remarkRemark hypothesisHypothesis claimClaim Compressible Leslie–Ericksen
equationsPatrick E. Farrell, Giovanni Russo and Umberto Zerbinati

# Kinetic derivation of a compressible Leslie–Ericksen equation for rarified
calamitic gases††thanks: Submitted to the editors DATE. PEF was supported by
the Engineering and Physical Sciences Research Council [EPSRC grants
EP/R029423/1 and EP/W026163/1].

Patrick E. Farrell Mathematical Institute, University of Oxford, Oxford,
United Kingdom<EMAIL_ADDRESS>Giovanni Russo Departement of
Mathematics, University of Catania, Catania CT, Italy<EMAIL_ADDRESS>Umberto Zerbinati Mathematical Institute, University of Oxford, Oxford, United
Kingdom<EMAIL_ADDRESS>
###### Abstract

Nematic ordering describes the phenomenon where anisotropic molecules tend to
locally align, like matches in a matchbox. This ordering can arise in solids
(as nematic elastomers), liquids (as liquid crystals), and in gases. In the
1940s, Onsager described how nematic ordering can arise in dilute colloidal
suspensions from the molecular point of view. However, the kinetic theory of
nonspherical molecules has not, thus far, accounted for phenomena relating to
the presence of nematic ordering.

In this work we develop a kinetic theory for the behavior of rarified
calamitic (rodlike) gases in the presence of nematic ordering. Building on
previous work by Curtiss, we derive from kinetic theory the rate of work
hypothesis that forms the starting point for Leslie–Ericksen theory. We
incorporate ideas from the variational theory of nematic liquid crystals to
create a moment closure that preserves the coupling between the laws of linear
and angular momentum. The coupling between these laws is a key feature of our
theory, in contrast to the kinetic theory proposed by Curtiss & Dahler, where
the couple stress tensor is assumed to be zero. This coupling allows the
characterization of anisotropic phenomena arising from the nematic ordering.
Furthermore, the theory leads to an energy functional that is a compressible
variant of the classical Oseen–Frank energy (with a pressure-dependent Frank
constant) and to a compressible analogue of the Leslie–Ericksen equations. The
emergence of compressible aspects in the theory for nematic fluids enhances
our understanding of these complex systems.

###### keywords:

Kinetic theory, non-spherical rarefied gases, nematic ordering, Oseen–Frank,
Leslie–Ericksen

82C40, 82D05, 82D30

##  1 Introduction

In recent years there has been much interest in the mathematical modeling of
liquid crystals [2], both due to their important applications and the
mathematical beauty of the arising theories. Liquid crystals are the core
technology of the billion-dollar liquid crystal display industry, and are
increasingly employed for the development of novel materials [24, 4, 21]. In
many liquid crystals, nematic ordering arises due to the calamitic (rodlike)
nature of the constituent molecules; the short-range interaction potential
causes nearby particles to be aligned, or the molecules are aligned by the
action of external fields.

In this work we develop a kinetic theory for rarified calamitic gases in the
presence of nematic ordering. This is a first step towards a kinetic theory of
nematic ordering in calamitic fluids, substances with very low shear modulus
whose constituent molecules are rodlike. This class encompasses both many
liquid crystals, and many polyatomic gases, such as N2, CO, and NO.

In particular, we focus on calamitic molecules, as a generic representative of
this class. A kinetic theory of nonspherical gases was developed by C. F.
Curtiss and his collaborators [7, 9, 27, 23], and was later extended to
general nonspherical molecules [8, 10, 30]. An overview of the theory of hard
convex body fluids is given by Allen et al. [1]. McCourt et al. [26] have
applied the Boltzmann–Curtiss and the Waldmann–Snider equations to study the
behavior of polyatomic gases under the effect of external fields.

We will build on the original theory proposed by Curtiss, relying on ideas
that have been developed for ordered fluids with a nematic nature.
Particularly useful will be ideas developed in the context of the variational
theory of nematic liquid crystals [38, 32, 33, 11]. These ideas will allow us
to develop a moment closure that preserves the coupling between the laws of
linear and angular momentum. The particular closure proposed here will allow
the characterization of anisotropic phenomena caused by the emergence of a
nematic ordering.

The theory leads to a new, kinetically-motivated energy functional for the
nematic configuration that is pressure-dependent. The energy functional is a
compressible variant of the familiar one-constant Oseen–Frank energy. This
represents a key difference with the standard variational theories of nematic
liquid crystals such as the Oseen–Frank and Landau–de Gennes theories [29, 14,
38, 11]. Compressible phenomena in liquid crystals have been studied from the
molecular point of view [15, 34], with a focus on calculating the dependence
of the Frank constants on the fluid density. Our theory differs from the
previous work on polyatomic gases of McCourt et al. in that we study the
regime where the nematic ordering is fully established by the action of an
external field.

We envision this theory as a first step towards the development of a kinetic
theory for dense calamitic fluids in the presence of nematic ordering, which
may inform the calculation of material constants required in the standard
variational theories.

##  2 The Boltzmann–Curtiss equation

Our starting point will be a reformulation of the Boltzmann equation for
calamitic molecules interacting by excluded volume, known as the
Boltzmann–Curtiss equation [7, 8]:

(1)
$\partial_{t}f+\nabla_{\bm{q}}\cdot(\bm{v}f)+\nabla_{\bm{\alpha}}\cdot(\dot{\bm{\alpha}}f)=C[f,f],$

where the usual configuration space of the Boltzmann equation, consisting of
position and velocity $(\bm{q},\bm{v})\in\mathbb{R}^{3}\times\mathbb{R}^{3}$,
is enlarged to also include the Euler angles describing the orientation of
each molecule and its angular velocity. We employ as configuration space the
position and Euler angles, and their time derivatives
$(\bm{q},\bm{v},\bm{\alpha},\dot{\bm{\alpha}})\in\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}\simeq\mathbb{R}^{12}$
(all notation is described in Table 1)111We use $(\bm{v},\dot{\bm{\alpha}})$
instead of $(\bm{p},\bm{\varsigma})$ as this choice is more convenient for our
calculations. The Hessian of the Lagrangian remains positive-definite in these
coordinates, ensuring the well-posedness of the Legendre transform. Details
are given in Appendix A..

The collision operator originally proposed for the Boltzmann–Curtiss equation
is of the form

(2)
$C[f,g]\coloneqq-\int\\!\\!\\!\\!\int\\!\\!\\!\\!\int\\!\\!\\!\\!\int(g^{\prime}f^{\prime}-gf)(\bm{k}\cdot\bm{\mathfrak{g}})dSd\bm{k}d\bm{p}_{2}d\bm{\alpha}_{2}d\bm{\varsigma}_{2},$

where $\mathfrak{g}$ is the relative velocity of the point of contact,
$dS\mathrm{d}\bm{k}$ is the surface element of the excluded volume, i.e. the
excluded volume of two convex rigid bodies is the volume by the first body
that can not be accessed by the second body by the presence of the first body,
as depicted in Figure 1.

Figure 1: A section of the excluded volume between two convex molecules is
represented by the shaded green region. We also draw the tangent plane to both
bodies at the point of contact in red.

The pre-collisional and post-collisional distribution functions are denoted by
$f,g$ and $f^{\prime},g^{\prime}$, respectively.

We know that the collision operator $C[f,f]$ is $L^{2}$-orthogonal to any
invariant preserved in the collision. In particular, it is known that the
collision operator $C[f,f]$ conserves [35]:

  1. 1.

$\psi_{1}=1$, the number of particles in the system;

  2. 2.

$\psi_{2}=m\bm{v}$, the linear momentum;

  3. 3.

$\psi_{3}=\mathbb{I}\,\bm{\omega}+\bm{q}\times m\bm{v}$, the angular momentum;

  4. 4.

$\psi_{4}=\frac{1}{2}m(\bm{v}\cdot\bm{v})+\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\,\bm{\omega}$,
the kinetic energy of the system.

Table 1: Table of notation. $\bm{q}$ | – | position of the center of mass
---|---|---
$\bm{v}$ | – | velocity of the center of mass
$\bm{\alpha}$ | – | Euler angles describing the orientation
$\bm{\varsigma}$ | – | conjugate moment to the Euler angles
$f(\bm{q},\bm{v},\bm{\alpha},\bm{\alpha},t)$ | – | solution of the Boltzmann–Curtiss equation
$\bm{\mathfrak{g}}$ | $\bm{p}_{1}-\bm{p}_{2}+\bm{\omega}_{1}\times\bm{g}_{1}-\bm{\omega}_{2}\times\bm{g}_{2}$ | relative velocity of the point of contact
$m$ | – | mass of a single molecule
$n(\bm{q},t)$ | $\int\\!\\!\\!\int\\!\\!\\!\int f(\bm{q},\bm{v},\bm{\alpha},\bm{\omega},t)d\bm{v}d\bm{\alpha}d\bm{\omega}$ | particle number density
$\rho(\bm{q},t)$ | $mn(\bm{q},t)$ | fluid mass density
$\langle\\!\langle\cdot\rangle\\!\rangle$ | $\frac{1}{n}\int\\!\\!\\!\int\\!\\!\\!\int\cdot\,f(\bm{q},\bm{v},\bm{\alpha},\bm{\omega},t)d\bm{v}d\bm{\alpha}d\bm{\omega}$ | –
$\bm{v}_{0}(\bm{q},t)$ | $\langle\\!\langle\bm{v}\rangle\\!\rangle$ | macroscopic stream velocity
$\bm{V}(\bm{q},t)$ | $\bm{v}-\bm{v}_{0}$ | peculiar velocity
$\bm{\omega}_{0}(\bm{q},t)$ | $\langle\\!\langle\bm{\omega}\rangle\\!\rangle$ | macroscopic stream angular velocity
$\bm{\Omega}(\bm{q},t)$ | $\bm{\omega}-\bm{\omega}_{0}$ | peculiar angular velocity
$\bm{\eta}(\bm{q},t)$ | $\langle\\!\langle\mathbb{I}\cdot\bm{\omega}\rangle\\!\rangle$ | macroscopic stream angular momentum
$\overline{\bm{\mathbb{I}}}(\bm{q},t)$ | $\langle\\!\langle\mathbb{I}\rangle\\!\rangle$ | macroscopic inertia tensor
$I_{1},\,I_{2},\,I_{3}$ |  – |  nondimensionalised microscopic principal moment of inertia
$I$ | – |  $3\times 3$ identity matrix
$\mathbb{P}$ | $\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle$ | stress tensor, also known as pressure tensor
$\Pi(\bm{q},t)$ | $\langle\\!\langle\bm{v}\otimes\bm{v}\rangle\\!\rangle$ | linear momentum flux tensor
$\mathbb{M}$ | $\langle\\!\langle\bm{V}\otimes(\mathbb{I}\bm{\omega})\rangle\\!\rangle$ | couple stress tensor
$\Pi_{c}(\bm{q},t)$ | $\langle\\!\langle\bm{v}\otimes(\mathbb{I}\,\bm{\omega})\rangle\\!\rangle$ | angular momentum flux tensor
$\bm{\xi}$ | $\langle\\!\langle\varepsilon_{lki}(nmv_{i}v_{k})\bm{e}_{l}\rangle\\!\rangle$ |  vector antisymmetric part of $\langle\\!\langle\bm{v}\otimes\bm{v}\rangle\\!\rangle$
$\bm{Q}$ | $\frac{1}{2}\langle\\!\langle\bm{V}\Big{(}m\lvert\bm{V}\rvert^{2}+\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\Big{)}\rangle\\!\rangle$ | heat flux
$\theta$ | $\frac{1}{2}m\lvert\bm{V}\rvert^{2}+\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}$ | peculiar kinetic energy
$\psi_{0}(\bm{q},t)$ | $\langle\\!\langle\theta\rangle\\!\rangle$ | internal energy
$\psi(\bm{q},t)$ | $\frac{1}{2}\Big{[}m\lvert\bm{v}\rvert^{2}+\bm{\omega}\cdot\mathbb{I}\bm{\omega}\Big{]}$ | total energy
$\psi_{K}(\bm{q},t)$ | $\frac{1}{2}\langle\\!\langle\Big{[}m\lvert\bm{v}_{0}\rvert^{2}+\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\Big{]}\rangle\\!\rangle$ | macroscopic kinetic energy
$p_{K}(\bm{q},t)$ | $tr[\mathbb{P}^{(0)}]=\frac{\rho}{m}I_{1}I_{2}I_{3}\langle\\!\langle\theta\rangle\\!\rangle$ | kinetic pressure
$\bm{\nu}$ |  (45) | nematic director
$k_{B}$ | $1.380649\times 10^{23}\;J\cdot K^{-1}$ | Boltzmann constant
$N_{A}$ | $6.02214076\times 10^{23}\;mol^{-1}$ | Avogadro constant
$R$ | $N_{A}k_{B}$ | universal gas constant
$\mathcal{N}$ |  – |  number of degrees of freedom in the microscopic Hamiltonian

Testing (1) against a collision invariant $\psi$ and integrating by parts,
since

(3) $\lim_{\lVert\dot{\bm{\alpha}}\rVert\to\infty}(\psi f)=0,$

we obtain

(4)
$\partial_{t}(n\langle\\!\langle\psi\rangle\\!\rangle)+\nabla_{\bm{q}}\cdot(n\langle\\!\langle\bm{v}\psi\rangle\\!\rangle)-n\langle\\!\langle\dot{\bm{\alpha}}\cdot\nabla_{\bm{\alpha}}\psi\rangle\\!\rangle=0,$

where the symbol $\langle\\!\langle\cdot\rangle\\!\rangle$ denotes the average
over $\bm{p}$, $\bm{\alpha}$ and $\bm{\varsigma}$, i.e.

(5)
$\langle\\!\langle\psi\rangle\\!\rangle\coloneqq\frac{1}{n}\int\\!\\!\\!\int\\!\\!\\!\int\psi\,f(\bm{r},\bm{v},\bm{\alpha},\bm{\omega},t)d\bm{v}d\bm{\alpha}d\bm{\omega}.$

Substituting the first collision invariant $\psi_{1}\equiv 1$ into (4), we
obtain the equation of conservation of number of particles:

(6) $\partial_{t}n+\nabla_{\bm{q}}\cdot(n\bm{v}_{0})=0.$

Multiplying by $m$ derives the well-known law of the conservation of mass,
i.e.

