Smoluchowski problem for degenerate Bose gases

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by  A. V. Latyshev

Smoluсhowski problem for degenerate Bose gases
Anatoly V. Latyshev and Alexander A. Yushkanov
Department of Mathematical Analysis and Department of Theoretical Physics,
Moscow State Regional University, 105005, Moscow, Radio st., 10–A
(Dated: 4 января 2010 г.)
We construct a kinetic equation simulating the behavior of degenerate quantum Bose gases with
the collision rate proportional to the molecule velocity. We obtain an analytic solution of the half–
spaceboundary–valueSmoluchowskiproblemofthetemperaturejumpattheinterfacebetweenthe
0 degenerate Bose gas and the condensed phase.
1 Keywords: degenerate quantum Bose gas, Bose — Einstein condensate, collision integral,
0
temperature jump, Kapitsa resistance.
2

n
a PACSnumbers:05.30.-d,05.70.Ce,65.80.+n,82.60.Qr
J

3
1. INTRODUCTION conditions at the surface are purely diffusive.
]
h
p
In recent years, the behavior of quantum gases
- 2. KINETIC EQUATION
h
has increased interest. In particular, this is related
t
a to the development of experimental procedure for
m Todescribethegasbehavior,weuseakineticequation
producing and studying quantum gases at extremely
with a model collision integral analogous to that used to
[
low temperatures [1]. In the majority of papers, bulk
describeaclassicalgas.Inthiscase,wetakeintoaccount
1
properties of quantum gases have been studied [2] and
v the quantum character of the Bose gas and the presence
7 [3].Atthesametime,itisobviousthatitisimportantto
of the Bose — Einstein condensate.
0
take boundary effects on the properties of such systems
4 For a rarefied Bose gas, the evolution of the molecule
0 into account. In particular, such a phenomenon as the distribution function f can be described by the kinetic
. temperature jump at the interface between a gas and
1 equation
0 a condensed (in particular, solid) body in the presence
0 ∂f ∂E
of a heat flux normal to the surface is important. Such + ∇f =I[f],
1 ∂t ∂p
: a temperature jump is frequently called the Kapitsa
v where E is the kinetic energy of molecules, p is the
i temperature jump [4]. We also mention a paper where
X
molecule momentum, and I[f] is the collision integral.
the problem of boundary conditions for the motion of
r
a superliquids was considered [5]. In the case of the kinetic description of a degenerate
Bosegas,wemusttakeintoaccountthatthepropertiesof
The problem of a temperature jump in a quantum
the Bose — Einstein condensate can change as functions
Fermi gas was studied in [6], where an analytic solution
of the space and time coordinates. In other words, we
for an arbitrary degree of gas degeneracy was obtained.
mustconsideratwo–liquidmodel(moreprecisely,atwo–
A similar problem for a Bose gas was considered in
fluid model, because we consider a gas rather than a
[7], where the gas was assumed to be nondegenerate,
liquid). We let ρ =ρ (r,t) and u=u(r,t) denote the
i.e., it was assumed that there was no Bose — Einstein c c
respectivedensityandvelocityoftheBosecondensate—
condensate.
Einstein. We then have the expressions [9]
Thepresentpaperisdevotedtosolvingtheproblemof
atemperaturejumpinadegenerateBosegasanalytically. j=ρ u, Q= ρcu2u,
c
The presence of a Bose — Einstein condensate [8] leads 2
to a considerable modification of both the problem
Π =ρ u u .
ik c i k
statementanditssolutionmethod.Inthiscase,akinetic
equation with a model collision integral is used to for the densities j and Q of the respective mass and
describekineticprocesses.Weassumethattheboundary energy fluxes and for the momentum flux tensor Π of
ik
2
the Bose condensate — Einstein (under the assumption The relation u2 ≪kT/m, where k is the Boltzmann
that the chemical potential is zero). constant and T is the gas temperature, holds for
The conservation laws for the number of particles, sufficiently small a. In this case, we can neglect the first
energy, and momentum require that the relations terminthebracketsin(1).Theexpressionfortheenergy
E(p) takesthesameformasinthecaseofnoninteracting
∂ρ
c +∇j=− I[f]dΩ ,
B molecules:
∂t
Z
p2
E(p)= .
