Table Of Content1 From adiabatic piston to non-equilibrium hydrodynamics
3
1 KEN SEKIMOTO1,2, ANTOINE FRULEUX2,1, RYOICHI KAWAI3 AND NATHAN RIDLING3
0
2
1Matie`resetSyste`mesComplexes,CNRS-UMR7057,Universite´Paris-Diderot,75205Paris,France,
n 2Gulliver,CNRS-UMR7083,ESPCI,75231Paris,France,
a
J 3DepartmentofPhysics,UniversityofAlabamaatBirmingham,
9
2
] Basedonthenewconceptofthemomentumtransferdeficiencyduetodissipation(MDD),the
h
physical basis of the mechanism of “adiabatic piston” is explained. The implication of MDD in
c
e termsofhydrodynamicsundernon-equilibriumsteadystatealsodiscussed.
m
- PACSnumbers:05.40.-a,05.70.Ln,05.20.Dd
t
a
t
s 1.Introduction
.
t
a Since the 1920’s a simple question associated to non-equilibrium statistical physics has been
m
addressed: IfweputaBrownianpistonofmass Mbetweenthetwosemi-infinitecylinderseachbe-
-
ingfilledwithanidealgasconsisting ofparticles withmassm( M),whatisthenon-equilibrium
d
≪
n steadystate? Herethetemperatureandpressureofthegasintheleftcylinderarepreparedat(T,p)
o while those in the right cylinder are at (T ,p). The surface area of the piston is the same on both
′
c
sides. Itisclearthat,ifthepistonwerefirmlyfixedandifthepistonis“adiabatic”,thentherewould
[
be no net force on the piston because the gas in each cylinder remains in equilibrium and presses
1
thepiston bythesamepressure butintheopposing directions. WhentheBrownianmotionofthis
v
5 adiabatic pistonisallowed,however,thismotionwillallowtheenergytransferfromthehottergas
3 (e.g. the side of T if T > T ) to the colder gas (ibid. T ) [Feynmanetal.(1963)(§46.1-§46.9)].
′ ′
0 The question is if this non-equilibrium precess leads to a non-vanishing net force on the piston.
7
Themacroscopic thermodynamics cannot answerthistype ofquestion [Callen(1965)],neither the
.
1
Langevindescriptioncangiveananswertothistypeofsetup[VandenBroecketal.(2004)]. While
0
the stochastic energetics [Sekimoto(1997), Sekimoto(2010)] can describe correctly the heat flow,
3
1 thenon-equilibrium forceisbeyondtheresolution ofthislevelofdescription.
: Many calculative studies have been reported in the past both on this problem and also on a
v
i class of Brownian ratchet models, which turned out to be essentially the same problem as adi-
X
abatic piston [Fruleuxetal.(2012)]. All these studies have been done using either by ad hoc
ar treatment ofMaster-Boltzmann equations withtruncated momenthierarchical expansion with ǫ =
√m/M as small parameter, or by molecular dynamics (MD) simulations, see the references cited
in[Fruleuxetal.(2012)]. Boththeperturbative approach andtheMDsimulation consistently con-
cluded that the Brownian piston will move steadily towards the hotter gas. Nevertheless a clear
Presentedat25thSmoluchowskisymposium,September,2012inKrakow.
∗
(1)
2 MDD-TO-HYDRO PRINTED ONJANUARY30,2013
physicalunderstanding wasmissing. Afrequentlygivenhand-wavingargumentwasthatthehotter
side losing the heat has locally lower pressure. But it is not a valid argument. If the ideal gas
is used, the cooled particle will never hit again the Brownian piston while the freshly colliding
particles are characterized by the equilibrium parameters, (T,p) or (T ,p). The adiabatic piston
′
has, therefore, remained among ”Some problems in statistical mechanics that I would like to see
solved”[Lieb(1999)].
It is only very recently [Fruleuxetal.(2012)] that the physical explanation to the above ques-
tionwasdefinitely given. Thekeyistotakeintoaccount theinterplay betweentheenergytransfer
and the momentum transfer at the gas-piston interfaces. Once this point is understood, the results
of elaborated perturbative calculations could be perfectly reproduced just by a few lines’ calcula-
tions, except for an overall numerical factor. The purpose of the present paper is to summary the
basicideaanddiscussitsgeneralization. Theorganizationofthepaperisthefollowing: Inthenext
section (§ 2) we recapitulate the main line of this mechanism. Especially the key concept of the
momentum transfer deficit due to dissipation (MDD) is explained using a simple argument. The
relation to the traditional calculative approach is also mentioned. As a prologue to the extension
to the dense gas case, we introduce in § 3.2 a toy model that shows how the energy flow and mo-
mentum flowhaving different symmetryinspace andtimecanmaketheuniform pressure andthe
heatconduction compatible. In§3.3weshowtheimplication oftheMDDtothenon-equilibrium
hydrodynamics ofdensehard-core gas.