(7) $\partial_{t}\rho+\nabla_{\bm{q}}\cdot(\rho\bm{v}_{0})=0.$

Substituting the second collision invariant $\psi_{2}=m\bm{v}$ into (4), we
obtain

(8)
$\partial_{t}(\rho\bm{v}_{0})+\nabla_{\bm{q}}\cdot\Big{[}\rho\langle\\!\langle\bm{v}\otimes\bm{v}\rangle\\!\rangle\Big{]}=0,$

where $\langle\\!\langle\bm{v}\otimes\bm{v}\rangle\\!\rangle$ is also known as
the linear momentum flux tensor and denoted as $\Pi$. The above equation is a
form of the law of linear momentum well known in continuum mechanics; to
rewrite it in a more standard form, we observe that introducing
$\bm{v}_{0}=\langle\\!\langle\bm{v}\rangle\\!\rangle$ we have

(9) $\displaystyle\Pi=\langle\\!\langle\bm{v}\otimes\bm{v}\rangle\\!\rangle$
$\displaystyle=\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle+\langle\\!\langle\bm{v}_{0}\otimes\bm{V}\rangle\\!\rangle+\langle\\!\langle\bm{V}\otimes\bm{v}_{0}\rangle\\!\rangle+\bm{v}_{0}\otimes\bm{v}_{0}$
$\displaystyle=\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle+\bm{v}_{0}\otimes\bm{v}_{0},$

where
$\langle\\!\langle\bm{v}_{0}\otimes\bm{V}\rangle\\!\rangle=\bm{v}_{0}\otimes\langle\\!\langle\bm{V}\rangle\\!\rangle=0$
because $\langle\\!\langle\bm{V}\rangle\\!\rangle=0$. Now we expand the time
derivative in (8) and use (7) to obtain

(10a)
$\displaystyle(\partial_{t}\rho)\bm{v}_{0}+\rho(\partial_{t}\bm{v}_{0})+\nabla_{\bm{q}}\cdot\Big{[}\rho\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle+\rho\langle\\!\langle\bm{v}_{0}\otimes\bm{v}_{0}\rangle\\!\rangle\Big{]}=0,$
(10b)
$\displaystyle-\Big{[}\nabla_{\bm{q}}\cdot(\rho\bm{v}_{0})\Big{]}\bm{v}_{0}+\rho(\partial_{t}\bm{v}_{0})+\nabla_{\bm{q}}\cdot\Big{[}\rho\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle+\rho\langle\\!\langle\bm{v}_{0}\otimes\bm{v}_{0}\rangle\\!\rangle\Big{]}=0,$
(10c)
$\displaystyle\rho(\partial_{t}\bm{v}_{0})+\rho\Big{[}(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}\rho\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle\Big{]}=0,$
(10d)
$\displaystyle\rho\Big{[}\partial_{t}\bm{v}_{0}+(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot(\rho\mathbb{P})=0,$

where the last equation is the usual law of linear momentum and
$\rho\mathbb{P}$ is the Cauchy stress tensor up to a change of sign.

Substituting the third collision invariant
$\psi_{3}=\mathbb{I}\cdot\bm{\omega}+m(\bm{q}\times\bm{v})$ into (4), we
obtain

(11)
$\partial_{t}(n\bm{\eta})+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}\otimes\bm{\eta}\rangle\\!\rangle\Big{]}+\partial_{t}(nm\langle\\!\langle\bm{q}\times\bm{v}\rangle\\!\rangle)+\nabla_{\bm{q}}\cdot\Big{[}nm\langle\\!\langle\bm{v}\otimes(\bm{q}\times\bm{v})\rangle\\!\rangle\Big{]}=0,$

where $\bm{\eta}=\mathbb{I}\bm{\omega}$ is the intrinsic angular momentum
field which measures the angular momentum of the fluid per unit mass. To
simplify the last equation we will use the following well-known lemma.

######  Lemma 2.1 (Result 3.5 in [17]).

Given a second order tensor $S=S_{ij}\bm{e}_{i}\otimes\bm{e}_{j}$, the
following identity holds:

(12)
$\varepsilon_{lki}\partial_{q_{j}}(S_{ij})r_{k}\bm{e}_{l}=-\varepsilon_{lki}S_{ik}\bm{e}_{l}+\partial_{q_{j}}(\varepsilon_{lki}q_{k}S_{ij})\bm{e}_{l}.$

We will also use the following lemma. It was used but not proven in [7], so we
include the proof here for completeness.

######  Lemma 2.2.

The following identity holds:

(13)
$\partial_{t}(nm\langle\\!\langle\bm{q}\times\bm{v}\rangle\\!\rangle)+\nabla_{\bm{q}}\cdot\Big{[}nm\langle\\!\langle\bm{v}\otimes(\bm{q}\times\bm{v})\rangle\\!\rangle\Big{]}=-\bm{\xi},$

where

(14)
$\bm{\xi}\coloneqq\langle\\!\langle\varepsilon_{lki}(mnv_{i}v_{k})\bm{e}_{l}\rangle\\!\rangle.$

######  Proof 2.3.

We begin by taking a cross product with $\bm{q}$ on both sides of (8) to
obtain

(15)
$\partial_{t}(\rho\bm{v}_{0})\times\bm{q}+\Big{[}\nabla_{\bm{q}}\cdot(\rho\Pi)\Big{]}\times\bm{q}=0.$

We then expand $\rho$, $\Pi$ and $\bm{v}_{0}$ in order to bring the cross
product with $\bm{q}$ inside $\langle\\!\langle\cdot\rangle\\!\rangle$ where
possible. Furthermore, since $\langle\\!\langle\cdot\rangle\\!\rangle$ does
not depend on time we can bring it inside the time derivative to obtain

(16)
$\langle\\!\langle\partial_{t}(nm\bm{v})\times\bm{q}\rangle\\!\rangle+\langle\\!\langle\nabla_{\bm{q}}\cdot(nm\bm{v}\otimes\bm{v})\times\bm{q}\rangle\\!\rangle=0.$

where the term $\langle\\!\langle
nm\bm{v}\times\partial_{t}\bm{q}\rangle\\!\rangle$ disappears because
$\partial_{t}\bm{q}\times\bm{v}=0$. Now we rewrite the divergence term using
tensor notation and apply Lemma 2.1:

(17)
$\partial_{r_{j}}(nmv_{i}v_{j})r_{k}\varepsilon_{lki}\bm{e}_{l}=-\varepsilon_{lki}(nmv_{i}v_{k})\bm{e}_{l}+\partial_{r_{j}}(nm\varepsilon_{lki}r_{k}v_{i}v_{j})\bm{e}_{l}.$

Defining $\bm{\xi}$ as in (14), the above expression can be rewritten as:

(18)
$\langle\\!\langle\nabla_{\bm{q}}\cdot(nm\bm{v}\otimes\bm{v})\times\bm{q}\rangle\\!\rangle=-\bm{\xi}+\nabla_{\bm{q}}\cdot\Big{[}nm\bm{v}\otimes(\bm{v}\times\bm{q})\Big{]}.$

Applying now Lemma 2.2 to (11) we get

(19)
$\partial_{t}(n\bm{\eta})+\nabla_{\bm{q}}\cdot\left(n\Pi_{c}\right)=\bm{\xi},$

where
$\Pi_{c}=\langle\\!\langle\bm{v}\otimes(\mathbb{I}\,\bm{\omega})\rangle\\!\rangle$
is the angular momentum flux tensor and the subscript $\cdot_{c}$ denotes its
connection with the couple stress tensor that we will show in a moment. As we
did for the linear momentum flux tensor we would like to rewrite the angular
momentum flux tensor in terms of the peculiar velocity, i.e.

(20)
$\Pi_{c}=\langle\\!\langle\bm{v}_{0}\otimes(\mathbb{I}\,\bm{\omega})\rangle\\!\rangle+\langle\\!\langle\bm{V}\otimes(\mathbb{I}\,\bm{\omega})\rangle\\!\rangle.$

Expanding the time derivative and using (7) we obtain

(21)
$\rho\Big{[}\partial_{t}\bm{\eta}+(\nabla_{\bm{q}}\bm{\eta})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot\left(\rho\mathbb{M}\right)=\bm{\xi},$

where $\mathbb{M}$ is defined as
$\langle\\!\langle\bm{V}\otimes(\mathbb{I}\,\bm{\omega})\rangle\\!\rangle$,
and it is the couple Cauchy stress tensor up to a change of sign.

In summary, systematically testing (1) against the first three collision
invariants retrieves the usual governing equations of continuum mechanics,
i.e. the conservation laws of mass, of linear momentum, and the balance law of
angular momentum:

(22a) $\displaystyle\partial_{t}\rho+\nabla_{\bm{q}}\cdot(\rho\bm{v}_{0})=0,$
(22b)
$\displaystyle\rho\Big{[}\partial_{t}\bm{v}_{0}+(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot(\rho\,\mathbb{P})=0,$
(22c)
$\displaystyle\rho\Big{[}\partial_{t}\bm{\eta}+(\nabla_{\bm{q}}\bm{\eta})\bm{v}_{0}\Big{]}+\nabla\cdot(\rho\,\mathbb{M})=\bm{\xi}.$

Now we notice that the law of angular momentum (22c) can be simplified. Since
$\Pi$ is by definition symmetric then $\bm{\xi}=0$, as formalised in the
following proposition.

######  Proposition 2.4.

Since $\Pi$ is symmetric by definition, we know that $\bm{\xi}=0$.

######  Proof 2.5.

Recall that $\Pi$ is defined as
$\langle\\!\langle\bm{v}\otimes\bm{v}\rangle\\!\rangle$, and since $\Pi$ is
symmetric we have that

(23)
$0=\varepsilon_{lki}nm\Big{[}\Pi_{ki}-\Pi^{T}_{ki}\Big{]}\bm{e}_{l}=\varepsilon_{lki}nm\Big{[}\langle\\!\langle
v_{i}v_{k}\rangle\\!\rangle-\langle\\!\langle
v_{k}v_{i}\rangle\\!\rangle\Big{]}\bm{e}_{l}=2\bm{\xi},$

where in the last equality we have used
$\varepsilon_{lki}=-\varepsilon_{lik}$.

######  Corollary 2.6.

Considering equation (22c) when $\mathbb{M}=0$ we have
$\rho\dot{\bm{\eta}}=\bm{\xi}$. Therefore since $\bm{\xi}=0$ we have that if
$\mathbb{M}=0$ then the angular momentum is conserved.

Notice that the same observation was also made in [8], after which Curtiss and
Dahler wrote:

> From this, it follows that in the absence of external torques the spin
> angular momentum density of a dilute gas is of no physical consequence since
> it is coupled neither to the surroundings of the system nor to the density,
> velocity, and temperature fields of the fluid. Because of this we may choose
> the $\mathbb{M}$ (and hence $\bm{\omega_{0}}$) to be identically zero.

The next sections of the chapter will be devoted to developing a theory that
will allow us to refute this conclusion.

##  3 Leslie–Ericksen rate of work hypothesis

To take into account the anisotropic behaviour that stems from the equation of
angular momentum, our objective will be to find a coupling between (22b) and
(22c). Indeed, in the absence of such coupling, one can assume $\mathbb{M}$ is
identically zero. If $\mathbb{M}$ vanishes this will result in the inability
of rarefied calamitic fluids to undergo reorientation under ultrasonic wave
propagation, a phenomenon that has been observed in other denser nematic
fluids [3, 31]. It is worth mentioning that such coupling exists in the
standard dynamic theory of liquid crystals, arising as the Ericksen tensor in
the Leslie–Ericksen equations.

For this reason, our first task will be to connect (22) with the
Leslie–Ericksen theory. In order to achieve this we consider the fourth
collision invariant
$\psi_{4}=\frac{1}{2}m(\bm{v}\cdot\bm{v})+\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}$
and substitute it into (4):

(24)
$\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}m(\bm{v}\cdot\bm{v})+\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}\frac{1}{2}m(\bm{v}\cdot\bm{v})+\bm{v}\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{]}=0.$

We now rewrite the previous expression using the notion of peculiar velocity.
In particular, using the identity
$\langle\\!\langle\bm{v}\cdot\bm{v}\rangle\\!\rangle=\langle\\!\langle\bm{V}\cdot\bm{V}\rangle\\!\rangle+\langle\\!\langle\bm{v}_{0}\cdot\bm{v}_{0}\rangle\\!\rangle$
yields

(25)
$\displaystyle\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{)}$
$\displaystyle+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{)}+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})\rangle\\!\rangle\Big{)}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}_{0}\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\langle\\!\langle\frac{1}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}nm\mathbb{P}\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}(\bm{v}_{0}\cdot\bm{v}_{0})\rangle\\!\rangle\Big{]}=0.$

First we notice that the last term vanishes because
$\langle\\!\langle\bm{V}(\bm{v}_{0}\cdot\bm{v}_{0})\rangle\\!\rangle=\langle\\!\langle\bm{V}\rangle\\!\rangle(\bm{v}_{0}\cdot\bm{v}_{0})=0$.
Next we develop a portion of the terms involved in the previous equation, i.e.

(26)
$\displaystyle\partial_{t}\Big{(}n\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})\Big{]}$
$\displaystyle=nm(\partial_{t}\bm{v}_{0})\cdot\bm{v}_{0}+m(\partial_{t}n)\frac{\bm{v}_{0}\cdot\bm{v}_{0}}{2}+\nabla_{\bm{q}}\cdot\Big{[}n{\bm{v}_{0}\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})}\Big{]}$
$\displaystyle=nm(\partial_{t}\bm{v}_{0})\cdot\bm{v}_{0}-\nabla_{\bm{q}}\cdot\Big{[}nm\bm{v}_{0}\Big{]}\frac{\bm{v}_{0}\cdot\bm{v}_{0}}{2}+\nabla_{\bm{q}}\cdot\Big{[}n{\bm{v}_{0}\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})}\Big{]}$
$\displaystyle=nm(\partial_{t}\bm{v}_{0})\cdot\bm{v}_{0}+nm\Big{[}(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}\cdot\bm{v}_{0}=-\bm{v}_{0}\cdot\Big{(}\nabla_{\bm{q}}\cdot(nm\mathbb{P})\Big{)},$

where we have employed the chain rule, Leibniz rule, (22a), and the scalar
product of (22b) with $\bm{v}_{0}$. Substituting the previous equation inside
(25) we obtain the following expression:

(27) $\displaystyle\partial_{t}\Big{(}n\langle\\!\langle$
$\displaystyle\frac{1}{2}m\bm{V}\cdot\bm{V}\rangle\\!\rangle\Big{)}\\!+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{)}-\bm{v}_{0}\\!\cdot\\!\Big{(}\nabla_{\bm{q}}\cdot(nm\mathbb{P})\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\frac{\bm{v}_{0}}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}nm\mathbb{P}\bm{v}_{0}\Big{]}=0.$

Combining (27) with the following well-known identity [17, Result 2.11]

(28) $\nabla_{\bm{q}}\cdot(S^{T}\bm{v}_{0})=(\nabla_{\bm{q}}\cdot
S)\cdot\bm{v}_{0}+S:\nabla_{\bm{q}}\bm{v}_{0},$

we obtain the following equation:

(29)
$\displaystyle\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}m\bm{V}\cdot\bm{V}\rangle\\!\rangle\Big{)}$
$\displaystyle+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{)}+\Big{(}nm\mathbb{P}\Big{)}:\nabla_{\bm{q}}\bm{v}_{0}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\frac{\bm{v}_{0}}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}m(\bm{V}\cdot\bm{V})\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle\Big{]}=0.$

We are now left to focus our attention on the terms involving the angular
velocity. In particular we begin rewriting the previous equations in terms of
the peculiar angular velocity, using the fact that
$\langle\\!\langle\mathbb{I}\bm{\Omega}\rangle\\!\rangle$ vanishes:

(30) $\displaystyle\partial_{t}$
$\displaystyle\Big{(}n\langle\\!\langle\frac{1}{2}m\bm{V}\cdot\bm{V}\rangle\\!\rangle\Big{)}+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{)}+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{)}+\Big{(}nm\mathbb{P}\Big{)}:\nabla\bm{v}_{0}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\frac{\bm{v}_{0}}{2}m\lvert\bm{V}\rvert^{2}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\frac{\bm{V}}{2}m\lvert\bm{V}\rvert^{2}\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}_{0}\frac{1}{2}\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}_{0}\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{(}n\mathbb{M}^{T}\bm{\omega}_{0}\Big{)}=0,$

where the second term in the last row vanishes as before because
$\langle\\!\langle\bm{V}\rangle\\!\rangle=0$. Now we develop a portion of the
terms involved in the previous equation, i.e.

