Geometrisation of Statistical Mechanics Dorje C. Brody∗ and Lane P. Hughston†
arXiv:gr-qc/9708032v1 15 Aug 1997
∗Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Silver Street, Cambridge CB3 9EW U.K. † Merrill Lynch International, 25 Ropemaker Street, London EC2Y 9LY U.K. and King’s College London, The Strand, London WC2R 2LS, U.K. (February 7, 2008) Abstract. Classical and quantum statistical mechanics are cast here in the language of projective geometry to provide a unified geometrical framework for statistical physics. After reviewing the Hilbert space formulation of classical statistical thermodynamics, we introduce projective geometry as a basis for analysing probabilistic aspects of statistical physics. In particular, the specification of a canonical polarity on RP n induces a Riemannian metric on the state space of statistical mechanics. In the case of the canonical ensemble, we show that equilibrium thermal states are determined by the Hamiltonian gradient flow with respect to this metric. This flow is concisely characterised by the fact that it induces a projective automorphism on the state manifold. The measurement problem for thermal systems is studied by the introduction of the concept of a random state. The general methodology is then extended to include the quantum mechanical dynamics of equilibrium thermal states. In this case the relevant state space is complex projective space, here regarded as a real manifold endowed with the natural FubiniStudy metric. A distinguishing feature of quantum thermal dynamics is the inherent multiplicity of thermal trajectories in the state space, associated with the nonuniqueness of the infinite temperature state. We are then led to formulate a geometric characterisation of the standard KMS-relation often considered in the context of C ∗ -algebras. The example of a quantum spin one-half particle in heat bath is studied in detail. Keywords: Hilbert space geometry, Projective geometry, Equilibrium statistical mechanics, Quantum dynamics I. INTRODUCTION
One of the most fascinating advances in the application of modern differential geometry is its use in statistical physics, including quantum and statistical mechanics. The purpose of this paper is to develop a unified geometrical framework that allows for a natural characterisation of both of these aspects of statistical physics. In quantum mechanics, one typically works with square-integrable wave functions, i.e., elements of a complex Hilbert space H. This space possesses natural geometrical structures induced by its norm. However, in order to seek a compelling axiomatic formulation of quantum mechanics, it may be reasonable to work with a space of more direct physical relevance [1,2]. This is not the Hilbert space H itself, but rather the manifold Σ of “instantaneous pure states” [3], which has the structure of a complex projective space CP n , possibly infinite dimensional, enriched with a Hermitian correspondence, i.e., a complex conjugation 1
operation that maps points to hyperplanes in CP n , and vice-versa. Equivalently, we think of CP n as being endowed with a natural Riemannian metric, the Fubini-Study metric. The space Σ is, in fact, the quantum analogue of the classical phase space of mechanical systems. Hence, one can interpret the Schr¨odinger equation as Hamilton’s equations on CP n , and the equation of motion for a general density matrix can be identified with the Liouville equation [4]. The advantage of working with the manifold Σ, rather than the Hilbert space of state vectors, above all, is that it can readily accommodate generalisations of quantum mechanics [5], including nonlinear relativistic models. Furthermore, the structure of Σ allows for a natural probabilistic interpretation even if the standard linear quantum theory is modified. As we discuss elsewhere [6], the statistical aspects of quantum measurement can be greatly clarified if we shift our view slightly, and regard the Hilbert space H of quantum mechanics not as a complex Hilbert space, but rather a real Hilbert space endowed with a real metric and a compatible complex structure. This would appear to be simply a change in formalism while keeping the same underlying physical structure. Indeed this is so, but once quantum theory is formulated this way its relation to other aspects of statistical physics becomes much more apparent. Statistical mechanics, in particular, can also be formulated concisely [7] in terms of the geometry of a real Hilbert space H. This can be seen by taking the square root of the Gibbs density function, which maps the space of probability distributions to vectors in a convex cone H+ in H. In this way, the various probabilistic and statistical operations of statistical mechanics can be given a transparent geometric meaning in H [8,9]. However, it can be argued that even at the classical level of statistical mechanics the ‘true’ state space is obtained by identifying all the pure states along the given ray through the origin of H. In this case, the space obtained is essentially the real projective space RP n . This is the view we take here, and we shall study properties of thermal states that become apparent only when the theory is developed in a fully geometric context. The present paper is organised as follows. In Section 2, we review the basics of the Hilbert space formulation of statistical mechanics. Since this formulation is perhaps not very widely appreciated, we can regard this section as an extended introduction which then paves the way to the approach in terms of projective geometry presented later. We begin with a brief review of statistical geometry, including the theory of the Fisher-Rao metric on the parameter space of a family of probability distributions. The Gibbs distribution when viewed in this way can be seen as a curve in Hilbert space, parameterised by the inverse temperature, and there is a striking formal resemblance to the Schr¨odinger equation, even though here we are working at a strictly classical level. A measurement theory for thermal states is developed by analogy with the standard density matrix theory used in quantum mechanics. We are then led to a set of uncertainty relations for the measurements of thermodynamic conjugate variables such as energy and inverse temperature. We also introduce an alternative approach to the measurement theory that is not based upon the density matrix description. Our approach, based on the introduction of random states, extends naturally also to quantum mechanics, where it can be seen to be more appealing in a probabilistic context than standard treatments, and indeed reduces to the conventional density matrix approach in special cases. In Section 3, we introduce a projective geometric framework for the probabilistic op2
erations involved in the representation of the canonical thermal state associated with the standard Gibbs measure. Thermal states are shown to lie on a trajectory in the real projective space RP n , which is endowed with the natural ‘spherical’ metric. In this connection we find it convenient to develop a number of useful differential geometric results characterising projective transformations on the state space RP n . We find that the equilibrium thermal trajectories, which are shown to be given by a Hamiltonian gradient flow, generate projective automorphisms of the state manifold. In Section 4, we then synthesise the approaches outlined in Sections 2 and 3, and consider the inter-relationship of the classical thermal state space RP n and the quantum phase space CP n , to study the quantum mechanical dynamics of equilibrium thermal states. First we examine the quantum state space from the viewpoint of complex algebraic geometry, which shows that this space is endowed with a natural Riemannian geometry given by the FubiniStudy metric, along with a natural symplectic structure. For thermal physics it is instructive to look at quantum mechanics from an entirely ‘real’ point of view as well, and this approach is developed in Section 4.B. Our formulation is then compared to the standard KMS-construction [10] for equilibrium states. In particular, once we pass to the mixed state description we recover the KMS-state. However, our quantum mechanical pure thermal state, which does not obey the KMScondition, can be viewed as a more fundamental construction. In Section 4.D we develop a theory of the quantum mechanical microcanonical ensemble, formulated entirely in terms of the quantum phase space geometry. This is set up in such a way as to admit generalisations to nonlinear quantum theories. Finally, we study more explicitly the case of a quantum mechanical spin one-half particle in heat bath. II. STATISTICAL STATES IN HILBERT SPACE A. Hilbert space geometry
Let us begin by demonstrating how classical statistical mechanics can be formulated in an appealing way by the use of a geometrical formalism appropriate for Hilbert space. Consider a real Hilbert space H equipped with an inner product gab . A probability density function p(x) can be mapped into H by taking the square-root ψ(x) = (p(x))1/2 , which is denoted R a by a vector ψ in H. The normalisation condition (ψ(x))2 dx = 1 is written gab ψ a ψ b = 1, indicating that ψ a lies on the unit sphere S in H. Since a probability density function is nonnegative, the image of the map f : p(x) → ψ(x) is the intersection S+ = S ∩ H+ of S with the convex cone H+ formed by the totality of quadratically integrable nonnegative functions. If we consider the space of all probability distributions as a metric space relative to the Hellinger distance [11], then f is an isometric embedding in H. We call ψ a the state vector of the corresponding probability density p(x). A typical random variable is represented on H by a symmetric tensor Xab , whose expectation in a normalised state ψ a is given by Eψ [X] = Xab ψ a ψ b .
(1)
Similarly, the expectation of its square is Xac Xbc ψ a ψ b . The variance of Xab in the state ψ a 3
˜ ac X ˜ c ψ a ψ b , where X ˜ ab = Xab − gab Eψ [X] represents the deviation of is therefore Varψ [X] = X b Xab from its mean in the state ψ a . We consider now the unit sphere S in H, and within this sphere a submanifold M given parametrically by ψ a (θ), where θi (i = 1, · · · , r) are local parameters. In particular, later on we have in mind the case where the parameter space spanned by θi represents the space of coupling constants in statistical mechanics associated with the given physical system. In the case of the canonical Gibbs measure there is a single such parameter, corresponding to the inverse temperature variable β = 1/kB T . We write ∂i for ∂/∂θi . Then, in local coordinates, there is a natural Riemannian metric Gij on the parameter space M, induced by gab , given by Gij = gab ∂i ψ a ∂j ψ b . This can be seen as follows. First, note that the squared distance between the endpoints of two vectors ψ a and η a in H is gab (ψ a − η a )(ψ b − η b ). If both endpoints lie on M, and η a is obtained by infinitesimally displacing ψ a in M, i.e., η a = ψ a + ∂i ψ a dθi , then the separation ds between the two endpoints on M is ds2 = Gij dθi dθj , where Gij is given as above. The metric Gij is, up to a conventional, irrelevant factor of four, the so-called Fisher-Rao metric on the space of the given family of distributions. The Fisher-Rao metric us usually defined in terms of a rather complicated expression involving the covariance matrix of the gradient of the log-likelihood function; but here we have a simple, transparent geometrical construction. The Fisher-Rao metric is important since it provides a geometrical basis for the key links between the statistical and physical aspects of the systems under consideration. B. Thermal trajectories
Now suppose we consider the canonical ensemble of classical statistical mechanics, in the case for which the system is characterised by a configuration space and an assignment of an energy value for each configuration. The parametrised family of probability distributions then takes the form of the Gibbs measure p(H, β) = q(x) exp [−βH(x) − W (β)] ,
(2)
where the variable x ranges over the configuration space, H(x) represents the energy, W (β) is a normalisation factor, and q(x) determines the distribution at β = 0, where β is the inverse temperature parameter. We now formulate a Hilbert space characterisation of this distribution. Taking the square-root of p(H, β), we find that the state vector ψ a (β) in H corresponding to the Gibbs distribution (2) satisfies the differential equation ∂ψ a 1 ˜a b ψ , = − H ∂β 2 b
(3)
˜ ab = Hab − gab Eψ [H]. Here the operator Hab in the Hilbert space H corresponds to where H the specified Hamiltonian function H(x) appearing in (2). The solution of this equation can be represented as follows: 1 ˜ (β)δ a ) q b , ψ (β) = exp − (βHba + W b 2 a
0015
0014
4
(4)
˜ (β) = W (β) − W (0) and q a = ψ a (0) is the prescribed distribution at β = 0. where W Since ψ a (β) respects the normalisation gab ψ a ψ b = 1, for each value of the temperature β we find a point on M in S+ . To be more specific, the thermal system can be described as follows. Consider a unit sphere S in H, whose axes label the configurations of the system, each of which has a definite energy. We let uak denote an orthonormal basis in H. Here, the index k labels all the points in the phase space of the given statistical system. In other words, for each point in phase space we have a corresponding basis vector uak in H for some value of k. With this choice of basis, a classical thermal state ψ a (β) can be expressed as a superposition 1
ψ a (β) = e− 2 W (β)
X
1
e− 2 βEk uak ,
(5)
k
where Ek is the energy for k-th configuration, and thus exp[W (β)] = k exp(−βEk ) is the partition function. We note that the states uak are, in fact, the energy eigenstates of the system, with eigenvalues Ek . That is to say, Hba ubk = Ek uak . The index k in these formulae is formal in the sense that the summation may, if appropriate, be replaced by an integration. By comparing equations (4) and (5), we find that the initial (β = 0) thermal state q a is P
1
q a = e− 2 W (0)
X
uak ,
(6)
k
which corresponds to the centre point in S+ . This relation reflects the fact that all configurations are equally probable likely at infinite temperature. Viewed as a function of β, the state trajectory ψ a (β) thus commences at the centre point q a , and follows a curve on S generated by the Hamiltonian Hab according to (3). It is interesting to note that the curvature of this trajectory, given by Kψ (β) =
˜ 4i ˜ 3 i2 hH hH − −1 , ˜ 2 i2 hH ˜ 2 i3 hH
(7)
arises naturally in a physical characterisation of the accuracy bounds for temperature measurements. This point is pursued further in Section 2.C below, and in [7]. In equation (7) the ˜ n i denotes the n-th central moment of the observable Hab . Here the curvature expression hH of the curve ψ a (β), which is necessarily positive, is the square of the ‘acceleration’ vector along the state trajectory ψ a (β), normalised by the square of the velocity vector. C. Measurement for thermal states
Given the thermal trajectory ψ a (β) above, we propose, in the first instance, to consider measurement and estimation by analogy with the von Neumann approach in quantum mechanics. According to this scheme the general state of a thermodynamic system is represented by a ‘density matrix’ ρab which in the present context should be understood to be a symmetric, semidefinite matrix with trace unity; that is to say, ρab ξa ξb ≥ 0 for any covector ξa , and ρab gab = 1. Then, for example, we can write Eρ [X] = Xab ρab 5
(8)
for the expectation of a random variable Xab in the state ρab , and Varρ [X] = Xab Xcb ρac − (Xab ρab )2
(9)
for the variance of Xab in that state. It should be evident that in the case of a pure state, for which ρab is of the form ρab = ψ a ψ b for some state vector ψ a , these formulae reduce to the expressions considered in Section 2.A. In particular, let us consider measurements made on a pure equilibrium state ψ a (β). Such measurements are characterised by projecting the prescribed state onto the ray in the Hilbert space corresponding to a specified point in the phase space. Hence, the probability of observing the k-th state, when the system is in the pure state ψ a , is given by the corresponding Boltzmann weight pk = (gab ψ a ubk )2 = e−βEk −W (β) .
(10)
In terms of the density matrix description, the state before measurement is given by the degenerate pure state matrix ρab = ψ a ψ b , for which the thermal development is 0011 1 0010 ˜ a bc dρab b ac ˜ , ρ + H ρ = − H c c dβ 2
(11)
˜ ρ}, where {A, B} denotes the symmetric product between or equivalently dρ/dβ = −{H, the operators A and B. After a measurement, ρab takes the form of a mixed state, characterised by a nondegenerate diagonal density operator for which the diagonal elements are the Boltzmann weights pk . In this state vector reduction picture, the von Neumann entropy −Tr[ρ ln ρ] changes from 0 to its maximum value S = βhHi + Wβ , which can be viewed as the quantity of information gained from the observation. More generally, suppose we consider the measurement of an arbitrary observable Xab in the state ψ a (β) in the situation when the spectrum of Xab admits a continuous component. In this case, we consider the spectral measure associated with the random variable Xab . Then, the probability density for the measurement outcome x is given by the expectation p(x, β) = Πab (X, x)ψ a ψ b of the projection operator Πab (X, x)
1 Z∞ = √ exp [iλ(Xba − xδba )] dλ . 2π −∞
(12)
In other words, we assign a projection-valued measure ΠX (x) on the real line associated with each symmetric operator X, so that for a given unit vector ψ a , the mapping x ∈ R 7→ Eψ [ΠX (x)] is a probability measure. This measure determines the distribution of values obtained when the observable X is measured while the system is in the state ψ. For a more refined view of the measurement problem we need to take into account some ideas from statistical estimation theory. Suppose that we want to make a measurement or series of measurements to estimate the value of the parameter characterising a given thermal equilibrium state. In this situation the observable we measure is called an ‘estimator’ for the given parameter. We are interested in the case for which the estimator is unbiased in the sense that its expectation gives the value of the required parameter. To be specific, we consider the case when we estimate the value of the temperature. Let Bab be an unbiased estimator for β, so that along the trajectory ψ a (β) we have: 6
Bab ψ a ψ b = β. gcd ψ c ψ d
(13)
As a consequence of this relation and the thermal state equation (3), we observe that the inverse temperature estimator B and the Hamiltonian H satisfy the ‘weak’ anticommutation ˜ = −1 along the state trajectory ψ. In statistical terms, this implies relation Eψ [{B, H}] that these conjugate variables satisfy the covariance relation Eψ [BH] − Eψ [B]Eψ [H] = −1 along the trajectory. D. Thermodynamic uncertainty relations
Equipped with the above definitions, one can easily verify that the variance in estimating the inverse temperature parameter β can be expressed by the geometrical relation 1 ab g ∇a β∇b β 4
Varψ [B] =
(14)
on the unit sphere S, where ∇a β = ∂β/∂ψ a is the gradient of the temperature estimate β. The essence of formula (14) can be understood as follows. First, recall that β is the expectation of the estimator Bab in the state ψ a (β). Suppose that the state changes rapidly as β changes. Then, the variance in estimating β is small—indeed, this is given by the squared magnitude of the ‘functional derivative’ of β with respect to the state ψ a . On the other hand, if the state does not change significantly as β changes, then the measurement outcome of an observable is less conclusive in determining the value of β. The squared length of the gradient vector ∇a β can be expressed as a sum of squares of orthogonal components. To this end, we choose a new set of orthogonal basis vectors given by the state ψ a and its higher derivatives. If we let ψna denote ψ a for n = 0, and for n > 0 the component of the derivative ∂ n ψ a /∂β n orthogonal to the state ψ a and its lower order derivatives, then our orthonormal vectors are given by ψˆna = ψna (gbc ψnb ψnc )−1/2 for n = 0, 1, 2, · · ·. With this choice of orthonormal vectors, we find that the variance of the estimator B satisfies Varψ [B] ≥
X n
˜ab ψ a ψ b )2 (B n , gcdψnc ψnd
(15)
for any range of the index n. This follows because the squared magnitude of the vector 1 ˜ab ψ b is greater than or equal to the sum of the squares of its projections onto the ∇ β=B 2 a basis vectors given by ψˆna for the specified range of n. In particular, for n = 1 we have Bab ψ1a ψ b = 12 on account of the relation Bab ψ a ψ b = β, and gab ψ1a ψ1b = 41 ∆H 2 , which follows from the thermal equation (3). Therefore, if we write Varψ [B] = ∆β 2 , we find for n = 1 that the inequality (15) implies the following thermodynamic uncertainty relation: ∆β 2 ∆H 2 ≥ 1 ,
(16)
valid along the trajectory consisting of the thermal equilibrium states. The variance ∆2 β here is to be understood in the sense of estimation theory. That is, although the variable 7
β does not actually fluctuate, as should be clear from the definition of canonical ensemble, there is nonetheless an inevitable lower bound for the variance of the measurement, given by (16), if we wish to estimate the value of the heat bath temperature β. It is worth pointing out that the exposition we have given here is consistent with the view put forward by Mandelbrot [12], who should perhaps be credited with first having introduced an element of modern statistical reasoning into the long-standing debate on the status of temperature fluctuations [13]. Note that, although we only considered the variance h(B − hBi)2 i here, the higher order central moments µn = h(B − hBi)n i can also be expressed geometrically. This can be seen as follows. First, recall that for any observable Fab with Eψ [F ] = f , we have ∇a f = 2F˜ab ψ b on ˜ n , we construct the higher central moments in the unit sphere S. Therefore, by letting F = B terms of the cosines of the angles between certain gradient vectors, e.g., 4µ3 = g ab ∇a µ2 ∇b β, 4µ4 = g ab ∇a µ2 ∇b µ2 − 4µ22 , and so on. In particular, the even order moments are expressible in terms of combinations of the squared lengths of normal vectors to the surfaces of constant central moments of lower order. E. Random states
Let us return to the consideration of measurements on thermal states, which we now pursue in greater depth. In doing so we shall introduce the idea of a ‘random state’, a concept that is applicable both in clarifying the measurement problem in statistical physics, as well as in providing a useful tool when we consider ensembles. It also turns out that the idea of a random state is helpful in the analysis of conceptual problems in quantum mechanics. Later on when we consider quantum statistical mechanics, we shall have more to say on this. Suppose we consider a pure thermal state ψ a (β) for some value β of the inverse temperature. We know that this state is given by ψ a (β) =
X
1/2
pk uak ,
(17)
k
where uak is a normalised energy eigenstate with eigenvalue Ek , and pk is the associated Gibbs probability. After measurement, it is natural to consider the outcome of the measurement to be a random state Ψa . Thus we consider Ψa to be a random variable (indicated by use of a bold font) such that the probability for taking a given eigenstate is Prob[Ψa = uak ] = pk .
(18)
This way of thinking about the outcome of the measurement process is to some extent complementary to the density matrix approach, though in what follows we shall make it clear what the relationship is. In particular, the expectation of an observable Xab in the random state Ψa is given by averaging over the random states, that is, EΨ [X] = Xab Ψa Ψb .
8
(19)
This relation should be interpreted as the specification of a conditional expectation, i.e., the conditional expectation of Xab in the random state Ψa . Then the associated unconditional expectation E[X] = E[EΨ [X]] is given by E[X] = Xab E[Ψa Ψb ] .
(20)
However, since Prob[Ψa = uak ] = pk , it should be evident that E[Ψa Ψb ] = ρab ,
(21)
where the density matrix ρab is defined by ρab =
X
pk uak ubk .
(22)
k
Thus, the unconditional expectation of the random variable Xab is given, as noted earlier, by E[X] = Xab ρab . It should be observed, however, that here we are not emphasising the role of the density matrix ρab as representing a ‘state’, but rather its role in summarising information relating to the random state Ψa . The feature that distinguishes the density matrix in this analysis is that it is fully sufficient for the characterisation of unconditional statistics relating to the observables and states under consideration. This point is clearly illustrated when we calculate the variance of a random variable Xab in a random state Ψa . Such a situation arises if we want to discuss the uncertainties arising in the measurement of an observable Xab for an ensemble. In this case the system we have in mind is a large number of identical, independent particles, each of which is in a definite energy eigenstate, where the distribution of the energy is given according to the Gibbs distribution. One might take this as an elementary model for a classical gas. Then the distribution of the ensemble can be described in terms of a random state Ψa . Note that here the interpretation is slightly different from what we had considered before (the random outcome of a measurement for an isolated system), though it will be appreciated that the relation of these two distinct interpretations is of considerable interest for physics and statistical theory alike. The conditional variance of the observable Xab in the random state Ψa is given by VarΨ [X] = Xab Xcb Ψa Ψc − (Xab Ψa Ψb )2 .
(23)
The average over the different values of Ψa then gives us E [VarΨ [X]] = Xab Xcb ρac − Xab Xcd ρabcd ,
(24)
where ρab is, as before, the density matrix (21), and ρabcd is a certain higher moment of Ψa , defined by ρabcd = E[Ψa Ψb Ψc Ψd ] .
(25)
The appearance of this higher order analogue of the density matrix may be surprising, though it is indeed a characteristic feature of conditional probability. However, the unconditional variance of X is not given simply by the expectation E [VarΨ [X]], but rather (see, e.g., [14]) by the conditional variance formula 9
Var[X] = E [VarΨ [X]] + Var[EΨ [X]] .
(26)
For the second term we have Var[EΨ [X]] = E[(Xab Ψa Ψb )2 ] − (E[Xab Ψa Ψb ])2 = Xab Xcd ρabcd − (Xab ρab )2 ,
(27)
which also involves the higher moment ρabcd . The terms in (26) involving ρabcd then cancel, and we are left with Var[X] = Xab Xcb ρac −(Xab ρab )2 for the unconditional variance, which, as indicated earlier, only involves the density matrix. It follows that the random state approach does indeed reproduce the earlier density matrix formulation of our theory, though the role of the density matrix is somewhat diminished. In other words, whenever conditioning is involved, it is the set of totally symmetric tensors ρab···cd = E[Ψa Ψb · · · Ψc Ψd ]
(28)
that plays the fundamental role, although it suffices to consider the standard density matrix ρab when conditioning is removed. All this is worth having in mind later when we turn to quantum statistical mechanics, where the considerations we have developed here in a classical context reappear in a new light. We want to de-emphasise the role of the density matrix, not because there is anything wrong per se with the use of the density matrix in an appropriate context, but rather for two practical reasons. First of all, when we want to consider conditioning, exclusive attention on the density matrix hampers our thinking, since, as we have indicated, higher moments of the random state vector also have a role to play. Second, when we go to consider generalisations of quantum mechanics, such as the nonlinear theories of the Kibble-Weinberg type [15,16], or stochastic theories of the type considered by Gisin, Percival, and others [17,18], the density matrix is either an ill formulated concept, or plays a diminished role. We shall return to this point for further discussion when we consider quantum statistical mechanics in Section 4. III. STATISTICAL PHASE SPACE A. Projective space and probabilities
To proceed further it will be useful to develop a formalism for the algebraic treatment of real projective geometry, with a view to its probabilistic interpretation in the context of classical statistical mechanics. Let Z a be coordinates for (n + 1)-dimensional real Hilbert space Hn+1 . Later, when we consider quantum theory from a real point of view we shall double this dimension. In the Hilbert space description of classical probabilities, the normalisation condition is written gab Z a Z b = 1. However, this normalisation is physically irrelevant since the expectation of an arbitrary operator Fab is defined by the ratio hF i =
Fab Z a Z b . gcd Z c Z d
10
(29)
Therefore, the physical state space is not the Hilbert space H, but the space of equivalence classes obtained by identifying the points {Z a } and {λZ a } for all λ ∈ R − {0}. In this way, we ‘gauge away’ the irrelevant degree of freedom. The resulting space is the real projective n-space RP n , the space of rays through the origin in Hn+1 . Thus, two points X a and Y a in Hn+1 are equivalent in RP n if they are proportional, i.e., X [a Y b] = 0. The coordinates Z a (excluding Z a = 0) can be used as homogeneous coordinates for points of RP n . Clearly Z a and λZ a represent the same point in RP n . In practice one treats the homogeneous coordinates as though they define points of Hn+1 , with the stipulation that the allowable operations of projective geometry are those which transform homogeneously under the rescaling Z a → λZ a . A prime, or (n − 1)-plane in RP n consists of a set of points Z a which satisfy a linear equation Pa Z a = 0, where we call Pa the homogeneous coordinates of the prime. Clearly, Pa and λPa determine the same prime. Therefore, a prime in RP n is an RP n−1 , and the set of all primes in RP n is itself an RP n (the ‘dual’ projective space). In particular, the metric gab on Hn+1 can be interpreted in the projective space as giving rise to a nonsingular polarity, that is, an invertible map from points to hyperplanes in RP n of codimension one. This map is given by P a → Pa := gab P b . See reference [19] for further discussion of some of the geometric operations employed here. If a point P a in RP n corresponds to a probability state, then its negation ¬P a is the hyperplane Pa Z a = 0. To be more precise, we take P a as describing the probability state for a set of events. Then, all the probability states corresponding to the complementary events lie in the prime Pa Z a = 0. Thus, the points Z a on this plane are precisely the states that are orthogonal to the original state P a . The intuition behind this is as follows. Two states P a and Qa are orthogonal if and only if any event which in the state P a (resp. Qa ) has a positive probability is assigned zero probability by the state Qa (resp. P a ). This is the sense in which orthogonal states are ‘complementary’. For any state P a , the plane consisting of all points Z a such that Pa Z a = 0 is the set of states complementary to P a in this sense. Distinct states X a and Y a are joined by a real projective line represented by the skew tensor Lab = X [a Y b] . The points on this line are the various real superpositions of the original two states. The intersection of the line Lab with the plane Ra is given by S a = Lab Rb . Clearly S a lies on the plane Ra , since Ra S a = 0 on account of the antisymmetry of Lab . The hyperplanes that are the negations (polar planes) of two points P a and Qa intersect ˜ a respectively. That is, if Pa = gab P b the joining line Lab = P [a Qb] at a pair of points P˜ a and Q is the coordinate of a plane and Lab represents a line in RP n , then the point of intersection ˜ a = Lab Qb . The projective cross ratio between these is given by P˜ a = Lab Pb , and similarly Q a a ˜a a b ˜ a = P a (Qb Qb ) − Qa (P bQb ), given by four points P , Q , P = P (Q Pb ) − Qa (P bPb ) and Q ˜ a Qb P˜b /P c P˜c Qd Q ˜ d , reduces, after some algebra, to the following simple expression: P aQ κ =
(P a Qa )2 , P b Pb Qc Qc
(30)
which can be interpreted as the transition probability between P a and Qa . It is interesting to note that this formula has an analogue in quantum mechanics [5]. ˜ a antipodal to the The projective line Lab can also be viewed as a circle, with P˜ a and Q points P a and Qa . In that case, the cross ratio κ is 21 (1 + cos θ) = cos2 (θ/2) where θ defines the angular distance between P a and Qa , in the geometry of RP n . We note that θ is, in 11
fact, twice the angle made between the corresponding Hilbert space vectors, so orthogonal states are maximally distant from one another. Now suppose we let the two states P a and Qa approach one another. In the limit the resulting formula for their second-order infinitesimal separation determines the natural line element on real projective space. This can be obtained by setting P a = Z a and Qa = Z a +dZ a in (30), while replacing θ with the small angle ds in the expression 21 (1 + cos θ), retaining terms of the second order in ds. Explicitly, we obtain ds
2
dZ a dZa (Z a dZa )2 = 4 − Z b Zb (Z b Zb )2 '
#
.
