Non-Maxwellian and Multi-Stream Flows in a Two-Level Gas
A. Muriel and L. Boot*
Data Transport Systems, 347 East 62nd St., New York, NY 10021
*National Institute of Physics, University of the Philippines,Diliman, Quezon City,
Philippines
Abstract. We use as motivation the study of nitrogen jets to study non-Maxwellian flows, adopting a quantum
kinetic model for a two-level system, to study the contribution of internal energy states in flow phenomena.
From a simple model of two low-energy states, we arrive at the following conclusions: (1) A jet of two-level
atoms or molecules starting with a displaced Maxwellian in the high-energy state will attain a non-Maxwellian
steady state, including a steady-state momentum distribution functions with two peaks in momentum. (2) Steady
state solutions of the model kinetic equations used are unique only for specific initial conditions. The solutions
admit the possibility of vortex motion. (3) For every equilibrium state, an equation of state may be derived, with
an effective temperature determined by the energy gap between the two internal energy states. A mixture of
such distribution functions is proposed to result in widely varying local pressure and temperature, characteristic
of a turbulent gas. We then review our other results, culminating in some speculation on the use of ultrasound to
control turbulence.
MOTIVATION
In several studies reported in the proceedings of the International Symposium on Rarefied Gas Dynamics
from 1959 to 1994 [1], it was shown that the flow characteristics of gases with internal degrees of freedom are
different from those derived in continuum dynamics. Some of the new results [2] include the following: (1)
slower relaxation to equilibrium due to transient behavior as molecules get excited and de-excited, (2) changes in
the values of transport coefficients due to the internal degrees of freedom, among which is the increased
significance of bulk viscosity, known to be associated with rotational angular momentum states as pointed out by
Landau [3], and (3) departures from Maxwell-Boltzmann local equilibrium in free jet experiments [4] in
nitrogen, which we use as inspiration for the model that follows. The nitrogen molecule is a rotating diatom, with
an infinite number of states. Not all of the states are occupied, it may suffice to study the low energy levels. In
particular, for our model, we look at only two such levels, a mathematical simplification that allows us to use
some recent results in non-equilibrium statistical mechanics.
From the theoretical point of view, the experimental results mentioned above should be interpreted using
the Wang-Chang-Uhlenbeck-de Boer modification of the one-specie Boltzmann transport equation [5]. In such a
case, the collision invariants are no longer the original ones due to the participation of internal degrees of
freedom, which become “unfrozen”, to use a description by Alves and Kremer [2]. But such an approach is quite
difficult, not only are there no known time-dependent solutions to the transport equations, the existence of
equilibrium states is not even proven. For this reason, there is some justification to use a much simpler model to
show the qualitative difference of the flows from classical fluid dynamics.
The basic new physics used in our model is the radiation produced by transitions between the two states.
This radiation results in energy dissipation. To study two features: the inelasticity of the collisions [6], and the
consequent radiation produced by the collisions [7], we adopt a molecular kinetic point of view. However,
discouraged by a long history of kinetic theory characterized by the absence of exact solutions of suggested
kinetic equations, such as the Pauli master equation [8], we propose a new simple model that allows exact
solutions. This is a traditional approach in non-equlibrium statistical mechanics [9,10], justifiable when new
physical insights are produced, as we report in this paper.
THE MODEL
In earlier papers [11,12], of which we will refer mostly to [12], we have suggested a kinetic equation that
displays new hydrodynamic flow features. These flow features are derived from coupled kinetic equations for
CP663, Rarefied Gas Dynamics: 23rd International Symposium, edited by A. D. Ketsdever and E. P. Muntz
© 2003 American Institute of Physics 0-7354-0124-1/03/$20.00
distribution functions for two states, f 2 ( r , p, t ) and f 1 ( r , p, t ) , which represent the single particle
distributions of each of the states 2 and 1, dependent on the coordinates r , the momentum p , and time t .
Our proposed kinetic equation is given by
γ 12 J
 f 2 
∂  f 2   D + γ 2 K − γ 21
 