(31)
$\displaystyle\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\langle\\!\langle\frac{1}{2}\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{]}$
$\displaystyle=(\partial_{t}n)\frac{1}{2}\langle\\!\langle\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle+n\partial_{t}\Big{(}\langle\\!\langle\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{)}\cdot\bm{\omega}_{0}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\frac{1}{2}\bm{\omega}_{0}\cdot\langle\\!\langle\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{]}$
$\displaystyle=-\frac{1}{2}\nabla_{r}\cdot(n\bm{v}_{0})\langle\\!\langle\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle+n\partial_{t}\Big{(}\langle\\!\langle\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{)}\cdot\bm{\omega}_{0}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\frac{1}{2}\bm{\omega}_{0}\cdot\langle\\!\langle\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{]}$
$\displaystyle=n\partial_{t}\Big{(}\langle\\!\langle\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\Big{)}\cdot\bm{\omega}_{0}+n\Big{[}\nabla_{\bm{q}}\big{(}\langle\\!\langle\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle\big{)}\bm{v}_{0}\Big{]}\cdot\bm{\omega}_{0}=-\bm{\omega}_{0}\cdot\Big{(}\nabla_{\bm{q}}\cdot(n\mathbb{M})\Big{)}$
$\displaystyle=\left(n\mathbb{M}\right):\nabla_{\bm{q}}\bm{\omega}_{0}-\nabla_{\bm{q}}\cdot\Big{(}n\mathbb{M}^{T}\bm{\omega}_{0}\Big{)},$

where we employ the chain rule, the Leibniz rule, (6), the scalar product of
(22c) with $\bm{\omega}_{0}$, and (28). We now substitute the previous
equation into (30) and obtain

(32)
$\displaystyle\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}m\bm{V}\cdot\bm{V}\rangle\\!\rangle\Big{)}$
$\displaystyle+\partial_{t}\Big{(}n\langle\\!\langle\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{)}+\Big{(}nm\mathbb{P}\Big{)}:\nabla\bm{v}_{0}$
$\displaystyle+\left(n\mathbb{M}\right):\nabla_{\bm{q}}\bm{\omega}_{0}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\frac{\bm{v}_{0}}{2}m\lvert\bm{V}\rvert^{2}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\frac{\bm{V}}{2}m\lvert\bm{V}\rvert^{2}\rangle\\!\rangle\Big{]}$
$\displaystyle+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{v}_{0}\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{]}+\nabla_{\bm{q}}\cdot\Big{[}n\langle\\!\langle\bm{V}\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{]}=0.$

Finally, we rewrite the previous equation using the heat flux and the kinetic
energy $\theta$, respectively

(33)
$\bm{\mathfrak{Q}}=\frac{1}{2}\langle\\!\langle\bm{V}\Big{(}m\lvert\bm{V}\rvert^{2}+\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\Big{)}\rangle\\!\rangle,\qquad\theta=\frac{1}{2}m\lvert\bm{V}\rvert^{2}+\frac{1}{2}\bm{\Omega}\cdot\mathbb{I}\bm{\Omega},$

to obtain the following theorem of power expended:

(34)
$\displaystyle\partial_{t}\Big{(}n\langle\\!\langle\theta\rangle\\!\rangle\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\langle\\!\langle\theta\rangle\\!\rangle\Big{]}+(\rho\mathbb{P}):\nabla_{\bm{q}}\bm{v}_{0}$
$\displaystyle+(n\mathbb{M}):\nabla_{\bm{q}}\bm{\omega}_{0}+\nabla_{\bm{q}}\cdot\bm{\mathfrak{Q}}=0.$

We will assume the heat flux has a constant sign. In particular for simplicity
we use the convention that $\nabla_{\bm{q}}\cdot\mathfrak{Q}\leq 0$ and
therefore

$\displaystyle\partial_{t}\Big{(}n\langle\\!\langle\theta\rangle\\!\rangle\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\langle\\!\langle\theta\rangle\\!\rangle\Big{]}$
$\displaystyle+(\rho\mathbb{P}):\nabla_{\bm{q}}\bm{v}_{0}+(n\mathbb{M}):\nabla_{\bm{q}}\bm{\omega}_{0}\geq
0.$

Using (6) together with the Leibniz rule we can rewrite (34) as:

$\displaystyle
n\Big{(}\partial_{t}\langle\\!\langle\theta\rangle\\!\rangle\Big{)}+\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\langle\\!\langle\theta\rangle\\!\rangle\Big{]}-\nabla_{\bm{q}}\cdot\Big{[}n\bm{v}_{0}\Big{]}\langle\\!\langle\theta\rangle\\!\rangle$
$\displaystyle+(\rho\mathbb{P}):\nabla_{\bm{q}}\bm{v}_{0}+(n\mathbb{M}):\nabla_{\bm{q}}\bm{\omega}_{0}\geq
0,$

which can be rewritten by making use of the definition of the total derivative
of an Eulerian field and the Leibniz rule as:

(35) $\rho\dot{\psi_{0}}+(\rho
m\mathbb{P}):\nabla_{\bm{q}}\bm{v}_{0}+(\rho\mathbb{M}):\nabla_{\bm{q}}\bm{\omega}_{0}\geq
0,$

where $\psi_{0}(\bm{q},t)$ is the internal energy and is defined as
$\psi_{0}(\bm{q},t)=\langle\\!\langle\theta\rangle\\!\rangle$.

A key observation of this work is the fact that (35), when integrated over a
domain $\Omega\subset\mathbb{R}^{3}$, is similar to the rate of work
hypothesis that forms the starting point of the modern development of
Leslie–Ericksen theory for the dynamics of liquid crystals. We will show in
the next sections that (35) relates to the rate of work hypothesis postulated
in [12] and is precisely the same rate of work hypothesis presented in [22].
Last we would like to point out that (35) is not only a reconciliation between
the well-established continuum mechanics theory of nematic fluids and the
kinetic theory approach here proposed but also a validation of the rate of
work hypothesis postulated in [22]. In fact rather than using the integral
version of (35) as an assumption, as in [12] and [22], the rate of work
hypothesis has been derived starting from the foundations of Boltzmann–Curtiss
kinetic theory.

##  4 Maxwellian distribution

In this section we derive constitutive relations for the stress tensor, the
couple stress tensor, and the internal energy from the Maxwellian distribution
for the Boltzmann–Curtiss equation (1), i.e. that $f^{(0)}$ satisfying
$C[f^{(0)},f^{0}]=0$. In seminal work by Curtiss [7, 8], this distribution was
obtained using formal computations from statistical mechanics. The Maxwellian
is given by

(36)
$f^{(0)}(\bm{\alpha},\bm{V},\bm{\Omega})=\frac{nQ\sin(\alpha_{2})}{\int_{\mathbb{R}^{3}}Q\sin(\alpha_{2})\,d\alpha}\frac{m^{\frac{3}{2}}(I_{1}I_{2}I_{3})^{\frac{1}{2}}}{(\frac{4}{\mathcal{N}}\pi\langle\\!\langle\theta\rangle\\!\rangle)^{3}}\exp\Big{[}-m\frac{\lvert\bm{V}\rvert^{2}}{\frac{4}{\mathcal{N}}\langle\\!\langle\theta\rangle\\!\rangle}-\frac{\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}}{\frac{4}{\mathcal{N}}\langle\\!\langle\theta\rangle\\!\rangle}\Big{]}$

where
$Q=\exp(\frac{\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}}{\frac{2}{3}\langle\\!\langle\theta\rangle\\!\rangle})$
and both $\omega_{0}$ and $\theta$ are assumed to remain constant at
equilibrium. Notice that the distribution here presented is the “absolute
Maxwellian” since it is spatially homogeneous, i.e. we are assuming that at
equilibrium $n,\bm{V},\bm{\Omega},\theta$ are spatially independent
quantities.

The only difference between the formulation of the Maxwellian considered here
and the one presented in [7, 8] is that (36) is expressed in terms of internal
energy $\langle\\!\langle\theta\rangle\\!\rangle$ rather than in terms of
kinetic temperature. If one is interested in obtaining (36) from [7, 8] one
only has to apply the equipartition of energy theorem, i.e.

(37) $\langle\\!\langle\theta\rangle\\!\rangle=\frac{\mathcal{N}}{2}k_{B}T,$

where $T$ is the kinetic temperature, $\mathcal{N}$ is the number of degrees
of freedom appearing in the Hamiltonian of a single molecule and $k_{B}$ is
the Boltzmann constant.

What is the correct value of $\mathcal{N}$ for a polyatomic gas? To answer
this question we consider linear diatomic gases, such as H2, N2 and O2. We can
think of these polyatomic gases as two spheres connected by a spring and
therefore the Hamiltonian of this model would have seven degrees of freedom:
the moments of the centers of mass of the two spheres, and the length of the
spring connecting the two centers of mass. Yet we know that vibration modes
for the previously mentioned diatomic molecules are excited only above
$2000K$, a temperature well above the regime of interest here. This suggests
that we can ignore the degrees of freedom corresponding to the spring
connecting the center of mass of each atom and model the polyatomic gas
molecule as a rigid body. Lastly, in a later section, the hypothesis that our
molecules are slender bodies will be essential to ignore some terms in the
expansion of the inertia tensor. In the slender body limit, we can regard our
molecules as segments in space, thus losing one rotational degree of freedom
in the kinetic energy; more details are provided in Appendix A. This suggests
that in our regime of interest, we have $\mathcal{N}=5$ and the correct
relation arising from the equipartition of energy theorem would be

(38) $\langle\\!\langle\theta\rangle\\!\rangle=\frac{5}{2}k_{B}T.$

We notice that the emergence of a nematic ordering implies that the remaining
five rotational degrees of freedom are not independent among molecules and
therefore the heat capacity of the nematic gas should be less than
$\frac{5}{2}R$ per mole, where $R$ denotes the gas constant. These
difficulties in understanding the correct way to apply the equipartition of
energy theorem are reflected in the experimental observation of the heat
capacity of nematic fluids such as liquid crystals [28]. In particular, in
[28] the author observes a sharp increase in the specific heat capacity of the
liquid crystal MBBA at constant pressure when the temperature is increased
from $319K$ to $324K$. Such behaviour is anomalous compared to what is
observed for polyatomic gases such as N2, for which the specific heat capacity
at constant pressure is constant between $123K$ and $573K$ [5]. For this
reason, we decided to treat non-spherical rarefied gases in the presence of
nematic alignment in a standard manner, i.e. assuming (38), and devote future
work to this particular issue.

We now proceed to use the Maxwellian distribution (36) to obtain constitutive
relations for the stress tensor, the couple stress tensor, and the internal
energy. We can compute the pressure tensor near the equilibrium by evaluating
$\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle$ using the Maxwellian
$f^{(0)}$ as a probability distribution. It turns out that the pressure tensor
near the equilibrium is nothing more than the variance of the Maxwellian
distribution, i.e.

(39)
$\mathbb{P}^{(0)}=\langle\\!\langle\bm{V}\otimes\bm{V}\rangle\\!\rangle_{\sim
f^{(0)}}=\frac{2(I_{1}I_{2}I_{3})^{\frac{1}{2}}}{5m}\langle\\!\langle\theta\rangle\\!\rangle
I.$

Following the same reasoning we can compute also the couple stress tensor,
which turns out to be zero:

(40)
$\mathbb{M}^{(0)}=\langle\\!\langle\bm{V}(\mathbb{I}\bm{\omega})\rangle\\!\rangle_{\sim
f^{(0)}}=\mathbb{E}_{f^{(0)}_{\bm{\Omega}}}\big{[}\mathbb{I}\bm{\omega}\big{]}\mathbb{E}_{f^{(0)}_{\bm{V}}}\big{[}\bm{V}\big{]}=0.$

This follows from the Fubini–Tonelli Theorem, the fact that
$f^{(0)}=f^{(0)}_{\bm{V}}f^{(0)}_{\bm{\Omega}}f^{(0)}_{\bm{\alpha}}$, and that
$f^{(0)}_{\bm{V}}$ has zero mean. In particular, since we know $\bm{\xi}$ is
also zero, we can use (22c) to deduce that $\dot{\bm{\eta}}=0$. In what
follows, we will be interested in weighting the pressure tensor and the couple
stress tensor by the fluid density, and for this reason, we introduce the
kinetic pressure

(41) $p_{K}\coloneqq
tr[\rho\mathbb{P}^{(0)}]=\frac{6}{5}\frac{\rho}{m}(I_{1}I_{2}I_{3})^{\frac{1}{2}}\langle\\!\langle\theta\rangle\\!\rangle.$

##  5 Oseen–Frank energy

Since we now know that energy stored in the particle ensemble is given by
$\psi=\langle\\!\langle\frac{1}{2}m\lvert\bm{v}\rvert^{2}+\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle$
we can proceed by rewriting,

$\displaystyle\psi=\frac{1}{2}\langle\\!\langle
m\lvert\bm{v}\rvert^{2}+\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle$
$\displaystyle=\frac{m}{2}\langle\\!\langle\lvert\bm{V}\rvert^{2}\rangle\\!\rangle+\frac{m}{2}\lvert\bm{v}_{0}\rvert^{2}+\frac{1}{2}\langle\\!\langle\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle+\frac{1}{2}\langle\\!\langle\bm{\omega}\cdot\mathbb{I}\rangle\\!\rangle\cdot\bm{\omega}_{0}$
$\displaystyle=\frac{1}{2}\Big{[}\langle\\!\langle
m\lvert\bm{V}\rvert^{2}\rangle\\!\rangle+\langle\\!\langle\bm{\Omega}\cdot\mathbb{I}\bm{\Omega}\rangle\\!\rangle\Big{]}+\frac{1}{2}\Big{[}m\lvert\bm{v}_{0}\rvert^{2}+\bm{\omega_{0}}\cdot\overline{\mathbb{I}}\bm{\omega}_{0}\Big{]}$
(42) $\displaystyle=:\psi_{0}+\psi_{K},$

where we make use of the fact that both $\langle\\!\langle V\rangle\\!\rangle$
and $\langle\\!\langle\mathbb{I}\bm{\Omega}\rangle\\!\rangle$ are zero, and
that $\bm{\omega}_{0}=\overline{\mathbb{I}}^{{}^{-1}}\bm{\eta}$. The first
term in (42) is the internal energy
$\langle\\!\langle\theta\rangle\\!\rangle$, from now on denoted $\psi_{0}$ to
avoid using the $\langle\\!\langle\cdot\rangle\\!\rangle$ notation. The second
term is the kinetic energy. It may be counterintuitive to identify
$\psi_{0}(\bm{q},t)$ as the internal energy, but from the decomposition (42)
we can obtain:

(43) $\psi_{0}=\psi-\psi_{K}$

which can be interpreted as the well-known decomposition of internal energy
into total energy minus macroscopic kinetic energy.