(31)
Note that this metric [20] is related to the metric on the sphere S n in Hn+1 , except that in the case of the sphere one does not identify opposite points. We now consider the case where the real projective space is the state space of classical statistical mechanics. If we write ψ a (β) for the trajectory of thermal state vectors, as discussed in Section 2, then we can regard ψ a (β), for each value of β, as representing homogeneous coordinates for points in the state space RP n . Since ψ a (β) satisfies (3) it ˜ 2 idβ 2, which can follows that the line element along the curve ψ a (β) is given by ds2 = hH be identified with the Fisher-Rao metric induced on the thermal trajectory by virtue of its embedding in RP n . This follows by insertion of the thermal equation (3) into expression (31) for the natural spherical metric on RP n . B. The projective thermal equations
Let us write Hab for the symmetric Hamiltonian operator, and ψ a (β) for the oneparameter family of thermal states. Then in our notation the thermal equation is dψ b = ˜ b ψ c dβ. However, we are concerned with this equation only inasmuch as it supplies − 12 H c information about the evolution of the state of the system, i.e., its motion in RP n . We are interested therefore primarily in the projective thermal equation, given by 1 ˜ b] ψ c dβ , ψ [a dψ b] = − ψ [a H c 2
(32)
obtained by skew-symmetrising the thermal equation (3) with ψ a . Equation (32) defines the equilibrium thermal trajectory of a statistical mechanical system in the proper state space. The thermal equation generates a Hamiltonian gradient flow on the state manifold. This can be seen as follows. First, recall that a physical observable F is associated with a symmetric operator Fab , and the set of such observables form a vector space of dimension 1 (n + 1)(n + 2). Such observables are determined by their diagonal matrix elements, which 2 are real valued functions on RP n of the form F (ψ a ) = Fab ψ a ψ b /ψ c ψc . In particular, we are interested in the Hamiltonian function H(ψ a ). Then, by a direct substitution, we find that the vector field H a = g ab ∂H/∂ψ b associated with the Hamiltonian function H takes the ˜ a ψ b /ψ c ψc . Therefore, we can write the differential equation for the thermal form H a = 2H b state trajectory in Hn+1 in the form 1 dψ a = − g ab ∇b H . dβ 4 12
(33)
By projecting this down to RP n , we obtain 1 ψ [a dψ b] = − ψ [a ∇b] Hdβ , 4 a ab where ∇ H = g ∇b H. From this we can then calculate the line element to obtain ds2 = 8ψ [a dψ b] ψ[a dψb] /(ψ c ψc )2 1 = ψ [a ∇b] Hψ[a ∇b] Hdβ 2 2 1 b = ∇ H∇b Hψ a ψa dβ 2 4 ˜ 2 idβ 2 , = hH
(34)
(35)
which establishes the result we noted earlier. The critical points of the Hamiltonian function are the fixed points in the state space associated with the gradient vector field g ab ∇b H. In the case of Hamilton’s equations the fixed points are called stationary states. In the Hilbert space Hn+1 these are the points corresponding to the energy eigenstates uak given by Hba ubk = Ek uak (in the general situation with distinct energy eigenvalues Ek , k = 0, 1, · · · , n). Therefore, in the projective space RP n we have a set of fixed points corresponding to the states uak , and the thermal states are obtained by superposing these points with an appropriate set of coefficients given by the Boltzmann weights. Since these coefficients are nonzero at finite temperature, it should be clear that the thermal trajectories do not intersect any of these fixed points. In particular, the infinite temperature (β = 0) thermal state ψ a (0) is located at the centre point of S+ in Hn+1 . The distances from ψ a (0) to the various energy eigenstates are all equal. This implies that in RP n the cross ratios between the fixed points and the state ψ a (0) are equal. Therefore, we can single out the point ψ a (0) as an initial point, and form a geodesic hypersphere in RP n . All the fixed points of the Hamiltonian vector field H a lie on this sphere. Since the cross ratios between the fixed points are also equal (i.e., maximal), these fixed points form a regular simplex on the sphere. The thermal trajectory thus commences at Z a (0), and asymptotically approaches a fixed point associated with the lowest energy eigenvalue E0 , as β → ∞. If we take the orthogonal prime of the thermal state for any finite β, i.e., ¬ψ a (β), then the resulting hyperplane clearly does not contain any of the fixed points. On the other hand, if we take the orthogonal prime of any one of the fixed points uak , then the resulting hyperplane includes a sphere of codimension one where all the other fixed points lie. This sphere is given by the intersection of the original hypersphere surrounding ψ a (0) with the orthogonal prime of the given excluded fixed point. There is a unique prime containing n general points in RP n . It is worth noting, therefore, that if we choose n general points given by uaj (j = 0, 1, · · · , n;j 6= k), then there is a unique solution, up to proportionality, of the n linear equations Xa ua0 = 0, · · ·, Xa uan = 0. The solution is then given by X a = uak , where uak = ǫabc···d ub0 uc1 · · · udn . Here, ǫab···c = ǫ[ab···c] , with n + 1 indices, is the totally skew tensor determined up to proportionality. It would be interesting to explore whether the orientability characteristics of RP n lead to any physical consequences. There may be a kind of purely classical ‘spin-statistics’ relation in the sense that the state space of half-integral spins are associated with a topological invariant, while the state space for even spins are not. 13
C. Hamiltonian flows and projective transformations
We have seen in Section 3.B how the thermal trajectory of a statistical mechanical system is generated by the gradient flow associated with the Hamiltonian function. In other words, for each point on the state manifold we form the expectation of the Hamiltonian in that state. This gives us a global function on the state manifold, which we call the Hamiltonian function. Next, we take the gradient of this function, and raise the index by use of the natural metric to obtain a vector field. This vector field is the generator of the thermal trajectories. It is interesting to note that there is a relation between the geometry of such vector fields and the global symmetries of the state manifold. In particular, we shall show below that gradient vector fields generated by observables on the state manifold can also be interpreted as the generators of projective transformations. A projective transformation on a Riemannian manifold is an automorphism that maps geodesics onto geodesics. In the case of a real projective space endowed with the natural metric, the general such automorphism is generated, as we shall demonstrate, by a vector field that is expressible in the form of a sum of a Killing vector and a gradient flow associated with an observable function. To pursue this point further we develop some differential geometric aspects of the state manifold. We consider RP n now to be a differential manifold endowed with the natural spherical metric gab . Here, bold upright indices signify local tensorial operations in the tangent space of this manifold. Thus we write ∇a for the covariant derivative associated with gab , and for an arbitrary vector field V a we define the Riemann tensor Rabc d according to the convention 1 ∇[a ∇b] V c = Rabdc V d . 2
(36)
It follows that Rabcd := Rabc e gde satisfies Rabcd = R[ab][cd], Rabcd = Rcdab , and R[abc]d = 0. In the case of RP n with the natural metric ds2 =
8Z [a dZ b] Z[a dZb] , λ(Z c Zc )2
(37)
where Z a are homogeneous coordinates and λ is a scale factor (set to unity in the preceding analysis), the Riemann tensor is given by Rabcd = λ(gac gbd − gbc gad) .
(38)
Now we turn to consider projective transformations on RP n . First we make a few general remarks about projective transformations on Riemannian manifolds [21]. Suppose we have a Riemannian manifold with metric gab and we consider the effect of dragging the metric along the integral curves of a vector field ξ a . For an infinitesimal transformation we have gab → gˆab = gab + ǫLξ gab ,
(39)
where Lξ gab = 2∇(a ξb) is the Lie derivative of the metric (ǫ 0. For the expectation of the energy E = Hβα Z β Z¯α we have E = h
e−βh − eβh e−βh + eβh
(93)
which, as expected, ranges from 0 to −h as β ranges from 0 to ∞. We note, in particular, that E is independent of the phase angle φ, and that the relation E = h tanh(βh) agrees with the result for a classical spin. 27
For other observables this is not necessarily the case, and we have to consider averaging over the random state Zα obtained by replacing φ with a random variable Φ, having a ¯ β ], uniform distribution over the interval (0, 2π). Then for the density matrix ραβ := E[Zα Z where E[−] is the unconditional expectation, we obtain ραβ =
e−βh P α P¯β − eβh P¯ αPβ . e−βh + eβh
(94)
The fact that ραβ has trace unity follows from the normalisation condition Z α Z¯α = 1, and the identity Z α Z¯α = −Zα Z¯ α . For the energy expectation ρ(H) = Hβα ρβα we then recover (93). Alternatively, we can consider the phase-space volume approach considered earlier, by assuming a microcanonical distribution for this system. Now, the phase space volume of the energy surface EE1 (a latitudinal circle) is given by V(E) = 2π sin θ, where θ is the angle measured from the pole of CP 1 ∼ S 2 . Hence, by use of (82), along with the energy expectation E = h cos θ, we deduce that the value of the system temperature is β(E) =
E . E 2 − h2
(95)
Since E ≤ 0 and E 2 ≤ h2 , the inverse temperature β is positive. Furthermore, we see that E = 0 implies θ = π/2, the equator of the sphere, which gives infinite temperature (β = 0), and E 2 = h2 corresponds to θ = π, which gives the zero temperature (β = ∞) state. The equation of state for this system can also be obtained by use of the standard relation βp = ∂S/∂V. In this case, we obtain the equation of state for an ideal gas, i.e., pV = β −1 . Explicitly, we have p(E) = −
h √ 2 h − E2 . 2πE
(96)
The pressure is minimised when the spin aligns with the external field, and is maximised at the equator. We note that for positive energies E > 0 the temperature takes negative values. If we take E < 0 and then flip the direction of the external field h, this situation can be achieved in practice Although the concept of such a negative temperature is used frequently in the study of Laser phenomena, it is essentially a transient phenomenon [28], and is thus not as such an objective of thermodynamics. We note, incidentally, that the distinct energy-temperature relationships obtained here in equation (93) for the canonical ensemble and in (95) for the microcanonical ensemble, have qualitatively similar behaviour. Indeed, for a system consisting of a large number of particles these two results are expected to agree in a suitable limit. V. DISCUSSION
The principal results of this paper are the following. First, we have formulated a projective geometric characterisation for classical probability states. By specialising then to the canonical ensemble of statistical mechanics, we have been able to determine the main features of thermal trajectories, which are expressed in terms of a Hamiltonian gradient 28
flow. This flow is then shown to be a special case of a projective automorphism on the state space when it is endowed with the natural RP n metric. It should be clear that the same formalism, and essentially the same results, apply also to the grand canonical and the pressure-temperature distributions. The quantum mechanical dynamics of equilibrium thermal states can be studied by consideration of the Hopf-type map RP 2n+1 → CP n , which in the present context allows one to regard the quantum mechanical state space as the base space in a fibre manifold which has the structure of an essentially classical thermal state space. The fact that a projective automorphism on a space of constant curvature can be decomposed into two distinct terms suggests the identification of the Killing term with the Schr¨odinger evolution of the Hamiltonian gradient flow with respect to the symplectic structure, and the other term with thermal evolution of the Hamiltonian gradient flow with respect to the metric— the former gives rise to a linear transformation, while the latter is nonlinear. There are a number of problems that still remain. First, much of our formulation has been based on the consideration of finite dimensional examples. The study of phase transitions, however, will require a more careful and extensive treatment of the infinite dimensional case. Our analysis of the projective automorphism group, on the other hand, suggests that the infinite dimensional case can also be handled comfortably within the geometric framework. Also, for most of the paper we have adopted the Schr¨odinger picture, which has perhaps the disadvantage of being inappropriate for relativistic covariance. It would be desirable to reformulate the theory in a covariant manner, in order to study the case of relativistic fields. Nevertheless, as regards the first problem noted above, the present formulation is sufficiently rich in order to allow us to speculate on a scenario for the spontaneous symmetry breaking of, say, a pure gauge group, in the infinite dimensional situation. In such cases the hypersurface of the parameter space (a curve in the one-parameter case considered here) proliferates into possibly infinite, thermally inequivalent hypersurfaces, corresponding to the multiplicity of the ground state degeneracy, at which the symmetry is spontaneously broken. The hyper-line characterising the proliferation should presumably be called the spinodal line (cf. [29]), along which the Riemann curvature of the parameter space manifold is expected to diverge. Furthermore, a pure thermal state in the ‘high temperature’ region should evolve into a mixed state, obtained by averaging over all possible surfaces, by passing through the geometrical singularity (the spinodal boundary) characterising phase transitions. By a suitable measurement that determines which one of the ground states the system is in, this mixed state will reduce back to a pure state. In a cosmological context this proliferation may correspond, for example, to different θ-vacuums [30]. It is interesting to note that the situation is analogous to the choice of a complex structure [31] for the field theory associated with a curved space-time, as the universe evolves. In any case, the remarkable advantages of the use of projective space should be stressed. As we have observed, the structure of projective space allows us to identify probabilistic operations with precise geometric relations. One of the problems involved in developing nonlinear (possibly relativistic) generalisations of quantum mechanics concerns their probabilistic interpretation. Formulated in a projective space, such generalisations can be obtained, for example, by replacing the Hamiltonian function by a more general function, or by introducing a more general metric structure. In this way, the assignment of a suitable probability theory can be approached in an appropriate way. In particular, as we have ob29
served, the canonical ensemble of statistical mechanics has an elegant characterisation in projective space—but this is an example of a theory that is highly nonlinear and yet purely probabilistic. It is also interesting to observe that the nonlinear generalisation of quantum mechanics considered by Kibble [3] and others can be applied to the thermal situation, in the sense that Hamiltonian function defined on the state manifold can be replaced by a general observable. For such generalisations it is not clear what physical interpretation can be assigned. Naively, one might expect that by a suitable choice of an observable the resulting trajectory characterises some kind of nonequilibrium process. One of the goals of this paper has been to formulate quantum theory at finite temperature in such a way as to allow for the possibility of various natural generalisations. These might include, for example, the stochastic approach to describe measurement theory, or nonlinear relativistic extensions of standard quantum theory, as noted above. These generalisations will be pursued further elsewhere. Acknowledgement. DCB is grateful to Particle Physics and Astronomy Research Council for financial support. ∗ Electronic address: [email protected] † Electronic address: [email protected] 1. Haag, R. and Kastler, D., J. Math. Phys. 5, 848 (1964). 2. Landsmann, N.P., Aspects of Classical and Quantum Mechanics (Springer, New York 1998). 3. Kibble, T.W.B., Commun. Math. Phys. 65, 189 (1979). 4. Gibbons, G.W., J. Geom. Phys. 8, 147 (1992). 5. Hughston, L.P., in Twistor Theory, ed. Huggett, S. (Marcel Dekker, Inc., New York 1995). 6. Brody, D.C. and Hughston, L.P., “Statistical Geometry”, gr-qc/9701051. 7. Brody, D.C. and Hughston, L.P., “Geometry of Thermodynamic States”, quantph/9706030. 8. Brody, D.C. and Hughston, L.P., Phys. Rev. Lett. 77, 2851 (1996). 9. Gibbons, G.W., Class. Quant. Grav. 14, A155 (1997). 10. Haag, R., Hugenholtz, N.M. and Winnink, M., Commun. math. Phys. 5, 215 (1967). 11. Xia, D.X., Measure and Integration Theory on Infinite-Dimensional Spaces (Academic Press, New York 1972). 12. Mandelbrot, B., Ann. Math. Stat. 33, 1021 (1962); Phys. Today, (January 1989). 13. Feshbach, H., Phys. Today, (November 1987); Kittel, C., Phys. Today, (May 1988). 30
14. Ross, S., Stochastic Process, (Wiley, New York 1996). 15. Kibble, T.W.B., Commun. Math. Phys. 64, 239 (1978). 16. Weinberg, S., Phys. Rev. Lett. 62, 485 (1989). 17. Gisin, N. and Percival, I., Phys. Lett. A167, 315 (1992). 18. Hughston, L.P., Proc. Roy. Soc. London 452, 953 (1996). 19. Hughston, L.P. and Hurd, T.R., Phys. Rep. 100, 273 (1983). 20. Kobayashi, S. and Nomizu, K. Foundations of differential geometry, vols. 1 and 2 (Wiley, New York, 1963 and 1969). 21. Tomonaga, Y., Riemannian Geometry, (Kyoritsu Publishing Co., Tokyo 1970). 22. Hughston, L.P. and Tod, K.P., An Introduction to General Relativity, (Cambridge University Press, Cambridge 1990). 23. Anandan, J. and Aharonov, Y., Phys. Rev. Lett. 65, 1697 (1990). 24. Ashtekar, A. and Schilling, T.A., “Geometrical Formulation of Quantum Mechanics”, gr-qc/9706069. 25. Bratteli, O. and Robinson, D.W., Operator algebras and quantum statistical mechanics, vols. I and II (Springer, New York 1979,1981). 26. Ruelle, D., Statistical Mechanics: Rigorous Results (Addison-Wesley, New York 1989). 27. Thompson, C.J., Mathematical Statistical Mechanics (Princeton University Press, Princeton 1972). 28. Kubo, R., Statistical Mechanics, (Kyoritsu Publishing Co., Tokyo 1951). 29. Brody, D. and Rivier, N., Phys. Rev. E 51, 1006 (1995). 30. Sacha, R.G., Phys. Rev. Lett. 78, 420 (1997). 31. Gibbons, G.W. and Pohle, H.J., Nucl. Phys. B410, 117 (1993).
31

Problems & Solutions
for
Statistical Physics of Particles
Updated July 2008 by Mehran Kardar Department of Physics Massachusetts Institute of Technology Cambridge, Massachusetts 02139, USA
Table of Contents
I.
Thermodynamics
II.
Probability
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
III.
Kinetic Theory of Gases
IV.
Classical Statistical Mechanics
V. Interacting Particles VI. VII.
1
. . . . . . . . . . . . . . . . . . . . . . . . 38 . . . . . . . . . . . . . . . . . . . . . 72
. . . . . . . . . . . . . . . . . . . . . . . . . . 93
Quantum Statistical Mechanics
. . . . . . . . . . . . . . . . . . . .
121
Ideal Quantum Gases . . . . . . . . . . . . . . . . . . . . . . . .
138
Problems for Chapter I - Thermodynamics 1. Surface tension:
Thermodynamic properties of the interface between two phases are
described by a state function called the surface tension S. It is defined in terms of the
work required to increase the surface area by an amount dA through d¯W = SdA.
(a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius R is larger than outside pressure by 2S/R. What is the air pressure inside a soap bubble of radius R? • The work done by a water droplet on the outside world, needed to increase the radius
from R to R + ∆R is
∆W = (P − Po ) · 4πR2 · ∆R, where P is the pressure inside the drop and Po is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy S∆A = S · 8πR · ∆R, where S is the surface tension, and
∆Wtotal = 0,
=⇒
∆Wpressure = −∆Wsurface ,
resulting in (P − Po ) · 4πR2 · ∆R = S · 8πR · ∆R,
=⇒
(P − Po ) =
2S . R
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and Pfilm − Po = Pinterior − Pfilm = leading to Pinterior − Po =
2S , R
4S . R
Hence, the air pressure inside the bubble is larger than atmospheric pressure by 4S/R. (b) A water droplet condenses on a solid surface. There are three surface tensions involved S aw , S sw , and S sa , where a, s, and w refer to air, solid and water respectively. Calculate
the angle of contact, and find the condition for the appearance of a water film (complete wetting).
• When steam condenses on a solid surface, water either forms a droplet, or spreads on
the surface. There are two ways to consider this problem: Method 1: Energy associated with the interfaces 1
In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore dE = Saw dAaw + Sas dAas + Sws dAws = 0. Since the total surface area of the solid is constant, dAas + dAws = 0. From geometrical considerations (see proof below), we obtain dAws cos θ = dAaw . From these equations, we obtain dE = (Saw cos θ − Sas + Sws ) dAws = 0,
=⇒
cos θ =
Sas − Sws . Saw
Proof of dAws cos θ = dAaw : Consider a droplet which is part of a sphere of radius R, which is cut by the substrate at an angle θ. The areas of the involved surfaces are Aws = π(R sin θ)2 ,
and
Aaw = 2πR2 (1 − cos θ).
Let us consider a small change in shape, accompanied by changes in R and θ. These variations should preserve the volume of water, i.e. constrained by V =
0001 πR3 cos3 θ − 3 cos θ + 2 . 3
Introducing x = cos θ, we can re-write the above results as 0001  2 2 A = πR 1 − x , ws     Aaw = 2πR2 (1 − x) ,  3    V = πR x3 − 3x + 20001 . 3
The variations of these quantities are then obtained from 0015 0014  dR  2  (1 − x ) − Rx dx, dAws = 2πR   dx    0015 0014  dR (1 − x) − R dx, dAaw = 2πR 2  dx   0014 0015   dR  2 3 2   dV = πR (x − 3x + 2) + R(x − x) dx = 0. dx 2
From the last equation, we conclude 1 dR x2 − 1 x+1 =− 3 =− . R dx x − 3x + 2 (x − 1)(x + 2) Substituting for dR/dx gives, dAws = 2πR2
dx , x+2
and
dAaw = 2πR2
x · dx , x+2
resulting in the required result of dAaw = x · dAws = dAws cos θ. Method 2: Balancing forces on the contact line Another way to interpret the result is to consider the force balance of the equilibrium surface tension on the contact line. There are four forces acting on the line: (1) the surface tension at the water–gas interface, (2) the surface tension at the solid–water interface, (3) the surface tension at the gas–solid interface, and (4) the force downward by solid–contact line interaction. The last force ensures that the contact line stays on the solid surface, and is downward since the contact line is allowed to move only horizontally without friction. These forces should cancel along both the y–direction x–directions. The latter gives the condition for the contact angle known as Young’s equation, S as = S aw · cos θ + S ws ,
=⇒
cos θ =
S as − S ws . S aw
The critical condition for the complete wetting occurs when θ = 0, or cos θ = 1, i.e. for cos θC =
S as − S ws = 1. S aw
Complete wetting of the substrate thus occurs whenever S aw ≤ S as − S ws .
(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is S o ≈ 7 × 10−2 N m−1 . Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important. 3
• Typical length scales at which the surface tension effects become significant are given
by the condition that the forces exerted by surface tension and relevant pressures become
comparable, or by the condition that the surface energy is comparable to the other energy changes of interest. Example 1: Size of water drops not much deformed on a non-wetting surface. This is given by equalizing the surface energy and the gravitational energy, S · 4πR2 ≈ mgR = ρV gR =
4π 4 R g, 3
leading to R≈
s
3S ≈ ρg
s
3 · 7 × 10−2 N/m ≈ 1.5 × 10−3 m = 1.5mm. 103 kg/m3 × 10m/s2
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel (about 1 atm) = surface tension of water, gives πgel ≈
2S N kB T ≈ , V R
where N is the number of counter ions within the gel. Thus, 0012 0013 2 × 7 × 10−2 N/m R≈ ≈ 10−6 m. 105 N/m2 ******** 2. Surfactants:
Surfactant molecules such as those in soap or shampoo prefer to spread
on the air-water surface rather than dissolve in water. To see this, float a hair on the surface of water and gently touch the water in its vicinity with a piece of soap. (This is also why a piece of soap can power a toy paper boat.) (a) The air-water surface tension S o (assumed to be temperature independent) is reduced
roughly by N kB T /A, where N is the number of surfactant particles, and A is the area. Explain this result qualitatively. • Typical surfactant molecules have a hydrophilic head and a hydrophobic tail, and prefer to go to the interface between water and air, or water and oil. Some examples are, CH3 − (CH2 )11 − SO3− · N a+ , 4
CH3 − (CH2 )11 − N + (CH3 )3 · Cl− , CH3 − (CH2 )11 − O − (CH2 − CH2 − O)12 − H. The surfactant molecules spread over the surface of water and behave as a two dimensional gas. The gas has a pressure proportional to the density and the absolute temperature, which comes from the two dimensional degrees of freedom of the molecules. Thus the surfactants lower the free energy of the surface when the surface area is increased. ∆Fsurfactant =
N kB T · ∆A = (S − So ) · ∆A, A
S = So −
=⇒
N kB T. A
(Note that surface tension is defined with a sign opposite to that of hydrostatic pressure.) (b) Place a drop of water on a clean surface. Observe what happens to the air-watersurface contact angle as you gently touch the droplet surface with a small piece of soap, and explain the observation. • As shown in the previous problem, the contact angle satisfies cos θ =
S as − S ws . S aw
Touching the surface of the droplet with a small piece of soap reduces S aw , hence cos θ
increases, or equivalently, the angle θ decreases.