  = 
γ 21 J
D + γ 1 K − γ 12  f1 
∂t  f 1  
(1)
p ∂
. Cartesian coordinates and dot products are implied. As we
m ∂r
remarked earlier, the operators J and K will be specified later, all that we stress now is that in general, they
do not commute. All s are constants, whose significance will be explained, but which are otherwise described
D is the streaming operator given by
−
γ
as well in [11,12]. We note that if the elements of the transition matrix in (1) are scalar, instead of noncommuting operators, the kinetic equation will be trivial.
An interesting result from [11,12] is an observed approach to steady states of some initial conditions,
using symbolic and numerical representations of the exact formal solution of Eq. (1). In [12], we stressed that we
needed to show rigorously that there exist equilibrium and steady states for the kinetic equation, and we
expressed the hope that this was doable in the future. In this review, we exhibit some equilibrium and steady
state solutions [13], in analogy with Boltzmann’s result for his transport equation. However, there is no unique
equilibrium solution, and neither are the equilibrium solutions Maxwellian.
EQUILIRIUM SOLUTION FOR A DISSIPATIVE SYSTEM
The system is dissipative when the jump operator J models the jump of one state to the other, radiating
energy equal to the energy gap between the two states.
The equilibrium solution is to be found by putting all time derivatives in Eq. (1) to zero:
 D + γ 2 K − γ 21

γ 21 J

γ 12 J
 f 2 
  = 0
D + γ 1 K − γ 12  f1 
(2)
or
(D + γ 2 K − γ 21 ) f 2 + γ 12 Jf1 = 0
(3)
γ 21 Jf 2 + (D + γ 1 K − γ 21 ) f1 = 0
(4)
Expressing
[− γ
f 2 in terms of f 1 from (3) and substituting into (4), we get
]
γ J (D + γ 2 K − γ 21 )−1 J + (D + γ 1 K − γ 21 ) f1 = 0
21 12
(5)
A similar equation may be derived for f 2 .
It turns out that if the system is uniform, or D = 0 , and if there is no external source of energy, or
kicks, that is , K = 0 , (so that we do not need to define it [11,12] for now ), the results are interesting enough
even if we put for the purpose of this report equal transition probabilities from 1 to 2 and vice versa.
γ
=γ
21 , which we interpret to mean as in [11,12], that it is equally probable for an atom to jump
For 12
from state 2 to 1 or state 1 to 2 . We then find the requirement that
J 2 =1
(6)
So, the question is: what momentum distributions will not change under the action of the “jump” operator J ?
In particular, for the one-dimensional system that we studied in [12], we defined the action of the jump operator
as
J m f ( p) =
[(
1
f
2
Assume that
)]
) (
p 2 + mδ + f − p 2 + mδ
f is an even function of
(
)
(p
,m > 0
)
(7)
+ 2δ , then
2
( )
J m f ( p 2 ) = f p 2 + mδ = f p 2
(8)
It is verifiable that the following distribution function satisfies Eq. (8):
f eq ( p 2 ) =
n =∞
∑
n = −∞
 2πinp 2 