We will next derive a slightly different decomposition in order to relate some
terms of $\psi(\bm{q},t)$ to the classical Oseen–Frank energy functional [14,
29]. It is a well-known fact that the inertia tensor of a needle-shaped
molecule, i.e. the vanishing-girth limit of a calamitic molecule, can be
decomposed as:

(44) $\mathbb{I}=\lambda_{1}\Big{(}I-\bm{\nu}\otimes\bm{\nu}^{T}\Big{)}$

with $\bm{\nu}\in\mathbb{S}^{2}$. Motivated by (44) we assume the following
decomposition holds for calamitic molecules:

(45)
$\mathbb{I}=\lambda_{1}\Big{(}I-\bm{\nu}\otimes\bm{\nu}\Big{)}+\mathcal{O}(\varepsilon),$

where now $\bm{\nu}(\bm{q},t)\in\mathbb{S}^{2}$ is the nematic vector, first
introduced in [12], and $\varepsilon\approx\left(\delta/\ell\right)^{2}$,
where $\delta$ and $\ell$ are respectively the diameter of the spherical
hemisphere at the end of the calamitic molecule and the height of the
calamitic molecule. Notice that $\bm{\nu}$ is normally defined in terms of the
orientation distribution function (ODF), i.e.
$\bm{\nu}=\langle\\!\langle\bm{\mathcal{N}}(\bm{\alpha})\rangle\\!\rangle$,
where $\bm{\mathcal{N}}(\bm{\alpha})$ is the unit vector corresponding to the
Euler angles $\bm{\alpha}$.

The introduction of a nematic director allows for substantial simplifications
of computations involving the average of the inertia tensor, since under the
hypothesis of the emergence of a nematic ordering we are allowed to move the
inertia tensor expressed as in (45) outside the
$\langle\\!\langle\cdot\rangle\\!\rangle$. Since by theorem A.4 the material
time derivative $\dot{\bm{\nu}}$ can be computed as
$\bm{\omega}_{0}\times\bm{\nu}=\dot{\bm{\nu}}$, we can obtain:

(46)
$\displaystyle\langle\\!\langle\bm{\omega}_{0}\cdot\mathbb{I}\bm{\omega}_{0}\rangle\\!\rangle=\lambda_{1}\Big{[}\bm{\omega}_{0}\cdot\bm{\omega}_{0}-\langle\\!\langle(\bm{\omega}_{0}\cdot\bm{\nu})(\bm{\omega_{0}}\cdot\bm{\nu})\rangle\\!\rangle\Big{]}+\mathcal{O}(\varepsilon)=\lambda_{1}\dot{\bm{\nu}}\cdot\dot{\bm{\nu}}+\mathcal{O}(\varepsilon),$

using the triple product identity
$\dot{\bm{\nu}}\cdot\dot{\bm{\nu}}=(\bm{\omega}_{0}\times\bm{\nu})\cdot(\bm{\omega}_{0}\times\bm{\nu})=(\bm{\omega}_{0}\cdot\bm{\omega}_{0})(\bm{\nu}\cdot\bm{\nu})-(\bm{\omega}_{0}\cdot\bm{\nu})(\bm{\nu}\cdot\bm{\omega}_{0})$
twice. This simplifies the kinetic energy

(47)
$\psi_{K}=\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})+\frac{\lambda_{1}}{2}(\dot{\bm{\nu}}\cdot\dot{\bm{\nu}})+\mathcal{O}(\varepsilon)$

which in the vanishing girth limit, i.e. $\varepsilon\searrow 0$, becomes
$\psi_{K}(\bm{q},t)=\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})+\frac{\lambda_{1}}{2}(\dot{\bm{\nu}}\cdot\dot{\bm{\nu}})$.
We proceed with our analysis of $\psi_{0}(\bm{q},t)$, making use of the fact
that by theorem A.4
$\bm{\omega}\times\bm{\nu}=\partial_{t}\bm{\nu}+(\nabla_{\bm{q}}\bm{\nu})\bm{v}$,
hence we can expand the
$\langle\\!\langle\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle$ in
$\psi$ as

$\displaystyle
2\lambda_{1}^{-1}\langle\\!\langle\bm{\omega}\cdot\mathbb{I}\bm{\omega}\rangle\\!\rangle$
$\displaystyle=\langle\\!\langle(\bm{\omega}\times\bm{\nu})\cdot(\bm{\omega}\times\bm{\nu})\rangle\\!\rangle=\langle\\!\langle\Big{(}\partial_{t}\bm{\nu}+(\nabla_{r}\bm{\nu})\bm{v}\Big{)}^{T}\Big{(}\partial_{t}\bm{\nu}+(\nabla_{r}\bm{\nu})\bm{v}\Big{)}\rangle\\!\rangle$
$\displaystyle=\langle\\!\langle\lvert\partial_{t}\bm{\nu}\rvert\rangle\\!\rangle^{2}+2\langle\\!\langle(\partial_{t}\bm{\nu})^{T}\Big{(}(\nabla_{\bm{q}}\bm{\nu})\bm{v}\Big{)}\rangle\\!\rangle+\operatorname{tr}\langle\\!\langle\nabla_{\bm{q}}{\bm{\nu}}(\bm{v}\otimes\bm{v})\nabla_{\bm{q}}\bm{\nu}^{T}\rangle\\!\rangle,$

where the last term has been obtained by
$tr(\bm{a}\bm{b}^{T})=\bm{a}^{T}\bm{b}$. Furthermore, we notice that in the
mixed-term

(48) $\langle\\!\langle
2(\partial_{t}\bm{\nu})^{T}\Big{(}(\nabla_{\bm{q}}\bm{\nu})\bm{v}\Big{)}\rangle\\!\rangle=-2\langle\\!\langle\lvert\partial_{t}\bm{\nu}\rvert^{2}\rangle\\!\rangle+\langle\\!\langle
2\partial_{t}{\bm{\nu}}\cdot\dot{\bm{\nu}}\rangle\\!\rangle,$

the last term vanishes because
$\partial_{t}{\bm{\nu}}\cdot\dot{\bm{\nu}}=\partial_{t}\bm{\nu}\cdot(\bm{\omega}\times\bm{\nu})=\bm{\omega}\cdot(\partial_{t}\bm{\nu}\times\bm{\nu})=0$,
as $\bm{\nu}\times\bm{\nu}=0$. We are now left with the following expression
for the internal energy:

$\displaystyle\psi_{0}$
$\displaystyle=\psi-\psi_{K}=\psi-\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})-\frac{\lambda_{1}}{2}(\dot{\bm{\nu}}\cdot\dot{\bm{\nu}})$
$\displaystyle=\frac{1}{2}m\langle\\!\langle\lvert\bm{v}\rvert^{2}\rangle\\!\rangle+\frac{\lambda_{1}}{2}\operatorname{tr}\langle\\!\langle\nabla_{\bm{q}}{\bm{\nu}}(\bm{v}\otimes\bm{v})\nabla_{\bm{q}}\bm{\nu}^{T}\rangle\\!\rangle-\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})$
$\displaystyle-\frac{\lambda_{1}}{2}(\dot{\bm{\nu}}\cdot\dot{\bm{\nu}})-\langle\\!\langle(\partial_{t}\bm{\nu}\cdot\partial_{t}\bm{\nu})\rangle\\!\rangle$
$\displaystyle=\frac{1}{2}\langle\\!\langle\lvert\bm{v}^{2}\rvert\rangle\\!\rangle+\frac{\lambda_{1}}{2}\operatorname{tr}\Big{[}\nabla_{\bm{q}}\bm{\nu}\langle\\!\langle\Pi\rangle\\!\rangle\nabla_{\bm{q}}\bm{\nu}^{T}\Big{]}-\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0})$
$\displaystyle-\frac{\lambda_{1}}{2}(\dot{\bm{\nu}}\cdot\dot{\bm{\nu}})-\langle\\!\langle(\partial_{t}\bm{\nu}\cdot\partial_{t}\bm{\nu})\rangle\\!\rangle$
(49)
$\displaystyle=\frac{1}{2}\langle\\!\langle\lvert\bm{v}^{2}\rvert\rangle\\!\rangle+\frac{\lambda_{1}}{2}\operatorname{tr}\Big{[}\nabla_{\bm{q}}\bm{\nu}\mathbb{P}\nabla_{\bm{q}}\bm{\nu}^{T}\Big{]}-\frac{1}{2}m(\bm{v}_{0}\cdot\bm{v}_{0}).$

To conclude this section we would like to highlight one term in the previous
decomposition of $\psi_{0}$, i.e (42), which we denote as

(50)
$\psi_{OF}(\bm{q},t)=\frac{\lambda_{1}}{2}\operatorname{tr}\Big{[}\nabla_{\bm{q}}\bm{\nu}\mathbb{P}\nabla_{\bm{q}}\bm{\nu}^{T}\Big{]}.$

Using the Maxwellian to compute $\mathbb{P}$, we have
$\mathbb{P}^{(0)}=\frac{2}{5m}(I_{1}I_{2}I_{3})^{\frac{1}{2}}\langle\\!\langle\theta\rangle\\!\rangle
I$. Since multiples of the identity matrix commute with all other matrices, we
get the one-constant approximation of the Oseen–Frank energy functional, with
the caveat that the Frank constant in this case depends on the pressure:

(51)
$\rho\psi_{OF}^{(0)}(\bm{q},t)=p_{K}\frac{\lambda_{1}}{2}tr\Big{[}\nabla_{\bm{q}}\bm{\nu}\nabla_{\bm{q}}\bm{\nu}^{T}\Big{]}.$

The dependence of the energy functional on the pressure is a striking
difference from the classical Oseen–Frank theory. Such a dependence is a
consequence of the fact that the principle of virtual work we started from and
the constitutive relation for the pressure tensor we used were derived from
the Boltzmann–Curtiss equation, which describes the dynamics of compressible
fluids.

##  6 Noll–Coleman procedure

We observe that (35) holds for all thermokinetic processes to which the
thermodynamic body representing our fluid can be subjected, in the sense of
[37]. We can therefore perform a Noll–Coleman procedure to derive from
$\psi^{(0)}_{OF}$ the nematic contribution to the stress tensor and the couple
stress tensor. The Noll–Coleman procedure is a method to derive the
constitutive equations for the moments from a principle of virtual work [6,
33].

First, we recall (35), where we have removed the dependence on the density:

(52)
$\displaystyle\dot{\psi}_{0}+m\mathbb{P}:\nabla_{\bm{q}}\bm{v}_{0}+\mathbb{M}:\nabla_{\bm{q}}\bm{\omega}_{0}\geq
0.$

We are interested in applying the Noll–Coleman procedure only to find the
relation between the pressure tensor and the nematic director, so we fix
$\psi_{0}=\psi_{OF}$. To begin with, we expand the total derivative of
$\psi_{0}$ in terms of $\dot{\bm{\nu}}$ and $\dot{\nabla_{\bm{q}}\bm{\nu}}$
and then adopt tensor notation. We then proceed using the identity
$\dot{\bm{\nu}}=\bm{\omega}_{0}\times\bm{\nu}$ and definition of material time
derivative to obtain
$\dot{\nabla_{\bm{q}}\bm{\nu}}=\nabla_{\bm{q}}\dot{\bm{\nu}}-\nabla_{\bm{q}}\bm{\nu}\nabla_{\bm{q}}\bm{v}_{0}$,
which we substitute into the expansion of $\dot{\psi}_{0}$:

$\displaystyle\dot{\psi_{0}}$
$\displaystyle=\frac{\partial\psi_{0}}{\partial\bm{\nu}}\dot{\bm{\nu}}+\frac{\partial\psi_{0}}{\partial(\nabla_{\bm{q}}\bm{\nu})}\dot{\nabla_{\bm{q}}\bm{\nu}}=\frac{\partial\psi_{0}}{\partial(\nu_{p})}\Big{[}\varepsilon_{iqp}\nu_{q}\omega_{0,i}\Big{]}+\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\partial_{k}\dot{\nu}_{p}$
$\displaystyle=\frac{\partial\psi_{0}}{\partial(\nu_{p})}\Big{[}\varepsilon_{iqp}\nu_{q}\omega_{0,i}\Big{]}+\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\Big{[}\varepsilon_{iqp}\partial_{k}(\nu_{q}\omega_{0,i})-\partial_{q}(\nu_{p})\partial_{k}(v_{0,q})\Big{]}$
$\displaystyle=\frac{\partial\psi_{0}}{\partial(\nu_{p})}\Big{[}\varepsilon_{iqp}\nu_{q}\omega_{0,i}\Big{]}+\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\Big{[}\varepsilon_{iqp}\partial_{k}(\nu_{q})\omega_{0,i}+\varepsilon_{iqp}\partial_{k}(\omega_{0,i})\nu_{q}-\partial_{q}(\nu_{p})\partial_{k}(v_{0,q})\Big{]}$
(53)
$\displaystyle=\varepsilon_{iqp}\Bigg{[}\Big{(}\nu_{q}\frac{\partial\psi_{0}}{\partial(\nu_{p})}+\partial_{k}\nu_{q}\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\Big{)}\omega_{0,i}+\nu_{q}\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\partial_{k}\omega_{0,i}\Bigg{]}-\frac{\partial\psi}{\partial(\partial_{k}\nu_{p})}\partial_{q}\nu_{p}\partial_{k}(v_{0,q}).$

We would like to express the previous equation using a vector product. To do
this we use the Ericksen identity stated in the following lemma [33, Appendix
B].

