial^{2}_{s\gamma}T+(\lambda_{0}^{d+1})^{\prime\prime}\partial_{s}T\partial_{\gamma}T)\int_{o}^{\infty}d\tau^{\prime}\int_{\tau^{\prime}}^{\infty}d\tau\int
d\xi\Big{\langle}\bar{w}^{\beta k}(0,0)({\mathscr{C}}^{\gamma s})(\xi,\tau),$

which becomes for $\beta=k$:

$\qquad\qquad\sum_{\gamma=1}^{d}\partial_{k}[((\lambda_{0}^{d+1})^{\prime}\partial^{2}_{\gamma\gamma}T+(\lambda_{0}^{d+1})^{\prime\prime}\partial_{\gamma}T\partial_{\gamma}T)g_{1}],$

with

$g_{1}:={1\over
d}\sum_{\gamma=1}^{d}\int_{0}^{\infty}d\tau^{\prime}\int_{\tau^{\prime}}^{\infty}d\tau\int
d\xi\Big{\langle}\bar{w}^{\beta
k}(0,0)({\mathscr{C}}^{\gamma\gamma})(\xi,\tau)$

$for\quad\beta\neq
k\qquad\qquad\partial_{k}[((\lambda_{0}^{d+1})^{\prime}\partial^{2}_{\beta
k}T+(\lambda_{0}^{d+1})^{\prime\prime}\partial_{k}T\partial_{\beta}T)g_{2}],$

with

$g_{2}:=\int_{0}^{\infty}d\tau^{\prime}\int_{\tau^{\prime}}^{\infty}d\tau\int
d\xi\Big{\langle}\bar{w}^{\beta
k}(0,0)({\mathscr{C}}^{\beta\beta})(\xi,\tau).$

and for the term ${\mathscr{N}}$: $\varepsilon^{-d}\varepsilon\int
dx{\mathscr{C}N}^{\ s\gamma}(x)\partial_{s}g_{\gamma}(x)$ in the equation for
$\beta\neq k$ is zero and for $\beta=k$:

$\qquad\qquad\sum_{\gamma=1}^{d}\partial_{k}[((\lambda_{0}^{d+1})^{\prime}\partial^{2}_{\gamma\gamma}T+(\lambda_{0}^{d+1})^{\prime\prime}\partial_{\gamma}T\partial_{\gamma}T)f_{1}],$

with

$f_{1}:={1\over
d}\sum_{\gamma=1}^{d}\int_{0}^{\infty}d\tau^{\prime}\int_{\tau^{\prime}}^{\infty}d\tau\int
d\xi\Big{\langle}\bar{w}^{\beta
k}(0,0)({\mathscr{N}}^{\gamma\gamma})(\xi,\tau)$

${\bf Important\ remark}$

This long calculation proves also that indeed $C$ is of order $1$ and hence
the assumption on the definition of $R_{1}^{s}$ is correct.

The transport coefficients in (3.35) in the stress thermal tensor $\tau^{(2)}$
have two different contributions. We can write $K_{i}=Y_{i}+X_{i},i=1,2$. We
have already given the expression of $Y_{i}$ and now we collect all the terms
to give the expression of $X_{i}$.

$X_{1}=\Big{[}[(\lambda_{0}^{0})^{\prime}(\lambda_{0}^{d+1})^{\prime}]4a_{2}(tz)+2b_{2}(\lambda_{0}^{0})^{\prime}(\lambda_{0}^{0})^{\prime}+2c_{2}(\lambda_{0}^{d+1})^{\prime}(\lambda_{0}^{d+1})^{\prime}+(\lambda_{0}^{d+1})^{\prime\prime}[h_{2}+2d_{2}+g_{2}]\Big{]}\quad\beta\neq
k,$

$X_{2}=2(\lambda_{0}^{d+1})^{\prime}d_{2}+2(\lambda_{0}^{0})^{\prime}T(z)h_{2}+(\lambda_{0}^{d+1})^{\prime}g_{2},\quad\beta\neq
k.$