an exp
 δ 
(9)
The most immediate question that comes to mind is the following: given that the possible equilibrium
states are not Maxwellian, would it still be possible to derive the thermodynamics of the system using the
conventional canonical ensemble approach? At this time, we think not, so that instead we fall back on
elementary physics definitions to derive such equilibrium concepts such as the equation of state and the adiabatic
constant. To do this, we need not restrict ourselves to one dimension as in [11,12].
To discuss the thermodynamics of the system, we follow these steps. First, choose an equilibrium state
from Eq. (9). Second, normalize the distribution function, using any maximum momentum obtained from the
requirement that the distribution function should always be positive. Third, calculate the pressure from the
elementary definition of pressure or momentum change against a fully reflecting wall. Fourth, calculate the
internal energy as a sum of the total kinetic energy of the atoms plus the total internal energy N∆ / 2 , where ∆
is the energy gap between the two states and N is the total number of particles. In the model that we now use,
half of the atoms are in the excited state.
We use the conventional thermodynamic relation PV = (γ − 1)U , where V is the volume of the
system, and P is the pressure, to get the adiabatic constant γ for each equilibrium state. From elementary
considerations,
P=
2N
Vm
pm
∫
dp p 2 f eq ( p )
0
U=
3N
2m
pm
∫ dp p
2
f eq ( p ) + N∆ / 2
(10)
p x2 f ( p x )
(11)
− pm
to give
γ = 1+
1
3 / 2 + m∆ / 4 I
pm
I=
∫ dp
x
0
From Eq. (9), we have constructed three normalized equilibrium states and calculated the pressure from
the transport of momentum across a unit area.
The results are given in tabular form :
Table 1. Possible Equilibrium States
Equilibrium Distribution
cos
(
2πp x2
/δ
Maximum momentum
pm
f ( px )
)
[ FresnelC (1) δ ]
(
)
(
cos 2πp x2 / δ + sin 2πp x2 / δ
)
PV / N =
δ /2
∆(1 − FresnelS (1)
= 0.229∆
2πFresnelC (1)
δ /2
∆( 2 + FresnelC ( 2 )
= 0.793∆
2πFresnelC (1)
δ /2
∆(1 − FresnelS (1) + FresnelC (1))
= 0.351∆
2π ( FresnelC (1) + FresnelS (1))
δ FresnelC (1)
sin( 2πp x2 / δ )
Equation of State
δ ( FresnelC (1) + FresnelS (1))
We have used the Fresnel sine and cosine functions. Each choice of the equilibrium state produces an equivalent
temperature, which is given by Teq = 0.229∆ / k for the first equilibrium state, where k is Boltzmann’s
constant .
We use the definition [12] δ = 2m∆ where ∆ is the energy gap between the two states and m
is the mass of the atom. There is a maximum momentum determined by the condition that the distribution
function should always be positive.
Why is there a maximum momentum for each of the equilibrium states? Is this not a major departure
from classical results? In fact, if we go back to continuum hydrodynamics, this is not at all new. For it is known
that in a vortex tube, the momentum of a fluid element increases linearly from the axis of the vortex, reaches a
maximum, then decays to zero as the distance from the axis increases [13]. Close to the axis of the vortex, the
fluid behaves like a rotating solid.
Furthermore, if a 3-d gas is characterized by vortex motion, it is quite easy to accept that there will be a
maximum momentum for the atoms, else, the vortex motion will be unstable. Our requirement for equilibrium
states thus admits the possibility of vortex motion. A stronger statement will be to interpret our discussion as an
indirect proof of the admissibility of vortex motion as a steady state. This is true for the first two examples
above. In the second case, we see the possibility of two vortices. The third case, we think, corresponds to three
vortices.
If we further consider the possibility that there are mixture of states, or if the system is perturbed to go
from one steady state to another, each state representing unstable equilibrium, then there will be fluctuations in
the measured temperature as well the pressure. Indeed we open up the possibility that there will be no steady
pressure nor local temperature, both of which are characteristics of turbulence, that subject which inspired our
suggestion of the proposed kinetic equation in the first place. Additionally, we also find that the adiabatic
constant cannot remain the same for a mixture of states. We have in the past [12] displayed some turbulent-like
behavior, we now stress as well the peculiar thermodynamics of the system under study.
We conclude this Section to highlight the importance of studying our new equilibrium and steady state
solutions for our proposed kinetic equation. More studies will help elucidate these states, which we suggest even
more strongly at this time are indicative of turbulence.
APPROACH TO EQUILIBRIUM
The main point of the presentation above is that there is theoretical justification to expect flows which are
non-Maxwellian. But how do Maxwellian initial conditions propagate|?
Let us now consider the initial condition
f 1 ( 0) =
β
exp(− β [ p − p o ] 2 )
π
f 2 ( 0) = 0
We present without proof [12] the time evolution of the above initial conditions for the jump operator given by
Eq. (8):
f2 =
e − gt
2
e − gt
2

2 exp − β [ p − p o ] 2 +


2
2  
 

β
2m
2
2




(gt ) exp − β  p + mδ + p o   + exp − β  p + mδ − p o   

π ∞

 

  

 


( 2m)!

 m =1
∑




(gt )2n +1 exp − β 
(
f1 =
2
2  

p 2 + (2n + 1)δ + p o   + exp − β  p 2 + (2n + 1)δ − p o   

  


 



(2n + 1)!