######  Lemma 6.1.

Given a rotationally invariant
$\psi_{0}=\psi_{0}(\bm{\nu},\nabla_{\bm{q}}\bm{\nu})$, so that
$\psi_{0}(\bm{\nu},\nabla_{\bm{q}}\bm{\nu})\allowbreak=\psi_{0}(Q\bm{\nu},Q\nabla_{\bm{q}}Q^{T})$
for all $Q\in SO(\mathbb{R}^{3})$, the following identity holds:

(54)
$\varepsilon_{iqp}\Big{[}\nu_{q}\frac{\partial\psi_{0}}{\partial\nu_{p}}+\partial_{k}\nu_{q}\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}+\partial_{q}\nu_{k}\frac{\partial\psi_{0}}{\partial(\partial_{p}\nu_{k})}\Big{]}=0.$

Using (54) to substitute for the coefficients of $\omega_{0,i}$ in (53) yields

(55)
$\dot{\psi_{0}}=\varepsilon_{iqp}\Big{(}\nu_{q}\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\partial_{k}\omega_{0,i}-\partial_{q}\nu_{k}\frac{\partial\psi_{0}}{\partial(\partial_{p}\nu_{k})}\omega_{0,i}\Big{)}-\frac{\partial\psi_{0}}{\partial(\partial_{k}\nu_{p})}\partial_{q}\nu_{p}\partial_{k}(v_{0,q}).$

Substituting (55) into (52) and overloading the cross product between a vector
and a tensor as $\bm{\nu}\times S=\varepsilon_{iqp}\nu_{q}S_{jp}$ we get, for
a thermokinetic process with $\bm{\omega}_{0}=0$, the following inequality

(56)
$\displaystyle\Big{(}\mathbb{P}-(\frac{\partial\psi_{0}}{\partial(\nabla\bm{\nu})})^{T}\nabla_{\bm{q}}\bm{\nu}\Big{)}:\nabla_{\bm{q}}\bm{v}_{0}+\Big{(}\mathbb{M}+\bm{\nu}\times(\frac{\partial\psi_{0}}{\partial\nabla\bm{\nu}})\Big{)}:\nabla_{\bm{q}}\bm{\omega}_{0}\geq
0.$

Furthermore, since we can choose the sign of $\nabla_{\bm{q}}\bm{v}_{0}$ and
$\nabla_{\bm{q}}{\bm{\omega}_{0}}$ arbitrarily to preserve the inequality in
(35), the following identities must always hold:

(57)
$\mathbb{P}^{(*)}=\left(\frac{\partial\psi_{OF}}{\partial(\nabla\bm{\nu})}\right)^{T}\nabla_{\bm{q}}\bm{\nu},\qquad\mathbb{M}^{(*)}=-\bm{\nu}\times\left(\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}\right),$

where the superscript (∗) denotes the additional component that arises from
the nematic ordering, via the Noll–Coleman procedure. This step is crucial: it
yields a closure that describes the coupling between the pressure stress
tensor and $\nabla\bm{\nu}$. Therefore, in contrast to the hypothesis of
Curtis and Dahler [7, 8], the intrinsic angular momentum is coupled to the
macroscopic velocity.

Furthermore, (22c) can be used to describe the evolution of the nematic
director. We proceed to substitute (57) inside (22) to obtain the following
set of equations, governing the dynamics of a polyatomic rarefied gas:

(58a) $\displaystyle\partial_{t}\rho+\nabla_{\bm{q}}\cdot(\rho\bm{v}_{0})=0,$
(58b)
$\displaystyle\rho\Big{[}\partial_{t}\bm{v}_{0}+(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot\left(\rho\,\mathbb{P}+\rho\left(\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}\right)^{T}\nabla_{\bm{q}}\bm{\nu}\right)=0,$
(58c)
$\displaystyle\rho\Big{[}\partial_{t}\bm{\eta}+(\nabla_{\bm{q}}\bm{\eta})\bm{v}_{0}\Big{]}-\nabla_{\bm{q}}\cdot\left(\rho\,\bm{\nu}\times\left(\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}\right)\right)=0.$

We focus our attention on the equation for the conservation of angular
momentum (58c) and notice that we have two variables $\bm{\nu}$ and
$\bm{\eta}$ but only one equation. Ideally, we would like to make one of the
two disappear or add an equation relating the two quantities. In deriving the
Leslie–Ericksen equations (see for example [33]), Leslie wrote [22]:

> The inertial term associated with local rotation of the material element is
> omitted because in general, it is negligible.

Given the symmetric nature of the stress tensor, in our model we can not
ignore the intrinsic angular momentum $\bm{\eta}$, since it is the only source
of anisotropy. In order to express the intrinsic angular momentum $\bm{\eta}$
in terms of the nematic director $\bm{\nu}$ we observe that if as in (45) we
assume $\varepsilon\searrow 0$ and use the two identities

(59) $\dot{\bm{\nu}}=\bm{\omega}\times\bm{\nu}$

and

(60)
$\dot{\bm{\nu}}\times\bm{\nu}=(\bm{\omega}\times\bm{\nu})\times\bm{\nu}=\Big{(}\bm{\omega}(\bm{\nu}^{T}\bm{\nu})-\bm{\nu}(\bm{\nu}^{T}\bm{\omega})\Big{)},$

we have:

(61)
$\dot{\bm{\eta}}=\langle\\!\langle\mathbb{I}\rangle\\!\rangle\bm{\omega}=\lambda_{1}\Big{(}I-\bm{\nu}\otimes\bm{\nu}\Big{)}\bm{\omega}=\lambda_{1}\Big{(}\bm{\omega}(\bm{\nu}^{T}\bm{\nu})-\bm{\nu}(\bm{\nu}^{T}\bm{\omega})\Big{)}=\lambda_{1}\dot{\bm{\nu}}\times\bm{\nu}.$

Substituting this last equation into (58c) we obtain

$\rho\dot{\bm{\nu}}\times\bm{\nu}-\nabla_{\bm{q}}\cdot\left(\rho\,\bm{\nu}\times\left(\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}\right)\right)=0.$

To massage the previous equation to a better-known form we subtract from the
previous equation the antisymmetric part of
$\rho(\frac{\partial\psi_{OF}}{\partial(\nabla\bm{\nu})})^{T}\nabla_{\bm{q}}\bm{\nu}$,
which we know will vanish since $\mathbb{P}$ is symmetric by definition. Using
the Ericksen identity once again yields, in tensor notation,

$\displaystyle\varepsilon_{ikq}\Big{[}\lambda_{1}\dot{\nu}_{q}\nu_{k}+\nu_{k}\partial_{j}\Big{(}\rho\frac{\partial\psi_{OF}}{\partial(\partial_{j}\nu_{q})}\Big{)}-\rho\frac{\partial\psi_{OF}}{\partial(\partial_{k}\nu_{p})}\partial_{q}\nu_{p}\Big{]}$
(62)
$\displaystyle=\varepsilon_{ikq}\nu_{k}\Big{[}\lambda_{1}\dot{\nu}_{q}+\partial_{j}\Big{(}\rho\frac{\partial\psi_{OF}}{\partial(\partial_{j}\nu_{q})}\Big{)}-\rho\frac{\partial\psi_{OF}}{\partial\nu_{q}}\Big{]}=0.$

We conclude by observing that the last equation prescribes that all the
components of

(63)
$\lambda_{1}\rho\dot{\bm{\nu}}+\nabla_{\bm{q}}\cdot\rho\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}-\rho\frac{\partial\psi_{OF}}{\partial\bm{\nu}},$

in the direction orthogonal to $\bm{\nu}$ vanish. We can express this
constraint using the Lagrange multiplier
$\tau(\bm{q}):\mathbb{R}^{3}\to\mathbb{R}$, i.e.

(64)
$\lambda_{1}\rho\dot{\bm{\nu}}+\nabla_{\bm{q}}\cdot\rho\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}-\rho\frac{\partial\psi_{OF}}{\partial\bm{\nu}}=\tau\bm{\nu},$

where the Lagrange multiplier $\tau(\bm{q})$ is determined by the unit length
constraint on $\bm{\nu}$. We have retrieved the balance law of angular
momentum presented in [22].

Next we evaluate the expression for
$\frac{\partial\psi_{OF}}{\partial\nabla\bm{\nu}}$ using the Maxwellian
Oseen–Frank energy $\psi_{OF}^{(0)}$, using the Maxwellian pressure tensor
$\mathbb{P}^{(0)}$ to obtain the following set of equations:

(65a) $\displaystyle\partial_{t}\rho+\nabla_{\bm{q}}\cdot(\rho\bm{v}_{0})$
$\displaystyle=0,$ (65b)
$\displaystyle\rho\Big{[}\partial_{t}\bm{v}_{0}+(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot\Big{(}p_{K}I+p_{K}\frac{\lambda_{1}}{2}\nabla_{\bm{q}}\bm{\nu}^{T}\nabla_{\bm{q}}\bm{\nu}\Big{)}$
$\displaystyle=0,$ (65c)
$\displaystyle\lambda_{1}\rho\dot{\bm{\nu}}+\nabla_{\bm{q}}\cdot\Big{(}p_{K}\frac{\lambda_{1}}{2}\nabla_{\bm{q}}\bm{\nu}\Big{)}$
$\displaystyle=\tau\bm{\nu},$ (65d) $\displaystyle\lVert\bm{\nu}\rVert$
$\displaystyle=1,$

where the last equation is the constraint on the nematic director length that
comes from (44). We observe that the previous system of equations has five
unknowns ($\rho$, $\bm{v}_{0}$, $p_{K}$, $\lambda$, $\bm{\nu}$) and only four
equations. This is because we still need to consider the equation describing
the evolution of $\psi_{0}$. To obtain this equation we start from (52), with
adiabatic conditions,

(66)
$\rho\Big{[}\partial_{t}\psi_{0}+\nabla_{\bm{q}}\psi_{0}\cdot\bm{v}_{0}\Big{]}+\Big{(}p_{K}I+p_{K}\frac{\lambda_{1}}{2}\nabla_{\bm{q}}\bm{\nu}^{T}\nabla_{\bm{q}}\bm{\nu}\Big{)}:\nabla_{\bm{q}}\bm{v}_{0}=0,$

and $\psi_{0}$ can be related to the other five variables in the system via
(41). We conclude now by observing that the combination of (65), (66) and (41)
form a closed system that describes the evolution of a rarefied gas subject to
a nematic ordering, near the thermodynamic equilibrium and in adiabatic
conditions. For completeness, we report the final system of six equations
determining the evolution of the six macroscopic unknowns $\rho$,
$\bm{v}_{0}$, $p_{K}$, $\bm{\nu},\tau$ and $\psi_{0}$:

(67a) $\displaystyle\partial_{t}\rho+\nabla_{\bm{q}}\cdot(\rho\bm{v}_{0})$
$\displaystyle=0,$ (67b)
$\displaystyle\rho\Big{[}\partial_{t}\bm{v}_{0}+(\nabla_{\bm{q}}\bm{v}_{0})\bm{v}_{0}\Big{]}+\nabla_{\bm{q}}\cdot\Big{(}p_{K}I+p_{K}\frac{\lambda_{1}}{2}\nabla_{\bm{q}}\bm{\nu}^{T}\nabla_{\bm{q}}\bm{\nu}\Big{)}$
$\displaystyle=0,$ (67c)
$\displaystyle\lambda_{1}\rho\dot{\bm{\nu}}+\nabla_{\bm{q}}\cdot\Big{(}p_{K}\frac{\lambda_{1}}{2}\nabla_{\bm{q}}\bm{\nu}\Big{)}$
$\displaystyle=\tau\bm{\nu},$ (67d)
$\displaystyle\rho\Big{[}\partial_{t}\psi_{0}+\nabla_{\bm{q}}\psi_{0}\cdot\bm{v}_{0}\Big{]}+\Big{(}p_{K}I+p_{K}\frac{\lambda_{1}}{2}\nabla_{\bm{q}}\bm{\nu}^{T}\nabla_{\bm{q}}\bm{\nu}\Big{)}:\nabla_{\bm{q}}\bm{v}_{0}$
$\displaystyle=0,$ (67e) $\displaystyle p_{K}$
$\displaystyle=\frac{6}{5}\frac{\rho}{m}(I_{1}I_{2}I_{3})^{\frac{1}{2}}\psi_{0},$
(67f) $\displaystyle\lVert\bm{\nu}\rVert$ $\displaystyle=1.$

These derived equations (67) are a compressible variant of the Leslie–Ericksen
equations. Under isothermal and isobaric conditions, this system reduces to
the Leslie–Ericksen equations.

##  Appendix A Rational mechanics of rigid bodies

This appendix aims to give a brief summary of the basic notions of the
rational mechanics of rigid bodies. In particular, we will discuss our choice
of generalized coordinates used to describe the configuration space of a rigid
body and what are the corresponding conjugate momenta. We will assume our
calamitic molecules can be modeled as a discrete rigid system of $N$ material
points and limit our focus to this specific case. It is trivial to extend the
notions here presented to the case $N\to\infty$. The presentation is primarily
drawn from [13]. We redirect the reader interested in a more in-depth study of
this topic to Goldstein et al. or Whittaker [16, 39].

######  Definition A.1.

We call a rigid body a set of material points $\\{(P_{1},m_{1})\\}_{i=1}^{N}$,
that at all time satisfy a set of so-called rigidity constraints:

(68) $\lVert P_{i}-P_{j}\rVert=R_{ij},\qquad 1\leq i\leq j\leq N.$

where $\lVert\cdot\rVert$ denotes the standard Euclidean norm in
$\mathbb{R}^{3}$. Furthermore, we require the rigidity constraints to satisfy
the Euclidean compatibility conditions, i.e.

  1. 1.

$R_{ij}\geq 0$ and $R_{ij}=0$ if and only if $P_{i}=P_{j}$,

  2. 2.

if $P_{i^{\prime}}=\lambda P_{i}$ and $P_{j^{\prime}}=\lambda P_{j}$ then
$R_{i^{\prime}j^{\prime}}=\lambda R_{ij}$,

  3. 3.

$R_{ij}\leq R_{ik}+R_{kj}$ for all $1\leq k<j\leq N$.

Lastly, we require that if $N>3$, not all $\frac{N(N-1)}{2}$ equations are
independent. In particular we require the system of the rigidity equations to
have rank at most $3$.

The system of rigidity constraints having at most rank 3 implies that once the
position of three points are determined, all other points of the material body
can be located using the rigidity constraints.

