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(with H = 1 2 kT and ( H
2 − H
2 )/ H
2 = 2).
Example 15.5 Consider the Hamiltonian of a harmonic oscillator with one degree of freedom: H(q, p) = p 2 2m + 1 2 mω 2 q 2 , and compute the canonical partition function Z and the average energy E .
Recalling the definition (15.61) of Z and Appendix 8, we have Z =
1 h +∞ −∞ dq +∞ −∞ dp e
−βH(q,p) = 2π hωβ .
642 Statistical mechanics: Gibbs sets 15.10 We arrive at the same result by considering the Hamiltonian expressed in terms of the angle-action variables: H(χ, J ) = ωJ. In this case, recalling that (χ, J ) ∈ T
1 × R
+ , we have Z = 1
2π 0 dχ +∞ 0 e −βωJ dJ =
2π hωβ
· The average value of the energy is equal to H = −
dβ log Z =
d dβ log β = 1 β , in agreement with what was predicted by the theorem of equipartition of the energy. It is also immediate to verify that q 2
q 2 e −βH(q,p) dp dq
e −βH(q,p)
dp dq = 1 mω 2 β , p 2 = p 2 e −βH(q,p)
dp dq e −βH(q,p) dp dq = m β , from which we arrive at the equipartition 1 2 mω 2 q 2 = 1 2 kT , p 2 /2m = 1 2 kT . It is interesting to compare these averages with the averages over one period of the deterministic motion. It is convenient to start from the transformation p =
√ 2ωmJ cos χ, q = 2J
sin χ, imposing for the energy the value ωJ = kT , and hence J = kT /ω. Taking the average over a period of χ for these quantities, we find exactly the same results. We now compute H 2 = (hωβ/2π) (1/h) 2q 0 dχ +∞ 0 ω 2 J 2 e −βωJ
dJ = 2(kT ) 2 . Note that ( H 2 − H 2 )/ H
2 = 1.
It is elementary to observe that the latter result is necessarily different from the deterministic motion, for which the energy is constant. Hence the average over a period of any power of the energy coincides with the same power of the average (in other words (H 2 ) = (H)
2 = ω
2 J 2 and there is no fluctuation). As an exercise, compare the average p 4 /4m
2 in the statistical motion with the average (p 4 /4m 2 ) over a period of the deterministic motion ( 3 4
2 T 2 and 3 8 k 2 T 2 , respectively). 15.10 Statistical mechanics: Gibbs sets 643 Example 15.6 Consider the Hamiltonian of a harmonic oscillator with two degrees of freedom: H(q
1 , q
2 , p
1 , p
2 ) =
p 2 1 + p 2 2 2m + 1 2 mω 2 q 2 1 + q 2 2 , and compute the canonical partition function Z and the average energy E . Introducing planar polar coordinates, the Hamiltonian becomes H(ϕ, r, p ϕ , p
r ) =
1 2m p 2 r + p 2 ϕ r 2 + 1 2 mω 2 r 2 , from which it follows that Z = 1
2 2π 0 dϕ +∞ 0 dr +∞ −∞ dp ϕ +∞ −∞ e −βH(ϕ,r,p ϕ ,p r ) dp r = 2π h +∞ 0 2πm β re −mω 2 βr 2 / 2 dr = 4π 2 h 2 ω 2 β 2 +∞ 0 e −mω 2 βr 2 / 2 d mω 2 βr 2 2 = 4π 2 h 2 ω 2 β 2 , and therefore H = −
dβ log Z =
d dβ 2 log β = 2 β . We would arrive at the same result considering the Hamiltonian expressed in terms of the action-angle variables: H(χ 1
2 , J
1 , J
2 ) = ω(J
1 + J
2 ), by simply recalling that (χ 1 , χ
2 , J
1 , J
2 ) ∈ T 2 × R
2 + . Example 15.7 Consider the Hamiltonian of a point particle of mass m constrained to move on the surface of a sphere of radius R under the action of weight: in spherical coordinates we have H(ϑ, ϕ, p ϑ , p ϕ ) =
1 2mR
2 p 2 ϑ + p 2 ϕ sin 2 ϑ + mgR cos ϑ, where 0 ≤ ϑ ≤ π and 0 ≤ ϕ ≤ 2π. The canonical partition function is given by Z = 1
2 2π 0 dϕ π 0 dϑ +∞ −∞ dp ϕ +∞ −∞ e −βH(ϑ,ϕ,p ϑ ,p ϕ ) dp ϑ = 4π 2 mR 2 h 2 β π 0 e −mgRβ cos ϑ sin ϑ dϑ = 8π 2 R gh 2 1 β 2 sinh(βmgR). 644 Statistical mechanics: Gibbs sets 15.10 The average energy is therefore given by E = − d dβ log Z =
2 β − mgR cotan h(βmgR), from which it follows that E ≈ ⎧ ⎨ ⎩ −mgR, for β → +∞, 1 β , for β
→ 0. Hence for low temperatures, the energy stabilises in correspondence with stable equilibrium, while at high temperatures the system tends to forget the gravitational potential energy and sees only the quadratic part of the Hamiltonian. Example 15.8 Consider a system of N rigid segments of length a and mass m, constrained to move on the semi-axis x > 0. Denoting by (x i , x
i + a) the interval occupied by the ith segment, the point x 1 is attracted with an elastic force of constant λ by the point x = 0, while the adjacent endpoints (x i −1 + a, x i ) of two consecutive segments are also subject to the same force. Defining ξ 1 = x 1 and ξ