Moreover, there are also new contributions to $\omega_{i}$. We write
$\omega_{i}=\bar{\omega}_{i}+{\mathscr{\phi}}_{i},i=1,2$. We get

${\mathscr{\phi}}_{1}=[(\lambda_{0}^{0})^{\prime\prime}h_{1}+(\lambda_{0}^{d+1})^{\prime\prime}d_{1}+a_{1}(\lambda_{0}^{d+1})^{\prime}(\lambda_{0}^{0})^{\prime}+b_{1}(\lambda_{0}^{0})^{\prime}(\lambda_{0}^{0})^{\prime}+c_{1}(\lambda_{0}^{d+1})^{\prime}(\lambda_{0}^{d+1})^{\prime}+(\lambda_{0}^{d+1})^{\prime\prime}(g_{1}+f_{1}),$
${\mathscr{\phi}}_{2}=(\lambda_{0}^{0})^{\prime}h_{1}+(\lambda_{0}^{d+1})^{\prime}d_{1}+(\lambda_{0}^{d+1})^{\prime}(g_{1}+f_{1}).$

###  A.2.  Compatibility conditions

We need to check that in the definitions of $R_{i}$ we can apply
${\mathscr{L}}^{-1}$ on the l.h.s.

$\bullet$ $R_{1}^{a}:\quad$ The simplest case is the definition of $R_{1}^{a}$
in (2.43). We need to know that ${\mathscr{L}}^{*}G_{0}$ has zero projection
on the null space. This is equivalent to prove

$\int dxz^{\mu}(x){\mathscr{L}}^{*}G_{0}=0,$

since $Z^{\mu}=\int dxz^{\mu}(x)$ are the total mass, the total momentum and
the total energy and they are conserved by the dynamics, namely
${\mathscr{L}}Z^{\mu}=0$. Hence for any observable $\phi$ $\int
dxz^{\mu}(x){\mathscr{L}}^{*}\phi=0$. We have

$\int dxz^{\mu}(x){\mathscr{L}}^{*}G_{0}=\int
dx{\mathscr{L}}z^{\mu}(x)G_{0}=\sum_{\gamma=1}^{d}{\partial_{\gamma}}\Big{\langle}\varepsilon^{d}\sum_{i}\delta(x-x_{i})w^{\mu\gamma}_{i}]\Big{\rangle}_{G_{0}}.$

For $\mu=0,d+1$, $\int dxz^{\mu}(x){\mathscr{L}}^{*}G_{0}$ is zero since
$\Big{\langle}v_{i}\Big{\rangle}_{G_{0}}=0$. For ${\mu=1\cdots d}$

$\int
dxz^{\mu}(x){\mathscr{L}}^{*}G_{0}=\sum_{\gamma=1}^{d}{\partial_{\gamma}}\Big{\langle}\varepsilon^{d}\sum_{i}w^{\mu\gamma}_{i}\Big{\rangle}_{G_{0}}=\sum_{\gamma=1}^{d}\delta_{\mu\gamma}{\partial_{\gamma}}\Big{\langle}\varepsilon^{d}\sum_{i}w^{\mu\mu}_{i}]\Big{\rangle}_{G_{0}}={\partial_{\mu}}P=0$

since $P$ is constant.