If we have a rigid body consisting of three unaligned material points,
$P_{1}$, $P_{2}$, and $P_{3}$, and a fixed reference frame
$(O,\bm{e}_{1},\bm{e}_{2},\bm{e}_{3})$ we can identify the position of the
three points assigning three coordinates to $P_{1}$, two coordinates to
$P_{2}$ and one coordinate to $P_{3}$. Therefore, any three unaligned points
can be identified by six coordinates. Notice also that this tells us that any
rigid body can be identified by six coordinates provided that it contains at
least three unaligned points. In the case where all points in the system are
aligned, we lose one degree of freedom because we only need five coordinates
to identify two key material points and then all the other material points of
the rigid body can be located using the rigidity constraints.

######  Proposition A.2.

The configuration space for a rigid body containing at least three unaligned
points is $\mathbb{R}^{3}\times SO(3)$.

Since we can represent an element of $SO(3)$ via an orthogonal matrix with
positive determinant, we can identify the configuration space of a rigid body
with the position of its center of mass $O$, and the three vectors
$\bm{e}_{1},\,\bm{e}_{2},\,\bm{e}_{3}$ representing the column of the
orthogonal matrix. We call the quadruplet
$(O,\bm{e}_{1},\bm{e}_{2},\bm{e}_{3})$ a body _reference frame_.

We now introduce Euler angles. Given two different body reference frames
$(O,\bm{e}_{1},\allowbreak\bm{e}_{2},\bm{e}_{3})$ and
$(O,\hat{\bm{e}_{1}},\hat{\bm{e}_{2}},\hat{\bm{e}}_{3})$ we can represent the
$SO(3)$ transformation that rotates one reference frame to the other making
use of three angles $\alpha_{1}$, $\alpha_{2}$ and $\alpha_{3}$ (see Fig. 2):

  1. 1.

the _precession angle_ $\alpha_{1}$, defined as the angle around $\bm{e}_{3}$
that aligns $\bm{e}_{1}$ with the intersection of the planes identified by the
normals $\bm{e}_{3}$ and $\hat{\bm{e}}_{3}$;

  2. 2.

the _nutation angle_ $\alpha_{2}$, defined as the angle between $\bm{e}_{3}$
and $\hat{\bm{e}}_{3}$;

  3. 3.

the _intrinsic rotation angle_ $\alpha_{3}$, defined as the angle around
$\hat{\bm{e}}_{3}$ that aligns $\hat{\bm{e}}_{1}$ with the intersection of the
planes identified by the normals $\bm{e}_{3}$ and $\hat{\bm{e}}_{3}$.

Figure 2: A depiction of the precession, nutation and intrinsic rotation
angles.

$\hat{\bm{e}}_{1}$$\hat{\bm{e}}_{2}$$\hat{\bm{e}}_{3}=\bm{e}_{3}$$\bm{e}_{1}$$\bm{e}_{2}$$\alpha_{1}$$\alpha_{1}$Precession
$\alpha_{1}$$\bm{e}_{2}$$\hat{\bm{e}_{2}}$$\hat{\bm{e}_{3}}$$\alpha_{2}$$\alpha_{2}$$\bm{e}_{1}=\hat{\bm{e}}_{1}$$\bm{e}_{3}$Nutation
$\alpha_{2}$$\hat{\bm{e}}_{1}$$\hat{\bm{e}}_{2}$$\bm{e}_{3}=\hat{\bm{e}}_{3}$${\bm{e}}_{1}$${\bm{e}}_{2}$$\alpha_{3}$$\alpha_{3}$
Intrinsic Rotation $\alpha_{3}$

The intersection of the planes identified by the normals $\bm{e}_{3}$ and
$\hat{\bm{e}}_{3}$ is called the _node line_ $\bm{N}$ and we denote
$\bm{\alpha}=(\alpha_{1},\alpha_{2},\alpha_{3})$.

We now proceed with some basic results.

######  Proposition A.3.

The Eulerian velocity field of a rigid body can be expressed by the formula

(69)
$v(\bm{x},t)=\bm{\omega}(t)\times\Big{(}\bm{x}-\bm{c}(t)\Big{)}+\dot{\bm{c}}(t),$

where $\bm{\omega}$ is referred to as the _angular velocity_ and $c(t)$ is the
velocity of the center of mass of the rigid body.

######  Theorem A.4.

Under any rigid motion, we can express the total derivative of any unit vector
$\bm{\nu}\in\mathbb{S}^{2}$, that expresses the direction connecting two
points making up the rigid body, as

(70) $\dot{\bm{\nu}}=\bm{\omega}\times\bm{\nu}.$

######  Proof A.5.

Let $\bm{x}_{1}$ and $\bm{x}_{2}$ denote two points of the rigid body aligned
along the direction $\bm{\nu}$, i.e.
$\bm{\nu}=(\bm{x}_{2}-\bm{x}_{1})/\lVert\bm{x}_{2}-\bm{x}_{1}\rVert$. The
velocity of the points can be expressed by relation (69):

(71)
$\displaystyle\dot{\bm{x}}_{1}=\bm{\omega}\times\Big{(}\bm{x}_{1}-\bm{c}\Big{)}+\dot{\bm{c}},\quad\dot{\bm{x}}_{2}=\bm{\omega}\times\Big{(}\bm{x}_{2}-\bm{c}\Big{)}+\dot{\bm{c}}.$

Subtracting the first equation from the second we obtain:

$\dot{\bm{x}}_{2}-\dot{\bm{x}}_{1}=\bm{\omega}\times\Big{(}\bm{x}_{2}-\bm{x}_{1}\Big{)}$

The result follows on division by $\lVert\bm{x}_{2}-\bm{x}_{1}\rVert$, which
remains constant under rigid motion.

We now would like to express $\bm{\omega}$ as the total derivative of the
Euler angles. To achieve this we decompose the angular velocity in terms of
the rotations mandated by the Euler angles,

(72)
$\bm{\omega}=\dot{\alpha_{1}}\bm{z}+\dot{\alpha_{2}}\bm{N}+\dot{\alpha_{3}}\hat{\bm{z}},$

where $(O,\bm{x},\bm{y},\bm{z})$ and
$(O,\hat{\bm{x}},\hat{\bm{y}},\hat{\bm{z}})$ are the body reference frame
before and after the rotation respectively, and $\bm{N}$ is the node line. Now
we introduce $\bm{N}_{\bot}=\bm{\hat{z}}\times\bm{N}$ and observe that the
following identities hold:

(73) $\displaystyle\bm{z}$
$\displaystyle=\sin(\alpha_{2})\bm{N}_{\bot}+\cos(\alpha_{2})\hat{\bm{z}},$
(74) $\displaystyle\bm{N}$
$\displaystyle=\cos(\alpha_{3})\hat{\bm{x}}-\sin(\alpha_{3})\hat{\bm{y}},$
(75) $\displaystyle\bm{N}_{\bot}$
$\displaystyle=\sin(\alpha_{3})\hat{\bm{x}}+\cos(\alpha_{3})\hat{\bm{y}}.$

Projecting $\bm{\omega}$ onto $(O,\hat{\bm{x}},\hat{\bm{y}},\hat{\bm{z}})$
yields

(76) $\displaystyle\bm{\omega}$
$\displaystyle=\Big{(}\dot{\alpha_{2}}\cos(\alpha_{3})+\dot{\alpha_{1}}\sin(\alpha_{2})\sin(\alpha_{3})\Big{)}\hat{\bm{x}}$
$\displaystyle+\Big{(}-\dot{\alpha_{2}}\sin(\alpha_{3})+\dot{\alpha_{1}}\sin(\alpha_{2})\cos(\alpha_{3})\Big{)}\hat{\bm{y}}+\Big{(}\dot{\alpha_{3}}+\dot{\alpha_{1}}\cos(\alpha_{2})\Big{)}\hat{\bm{z}}.$

We can rewrite the previous expression in matrix form as follows,

(77)
$\bm{\omega}=\begin{bmatrix}\sin(\alpha_{2})\sin(\alpha_{3})&\cos(\alpha_{3})&0\\\
\sin(\alpha_{2})\cos(\alpha_{3})&-\sin(\alpha_{3})&0\\\
\cos(\alpha_{2})&0&1\end{bmatrix}\begin{bmatrix}\dot{\alpha_{1}}\\\
\dot{\alpha_{2}}\\\ \dot{\alpha_{3}}\end{bmatrix}=\Xi\dot{\bm{\alpha}}.$

We now introduce notions related to the dynamic aspects of rigid bodies,
rather than the kinematic aspects.

######  Definition A.6.

We define the _center of mass_ of a rigid body as

(78) $O_{M}=\frac{1}{M}\sum_{i=1}^{N}m_{i}P_{i},$

where $M$ is the total mass of the rigid body, i.e. $M=\sum_{i=1}^{N}m_{i}$.

Once the center of mass has been defined we can introduce the notion of
_inertia moment_ with respect to the planes identified respectively by the
orthogonal normals $\bm{n}$ and $\bm{n}^{\prime}$, i.e.

(79)
$\Gamma_{\bm{n}\bm{n}^{\prime}}=\sum_{i=1}^{N}m_{i}\Big{[}(P_{i}-O_{M})\cdot\bm{n}\Big{]}\Big{[}(P_{i}-O_{M})\cdot\bm{n}^{\prime}\Big{]}.$

Once we are equipped with this definition we can introduce the notion of
inertia tensor with respect to a reference frame $(O,\bm{x},\bm{y},\bm{z})$,
i.e.

(80)
$\mathbb{I}=\begin{bmatrix}\Gamma_{\bm{x}\bm{x}}&\Gamma_{\bm{x}\bm{y}}&\Gamma_{\bm{x}\bm{z}}\\\
\Gamma_{\bm{y}\bm{x}}&\Gamma_{\bm{y}\bm{y}}&\Gamma_{\bm{y}\bm{z}}\\\
\Gamma_{\bm{z}\bm{x}}&\Gamma_{\bm{z}\bm{y}}&\Gamma_{\bm{z}\bm{z}}\end{bmatrix}.$

In particular, we notice that since $\mathbb{I}$ is a symmetric matrix it can
be orthogonally diagonalised, i.e.

(81) $\mathbb{I}=Q\begin{bmatrix}\Gamma_{1}&&\\\ &\Gamma_{2}&\\\
&&\Gamma_{3}\end{bmatrix}Q^{T},\qquad Q\in SO(3).$

Notice that $\Gamma_{i}$ are positive because they can be expressed as
$\Gamma_{\bm{n}\bm{n}}$ for a certain $\bm{n}$. We will refer to $\Gamma_{1}$,
$\Gamma_{2}$ and $\Gamma_{3}$ as the _principal moments of inertia_. We call
the principal reference frame the reference frame obtained by applying to
$(O,\bm{x},\bm{y},\bm{z})$ the rotation $Q$.

Lastly, we would like to apply the Hamiltonian formalism to the study of rigid
motion. To achieve this we first observe that the Lagrangian corresponding to
a rigid motion is only composed by the kinetic energy term,

(82)
$T=\frac{1}{2}\lvert\bm{v}\rvert^{2}+\frac{1}{2}\bm{\omega}\cdot\mathbb{I}\bm{\omega}.$

We chose as Lagrangian coordinates the position in space and the Euler angles.
We therefore need to express the above kinetic energy in terms of the total
derivatives of the position and of the Euler angles. To achieve this we resort
to (77),

(83)
$\mathcal{L}(\bm{q},\bm{\alpha},\dot{\bm{q}},\dot{\bm{\alpha}})=\frac{1}{2}m\lvert\dot{\bm{q}}\rvert^{2}+\frac{1}{2}\dot{\bm{\alpha}}\cdot\Xi^{T}\mathbb{I}\Xi\dot{\bm{\alpha}}.$

By Sylvester’s law of inertia [36], the congruent transformation
$D_{\dot{\bm{\alpha}}}^{2}\mathcal{L}=\Xi^{T}\mathbb{I}\Xi$ preserves the sign
of the eigenvalues, i.e. the eigenvalues of $\Xi^{T}\mathbb{I}\Xi$ have the
same signs as the eigenvalues of $\mathbb{I}$, which are all positive. The
Hessian $D^{2}\mathcal{L}$ of the Lagrangian $\mathcal{L}$ expressed in terms
of position, Euler angles and their total time derivative therefore remains
positive definite. Thus the Legendre transformation that defines the
Hamiltonian coordinates is well-posed:

(84)
$\bm{p}=\frac{\partial\mathcal{L}}{\partial\dot{\bm{q}}}=m\dot{\bm{q}},\qquad\bm{\varsigma}=\frac{\partial\mathcal{L}}{\partial\dot{\bm{\alpha}}}=\Xi^{T}\mathbb{I}\Xi\dot{\bm{\alpha}}.$

Notice now that given the fact that $D^{2}\mathcal{L}$ is positive definite
then $\varsigma$ is well-defined even if it depends on both $\alpha$ and
$\dot{\alpha}$, and we can proceed to define the Hamiltonian of a rigid motion
as

(85)
$\mathcal{H}(\bm{q},\bm{\alpha},\bm{p},\bm{\varsigma})=\frac{1}{2m}\lvert\bm{p}\rvert^{2}+\frac{1}{2}\bm{\varsigma}\cdot(\Xi^{T}\mathbb{I}\Xi)^{-1}\bm{\varsigma}.$

Hence our choice of coordinates in (1) is perfectly admissible.