2.Reviewofthephysicsofadiabaticpiston
We outline the concept of MDD as the underlining mechanism of the adiabatic piston. The
readers might refer to [Fruleuxetal.(2012)] [Kawaietal.(2012)] for the technical details and its
generalization toinelastic case.
The essential point of the adiabatic piston is more clearly grasped when the Brownian piston
separatingtheidealgasesistrappedbyapotentialforcesuchasanelasticspring(Fig.1)sothatthe
meanvelocity ofthepistonvanishes ( V = 0). Ifthereappears anon-equilibrium force F on
NESS
h i
thistrapped Brownianpiston, thesteady statevelocity V ofthepistonintheabsence oftrapping
isgiven bythe balance withthe passive frictional forceh, Fi (γ+γ ) V = 0, where γ and γ
NESS ′ ′
− h i
arethefrictioncoefficientoftheBrownianpistonagainsttherespective gas.
T, p T’, p
Fig.1. Trappedadiabaticpiston.
The first step is to realize that the Brownian motion of the piston serves merely as the medi-
ator of the energy transfer, or heat, from hotter gas to the cooler gas. While correlation between
collisions withthepiston bythehotgasparticles andbythe coldgasparticles areessential forthe
irreversibility of this purely mechanical problem, we can bypass all the details for the purpose of
understanding thenon-equilibrium force F . Therateoftheenergy transferperunitsurface of
NESS
MDD-to-hydro printedonJanuary30,2013 3
thepiston, j(e)canbefoundusingthestochasticenergetics[Sekimoto(2010)]orevenbyaheuristic
argument [ParrondoandEspan˜ol(1996)]. The result reads, j(e) = (k T k T )/[M(γ˜ 1+γ˜ 1)],
B B ′ − ′−
whereγ˜ = γ/Aandγ˜ = γ /Awith Abeingtheareaofeachpistonsurfac−e.
′ ′
Once we know the energy transfer rate across the gas-piston interface, we can concentrate on
the following problem: When an ideal gas prepared in the equilibrium characterized by (T,p) is
brought into contact with the wall that absorbs (or injects) energy at the rate j(e) per unit surface
(Fig. 2), how the pressure on the wall is modified from p? Consider that gas particles with a
ρ
(0)
T (x ) p
neq ’
j(e)
Fig.2. Gasincontactwithanenergy-transferringwallat x = 0. Thecontactdensityisdenotedbyρ(0). In
generaltheeffectivetemperatureT (x)shoulddependonthepositionx.
new
typical velocity component normal to the wall v collides the energy-transferring wall. They are
⊥in
reflected back with a velocity v . While the typical incoming velocity should be the thermal
⊥out
velocityv = v = √k T/m,exceptforthenumericalfactor,thetypicaloutgoingvelocitydepends
⊥in th B
on the energy transfer rate j(e) and the collision rate ν per unit area on the energy-transferring
col
wallthrough theenergybalance condition:
j(e) m m m m
= v 2 v 2 = v 2 v 2.
ν 2 ⊥in − 2 ⊥out 2 th − 2 ⊥out
col
Assuming that the energy transfer is sufficient small, the right hand side (r.h.s.) of the above
equation isapproximated as
v + v
(mv mv ) ⊥in | ⊥out| (mv mv )v .
⊥in − | ⊥out| 2 ≃ ⊥in− | ⊥out| th
Substituting thisresultintotheaboveequation, wefind
j(e)
(mv mv )ν (1)
⊥in− | ⊥out| col ≃ v
th
The left hand side (l.h.s.) of this relation gives the momentum transfer deficit due to dissipation
(MDD). In other words, upon the collision, the gas particles kicks the wall less strongly in non-
equilibriumthanequilibriumifapartoftheirincomingkineticenergywastakenoutbytheenergy-
transferring wall. Intermsofthenetmomentumtransferrateperunitsurface, j(p) = ν (mv +
⊥⊥ col th
v ),theequilibrium value, p = 2mv ν ,iscorrected bythisMDDtogive
| ⊥out| th col
j(e)
j(p) = p .