$\bullet$ $R_{2}:\quad$ $R_{2}$ is defined by (2.46). It is not difficult to
see that the condition in this case amount to say that $\rho$ and $e$ are
solutions of the continuity equation and the energy equation. We have the
condition to be verified for each $\mu$

$0=\int
dxz^{\mu}(x)[{\mathscr{L}}^{*}G_{0}g_{1}+\partial_{t}G_{0}]=\sum_{k=1}^{d}\int
dx{u^{k}\over
T}\Big{\langle}{\mathscr{L}}z^{\mu}(x)z^{k}\Big{\rangle}_{G_{0}}+\int
dx\partial_{t}\Big{\langle}z^{\mu}\Big{\rangle}_{G_{0}}.$

This is naturally true for $\mu=1\cdots d$. Then,

$\mu=0:\quad$

$\int dx\sum_{k=1}^{d}\partial_{\mu}[{u^{k}\over
T}\Big{\langle}z^{\mu}z^{k}\Big{\rangle}_{G_{0}}]+\int
dx\partial_{t}\Big{\langle}z^{\mu}\Big{\rangle}_{G_{0}}=\int
dx[\partial_{\mu}(\rho u^{\mu})+\partial_{t}\rho]=0.$ (A. 2.1)

For $\mu=d+1$ we can add in the condition ${\mathscr{L}}^{*}G_{0}R_{1}$ since
we have already proven that it is orthogonal to the invariants.

$\mu=d+1:\qquad\qquad\qquad$

$\int
dxz^{d+1}(x)[{\mathscr{L}}^{*}G_{0}g_{1}+\partial_{t}G_{0}+{\mathscr{L}}^{*}G_{0}R_{1}]=$
$\int dx\sum_{k=1}^{d}\partial_{\nu}[{u^{k}\over
T}\Big{\langle}w^{d+1\nu}z^{k}\Big{\rangle}_{G_{0}}]+\int
dx\partial_{t}\Big{\langle}z^{d+1}\Big{\rangle}_{G_{0}}+\int
dx\Big{\langle}w^{d+1\nu}G_{0}R_{1}\Big{\rangle}_{G_{0}}$ $=\int
dx[\partial_{\mu}((\rho e+P)u^{\mu})+\partial_{t}(\rho e)-\nabla(k\nabla
T)]=0.$

$\bullet$ $R_{1}^{s}:\quad$ The most difficult case is the definition of
$R_{1}^{s}$ in (2.44). We need to show that
${\mathscr{L}}^{*}{\mathscr{L}}^{*}G_{0}$ has zero projection on the null. We
can write

${\mathscr{L}}^{*}{\mathscr{L}}^{*}G_{0}={\mathscr{L}}^{*}[G_{0}\sum_{i=1}^{N}\sum_{\gamma=1}^{d}\sum_{\mu=0,4}{\partial_{\gamma}\lambda_{0}^{\mu}}(x_{i})w^{\mu\gamma}_{i}]=\varepsilon^{-d}{\mathscr{L}}^{*}[\sum_{\mu=0,4}\sum_{\gamma=1}^{d}\int
dyG_{0}{\partial_{\gamma}\lambda_{0}^{\mu}}(y)w^{\mu\gamma}(y)]\Big{\rangle}.$

The condition to be satisfied is

$\varepsilon^{-d}\int
dx\Big{\langle}z^{\alpha}(x){\mathscr{L}}^{*}[\sum_{\mu=0,4}\sum_{\gamma=1}^{d}\int
dyG_{0}{\partial_{\gamma}\lambda_{0}^{\mu}}(y)w^{\mu\gamma}(y)]\Big{\rangle}=0,\quad\
for\ \alpha=0,\cdots,d+1.$

The lhs is

$\varepsilon^{-d}\int
dx\Big{\langle}{\mathscr{L}}z^{\alpha}(x)[\sum_{\mu=0,4}\sum_{\gamma=1}^{d}\int
dyG_{0}{\partial_{\gamma}\lambda_{0}^{\mu}}(y)w^{\mu\gamma}(y)]\Big{\rangle}$
$=\varepsilon^{-d}\int dx\Big{\langle}\nabla\cdot
w^{\alpha}(x)[\sum_{\mu=0,4}\sum_{\gamma=1}^{d}\int
dyG_{0}{\partial_{\gamma}\lambda_{0}^{\mu}}(y)w^{\mu\gamma}(y)]\Big{\rangle}$
$=\varepsilon^{-d}\int
dx\sum_{\nu,\gamma=1}^{d}\sum_{\mu=0,4}\partial_{\nu}\int
dy{\partial_{\gamma}\lambda_{0}^{\mu}}(y)\Big{\langle}w^{\alpha\nu}(x)w^{\mu\gamma}(\xi)\Big{\rangle}_{G_{0}}$
$=\sum_{\nu,\gamma=1}^{d}\sum_{\mu=0,4}\partial_{\nu}\int
dy{\partial_{\gamma}\lambda_{0}^{\mu}}(y)\int
d\xi\Big{\langle}w^{\alpha\nu}(0)w^{\mu\gamma}(\xi)\Big{\rangle}_{G_{0}}.$