i = x
i −x i −1 −a,
i = 2, . . . , N and ξ i ≥ 0, we have the following expression for the Hamiltonian: H(ξ 1 , . . . , ξ N , p
1 , . . . , p N ) =
N i =1 p 2 i 2m + λ 2 ξ 2 i . If the system is kept at temperature T , its partition function has the value Z = 1 N !h N +∞ −∞ e −βp
2 / 2m dp +∞ 0 e −βλξ
2 / 2 dξ N = 1 N !h
N 2mπ
β N/ 2 π 2βλ
N/ 2 = π N N !h N m λ N/ 2 β −N , from which we deduce the average energy H = N kT , in agreement with the equipartition theorem. We now want to compute the average length of the system, given by
L = x N + a − ξ 1 = N a + N i =2 ξ i . The averages ξ i are all equal to ξ = ∞ 0 ξe −βλξ
2 / 2 dξ ∞ 0 e −βλξ
2 / 2 dξ = 2 πβλ .
15.11 Statistical mechanics: Gibbs sets 645 Hence L = N a + (N −1) 2/πβλ. The average x-coordinate of the last endpoint is x
N + a = N a + N 2/πβλ. The result is consistent with physical intuition: when the intensity of the elastic interactions grows, L decreases; when the temperature rises, L increases. In addition, we can compute the coefficient of thermal dilation δ of the system, which we can define as the derivative with respect to T of the average elongation : x
N + a
− Na , with respect to the extension x N + a :
δ = 1 x N + a
∂ ∂T x N + a
− Na = kβ a √ 2πβλ + 2
. 15.11
Helmholtz free energy and orthodicity of the canonical set In the microcanonical set, given the parameters E, V it is natural to choose the entropy S(E, V ) as the fundamental thermodynamical quantity. In the canonical set, the independent variables are rather T, V and the connection with thermodynamics is established through the Helmholtz free energy Ψ (V, T ), defined by T ∂ Ψ ∂T − Ψ = −U, (15.65) where U is the internal energy (hence H ). This expression highlights the fact that −U is the Legendre transform of Ψ (hence that Ψ is the Legendre transform of −U). A more familiar expression is U = Ψ
(15.66) from which, differentiating with respect to T , we find S = −∂ Ψ
can write ∂S ∂T = ∂S ∂E ∂E ∂T = 1 T ∂U ∂T . If H is independent of V , as for the equipartition case, by differentiating with respect to V we deduce P = −∂ Ψ /∂V (it is enough to observe that T ∂S/∂V = P ). We show now that a correct definition of the function Ψ (V, T ) starting from the canonical partition function is the following: Z(V, T ) = exp [ −β Ψ
(15.67) Note that by altering the numerical factors in the definition of Z the free energy is modified by the addition of a term linear in the temperature (which does not contribute to the expression Ψ −T (∂
Ψ /∂T ) for the internal energy). P roposition 15.3 The free energy Ψ (V, N, T ) = −kT log Z(V, N, T ) satisfies equation (15.65), and hence it coincides with the Helmholtz free energy. 646 Statistical mechanics: Gibbs sets 15.12 Proof
Using the expression (15.67) in the computation of H through equation (15.63), we obtain H = Ψ
∂ Ψ ∂T . (15.68)
Equation (15.68) reproduces the relation between the free energy and the internal energy (identified with the average value of the Hamiltonian). Example 15.9 Recalling the expression (15.64) for the partition function Z(V, T ) for a monatomic perfect gas, we easily find Ψ (V, T ) = −NkT log V N (2πmkT ) 3/2
h −3 . (15.69) From this it is immediate to check that ∂ Ψ /∂T =
−S, ∂ Ψ /∂V = −P . Note also that equation (15.68) gives H = 3 2
It is important to remark that the derivation of these results does not require the hypothesis N 1 (explicitly used in the microcanonical set), i.e. there is no need to invoke the thermodynamical limit for their justification, and the canonical set is naturally orthodic. 15.12
Canonical set and energy fluctuations In the discussion introducing the structure of the canonical set, we anticipated that for N 1 this clusters around the microcanonical set, and hence the manifold H = H . We can now verify this fact, studying the energy fluctuations. T heorem 15.11 In the limit N → ∞, the mean quadratic energy fluctuation in the canonical set H 2 − H 2 H 2 (15.70)
tends to zero. More precisely: H 2 − H 2 H 2 ≈ 1 N , f or N → ∞.