∂E p2 2m
c +∇Q=− I[f]dΩ ,
B
∂t 2m
Z We now consider the widely used kinetic equation
in the Boltzmann — Krook — Welander form with
∂(ρ u)
c +∇Π=− pI[f]dΩ the molecule collision rate proportional to the molecule
B
∂t
Z velocity [6, 7, 10]
be satisfied, where s is the molecule spin,
∂f
+(v∇)f =ν w(f∗ −f). (2)
(2s+1)d3p ρ u2 ∂t 0 M
c
dΩ = , E = ,
B (2π~)3 c 2
Here, f isthedistributionfunction, v isthemolecule
where ~ is the Planck constant. velocity, w =|v−v |, v isthemean–massgasvelocity,
0 0
In what follows, we are interested in the case of f∗ is the Maxwell distribution function,
M
stationary motion with small velocities (compared with
3/2
m m
the thermal velocities). We note that for the Bose fM∗ =n∗ 2πkT∗ exp −2kT∗(v−u∗)2 ,
condensate, the quantities Q and Π , are depend (cid:18) (cid:19) (cid:20) (cid:21)
ik
nonlinearly on the velocity (they are proportional to and ν isamodelparametercorrespondingtotheinverse
0
the respective third and second powers of the velocity). mean free path l of a molecule, ν ∼1/l.
0
Therefore, in the approximation linear in the velocity The parameters in the formula for fM∗ , namely, n∗,
u, the energy and momentum conservation laws can be T∗ and u∗, can be determined from the conservation
written as lawsforthenumberofmolecules,momentum,andenergy
p2
2mI[f]dΩB wfdΩM = wfM∗ dΩM, (3a)
Z Z Z
and
wvfdΩ = wvf∗ dΩ , (3b)
pI[f]dΩ =0. M M M
B
Z Z
Z
According to the Bogolyubov theory, the relation for
m m
the excitation energy ε(p) [8] w 2 (v−u)2fdΩM = w 2 (v−u)2fM∗ dΩM, (3c)
Z Z
p2 2 1/2 where
ε(p)= u2p2+ , (1)
2m
" (cid:18) (cid:19) # dΩ =d3v.
M
where
The conservation law for the number of particles in
4π~2an 1/2
u= . thenormalstateisinapplicablebecausethetransitionof
m2
(cid:18) (cid:19) particles to the Bose — Einstein condensate can occur.
Here a is the scattering length for gas molecules, As mentioned above, the effect of the condensate on
n is the concentration, m is the mass, and p is the the energy and momentum conservation laws can be
momentum of the gas molecule, holds for a weakly neglected in the approximation linear in u.
interacting Bosegas.Theparameter a characterizes the We note that Eq. (2) corresponds to the assumption
interaction force of gas molecules and can be assumed to that the mean free path of molecules is constant (it is
be small for a weakly interacting gas. independentofmoleculevelocities).Ithencefollowsthat
3
Eq. (2) to a greater extent corresponds to the model in WelinearizethelocalBose—Einsteindistribution f∗ ,
B
which the molecules are regarded as solid spheres. passing to dimensionless quantities. We note that
We consider a generalization of Eq. (2) to the case
T T
of a degenerate Bose gas. We assume that the general ε∗ = s βs(v−u∗)2 = s (C−W)2 ,
T∗ T∗
structure of Eq. (2) is preserved, but by a function fM∗ , (cid:2) (cid:3) (cid:2) (cid:3)
where
wemustmeantheBose—Einsteindistribution(Bosean)
with a zero chemical potential [8]
W= βsu∗.
−1
fB∗ = exp 2kmT∗(v−u∗)2 −1 . Taking T∗ =Ts+δTs inpto account, we obtain
(cid:20) (cid:18) (cid:19) (cid:21)
Here, the parameters T∗ and u∗ , are determined by ε∗ =C2− δT∗C2−2CW,
T
the second and third conditions in (3). In this case, we s
have whence we find
dΩB = (22sπ+~)13d3p δε∗ =−2CW−C2δTT∗,
s
instead dΩ . where
M
We assume that the mass velocity of the gas is much
less than the mean thermal velocity of the molecules δε∗ =ε∗−εs, εs =C2.
and the typical temperature variations along the mean
Consequently,
free path of molecules are small compared with the gas
taesmsupmeprattiounres.. The problem can be linearized under these fB∗ =fBs +(cid:18)∂∂fεB∗∗(cid:19)ε∗=εsδε∗,
We seek the distribution function in the form or
f =fBs(v)+ϕ(t,r,v)g(v), f∗ =fs +g(C) 2CW+C2δT∗ .