##  Appendix B The Boltzmann–Curtiss equation

In this section we provide a detailed derivation of the Boltzmann–Curtiss
equation for convex non-spherical symmetric-top rigid molecules, using the
modern BBGKY hierarchy approach [18]. We assume the calamitic rarefied gas
under consideration is an ensemble of $N$ convex non-spherical symmetric-top
rigid bodies. The state of each particle can be identified by the coordinates
$\bm{\Gamma}=(\bm{q},\bm{\alpha},\bm{p},\bm{\varsigma})$, where $\bm{q}$
represents the position of the center of mass of the particle, $\bm{\alpha}$
are the Euler angles representing its orientation while $\bm{p}$ and
$\bm{\varsigma}$ are the conjugate moments to $\bm{q}$ and $\bm{\alpha}$
respectively. The configuration of the rarefied calamitic fluid at time $t$ is
given by
$\\{\bm{\Gamma}_{i}^{*}(t)\\}_{i=1}^{N}=\\{(\bm{q}_{i}(t),\bm{\alpha}_{i}(t),\bm{p}_{i}(t),\bm{\varsigma}_{i}(t))\\}_{i=1}^{N}$.
We can now introduce the Klimontovich distribution $\pi$ and the reduced
particle distribution $f_{s}$ as follows,

(86)
$\displaystyle\pi\Big{(}\\{\bm{\Gamma}_{i}\\}_{i=1}^{N},t\Big{)}=\sum_{i=1}^{N}\delta\Big{(}\bm{\Gamma}_{i}-\bm{\Gamma}_{i}^{*}(t)\Big{)},$
(87) $\displaystyle
f_{s}(\bm{\Gamma}_{1},\dots,\bm{\Gamma}_{s},t)=\int\pi(\bm{\Gamma}_{1},\dots,\bm{\Gamma}_{s},\bm{\Gamma}^{(s)},t)\,d\bm{\Gamma}^{(s)},$

where $d\bm{\Gamma}^{(s)}=d\bm{\Gamma}_{s+1}\dots d\bm{\Gamma}_{n}$. From (85)
we know that the Hamiltonian of the particle ensemble is given by

(88)
$\mathcal{H}=\left(\sum_{i=1}^{N}\frac{\lvert\bm{p}_{i}\rvert^{2}}{2m}+\frac{1}{2}\bm{\varsigma}_{i}\cdot(\Xi^{T}\mathbb{I}\Xi)^{-1}\bm{\varsigma}_{i}\right)+\sum_{1\leq
i<j\leq N}\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j}),$

where $\mathbb{I}$ is the inertia tensor of a single molecule, $\Xi$ is the
tensor used to pass from angular velocity to the total time derivative of the
Euler angles in (77), and $\phi:\mathbb{R}^{3}\to\mathbb{R}$ is the
interaction potential we are considering in addition to the Hamiltonian
corresponding to the rigid motion described in (85). We decompose the
Hamiltonian into three terms: two containing respectively contributions from
particles $1,\dots,s$ and $s+1,\dots,N$ and a third term containing mixed
terms,

(89)
$\displaystyle\mathcal{H}_{s}=\left(\sum_{i=1}^{s}\frac{\lvert\bm{p}_{i}\rvert^{2}}{2m}+\frac{1}{2}\bm{\varsigma}_{i}\cdot(\Xi^{T}\mathbb{I}\Xi)^{-1}\bm{\varsigma}_{i}\right)+\sum_{1\leq
i<j\leq s}\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j}),$ (90)
$\displaystyle\mathcal{H}_{N-s}=\left(\sum_{i=s+1}^{N}\frac{\lvert\bm{p}_{i}\rvert^{2}}{2m}+\frac{1}{2}\bm{\varsigma}_{i}\cdot(\Xi^{T}\mathbb{I}\Xi)^{-1}\bm{\varsigma}_{i}\right)+\sum_{s+1\leq
i<j\leq N}\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j}),$ (91)
$\displaystyle\hat{\mathcal{H}}_{s}=\sum_{i=1}^{s}\sum_{j=s+1}^{N}\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j}).$

We then differentiate $f_{s}$ with respect to time and use Liouville’s theorem

(92) $\frac{\partial\pi}{\partial t}=-\\{\pi,\mathcal{H}\\}$

to obtain the following integro-differential equation for the reduced particle
distribution $f_{s}$,

(93) $\frac{\partial f_{s}}{\partial t}=\int\frac{\partial\pi}{\partial
t}\,d\bm{\Gamma}^{(s)}=-\int\left(\\{\pi,\mathcal{H}_{s}\\}+\\{\pi,\mathcal{H}_{n-s}\\}+\\{\pi,\hat{\mathcal{H}}_{s}\\}\right)\,d\Gamma^{(s)}.$

We notice that since $\mathcal{H}_{s}$ is independent of
$\bm{\Gamma}_{s+1},\dots,\bm{\Gamma}_{n}$ we can bring the integral inside the
Poisson brackets to obtain the following equation,

(94) $\frac{\partial f_{s}}{\partial t}=\int\frac{\partial\pi}{\partial
t}\,d\bm{\Gamma}^{(s)}=-\\{f_{s},\mathcal{H}_{s}\\}-\int\\{\pi,\mathcal{H}_{n-s}\\}\,d\Gamma^{(s)}-\int\\{\pi,\hat{\mathcal{H}}_{s}\\}\,d\Gamma^{(s)}.$

We then notice that the second term on the right-hand side vanishes given the
fact it is an exact divergence and we have appropriate boundary conditions:

(95) $\displaystyle\int\\!\\!\\{\pi$
$\displaystyle,\mathcal{H}_{N-s}\\}d\bm{\Gamma}^{(s)}\\!=\\!\\!\\!\int\\!\\!\sum_{i=1}^{n}\\!\Big{(}\\!\frac{\partial\pi}{\partial\bm{q}_{i}}\frac{\partial\mathcal{H}_{N-s}}{\partial\bm{p}_{i}}\\!+\\!\frac{\partial\pi}{\partial\bm{\alpha}_{i}}\frac{\partial\mathcal{H}_{N-s}}{\partial\bm{\varsigma}_{i}}\\!-\\!\frac{\partial\pi}{\partial\bm{p}_{i}}\frac{\partial\mathcal{H}_{N-s}}{\partial\bm{q}_{i}}\\!-\\!\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\frac{\partial\mathcal{H}_{N-s}}{\partial\bm{\alpha}_{i}}\\!\Big{)}d\bm{\Gamma}^{(s)}$
$\displaystyle=\int\sum_{i=s+1}^{N}\Big{(}\frac{\partial\pi}{\partial\bm{q}_{i}}\frac{\bm{p}_{i}}{m}+\frac{\partial\pi}{\partial\bm{\alpha}}(\Xi^{T}\mathbb{I}\Xi)^{-1}\bm{\varsigma}\Big{)}\,d\bm{\Gamma}^{(s)}$
$\displaystyle\;\;\;\;-\int\sum_{i=s+1}^{N}\Big{(}\frac{\partial\pi}{\partial\bm{p}_{i}}\sum_{j=i+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{q}_{i}}+\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\sum_{j=i+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{\alpha}_{i}}\Big{)}\,d\bm{\Gamma}^{(s)}$
$\displaystyle=\int\sum_{i=s+1}^{N}\Bigg{(}\frac{\partial}{\partial\bm{q}_{i}}\cdot\Big{(}\pi\frac{\bm{p}_{i}}{m}\Big{)}+\frac{\partial}{\partial\bm{\alpha}}\cdot\Big{(}\pi(\Xi^{T}\mathbb{I}\Xi)^{-1}\bm{\varsigma}\Big{)}\Bigg{)}d\bm{\Gamma}^{(s)}$
$\displaystyle\;\;\;\;\;\;\;-\int\sum_{i=s+1}^{N}\frac{\partial}{\partial\bm{p}_{i}}\cdot\Big{(}\pi\sum_{j=i+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{q}_{i}}\Big{)}\,d\bm{\Gamma}^{(s)}$
$\displaystyle\;\;\;\;\;\;\;-\int\sum_{i=s+1}^{N}\frac{\partial}{\partial\bm{\varsigma}_{i}}\cdot\Big{(}\pi\sum_{j=i+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{\alpha}_{i}}\Big{)}\,d\bm{\Gamma}^{(s)}=0.$

It remains to study the last term in (94):

(96) $\displaystyle\int\\!\\{\pi,\hat{\mathcal{H}}_{s}\\}d\bm{\Gamma}^{(s)}$
$\displaystyle=\\!\int\\!\sum_{i=1}^{N}\\!\Big{(}\frac{\partial\pi}{\partial\bm{q}_{i}}\frac{\partial\hat{\mathcal{H}}_{s}}{\partial\bm{p}_{i}}\\!+\\!\frac{\partial\pi}{\partial\bm{\alpha}_{i}}\frac{\partial\hat{\mathcal{H}}_{s}}{\partial\bm{\varsigma}_{i}}\\!-\\!\frac{\partial\pi}{\partial\bm{p}_{i}}\frac{\partial\hat{\mathcal{H}}_{s}}{\partial\bm{q}_{i}}\\!-\\!\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\frac{\partial\hat{\mathcal{H}}_{s}}{\partial\bm{\alpha}_{i}}\Big{)}\,d\bm{\Gamma}^{(s)}$
$\displaystyle=\int\sum_{i=1}^{s}\frac{\partial\pi}{\partial\bm{p}_{i}}\cdot\sum_{j=s+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{q}_{i}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle+\int\sum_{j=s+1}^{N}\frac{\partial\pi}{\partial\bm{p}_{j}}\cdot\sum_{i=1}^{s}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{q}_{j}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle+\int\sum_{i=1}^{s}\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\cdot\sum_{j=s+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{\alpha}_{i}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle+\int\sum_{j=s+1}^{N}\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\cdot\sum_{i=1}^{s}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{\alpha}_{j}}\,d\bm{\Gamma}^{(s)}.$

We notice that the second and last term in the previous equation vanish since
they are exact divergences, which leaves us with

(97)
$\displaystyle\int\\!\\{\pi,\hat{\mathcal{H}}_{s}\\}d\bm{\Gamma}^{(s)}\\!$
$\displaystyle=\int\sum_{i=1}^{s}\frac{\partial\pi}{\partial\bm{p}_{i}}\cdot\sum_{j=s+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{q}_{i}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle+\int\sum_{i=1}^{s}\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\cdot\sum_{j=s+1}^{N}\frac{\partial\phi(\bm{q}_{i},\bm{q}_{j},\bm{\alpha}_{i},\bm{\alpha}_{j})}{\partial\bm{\alpha}_{i}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle=(N-s)\int\sum_{i=1}^{s}\frac{\partial\pi}{\partial\bm{p}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{q}_{i}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle+(N-s)\int\sum_{i=1}^{s}\frac{\partial\pi}{\partial\bm{\varsigma}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{\alpha}_{i}}\,d\bm{\Gamma}^{(s)}$
$\displaystyle=(N-s)\int\sum_{i=1}^{s}\frac{\partial
f_{s+1}}{\partial\bm{p}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{q}_{i}}\,d\bm{\Gamma}_{s+1}$
$\displaystyle+(N-s)\int\sum_{i=1}^{s}\frac{\partial
f_{s+1}}{\partial\bm{\varsigma}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{\alpha}_{i}}\,d\bm{\Gamma}_{s+1}.$

We have the BBGKY hierarchy for the Boltzmann–Curtiss equation, i.e.

(98) $\displaystyle\frac{\partial f_{s}}{\partial
t}+\\{f_{s},\mathcal{H}_{s}\\}$
$\displaystyle=(N-s)\int\sum_{i=1}^{s}\frac{\partial
f_{s+1}}{\partial\bm{p}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{q}_{i}}\,d\bm{\Gamma}_{s+1}$
$\displaystyle+(N-s)\int\sum_{i=1}^{s}\frac{\partial
f_{s+1}}{\partial\bm{\varsigma}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{\alpha}_{i}}\,d\bm{\Gamma}_{s+1}.$

By abuse of notation, we will denote $f_{s}$ the function $f_{s}$ multiplied
by the factor $\frac{N!}{(N-s)!}$ to obtain the following expression for the
BBGKY hierarchy, i.e.

(99) $\displaystyle\frac{\partial f_{s}}{\partial
t}+\\{f_{s},\mathcal{H}_{s}\\}$
$\displaystyle=\int\sum_{i=1}^{s}\frac{\partial
f_{s+1}}{\partial\bm{p}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{q}_{i}}\,d\bm{\Gamma}_{s+1}$
$\displaystyle+\int\sum_{i=1}^{s}\frac{\partial
f_{s+1}}{\partial\bm{\varsigma}_{i}}\cdot\frac{\partial\phi(\bm{q}_{i},\bm{q}_{s+1},\bm{\alpha}_{i},\bm{\alpha}_{s+1})}{\partial\bm{\alpha}_{i}}\,d\bm{\Gamma}_{s+1}.$

In particular, we focus our attention on the first two terms of the BBGKY
hierarchy, i.e.

(100) $\displaystyle\frac{\partial f_{1}}{\partial
t}+\frac{\bm{p}_{1}}{m}\cdot\frac{\partial
f_{1}}{\partial\bm{q}_{1}}+(\Xi^{T}_{1}\mathbb{I}\Xi_{1})^{-1}\bm{\varsigma}_{1}\cdot\frac{\partial
f_{1}}{\partial\bm{\alpha}_{1}}$
$\displaystyle=\int\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{1}}\frac{\partial
f_{2}}{\partial\bm{p}_{1}}\,d\bm{\Gamma}_{2}$
$\displaystyle+\int\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{1}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{1}}\,d\bm{\Gamma}_{2}$ (101)
$\displaystyle\frac{\partial f_{2}}{\partial
t}+\frac{\bm{p}_{1}}{m}\cdot\frac{\partial f_{2}}{\partial\bm{q}_{1}}$
$\displaystyle+(\Xi^{T}_{1}\mathbb{I}\Xi_{1})^{-1}\bm{\varsigma}_{1}\cdot\frac{\partial
f_{2}}{\partial\bm{\alpha}_{1}}+\frac{\bm{p}_{2}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}_{2}}+(\Xi^{T}_{2}\mathbb{I}\Xi_{2})^{-1}\bm{\varsigma}_{2}\cdot\frac{\partial
f_{2}}{\partial\bm{\alpha}_{2}}$
$\displaystyle-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{1}}\frac{\partial
f_{2}}{\partial\bm{p}_{1}}-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{1}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{1}}=0$

where the right-hand side of the last equation vanishes because we have
assumed $f_{3}\equiv 0$ since we are working under the assumption that we only
have binary collisions. Indeed under the Boltzmann–Grad limit it is possible
to justify more precisely that $f_{3}$ is negligible; we redirect the
interested reader to [19, 20, 18]. Notice now that we can add under the
integral appearing on the right-hand side of (100) and (101) terms of the form
$\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{2}}\frac{\partial
f_{2}}{\partial\bm{p}_{2}}$ and
$\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{2}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{2}}$ respectively, since they will vanish due
to the fact that they are exact divergences. We focus our attention on the
first two terms of the BBGKY hierarchy, i.e.