(c) More careful observations show that at higher surfactant densities 0012 00132 ∂S 2a N N kB T − = ∂A T (A − N b)2 A A
,
∂T A − Nb ; =− ∂S A N kB
and
where a and b are constants. Obtain the expression for S(A, T ) and explain qualitatively the origin of the corrections described by a and b.
• When the surfactant molecules are dense their interaction becomes important, resulting
in
and
0012 00132 N kB T 2a N ∂S = , − ∂A T (A − N b)2 A A
Integrating the first equation, gives
∂T A − Nb . =− ∂S A N kB
N kB T +a S(A, t) = f (T ) − A − Nb 5
0012
N A
00132
,
where f (T ) is a function of only T , while integrating the second equation, yields S(A, T ) = g(A) −
N kB T , A − Nb
with g(A) a function of only A. By comparing these two equations we get N kB T +a S(A, T ) = S o − A − Nb
0012
N A
00132
,
where S o represents the surface tension in the absence of surfactants and is independent
of A and T . The equation resembles the van der Waals equation of state for gas-liquid systems. The factor N b in the second term represents the excluded volume effect due to the finite size of the surfactant molecules. The last term represents the binary interaction between two surfactant molecules. If surfactant molecules attract each other the coefficient a is positive the surface tension increases. (d) Find an expression for CS − CA in terms of ∂E . ∂T S
∂E ∂A T ,
S,
• Taking A and T as independent variables, we obtain δQ = dE − S · dA,
=⇒
and δQ =
0012
∂S ∂A
T
, and
∂E ∂E δQ = dA + dT − S · dA, ∂A T ∂T A
0013 ∂E ∂E − S dA + dT. ∂A T ∂T A
From the above result, the heat capacities are obtained as
resulting in
 ∂E δQ    CA ≡ δT = ∂T A 0012 A , 0013  ∂E ∂A ∂E δQ   CS ≡ = −S + δT S ∂A T ∂T S ∂T S CS − CA =
Using the chain rule relation
∂T , ∂S A
0012
0013 ∂A ∂E −S . ∂A T ∂T S
∂T ∂S ∂A · · = −1, ∂S A ∂A T ∂T S 6
for
∂E ∂T A
=
we obtain CS − CA =
0012
 0013 ∂E −S · ∂A T ********
3. Temperature scales:
−1
∂T ∂S A
·

.
∂S ∂A T
Prove the equivalence of the ideal gas temperature scale Θ, and
the thermodynamic scale T , by performing a Carnot cycle on an ideal gas. The ideal gas satisfies P V = N kB Θ, and its internal energy E is a function of Θ only. However, you may not assume that E ∝ Θ. You may wish to proceed as follows: (a) Calculate the heat exchanges QH and QC as a function of ΘH , ΘC , and the volume expansion factors. • The ideal gas temperature is defined through the equation of state θ=
PV . N kB
The thermodynamic temperature is defined for a reversible Carnot cycle by Qhot Thot = . Tcold Qcold For an ideal gas, the internal energy is a function only of θ, i.e. E = E(θ), and d¯Q = dE − d¯W =
dE · dθ + P dV. dθ
adiabatics (∆Q = 0)
pressure P
1 Q hot 2
θ hot isothermals
4 Q cold 3 volume V 7
θ cold
Consider the Carnot cycle indicated in the figure. For the segment 1 to 2, which undergoes an isothermal expansion, we have dθ = 0,
=⇒ d¯Qhot = P dV,
and P =
N kB θhot . V
Hence, the heat input of the cycle is related to the expansion factor by 0012 0013 Z V2 dV V2 N kB θhot Qhot = . = N kB θhot ln V V1 V1 A similar calculation along the low temperature isotherm yields 0012 0013 Z V3 V3 dV , = N kB θcold ln Qcold = N kB θcold V V4 V4 and thus
Qhot θhot ln (V2 /V1 ) = . Qcold θcold ln (V3 /V4 )
(b) Calculate the volume expansion factor in an adiabatic process as a function of Θ. • Next, we calculate the volume expansion/compression ratios in the adiabatic processes. Along an adiabatic segment d¯Q = 0,
=⇒
0=
dE N kB θ · dθ + · dV, dθ V
=⇒
dV 1 dE =− · dθ. V N kB θ dθ
Integrating the above between the two temperatures, we obtain  0012 0013 Z θhot 1 dE 1 V3   =− · dθ, and   ln V N kB θcold θ dθ 2 0012 0013 Z θhot  1 dE 1 V4   =−  ln · dθ. V1 N kB θcold θ dθ
While we cannot explicitly evaluate the integral (since E(θ) is arbitrary), we can nonetheless conclude that
V2 V1 = . V4 V3
(c) Show that QH /QC = ΘH /ΘC . • Combining the results of parts (a) and (b), we observe that Qhot θhot = . Qcold θcold 8
Since the thermodynamic temperature scale is defined by Qhot Thot = , Qcold Tcold we conclude that θ and T are proportional. If we further define θ(triple pointH2 0 ) = T (triple pointH2 0 ) = 273.16, θ and T become identical. ******** 4. Equations of State:
The equation of state constrains the form of internal energy as
in the following examples. (a) Starting from dE = T dS − P dV , show that the equation of state P V = N kB T , in fact
implies that E can only depend on T .
• Since there is only one form of work, we can choose any two parameters as independent
variables. For example, selecting T and V , such that E = E(T, V ), and S = S(T, V ), we obtain
∂S ∂S dT + T dV − P dV, dE = T dS − P dV = T ∂T V ∂V T
resulting in
Using the Maxwell’s relation†
we obtain
Since T
∂P ∂T V
E = E(T ).
B = T Nk V
∂S ∂E =T − P. ∂V T ∂V T ∂S ∂P = , ∂V T ∂T V
∂P ∂E =T − P. ∂V T ∂T V ∂E = 0. Thus E depends only on T , i.e. = P , for an ideal gas, ∂V T
(b) What is the most general equation of state consistent with an internal energy that depends only on temperature? • If E = E(T ),
∂E = 0, ∂V T
=⇒
∂P T = P. ∂T V
The solution for this equation is P = f (V )T, where f (V ) is any function of only V . ∂Y ∂2L † dL = Xdx + Y dy + · · · , =⇒ ∂X ∂y = ∂x y = ∂x·∂y . x
9
(c) Show that for a van der Waals gas CV is a function of temperature alone. • The van der Waals equation of state is given by '
P −a
0012
N V
00132 #
· (V − N b) = N kB T,
or N kB T P = +a (V − N b)
0012
N V
00132
.
From these equations, we conclude that ∂E CV ≡ , ∂T V
=⇒
001b 001a ∂CV ∂ 2E ∂ 2 P ∂P ∂ = −P =T T = 0. = ∂V T ∂V ∂T ∂T ∂T V ∂T 2 V ********
5. Clausius–Clapeyron equation describes the variation of boiling point with pressure. It is usually derived from the condition that the chemical potentials of the gas and liquid phases are the same at coexistence. • From the equations
µliquid (P, T ) = µgas (P, T ),
and µliquid (P + dP, T + dT ) = µgas (P + dP, T + dT ), we conclude that along the coexistence line dP = dT coX
∂µg ∂T P ∂µl ∂P T
− −
∂µl ∂T P ∂µg ∂P T
.
The variations of the Gibbs free energy, G = N µ(P, T ) from the extensivity condition, are given by ∂G V = , ∂P T
In terms of intensive quantities
∂µ V , = v= N ∂P T
∂G S=− . ∂T P S ∂µ s= , =− N ∂T P
10
where s and v are molar entropy and volume, respectively. Thus, the coexistence line satisfies the condition dP Sg − Sl sg − sl = = . dT coX Vg − Vl vg − vl
For an alternative derivation, consider a Carnot engine using one mole of water. At the source (P, T ) the latent heat L is supplied converting water to steam. There is a volume increase V associated with this process. The pressure is adiabatically decreased to P − dP . At the sink (P − dP, T − dT ) steam is condensed back to water. (a) Show that the work output of the engine is W = V dP + O(dP 2 ). Hence obtain the Clausius–Clapeyron equation dP L = . dT boiling T V
(1)
• If we approximate the adiabatic processes as taking place at constant volume V (vertical lines in the P − V diagram), we find P dV = P V − (P − dP )V = V dP.
pressure
W =
I
P
liq.
1
Q hot
2 T
gas
V
P-dP 4 Q cold
T-dT
3
volume 11
Here, we have neglected the volume of liquid state, which is much smaller than that of the gas state. As the error is of the order of ∂V dP · dP = O(dP 2 ), ∂P S
we have
W = V dP + O(dP 2 ). The efficiency of any Carnot cycle is given by η=
W TC =1− , QH TH
and in the present case, QH = L,
W = V dP,
TH = T,
TC = T − dT.
Substituting these values in the universal formula for efficiency, we obtain the ClausiusClapeyron equation dT V dP = , L T
L dP = . dT coX T ·V
or
(b) What is wrong with the following argument: “The heat QH supplied at the source to convert one mole of water to steam is L(T ). At the sink L(T − dT ) is supplied to condense
one mole of steam to water. The difference dT dL/dT must equal the work W = V dP , equal to LdT /T from eq.(1). Hence dL/dT = L/T implying that L is proportional to T !” • The statement “At the sink L(T − dT ) is supplied to condense one mole of water” is
incorrect. In the P − V diagram shown, the state at “1” corresponds to pure water, “2”
corresponds to pure vapor, but the states “3” and “4” have two phases coexisting. In
going from the state 3 to 4 less than one mole of steam is converted to water. Part of the steam has already been converted into water during the adiabatic expansion 2 → 3, and
the remaining portion is converted in the adiabatic compression 4 → 1. Thus the actual
latent heat should be less than the contribution by one mole of water.
(c) Assume that L is approximately temperature independent, and that the volume change is dominated by the volume of steam treated as an ideal gas, i.e. V = N kB T /P . Integrate equation (1) to obtain P (T ). 12
• For an ideal gas N kB T V = , P
=⇒
LP dP = , dT coX N kB T 2
dP L = dT. P N kB T 2
or
Integrating this equation, the boiling temperature is obtained as a function of the pressure P , as 0012 P = C · exp −
L kB TBoiling
0013
.
(d) A hurricane works somewhat like the engine described above. Water evaporates at the warm surface of the ocean, steam rises up in the atmosphere, and condenses to water at the higher and cooler altitudes. The Coriolis force converts the upwards suction of the air to spiral motion. (Using ice and boiling water, you can create a little storm in a tea cup.) Typical values of warm ocean surface and high altitude temperatures are 800 F and −1200 F respectively. The warm water surface layer must be at least 200 feet thick to provide sufficient water vapor, as the hurricane needs to condense about 90 million tons of water vapor per hour to maintain itself. Estimate the maximum possible efficiency, and power output, of such a hurricane. (The latent heat of vaporization of water is about 2.3 × 106 Jkg −1 .)
• For TC = −120o F = 189o K, and TH = 80o F = 300o K, the limiting efficiency, as that
of a Carnot engine, is
ηmax =
TH − TC = 0.37. TH
The output power, is equal to (input power) x (efficiency). The input in this case is the energy obtained from evaporation of warm ocean temperature; hence dQc TH − TC dW = × dt dt TC 6 1hr 1000kg 2.3 × 106 J 90 × 10 tons · · · × 0.67 ≈ 4 × 1013 watts. = hr 3600sec ton kg
Power output =
(e) Due to gravity, atmospheric pressure P (h) drops with the height h. By balancing the forces acting on a slab of air (behaving like a perfect gas) of thickness dh, show that P (h) = P0 exp(−mgh/kT ), where m is the average mass of a molecule in air. 13
• Consider a horizontal slab of area A between heights h and h + dh. The gravitational force due to mass of particles in the slab is dFgravity = mg
N P Adh = mg Adh, V kB T
where we have used the ideal gas law to relate the density (N/V ) to the pressure. The gravitational force is balanced in equilibrium with the force due to pressure ∂P Adh. dFpressure = A [P (h) − P (h + dh)] = − ∂h
Equating the two forces gives ∂P P = −mg , ∂h kB T
0013 0012 mgh , P (h) = p0 exp − kB T
=⇒
assuming that temperature does not change with height. (f) Use the above results to estimate the boiling temperature of water on top of Mount Everest (h ≈ 9km). The latent heat of vaporization of water is about 2.3 × 106 Jkg −1 .
• Using the results from parts (c) and (e), we conclude that 0012 00130015 0012 0013 0014 PEverest 1 mg L 1 . ≈ exp − (hEverest − hsea ) ≈ exp − − Psea kB T kB TEverest (boil) Tsea (boil)
Using the numbers provided, we find TEverest (boil) ≈ 346o K (74o C≈ 163o F). ********
6. Glass:
Liquid quartz, if cooled slowly, crystallizes at a temperature Tm , and releases
latent heat L. Under more rapid cooling conditions, the liquid is supercooled and becomes glassy. (a) As both phases of quartz are almost incompressible, there is no work input, and changes in internal energy satisfy dE = T dS + µdN . Use the extensivity condition to obtain the expression for µ in terms of E, T , S, and N . • Since in the present context we are considering only chemical work, we can regard
entropy as a function of two independent variables, e.g. E, and N , which appear naturally from dS = dE/T − µdN/T . Since entropy is an extensive variable, λS = S(λE, λN ).
Differentiating this with respect to λ and evaluating the resulting expression at λ = 1, gives ∂S E ∂S Nµ E + N = S(E, N ) = − , ∂E N ∂N E T T 14
leading to µ=
E − TS . N
(b) The heat capacity of crystalline quartz is approximately CX = αT 3 , while that of glassy quartz is roughly CG = βT , where α and β are constants. Assuming that the third law of thermodynamics applies to both crystalline and glass phases, calculate the entropies of the two phases at temperatures T ≤ Tm .
• Finite temperature entropies can be obtained by integrating d¯Q/T , starting from S(T =
0) = 0. Using the heat capacities to obtain the heat inputs, we find  T dScrystal   Ccrystal = αT 3 = , N dT  T dSglass  Cglass = βT = , N dT
N αT 3 , 3
=⇒
Scrystal =
=⇒
Sglass = βN T.
(c) At zero temperature the local bonding structure is similar in glass and crystalline quartz, so that they have approximately the same internal energy E0 . Calculate the internal energies of both phases at temperatures T ≤ Tm .
• Since dE = T dS + µdN , for dN = 0, we have (
dE = T dS = αN T 3 dT dE = T dS = βN T dT
(crystal), (glass).
Integrating these expressions, starting with the same internal energy Eo at T = 0, yields  αN 4   E = Eo + T 4   E = E + βN T 2 o 2
(crystal), (glass).
(d) Use the condition of thermal equilibrium between two phases to compute the equilibrium melting temperature Tm in terms of α and β. • From the condition of chemical equilibrium between the two phases, µcrystal = µglass , we obtain
0012
1 1 − 3 4
0013
0012 0013 1 · αT = 1 − · βT 2 , 2 4
15
=⇒
αT 4 βT 2 = , 12 2
resulting in a transition temperature Tmelt =
r
6β . α
(e) Compute the latent heat L in terms of α and β. • From the assumptions of the previous parts, we obtain the latent heats for the glass to
crystal transition as
0012 0013 3 αTmelt L = Tmelt (Sglass − Scrystal ) = N Tmelt βTmelt − 3 0013 0012 2 αTmelt 2 2 2 = N Tmelt (β − 2β) = −N βTmelt < 0. = N Tmelt β− 3 (f) Is the result in the previous part correct? If not, which of the steps leading to it is most likely to be incorrect? • The above result implies that the entropy of the crystal phase is larger than that of
the glass phase. This is clearly unphysical, and one of the assumptions must be wrong.
The questionable step is the assumption that the glass phase is subject to the third law of thermodynamics, and has zero entropy at T = 0. In fact, glass is a non-ergodic state of matter which does not have a unique ground state, and violates the third law. ******** 7. Filament: For an elastic filament it is found that, at a finite range in temperature, a displacement x requires a force J = ax − bT + cT x, where a, b, and c are constants. Furthermore, its heat capacity at constant displacement is proportional to temperature, i.e. Cx = A(x)T . (a) Use an appropriate Maxwell relation to calculate ∂S/∂x|T . • From dF = −SdT + Jdx, we obtain ∂S ∂J =− = b − cx. ∂x T ∂T x (b) Show that A has to in fact be independent of x, i.e. dA/dx = 0. 16
∂S = A(x)T , where S = S(T, x). Thus • We have Cx = T ∂T
∂ ∂S ∂ ∂S ∂A = = =0 ∂x ∂x ∂T ∂T ∂x from part (a), implying that A is independent of x. (c) Give the expression for S(T, x) assuming S(0, 0) = S0 . • S(x, T ) can be calculated as T ′ =T
∂S(T ′ , x = 0) ′ S(x, T ) = S(0, 0) + dT + ∂T ′ T ′ =0 Z x Z T ′ (b − cx′ )dx′ AdT + = S0 + Z
Z
x′ =x x′ =0
∂S(T, x′ ) dx ∂x′
0
0
c = S0 + AT + (b − x)x. 2
(d) Calculate the heat capacity at constant tension, i.e. CJ = T ∂S/∂T |J as a function of
T and J.
• Writing the entropy as S(T, x) = S(T, x(T, J)), leads to ∂S ∂S ∂x ∂S = + . ∂T J ∂T x ∂x T ∂T J = b − cx and ∂S x = A. Furthermore, From parts (a) and (b), ∂S ∂x T ∂T ∂x ∂x − b + cx + cT ∂T = 0, i.e. a ∂T
Thus
Since x =
b − cx ∂x = . ∂T a + cT
∂x ∂T J
is given by
0015 0014 (b − cx)2 . CJ = T A + (a + cT ) J+bT a+cT
, we can rewrite the heat capacity as a function of T and J, as '
CJ = T A +
2 (b − c J+bT a+cT )
(a + cT ) 0015 0014 (ab − cJ)2 . =T A+ (a + cT )3
#
******** 8.
Hard core gas:
A gas obeys the equation of state P (V − N b) = N kB T , and has a
heat capacity CV independent of temperature. (N is kept fixed in the following.) 17
(a) Find the Maxwell relation involving ∂S/∂V |T,N . • For dN = 0,
d(E − T S) = −SdT − P dV,
=⇒
∂P ∂S = . ∂V T,N ∂T V,N
(b) By calculating dE(T, V ), show that E is a function of T (and N ) only. • Writing dS in terms of dT and dV , dE = T dS − P dV = T
! ∂S ∂S dT + dV − P dV. ∂T V,N ∂V T,N
Using the Maxwell relation from part (a), we find ∂S dT + dE(T, V ) = T ∂T V,N
But from the equation of state, we get N kB T P = , (V − N b)
=⇒
P ∂P = , ∂T V,N T
! ∂P T − P dV. ∂T V,N =⇒
i.e. E(T, N, V ) = E(T, N ) does not depend on V .
∂S dT, dE(T, V ) = T ∂T V,N
(c) Show that γ ≡ CP /CV = 1 + N kB /CV (independent of T and V ). • The hear capacity is
∂Q ∂E + P V = CP = ∂T P ∂T
But, since E = E(T ) only,
= ∂E + P ∂V . ∂T P ∂T P P
∂E ∂E = = CV , ∂T P ∂T V
and from the equation of state we get N kB ∂V = , ∂T P P
=⇒
CP = CV + N kB ,
=⇒
γ =1+
N kB , CV
which is independent of T , since CV is independent of temperature. The independence of CV from V also follows from part (a). 18
(d) By writing an expression for E(P, V ), or otherwise, show that an adiabatic change satisfies the equation P (V − N b)γ =constant. • Using the equation of state, we have dE = CV dT = CV d
0012
P (V − N b) N kB
0013
=
CV (P dV + (V − N b)dP ) . N kB
The adiabatic condition, dQ = dE + P dV = 0, can now be written as 0 = dQ =
0012
CV 1+ N kB
0013
P d(V − N b) +
CV (V − N b)dP. N kB
Dividing by CV P (V − N b)/(N kB ) yields d(V − N b) dP +γ = 0, P (V − N b)
ln [P (V − N b)γ ] = constant.
=⇒
******** 9. Superconducting transition: Many metals become superconductors at low temperatures T , and magnetic fields B. The heat capacities of the two phases at zero magnetic field are approximately given by (
Cs (T ) = V αT 3 0002 0003 Cn (T ) = V βT 3 + γT
in the superconducting phase in the normal phase
,
where V is the volume, and {α, β, γ} are constants. (There is no appreciable change in volume at this transition, and mechanical work can be ignored throughout this problem.)
(a) Calculate the entropies Ss (T ) and Sn (T ) of the two phases at zero field, using the third law of thermodynamics. • Finite temperature entropies are obtained by integrating dS = d¯Q/T , starting from S(T = 0) = 0. Using the heat capacities to obtain the heat inputs, we find    
dSs , dT 0002 0003 dS    Cn = V βT 3 + γT = T n , dT Cs = V αT 3 = T
19
=⇒ =⇒
αT 3 , 00143 3 0015 . βT Sn = V + γT 3 Ss = V
(b) Experiments indicate that there is no latent heat (L = 0) for the transition between the normal and superconducting phases at zero field. Use this information to obtain the transition temperature Tc , as a function of α, β, and γ. • The Latent hear for the transition is related to the difference in entropies, and thus L = Tc (Sn (Tc ) − Ss (Tc )) = 0. Using the entropies calculated in the previous part, we obtain r αTc3 3γ βTc3 = + γTc , =⇒ Tc = . 3 3 α−β (c) At zero temperature, the electrons in the superconductor form bound Cooper pairs. As a result, the internal energy of the superconductor is reduced by an amount V ∆, i.e. En (T = 0) = E0 and Es (T = 0) = E0 −V ∆ for the metal and superconductor, respectively. Calculate the internal energies of both phases at finite temperatures.
• Since dE = T dS + BdM + µdN , for dN = 0, and B = 0, we have dE = T dS = CdT .
Integrating the given expressions for heat capacity, and starting with the internal energies E0 and E0 − V ∆ at T = 0, yields h  α 4i   Es (T ) = E0 + V −∆ + 4 T 0014 0015 β 4 γ 2 .   En (T ) = E0 + V T + T 4 2
(d) By comparing the Gibbs free energies (or chemical potentials) in the two phases, obtain an expression for the energy gap ∆ in terms of α, β, and γ. • The Gibbs free energy G = E − T S − BM = µN can be calculated for B = 0 in each
phase, using the results obtained before, as h h  α 3 α 4i α 4i  T − T V T = E − V ∆ + T G (T ) = E + V −∆ + 0 0  s 4 3 12 0014 0015 0014 0015 0015 0014 β 4 γ 2 β 3 β 4 γ 2 .   Gn (T ) = E0 + V T + T − TV T + γT = E0 − V T + T 4 2 3 12 2
At the transition point, the chemical potentials (and hence the Gibbs free energies) must be equal, leading to ∆+
α 4 β 4 γ 2 Tc = T + Tc , 12 12 c 2
=⇒ 20
∆=
γ 2 α−β 4 T − T . 2 c 12 c
Using the value of Tc =
p
3γ/(α − β), we obtain ∆=
3 γ2 . 4α−β
(e) In the presence of a magnetic field B, inclusion of magnetic work results in dE = T dS +BdM +µdN , where M is the magnetization. The superconducting phase is a perfect diamagnet, expelling the magnetic field from its interior, such that Ms = −V B/(4π) in
appropriate units. The normal metal can be regarded as approximately non-magnetic,
with Mn = 0. Use this information, in conjunction with previous results, to show that the superconducting phase becomes normal for magnetic fields larger than 0013 0012 T2 Bc (T ) = B0 1 − 2 , Tc giving an expression for B0 . • Since dG = −SdT − M dB + µdN , we have to add the integral of −M dB to the Gibbs
free energies calculated in the previous section for B = 0. There is no change in the metallic phase since Mn = 0, while in the superconducting phase there is an additional R R contribution of − Ms dB = (V /4π) BdB = (V /8π)B 2 . Hence the Gibbs free energies
at finite field are
 h B2 α 4i   + V T G (T, B) = E − V ∆ +  s 0 12 0014 0015 8π . β 4 γ 2    Gn (T, B) = E0 − V T + T 12 2
Equating the Gibbs free energies gives a critical magnetic field
γ α−β 4 3 γ2 γ α−β 4 Bc2 = ∆ − T2 + T = − T2 + T 8π 2 12 4α−β 2 12 '0012 # 00132 00012 α−β 3γ 6γT 2 α−β = − Tc2 − T 2 , + T4 = 12 α−β α−β 12 where we have used the values of ∆ and Tc obtained before. Taking the square root of the above expression gives 0013 0012 T2 Bc = B0 1 − 2 , Tc
where
B0 =
r
21
2π(α − β) 2 Tc = 3
s
p 6πγ 2 = Tc 2πγ. α−β
******** 10. Photon gas Carnot cycle:
The aim of this problem is to obtain the blackbody 4
radiation relation, E(T, V ) ∝ V T , starting from the equation of state, by performing an
infinitesimal Carnot cycle on the photon gas.
P P+dP
T+dT
P V
T
V+dV
V (a) Express the work done, W , in the above cycle, in terms of dV and dP . • Ignoring higher order terms, net work is the area of the cycle, given by W = dP dV . (b) Express the heat absorbed, Q, in expanding the gas along an isotherm, in terms of P , dV , and an appropriate derivative of E(T, V ). • Applying the first law, the heat absorbed is Q = dE + P dV =
00140012
∂E ∂T
0013
dT +
V
0012
∂E ∂V
0013
dV + P dV
T
0015
= isotherm
00140012
∂E ∂V
0013
T
0015
+ P dV.
(c) Using the efficiency of the Carnot cycle, relate the above expressions for W and Q to T and dT . • The efficiency of the Carnot cycle (η = dT /T ) is here calculated as η=
dP dT W = = . Q [(∂E/∂V )T + P ] T
(d) Observations indicate that the pressure of the photon gas is given by P = AT 4 , 3
4 where A = π 2 kB /45 (¯ hc) is a constant. Use this information to obtain E(T, V ), assuming
E(T, 0) = 0. • From the result of part (c) and the relation P = AT 4 , 4
4AT =
0012
∂E ∂V
0013
4
+ AT ,
T
22
or
0012
∂E ∂V
0013
= 3AT 4 , T
so that E = 3AV T 4 .