B B T
(cid:20) s (cid:21)
where
Wenotethatthequantity w inEq.(2)canbereplaced
1 m
fs(v)= , β = , with v in the approximation under consideration. We
B exp(β v2)−1 s 2kT
s s introduce the dimensionless quantities t∗ = tν /β and
0 s
ϕ is a new unknown function, T is the surface r∗ = rν and omit the asterisks on these quantities
s 0
temperature, and below. It is now clear that Eq. (2) (in the dimensionless
variables) becomes
∂
g(v)=− fs, ε =β v2.
∂εs B s s ∂ϕ +(C∇)ϕ=2C(CW)+C3δT∗ −Cϕ. (4)
∂t T
We introduce the notation s
m Theparametersofthisequationcanbefoundfromthe
C= βsv, ε∗ = 2kT∗(v−u∗)2. momentum and energy conservation laws (relations (3)),
p which now become
Taking this notation into account, we have
∗ 1 (Lϕ)C2g(C)d3C =0,
fB(ε∗)= exp(ε∗)−1, Z
1 (Lϕ)Cg(C)d3C =0,
fs(C)= ,
B exp(C2)−1 Z
where
g(C)= exp(C2) . Lϕ=2CCW+C3δT∗ −Cϕ.
(exp(C2)−1)2 T
s
4
From this system, we find temperatures.Thequantity ∆ iscalledthetemperature
jump (the Kapitsa temperature jump in the case of low
3
W= ϕCg(C)Cd3C, temperatures).
8πg
0 Z
The problem is to find ∆T as a function of the heat
flux Q. Taking into account that the problem is linear,
δT∗ = 1 ϕg(C)C3d3C,
we can write
T 4πg
s 2 Z
where ∆T =T Q.
Q
∞
The quantity T is called the coefficient of the
g = C5+ng(C)dC, n=0,1,2, Q
n
temperature jump. Another notation can be used:
Z
0
In this case, we have Q=R∆T,
∞ ∞
where R is called the Kapitsa resistance.
exp(C2)C5dC
g0 = g(C)C5dC = 2 = Takingintoaccountthattheproblemisstationaryand
Z Z exp(C2)−1
0 0 that the function ϕ is independent of the coordinates
(cid:16) (cid:17)
and z, we simplify Eq. (5):
∞
π2 ∂ϕ
=−2 Cln(1−exp(−C2))dC = =1.64493, µ +ϕ(x,µ,C)=
6 ∂x
Z
0
1 ∞
∞ 1 ′ ′ ′ ′ ′
= k(µ,C;µ,C )ϕ(x,µ,C )dΩ(C ), (6)
g = g(C)C6dC =2.22912, 2
1 −Z1 Z0
Z
0
where
∞ µ= Cx, dΩ(C′)=g(C′)C′3dµ′dc′,
g = g(C)C7dC =3.60617. C
2
Z
0 3 1
k(C,µ;C′,µ′)= CµC′µ′+ C2C′2.
We represent Eq. (4) in the standard form g0 g2
∂ϕ It is easy to verify that Eq. (6) has the particular
+(C∇)ϕ+Cϕ(t,r,C)=
∂t solutions
ϕ =µC, ϕ =C2.
C 1 2
= k(C,C′)ϕ(t,r,C′)C′g(C′)d3C′, (5)
4π
Z Consequently, the function
где
ϕ (x,µ,C)=BµC+ε C2, (7)
as T
3 1
k(C,C′)= CC′+ C2C′2.
g0 g2 where isproportionaltotheheatflux,isanasymptotic
distribution function (as x→+∞).
3. PROBLEM STATEMENT Assumingthatthereflectionofthemoleculesfromthe
wall is purely diffusive, we now formulate the boundary
A degenerate Bose gas occupies the half-space x > conditions
0 above the plane surface in the problem under
ϕ(0,µ,C)=0, 0<µ<1, (8)
consideration. A heat flux Q normal to the surface
is maintained in the gas. We let T denote the gas
0
ϕ(x,µ,C)=
temperaturefarfromthesurface.Thequantity T differs
0
from the surface temperature T if there is a heat flux.
s
We let ∆ = T0 − Ts be the difference between these =ϕas(x,µ,C)+o(1), x→+∞, −1<µ<0. (9)
5
4. REDUCTION TO THE ONE–VELOCITY where h (x,µ) is the asymptotic distribution function
as
PROBLEM AND THE SEPARATION OF
VARIABLES Bµ
h (x,µ)= .