We know that [28]

$\int
d\xi\Big{\langle}w^{\alpha\nu}(0)w^{\mu\gamma}(\xi)\Big{\rangle}_{G_{0}}=0\quad
if\ \alpha=1\cdots d,\quad\mu=0,d+1;\quad\int
d\xi\Big{\langle}w^{0\nu}(0)w^{0\gamma}(\xi)\Big{\rangle}_{G_{0}}=\rho
T\delta_{\nu\gamma};$ $\int
d\xi\Big{\langle}w^{d+1\nu}(0)w^{d+1\gamma}(\xi)\Big{\rangle}_{G_{0}}={T(\rho
e+P)^{2}\over\rho}\delta_{\nu\gamma};\quad\int
d\xi\Big{\langle}w^{0\nu}(0)w^{d+1\gamma}(\xi)\Big{\rangle}_{G_{0}}=T(\rho
e+P)\delta_{\nu\gamma},$

so that we have two conditions:

$\int dy[{\partial_{\gamma}\lambda_{0}^{0}}(y)\rho
T(y)+{\partial_{\gamma}\lambda_{0}^{d+1}}T(\rho e+P)]=0,$ $\int
dy[{\partial_{\gamma}\lambda_{0}^{0}}(y)T(\rho
e+P)(y)+{\partial_{\gamma}\lambda_{0}^{d+1}}{T(e+P)^{2}\over\rho}]=0.$

Now we prove that

${\partial_{\gamma}\lambda_{0}^{0}}(y)\rho(y)+{\partial_{\gamma}\lambda_{0}^{d+1}}(\rho
e+P)=0,$

so that both conditions are satisfied.

$\rho\partial_{\gamma}\log z-P\partial_{\gamma}\beta=$ $=\rho{\delta(\log
z)\over\delta P}\partial_{\gamma}P+\rho{\delta(\log
z)\over\delta\beta}\partial_{\gamma}\beta+\log z{\delta\rho\over\delta
P}\partial_{\gamma}P-(\rho e+P)\partial_{\gamma}\beta.$

Since $P$ is constant we are left with

$[\rho{\delta(\log z)\over\delta\beta}-(\rho e+P)]=0$

by thermodynamic relations.

###  A.3. Galileian invariance

We write (3.23)+second term in (3.26) as

$\displaystyle\begin{split}&\sum_{\mu=1}^{d}\sum_{\gamma=1}^{d}\int
dy\Big{[}\lambda_{1}^{\mu}(y){\partial_{\gamma}\lambda_{0}^{0}}(y)\int
d\xi\Big{\langle}z(\xi)^{\mu}w^{0\gamma}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}_{G_{0}}\\\ +&\sum_{\mu=1}^{d}\sum_{\gamma=1}^{d}\int
dy\Big{[}\lambda_{1}^{\mu}(y){\partial_{\gamma}\lambda_{0}^{d+1}}(y)\int
d\xi\Big{\langle}z(\xi)^{\mu}w^{d+1\gamma}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}_{G_{0}}\\\
&-\sum_{\mu=1}^{d}\sum_{\ell=1}^{d}u^{\mu}{1\over T^{2}}\int
dy{\partial_{l}T}(y)]\int d\xi\Big{\langle}\bar{w}^{\mu
l}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)(0)\Big{\rangle}_{G_{0}}\Big{]}=0\end{split}$ (A. 3.1)

and we want to prove that (A. 3.1) is zero.