(e) Find the relation describing the adiabatic paths in the above cycle. • Adiabatic curves are given by dQ = 0, or 0=
0012
∂E ∂T
0013
0012
dT +
V
∂E ∂V
0013
dV + P dV = 3V dP + 4P dV, T
i.e. P V 4/3 = constant.
******** 11. Irreversible Processes: (a) Consider two substances, initially at temperatures T10 and T20 , coming to equilibrium at a final temperature Tf through heat exchange. By relating the direction of heat flow to the temperature difference, show that the change in the total entropy, which can be written as Z
∆S = ∆S1 + ∆S2 ≥
Tf
T10
d¯Q1 + T1
Z
Tf
T20
d¯Q1 = T2
Z
T1 − T2 d¯Q, T1 T2
must be positive. This is an example of the more general condition that “in a closed system, equilibrium is characterized by the maximum value of entropy S.” • Defining the heat flow from substance 1 to 2 as, d¯Q1→2 , we get, ∆S = ∆S1 + ∆S2 ≥
Z
Tf
T10
d¯Q1 + T1
Z
Tf
T20
d¯Q2 = T2
Z
T1 − T2 d¯Q1→2 . T1 T2
But according to Clausius’ statement of the second law d¯Q1→2 > 0, if T1 > T2 and d¯Q1→2 < 0, if T1 < T2 . Hence, (T1 − T2 )d¯Q1→2 ≥ 0, resulting in ∆S ≥
Z
T1 − T2 d¯Q1→2 ≥ 0. T1 T2
23
(b) Now consider a gas with adjustable volume V , and diathermal walls, embedded in a heat bath of constant temperature T , and fixed pressure P . The change in the entropy of the bath is given by ∆Sbath =
∆Qbath ∆Qgas 1 =− = − (∆Egas + P ∆Vgas ) . T T T
By considering the change in entropy of the combined system establish that “the equilibrium of a gas at fixed T and P is characterized by the minimum of the Gibbs free energy G = E + P V − T S.”
• The total change in entropy of the whole system is, 1 1 ∆S = ∆Sbath + ∆Sgas = − (∆Egas + P ∆Vgas − T ∆Sgas ) = − ∆Ggas . T T By the second law of thermodynamics, all processes must satisfy, 1 − ∆Ggas ≥ 0 ⇔ ∆Ggas ≤ 0, T that is, all processes that occur can only lower the Gibbs free energy of the gas. Therefore the equillibrium of a gas in contact with a heat bath of constant T and P is established at the point of minimum Gibbs free energy, i.e. when the Gibbs free energy cannot be lowered any more. ******** 12. The solar system originated from a dilute gas of particles, sufficiently separated from other such clouds to be regarded as an isolated system. Under the action of gravity the particles coalesced to form the sun and planets. (a) The motion and organization of planets is much more ordered than the original dust cloud. Why does this not violate the second law of thermodynamics? • The formation of planets is due to the gravitational interaction. Because of the attrac-
tive nature of this interaction, the original dust cloud with uniform density has in fact
lower entropy. Clumping of the uniform density leads to higher entropy. Of course the gravitational potential energy is converted into kinetic energy in the process. Ultimately the kinetic energy of falling particles is released in the form of photons which carry away a lot of entropy. (b) The nuclear processes of the sun convert protons to heavier elements such as carbon. Does this further organization lead to a reduction in entropy? 24
• Again the process of formation of heavier elements is accompanied by the release of large amounts of energy which are carry away by photons. The entropy carry away by these
photons is more than enough to compensate any ordering associated with the packing of nucleons into heavier nuclei. (c) The evolution of life and intelligence requires even further levels of organization. How is this achieved on earth without violating the second law? • Once more there is usage of energy by the organisms that converts more ordered forms of energy to less ordered ones.
********
25
Problems for Chapter II - Probability 1. Characteristic functions:
Calculate the characteristic function, the mean, and the
variance of the following probability density functions: (a) Uniform
1 2a
p(x) =
−a < x < a , and
for
• A uniform probability distribution,   1 p(x) = 2a  0
p(x) = 0
for − a < x < a
otherwise;
,
otherwise
for which there exist many examples, gives 1 f (k) = 2a = Therefore,
a 1 1 exp(−ikx)dx = exp(−ikx) 2a −ik −a −a
Z
a
∞ X 1 (ak)2m sin(ka) = (−1)m . ak (2m + 1)! m=0
m1 = hxi = 0, (b) Laplace
p(x) =
• The Laplace PDF,
1 2a
exp
0010
− |x| a
0011
and
m2 = hx2 i =
1 2 a . 3
;
0013 0012 1 |x| p(x) = , exp − 2a a
for example describing light absorption through a turbid medium, gives 0012 0013 Z ∞ |x| 1 dx exp −ikx − f (k) = 2a −∞ a Z ∞ Z 0 1 1 dx exp(−ikx − x/a) + dx exp(−ikx + x/a) = 2a 0 2a −∞ 0014 0015 1 1 1 1 = = − 2a −ik + 1/a −ik − 1/a 1 + (ak)2 = 1 − (ak)2 + (ak)4 − · · · .
Therefore, m1 = hxi = 0,
and
26
m2 = hx2 i = 2a2 .
(c) Cauchy
p(x) =
a π(x2 +a2 )
.
• The Cauchy, or Lorentz PDF describes the spectrum of light scattered by diffusive
modes, and is given by
p(x) =
π(x2
a . + a2 )
For this distribution, ∞
a exp(−ikx) dx 2 π(x + a2 ) −∞ 0014 0015 Z ∞ 1 1 1 exp(−ikx) dx. − = 2πi −∞ x − ia x + ia
f (k) =
Z
The easiest method for evaluating the above integrals is to close the integration contours in the complex plane, and evaluate the residue. The vanishing of the integrand at infinity determines whether the contour has to be closed in the upper, or lower half of the complex plane, and leads to Z  1 exp(−ikx)   dx = exp(−ka)  − 2πi x + ia C f (k) = Z  exp(−ikx) 1   dx = exp(ka) 2πi B x − ia
  for k ≥ 0  
= exp(−|ka|).
  for k < 0 
Note that f (k) is not an analytic function in this case, and hence does not have a Taylor expansion. The moments have to be determined by another method, e.g. by direct evaluation, as m1 = hxi = 0,
and
2
m2 = hx i =
Z
dx
π x2 · 2 → ∞. a x + a2
The first moment vanishes by symmetry, while the second (and higher) moments diverge, explaining the non-analytic nature of f (k). The following two probability density functions are defined for x ≥ 0. Compute only
the mean and variance for each. (d) Rayleigh
p(x) =
x a2
2
x exp(− 2a , 2)
• The Rayleigh distribution, 0012 0013 x2 x p(x) = 2 exp − 2 , a 2a 27
for
x ≥ 0,
can be used for the length of a random walk in two dimensions. Its characteristic function is 0012 0013 Z ∞ x2 x f (k) = exp(−ikx) 2 exp − 2 dx a 2a 0 0012 0013 Z ∞ x x2 = [cos(kx) − i sin(kx)] 2 exp − 2 dx. a 2a 0 The integrals are not simple, but can be evaluated as 0012 0013 Z ∞ ∞ X 0001n (−1)n n! x2 x 2a2 k 2 , cos(kx) 2 exp − 2 dx = a 2a (2n)! 0 n=0 and
Z
0
resulting in
∞
0012 0013 0012 0013 Z x x x2 1 ∞ x2 sin(kx) 2 exp − 2 dx = sin(kx) 2 exp − 2 dx a 2a 2 −∞ a 2a r 0012 2 20013 π k a , ka exp − = 2 2 r 0012 2 20013 ∞ X 0001 (−1)n n! k a π 2 2 n 2a k −i . f (k) = ka exp − (2n)! 2 2 n=0
The moments can also be calculated directly, from r 0013 0013 0012 0012 Z ∞ 2 Z ∞ 2 x π x x2 x2 m1 = hxi = dx = dx = exp − exp − a, 2 a2 2a2 2a2 2 0 −∞ 2a 0012 0013 0012 0013 0012 20013 Z ∞ 2 Z ∞ 3 x x2 x2 x x 2 2 exp − 2 dx = 2a exp − 2 d m2 = hx i = 2 2 a 2a 2a 2a 2a2 0 0 Z ∞ y exp(−y)dy = 2a2 . = 2a2 0
q
2
2
x (e) Maxwell p(x) = π2 xa3 exp(− 2a . 2) • It is difficult to calculate the characteristic function for the Maxwell distribution r 0013 0012 2 x2 x2 exp − 2 , p(x) = π a3 2a
say describing the speed of a gas particle. However, we can directly evaluate the mean and variance, as r Z ∞ 0012 0013 x3 2 x2 exp − 2 dx m1 = hxi = π 0 a3 2a r 0012 0013 0012 20013 Z ∞ 2 x x2 x 2 =2 a exp − 2 d 2 π 0 2a 2a 2a2 r r Z ∞ 2 2 a y exp(−y)dy = 2 a, =2 π 0 π 28
and 2
m2 = hx i =
r
2 π
Z
∞
o
0012 0013 x4 x2 exp − 2 dx = 3a2 . a3 2a
******** 2. Directed random walk:
The motion of a particle in three dimensions is a series
of independent steps of length ℓ. Each step makes an angle θ with the z axis, with a probability density p(θ) = 2 cos2 (θ/2)/π; while the angle φ is uniformly distributed between 0 and 2π. (Note that the solid angle factor of sin θ is already included in the definition of p(θ), which is correctly normalized to unity.) The particle (walker) starts at the origin and makes a large number of steps N .
(a) Calculate the expectation values hzi, hxi, hyi, z 2 , x2 , and y 2 , and the covariances hxyi, hxzi, and hyzi.
• From symmetry arguments,
hxi = hyi = 0,
while along the z-axis, hzi =
X i
hzi i = N hzi i = N a hcos θi i =
Na . 2
The last equality follows from Z Z π 1 hcos θi i = p(θ) cos θdθ = cos θ · (cos θ + 1)dθ 0 π Z π 1 1 (cos 2θ + 1)dθ = . = 2 0 2π The second moment of z is given by X XX
2 X zi2 hzi zj i = hzi zj i + z = i
i,j
=
XX i
Noting that
i6=j
i
i6=j
X hzi i hzj i + zi2 i
2
= N (N − 1) hzi i + N zi2 .
2 Z π Z π zi 1 1 1 2 = cos θ(cos θ + 1)dθ = (cos 2θ + 1)dθ = , 2 a 2 0 π 0 2π 29
we find
0010 a 00112
2 a2 a2 +N z = N (N − 1) = N (N + 1) . 2 2 4
The second moments in the x and y directions are equal, and given by XX X
2 X x2i = N x2i . hxi xj i = hxi xj i + x = i,j
i
i
i6=j
Using the result
2
xi = sin2 θ cos2 φ 2 a Z π Z 2π 1 1 2 dθ sin2 θ(cos θ + 1) = , dφ cos φ = 2 2π 0 4 0 we obtain
2 2 N a2 x = y = . 4
While the variables x, y, and z are not independent because of the constraint of unit length, simple symmetry considerations suffice to show that the three covariances are in fact zero, i.e. hxyi = hxzi = hyzi = 0. (b) Use the central limit theorem to estimate the probability density p(x, y, z) for the particle to end up at the point (x, y, z). • From the Central limit theorem, the probability density should be Gaussian. However,
for correlated random variable we may expect cross terms that describe their covariance. However, we showed above that the covarainces between x, y, and z are all zero. Hence we can treat them as three independent Gaussian variables, and write 0014 0015 (x − hxi)2 (y − hyi)2 (z − hzi)2 p(x, y, z) ∝ exp − − − . 2σx2 2σy2 2σz2 (There will be correlations between x, y, and z appearing in higher cumulants, but all such cumulants become irrelevant in the N → ∞ limit.) Using the moments hxi = hyi = 0,
and
a hzi = N , 2
a2 2 = σy2 , σx2 = x2 − hxi = N 4 30
σz2
and we obtain
a2 2 = z 2 − hzi = N (N + 1) − 4
p(x, y, z) =
0012
2 πN a2
00133/2
0012
'
Na 2
00132
=N 2
x2 + y 2 + (z − N a/2) exp − N a2 /2
a2 , 4
#
.
******** Consider any probability density p(x) for (−∞ < x < ∞),
3. Tchebycheff inequality:
with mean λ, and variance σ 2 . Show that the total probability of outcomes that are more than nσ away from λ is less than 1/n2 , i.e. Z |x−λ|≥nσ
dxp(x) ≤
1 . n2
Hint: Start with the integral defining σ 2 , and break it up into parts corresponding to |x − λ| > nσ, and |x − λ| < nσ.
• By definition, for a system with a PDF p(x), and average λ, the variance is Z 2 σ = (x − λ)2 p(x)dx. Let us break the integral into two parts as Z Z 2 2 (x − λ) p(x)dx + σ = |x−λ|≥nσ
|x−λ| {x1 , x2 , · · · , xn−1 }. 45
We can then define an associated sequence of indicators {R1 , R2 , · · · , Rn , · · ·} in which Rn = 1 if xn is a record, and Rn = 0 if it is not (clearly R1 = 1).
(a) Assume that each entry xn is taken independently from the same probability distribution p(x). [In other words, {xn } are IIDs (independent identically distributed).] Show
that, irrespective of the form of p(x), there is a very simple expression for the probability
Pn that the entry xn is a record. • Consider the n–entries {x1 , x2 , · · · , xn }. Each one of them has the same probability to
be the largest one. Thus the probability that xn is the largest, and hence a record, is Pn = 1/n. (b) The records are entered in the Guinness Book of Records. What is the average number hSN i of records after N attempts, and how does it grow for, N ≫ 1? If the number of trials, e.g. the number of participants in a sporting event, doubles every year, how does the number of entries asymptotically grow with time. •
N X
0012 0013 N X 1 1 hSN i = Pn = ≈ ln N + γ + O n N n=1 n=1
for
N ≫ 1,
where γ ≈ 0.5772 . . . is the Euler number. Clearly if N ∝ 2t , where t is the number of
years,
hSt i ≈ ln N (t) = t ln 2. (c) Prove that the record indicators {Rn } are independent random variables (though not
identical), in that hRn Rm ic = 0 for m 6= n.
• hRn Rm i = 1 · Pn · Pm while hRn i = 1 · Pn , hence yielding, hRn Rm ic = Pn Pm − Pn · Pm = 0. This is by itself not sufficient to prove that the random variables Rn and Rm are independent variables. The correct proof is to show that the joint probability factorizes, i.e. p(Rn , Rm ) = p(Rn )p(Rm ). Let us suppose that m > n, the probability that xm is the largest of the random variables up to m is 1/m, irrespective of whether some other random number in the set was itself a record (the largest of the random numbers up to n). The equality of the conditional and unconditional probabilities is a proof of independence. (d) Compute all moments, and the first three cumulants of the total number of records SN after N entries. Does the central limit theorem apply to SN ? 46
• We want to compute he−ikSN i. This satisfies, he−ikSN i =
e−ik −ikSN −1 N − 1 −ikSN −1 e−ik + N − 1 −ikSN −1 he i+ he i= he i, N N N
since the probability for aquiring an additional phase −ik is PN = 1/N . Since he−ikS1 i =
e−ik , by recursion we obtain, he−ikSN i =
N N 1 X 1 Y −ik (e + n − 1) = S1 (N, n)e−ikn , N ! n=1 N ! n=0
where S1 (n, m) denotes the unsigned Stirling number of the first kind. Expanding the exponent, we obtain, −ikSN
he
N ∞ ∞ X X 1 X nm (−ik)m (−ik)m i= S1 (N, n) = N ! n=0 m! m! m=0 m=0
'
# N 1 X m S1 (N, n)n . N ! n=0
Hence we obtain the mth moment, m hSN i
N 1 X S1 (N, n)nm . = N ! n=0
The first three cumulants can be obtained using the moments, but the easier way to obtain them is by using the fact that SN = R1 + · · · + RN and that Rn are independent variables.
Due to the independence of Rn ,
m hSN ic
=
N X
n=1
hRnm ic .
From this, and the fact that for any m, hRnm i = 1/n, we easily obtain, N X
N X 1 hSN ic = hRn ic = , n n=1 n=1 2 hSN ic
3 hSN ic
=
=
N X
n=1 N X
n=1
hRn2 ic
hRn3 ic
N X 1 1 ( − 2 ), = n n n=1
N X 3 2 1 = ( − 2 + 3 ). n n n n=1
47
The central limit theorem applies to SN since Rn are independent variables. Seeing this in another way is to observe that for large N , m hSN ic =
N X
n=1
hRnm ic
0. Since the same ω can be obtained for different choices of W , we have Z p(ω) = dW pf (W − ω)pb (−W ). By the relation obtained in (a), we get,
pf (W − ω)pb (−W ) exp(β(W − ω + F − F ′ )) = , pb (−W + ω)pf (W ) exp(β(W + F − F ′ ))
which yields,
pf (W − ω)pb (−W ) = pb (−W + ω)pf (W ) exp(−βω), hence, p(ω) =
Z
dW pb (−W + ω)pf (W ) exp(−βω) = p(−ω) exp(−βω).
The cumulative probability of violating the second law by ω or more is now obtained as Z ∞ Z ∞ Z ∞ ′ ′ ′ ′ ′ P (ω) = ω p(ω ) = ω p(−ω ) exp(−βω ) < exp(−βω) ω ′ p(−ω ′ ) < exp(−βω). ω
ω
ω
′
The first inequality is because for all ω in the integrand, exp(−βω ′ ) < exp(−βω), and the second since any cumulative probability is less than one. ******** 51
Problems for Chapter III - Kinetic Theory of Gases 1. One dimensional gas:
A thermalized gas particle is suddenly confined to a one–
dimensional trap. The corresponding mixed state is described by an initial density function √ ρ(q, p, t = 0) = δ(q)f (p), where f (p) = exp(−p2 /2mkB T )/ 2πmkB T . (a) Starting from Liouville’s equation, derive ρ(q, p, t) and sketch it in the (q, p) plane. • Liouville’s equation, describing the incompressible nature of phase space density, is ∂ρ ∂ρ ∂ρ ∂H ∂ρ ∂H ∂ρ = −q˙ − p˙ =− + ≡ −{ρ, H}. ∂t ∂q ∂p ∂p ∂q ∂q ∂p For the gas particle confined to a 1–dimensional trap, the Hamiltonian can be written as H=
p2 p2 + V (qx ) = , 2m 2m
since Vqx = 0, and there is no motion in the y and z directions. With this Hamiltonian, Liouville’s equation becomes p ∂ρ ∂ρ =− , ∂t m ∂q whose solution, subject to the specified initial conditions, is 0011 0010 0010 p 0011 p ρ(q, p, t) = ρ q − t, p, 0 = δ q − t f (p). m m
p
t=0
q
slope m/ t
52
(b) Derive the expressions for the averages q 2 and p2 at t > 0. • The expectation value for any observable O is Z hOi = dΓOρ(Γ, t),
and hence
Z 0010 p 0011 p f (p)δ q − t dp dq = p2 f (p)dp m 0012 0013 Z ∞ 1 p2 2 dp p √ exp − = = mkB T. 2mkB T 2πmkB T −∞
2 p =
Z
2
Likewise, we obtain 0012 00132 Z Z 0010 00112 Z 0010
2 kB T 2 p t p 0011 2 p2 f (p)dp = t f (p)dp = t . q = q f (p)δ q − t dp dq = m m m m (c) Suppose that hard walls are placed at q = ±Q. Describe ρ(q, p, t ≫ τ ), where τ is an appropriately large relaxation time. • Now suppose that hard walls are placed at q = ±Q. The appropriate relaxation time τ , is the characteristic length between the containing walls divided by the characteristic velocity of the particle, i.e. r 2Q 2Qm m = 2Q τ∼ =p . |q| ˙ kB T hp2 i Initially ρ(q, p, t) resembles the distribution shown in part (a), but each time the particle hits the barrier, reflection changes p to −p. As time goes on, the slopes become less, and ρ(q, p, t) becomes a set of closely spaced lines whose separation vanishes as 2mQ/t.
p
q -Q
+Q
53
(d) A “coarse–grained” density ρ˜, is obtained by ignoring variations of ρ below some small resolution in the (q, p) plane; e.g., by averaging ρ over cells of the resolution area. Find ρ˜(q, p) for the situation in part (c), and show that it is stationary. • We can choose any resolution ε in (p, q) space, subdividing the plane into an array of pixels of this area. For any ε, after sufficiently long time many lines will pass through this area. Averaging over them leads to ρ˜(q, p, t ≫ τ ) =
1 f (p), 2Q
as (i) the density f (p) at each p is always the same, and (ii) all points along q ∈ [−Q, +Q] are equally likely. For the time variation of this coarse-grained density, we find ∂ ρ˜ p ∂ ρ˜ =− = 0, i.e. ρ˜ is stationary. ∂t m ∂q
******** 2. Evolution of entropy:
The normalized ensemble density is a probability in the phase R space Γ. This probability has an associated entropy S(t) = − dΓρ(Γ, t) ln ρ(Γ, t).
(a) Show that if ρ(Γ, t) satisfies Liouville’s equation for a Hamiltonian H, dS/dt = 0. • A candidate “entropy” is defined by S(t) = −
Z
dΓρ(Γ, t) ln ρ(Γ, t) = − hln ρ(Γ, t)i .
Taking the derivative with respect to time gives dS =− dt
Z
dΓ
0012
∂ρ 1 ∂ρ ln ρ + ρ ∂t ρ ∂t
0013
=−
Z
dΓ
∂ρ (ln ρ + 1) . ∂t
Substituting the expression for ∂ρ/∂t obtained from Liouville’s theorem gives dS =− dt
Z
0013 3N 0012 X ∂ρ ∂H ∂ρ ∂H − (ln ρ + 1) . dΓ ∂pi ∂qi ∂qi ∂pi i=1 54
(Here the index i is used to label the 3 coordinates, as well as the N particles, and hence runs from 1 to 3N .) Integrating the above expression by parts yields†
0012 0013 0012 00130015 3N 0014 X ∂ ∂H ∂ ∂H ρ (ln ρ + 1) − ρ (ln ρ + 1) dΓ ∂pi ∂qi ∂qi ∂pi i=1 0015 Z 3N 0014 X ∂2H ∂H 1 ∂ρ ∂2H ∂H 1 ∂ρ ρ = dΓ (ln ρ + 1) + ρ −ρ (ln ρ + 1) − ρ ∂p ∂q ∂q ρ ∂p ∂q ∂p ∂pi ρ ∂qi i i i i i i i=1 0015 Z 3N 0014 X ∂H ∂ρ ∂H ∂ρ = dΓ − . ∂qi ∂pi ∂pi ∂qi i=1
dS = dt
Z
Integrating the final expression by parts gives dS =− dt
Z
0015 3N 0014 X ∂ 2H ∂ 2H −ρ +ρ = 0. dΓ ∂pi ∂qi ∂qi ∂pi i=1
(b) Using the method of Lagrange multipliers, find the function ρmax (Γ) which maximizes R the functional S[ρ], subject to the constraint of fixed average energy, hHi = dΓρH = E. • There are two constraints, normalization and constant average energy, written respectively as
Z †
dΓρ(Γ) = 1,
hHi =
and
This is standard integration by parts, i.e.
Ra b
Z
dΓρ(Γ)H = E. a
F dG = F G|b −
Ra b
GdF . Looking
explicitly at one term in the expression to be integrated in this problem, Z Y 3N
∂ρ ∂H = dVi ∂qi ∂pi i=1
Z
dq1 dp1 · · · dqi dpi · · · dq3N dp3N
∂ρ ∂H , ∂qi ∂pi
∂ρ we identify dG = dqi ∂q , and F with the remainder of the expression. Noting that i
ρ(qi ) = 0 at the boundaries of the box, we get Z Y 3N
∂ρ ∂H =− dVi ∂qi ∂pi i=1
55
Z Y 3N
i=1
dVi ρ
∂ ∂H . ∂qi ∂pi
Rewriting the expression for entropy, S(t) =
Z
dΓρ(Γ) [− ln ρ(Γ) − α − βH] + α + βE,
where α and β are Lagrange multipliers used to enforce the two constraints. Extremizing the above expression with respect to the function ρ(Γ), results in ∂S = − ln ρmax (Γ) − α − βH(Γ) − 1 = 0. ∂ρ(Γ) ρ=ρmax
The solution to this equation is
ln ρmax = −(α + 1) − βH, which can be rewritten as ρmax = C exp (−βH) ,
where
C = e−(α+1) .
(c) Show that the solution to part (b) is stationary, i.e. ∂ρmax /∂t = 0. • The density obtained in part (b) is stationary, as can be easily checked from n o ∂ρmax −β H = − {ρmax , H} = − Ce ,H ∂t ∂H ∂H −β H ∂H ∂H −β H = C(−β) e − C(−β) e = 0. ∂p ∂q ∂q ∂p
(d) How can one reconcile the result in (a), with the observed increase in entropy as the system approaches the equilibrium density in (b)? (Hint: Think of the situation encountered in the previous problem.) • Liouville’s equation preserves the information content of the PDF ρ(Γ, t), and hence S(t)
does not increase in time. However, as illustrated in the example in problem 1, the density
becomes more finely dispersed in phase space. In the presence of any coarse-graining of phase space, information disappears. The maximum entropy, corresponding to ρ, ˜ describes equilibrium in this sense. ******** 56
3. The Vlasov equation is obtained in the limit of high particle density n = N/V , or large inter-particle interaction range λ, such that nλ3 ≫ 1. In this limit, the collision terms are
dropped from the left hand side of the equations in the BBGKY hierarchy. The BBGKY hierarchy ' s s X X ∂ ~pn ∂ − + · ∂t n=1 m ∂~qn n=1
! # X ∂V(~qn − ~ql ) ∂ ∂U + · fs ∂~qn ∂~qn ∂~pn l s Z X ∂V(~qn − ~qs+1 ) ∂fs+1 · , = dVs+1 ∂~qn ∂~pn n=1
has the characteristic time scales  ∂U ∂ v 1  ∼ · ∼ ,    τU ∂~q ∂~p L    1 ∂V ∂ v ∼ · ∼ ,  τc ∂~q ∂~p λ   Z     1 ∼ dx ∂V · ∂ fs+1 ∼ 1 · nλ3 , τX ∂~q ∂~p fs τc
where nλ3 is the number of particles within the interaction range λ, and v is a typical velocity. The Boltzmann equation is obtained in the dilute limit, nλ3 ≪ 1, by disregarding terms of order 1/τX ≪ 1/τc . The Vlasov equation is obtained in the dense limit of nλ3 ≫ 1 by ignoring terms of order 1/τc ≪ 1/τX .
