Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Moderne Theoretische Physik III SS 2015
Prof. Dr. A. Mirlin Blatt 03, 100 Punkte
Dr. U. Karahasanovic, Dr. I. Protopopov Besprechung, 15.05.2015
1. Microcanonical ensemble for the ideal Boltzmann gas.
(15 + 15 + 5 + 5 = 40 Punkte, m¨undlich) (a) We write (p, r) for the full set of coordinates and momenta of the gas particles (DN
coordinates and DN momenta). We also denote by H(p, r) =X
i
p2i
2m (1)
the Hamiltonian of the system.
We start from the microcanonical distribution in the phase space, which is a uniform distribution on the surfaceH(p, r) =P
p2i/2m=E
ρ(p, r) =Aδ X
i
p2i/2m−E
!
. (2)
Hereδ(x) is the Dirac δ function andAis the normalization constant which we will find next from the normalization condition (ri demote the coordinates of particles)
1 = 1 N!
Z N Y
i=1
dDpidDri
(2π~)D ρ(p, r). (3) The factor 1/N! appears here because of the indistinguishability of particles.
The integral over coordinates ri produces simply VN. We now note that the hy- persurface of constant energy in momentum space is in our case just a sphere of dimension d = N D−1 with radius R = √
2mE so it is natural to compute the integral in spherical coordinates. We introduceX2 =PN
i=1p2i and write the norma- lization integral as
1 = A N!
VN (2π~)N Dσd
Z ∞
0
dXXdδ(X2/2m−E) = Amσd N!
VN
(2π~)N DRd−1 (4) Here σd= 2πN D/2/Γ (N D/2) is the surface of ad=N D−1 dimensional sphere.
We find for the normalization constant A=N! Γ
N D 2
2π~2 mE
N D/2
E
VN. (5)
(b) Let us now compute the entropyS(E, V, N). We shall use the definition of entropy in statistical mechanics
S =−kBhlnρi. (6)
Before we do so we regularize our microcanonical ensemble by giving the delta function in Eq.(2) an infinitesimal width ∆E (cf. lectures).
ρ(p, r) = A
∆E, E < H(p, r)< E+ ∆E
0, otherwise. (7)
The entropy now reads S =−kBln (A/∆E) =−kBln
N! Γ
N D 2
+kBN D
2 ln mE
2π~2+kBNlnV−kBln E
∆E. (8) We are interested only in the terms proportional to the Number of particles in the limitN ≫1. We apply the asymptotic expansion of the gamma function and get.
S
kB =−NlnN−N D
2 lnN D
2 +N+N D
2 +N D
2 ln mE
2π~2 +NlnV
= N D
2 ln mEe
π~2N D +NlnV e N (9) As it should be, at given concentration N/V and energy per particle E/N the entropy is proportional to the number of particles.
(c) We now remind that (the energyE is just the average energy of the system)
dE=T dS−pdV +µdN. (10)
Correspondingly
dS = 1
T (dE+pdV −µdN). (11) We thus find
T = 2E
N D, E = N D
2 kBT, (12)
p= N kBT
V , p= 2E
V D, (13)
µ=− 2E N DkB
D
2 ln mE
π~2N D + lnV N
, (14)
µ=−T kB D
2 lnmT kB
2π~2 + lnV N
. (15)
(d) We find the entropy as a function of temperature S(T, V, N) = N DkB
2 lnmT kBe
2π~2 +N kBlnV e
N (16)
This result coincides with that derived in exercise sheet 1 up to the dependence of the entropy on the number of particles (which was assumed to be constant there).
The entropy (16) does not satisfy the third law of thermodynamics. It even beco- mes negative when the mean inter-particle distance in the gas (V /N)1/D becomes of the order of the thermal de Broglie wavelength λTp
2π~2/kBmT. This is the manifestation of the fact that the classical description of the particle motion is not applicable to at low temperatures.
2. Moving gas and generalizations of microcanonical distribution.
(a) We start from the consideration of the stationary solutions of the Liouville equation
∂ρ
∂t =− {H, ρ}, (17)
where the Poisson brackets {,} are given by {f(x), g(x)} =X
i
∂f
∂pi
∂g
∂ri − ∂f
∂ri
∂g
∂pi. (18)
Here,x stands for the full set of coordinates in the phase space.
Let us consider the chain rule for the Poisson brackets. Specifically, let f(x) be expressed as a function of some variableJα(x),α = 1, . . . m
f(x) =f(J1(x), J2(x), . . . Jm(x)). (19) We have
{f(x), g(x)}=
N
X
i=1
∂f
∂pi
∂g
∂ri − ∂f
∂ri
∂g
∂pi =
N
X
i=1 m
X
α=1
∂f
∂Jα
∂Jα
∂pi
∂g
∂ri − ∂f
∂Jα
∂Jα
∂ri
∂g
∂pi
=
m
X
α=1
∂f
∂Jα {Jα, g}. (20) In particular, for ρ(x) =ρ(H(x), Q(x)) we get
ρ(x), H(x) = ∂ρ
∂H{H, H}+ ∂ρ
∂Q{Q, H}. (21) The Poisson bracketH, Hvanishes trivially, while the Poisson bracketQ, Hvanishes in any system where Qis conserved quantity.
(b) To find the entropy S(E, Q, N, V) we proceed along the lines of exercises 1a and 1b. We density of the microcanonical reads now
ρ(p, r) =Aδ X
i
p2i/2m−E
!
δ X
i
pi−Q
!
. (22)
The normalization constantA can be obtained from the equation 1 = 1
N! Z N
Y
i=1
dpidri
(2π~)ρ(p, r) = A N!