⊥⊥
− v
th
4 MDD-TO-HYDRO PRINTED ONJANUARY30,2013
Inretrospect,thetraditionalapproachthroughtheMaster-Boltzmannequationcouldhavegiven
thesameinsight. ForthesetupofFig.2,thisequation canbewrittenasfollows:
∂ P(X,V,t)= V∂ P(X,V,t) [ γ V ∂ U(X)]∂ P(X,V,t)
t X b X V
− − − − k T
W(V V)P(X,V,t)+ W(V V )P(X,V ,t)+ B b∂2P(X,V,t),
−ZV ′| ZV | ′ ′ γb X
′ ′
whereP(X,V,t)istheprobabilitydensityofthepositionXandvelocityV ofthewallasaBrownian
piston, and U(X) represents the trapping potential energy. The heat absorption by the wall is
modeled by the coupling to a Langevin bath [Sekimoto(2010)] at the temperature T with the
b
coupling, i.e. friction, constant γ .Thecollision ofthegasparticles isrepresented bythevelocity
b
transition rate, W(V V),givenby
′
|
W(V′V)dV′dt = H(v⊥ V) ρA(v⊥ V)dt m e−2kmBTv⊥2 m+ M dV′
| − × − r2πkBT (cid:18) 2m (cid:19)
whereAisthesurfaceareaofthewall,v isthenormalcomponentoftheincomingvelocityofgas
⊥
particle, and H(z)istheHeaviside unitstepfunction. v isthefunction ofwall’svelocities before
⊥
(V)andafter(V )thecollision, respectively, throughthemomentumconservation rule,
′
2m
V = V + (v V).
′ m+ M ⊥ −
The truncated equations for the first two moments of V can be derived from the above Master-
Boltzmannequation, andtheresultsread
d V j(e)
M h i = U (X) (γ+γ ) V + p c A
′ b
dt −h i− h i − v !
th
d V2 γ γ
M h i = VU (X) [ V2 k T] b[ V2 k T ]+c V
′ B B b ′
dt −h i− M h i− − M h i− h i
where c = √π/8 and the other constant c is irrelevant as we shall see immediately below. In
′
the steady state, not only V = d/dt = 0 but also VU (X) vanishes. Then the second moment
′
h i h i
equation tells that the kinetic temperature of the Brownian piston, T M V2 , is given by the
kin
≡ h i
wellknownformulaofLangevindynamics,
γk T +γ k T
k T = B b B b
B kin γ+γ
b
Moreover, the second and the third terms on the r.h.s. of the second moment equation gives the
energytransferto[from]thewall,respectively:
γ γ
j(e) = [ V2 k T]= b[ V2 k T ].
B B b
−M h i− M h i−
MDD-to-hydro printedonJanuary30,2013 5
With j(e) thus known, the first moment equation in the steady state is nothing but the momentum
balancecondition:
j(e)
U (X) + p c A = 0.
′
−h i − v !
th
Our physical reasoning, therefore, reproduces completely the traditional result except for the nu-
merical factor c. Moreover, our explanation allows to treat the adiabatic piston, Brownian ratchet
models[VandenBroecketal.(2004)],orinelasticpiston[Costantini etal.(2008)]onthesamefoot-
ing[Fruleuxetal.(2012)].
3.Momentumtransferofagaswithheattransport
3.1.Preliminaryargument
Themean free path ℓ ofanideal gasisinfinite because the particles undergo no collisions.
map
Knudsen numberKn ℓ /L istherefore infinite withanysystem size, L .Themacroscopic
map sys sys
≡
thermo-hydrodynamics [LandauandLifshitz(2000)] supposes the opposite limit, Kn 1. When
≪
we study the thermo-hydrodynamics with energy-transferring boundaries, the physical ideas ob-
tained in the previous section should, therefore, be applicable only to the vicinities of those walls
probablywithsomemodifications. ThemainquestionishowtoreconciletheformulaEq.1forthe
idealgaswiththemacroscopicdescriptionofthermo-hydrodynamics withnon-equilibrium bound-
ary condition. In this paper we limit ourselves to the steady states with vanishing macroscopic
velocity of the gas. The conservation laws of mass, momentum and energy then impose the con-
stancyofthosefluxes. Belowwestudyfirstbyapurelymechanicaltoymodelthatshowsthebasic
compatibilitybetweenthesefluxesandtheirnatureofsymmetryinspaceandintime(§§3.2). Then
wegoontothedensehard-core gaswith Kn 1(§§3.3).
≪
3.2.Toymodel
Webeginbyaveryelementarykineticmodeltodiscusstheinterplayoftheenergyandmomen-
tum transfer1. We take up a single gas particle on the x-axis bounded by the energy-transferring
walls at x = 0 and at x = L , which are macroscopically fixed in space, see Fig. 3. We further
sys
simplify that the hot wall (on the left) receives the particle of velocity v and returns with the
velocityv+ with0 < v < v+. Thecoldwall(ontheright)doestheoppos−ite−operation. Themicro-
scopicmechanism und−erlying thesereflectionsareirrelevantforourargument. (Onmightimagine
thetwotennisplayerengaginginarally.)