(a) Assume that the N body density is a product of one particle densities, i.e. ρ = QN pi , ~qi ). Calculate the densities fs , and their normalizations. i=1 ρ1 (xi , t), where xi ≡ (~ • Let bf xi denote the coordinates and momenta for particle i. Starting from the joint QN probability ρN = i=1 ρ1 (xi , t), for independent particles, we find N! fs = (N − s)!
Z
N Y
α=s+1
The normalizations follow from Z dΓρ = 1, and
Z Y s
n=1
dVα ρN =
dVn fs =
=⇒
Z
dV1 ρ1 (x, t) = 1,
N! ≈ Ns (N − s)! 57
s Y N! ρ1 (xn , t). (N − s)! n=1
for
s ≪ N.
(b) Show that once the collision terms are eliminated, all the equations in the BBGKY hierarchy are equivalent to the single equation 0014 0015 ∂ ~p ∂ ∂Ueff ∂ f1 (~ p, ~q, t) = 0, + · − · ∂t m ∂~q ∂~q ∂~p where Ueff (~q, t) = U (~q ) +
• Noting that
Z
dx′ V(~q − ~q ′ )f1 (x′ , t).
(N − s)! fs+1 = ρ1 (xs+1 ), fs (N − s − 1)!
the reduced BBGKY hierarchy is ' ≈
s Z X
s
X ∂ + ∂t n=1
0012
∂U ∂ p~ n ∂ − · · m ∂~qn ∂~qn ∂~pn
0013#
fs
∂V(~qn − ~qs+1 ) ∂ · [(N − s)fs ρ1 (xs+1 )] ∂~qn ∂~pn n=1 0014Z 0015 s X ∂ ∂ ≈ dVs+1 ρ1 (xs+1 )V(~qn − ~qs+1 ) · N fs , ∂~qn ∂~pn n=1 dVs+1
where we have used the approximation (N − s) ≈ N for N ≫ s. Rewriting the above
expression,
' where
s
X ∂ + ∂t n=1
0012
p~n ∂ ∂Uef f ∂ · − · m ∂~qn ∂~qn ∂~pn
Uef f = U (~q ) + N
Z
0013#
fs = 0,
dV ′ V(~q − ~q ′ )ρ1 (x′ , t).
(c) Now consider N particles confined to a box of volume V , with no additional potential. Show that f1 (~q, p~ ) = g(~ p )/V is a stationary solution to the Vlasov equation for any g(~ p ). Why is there no relaxation towards equilibrium for g(~ p )? • Starting from
ρ1 = g(~ p )/V,
we obtain Hef f =
N 0014 X p~i 2 i=1
2m 58
0015
+ Uef f (~qi ) ,
with Z N 1 d3 qV(~q ). p) = Uef f = 0 + N dV V(~q − ~q ) g(~ V V R 3 (We have taken advantage of the normalization d pg(~ p ) = 1.) Substituting into the Z
′
′
Vlasov equation yields
0012
∂ p~ ∂ + · ∂t m ∂~q
0013
ρ1 = 0.
There is no relaxation towards equilibrium because there are no collisions which allow g(~ p ) to relax. The momentum of each particle is conserved by Hef f ; i.e. {ρ1 , Hef f } = 0, preventing its change. ******** 4. Two component plasma:
Consider a neutral mixture of N ions of charge +e and mass
m+ , and N electrons of charge −e and mass m− , in a volume V = N/n0 . (a) Show that the Vlasov equations for this two component system are 0015 0014 p~ ∂ ∂Φeff ∂ ∂   p, ~q, t) = 0  ∂t + m · ∂~q + e ∂~q · ∂~p f+ (~ + 0015 0014  ~p ∂ ∂Φeff ∂ ∂   f− (~ p, ~q, t) = 0 + · −e · ∂t m− ∂~q ∂~q ∂~p
,
where the effective Coulomb potential is given by Φeff (~q, t) = Φext (~q ) + e
Z
dx′ C(~q − ~q ′ ) [f+ (x′ , t) − f− (x′ , t)] .
Here, Φext is the potential set up by the external charges, and the Coulomb potential C(~ q) satisfies the differential equation ∇2 C = 4πδ 3 (~q ). • The Hamiltonian for the two component mixture is 0015 X N 0014 2N 2N X X p~i 2 1 ~pi 2 H= ei ej ei Φext (~qi ), + + + 2m 2m |~ q − ~ q | + − i j i=1 i,j=1 i=1 where C(~qi − ~qj ) = 1/|~qi − ~qj |, resulting in X ∂Φext ∂ ∂H = ei + ei ej C(~qi − ~qj ). ∂~qi ∂~qi ∂~qi j6=i
59
Substituting this into the Vlasov equation, we obtain 0015 0014 p ~ ∂ ∂Φ ∂ ∂ ef f   f+ (~ p, ~q, t) = 0, + · +e ·  ∂t m+ ∂~q ∂~q ∂~p 0014 0015  ∂ p~ ∂ ∂Φef f ∂   f− (~ p, ~q, t) = 0. + · −e · ∂t m− ∂~q ∂~q ∂~p (b) Assume that the one particle densities have the stationary forms f± = g± (~ p )n± (~ q ). Show that the effective potential satisfies the equation ∇2 Φeff = 4πρext + 4πe (n+ (~q ) − n− (~q )) , where ρext is the external charge density. • Setting f± (~ p, ~q ) = g± (~ p )n± (~q ), and using potential simplify to
Φef f (~q, t) = Φext (~q ) + e
Z
R
d3 pg± (~ p ) = 1, the integrals in the effective
d3 q ′ C(~q − ~q ′ ) [n+ (~q ′ ) − n− (~q ′ )] .
Apply ∇2 to the above equation, and use ∇2 Φext = 4πρext and ∇2 C(~q−~q ′ ) = 4πδ 3 (~q−~ q ′ ), to obtain
∇2 φef f = 4πρext + 4πe [n+ (~q ) − n− (~q )] . (c) Further assuming that the densities relax to the equilibrium Boltzmann weights n± (~q ) = n0 exp [±βeΦeff (~q)], leads to the self-consistency condition 0002 00010003 ∇2 Φeff = 4π ρext + n0 e eβeΦeff − e−βeΦeff ,
known as the Poisson–Boltzmann equation. Due to its nonlinear form, it is generally not possible to solve the Poisson–Boltzmann equation. By linearizing the exponentials, one obtains the simpler Debye equation ∇2 Φeff = 4πρext + Φeff /λ2 . Give the expression for the Debye screening length λ. • Linearizing the Boltzmann weights gives n± = no exp[∓βeΦef f (~q )] ≈ no [1 ∓ βeΦef f ] , 60
resulting in ∇2 Φef f = 4πρext +
1 Φef f , λ2
with the screening length given by λ2 =
kB T . 8πno e2
(d) Show that the Debye equation has the general solution Z Φeff (~q ) = d3 ~q′ G(~q − ~q ′ )ρext (~q ′ ), where G(~q ) = exp(−|~q |/λ)/|~q | is the screened Coulomb potential.
• We want to show that the Debye equation has the general solution Z Φef f (~q ) = d3 ~qG(~q − ~q ′ )ρext (~q ′ ), where G(~q ) =
exp(−|q|/λ) . |q|
Effectively, we want to show that ∇2 G = G/λ2 for ~q 6= 0. In spherical coordinates,
G = exp(−r/λ)/r. Evaluating ∇2 in spherical coordinates gives
0014 0012 0013 0015 1 e−r/λ 1 ∂ 2 1 ∂ e−r/λ 2 ∂G r = 2 r − ∇ G= 2 − r ∂r ∂r r ∂r λ r r2 0015 0014 1 1 −r/λ 1 −r/λ 1 −r/λ 1 e−r/λ G = 2 = re − e + e = 2. 2 2 r λ λ λ λ r λ 2
(e) Give the condition for the self-consistency of the Vlasov approximation, and interpret it in terms of the inter-particle spacing? • The Vlasov equation assumes the limit no λ3 ≫ 1, which requires that (kB T )3/2 1/2
no e3
≫ 1,
where ℓ is the interparticle spacing. consistency condition is
=⇒
e2 ≪ no−1/3 ∼ ℓ, kB T
In terms of the interparticle spacing, the self– e2 ≪ kB T, ℓ 61
i.e. the interaction energy is much less than the kinetic (thermal) energy. (f) Show that the characteristic relaxation time (τ ≈ λ/c) is temperature independent.
What property of the plasma is it related to?
• A characteristic time is obtained from r r r λ 1 kB T m m τ∼ ∼ · ∼ ∼ , 2 2 c no e kB T no e ωp where ωp is the plasma frequency. ******** 5. Two dimensional electron gas in a magnetic field: When donor atoms (such as P or As) are added to a semiconductor (e.g. Si or Ge), their conduction electrons can be thermally excited to move freely in the host lattice. By growing layers of different materials, it is possible to generate a spatially varying potential (work–function) which traps electrons at the boundaries between layers. In the following, we shall treat the trapped electrons as a gas of classical particles in two dimensions. If the layer of electrons is sufficiently separated from the donors, the main source of scattering is from electron–electron collisions. (a) The Hamiltonian for non–interacting free electrons in a magnetic field has the form 0010
~ X  ~pi − eA H=  2m i
00112

~  ± µB |B| .
(The two signs correspond to electron spins parallel or anti-parallel to the field.) The ~=B ~ × ~q/2 describes a uniform magnetic field B. ~ Obtain the classical vector potential A
equations of motion, and show that they describe rotation of electrons in cyclotron orbits ~ in a plane orthogonal to B. • The Hamiltonian for non–interacting free electrons in a magnetic field has the form 0010
~ X  ~pi + eA H=  2m i or in expanded form H=
00112

~  ± µB |B| ,
2 e p2 ~ 2 ± µB |B|. ~ ~+ e A + p~ · A 2m m 2m
62
~=B ~ × ~q/2, results in Substituting A e p2 ~ × ~q + + ~p · B 2m 2m p2 e ~ · ~q + = + ~ ×B p 2m 2m
H=
00112 e2 0010 ~ ~ B × ~q ± µB |B| 8m 0011 e2 0010 2 2 ~ · ~q )2 ± µB |B|. ~ B q − (B 8m
Using the canonical equations, ~q˙ = ∂H/~ p and ~p˙ = −∂H/~q , we find  ∂H e~ ~p e ~  ˙  × ~q , =⇒ p~ = m~q˙ − B × ~q ,  ~q = ∂~p = m + 2m B 2 0011 2 2 0010  e ∂H  ~ · ~q B. ~ ~ − e B 2 ~q + e  ~p˙ = − B =− p~ × B ∂~q 2m 4m 4m
Differentiating the first expression obtained for p~, and setting it equal to the second expression for p~˙ above, gives 0011 0011 e~ ˙ e 0010 ˙ e~ e2 ~ 2 e2 0010 ~ ¨ ~ ~ m~q − B × ~q = − m~q − B × ~q × B − B · ~q B. |B| ~q + 2 2m 2 4m 4m 0010 0011 ~ × ~q × B ~ = B 2 ~q − B ~ · ~q B, ~ leads to Simplifying the above expression, using B ~ × ~q˙ . m¨~q = eB This describes the rotation of electrons in cyclotron orbits, ¨~q = ω ~ c × ~q˙ , ~ ~ where ~ωc = eB/m; i.e. rotations are in the plane perpendicular to B. (b) Write down heuristically (i.e. not through a step by step derivation), the Boltzmann equations for the densities f↑ (~ p, ~q, t) and f↓ (~ p, ~q, t) of electrons with up and down spins, in terms of the two cross-sections σ ≡ σ↑↑ = σ↓↓ , and σ× ≡ σ↑↓ , of spin conserving collisions.
• Consider the classes of collisions described by cross-sections σ ≡ σ↑↑ = σ↓↓ , and σ× ≡ σ↑↓ . We can write the Boltzmann equations for the densities as ∂f↑ − {H↑ , f↑ } = ∂t
Z
2
d p2 dΩ|v1 − v2 |
001a
dσ [f↑ (p~1 ′ )f↑ (p~2 ′ ) − f↑ (p~1 )f↑ (p~2 )] + dΩ
001b dσ× ′ ′ [f↑ (p~1 )f↓ (p~2 ) − f↑ (p~1 )f↓ (p~2 )] , dΩ 63
and ∂f↓ − {H↓ , f↓ } = ∂t
001a
dσ [f↓ (p~1 ′ )f↓ (p~2 ′ ) − f↓ (p~1 )f↓ (p~2 )] + dΩ 001b dσ× ′ ′ [f↓ (p~1 )f↑ (p~2 ) − f↓ (p~1 )f↑ (p~2 )] . dΩ
Z
2
d p2 dΩ|v1 − v2 |
(c) Show that dH/dt ≤ 0, where H = H↑ + H↓ is the sum of the corresponding H functions.
• The usual Boltzmann H–Theorem states that dH/dt ≤ 0, where Z H = d2 qd2 pf (~q, ~p, t) ln f (~q, ~p, t).
For the electron gas in a magnetic field, the H function can be generalized to Z H = d2 qd2 p [f↑ ln f↑ + f↓ ln f↓ ] , where the condition dH/dt ≤ 0 is proved as follows: 0014 0015 Z ∂f↑ ∂f↓ dH 2 2 = d qd p (ln f↑ + 1) + (ln f↓ + 1) dt ∂t ∂t Z = d2 qd2 p [(ln f↑ + 1) ({f↑ , H↑ } + C↑↑ + C↑↓ ) + (ln f↓ + 1) ({f↓ , H↓ } + C↓↓ + C↓↑ )] , with C↑↑ , etc., defined via the right hand side of the equations in part (b). Hence Z dH = d2 qd2 p (ln f↑ + 1) (C↑↑ + C↑↓ ) + (ln f↓ + 1) (C↓↓ + C↓↑ ) dt Z = d2 qd2 p (ln f↑ + 1) C↑↑ + (ln f↓ + 1) C↓↓ + (ln f↑ + 1) C↑↓ + (ln f↓ + 1) C↓↑ ≡
d dH↑↑ dH↓↓ + + (H↑↓ + H↓↑ ) , dt dt dt
where the H′ s are in correspondence to the integrals for the collisions. We have also made R R use of the fact that d2 pd2 q {f↑ , H↑ } = d2 pd2 q {f↓ , H↓ } = 0. Dealing with each of the
terms in the final equation individually, Z dσ dH↑↑ = d2 qd2 p1 d2 p2 dΩ|v1 − v2 | (ln f↑ + 1) [f↑ (~ p1 ′ )f↑ (~ p2 ′ ) − f↑ (~ p1 )f↑ (~ p2 )] . dt dΩ After symmetrizing this equation, as done in the text, Z dσ 1 dH↑↑ d2 qd2 p1 d2 p2 dΩ|v1 − v2 | =− [ln f↑ (~ p1 )f↑ (~ p2 ) − ln f↑ (~ p1 ′ )f↑ (~ p2 ′ )] dt 4 dΩ 64
· [f↑ (~ p1 )f↑ (~ p2 ) − f↑ (~ p1 ′ )f↑ (~ p2 ′ )] ≤ 0. Similarly, dH↓↓ /dt ≤ 0. Dealing with the two remaining terms, dH↑↓ = dt
dσ× [f↑ (~ p1 ′ )f↓ (~ p2 ′ ) − f↑ (~ p1 )f↓ (~ p2 )] dΩ Z dσ× [f↑ (~ p1 )f↓ (~ p2 ) − f↑ (~ p1 ′ )f↓ (~ p2 ′ )] , = d2 qd2 p1 d2 p2 dΩ|v1 − v2 | [ln f↑ (~ p1 ′ ) + 1] dΩ Z
d2 qd2 p1 d2 p2 dΩ|v1 − v2 | [ln f↑ (~ p1 ) + 1]
where we have exchanged (~ p1 , ~p2 ↔ ~p1 ′ , ~p2 ′ ). Averaging these two expressions together, dH↑↓ 1 =− dt 2
Z
d2 qd2 p1 d2 p2 dΩ|v1 − v2 |
dH↓↑ 1 =− dt 2
Z
d2 qd2 p1 d2 p2 dΩ|v1 − v2 |
dσ× [ln f↑ (~ p1 ) − ln f↑ (~ p1 ′ )] dΩ · [f↑ (~ p1 )f↓ (~ p2 ) − f↑ (~ p1 ′ )f↓ (~ p2 ′ )] .
Likewise dσ× [ln f↓ (~ p2 ) − ln f↓ (~ p2 ′ )] dΩ · [f↓ (~ p2 )f↑ (~ p1 ) − f↓ (~ p2 ′ )f↑ (~ p1 ′ )] .
Combining these two expressions, Z dσ× 1 d d2 qd2 p1 d2 p2 dΩ|v1 − v2 | (H↑↓ + H↓↑ ) = − dt 4 dΩ ′ ′ [ln f↑ (~ p1 )f↓ (~ p2 ) − ln f↑ (~ p1 )f↓ (~ p2 )] [f↑ (~ p1 )f↓ (~ p2 ) − f↑ (~ p1 ′ )f↓ (~ p2 ′ )] ≤ 0. Since each contribution is separately negative, we have dH dH↑↑ dH↓↓ d = + + (H↑↓ + H↓↑ ) ≤ 0. dt dt dt dt
(d) Show that dH/dt = 0 for any ln f which is, at each location, a linear combination of quantities conserved in the collisions. • For dH/dt = 0 we need each of the three square brackets in the previous derivation to be zero. The first two contributions, from dH↑↓ /dt and dH↓↓ /dt, are similar to those discussed
in the notes for a single particle, and vanish for any ln f which is a linear combination of quantities conserved in collisions ln fα =
X
aα q )χi (~ p), i (~
i
65
where α = (↑ or ↓) . Clearly at each location ~q, for such fα , ln fα (~ p1 ) + ln fα (~ p2 ) = ln fα (~ p1 ′ ) + ln fα (~ p2 ′ ). If we consider only the first two terms of dH/dt = 0, the coefficients aα q) can vary i (~ with both ~q and α = (↑ or ↓).
This changes when we consider the third term
d (H↑↓ + H↓↑ ) /dt. The conservations of momentum and kinetic energy constrain the corresponding four functions to be the same, i.e. they require a↑i (~q) = a↓i (~q). There is, however, no similar constraint for the overall constant that comes from particle number conservation, as the numbers of spin-up and spin-down particles is separately conserved, i.e. a↑0 (~q) = a↓0 (~q). This implies that the densities of up and down spins can be different in the final equilibrium, while the two systems must share the same velocity and temperature. (e) Show that the streaming terms in the Boltzmann equation are zero for any function that depends only on the quantities conserved by the one body Hamiltonians. • The Boltzmann equation is ∂fα = − {fα , Hα } + Cαα + Cαβ , ∂t where the right hand side consists of streaming terms {fα , Hα }, and collision terms C.
Let Ii denote any quantity conserved by the one body Hamiltonian, i.e. {Ii , Hα } = 0.
Consider fα which is a function only of the Ii′ s
fα ≡ fα (I1 , I2 , · · ·) . Then {fα , Hα } =
X ∂fα j
∂Ij
{Ij , Hα } = 0.
~ = ~q × p~, is conserved during, and away from collisions. (f) Show that angular momentum L • Conservation of momentum for a collision at ~q
(~ p1 + p~2 ) = (~ p1 ′ + p~2 ′ ), implies ~q × (~ p1 + p~2 ) = ~q × (~ p1 ′ + p~2 ′ ), or ~1 + L ~2 = L ~1 ′ + L ~ 2 ′, L 66
~ i = ~q × p~i . Hence angular momentum L ~ is conserved during colliwhere we have used L
sions. Note that only the z–component Lz is present for electrons moving in 2–dimensions,
~q ≡ (x1 , x2 ), as is the case for the electron gas studied in this problem. Consider the Hamiltonian discussed in (a)
0011 e e2 0010 2 2 p2 2 ~ ~ ~ B q − (B · ~q ) ± µB |B|. + p~ × B · ~q + H= 2m 2m 8m Let us evaluate the Poisson brackets of the individual terms with Lz = ~q × p~ |z . The first
term is
2 ∂ |~ p | , ~q × ~p = εijk {pl pl , xj pk } = εijk 2pl (xj pk ) = 2εilk pl pk = 0, ∂xl
where we have used εijk pj pk = 0 since pj pk = pk pj is symmetric. The second term is proportional to Lz , n
~ · ~q, Lz p~ × B
o
= {Bz Lz , Lz } = 0. The final terms are proportional to q 2 , and q 2 , ~q × p~ = 0 for the same reason that 2 p , ~q × p~ = 0, leading to {H, ~q × p~ } = 0.
Hence angular momentum is conserved away from collisions as well. (g) Write down the most general form for the equilibrium distribution functions for particles confined to a circularly symmetric potential. • The most general form of the equilibrium distribution functions must set both the
collision terms, and the streaming terms to zero. Based on the results of the previous parts, we thus obtain fα = Aα exp [−βHα − γLz ] . The collision terms allow for the possibility of a term −~u · p~ in the exponent, corresponding
to an average velocity. Such a term will not commute with the potential set up by a stationary box, and is thus ruled out by the streaming terms. On the other hand, the angular momentum does commute with a circular potential {V (~q), L} = 0, and is allowed by the streaming terms. A non–zero γ describes the electron gas rotating in a circular box. (h) How is the result in part (g) modified by including scattering from magnetic and non-magnetic impurities? 67
~ in collisions. • Scattering from any impurity removes the conservation of p~, and hence L, The γ term will no longer be needed. Scattering from magnetic impurities mixes populations of up and down spins, necessitating A↑ = A↓ ; non–magnetic impurities do not have this effect. (i) Do conservation of spin and angular momentum lead to new hydrodynamic equations? • Conservation of angular momentum is related to conservation of p~, as shown in (f), and hence does not lead to any new equation. In contrast, conservation of spin leads to an additional hydrodynamic equation involving the magnetization, which is proportional to (n↑ − n↓ ). ********
6. The Lorentz gas describes non-interacting particles colliding with a fixed set of scatterers. It is a good model for scattering of electrons from donor impurities. Consider a uniform two dimensional density n0 of fixed impurities, which are hard circles of radius a. (a) Show that the differential cross section of a hard circle scattering through an angle θ is dσ =
θ a sin dθ, 2 2
and calculate the total cross section. • Let b denote the impact parameter, which (see figure) is related to the angle θ between p~ ′ and ~p by
68
π−θ θ = a cos . 2 2 The differential cross section is then given by b(θ) = a sin
θ dσ = 2|db| = a sin dθ. 2 Hence the total cross section σtot =
Z
π 0
0015π 0014 θ θ = 2a. dθa sin = 2a − cos 2 2 0
(b) Write down the Boltzmann equation for the one particle density f (~q, ~p, t) of the Lorentz gas (including only collisions with the fixed impurities). (Ignore the electron spin.) • The corresponding Boltzmann equation is ∂f p~ ∂f ~ · ∂f + · +F ∂t m ∂~q ∂~p Z Z dσ p| no |~ p| dσ |~ ′ dθ n0 [−f (~ p ) ) + f (~ p )] = [f (~ p ′ ) − f (~ p )] ≡ C [f (~ p )] . = dθ dθ m m dθ (c) Find the eigenfunctions and eigenvalues of the collision operator. ~ ≡ −∂U/∂~q, and (d) Using the definitions F Z n(~q, t) = d2 ~pf (~q, ~p, t), and hg(~q, t)i =
1 n(~q, t)
Z
d2 p~f (~q, ~p, t)g(~q, t),
show that for any function χ(|~ p|), we have 0012 001c 001d0013 0012 001c 001d0013 ∂ ∂ p~ ∂χ ~ . (n hχi) + · n χ =F · n ∂t ∂~q m ∂~p • Using the definitions F~ ≡ −∂U/∂~q, Z n(~q, t) = d2 ~pf (~q, ~p, t), and
1 hg(~q, t)i = n(~q, t)
Z
d2 p~f (~q, ~p, t)g(~q, t),
we can write 0014 0015 Z p~ ∂f ∂f dσ |~ p | ′ ~· d pχ(|~ p |) − · −F + dθ no (f (~ p) − f (~ p )) m ∂~q ∂~p dθ m 0012 001c 001d0013 0012 001c 001d0013 ∂ p~ ∂χ . =− · n χ + F~ · n ∂~q m ∂~p
d (n hχ(|~ p |)i) = dt
Z
2
69
Rewriting this final expression gives the hydrodynamic equation 0012 001c 001d0013 0012 001c 001d0013 ∂ p~ ∂χ ∂ ~ . (n hχi) + · n χ =F · n ∂t ∂~q m ∂~p (e) Derive the conservation equation for local density ρ ≡ mn(~q, t), in terms of the local velocity ~u ≡ h~p/mi.