as
" εT #
We seek the solution of problem (6)–(9) in the form
We note that the equation kernel can be represented
ϕ(x,µ,C)=Ch (x,µ)+C2h (x,µ). (10) in the form
1 2
′ ′
K(µ,µ)=K +3µµK ,
We obtain the system of equations 0 1
µ∂h1 +h1(x,µ)= 0 0 1 g1
∂x K0 = g1 1, K1 = g0 .
g 0 0
2
1 1    
3 3g The general method for the separation of variables
′ ′ ′ 1 ′ ′ ′
= µ µh (x,µ)dµ + µ µh (x,µ)dµ,
1 2
2 2g yields the expression
−Z1 0 −Z1
x
h (x,µ)=exp(− )Φ(η,µ). (14)
η
η
∂h
2
µ +h (x,µ)=
∂x 2 Substituting (14) in (11), we obtain the characteristic
equation
1 1 η 3
= g1 h1(x,µ′)dµ′+ h2(x,µ′)dµ′. (η−µ)Φ(η,µ)= 2K0n(0)(η)+ 2µηK1n(1)(η),
2g
2−Z1 −Z1 where
We represent this system of equations with respect to 1
the column vector n(k)(η)= µkΦ(η,µ)dµ, k =0,1.
−Z1
h (x,µ)
1
h(x,µ)=
"h2(x,µ) # It is obvious, that
in the vector form n(1)(η)=η(E−K0)n(0)(η),
1 where E is the unit matrix of the second order.
∂h 1
′ ′ ′
µ +h(x,µ)= K(µ,µ)h(x,µ)dµ, (11)
We obtain now the characteristic equation
∂x 2
−Z1
1
(η−µ)Φ(η,µ)= ηD(µη)n(η), (15)
where K(µ,µ′) is the kernel of Eq. (11), 2
g
3µµ′ 3 1µµ′ 1
K(µ,µ′)= g1 g01 . n(η)≡n(0)(η)= Φ(η,µ)dµ,
g2 −Z1
 
Using (10), we transform boundary conditions (8) and where
(9) into
D(µη)=K +3µηK (E−K )=
0 1 0
0
h(0,µ)=0, 0<µ<1, 0= , (12)
"0#
3gµη 0
= g1 ,
 1
g
2
h(x,µ)=
 
g2
g =1− 1 =0.96288.
=has(x,µ)+o(1), x→+∞, −1<µ<0, (13) g0g2
6
For η ∈ (−1,1), we use Eqs. (15) in the class of 5. EIGENVECTORS OF THE DISCRETE
generalized functions [11] to find the eigenvectors SPECTRUM
Φ(η,µ)=F(η,µ)n(η) By definition [7], the set of zeros of the dispersion
function is called the discrete spectrum.
of the continuous spectrum, where
The function λ (z) has a single double zero at the
0
1 1 point at infinity z = ∞. Two eigensolutions of Eq.
F(η,µ)= ηD(µη)P +Λ(η)δ(η−µ), i
2 η−µ (11) correspond to this zero (they coincide with the
eigenvectors of the characteristic equation):
1
n1(η) 1 0
n(η)= = Φ(η,µ)dµ, (16) h (x,µ)=µ , h (x,µ)= .
"n2(η)# −Z1 0 "0# 1 "1#
F(η,µ) is the eigen matrix–function, the symbol Px−1 We find zeros of λ1(z) outside the cut [−1,1]. It is
obvious that λ (∞)=1−g =0.03712>0. Wetake the
denotes the principal value of the integral in the 1
integrationof x−1, δ(x) isthedeltafunction,and Λ(z) contour γε (seeFig.1)thatencompassesthecut [−1,1]
at a distance ε from it such that there are no zeros
is the dispersion matrix,
of the function λ (z) inside the contour. The number
1
1
N of zeros of λ (z) outside the contour is equal to the
z D(µz) 1
Λ(z)=E+ dµ.