Before counting the momentum and energy flux, j(e) and j(p), weimpose the vanishing of the
massflux, j(m) inthesteadystate:
j(m) = (ρ+mv+ ρ mv )xˆ = 0.
− − −
Thisissatisfiedbythedensities ofrightward andleftwardparticle, respectively,
v
ρ± = (v++v∓ )Lsys.
−
1ThisisasimplifiedversionofKnudssenheattransfer,see,forexample,[Struchtrup(2005)],page25.
6 MDD-TO-HYDRO PRINTED ONJANUARY30,2013
time
x
0 Lsys
Fig.3. Space(x)-timetrajectoryofaparticlebetweenthehot(x=0)andcold(x= L )walls.
sys
Alsothecollision frequency oneachwall,ν isfoundtobe
col
νcol = (v+ +v+vv−)Lsys.
−
Withthisν ,theenergy transferrate j(e) is
col
j(e) = m2(v2+ −v2−)(v+ +v+vv−)Lsysxˆ = m2(v+−v−)vL+svy−s xˆ
−
whilethemomentumflux j(p) reads
j(p) = m(v++v−)(v++v+vv−)Lsysxˆxˆ = mvL+svy−s xˆxˆ.
−
We verify that j(e) is odd under time or space-inversion, while j(p) is even under these operation.
The symmetry of j(p) can also be seen from Fig 3. This model, although simple, shows how the
directed energy transfer is established without gradient of momentum flux. In other words, the
pressure onthehotandcoldwallsarethesame.
To see more in detail the process at the walls, we refer to the contact value theorem, p =
k Tρ(0) [Henderson etal.(1979)], which give the equilibrium momentum transfer to a hard wall
B
by a hard core gas in terms of the equilibrium temperature T and the gas particle density at the
closestcontactsurfaceofthehardwall. Ourinterestisthe casewithenergy-transferring walls,see
Fig.2. Whenthewallstransfertheenergy, the pshouldindicatethetotalmomentumflux j(p),i.e.,
pneq = j(p) = m(v++v )νcol
| | −
andk T /2thekineticenergy perparticle,
B neq
kBT2neq = m2 ρ+vρ++2++ρρ−v−2 = m2v+v−.
−
Sincethetotaldensityρ(0)onthewallis,bythehomogeneity,
1 1 1
ρ(0) = + ν =
col
v+ v ! Lsys
−
MDD-to-hydro printedonJanuary30,2013 7
wearriveataformofthecontactvaluetheoreminnon-equilibrium.
p = k T ρ(0). (2)
neq B neq
In other words, while the symmetry allows the correction to the r.h.s. of the form (v+ v )2 or
j(e)2,thecontact valuetheorem holdsuptotheorderof (j(e)2)if p andT ar∝eappr−opr−iately
eq eq
∝ O
chosen(cf.[Komatsuetal.(2008)]).
3.3.Non-equilibrium hydrodynamics
Inthenon-equilibriumsteadystatewithheatfluxofadensehard-coregaswithKn 1,theen-
ergyfluxvectorfield, j(e),andmomentumfluxtensorfield,j(p),mustsatisfythebasicc≪onservation
laws:
j(e) = 0, j(p) = 0.
∇· ∇·
Ifthe wall isperpendicular to the x-axis, thesystem ishomogeneous in yand zdirections and the
aboveconditions arereduces to
j(e) = const. j(p) = const.
x xx
Ifthe heat conduction obeys approximately the Fourier’s law, j(e) = k T, withk being the heat
T T
∇
conductivity, the temperature gradient keeps constancy of the energy flux. As for the momentum
flux j(p), the symmetry argument or Curie principle [Curie(1894)] allows the anisotropy of the
type j(p) = p1 +a(xˆxˆ 11) with a characterizing the deviatoric part of the flux due to heat flux
− 3
x. However, seeing that the (j(e)2) contribution was missing in the above simple model Eq. 2,
wk e simply identify j(p) = p O1 to be the pressure in the present approximation. If p obeys
neq neq
approximately the equilibrium equation of state, p = p(ρ,T), among the pressure p, temperature
T andthedensity ρ,thedensity ρ(x)variesinamannerlocally compensating theheterogeneity of
thetemperature T(x)sothatthe p remainshomogeneous.
neq
Our concern is how we can relate the momentum flux and the energy flux in the dense hard-
core gas where p reflects already both the incoming and outgoing particles. Below we will
neq
indicate that the relation like Eq. 1 corresponds to the skewness of the velocity distribution of
particles (especially) at the energy-transferring wall. To be concrete we imagine the dense hard-
coregaswhichisconductingtheheatrightwardsuptotheenergy-transferringwallat x = 0without
convection (Fig.2). Wealsoassumethatthewallexchanges onlythe x-component ofmomentum.