(102) $\displaystyle\frac{\partial f_{1}}{\partial
t}+\frac{\bm{p}_{1}}{m}\cdot\frac{\partial
f_{1}}{\partial\bm{q}_{1}}+(\Xi^{T}_{1}\mathbb{I}\Xi_{1})^{-1}\bm{\varsigma}_{1}\frac{\partial
f_{1}}{\partial\bm{\alpha}_{1}}$
$\displaystyle=\int\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{1}}\Big{(}\frac{\partial
f_{2}}{\partial\bm{p}_{1}}-\frac{\partial
f_{2}}{\partial\bm{p}_{2}}\Big{)}\,d\bm{\Gamma}_{2}$
$\displaystyle+\int\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{1}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{1}}d\bm{\Gamma}_{2}$
$\displaystyle+\int\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{2}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{2}}d\bm{\Gamma}_{2},$ (103)
$\displaystyle\frac{\partial f_{2}}{\partial
t}+\frac{\bm{p}_{1}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}_{1}}+(\Xi^{T}_{1}\mathbb{I}\Xi_{1})^{-1}\cdot\bm{\varsigma}_{1}\frac{\partial
f_{2}}{\partial\bm{\alpha}_{1}}$
$\displaystyle+\frac{\bm{p}_{2}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}_{2}}+(\Xi^{T}_{2}\mathbb{I}\Xi_{2})^{-1}\bm{\varsigma}_{2}\cdot\frac{\partial
f_{2}}{\partial\bm{\alpha}_{2}}$
$\displaystyle-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{1}}\Big{(}\frac{\partial
f_{2}}{\partial\bm{p}_{1}}-\frac{\partial f_{2}}{\partial\bm{p}_{2}}\Big{)}$
$\displaystyle-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{1}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{1}}$
$\displaystyle-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{2}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{2}}=0,$

We notice that there are two different length scales in both (100) and (101).
In particular, the left-hand side of (100) acts on the macroscopic length
scale while the right-hand side acts on the microscopic length scale. We
introduce fast varying coordinates, i.e. $\bm{q}=\bm{q}_{2}-\bm{q}_{1}$ and
$\bm{\sigma}=\bm{\alpha}_{2}-\bm{\alpha}_{1}$ together with slowly varying
coordinates, i.e. $\bm{Q}=\frac{\bm{q}_{2}+\bm{q}_{1}}{2}$ and
$\bm{\Sigma}=\frac{\bm{\alpha}_{2}+\bm{\alpha}_{1}}{2}$. Now we can rewrite
(101) in terms of the new coordinates as,

(104) $\displaystyle\frac{\partial f_{2}}{\partial
t}\\!+\frac{1}{2}\frac{\bm{p}_{2}\\!+\\!\bm{p}_{1}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{Q}}+A(\bm{\varsigma}_{2}\\!+\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\Sigma}_{1}}\boxed{\\!+\frac{\bm{p}_{2}\\!-\\!\bm{p}_{1}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}}\\!+\\!B(\bm{\varsigma}_{2}\\!-\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}}$
$\displaystyle\boxed{-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{1}}\Big{(}\frac{\partial
f_{2}}{\partial\bm{p}_{1}}-\frac{\partial f_{2}}{\partial\bm{p}_{2}}\Big{)}}$
$\displaystyle\boxed{-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{1}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{1}}-\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{2}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{2}}}\\!$ $\displaystyle=\\!0,$

where
$A=(\Xi_{1}^{T}\mathbb{I}\Xi_{1})^{-1}+(\Xi_{2}^{T}\mathbb{I}\Xi_{2})^{-1}$,
$B=\left((\Xi_{2}^{T}\mathbb{I}\Xi_{2})^{-1}-(\Xi_{1}^{T}\mathbb{I}\Xi_{1})^{-1}\right)(\Xi_{1}^{T}\mathbb{I}\Xi_{1})(\Xi_{2}^{T}\mathbb{I}\Xi_{2})$
and we have boxed all the terms acting on the microscopic scale length. We
proceed to treat the terms acting on different scale lengths separately, and
we notice that the terms in the previous equation corresponding to the
microscopic scale length can be treated as steady with respect to the
collisional time scale,

(105) $\displaystyle\frac{\bm{p}_{2}\\!-\\!\bm{p}_{1}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}}+B(\bm{\varsigma}_{2}\\!-\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}=\frac{\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{q}_{1}}\Big{(}\frac{\partial
f_{2}}{\partial\bm{p}_{1}}-\frac{\partial f_{2}}{\partial\bm{p}_{2}}\Big{)}$
$\displaystyle+\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{1}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{1}}+\frac{\partial\phi(\bm{q}_{1},\bm{q}_{2},\bm{\alpha}_{1},\bm{\alpha}_{2})}{\partial\bm{\alpha}_{2}}\frac{\partial
f_{2}}{\partial\bm{\varsigma}_{2}}.$

Substituting (105) into (100) we obtain,

(106) $\displaystyle\frac{\partial f_{1}}{\partial
t}+\frac{\bm{p}_{1}}{m}\cdot\frac{\partial
f_{1}}{\partial\bm{q}_{1}}+(\Xi^{T}_{1}\mathbb{I}\Xi_{1})^{-1}\bm{\varsigma}_{1}\frac{\partial
f_{1}}{\partial\bm{\alpha}_{1}}=\left.\frac{d\,f_{1}}{d\,t}\right\lvert_{\vspace{0.5pt}\text{collision}},$
(107)
$\displaystyle\left.\frac{d\,f_{1}}{d\,t}\right\lvert_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!=\int\left(\frac{\bm{p}_{2}\\!-\\!\bm{p}_{1}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}}+B(\bm{\varsigma}_{2}\\!-\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}\,d\bm{\Gamma}_{2}\right),$

where the right-hand side of (107) represents the total variation of $f_{1}$
due to collisions between two molecules.

###  B.1 The rigid collision operator

In the previous section, we derived a hierarchy of equations to describe the
evolution of the reduced distribution functions $f_{s}$. In particular, each
equation in the hierarchy depended on a higher-order reduced particle
distribution, hence the system of equations was not closed. To solve this
issue we removed the dependence on $f_{3}$ from the second equation in the
hierarchy, i.e. (101), by using the physical assumption that we are only
considering binary collisions. We will now focus our attention on a way to
remove the dependence on $f_{2}$ from the first equation in the hierarchy,
i.e. (100) when considering the collision between two hard convex molecules.

Notice that since $f_{3}$ vanishes we are only considering collisions between
two molecules. We will from now on label these two molecules as molecule 1 and
molecule 2 and use the notation $\bm{\Gamma}_{1}$ and $\bm{\Gamma}_{2}$ to
denote the phase space coordinates of the two molecules respectively. In
particular, we will consider a reference frame having origin in the center of
mass of molecule 1 and imagine that molecule 2 is approaching, colliding and
then moving away from molecule 1.

We begin expanding $d\bm{\Gamma}_{2}$ in (107) and making use of the fact that
since we are only considering collisions between rigid bodies $f_{2}$ will
vanish outside of the translated excluded volume
$V=V_{ex}^{1,2}(\sigma)+\bm{q}_{1}$, hence we can restrict the domain of
integration to $V$, i.e.

(108)
$\displaystyle\left.\frac{d\,f_{1}}{d\,t}\right\lvert_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!\\!\\!\\!$
$\displaystyle=\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{V}\frac{\bm{p}_{2}\\!-\\!\bm{p}_{1}}{m}\cdot\frac{\partial
f_{2}}{\partial\bm{q}}\,d\bm{q}_{2}+\\!\\!\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\\!\\!\int_{V}\\!\\!\\!B(\bm{\varsigma}_{2}\\!-\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}\,d\bm{q}_{2}$ (109) $\displaystyle=\\!\\!\\!\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\partial
V}\frac{\bm{p}_{2}\\!-\\!\bm{p}_{1}}{m}\cdot\bm{k}f_{2}d\bm{S}+\\!\\!\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{V}\\!\\!\\!B(\bm{\varsigma}_{2}\\!-\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}\,d\bm{q}_{2}.\;\;$

We now observe that the translated excluded volume $V$ is a strictly convex
set with smooth boundary, hence we can apply results from differential
geometry of surfaces with positive curvature. In particular, we can transform
the integral on the boundary of $V$ into an integral on the surface of the
unit sphere $\mathbb{S}^{2}$, making use of the element of surface area per
solid angle $dS$ that can be computed explicitly [8], and rewrite (109) as

(110)
$\displaystyle\left.\frac{d\,f_{1}}{d\,t}\right\lvert_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!$
$\displaystyle=\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\mathbb{S}^{2}}\frac{\bm{p}_{2}\\!-\\!\bm{p}_{1}}{m}\cdot\bm{k}f_{2}dS\,d\bm{k}$
(111) $\displaystyle+\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{V}B(\bm{\varsigma}_{2}\\!-\\!\bm{\varsigma}_{1})\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}\,d\bm{q}_{2}.$

We now need to focus on the second term in the previous equation. In
particular we need to rewrite it in terms of the surface element of the unit
sphere $\mathbb{S}^{2}$. Notice that
$B(\varsigma_{2}-\varsigma_{1})=\dot{\sigma}$, and so we can rewrite the
second term in the previous equation as

(112)
$\displaystyle\int\\!\\!d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{V}\dot{\bm{\sigma}}\cdot\frac{\partial
f_{2}}{\partial\bm{\sigma}}\,d\bm{q}_{2}$ $\displaystyle=\\!\\!\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\mathbb{S}^{2}}\dot{\bm{\sigma}}\frac{\bm{q}}{\bm{\sigma}}\cdot\frac{\partial
f_{2}}{\partial\bm{q}}\,d\bm{q}_{2}$ (113)
$\displaystyle=\\!\\!\int\\!\\!d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\mathbb{S}^{2}}\dot{\bm{q}}\cdot\bm{k}f_{2}\,dS\,d\bm{k}.$

By virtue of proposition A.4 we can express $\dot{\bm{q}}$ as
$\bm{\omega}_{1}\times\bm{g}_{1}-\bm{\omega}_{2}\times\bm{g}_{2}$, where
$\bm{g}_{1}$ and $\bm{g}_{2}$ are the vectors connecting $\bm{q}_{1}$ and
$\bm{q}_{2}$ with the contact point $\bm{\zeta}$. In particular we can rewrite
(109) as

(114)
$\left.\frac{d\,f_{1}}{d\,t}\right\lvert_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!=\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\mathbb{S}^{2}}\bm{\mathfrak{g}}\cdot\bm{k}f_{2}\,dS\,d\bm{k},$

where $\bm{\mathfrak{g}}$ is the relative velocity of the contact point
$\bm{\zeta}$, i.e.

(115)
$\bm{\mathfrak{g}}=\bm{p}_{1}-\bm{p}_{2}+\bm{\omega}_{1}\times\bm{g}_{1}-\bm{\omega}_{2}\times\bm{g}_{2}.$

We decompose $\mathbb{S}^{2}$ into two surfaces: $\mathbb{S}_{+}$ is the
portion of the surface $\mathbb{S}^{2}$ such that along the collision
trajectory molecule 2 is coming towards molecule 1, i.e.
$\mathfrak{g}\cdot\bm{k}$ is positive, while $\mathbb{S}_{-}$ is the portion
of the surface $\mathbb{S}^{2}$ such that along the collision trajectory
molecule 2 is moving away from molecule 1, i.e. $\mathfrak{g}\cdot\bm{k}$ is
negative. With this we split the integral in (114) into two integrals on
$\mathbb{S}_{+}$ and $\mathbb{S}_{-}$ respectively, i.e.

(116)
$\displaystyle\left.\frac{d\,f_{1}}{d\,t}\right\lvert_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!$
$\displaystyle=\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\mathbb{S}_{+}}\bm{\mathfrak{g}}\cdot\bm{k}f_{2}\,dS\,d\bm{k}+\int
d\bm{\varsigma}_{2}d\bm{p}_{2}d\bm{\alpha}_{2}\int_{\mathbb{S}_{-}}\bm{\mathfrak{g}}\cdot\bm{k}f_{2}\,dS\,d\bm{k}.$

The first term in the previous equation is known as the _gain term_ , while
the second term is known as the _loss term_. Notice now that we can make use
of the _molecular chaos assumption_ [25], i.e.
$f_{2}(\bm{\Gamma}_{1},\bm{\Gamma}_{2},t)=f_{1}(\bm{\Gamma}_{1},t)f_{1}(\bm{\Gamma}_{2},t)$
on the loss term to obtain

(117)
$\displaystyle\left.\frac{df_{1}}{dt}\right|_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!$
$\displaystyle=\int\\!d\bm{p}_{2}\,d\bm{\varsigma}_{2}\,d\bm{\alpha}\\!\int_{\mathbb{S}_{+}}\\!\\!(\bm{k}\cdot\bm{\mathfrak{g}})f_{2}\,dSd\bm{k}$
$\displaystyle+\int\\!d\bm{p}_{2}\,d\bm{\varsigma}_{2}\,d\bm{\alpha}\\!\int_{\mathbb{S}_{-}}\\!\\!(\bm{k}\cdot\bm{g})f_{1}(\bm{q}_{1},\bm{\alpha}_{1},\bm{p}_{1},\bm{\varsigma}_{1})f_{1}(\bm{q}_{2},\bm{\alpha}_{2},\bm{p}_{2},\bm{\varsigma}_{2})\,S(\bm{k})d\bm{k}.$

As discussed in [35], since energy and linear and angular momentum are
conserved after the collision we can trace back the post-collisional
coordinates in the gain term of (116) to the corresponding pre-collisional
coordinates. Applying the molecular chaos assumption once again we obtain,

(118)
$\displaystyle\left.\frac{df_{1}}{dt}\right|_{\vspace{0.5pt}\text{collision}}\\!\\!\\!\\!$
$\displaystyle=\int\\!d\bm{p}_{2}\,d\bm{\varsigma}_{2}\,d\bm{\alpha}\\!\int_{\mathbb{S}_{+}}\\!\\!(\bm{k}\cdot\bm{\mathfrak{g}})f_{1}(\bm{q}_{1}^{\prime},\bm{\alpha}_{1}^{\prime},\bm{p}_{1}^{\prime},\bm{\varsigma}_{1}^{\prime},t)f_{1}(\bm{q}_{2}^{\prime},\bm{\alpha}_{2}^{\prime},\bm{p}_{2}^{\prime},\bm{\varsigma}_{2}^{\prime},t)\,dSd\bm{k}$
$\displaystyle+\int\\!d\bm{p}_{2}\,d\bm{\varsigma}_{2}\,d\bm{\alpha}\\!\int_{\mathbb{S}_{-}}\\!\\!(\bm{k}\cdot\bm{\mathfrak{g}})f_{1}(\bm{q}_{1},\bm{\alpha}_{1},\bm{p}_{1},\bm{\varsigma}_{1},t)f_{1}(\bm{q}_{2},\bm{\alpha}_{2},\bm{p}_{2},\bm{\varsigma}_{2},t)\,dSd\bm{k}.$

Lastly, we shorten the notation by writing the collision operator in the
familiar way:

(119)
$\displaystyle\left.\frac{df_{1}}{dt}\right|_{\vspace{0.5pt}\text{collision}}=C[f_{1},f_{1}],$

with

(120)
$C[f,g]\coloneqq\int\\!\\!\\!\\!\int\\!\\!\\!\\!\int\\!\\!\\!\\!\int(g^{\prime}f^{\prime}-gf)(\bm{k}\cdot\bm{\mathfrak{g}})dSd\bm{k}d\bm{p}_{2}d\bm{\alpha}_{2}d\bm{\varsigma}_{2},$

where the arguments of $f$ are
$(\bm{q}_{1},\bm{\alpha}_{1},\bm{p}_{1},\bm{\varsigma}_{1},t)$ and the
arguments of $g$ are
$(\bm{q}_{2},\bm{\alpha}_{2},\bm{p}_{2},\allowbreak\bm{\varsigma}_{2},t)$.

## Acknowledgments

The authors would like to acknowledge Professor E. Virga, whose insights made
this work possible. PEF acknowledges support from the UK Engineering and
Physical Sciences Research Council [EPSRC grants EP/R029423/1 and
EP/W026163/1].

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