2 µ−z increment of the argument of this function according to
−Z1 the argument principle [12], i.e.,
The dispersion matrix has following elements
N =(2π)−1∆ argλ (z),
γε 1
Λ (z)=1+3gz2λ (z)≡λ (z),
11 0 1 wherethesymbol ∆ f meanstheincrementof f along
γε
γ . Passing to the limit as ε → 0 in this equality, we
ε
Λ (z)≡0, obtain
12
1 λ+(µ)
Λ21(z)= g1λ0(z)− g1, N = 2π∆(−1,1)argλ−11(µ),
g g
2 2
±
where λ (µ) are the boundary values of λ (z) on the
1 1
interval (−1,1) from above and from below,
Λ (z)=λ (z).
22 0
λ±(µ)=λ (µ)±3iπgµ3.
Werepresentthedispersionmatrixintheexplicitform 1 1
Taking into account that Reω+(µ)=ω(µ) is an even
Λ(z)=λ (z)D(z2)+D ,
0 0 function and Imω+(µ) is odd, we have
where 1
N = ∆ argG(µ),
π (0,1)
1 0
D0 =−gg21 0, G(µ)= λ+1(µ).
−
  λ (µ)
1
1 Let θ (µ) = argλ+(µ) be the principle value of the
z du 1 1
λ0(z)=1+ 2 u−z. argument specified by the condition θ1(0)=0. Because
−Z1
λ+(µ)=λ−(µ), |λ+(µ)|=|λ−(µ)|,
1 1 1 1
Itsdeterminant,calledthedispersionfunction,isgiven
by we have argG(µ)=2θ1(µ) and therefore
2
λ(z)=detΛ(z)=λ (z)λ (z). N = ∆ θ (µ).
0 1 π (0,1) 1
7
has a nonzero solution because
detΛ(η )≡λ(η )=0.
0 0
We represent Eq. (17) in the form of two scalar
equations
λ (η )n (η )=0, (18)
1 0 1 0
g
1
[λ (η )−1]n (η )+λ (η )n (η )=0. (19)
0 0 1 0 0 0 2 0
g
2
In view of the condition λ (η ) = 0, it follows from
1 0
Eq. (18) that the upper element n (η ) is arbitrary and
1 0
nonzero. From Eq. (19), we now find the lower element
of the vector n(η ),
0
y
g 1
z 1
n (η )=− 1− n (η ).
2 0 1 0
g λ (η )
2 0 0
(cid:16) (cid:17)
ΓR
From the equation
DR λ (η )=1+3gη2λ (η )=0
1 0 0 0 0
γε we find
• •• •
x 1
−1 O γε 1 λ0(η0)=−3gη2.
0
R=1
ε
Hence, the vector n(η ) has the form
0
ΓR 1
n(η0)=−g1(1+3gη2) n1(η0).
g 0
Fig.1. 2
 
We substitute this vector in the second condition in
Fig. 1. Dependence of the temperature jump coefficient on
(17) and obtain
the parameter 7: the specular reflection coefficient is q = 0.3
for curve 1, q = 0.5 for curve 2, and q = 0.8 for curve 3.
1 µ
Φ(η0)= η0−µ−gg1η0 ,
2
It is easy to see that the angle θ (µ) on the cut
1  
[0,1] has an increment equal to π; consequently, we for
have N = 2 (the number of zeros is equal to two). We 2
n (η )= =−2λ (η ).
let ±η (η = 1.27573) denote these zeros. In view of 1 0 3gη2 0 0
0 0 0
the equality λ (z) = λ (z¯), these zeros are real. Two
1 1
eigensolutions h±η0(x,µ), where 6. DECOMPOSITION OF THE SOLUTION INTO
x EIGENVECTORS: THE RIEMANN — HILBERT
h (x,µ)=Φ(η ,µ)exp(− ), (17a)
η0 0 η0 BOUNDARY VALUE PROBLEM
η D(µη )
0 0
Φ(η ,µ)= n(η ), (17b) We show that the solution of boundary value
0 0
2 η −µ
0
problem (11)–(13) can be represented in the form of a
correspond to them.
decomposition, namely,
We note that the homogeneous equation
x
h(x,µ)=h (x,µ)+A exp(− )Φ(η ,µ)+
Λ(η0)n(η0)=0 as 0 η0 0
8
∞
x Let’s lead reduction to the diagonal form of the
+ exp(− )F(η,µ)A(η)dη, (20)
η problem(25).Forthispurposewewillleadtoadiagonal
Z
0 kind the matrix
where the unknowns are the constants εT and A0 and λ1(z)
0
a vector function A(η) with the elements Aj(η),j = P(z)= 3gz2 .
 g 
1,2,3. Decomposition (20) can be represented in the − 1 λ (z)
3gg z2 0
form  2 
 
The matrix bringing the matrix P(z) to a diagonal
x
h(x,µ)=has(x,µ)+A0exp(− )Φ(η0,µ)+ kind, is that
η
0
g
0 −1 1 1
x S = g1 , detS =1, S−1 = g2 .