β  ∞

π  n =0


(12)
)
∑
We may plot Eqs. (12) and (13 with the parameters
β = 1,
δ = 0.02.
p o = 2.0,
(13)
g = 0.001 using 40
terms of the two infinite series. We find steady-state distributions, which look like a linear combination of
solutions of the form given by Eq. (9). For lack of space, we simply show the plot for the momentum distribution
for the ground state. The momentum distribution for the excited state starts with zero, as all atoms are initially in
the ground state. Then two bumps, approximately Gaussian, grow slowly, symmetric about zero momentum,
and eventually saturate. The behavior of the ground state distribution function is more interesting. First there is a
displaced Maxwellian peaked at the mean velocity. This peak diminishes, while another peak opposite the first
peak grows. This is shown in Fig. 1. Finally the peaks in the momentum equalize. In the meantime, the excited
state distribution becomes identical to that of the ground state, with two peaks each. We demonstrate an
unmistakable approach to a steady-state, or equilibrium, for an initial condition that is a displaced Maxwellian .
But we see that the steady-state solution is not Maxwellian.
FIGURE 1. Time evolution of the momentum distribution function for the low-energy state. Refer to the
narrative for the analogous figure for the excited state distribution.
The physical interpretation of the above result is the following. We start with a uniform jet of atoms in
the ground state. As the system evolves, by collisional transitions described by the jump operator, with equal
likelihood of jumping from one state to the other that results in radiation, the kinetic energy of the atoms is
eventually dissipated, while the two states become equally populated with atoms described by identical
distributions. There will be no more mean motion of the atoms, the entire system relaxes to a symmetric
momentum distribution representing two opposing streams. We conjecture that were the transition probabilities
unequal, the two states will no longer be equally populated. This point still remains as an analytical challenge.
Although quite different, it is interesting to compare this simple analytic result with transient
characteristics observed in a real system such as a nitrogen free jet, as done by R.G. Sharafutdinov and P. A.
Skovorodko [5], who concluded that the hydrodynamic approximation must be refined due to rotational
relaxation.
We now make the following qualitative prediction from our theoretical results. Suppose that the
longtitudinal momentum distribution function of a jet of nitrogen were measured as a function of distance from
the nozzle, thus effectively studying its time evolution. According to our theoretical result the distribution will
first be a displaced Maxwellian, symmetric about the mean velocity. It will then become an asymmetric
distribution, which becomes symmetric about zero momentum. But given that it may not be possible to study the
jet too far from the nozzle, only the approach to a symmetric, but non-Maxwellian distribution will be seen. If
confirmed, it will attest to the relevancy of the kinetic equation used in our work, and perhaps render plausible
other conclusions we have made in regard to turbulence [ 11,12,15-19].
SOME OTHER RELATED RESULTS
In every reported experiment on turbulence, the mechanical energy characterizing the energy per
molecule exceeds the accessible lowest energies arising from the internal degrees of freedom, like the rotational
energy of the nitrogen molecule, for example, which is of the order of kT / 40 , where k is the Boltzmann
constant and T is the temperature. For emphasis, we note that the Navier-Stokes equation excludes molecular
excitation energies. Furthermore, it is interesting to note that an apparently successful analysis of the probability
distribution functions for momentum differences between two points in a fluid does not make any direct use of
the Navier-Stokes equation [14].
We may ask: would the onset of turbulence be due to the unfreezing of the internal degrees of freedom
of molecules, as has been seen in aerodynamic flows (Karelov, et.al., 1981)? And, if indeed, the onset of
turbulence is associated with the unfreezing of the internal degrees of freedom, any external mechanism for such
unfreezing or excitation will affect the onset of turbulence. We thus make two other predictions, again testable
by experiments, and in fact, proposed here not without experimental hints from unpublished data.
First, a precise free efflux experiment to study the transition from turbulent flow to laminar flow, much
like a controlled efflux, as in a medical sphigmanometer, pressure versus time, will show distinctive signatures
of relaxation due to molecular properties like their energy states. Second, any mechanical, electromagnetic, or
thermal perturbation will effectively increase, or even decrease -- and therefore control -- the critical Reynolds
number as a function of the amplitude of the external perturbation, as when ultrasound is applied to a gas. This
effect will be independent of the mere change of density and viscosity due to the perturbation, it will be a new
qualitative effect. The decrease, or increase of the critical Reynolds number is predicted due to the following
reasoning [15]. When the internal degrees of freedom of the molecules are frozen, or unexcited, laminar flow is
observed. As the internal degrees of freedom are excited, or unfrozen, for simplicity, from the ground state to the
first excited state, turbulence begins. Finally, if the intensity of the external perturbation, such as ultrasound, is
increased to the point that the molecules are uniformly refrozen into another (higher) state, making the molecules
identical once again, all in the next accessible excited state, then an already turbulent gas, earlier unfrozen first
from its ground state to render it turbulent, can then re-laminarize as a “new” ground state is defined. These last
two speculations are based on preliminary experimental results, which we feel should be rigorously verified by
laboratories equipped to combine molecular physics with flow experiments. A new field of turbulence research
combining molecular physics and hydrodynamics may yet open up. These ideas are theoretically justified in the
physics literature published over the last six years [11,12, 15-19], outside the more familiar aerodynamics
literature.
ACKNOWLEDGEMENTS
We acknowledge the assistance of M. A. Sereno, F. N. Paraan, C. Catalan, J. Afalla, L. Alejan and L.
Jirkovsky. This research was supported in part by the World Laboratory, Lausanne.
*Email: [email protected]
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