• Using χ = 1 in the above expression 0012 001c 001d0013 ∂ ∂ p~ = 0. n+ · n ∂t ∂~q m In terms of the local density ρ = mn, and velocity ~u ≡ h~ p/mi, we have ∂ ∂ ρ+ · (ρ~u ) = 0. ∂t ∂~q
(f) Since the magnitude of particle momentum is unchanged by impurity scattering, the Lorentz gas has an infinity of conserved quantities |~ p|m . This unrealistic feature is removed
upon inclusion of particle–particle collisions. For the rest of this problem focus only on p2 /2m as a conserved quantity. Derive the conservation equation for the energy density p~ ρ 2 c , where ~c ≡ − ~u, 2 m
in terms of the energy flux ~h ≡ ρ ~c c2 /2, and the pressure tensor Pαβ ≡ ρ hcα cβ i. ǫ(~q, t) ≡
• With the kinetic energy χ = p2 /2m as a conserved quantity, the equation found in (c) gives
001d0013 0012 0013 0012 001c ∂ 0010 n 2 0011 ∂ h~ p i n p~ p2 ~· n + =F . |~ p| · ∂t 2m ∂~q 2 mm m
Substituting ~p/m = ~u + ~c , where h~c i = 0, and using ρ = nm,
000bi ∂ hρ
ρ ∂ h ρ 2 ρ 2 i 2 2 + c (~u + ~c )(u + c + 2~u · ~c ) = F~ · ~u. u + · ∂t 2 2 ∂~q 2 m
From the definition ε = ρ c2 /2, we have
i
0001i ∂ hρ 2 ρ ∂ hρ ~uu2 + ~u c2 + ~cc2 + 2~u · h~c ~c i = F~ · ~u. u +ε + · ∂t 2 ∂~q 2 m 70
Finally, by substituting ~h ≡ ρ ~c c2 /2 and Pαβ = ρ hcα cβ i , we get
i 0011 i ∂ hρ 2 ∂ h 0010ρ 2 ∂ ρ u +ε + · ~u u + ε + ~h + (uβ Pαβ ) = F~ · ~u. ∂t 2 ∂~q 2 ∂qα m
(g) Starting with a one particle density 0015 1 p2 , f (~ p, ~q, t) = n(~q, t) exp − 2mkB T (~q, t) 2πmkB T (~q, t) 0014
0
reflecting local equilibrium conditions, calculate ~u, ~h, and Pαβ . Hence obtain the zeroth order hydrodynamic equations. • There are only two quantities, 1 and p2 /2m, conserved in collisions. Let us start with the one particle density
0015 p2 1 f (~ p, ~q, t) = n(~q, t) exp − . 2mkB T (~q, t) 2πmkB T (~q, t) 0014
0
Then ~u =
001c
p~ m
001d
= 0,
and
0
~h =
001c
p~ p2 mm
001d
0
ρ = 0, 2
o
since both are odd functions of p~, while f is an even function of p~, while Pαβ = ρ hcα cβ i =
n n hpα pβ i = δαβ · mkB T = nkB T δαβ . m m
Substituting these expressions into the results for (c) and (d), we obtain the zeroth–order hydrodynamic equations  ∂ρ   = 0, ∂t   ∂ ε = ∂ ρ c2 = 0. ∂t ∂t 2
The above equations imply that ρ and ε are independent of time, i.e. ρ = n(~q ), or
and
ε = kB T (~q ),
0015 0014 p2 n(~q ) . exp − f = 2πmkB T (~q ) 2mkB T (~q ) 0
71
(h) Show that in the single collision time approximation to the collision term in the Bolzmann equation, the first order solution is ' !# 2 ~ p~ p ∂ln ρ ∂ln T ∂T F f 1 (~ p, ~q, t) = f 0 (~ p, ~q, t) 1 − τ · . − + − 2 m ∂~q ∂~q 2mkB T ∂~q kB T • The single collision time approximation is f0 − f . C [f ] = τ The first order solution to Boltzmann equation f = f 0 (1 + g) , is obtained from
as
Noting that
0002 0003 f 0g L f0 = − , τ
0015 0014 ∂ p~ ∂ ∂ 1 0002 00003 0 0 0 ~ ln f + · ln f + F · ln f . g = −τ 0 L f = −τ f ∂t m ∂~q ∂~p ln f 0 = −
p2 + ln n − ln T − ln(2πmkB ), 2mkB T
where n and T are independent of t, we have ∂ ln f 0 /∂t = 0, and 0013 0015001b 0014 001a 0012 2 p ~ ∂n 1 ∂T p 1 ∂T −~ p + · − + g = −τ F~ · mkB T m n ∂~q T ∂~q 2mkB T 2 ∂~q !) ( ~ F 1 ∂T p2 1 ∂ρ ∂T p~ . · − + − = −τ m ρ ∂~q T ∂~q 2mkB T 2 ∂~q kB T (i) Show that using the first order expression for f , we obtain h i ρ~u = nτ F~ − kB T ∇ ln (ρT ) . R d2 qf 0 (1 + g) = d2 qf 0 = n, and Dp E 1 Z pα α = d2 p f 0 (1 + g) uα = m n m 0014 0012 00130015 Z pβ ∂ ln ρ ∂ ln T 1 Fβ p2 ∂T 2 pα d p −τ f 0. = − − + 2 n m m ∂qβ ∂qβ kB T 2mkB T ∂qβ
• Clearly
R
72
Wick’s theorem can be used to check that
resulting in
hpα pβ i0 = δαβ mkB T,
2 p pα pβ 0 = (mkB T )2 [2δαβ + 2δαβ ] = 4δαβ (mkB T )2 , 0014 0012 0015 0013 0010ρ0011 ∂ 1 ∂T nτ δαβ kB T ln − δαβ . Fβ + 2kB uα = − ρ ∂qβ T kB T ∂qβ
Rearranging these terms yields 0015 ∂ ln (ρT ) . ρuα = nτ Fα − kB T ∂qα 0014
(j) From the above equation, calculate the velocity response function χαβ = ∂uα /∂Fβ . • The velocity response function is now calculated easily as χαβ =
∂uα nτ = δαβ . ∂Fβ ρ
(k) Calculate Pαβ , and ~h, and hence write down the first order hydrodynamic equations. • The first order expressions for pressure tensor and heat flux are ρ hpα pβ i = δαβ nkB T, and δ 1 Pαβ = 0, m2 0001E τρ D ρ
2 pi 2 2 ai + bi p pα p = − 3 pα p . hα = 2m3 2m m 0
Pαβ =
The latter is calculated from Wick’s theorem results
2 pi pα p2 = 4δαi (mkB T ) , and
3 pi pα p2 p2 = (mkB T ) [δαi (4 + 4) + 4 × 2δαi + 4 × 2δαi ] = 24δαi , as
0012 0014 0013 0015 ∂ ρ 24(mkB T )3 ∂ ρτ 2 ~ ln − F + ln T hα = − 3 (mkB T ) 2m ∂qα T 2mkb T ∂qα ∂T 2 = −12nkB Tτ . ∂qα
Substitute these expressions for Pαβ and hα into the equation obtained in (e) 0015 0014 0010 i 0011 ∂ hρ 2 ∂ ∂ ρ 2 ∂T ρ 2 + u +ǫ + · ~u u + ǫ − 11nkB T τ (~u nkB T ) = F~ · ~u. ∂t 2 ∂~q 2 ∂~q ∂~q m 73
******** 7. Thermal conductivity:
Consider a classical gas between two plates separated by a
distance w. One plate at y = 0 is maintained at a temperature T1 , while the other plate at y = w is at a different temperature T2 . The gas velocity is zero, so that the initial zeroth order approximation to the one particle density is, f10 (~ p, x, y, z)
=
n(y) 3/2
[2πmkB T (y)]
0015 p~ · p~ . exp − 2mkB T (y) 0014
(a) What is the necessary relation between n(y) and T (y) to ensure that the gas velocity ~u remains zero? (Use this relation between n(y) and T (y) in the remainder of this problem.) • Since there is no external force acting on the gas between plates, the gas can only flow
locally if there are variations in pressure. Since the local pressure is P (y) = n(y)kB T (y), the condition for the fluid to be stationary is n(y)T (y) = constant.
(b) Using Wick’s theorem, or otherwise, show that
2 0 0 p ≡ hpα pα i = 3 (mkB T ) ,
and
4 0 0 2 p ≡ hpα pα pβ pβ i = 15 (mkB T ) ,
0 0 where hOi indicates local averages with the Gaussian weight f10 . Use the result p6 =
105(mkB T )3 in conjunction with symmetry arguments to conclude
2 4 0 3 py p = 35 (mkB T ) . 0
• The Gaussian weight has a covariance hpα pβ i = δαβ (mkB T ). Using Wick’s theorem
gives
Similarly
2 0 0 p = hpα pα i = (mkB T ) δαα = 3 (mkB T ) .
4 0 0 2 2 p = hpα pα pβ pβ i = (mkB T ) (δαα + 2δαβ δαβ ) = 15 (mkB T ) . 74
The symmetry along the three directions implies
2 4 0 2 4 0 2 4 0 1 2 4 0 1 3 3 px p = py p = pz p = p p = × 105 (mkB T ) = 35 (mkB T ) . 3 3 (c) The zeroth order approximation does not lead to relaxation of temperature/density variations related as in part (a). Find a better (time independent) approximation f11 (~ p, y), by linearizing the Boltzmann equation in the single collision time approximation, to 0002 0003 L f11 ≈
0014
0015 ∂ f 1 − f10 py ∂ f10 ≈ − 1 , + ∂t m ∂y τK
where τK is of the order of the mean time between collisions. • Since there are only variations in y, we have 0014
0015 0014 0015 ∂ py ∂ 3 p2 3 0 0 py 0 0 py f = f1 ∂y ln f1 = f1 ∂y ln n − ln T − + − ln (2πmkB ) ∂t m ∂y 1 m m 2 2mkB T 2 0014 0015 0014 0015 2 5 ∂y T 3 ∂y T p p2 ∂T 0 py 0 py ∂y n = f1 − + − + , = f1 m n 2 T 2mkB T T m 2 2mkB T T
where in the last equality we have used nT = constant to get ∂y n/n = −∂y T /T . Hence
the first order result is
f11 (~ p, y)
=
f10 (~ p, y)
0014 0012 0013 0015 py p2 5 ∂y T 1 − τK . − m 2mkB T 2 T
(d) Use f11 , along with the averages obtained in part (b), to calculate hy , the y component of the heat transfer vector, and hence find K, the coefficient of thermal conductivity. • Since the velocity ~u is zero, the heat transfer vector is 001d1 001c n 2 1 mc2 = py p . hy = n cy 2 2m2 In the zeroth order Gaussian weight all odd moments of p have zero average. The corrections in f11 , however, give a non-zero heat transfer n ∂y T hy = −τK 2m2 T
001c
py m
0012
75
p2 5 − 2mkB T 2
0013
2
py p
001d0
.
000b0
000b0 Note that we need the Gaussian averages of p2y p4 and p2y p2 . From the results of part (b), these averages are equal to 35(mkB T )3 and 5(mkB T )2 , respectively. Hence n ∂y T 2 hy = −τK (mkB T ) 3 2m T
0012
35 5 × 5 − 2 2
0013
=−
2 5 nτK kB T ∂y T. 2 m
The coefficient of thermal conductivity relates the heat transferred to the temperature gradient by ~h = −K∇T , and hence we can identify K=
2 5 nτK kB T . 2 m
(e) What is the temperature profile, T (y), of the gas in steady state? • Since ∂t T is proportional to −∂y hy , there will be no time variation if hy is a constant.
But hy = −K∂y T , where K, which is proportional to the product nT , is a constant in
the situation under investigation. Hence ∂y T must be constant, and T (y) varies linearly between the two plates. Subject to the boundary conditions of T (0) = T1 , and T (w) = T2 , this gives T (y) = T1 +
T2 − T1 y. w
******** 8. Zeroth-order hydrodynamics: The hydrodynamic equations resulting from the conservation of particle number, momentum, and energy in collisions are (in a uniform box):  ∂t n + ∂α (nuα ) = 0      1 ∂β Pαβ ∂ t uα + uβ ∂ β uα = − , mn      ∂t ε + uα ∂α ε = − 1 ∂α hα − 1 Pαβ uαβ n n
where n is the local density, ~u = h~p/mi, uαβ = (∂α uβ + ∂β uα ) /2, and ε = mc2 /2 , with
~c = p~/m − ~u.
(a) For the zeroth order density f10 (~ p, ~q, t) =
n(~q, t) 3/2
(2πmkB T (~q, t))
'
2
(~ p − m~u(~q, t)) exp − 2mkB T (~q, t)
#
,
0 0 0 calculate the pressure tensor Pαβ = mn hcα cβ i , and the heat flux h0α = nm cα c2 /2 . 76
0001 • The PDF for ~c is proportional to the Gaussian exp −mc2 /(2kB T ) , from which we
immediately get
0
hcα cβ i =
kB T δαβ m
=⇒
0 Pαβ = nkB T δαβ ,
and ε =
3 kB T. 2
All odd expectation values of the symmetric weight are zero, specifically h~ci = 0, and D E ~h0 = 0.
(b) Obtain the zeroth order hydrodynamic equations governing the evolution of n(~ q, t),
~u(~q, t), and T (~q, t). •
Substituting
0 Pαβ
equations gives:
D E = nkB T δαβ , ε = 3kB T /2, and ~h0 = 0 in the hydrodynamic  ∂t n + uα ∂α n = −n∂α uα      kB ∂α (nT ) ∂ t uα + uβ ∂ β uα = − mn      3 (∂t T + uα ∂α T ) = −T ∂α uα 2
.
0001 (c) Show that the above equations imply Dt ln nT −3/2 = 0, where Dt = ∂t + uβ ∂β is the material derivative along streamlines.
• Using Dt = ∂t + uβ ∂β , the above equations can be written as  Dt ln n = −∂α uα      kB Dt uα = − ∂α (nT ) mn      3 Dt ln T = −∂α uα 2
.
Eliminating ∂α uα between the first and third equations gives the required result of 0001 Dt ln nT −3/2 = 0. R (d) Write down the expression for the function H0 (t) = d3 ~qd3 p~f10 (~ p, ~q, t) ln f10 (~ p, ~q, t),
after performing the integrations over p~, in terms of n(~q, t), ~u(~q, t), and T (~q, t). • Using the expression for f10 , H0 (t) =
Z
d3 ~qd3 p~
n 3/2
'
exp −
(~ p − m~u) 2mkB T
2
#
(2πmkB T ) ' #. 2 0010 0011 3 (~ p − m~ u ) × ln nT −3/2 − ln (2πmkB ) − 2 2mkB T 77
The Gaussian averages over ~p are easily performed to yield 0014 0010 0015 Z 0011 3 3 0 3 −3/2 − ln (2πmkB ) − H (t) = d ~q n ln nT . 2 2 (e) Using the hydrodynamic equations in (b) calculate dH0 /dt. • Taking the time derivative inside the integral gives Z h 0010 0011 0010 0011i dH0 3 −3/2 −3/2 = d ~q ∂t n ln nT + n∂t ln nT . dT 0001 Use the results of parts (b) and (c) to substitute for ∂t n and ∂t ln nT −3/2 , to get dH0 =− dT
=−
Z
Z
0011i 0011 0010 h 0010 −3/2 −3/2 ∂α (nuα ) + nuα ∂α ln nT d ~q ln nT 3
h 0010 0011i d3 ~q ∂α nuα ln nT −3/2 = 0,
since the integral of a complete derivative is zero.
(f) Discuss the implications of the result in (e) for approach to equilibrium. • The expression for −H0 is related to the entropy of the gas. The result in (f) implies
that the entropy of the gas does not change if its n ~u, and T vary according to the zeroth order equations. The corrections due to first order hydrodynamics are necessary in order to describe the increase in entropy. ******** 9. Viscosity:
Consider a classical gas between two plates separated by a distance w.
One plate at y = 0 is stationary, while the other at y = w moves with a constant velocity vx = u. A zeroth order approximation to the one particle density is, 0014 0015 0001 1 n 0 2 2 2 exp − f1 (~ p, ~q) = (px − mαy) + py + pz , 3/2 2mkB T (2πmkB T )
obtained from the uniform Maxwell–Boltzmann distribution by substituting the average value of the velocity at each point. (α = u/w is the velocity gradient.) (a) The above approximation does not satisfy the Boltzmann equation as the collision term vanishes, while df10 /dt 6= 0. Find a better approximation, f11 (~ p), by linearizing the Boltzmann equation, in the single collision time approximation, to 0015 0014 0002 10003 f 1 − f10 p~ ∂ ∂ f10 ≈ − 1 , + · L f1 ≈ ∂t m ∂~q τ× 78
where τ× is a characteristic mean time between collisions. • We have
0012
p~ ∂ ∂ + · ∂t m ∂~q
whence f11
=
f10
0013
f10 =
001a 1 − τx
α py (px − mαy)f10 , mkB T
001b α py (px − mαy) . mkB T
(b) Calculate the net transfer Πxy of the x component of the momentum, of particles passing through a plane at y, per unit area and in unit time. • The transfer of x-momentum in the y direction, across a plane at y, per unit area and
per unit time, is calculated as Z Z (−py ) 3 py 1 Πxy = d p px f1 (y) − d3 p px f11 (y) m m p >0 py 0, i.e. f (px , x > 0) = A (px ) e−ax/px + feq. (px ) . The constant A (px ) can be determined by matching to solution for x < 0 at x = 0, and is related to the incoming flux. The penetration depth d is the inverse of the decay parameter, and given by d=
px , a
with
a = σst c (n0 − n1 ) > 0. 81
******** 11. Equilibrium density: Consider a gas of N particles of mass m, in an external potential U (~q ). Assume that the one body density ρ1 (~p, ~q, t), satisfies the Boltzmann equation. For a stationary solution, ∂ρ1 /∂t = 0, it is sufficient from Liouville’s theorem for ρ1 to satisfy 0002 00010003 ρ1 ∝ exp −β p2 /2m + U (~q ) . Prove that this condition is also necessary by using the
H-theorem as follows.
(a) Find ρ1 (~ p, ~q ) that minimizes H = N
R
d3 p~d3 ~qρ1 (~ p, ~q ) ln ρ1 (~ p, ~q), subject to the con-
straint that the total energy E = hHi is constant. (Hint: Use the method of Lagrange
multipliers to impose the constraint.)
• Using Lagrange multipliers to impose the constraints, hHi = E and
minimizing H with the given constraints reduces to minimizing, N
Z
R
d3 p~d3 ~q ρ1 = 1,
d3 ~pd3 ~q(ρ1 ln ρ1 + βρ1 H + αρ1 ) − βE − αN.
Differentiating with respect to α, β, and the function ρ1 we get, N Z
Z
3
3
d p~d ~q ρ1 = N 3
3
d ~pd ~q ρ1 H = E
→ →
Z
Z
d3 p~d3 ~q ρ1 = 1, d3 p~d3 ~q ρ1 H = E/N,
ln ρ1 + βH + α = 0 → ρ1 = exp(−βH − α),
respectively. Hence we conclude, ρ1 = R
where β is determined by, R
exp(−βH) d3 p~d3 ~q exp(−βH)
,
d3 ~pd3 ~q H exp(−βH) E R = . 3 3 N d ~pd ~q exp(−βH) (a)
(b) For a mixture of two gases (particles of masses ma and mb ) find the distributions ρ1 (b)
and ρ1 that minimize H = H(a) + H(b) subject to the constraint of constant total energy. Hence show that the kinetic energy per particle can serve as an empirical temperature. 82
• If we have Na and Nb of each particle type with total energy E, then H is minimized with the total energy constraint by extremizing, Z (a) (a) (b) (b) (a) (b) d3 p~d3 ~q(Na ρ1 ln ρ1 + Nb ρ1 ln ρ1 + β(Na Ha ρ1 + Nb Hb ρ1 ) . (a) ′ (b) ′ +Na αρ1 + Nb α ρ1 ) − βE − αNa − α Nb (a)
(b)
Differentiating this expression with respect to α, α′ , β, ρ1 , and ρ1 , we get, Z (a) d3 p~d3 ~q ρ1 = 1, Z (b) d3 p~d3 ~q ρ1 = 1, Z 0010 0011 (a) (b) 3 3 d p~d ~q Na Ha ρ1 + Nb Hb ρ1 = E, (a)
ln ρ1 + βHa + α = 0, (b)
ln ρ1 + βHb + α′ = 0. So we get, (a) ρ1 = R (a) ρ2 = R
exp(−βHa ) 3 d ~pd3 ~q exp(−βH
a)
,
exp(−βHb ) . d3 ~pd3 ~q exp(−βHb )
where β is obtained by, R 3 3 R 3 3 d p~d ~qHa exp(−βHa ) d p~d ~qHb exp(−βHb ) Na R 3 3 + Nb R 3 3 = E. d p~d ~q exp(−βHa ) d p~d ~q exp(−βHb )
Note that β is a value defined for both gases a and b, and hence can serve as an empirical temperature. For the specific case of Ha =
p2 + Ua (~q), 2ma
Hb =
p2 + Ub (~q), 2ma
the kinetic energy per particle in a distribution with equal β is also equal, since R 3 3 p2 0002 0003 R 3 R p2 d p~d ~q 2ma exp −β(p2 /2ma + Ua (~q)) exp(−βp2 /2ma ) d ~q exp(−βUa ) d3 p~ 2m a R R R = d3 p~d3 ~q exp(−βHa ) d3 ~q exp(−βUa ) d3 p~ exp(−βp2 /2ma ) R∞ p4 4π 0 dp 2m exp(−βp2 /2ma ) a . R = ∞ 4π 0 dp p2 exp(−βp2 /2ma ) R∞ 2 3 1 0 dt t4 e−t = = R∞ 2 β 0 dt t2 e−t 2β 83
So we see that the kinetic energy per particle for the gas can also serve as an empirical temperature in this case. ******** 12. Moments of momentum: Consider a gas of N classical particles of mass m in thermal equilibrium at a temperature T , in a box of volume V . (a) Write down the equilibrium one particle density feq. (~p, ~q ), for coordinate ~q, and momentum p~. • The equilibrium Maxwell-Boltzmann distribution reads f (~ p, ~q ) =
n (2πmkB T )
3/2
0012 exp −
D
0010
p2 2mkB T
(b) Calculate the joint characteristic function, exp −i~k · p~
• Performing the Gaussian average yields
0013
0011E
.
, for momentum.
0013 0012 0010 0011 D E mkB T 2 −i~ k·~ p ~ k . p˜ k = e = exp − 2
n (c) Find all the joint cumulants pℓx pm y pz c .
• The cumulants are calculated from the characteristic function, as
ℓ m n px py pz c =
0014
∂ ∂ (−ikx )
0015ℓ 0014
∂ ∂ (−iky )
0015m 0014
∂ ∂ (−ikz )
0015n
0010 0011 ln p˜ ~k ~
= mkB T (δℓ2 δm0 δn0 + δℓ0 δm2 δn0 + δℓ0 δm0 δn2 ) ,
i.e., there are only second cumulants; all other cumulants are zero. (d) Calculate the joint moment hpα pβ (~ p · p~ )i. • Using Wick’s theorem
hpα pβ (~p · p~ )i = hpα pβ pγ pγ i = hpα pβ i hpγ pγ i + 2 hpα pγ i hpβ pγ i 2
2
= (mkB T ) δαβ δγγ + 2 (mkB T ) δαγ δβγ 2
= 5 (mkB T ) δαβ . 84
k=0
Alternatively, directly from the characteristic function, 0010 0011 ∂ ∂ ∂ ∂ hpα pβ (~ p · p~ )i = p˜ ~k ∂ (−ikα ) ∂ (−ikβ ) ∂ (−ikγ ) ∂ (−ikγ ) ~ k=0 i h mk T 2 ∂ ∂ 2 ~2 − 2B ~ k 3mkB − (mkB T ) k e = ~ ∂ (−ikα ) ∂ (−ikβ ) k=0 2
= 5 (mkB T ) δαβ .
******** 13. Generalized ideal gas: Consider a gas of N particles confined to a box of volume V in d-dimensions. The energy, ǫ, and momentum, p, of each particle are related by ǫ = ps , where p = |p|. (For classical particles s = 2, while for highly relativistic ones s = 1.) Let
f (v)dv denote the probability of finding particles with speeds between v and v + dv, and n = N/V .
(a) Calculate the number of impacts of gas molecules per unit area of the wall of the box, and per unit time as follows: (i) Show that the number of particles hitting area A in a time dt arriving from a ~ with a speed v, is proportional to A · vdt cos θ · nf (v)dv , specific direction Ω, ~ and the normal to the wall. where θ is the angle between the direction Ω ~ with the same polar angle θ, demonstrate that (ii) Summing over all directions Ω dN (θ, v) = A · vdt cos θ · nf (v)dv ·
Sd−1 sind−2 θ dθ Sd
,
where Sd = 2π d/2 /(d/2 − 1)! is the total solid angle in d dimensions. (iii) By averaging over v and θ show that Sd−1 N · nv , = A dt (d − 1)Sd
where
v=
Z
vf (v)dv
is the average speed.
(b) Each (elastic) collision transfers a momentum 2p cos θ to the wall. By examining the net force on an element of area prove that the pressure P equals
s d
.
E V ,
where E is the
average (kinetic) energy. (Note that the velocity v is not p/m but ∂ǫ/∂p.) Hint: Clearly
upon averaging over all directions cos2 θ = 1/d.
(c) Using thermodynamics and the result in (b) show that along an adiabatic curve P V γ is constant, and calculate γ. 85
(d) According to the equipartition theorem, each degree of freedom which appears quadratically in the energy has an energy kB T /2. Calculate the value of γ if each gas particle has ℓ such quadratic degrees of freedom in addition to its translational motion. What values of γ are expected for helium and hydrogen at room temperature? (e) Consider the following experiment to test whether the motion of ants is random. 250 ants are placed inside a 10cm × 10cm box. They cannot climb the wall, but can escape
through an opening of size 5mm in the wall. If the motion of ants is indeed random, and they move with an average speed of 2mms−1 , how may are expected to escape the box in the first 30 seconds? ******** 14. Effusion:
A box contains a perfect gas at temperature T and density n.
(a) What is the one-particle density, ρ1 (~v), for particles with velocity ~v ? A small hole is opened in the wall of the box for a short time to allow some particles to escape into a previously empty container. (b) During the time that the hole is open what is the flux (number of particles per unit time and per unit area) of particles into the container? (Ignore the possibility of any particles returning to the box.) (c) Show that the average kinetic energy of escaping particles is 2kB T . (Hint: calculate contributions to kinetic energy of velocity components parallel and perpendicular to the wall separately.) (d) The hole is closed and the container (now thermally insulated) is allowed to reach equilibrium. What is the final temperature of the gas in the container? (e) A vessel partially filled with mercury (atomic weight 201), and closed except for a hole of area 0.1mm2 above the liquid level, is kept at 00 C in a continuously evacuated enclosure. After 30 days it is found that 24mg of mercury has been lost. What is the vapor pressure of mercury at 00 C? ******** 15. Adsorbed particles: Consider a gas of classical particles of mass m in thermal equilibrium at a temperature T , and with a density n. A clean metal surface is introduced into the gas. Particles hitting this surface with normal velocity less than vt are reflected back into the gas, while particles with normal velocity greater than vt are absorbed by it. 86
(a) Find the average number of particles hitting one side of the surface per unit area and per unit time. (b) Find the average number of particles absorbed by one side of the surface per unit area and per unit time. ******** 16. Electron emission:
When a metal is heated in vacuum, electrons are emitted from
its surface. The metal is modeled as a classical gas of noninteracting electrons held in the solid by an abrupt potential well of depth φ (the work function) relative to the vacuum. (a) What is the relationship between the initial and final velocities of an escaping electron? (b) In thermal equilibrium at temperature T , what is the probability density function for the velocity of electrons? (c) If the number density of electrons is n, calculate the current density of thermally emitted electrons. ********
87
Problems for Chapter IV - Classical Statistical Mechanics 1. Classical harmonic oscillators:
Consider N harmonic oscillators with coordinates and
momenta {qi , pi }, and subject to a Hamiltonian 0015 N 0014 2 X pi mω 2 qi2 . + H({qi , pi }) = 2m 2 i=1 (a) Calculate the entropy S, as a function of the total energy E. (Hint: By appropriate change of scale, the surface of constant energy can be deformed into a sphere. You may then ignore the difference between the surface area and volume for N ≫ 1. A more elegant method is to implement this deformation through a canonical
transformation.)