+exp(− )Λ(µ)A(µ)θ (µ)+ 1   
µ + g2 −1 0
   
It is obvious that
1
+21 exp(−xη)ηD(µη)A(η)ηd−ηµ, (21) S−1P(z)S ≡Ω(z)=diag λ31g(zz2), λ0(z) .
Z n o
0
We first solve the homogeneous boundary value
where
problem
θ+(µ)=1,µ∈(0,1); θ+(µ)=0,µ∈/ (−1,0). P+(µ)X+(µ)=P−(µ)X−(µ), 0<µ<1, (25)
Substitutingdecomposition(21)inboundarycondition corresponding to (24).
(12), we obtain a singular integral equation with the Clearly, that it is necessary to search for matrix X(z)
Cauchy kernel [12] in the form
h (0,µ)+A Φ(η ,µ)+Λ(µ)A(µ)+ X(z)=Sdiag U (z), U (z) S−1.
as 0 0 1 2
n o
The method for solving such problems was developed
1
1 dη in [8]; therefore, we give the solution of problem (25)
+ ηD(µη)A(η) =0, 0<µ<1. (22)
2 η−µ without a derivation,
Z
0
U (z) 0
We introduce an auxiliary vector function 1
X(z)= g1 ,
 [U0(z)−U1(z)] U0(z)
g
1 2
1 dη  
N(z)= ηD(zη)A(η) (23) where
2 η−z
Z
0
U (z)=zexp(−V (z)),
0 0
and the matrix
P(z)=Λ(z)D−1(z2). U (z)=zexp(−V (z)),
1 1
Using the boundary values N(z), Λ(z) and P(z)
1
and the corresponding Sokhotski formulas, we reduce 1 ζ0(u)du
V (z)= ,
0
π u−z
Eq.(22)toaninhomogeneousvectorRiemann—Hilbert
Z
0
boundary value problem
1
P+(µ)[N+(µ)+h (0,µ)+A Φ(η ,µ)]= 1 ζ (u)du
as 0 0 1
V (z)= ,
1
π u−z
Z
0
− −
=P (µ)[N (µ)+h (0,µ)+A Φ(η ,µ)], 0<µ<1.
as 0 0
(24) ζ (u)=θ (u)−π, ζ (0)=−π, ζ (1−0)=0,
0 0 0 0
9
ζ (u)=θ (u)−π, ζ (0)=−π, ζ (1−0)=0. C 1 d
1 1 1 1 1 1
+X(z) + ,
0 η −z d
In this case, the angles have the form (cid:20)h i 0 h 2 i(cid:21)
or, in scalar form,
π 2λ (u)
0
ζ (u)=− −arctg ,
0 2 πu A0z d1
N (z)=−Bz− +U (z) C + , (27)
1 1 1
η −z η −z
0 0
h i
π 2λ (u)
1
ζ (u)=− −arctg .
1 2 3gπu3 g A η
1 0 0
N (z)=−ε + +
2 T
g η −z
We now return to the solution of inhomogeneous 2 0
problem (24). Substituting the matrix P+(µ) found
g d d U (z)
using (25) in (24), we obtain the problem of determining + 1 U (z)−U (z) C + 1 + 2 0 . (28)
0 1 1
g η −z η −z
an analytic vector function from its jump 2 0 0
h ih i
Here C , d and d areconstantsthatcanbefound
−1 1 1 2
X+(µ) N+(µ)+has(0,µ)+A0Φ(η0,µ) = from the solvability conditions for the boundary value
h i h i problem.
−1 We note that
− −
= X (µ) N (µ)+h (0,µ)+A Φ(η ,µ) . (26)
as 0 0
h i h i U0(z)=z−V0−1+o(1), z →∞,
We note that
1
0 U (z)=z−V−1+o(1), z →∞,
X−1(z)= U1(z) , 1 1
 g 1 1 1 
1
− where
g U (z) U (z) U (z)
 2 0 1 0 
 (cid:16) (cid:17)  1
V(−1) =−1 ζ (u)du=0.84188,
N (z) 1 π 1
N(z)= 1 . Z
0
"N2(z)#
The asymptotic behavior of these functions in a 1
neighborhood of the point at infinity can be described V0(−1) =−π1 ζ0(u)du=0.71045,
as Z0
X−1(z)∼ z−1 0 , z →∞, Eliminating the pole of the function N1(z) given by
"z−2 z−1 # equality (27) at thepoint atinfinity, weobtain C1 =B.