(15.71) Proof
We start from the known relation H = −(∂/∂β) log Z. Differentiating with respect to β, we obtain ∂ H ∂β
1 Z 2 ∂Z ∂β 2 − 1 Z ∂ 2 Z ∂β 2 . 15.13 Statistical mechanics: Gibbs sets 647 Since the first term on the right-hand side is just H 2 , while ∂ 2 Z/∂β
2 = Z H
2 , we conclude that H 2 − H 2 = − ∂ H ∂β . Note now that −(∂ H /∂β) = kT 2 (∂ H /∂T ) = kT 2 C V is, like H , proportional to N . Hence (15.71) is proved. Example 15.10 The explicit computation for a perfect monatomic gas gives the result H 2
2 H 2 = 2 3N , independent of the temperature. We can therefore confirm that the states of the canonical set are concentrated in a ‘thin’ region around the surface H = H , and therefore the canonical set is not different from the microcanonical set in the limit N → ∞, and the two formalisms give rise to thermodynamical descriptions which are essentially equivalent for sufficiently large values of N . Thanks to the results of the previous section we again find that the microcanonical set is orthodic in the limit N → ∞.
15.13 Open systems with fixed temperature. Grand canonical set Our discussion of statistical mechanics, and of Gibbs’ statistical sets, has so far led us to consider two distributions, the microcanonical and the canonical. In the former, we consider closed and isolated systems, with a fixed number of particles N in a region of fixed volume V and with total energy E fixed as well. In the latter, we studied closed but not isolated systems, and in contact with a thermostat at constant temperature T ; hence with a fixed number of particles N and volume V but with variable energy E. In many cases—for example when a chemical reaction is taking place—one needs to consider open systems in which the temperature T and the volume V are fixed, but the energy E and the number of particles N are variable. To model such a situation we can simply imagine eliminating the separation between the system and the thermostat, letting the two systems exchange their particles. We then have a fluctuation of the number of particles in the two compartments. The statistical set corresponding to this situation is the grand canonical set. In what follows we discuss briefly some properties of its density and its orthodicity in the thermodynamical limit. Assume that N 1 , N 2 are the number of particles in the system under consid- eration and in the thermostat, respectively, and that the sum N 1 + N 2 = N is
constant. Denote by V 1 , V 2 (V 1 +V 2 = V and V 1 V 2 ) the respective volumes, and by T the temperature. Naturally the two systems must have identical particles 648 Statistical mechanics: Gibbs sets 15.13 and we must assume that the respective Hamiltonians are subject to fluctu- ations also because of the exchange of particles. The space Γ 1 has the variable dimension 6N 1 , but for fixed N 1 we can consider the canonical partition function Z 1
1 , N
1 , T ) of the first system and simultaneously the one Z 2 (V
, N 2 , T ) of the thermostat. In addition we can consider that the union of the two systems is in turn kept at temperature T and therefore also consider the total partition function Z(V, N, T ). If Z 1 , Z 2 , Z count the number of states in the respective systems, then the following relation must necessarily hold: Z(V, N, T ) = N N
=0 Z 1 (V 1 , N 1 , T )Z
2 (V 2 , N 2 , T ). (15.72) This suggests that we assume as density of the grand canonical set associated with system, the usual one of a canonical set with N 1 particles, corrected by the factor Z 2 /Z, i.e. ρ G (X 1 , N
1 , T ) =
Z 2 Z 1 N 1 !h 3N 1 exp[ −βH(X
1 , N
1 )],
(15.73) and hence with integral Γ 1
G (X 1 , N 1 , T ) dX 1 = Z 1 Z 2 Z , (15.74) in such a way that ρ G is normalised to 1 when summing over all possible states of system 1: N N 1 =0 Γ
1 ρ G (X 1 , N 1 , T ) dX
1 = 1.
We now seek an expression for the correction factor Z 2 /Z that is valid in the situation in which not only V 1 V 2 , but also N 1 N
, as it is natural to expect. In this case we can consider Z 2 as a perturbation of Z. It is convenient to refer to the Helmholtz free energy and write Z 2 (V 2 , N 2 , T )
Z(V, N, T ) = exp[
−β Ψ (V − V 1 , N − N 1 , T ) + β Ψ (V, N, T )]. (15.75) Using the expansion to first order we obtain Ψ (V − V 1 , N
− N 1 , T ) − Ψ (V, N, T ) = − ∂ Ψ ∂V V 1 − ∂ Ψ ∂N N 1 . (15.76)
We already know that −∂ Ψ /∂V = P . We now introduce two new quantities: the chemical potential : µ = ∂
∂N , (15.77) 15.13 Statistical mechanics: Gibbs sets 649 and the fugacity: z = e βµ , (15.78) through which we finally arrive at the expression Z 2
2 , N
2 , T )
Z(V, N, T ) = e
−βP V 1 z N 1 . (15.79) The condition N 1 N
can be verified on average a posteriori. Note that the Download 10.87 Mb. Do'stlaringiz bilan baham: |
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