• The volume of accessible phase space for a given total energy is proportional to Z 1 dq1 dq2 · · · dqN dp1 dp2 · · · dpN , Ω= N h H=E where the integration is carried out under the condition of constant energy, 0015 N 0014 2 X pi mω 2 qi2 . E = H ({qi , pi }) = + 2m 2 i=1 Note that Planck’s constant h, is included as a measure of phase space volume, so as to make the final result dimensionless. The surface of constant energy is an ellipsoid in 2N dimensions, whose area is difficult to calculate. However, for N → ∞ the difference between volume and area is subleading in N , and we shall instead calculate the volume of the ellipsoid, replacing the constraint H = E by H ≤ E. The ellipsoid can be distorted into a sphere by a canonical transformation, changing coordinates to
qi′ ≡
√
mωqi ,
and
pi . p′i ≡ √ mω
The Hamiltonian in this coordinate system is E=
H ({qi′ , p′i })
N 0001 ω X ′2 pi + qi′2 . = 2 i=1
88
Since the canonical transformation preserves volume in phase space (the Jacobian is unity), we have Z 1 ′ dq1′ · · · dqN dp′1 · · · dp′N , Ω≈ N h H≤E where the integral is now over the 2N -dimensional (hyper-) sphere of radius R = As the volume of a d-dimensional sphere of radius R is Sd Rd /d, we obtain 0013N
=
0012
S ≡ kB ln Ω ≈ N kB ln
0012
2πeE N hω
2π N 1 Ω≈ · (N − 1)! 2N
0012
2E hω
2πE hω
0013N
p
2E/ω.
1 . N!
The entropy is now given by 0013
.
(b) Calculate the energy E, and heat capacity C, as functions of temperature T , and N . • From the expression for temperature, 1 N kB ∂S ≈ ≡ , T ∂E N E
we obtain the energy
E = N kB T, and the heat capacity C = N kB .
(c) Find the joint probability density P (p, q) for a single oscillator. Hence calculate the mean kinetic energy, and mean potential energy for each oscillator. • The single particle distribution function is calculated by summing over the undesired
coordinates and momenta of the other N − 1 particles. Keeping track of the units of h used to make phase space dimensionless, gives p(p1 , q1 )dp1 dq1 =
1 dq2 · · · dqN dp2 · · · dpN (H≤EN −1 ) hN −1 R 1 dq1 dq2 · · · dqN dp1 dp2 · · · dpN (H≤E) hN
R
89
×
dp1 dq1 , h
where EN−1 = E − p21 /2m − mω 2 q12 /2. Using the results from part (a), p(p1 , q1 ) =
=
=
Ω (N − 1, EN−1 ) hΩ(N, E) 0010 0001 2 N−1 π N −1 hω
ω N 2π E
p2
2
1 E − 2m − mω q12 (N−1)! 2 0001 2 N πN N h hω N! E !N−1 p21 mω 2 2 + q 1 2 . 1 − 2m E
0011N−1
Using the approximation (N − 1) ∼ N , for N ≫ 1, and setting E = N kB T , we have ω N p(p1 , q1 ) = 2π N kB T
1−
!N 2 2 + mω q 1 2 N kB T ! 2 2 + mω q 1 2 . kB T
p21 2m
p2
1 ω ≈ exp − 2m 2πkB T
Let us denote (p1 , q1 ) by (p, q), then 0013 0012 ω p2 mω 2 q 2 p(p, q) = , exp − − 2πkB T 2mkB T 2kB T is a properly normalized product of two Gaussians. The mean kinetic energy is 001c
p2 2m
001d
=
Z
p(p, q)
p2 kB T dqdp = , 2m 2
p(p, q)
mω 2 q 2 kB T dqdp = . 2 2
while the mean potential energy is also 001c
mω 2 q 2 2
001d
=
Z
******** 2. Quantum harmonic oscillators: Consider N independent quantum oscillators subject to a Hamiltonian
N X
0012 0013 1 ¯hω ni + H ({ni }) = , 2 i=1 where ni = 0, 1, 2, · · ·, is the quantum occupation number for the ith oscillator. 90
(a) Calculate the entropy S, as a function of the total energy E. (Hint: Ω(E) can be regarded as the number of ways of rearranging M = N − 1 partitions along a line.)
P
i
ni balls, and
• The total energy of the set of oscillators is E=h ¯ω
N X
N ni + 2 i=1
!
.
Let us set the sum over the individual quantum numbers to M≡
N X
ni =
i=1
E N − . ¯hω 2
The number of configurations {ni } for a given energy (thus for a given value of M ) is
equal to the possible number of ways of distributing M energy units into N slots, or of
partitioning M particles using N − 1 walls. This argument gives to the number of states as
Ω=
(M + N − 1)! , M ! (N − 1)!
and a corresponding entropy 0013 00130015 0014 0012 0012 M M +N −1 N −1 − M ln . S = kB ln Ω ≈ kB (M + N − 1) ln − (N − 1) ln e e e (b) Calculate the energy E, and heat capacity C, as functions of temperature T , and N . • The temperature is calculated by 0013 0012 1 kB kB ∂S M +N −1 ≈ = ≡ ln ln T ∂E N ¯ω h M ¯hω
E h ¯ω
+
E h ¯ω
N 2
−
−1
N 2
!
kB ln ≈ ¯hω
E+ E−
By inverting this equation, we get the energy
0015 0014 N exp (¯ hω/kB T ) + 1 1 1 E = ¯hω , = N ¯hω + 2 exp (¯ hω/kB T ) − 1 2 exp (¯ hω/kB T ) − 1 and a corresponding heat capacity 0012 00132 ¯hω exp (¯ hω/kB T ) ∂E = N kB C≡ 2. ∂T N kB T [exp (¯ hω/kB T ) − 1] 91
N ¯hω 2 N ¯hω 2
!
.
(c) Find the probability p(n) that a particular oscillator is in its nth quantum level. • The probability that a particular oscillator is in its nth quantum level is given by summing the joint probability over states for all the other oscillators, i.e. P 0001 X Ω N − 1, E − (n + 12 )¯ hω (E−(n+1/2)¯ hω) 1 P = p(n) = p(ni ) = Ω(N, E) (E) 1 {ni6=1 }
[(M − n) + N − 2]! M ! · (N − 1)! · (M − n)! · (N − 2)! (M + N − 1)! M (M − 1) · · · (M − n + 1) · N ≈ (M + N − 1)(M + N − 2) · · · (M + N − n − 1) =
≈ N (M + N )−n−1 M n ,
where the approximations used are of the from (I − 1) ≈ I, for I ≫ 1. Hence, p(n) = N
0012
E N − +N ¯hω 2
N = E h ¯ω +
N 2
E h ¯ω E h ¯ω
0013−n−1 0012
N 2 N 2
− +
!n
N E − ¯hω 2
0013n
,
which using 1 kB = ln T ¯ω h
E+ E−
N¯ hω 2 N¯ hω 2
!
,
E h ¯ω E h ¯ω
=⇒
+ −
N 2 N 2
!n
0013 0012 n¯hω , = exp − kB T
leads to the probability 0012 00130014 0012 00130015 ¯ω h ¯hω p(n) = exp −n 1 − exp − . kB T kB T (d) Comment on the difference between heat capacities for classical and quantum oscillators. • As found in part (b), Cquantum = N kB
0012
¯ω h kB T
00132
h
0010
h ¯ω kB T
0011
exp 0010 0011 i2 . h ¯ω exp kB T − 1
In the high temperature limit, h ¯ ω/(kB T ) ≪ 1, using the approximation ex ≈ 1 + x for
x ≪ 1, gives
Cquantum = N kB = Cclassical . 92
At low temperatures, the quantized nature of the energy levels of the quantum oscillators becomes noticeable. In the limit T → 0, there is an energy gap between the ground state
and the first excited state. This results in a heat capacity that goes to zero exponentially, as can be seen from the limit ¯hω/(kB T ) ≫ 1, Cquantum = N kB
0012
¯ω h kB T
00132
0013 0012 ¯ω h . exp − kB T
******** 3. Relativistic particles: N indistinguishable relativistic particles move in one dimension subject to a Hamiltonian H ({pi , qi }) =
N X i=1
[c|pi | + U (qi )] ,
with U (qi ) = 0 for 0 ≤ qi ≤ L, and U (qi ) = ∞ otherwise. Consider a microcanonical ensemble of total energy E.
(a) Compute the contribution of the coordinates qi to the available volume in phase space Ω(E, L, N ). •
Each of N coordinates explores a length L, for an overall contribution of LN /N !.
Division by N ! ensures no over-counting of phase space for indistinguishable particles. (b) Compute the contribution of the momenta pi to Ω(E, L, N ). Pd (Hint: The volume of the hyper-pyramid defined by i=1 xi ≤ R, and xi ≥ 0, in d dimensions is Rd /d! .)
• The N momenta satisfy the constraint
PN
i=1
|pi | = E/c. For a particular choice of the
signs of {pi }, this constraint describes the surface of a hyper-pyramid in N dimensions. If
we ignore the difference between the surface area and volume in the large N limit, we can calculate the volume in momentum space from the expression given in the hint as 1 · Ωp = 2 · N! N
0012
E c
0013N
.
The factor of 2N takes into account the two possible signs for each pi . The surface area √ √ of the pyramid is given by dRd−1 /(d − 1)!; the additional factor of d with respect to 93
dvolume/dR is the ratio of the normal to the base to the side of the pyramid. Thus, the volume of a shell of energy uncertainly ∆E , is Ω′p
√
N · =2 · (N − 1)! N
0012
E c
0013N−1
·
∆E . c
We can use the two expressions interchangeably, as their difference is subleading in N . (c) Compute the entropy S(E, L, N ). • Taking into account quantum modifications due to anisotropy, and phase space measure, we have
√ 0012 0013N−1 1 LN N N E ∆E Ω(E, L, N ) = N · ·2 · · · . h N! (N − 1)! c c
Ignoring subleading terms in the large N limit, the entropy is given by S(E, L, N ) = N kB ln
0012
2e2 L E · · hc N N
0013
.
(d) Calculate the one dimensional pressure P . • From dE = T dS − P dV + µdN , the pressure is given by N kB T ∂S = . P =T ∂L E,N L (e) Obtain the heat capacities CL and CP . • Temperature and energy are related by N kB ∂S 1 = = , T ∂E L,N E
=⇒
E = N kB T,
=⇒
∂E = N kB . CL = ∂T L,N
Including the work done against external pressure, and using the equation of state, ∂E ∂L CP = +P = 2N kB . ∂T P,N ∂T P,N (f) What is the probability p(p1 ) of finding a particle with momentum p1 ? 94
• Having fixed p1 for the first particle, the remaining N − 1 particles are left to share
an energy of (E − c|p1 |). Since we are not interested in the coordinates, we can get the
probability from the ratio of phase spaces for the momenta, i.e.
Ωp (E − c|p1 |, N − 1) Ωp (E, N ) ' 0013N−1 # 0014 0015 0012 N ! 0010 c 0011N E − c|p1 | 2N−1 × N · = · (N − 1)! c 2 E 0013N 0013 0012 0012 cN c|p1 | cN |p1 | cN ≈ . · 1− · exp − ≈ 2E E 2E E
p(p1 ) =
Substituting E = N kB T , we obtain the (property normalized) Boltzmann weight 0013 0012 c|p1 | c . · exp − p(p1 ) = 2kB T kB T ******** 4. Hard sphere gas: Consider a gas of N hard spheres in a box. A single sphere occupies volume ω, while its center of mass can explore a volume V (if the box is otherwise empty). There are no other interactions between the spheres, except for the constraints of hard-core exclusion. (a) Calculate the entropy S, as a function of the total energy E. 0001 0001 00012 (Hint: V − aω V − (N − a)ω ≈ V − N ω/2 .)
• The available phase space for N identical particles is given by Z 1 Ω= d3 ~q1 · · · d3 ~qN d3 p~1 · · · d3 ~pN , N !h3N H=E where the integration is carried out under the condition, N X p2i E = H(~qi , ~pi ) = , 2m i=1
N X
or
p2i = 2mE.
i=1
The momentum integrals are now performed as in an ideal gas, yielding (2mE)3N/2−1 · Ω= N !h3N
2π 3N 0001 · 3N 2 −1 ! 95
Z
d3 ~q1 · · · d3 ~qN .
The joint integral over the spacial coordinates with excluded volume constraints is best performed by introducing particles one at a time. The first particle can explore a volume V , the second V − ω, the third V − 2ω, etc., resulting in Z
d3 ~q1 · · · d3 ~qN = V (V − ω)(V − 2ω) · · · (V − (N − 1)ω).
Using the approximation (V − aω)(V − (N − a)ω) ≈ (V − N ω/2)2 , we obtain Z
0012 0013N Nω d ~q1 · · · d q~N ≈ V − . 2 3
3
Thus the entropy of the system is '
e S = kB ln Ω ≈ N kB ln N
0012 00130012 00133/2 # Nω 4πmEe V − . 2 3N h2
(b) Calculate the equation of state of this gas. • We can obtain the equation of state by calculating the expression for the pressure of the gas,
which is easily re-arranged to,
N kB ∂S P ≈ = , T ∂V E,N V − Nω 2 P
0012
Nω V − 2
0013
= N kB T.
Note that the joint effective excluded volume that appears in the above expressions is one half of the total volume excluded by N particles. (c) Show that the isothermal compressibility, κT = −V −1 ∂V /∂P |T , is always positive. • The isothermal compressibility is calculated from N kB T 1 ∂V = κT = − > 0, V ∂P T,N P 2V
and is explicitly positive, as required by stability constraints. ******** 96
5. Non-harmonic gas:
Let us reexamine the generalized ideal gas introduced in the
previous section, using statistical mechanics rather than kinetic theory. Consider a gas of N non-interacting atoms in a d-dimensional box of “volume” V , with a kinetic energy H=
N X i=1
s
A |~ pi | ,
where p~i is the momentum of the ith particle. (a) Calculate the classical partition function Z(N, T ) at a temperature T . (You don’t have to keep track of numerical constants in the integration.) • The partition function is given by 1 Z(N, T, V ) = N !hdN 1 = N !hdN
Z
···
0014Z Z
Z
'
dd ~q1 · · · dd ~qN dd p~1 · · · dd ~pN exp −β d
s
d
d ~qd p~ exp (−βA |~ p| )
0015N
N X i=1
A |~ pi |
s
#
.
Ignoring hard core exclusion, each atom contributes a d–dimensional volume V to the integral over the spatial degrees of freedom, and 0014Z 0015N VN s d d p~ exp (−βA |~p | ) . Z(N, T, V ) = N !hdN Observing that the integrand depends only on the magnitude |~ p | = p, we can evaluate the R d R d−1 integral in spherical coordinates using d p~ = Sd dpp , where Sd denotes the surface area of a unit sphere in d–dimensions, as 0015N 0014 Z ∞ VN d−1 s dpp exp (−βAp ) . Z(N, T, V ) = Sd N !hdN 0
Introducing the variable x ≡ (βA)1/s p, we have 0013−dN /s 0014Z ∞ 0015N 0012 A V N SdN d−1 s dxx exp(−x ) Z(N, T, V ) = N !hdN kB T 0 0012 0013−dN /s 0013N 0012 1 V Sd A N = C (d, s) , N! hd kB T where C denotes the numerical value of the integral. (We assume that A and s are both real and positive. These conditions ensure that energy increases with increasing |~ p |.) The integral is in fact equal to
C(d, s) =
Z
∞ 0
dxx
d−1
1 exp(−x )dx = Γ s s
97
0012 0013 d , s
and the partition function is 1 Z= N!
0012
V Sd hd s
0013N 0012
A kB T
0013−dN /s 0014 0012 00130015N d Γ . s
(b) Calculate the pressure and the internal energy of this gas. (Note how the usual equipartition theorem is modified for non-quadratic degrees of freedom.) • To calculate the pressure and internal energy, note that the Helmholtz free energy is F = E − T S = −kB T ln Z, and that
∂F P =− , ∂V T
while
First calculating the pressure:
∂ ln Z E=− . ∂β V
∂ ln Z N kB T ∂F . = kB T = P =− ∂V T ∂V T V
Now calculating the internal energy:
0014 00130015 0012 ∂ dN d ∂ ln Z A = − − = E=− ln N kB T. ∂β V ∂β s kB T s
Note that for each degree of freedom with energy A|~ pi |s , we have the average value,
hA|~ pi |s i = ds kB T. This evaluates to 32 kB T for the 3–dimensional ideal gas. (c) Now consider a diatomic gas of N molecules, each with energy 0011 t 0010 (1) s (2) s (1) (2) Hi = A p~i + p~i + K ~qi − ~qi ,
where the superscripts refer to the two particles in the molecule. (Note that this unrealistic potential allows001dthe two atoms to occupy the same point.) Calculate the expectation value 001c t (1) (2) ~qi − ~qi , at temperature T .
• Now consider N diatomic molecules, with H=
N X i=1
Hi ,
where
0011 t 0010 (1) s (2) s (1) (2) Hi = A ~p i + p~ i + K ~q i − ~q i . 98
The expectation value 001c t 001d (1) (2) ~q i − ~q i =
1 N!
t P (1) (2) ~q i − ~q i exp [−β i Hi ] , R QN P (1) d (2) d (1) d (2) 1 d~ d q d ~ q d p ~ d p ~ exp [−β H ] i i i i i i=1 i N! d (1) d (2) d (1) d (2) q i d ~q i d p~ i d p~ i i=1 d ~
R QN
is easily calculated by changing variables to ~x ≡ ~q (1) − ~q (2) ,
and
~y ≡
q~ (1) + ~q (2) , 2
as (note that the Jacobian of the transformation is unity) 001c t 001d R dd ~xdd ~y · |~x |t · exp [−βK|~x |t ] (1) R ~q − ~q (2) = dd ~xdd ~y · exp [−βK|~x |t ] R d d ~x · |~x |t · exp [−βK|~x |t ] R . = dd ~x · exp [−βK|~x |t ]
Further simplifying the algebra by introducing the variable ~z ≡ (βK)1/t~x, leads to 001c t 001d (βK)−t/t R dd ~z · |~z |t · exp [−|~z |t ] d kB T (1) (2) R . = · − ~q = ~q d t t K d ~z · exp [−|~z | ]
Here we have assumed that the volume is large enough, so that the range of integration over the relative coordinate can be extended from 0 to ∞. Alternatively, note that for the degree of freedom ~x = ~q (1) − ~q (2) , the energy is K|~x |t .
Thus, from part (b) we know that
d kB T K|~x |t = · , t K
i.e.
t |~x | =
001c t 001d d k T (1) B (2) . − ~q = · ~q t K
And yet another way of calculating the expectation value is from N 001c t 001d X 1 ∂ ln Z N d kB T (1) (2) = · , ~q i − ~q i = − β ∂K t K i=1
(note that the relevant part of Z is calculated in part (d) below).
(d) Calculate the heat capacity ratio ratio γ = CP /CV , for the above diatomic gas. 99
• For the ideal gas, the internal energy depends only on temperature T . The gas in part
(c) is ideal in the sense that there are no molecule–molecule interactions. Therefore, dE + P dV dE(T ) d¯Q = = , CV = dT V dT dT V and
dE + P dV d¯Q = CP = dT P dT
Since P V = N kB T,
CP =
dE(T ) ∂V (T ) = . + P dT ∂T P P
dE(T ) + N kB . dT
We now calculate the partition function Z= =
1 N !hdN
Z Y N
i=1
(1)
(2)
(1)
(2)
'
dd ~q i dd ~q i dd p~ i dd p~ i exp −β
1 z1N , dN N !h
X i
Hi
#
where z1 = =
Z
0014Z
d (1) d (2) d
d ~q
d ~q
d p~
d (1) d (2)
d ~q
d ~q
(1) d
d p~
(2)
0014
0012 t 00130015 (1) s (2) s (1) (2) exp −β · A ~p + A p~ + K ~q − ~q
0012 t 00130015 0014Z 001100152 0010 (1) (1) s (2) d (1) exp −βK ~q − ~q · d p~ exp −βA p~ .
Introducing the variables, ~x, ~y , and ~z, as in part (c), 0014 0015N 0014Z ∞ 00152N Z ∞ VN −d/t d−1 t d−1 s (βK) z exp(−z )dz · p exp(−βAp )dp Z∝ N! 0 0 0014 0012 00130015N 0014 0012 001300152N 1 d d VN −d/t 1 −d/s (βK) · (βA) Γ Γ = N! t t s s ∝
VN −dN/t −2Nd/s (βK) (βA) . N!
Now we can calculate the internal energy as ∂ ln Z d 2d hEi = − = N kB T + N kB T = dN kB T ∂β t s
0012
1 2 + t s
From this result, the heat capacities are obtained as 0012 0013 ∂V 2d d ∂E +P = N kB + +1 , CP = ∂T P ∂T P s t 0012 0013 2 1 ∂E = dN k . CV = + B ∂T V s t 100
0013
.
resulting in the ratio γ=
CP 2d/s + d/t + 1 st = =1+ . CV 2d/s + d/t d(2t + s)
******** 6. Surfactant adsorption:
A dilute solution of surfactants can be regarded as an ideal
three dimensional gas. As surfactant molecules can reduce their energy by contact with air, a fraction of them migrate to the surface where they can be treated as a two dimensional ideal gas. Surfactants are similarly adsorbed by other porous media such as polymers and gels with an affinity for them. (a) Consider an ideal gas of classical particles of mass m in d dimensions, moving in a uniform attractive potential of strength εd . By calculating the partition function, or otherwise, show that the chemical potential at a temperature T and particle density nd , is given by 0002 0003 µd = −εd + kB T ln nd λ(T )d ,
where
λ(T ) = √
h . 2πmkB T
• The partition function of a d-dimensional ideal gas is given by ( ' 0013#) Z Z Y Nd Nd 0012 2 X p ~ 1 i ··· dd ~qi dd ~p i exp −β Nd εd + Zd = Nd !hdNd 2m i=1 i=1 0012 0013Nd 1 Vd = e−βNd εd , Nd ! λd where λ≡ √
h . 2πmkB T
The chemical potential is calculated from the Helmholtz free energy as ∂ ln Zd ∂F = −kB T µd = ∂N V,T ∂Nd V,T 0013 0012 Vd . = −εd + kB T ln Nd λd 101
(b) If a surfactant lowers its energy by ε0 in moving from the solution to the surface, calculate the concentration of floating surfactants as a function of the solution concentration n (= n3 ), at a temperature T . •
The density of particles can also be calculated from the grand canonical partition
function, which for particles in a d–dimensional space is Ξ(µ, Vd , T ) = =
∞ X
Nd =0 ∞ X
Nd =0
Z(Nd , Vd , T )eβNd µ 1 Nd !
0012
Vd λd
0013Nd
−βNd εd βNd µ
e
e
= exp
00140012
Vd λd
0013
β(µ−εd )
·e
0015
.
The average number of particles absorbed in the space is 1 ∂ 1 ∂ hNd i = ln Ξ = β ∂µ β ∂µ
00140012
Vd λd
0013
β(µ−εd )
·e
0015
=
0012
Vd λd
0013
· eβ(µ−εd ) .
We are interested in the coexistence of surfactants between a d = 3 dimensional solution, and its d = 2 dimensional surface. Dividing the expressions for hN3 i and hN2 i, and taking
into account ε0 = ε3 − ε2 , gives
Aλ βε0 hN2 i = e , hN3 i V
which implies that n2 =
hN2 i = nλeβε0 . A
(c) Gels are formed by cross-linking linear polymers. It has been suggested that the porous gel should be regarded as fractal, and the surfactants adsorbed on its surface treated as a gas in df dimensional space, with a non-integer df . Can this assertion be tested by comparing the relative adsorption of surfactants to a gel, and to the individual polymers (presumably one dimensional) before cross-linking, as a function of temperature? • Using the result found in part (b), but regarding the gel as a df –dimensional container,
the adsorbed particle density is
hngel i = nλ3−df exp [β(ε3 − εgel )] . Thus by studying the adsorption of particles as a function of temperature one can determine the fractal dimensionality, df , of the surface. The largest contribution comes from the 102
difference in energies. If this leading part is accurately determined, there is a subleading dependence via λ3−df which depends on df . ******** 7. Molecular adsorption:
N diatomic molecules are stuck on a metal surface of square
symmetry. Each molecule can either lie flat on the surface in which case it must be aligned to one of two directions, x and y, or it can stand up along the z direction. There is an energy cost of ε > 0 associated with a molecule standing up, and zero energy for molecules lying flat along x or y directions. (a) How many microstates have the smallest value of energy? What is the largest microstate energy? • The ground state energy of E = Emin = 0 is obtained for 2N configurations. The largest microstate energy is N ǫ is unique.
(b) For microcanonical macrostates of energy E, calculate the number of states Ω(E, N ), and the entropy S(E, N ). • Let Nz = E/ǫ. Expressing Ω as the number of ways to choose the Ng excited molecules,
multiplied by the number of possible configurations Ω(E, N ) =
N! · 2N−Nz , Nz !(N − Nz )!
and S(E, N ) = Stwo−level system + kb (N − Nz ) ln 2 001a 001b E E E E E = −N kB ln + (1 − ) ln(1 − ) + kb (N − ) ln 2. Nǫ Nǫ Nǫ Nǫ ǫ (c) Calculate the heat capacity C(T ) and sketch it. • The temperature dependence of the energy is obtained from the relation kB ∂S E kB 1 =− = ln( )− ln 2, T ∂E N ǫ Nǫ − E ǫ
whence
E= and
exp( kǫB ( T1
Nǫ Nǫ = , kB 2 exp( kBǫ T ) + 1 + ǫ ln 2)) + 1
2 exp( kBǫ T ) ǫ 2 dE = N kB ( ) . C= dT kB T (1 + 2 exp( kBǫ T ))2 103
C/kBN
(kεT)2e- ε/k T B
1 ε 2 4 kB T
( )
B
T
(d) What is the probability that a specific molecule is standing up? • The probability that a specific molecule is standing up is Ω(E − ǫ, N − 1) Ω(E, N ) (N − 1)! Nz !(N − Nz )! 1 = 2(N−1)−(Nz −1) (Nz − 1)!((N − 1) − (Nz − 1))! N! 2N−Nz E Nz = = N Nǫ 1 . = 2 exp kBǫ T + 1
p(~r1 = zˆ) =
(e) What is the largest possible value of the internal energy at any positive temperature? • Since
dE dT
> 0 for all T > 0, the energy is largest for → ∞, i.e. Emax =
Nǫ . 3
******** 8. Curie susceptibility: Consider N non-interacting quantized spins in a magnetic field ~ = B zˆ, and at a temperature T . The work done by the field is given by BMz , with a B PN magnetization Mz = µ i=1 mi . For each spin, mi takes only the 2s + 1 values −s, −s + 1, · · · , s − 1, s.