Equating the limits on the right and on the left at the
point z =∞ in equality (27) for N (z) and using (23),
1
N(0)
N(z)= 1 +o(1), z →∞, we obtain the equation
" 0 #
1
where 3g η2A (η)dη =−A +d +BV(−1). (29)
2 1 0 1 1
Z
1 0
3g
N(0) =− η2A (η)dη
1 2 1 From the equation N2(∞) = 0, where the function
Z
0 N (z) is given by equality (28), we find
2
according to (23). ε =−d +Bg1 V(−1)−V(−1) . (30)
Taking the behavior of X−1(z) and N(z) at finite T 2 g 1 0
2
h i
points of the complex plane and in the neighborhood of
Eliminating the poles of the functions N (z) and
1
the point at infinity into account, we obtain a general
N (z) at the point η , we obtain
2 0
solution of problem (26)
A η
0 0
d =
N(z)=−has(0,z)−A0Φ(η0,z)+ 1 U1(η0)
10
and Substituting these equalities in (34), we obtain the
g g A η equation
1 1 0 0
d =− d =− .
2 1
g g U (η )
2 2 1 0 1
Taking these relations into account, we can rewrite 32g η2A1(η)dη = C1+ dη1 U1(0)+V1(−1) −
equality (30) as Z0 (cid:16) 0(cid:17)(cid:16) (cid:17)
εT = gg21 UA10(ηη00) +B(V1(−1)−V0(−1)) . (31) −dη1 U1(η0)−η0+V1(−1) . (35)
h i 0
(cid:16) (cid:17)
From definition (23) of the function N(z), we have
From Eqs. (29) and (35), we now obtain an equation
1 from which we find
3g η2A (η)dη
1
N (z)= z ,
1
2 η−z A =−BU (η ).
Z 0 1 0
0
whence Substituting the found value A0 in (31), we find that
the temperature jump is given by
N+(µ)−N−(µ)=3gπiµ3A (µ) (32)
1 1 1 ε =Bg1 η +V(−1)−V(−1) . (36)
T g 0 1 0
according to the Sokhotsli formulas. 2
h i
Substituting solution (27) in (32), we find
7. TEMPERATURE JUMP
3g
zη2A (η)=
1
2
Weexpressthetemperaturejumpintermsoftheheat
flux, which is only transferred by the normal component
1 d 1 d
+ C + 1 − 1 U+(η)−U−(η) . (33)
2πi 1 η η η (η−η ) 1 1 and is proportional to its velocity [9]. We note that the
0 0 0
h(cid:16) (cid:17) ih i mean velocity of the gas (the normal component and
Integrating equality (33) over η from 0 to 1, we
Bose condensate) is equal to zero in the gas volume. We
obtain
calculate the heat flux using the formula
1
32g Z0 η2A1(η)dη =(cid:16)C1+ dη01(cid:17)J(0)− dη01J(η0), (34) Q=Z fvm2v2 (2s(2+π~1))3d3p, p=mv. (37)
Passing to dimensionless quantities in the integral in
where
(37), we obtain
1 1 U+(τ)−U−(τ)
J(z)= 2πiZ0 1 τ −z1 dτ. Q= (22(s2π+~1))3mβs64 Z [fBs +ϕ(x,C)g(C)]CC2d3C,
or
We give the integral representation of the function
U (z) without a derivation, (2s+1)mk3T3
1 Q= s ×
2π3~3
U (z)−z+V(−1) =
1 1
× Ch (x,µ)+C2h (x,µ) CC2g(C)d3C.
1 2
1 1 U+(τ)−U−(τ) Z h i
= 1 1 dτ, z ∈/ [0,1]. Taking into account that
2πi τ −z
Z
0
C=(Cµ,Csinθcosχ,Csinθsinχ),
According to this representation, we have
J(η0)=U1(η0)−η0+V1(−1) d3C =C2dµdCdχ,
and we obtain
J(0)=U (0)+V(−1). Q (x)=
1 1 x