Z N
Y
i=1
dpidri
(2π~)δ X
i
p2i/2m−E
!
δ X
i
pi−Q
!
= AVN N!(2π~)N
Z N Y
i=1
dpiδ X
i
p2i/2m−E
!
δ X
i
pi−Q
! (23) To compute the normalization integral we first make a Galilean transformation to the center of mass frame, ˜pi=pi−Q/N. We get
1 = AVN N!(2π~)N
Z N Y
i=1
d˜piδ X
i
˜
p2i/2m−E˜
!
δ X
i
˜ pi
!
, (24)
where ˜E =E−Q2/2mN is the energy of the system in the moving frame.
Geometrically, the integration in Eq. (24) runs over section of the (N−1)-dimensional sphere (the first delta-function) by the hyperplane P
ip˜i which is obviously an (N−2)-dimensional sphere. The unit vector orthogonal to the hyperplane has coor- dinates
n= 1
√N(1,1, . . .1). (25)
We can now go to the new orthogonal coordinate system ˜p˜i, such that the coordinate
˜˜
pN is the coordinate in the direction perpendicular to the hyperplaneP
ip˜i = 0 (so that ˜p˜N = 0 at the hyperplane)
˜˜
pN =γX
i
˜
pi (26)
The proportionality coefficientγ can be found by noting that the point (˜p˜1 = 0,p˜˜2= 0, . . .p˜˜N−1 = 0,p˜˜N = 1) has in the old coordinate system the coordinates n, Eq.
(25). Finally
˜˜
pN = 1
√N X
i
˜
pi (27)
We now introduce the spherical coordinates in the hyperplane P
ip˜i = 0 (equiva- lently, ˜p˜N = 0), and also the notation X2 =PN−1
i=1 p˜˜2i. The normalization integral can now be presented as
1 = AVN
N!(2π~)NσN−2
Z
dXdp˜˜NXN−2δ X2 2m + p˜˜2N
2m −E˜
! δ√
Np˜˜N
= AVNσN−2
√N N!(2π~)N m p2mE˜
(2mE)˜ N/2−1. (28) To compute the entropy we broaden theδ-functions ensuring energy and momentum conservation, use the definition
S =−kBhlnρi (29)
and take the limit N ≫1.
We find
S(E, Q, N, V) kB = N
2 ln mEe˜
π~2N +NlnV e
N , E˜ =E− Q2
2N. (30)
This result differs from the one obtained in the standard microcanonical ensemble just by replacement E→E. All traces of the funny integration we had to perform˜ to obtain the finite-N result are washed out in this limit. In particular,
S(E, Q= 0, N, V) =S(E, N, V). (31) We can understand this as follows. Consider a gas in box. This is a system where the momentum is not conserved and which is described by the standard microcano- nical distribution. The average value of the momentum of the gas is zero. As with all extensive quantities in thermodynamics, the fluctuations of the momentum Q, (∆Q)2 ∝ N. The momentum is not conserved but deviates only slightly from the its average value 0. In particular, the typical value for the velocity of the center of mass of the gas isp
(∆Q)2/N m∝1/√
N. This is the reason why microcanonical en- semble without momentum conservation is equivalent to the momentum-conserving ensemble with Q= 0 (for N ≫1).
As for the derivative of the entropy with respect toQ, we find
∂QS=− ∂E˜S Q
N m =−v
T (32)
Here v is the macroscopic velocity of the gas.
(c) The consideration of the exercise 2b tells us that
S(E, Q= 0, N, V) =S(E, N, V) (33) We can now use the Galilean invariance to show that
S(E, Q, N, V) =S(E−Q2/2mN, N, V) (34)
3. Grand canonical ensemble and generalized Gibbs ensemble
(a) Let us consider a system A attached to the reservoir R. A and R can exchange energy and particles. We need to compute the probability Wn that A is in a state n with energy En and number of particles Nn. The energy of the R+A and the number of particles in R+A are fixed. We denote them by E and N respectively.
The full system of A+R obeys the microcanonical ensemble. Let us denote (in accord with the notations on the lectures) the number of states in some system S with energy in the interval (E, E+dE) and number of particles in the interval (N, N +dN) by dNS(N, E).
Among dNR+A(E, N) the states of R +A with in (E, E +dE) and the particle number in (N, N+dN) (which the systemR+Acan occupy with equal probabilities) there are exactly NR(E−En, N−Nn) states where the system Ais in the state n.
So the probability
Wn= dNR(E−En, N −Nn)
dNA+R(E, N) (35)
We now compute the logarithm of Wn taking into account that the system A is small;l compared to the reservoirR
kBlnWn=kBlndNR(E−En, N −Nn)−kBlndNR+A(E, N)
=SR(E−En, N−Nn)−SR+A(E, N) =SR(E, N)−SR+A(E, N)−∂SR
∂E En−∂SR
∂N Nn (36) Taking into account the thermodynamic relations∂S/∂E = T1 and∂S/∂N =−µ/T we get
lnWn= const− En
kBT +µNn
kBT. (37)
Here const denotes the terms which do not depend onEn and Nn.
(b) We follow the same route as in the previous exercise. For the probability to find the system in the state nwe find
kBlnWn=SR(E−En, N−Nn, Q−Qn)−SR+A(E, N, Q)
=SR(E, N, Q)−SR+A(E, N, Q)− ∂SR
∂E En− ∂SR
∂N Nn−∂SR
∂Q Qn
=−const− En
kBT +µNn
kBT + vQn kBT (38) Here we putv/T ≡ −∂QSR. The parameter v has the meaning of the macroscopic velocity of the reservoir.