We start from the identity, for any $\gamma,\beta,k=1,\cdots,d$,

${\mathscr{D}}:=\int
d\xi\Big{\langle}w^{0\gamma}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}_{G_{0}}=0$

and define ${\mathscr{D}}_{s}$ and ${\mathscr{D}}^{1}_{s}$ below the same
quantity where all the velocity $v$ are changed to $v-su$. We have also

${d\over ds}{\mathscr{D}}_{s}|{{}_{s=0}}=0.$

We compute the derivative

${d\over ds}{\mathscr{D}}_{s}=\int d\xi\Big{\langle}{G_{0}}{d\over
ds}(w^{0\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta k}(0)\Big{\rangle}+\int
d\xi\Big{\langle}{d\over
ds}{G_{0}}(w^{0\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}$ $+\int d\xi\Big{\langle}w^{0\gamma}(\xi){d\over
ds}[{\mathscr{L}}^{-1}]\bar{w}^{\beta k}(0)+\int
d\xi\Big{\langle}w^{0\gamma}(\xi){\mathscr{L}}^{-1}[{d\over ds}\bar{w}^{\beta
k}(0)].$

We use the identity

${d\over ds}[{\mathscr{L}}^{-1}]=-{\mathscr{L}}^{-1}{d\over
ds}[{\mathscr{L}}]{\mathscr{L}}^{-1}.$

It is not difficult to see that the only surviving term evaluated in $s=0$ is

$\int d\xi\Big{\langle}{d\over
ds}{G_{0}}(w^{0\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}=-\sum_{\mu=1}^{d}{u^{\mu}\over T}\int
d\xi\Big{\langle}{G_{0}}(z^{\mu}w^{0\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}$

and this implies that the first term in (A. 3.1) is zero.

Now we start from the identity for any $\gamma,\beta,k=1,\cdots,d$

${\mathscr{D}}^{1}:=\int
d\xi\Big{\langle}w^{d+1\gamma}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}_{G_{0}}=0$

and

${d\over ds}{\mathscr{D}^{1}}_{s}|{{}_{s=0}}=0.$

We have

$\displaystyle\begin{split}{d\over
ds}{\mathscr{D}^{1}}_{s}|{{}_{s=0}}&=-\sum_{\mu=1}^{d}u^{\mu}\int
d\xi\Big{\langle}{G_{0}}(z^{\mu}w^{d+1\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}+\Big{\langle}{G_{0}}{d\over
ds}(w^{d+1\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta k}(0)\Big{\rangle}\\\
&=\sum_{\mu=1}^{d}\Big{[}-{u^{\mu}\over T}\int
d\xi\Big{\langle}{G_{0}}(z^{\mu}w^{d+1\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}-u^{\mu}\Big{\langle}{G_{0}}(w^{\mu\gamma})(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}\Big{]}=0.\end{split}$ (A. 3.2)

The sum of the last two terms in (A. 3.1) is

$\displaystyle\sum_{\mu=1}^{d}\Big{[}-\int dy{u^{\mu}\over T}(y){1\over
T^{2}}{\partial_{\gamma}T}(y)\int
d\xi\Big{\langle}z(\xi)^{\mu}w^{d+1\gamma}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta
k}(0)\Big{\rangle}_{G_{0}}$ (A. 3.3) $\displaystyle-u^{\mu}{1\over T^{2}}\int
dy{\partial_{l}T}(y)]\int d\xi\Big{\langle}\bar{w}^{\mu
l}(\xi){\mathscr{L}}^{-1}\bar{w}^{\beta k}(0)(0)\Big{\rangle}_{G_{0}}\Big{]}$
(A. 3.4)

and is equal to zero by using the identity (A. 3.2).

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