Nowweintroducethevelocitydistribution function f(v ;x)perunitvolumeofgasparticles. Then
x
theparticledensity ρ(x)isgivenby
ρ(x)= f(v ;x)dv .
x x
Z
The conditions on the fluxes of mass, momentum and energy along the x axis are given, respec-
tively,as
0 = j(m)(x) xˆ = mv f(v ;x)dv
x x x
· Z
8 MDD-TO-HYDRO PRINTED ONJANUARY30,2013
p = xˆ j(p)(x) xˆ = mv2f(v ;x)dv
neq · · Z x x x
m
j(e) = j(e)(x) xˆ = v3f(v ;x)dv , (3)
· Z 2 x x x
where p and j(e) areindependent oftheposition x.
neq
Nowwefocusonthethinslabofthedistance ℓ fromtheenergy-transferring wall. Inthis
mfp
≪
slabweassumethatthegasparticlesundergopracticallynocollisionsexceptforwiththewall. We
(p)
introducethepartialmomentumfluxesassociatedtotheincomingparticles, j andtotheoutgoing
in
particles, j(p),rightbeforethewall:
out
j(p) mv2H(+v )f(v ;x)dv , j(p) mv2H( v )f(v ;x)dv .
in (cid:12)x=0 ≡ Z x x x x out(cid:12)x=0 ≡ Z x − x x x
(cid:12) − (cid:12) −
(cid:12) (cid:12)
Inequilibrium(cid:12) where f(v ;x)aresymmetricwithrespec(cid:12)ttov ,thebothpartialfluxesarethesame.
x x
In the presence of the heat fluxit is no more the case. While the asymmetric velocity distribution
for Knudsen heat transfer, i.e. the above toy model, is usually singular and far from Maxwellian,
thecollisionsmakethevelocitydistributionlookslikeskewedMaxwelldistribution. We,therefore,
assumeanapproximate form2:
f(vx;x)|x=0− = "c0+c1(cid:18)vσx(cid:19)+c2(cid:18)vσx(cid:19)2+c3(cid:18)vσx(cid:19)3#e−2vσx22.
For a week energy flux, the terms containing c , c and c are regarded to be small perturbations
1 2 3
withrespect tothe mainterm c . Theexpression ofthedensity andthe fluxconditions mentioned
0
aboveimpose
ρ
|x=0− = (c0+c2)σ
√2π
0 = c +3c
1 3
p
neq = (c +3c )σ3
0 2
√2πm
2j(e)
= (c +5c )σ5 = 2c σ5, (4)
1 3 3
3√2πm
where in the last line we used the vanishing mass flux condition in the second line; c = 3c .
1 3
−
(p) (p)
Finallythedifference betweenthepartialmomentumfluxes j and j read
in out
(p) (p)
(jin − jout)|x=0− = (c +4c )σ4 = c σ4,
1 3 3
2m
whereagainweused thevanishing massfluxcondition. Thenifweintroduce thesquared average
oftheparticle velocity(noting c /c 1forweaknon-equilibrium),
2 0
| | ≪
p 1+3c /c
v2x|x=0− = ρ|xn=e0q− = 1+c22/c00σ2 ≃ σ2,
2Onthewall, x = 0,theveryMDDimpliesthediscontinuityin f(v ;x)atv = 0. Here,however, weshallusea
x x
smoothedformasqualitativemodel.Seealso,forexample,[Struchtrup(2005)],page202.
MDD-to-hydro printedonJanuary30,2013 9
weobtaintherelationofMDDandtheenergy flux,reminiscent ofEq.1:
j(e)
j(p) j(p) =C
(cid:16) in − out(cid:17)(cid:12)(cid:12)(cid:12)(cid:12)x=0− (cid:20)v2x|x=0−(cid:21)1/2
withanumerical factorC = √2/(3√π),whichissubject toourapproximations.
In onclusion, the dense hard-core gas conducting the heat carries also momentum through the
asymmetricdistribution ofparticle’svelocity, andtheconceptofMDDisaneatwaytoexplainthe
relation between the energy flux and the partial momentum fluxes in the non-equilibrium steady
state.
Wewouldliketoacknowledge theorganizers ofthe25thSmoluchowskiSymposium.
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