(a) Calculate the Gibbs partition function Z(T, B). (Note that the ensemble corresponding
to the macrostate (T, B) includes magnetic work.) • The Gibbs partition function is Z=
X
{mi }
0010
0011
~ ·M ~ = exp β B
X
exp βBµ
N X i=1
{mi }
104
mi
!
=
'
m i =s X
mi =−s
exp(βµB · mi )
#N
.
Thus we obtain the series N
Z = [exp(−βBµs) + exp(−βBµ(s − 1)) + · · · + exp(βBµ(s − 1)) + exp(βBµs)] . In general, to evaluate a geometrical series of the form S = x−s + x−(s−1) + · · · + xs−1 + xs , increase the order of the series by one, Sx = x−s+1 + · · · + xs + xs+1 , and subtract from the original series: (1 − x)S = x−s − xs+1 ,
=⇒
S=
x−s − xs+1 . 1−x
(Note that the same result is obtained whether s is an integer or half–integer quantity.) Using this expression, we get Z=
0012
=
0012
exp(−βBµs) − exp(βBµ(s + 1)) 1 − exp(βBµ)
0013N
exp(−βBµ(s + 1/2)) − exp(−βBµ(s + 1/2)) exp(−βBµ/2) − exp(−βBµ/2)
0013N
.
Substituting in the proper trigonometric identity, 0014
sinh (βµB(s + 1/2)) Z= sinh(βµB/2)
0015N
.
(b) Calculate the Gibbs free energy G(T, B), and show that for small B, G(B) = G(0) −
N µ2 s(s + 1)B 2 + O(B 4 ). 6kB T
• The Gibbs free energy is G = E − BM = −kB T ln Z = −N kB T ln[sinh(βµB(s + 1/2))] + N kB T ln[sinh(βµB/2)]. 105
Using an approximation of sinh θ for small θ, 0001 1 1 θ e − e−θ ≈ sinh θ = 2 2
0012 0013 θ3 2θ + 2 + O(θ 5 ), 3!
for
θ ≪ 1,
we find (setting α = βµB), G ≈ −N kB T
(
' 0012 0013 0012 00132 !# 0014 0012 00130015 001b 1 α2 1 α α2 4 ln α s + 1+ s+ − ln 1+ + O(α ) . 2 6 2 2 24
Using the expansion ln(1 + x) = x − x2 /2 + x3 /3 − · · ·, we find 0014
1 1 2 2 G ≈ −N kB T ln(2s + 1) + (α(s + 1/2)) − (α/2) + O(α4 ) 6 6 2 (s + s) ≈ −N kB T ln(2s + 1) − N kB T α2 6 2 2 N µ B s(s + 1) = G0 − + O(B 4 ). 6kB T
0015
(c) Calculate the zero field susceptibility χ = ∂Mz /∂B|B=0 , and show that is satisfies Curie’s law χ = c/T.
• The magnetic susceptibility, χ = ∂Mz /∂B, is obtained by noting that the average
magnetization is
hMz i = kB T Thus χ=
∂G ∂ ln Z =− . ∂B ∂B
∂ hMz i ∂ ∂G N µ2 ts(s + 1) =− = , ∂B ∂B ∂B 3kB T
which obeys Curie’s law, χ = c/T , with c = N µ2 s(s + 1)/3kB . (d) Show that CB − CM = cB 2 /T 2 where CB and CM are heat capacities at constant B and M respectively.
******** 9. Langmuir isotherms:
An ideal gas of particles is in contact with the surface of a
catalyst. 106
(a) Show that the chemical potential of the gas particles is related to their temperature 0002 0001 0003 and pressure via µ = kB T ln P/T 5/2 + A0 , where A0 is a constant. • For convenience, we begin by defining the characteristic length, λ= √
h , 2πmkB T
in terms of which the free energy of ideal gas is given by, F = −N kB T ln
0012
Ve N λ3
0013
.
Then using the equation of state of ideal gas we get, ∂F µ= = −kB T ln ∂N '
0012
V N λ3
0013
= kB T ln(P T −5/2 ) + ln
= −kB T ln h3
0012
kB T P λ3
5/2
kB 23/2 π 3/2 m3/2
0013
!#
.
(b) If there are N distinct adsorption sites on the surface, and each adsorbed particle gains an energy ǫ upon adsorption, calculate the grand partition function for the two dimensional
gas with a chemical potential µ. • We have, Q(T, µ) =
0013 N 0012 X N
n=0
n
0010 0011N enβµ e−nβǫ = 1 + eβ(µ−ǫ) .
(c) In equilibrium, the gas and surface particles are at the same temperature and chemical potential. Show that the fraction of occupied surface sites is then given by f (T, P ) = 0001 P/ P + P0 (T ) . Find P0 (T ). • Since the average number of absorption sites occupied is, hN i = kB T
eβ(µ−ǫ) ∂ ln Q =N , ∂µ 1 + eβ(µ−ǫ)
and the fraction of occupied sites is given by, f (T, P ) =
eβ(µ−ǫ) . 1 + eβ(µ−ǫ)
107
Since the gas and surface particles have the same temperature and chemical potential, the realtion in (a), namely, P λ3 kB T
eβµ =
holds. Plugging this into the formula for f we obtain, f (T, P ) =
P , P + P0 (T )
with P0 (T ) =
kB T βǫ e . λ3
(d) In the grand canonical ensemble, the particle number N is a random variable. Calculate its characteristic function hexp(−ikN )i in terms of Q(βµ), and hence show that m
hN ic = −(kB T )
m−1
where G is the grand potential.
∂ m G , ∂µm T
• Note that in calculating Q(βµ) the term for N particles is proportional to eβµN . In calculating the average of e−ikN , we just replace the initial factor with e(βµ−ik)N , and hence
−ikN Q(βµ − ik) . e = Q(βµ)
For the cumulant generating function, we get
Hence,
ln e−ikN = ln Q(βµ − ik) − ln Q(βµ) = −βG(βµ − ik) + βG(βµ). ∂m (−βG(βµ − ik))|T ∂(−ik)m m ∂ m G 1−m ∂ G = −β = −β ∂(βµ)m T ∂µm T ∂ m G . = −(kB T )m−1 ∂µm T
hN m ic =
(e) Using the characteristic function, show that
2 ∂hN i . N c = kB T ∂µ T 108
•
2 ∂ G ∂hN i . = kB T hN 2 ic = −kB T ∂µ2 T ∂µ T
(f) Show that fluctuations in the number of adsorbed particles satisfy
2 N c 2
hN ic
• By definition of f ,
and since,
=
1−f . Nf
∂G hN ic = − = N f, ∂µ T
−βµ −βǫ ∂f e e−βǫ ∂ = βe = 2 = βf (1 − f ), ∂µ T ∂µ e−βµ + e−βǫ T (e−βµ + e−βǫ )
we get,
hN 2 ic =
1 ∂f 1 ∂N = N = N f (1 − f ), β ∂µ β ∂µ
leading to the equality,
hN 2 ic 2 hN ic
=
1−f . Nf
******** ~ of unity, i.e. S z is quantized to -1, 0, 10. Molecular oxygen has a net magnetic spin, S, ~ k zˆ is or +1. The Hamiltonian for an ideal gas of N such molecules in a magnetic field B H=
N 0014 X p~i i=1
2
2m
−
µBSiz
0015
,
where {~ pi } are the center of mass momenta of the molecules. The corresponding coordi-
nates {~qi } are confined to a volume V . (Ignore all other degrees of freedom.)
(a) Treating {~ pi , ~qi } classically, but the spin degrees of freedom as quantized, calculate the ˜ partition function, Z(T, N, V, B). • Z=
X µs
−βH(µs )
e
=
X µs
−β
e
P
(~ p2i /2m−µBSiz )
109
1 = N!
0012
V βµB (e + 1 + e−βµB ) λ3
0013N
,
where, λ= √
h . 2πmkB T
(b) What are the probabilities for Siz of a specific molecule to take on values of -1, 0, +1 at a temperature T ? • The probabilities for Siz of a given molecule to take values -1, 0, +1, are, e−βµB , 2 cosh βµB + 1
eβµB , 2 cosh βµB + 1
1 , 2 cosh βµB + 1
respectively. (c) Find the average magnetic dipole moment, hM i /V , where M = µ
•
1 ∂ ln Z 1 ∂N ln(2 cosh βµB + 1) = β ∂B β ∂B 0013 0012 2 sinh βµB . = Nµ 2 cosh βµB + 1
PN
i=1
Siz .
hM i =
(d) Calculate the zero field susceptibility χ = ∂ < M > /∂B|B=0 . •
2 cosh βµB(2 cosh βµB + 1) − 4 sinh βµB sinh βµB 2 χ = N βµ = N βµ2 . 2 (2 cosh βµB + 1) 3 B=0 2
********
~ of unity, i.e. S z is quantized to -1, 0, 11. Molecular oxygen has a net magnetic spin, S, ~ k zˆ is or +1. The Hamiltonian for an ideal gas of N such molecules in a magnetic field B H=
N 0014 X p~i i=1
2
2m
−
µBSiz
0015
,
where {~ pi } are the center of mass momenta of the molecules. The corresponding coordi-
nates {~qi } are confined to a volume V . (Ignore all other degrees of freedom.)
(a) Treating {~ pi , ~qi } classically, but the spin degrees of freedom as quantized, calculate the ˜ partition function, Z(T, N, V, B). • Z=
X µs
−βH(µs )
e
=
X µs
−β
e
P
(~ p2i /2m−µBSiz )
110
1 = N!
0012
V βµB (e + 1 + e−βµB ) λ3
0013N
,
where, λ= √
h . 2πmkB T
(b) What are the probabilities for Siz of a specific molecule to take on values of -1, 0, +1 at a temperature T ? • The probabilities for Siz of a given molecule to take values -1, 0, +1, are, e−βµB , 2 cosh βµB + 1
1 , 2 cosh βµB + 1
eβµB , 2 cosh βµB + 1
respectively. (c) Find the average magnetic dipole moment, hM i /V , where M = µ
•
1 ∂N ln(2 cosh βµB + 1) 1 ∂ ln Z = β ∂B β ∂B 0013 0012 2 sinh βµB . = Nµ 2 cosh βµB + 1
PN
i=1
Siz .
hM i =
(d) Calculate the zero field susceptibility χ = ∂ < M > /∂B|B=0 . •
2 2 cosh βµB(2 cosh βµB + 1) − 4 sinh βµB sinh βµB = N βµ2 . χ = N βµ 2 (2 cosh βµB + 1) 3 B=0 2
******** 12. Polar rods:
Consider rod shaped molecules with moment of inertia I, and a dipole
moment µ. The contribution of the rotational degrees of freedom to the Hamiltonian is given by Hrot.
1 = 2I
p2θ +
p2φ sin2 θ
!
− µE cos θ
,
where E is an external electric field. (φ ∈ [0, 2π], θ ∈ [0, π] are the azimuthal and polar
angles, and pφ , pθ are their conjugate momenta.)
(a) Calculate the contribution of the rotational degrees of freedom of each dipole to the classical partition function. 111
• The classical partition function is obtained by integrating over the angles θ and φ, and
the corresponding momenta as ' Z π Z 2π Z ∞ 1 β Zrot = 2 dθ dφ dpθ dpφ exp − h 0 2I 0 −∞
p2θ +
p2φ sin2 θ
!
#
+ βµ cos θ .
The Gaussian integrations over momenta are easily carried out, and after the change of variables to x = cos θ, we find 0012 0012 2 0013 0013 Z +1 2πI 8π I sinh(βµE) βµEx Zrot = dxe = (2π) . 2 βh βh2 βµE −1 (b) Obtain the mean polarization P = hµ cos θi, of each dipole.
• From the form of the partition function, we note that 0015 0014 ∂ ln Zrot 1 P = hµ cos θi = . = µ coth(βµE) − ∂βE βµE (c) Find the zero–field polarizability
• The susceptibility is
∂P χT = ∂E E=0
.
0014 0015 ∂P 1 1 µ2 1 2 χT = = + = βµ − ∂E 3 kB T sinh2 (βµE) (βµE)2
for E = 0 .
(d) Calculate the rotational energy per particle (at finite E), and comment on its high and low temperature limits. • The energy stored in this degree of freedom is hErot i = −
∂ ln Zrot = 2kB T − µE coth(βµE). ∂β
At high temperature hErot i ≈ kB T coming entirely from the kinetic energies. At temper-
atures kB T ≪ µE, the polarization is saturated, and hErot i ≈ −µE + 2kB T . (e) Sketch the rotational heat capacity per dipole.
• The classical heat capacity is 2kB at low temperatures, and decreases to 1kB at high temperatures, with a ‘knee’ at kB T ≈ µE.
******** 112
Problems for Chapter V - Interacting Particles 1. Debye–H¨ uckel theory and ring diagrams:
The virial expansion gives the gas pressure
as an analytic expansion in the density n = N/V . Long range interactions can result in non-analytic corrections to the ideal gas equation of state. A classic example is the Coulomb interaction in plasmas, whose treatment by Debye–H¨ uckel theory is equivalent to summing all the ring diagrams in a cumulant expansion. For simplicity consider a gas of N electrons moving in a uniform background of positive charge density N e/V to ensure overall charge neutrality. The Coulomb interaction takes the form UQ =
X i Tc , T = Tc , and T < Tc . For T < Tc there is a range of compositions x < |xsp (T )| where F (x) is not convex and hence the composition is locally unstable. Find xsp (T ).
• The function F (x) is concave if ∂ 2 F /∂x2 < 0, i.e. if 2
x
<=' tc=' ,=' at=' the=' spinodal=' line=' given=' by=' xsp=' (t=' )='br' rbr=' −=' 1,=' 144br=' f(x)=' nj=' x=' sp(t)=' ttcbr=' ttc=' +1br=' -1br=' xbr=' as=' indicated=' dashed=' in=' figure=' below.=' (f)=' alloy=' globally=' minimizes=' its=' free=' energy=' separating=' into=' a=' rich=' and=' b=' phases=' of=' compositions=' ±xeq=' ),=' where=' xeq=' function=' f=' (x).=' find=' ).=' ='="" •=' setting=' first=' derivative=' df=' (x)=' dx='N' (kb=' 6j)=' +=' x2=' 3=' to=' zero=' yieldsbr=' equilibrium=' value=' ofbr=' =' r=' =' =' ±√3=' 1=' =' 0='> Tc (g) In the (T, x) plane sketch the phase separation boundary ±xeq (T ); and the so called
spinodal line ±xsp (T ). (The spinodal line indicates onset of metastability and hysteresis
effects.)
• The spinodal and equilibrium curves are indicated in the figure above. In the interval
between the two curves, the system is locally stable, but globally unstable. The formation of ordered regions in this regime requires nucleation, and is very slow. The dashed area is locally unstable, and the system easily phase separates to regions rich in A and B. ********
145
T Tc x eq(T) x sp(T)
unstable
metastable
metastable
1
1
x
Problems for Chapter VI - Quantum Statistical Mechanics 1. One dimensional chain: A chain of N +1 particles of mass m is connected by N massless springs of spring constant K and relaxed length a. The first and last particles are held fixed at the equilibrium separation of N a. Let us denote the longitudinal displacements of the particles from their equilibrium positions by {ui }, with u0 = uN = 0 since the end particles are fixed. The Hamiltonian governing {ui }, and the conjugate momenta {pi }, is H=
N−1 X i=1
' # N−2 X p2i K 2 u21 + (ui+1 − ui ) + u2N−1 . + 2m 2 i=1
(a) Using the appropriate (sine) Fourier transforms, find the normal modes {˜ uk }, and the corresponding frequencies {ωk }.
• From the Hamiltonian H=
N−1 X i=1
# ' N−1 X K p2i 2 (ui − ui−1 ) + u2N−1 , u21 + + 2m 2 i=2
the classical equations of motion are obtained as m
d 2 uj = −K(uj − uj−1 ) − K(uj − uj+1 ) = K(uj−1 − 2uj + uj+1 ), dt2 146
for j = 1, 2, · · · , N − 1, and with u0 = uN = 0. In a normal mode, the particles oscillate in phase. The usual procedure is to obtain the modes, and corresponding frequencies,
by diagonalizing the matrix of coefficeints coupling the displacements on the right hand side of the equation of motion. For any linear system, we have md2 ui /dt2 = Kij uj , and
we must diagonalize Kij . In the above example, Kij is only a function of the difference
i − j. This is a consequence of translational symmetry, and allows us to diagonalize the
matrix using Fourier modes. Due to the boundary conditions in this case, the appropriate transformation involves the sine, and the motion of the j-th particle in a normal mode is given by r
2 ±iωn t e sin (k(n) · j) . N The origin of time is arbitrary, but to ensure that uN = 0, we must set u ˜k(n) (j) =
k(n) ≡
nπ , N
n = 1, 2, · · · , N − 1.
for
Larger values of n give wave-vectors that are simply shifted by a multiple of π, and hence coincide with one of the above normal modes. The number of normal modes thus equals the number of original displacement variables, as required. Furthermore, the amplitudes are chosen such that the normal modes are also orthonormal, i.e. N−1 X j=1
u ˜k(n) (j) · u ˜k(m) (j) = δn,m .
By substituting the normal modes into the equations of motion we obtain the dispersion relation
where ω0 ≡
0010 nπ 0011 h 0010 nπ 0011i = ω02 sin2 , ωn2 = 2ω02 1 − cos N 2N
p K/m.
The potential energy for each normal mode is given by N N h nπ io2 KX K X n 0010 nπ 0011 2 Un = |ui − ui−1 | = sin i − sin (i − 1) 2 i=1 N i=1 N N 0014 0012 00130015 N 0011X 0010 1 4K 2 nπ 2 nπ i− . cos sin = N 2N i=1 N 2
Noting that N X i=1
cos
2
0014
nπ N
0012
1 i− 2
00130015
N
h nπ io N 1 Xn 1 + cos = (2i − 1) = , 2 i=1 N 2 147
we have 2
Uk(n) = 2K sin
0010 nπ 0011 2N
.
(b) Express the Hamiltonian in terms of the amplitudes of normal modes {˜ uk }, and evaluate the classical partition function. (You may integrate the {ui } from −∞ to +∞). • Before evaluating the classical partition function, lets evaluate the potential energy by first expanding the displacement using the basis of normal modes, as uj =
N−1 X n=1
an · u ˜k(n) (j).
The expression for the total potential energy is (N−1 )2 N N X X 0002 0003 K KX . (ui − ui−1 )2 = an u ˜k(n) (j) − u ˜k(n) (j − 1) U= 2 i=1 2 i=1 n=1
Since N−1 X j=1
u ˜k(n) (j) · u ˜k(m) (j − 1) =
N−1 X 1 δn,m {− cos [k(n)(2j − 1)] + cos k(n)} = δn,m cos k(n), N j=1
the total potential energy has the equivalent forms U=
=
N−1 N X KX 2 (ui − ui−1 ) = K a2n (1 − cos k(n)) , 2 i=1 n=1
N−1 X i=1
a2k(n) ε2k(n)
= 2K
N−1 X i=1
a2k(n)
2
sin
0010 nπ 0011 2N
.
The next step is to change the coordinates of phase space from uj to an . The Jacobian associated with this change of variables is unity, and the classical partition function is now obtained from # ' Z ∞ Z ∞ N−1 0011 0010 X 1 2 nπ 2 , da1 · · · daN−1 exp −2βK an sin Z = N−1 λ 2N −∞ −∞ n=1 √ where λ = h/ 2πmkB T corresponds to the contribution to the partition function from each momentum coordinate. Performing the Gaussian integrals, we obtain N−1 001aZ ∞ 0010 0011i001b h 1 Y 2 nπ 2 , dan exp −2βKan sin Z = N−1 λ 2N −∞ n=1 0013 N2−1 N−1 0012 Y h 0010 nπ 0011i−1 1 πkB T = N−1 . sin λ 2K 2N n=1 148
D
2
(c) First evaluate |˜ uk |
E
, and use the result to calculate u2i . Plot the resulting squared
displacement of each particle as a function of its equilibrium position. • The average squared amplitude of each normal mode is R∞
00010003 0002 2 nπ 2 2 da (a ) exp −2βKa sin n n n 2N −∞ 00010003 0002 R∞ 2 sin2 nπ da exp −2βKa n n 2N −∞ h 0010 nπ 0011i−1 kB T 1 0001 = 4βK sin2 = 2 nπ . 2N 4K sin 2N
2 an =
The variation of the displacement is then given by
2 uj =
*'N−1 X
2 = N
n=1
an u ˜n (j)
#2 +
=
N−1 X
N−1 X n=1
2 2 an u ˜n (j)
N−1
2 2 0010 nπ 0011 kB T X sin2 j = an sin N 2KN n=1 sin2 n=1
nπ j N 0001 nπ 2N
0001
.
The evaluation of the above sum is considerably simplified by considering the combination N−1
2 2
2 kB T X 2 cos uj+1 + uj−1 − 2 uj = 2KN n=1
0002 2nπ 0003 0002 2nπ 0003 0002 2nπ 0003 j − cos (j + 1) − cos (j − 1) N N N 0001 1 − cos nπ N 0001 0002 00010003 N−1 1 − cos nπ kB T X 2 cos 2nπ kB T N j 0001 N = , =− nπ 2KN n=1 KN 1 − cos N
PN−1
cos(πn/N ) = −1. It is easy to check that subject to the
n=1
boundary conditions of u20 = u2N = 0, the solution to the above recursion relation is
where we have used
2 kB T j(N − j) . uj = K N
(d) How are the results modified if only the first particle is fixed (u0 = 0), while the other end is free (uN 6= 0)? (Note that this is a much simpler problem as the partition function can be evaluated by changing variables to the N − 1 spring extensions.)
• When the last particle is free, the overall potential energy is the sum of the contributions PN−1 of each spring, i.e. U = K j=1 (uj − uj−1 )2 /2. Thus each extension can be treated
independently, and we introduce a new set of independent variables ∆uj ≡ uj − uj−1 . (In 149
Amplitude squared
NkT/K B
free end
fixed end 0
N/2 Position j
0
N
the previous case, where the two ends are fixed, these variables were not independent.) The partition function can be calculated separately for each spring as   Z ∞ Z ∞ N−1 X K 1 (uj − uj−1 )2  du1 · · · duN−1 exp − Z = N−1 λ 2k T B −∞ −∞ j=1   0013(N−1)/2 0012 Z ∞ Z ∞ N−1 X K 2πk T 1 B 2 d∆uN−1 exp − . d∆u1 · · · ∆uj  = = N−1 λ 2kB T λ2 K −∞ −∞ j=1
For each spring extension, we have
2
kB T ∆uj = (uj − uj−1 )2 = . K The displacement
uj =
j X
∆ui ,
i=1
is a sum of independent random variables, leading to the variance * j !2 + j X X
2 kB T 2 uj = (∆ui ) = ∆ui = j. K i=1
i=1
The results for displacements of open and closed chains are compared in the above figure. ******** 2. Black hole thermodynamics: According to Bekenstein and Hawking, the entropy of a black hole is proportional to its area A, and given by S=
kB c3 A . 4G¯h 150
(a) Calculate the escape velocity at a radius R from a mass M using classical mechanics. Find the relationship between the radius and mass of a black hole by setting this escape velocity to the speed of light c. (Relativistic calculations do not modify this result which was originally obtained by Laplace.) • The classical escape velocity is obtained by equating the gravitational energy and the
kinetic energy on the surface as,
2 Mm mvE G = , R 2
leading to vE =
r
2GM . r
Setting the escape velocity to the speed of light, we find R=
2G M. c2
For a mass larger than given by this ratio (i.e. M > c2 R/2G), nothing will escape from distances closer than R. (b) Does entropy increase or decrease when two black holes collapse into one? What is the entropy change for the universe (in equivalent number of bits of information), when two solar mass black holes (M⊙ ≈ 2 × 1030 kg) coalesce?
• When two black holes of mass M collapse into one, the entropy change is 0001 kB c3 kB c3 ∆S = S2 − 2S1 = (A2 − 2A1 ) = 4π R22 − 2R12 4G¯h '0012 4G¯h 0013 00132 # 0012 2 3 πkB c 2G 8πGkB M 2 2G = 2M M > 0. − 2 = G¯h c2 c2 c¯h Thus the merging of black holes increases the entropy of the universe. Consider the coalescence of two solar mass black holes. The entropy change is 2 8πGkB M⊙ c¯h 8π · 6.7 × 10−11 (N · m2 /kg 2 ) · 1.38 × 10−23 (J/K) · (2 × 1030 )2 kg 2 ≈ 3 × 108 (m/s) · 1.05 × 10−34 (J · s)
∆S =
≈ 3 × 1054 (J/K).
151
In units of bits, the information lost is NI =
∆S ln 2 = 1.5 × 1077 . kB
(c) The internal energy of the black hole is given by the Einstein relation, E = M c2 . Find the temperature of the black hole in terms of its mass. • Using the thermodynamic definition of temperature
1 T
=
∂S ∂E ,
and the Einstein relation
2
E = Mc ,
1 ∂ 1 = 2 T c ∂M
'
3
kB c 4π 4G¯h
0012
2G M c2
00132 #
8πkB G = M, ¯hc3
=⇒
¯ c3 1 h T = . 8πkB G M
(d) A “black hole” actually emits thermal radiation due to pair creation processes on its event horizon. Find the rate of energy loss due to such radiation. • The (quantum) vacuum undergoes fluctuations in which particle–antiparticle pairs are
constantly created and destroyed. Near the boundary of a black hole, sometimes one member of a pair falls into the black hole while the other escapes. This is a hand-waving explanation for the emission of radiation from black holes. The decrease in energy E of a black body of area A at temperature T is given by the Stefan-Boltzmann law, 1 ∂E = −σT 4 , A ∂t
where
4 π 2 kB σ= . 60¯ h3 c2
(e) Find the amount of time it takes an isolated black hole to evaporate. How long is this time for a black hole of solar mass? • Using the result in part (d) we can calculate the time it takes a black hole to evaporate.
For a black hole 2
A = 4πR = 4π Hence
which implies that
0012
2G M c2
00132
=
16πG2 2 M , c4
4 0001 d π 2 kB M c2 = − dt 60¯ h3 c2
M2
0012
E = M c2 ,
16πG2 2 M c4
00130012
¯ c3 1 h 8πkB G M
dM ¯hc4 =− ≡ −b. dt 15360G2 152
and T =
00134
,
¯ c3 1 h . 8πkB G M
This can be solved to give M (t) = M03 − 3bt
00011/3
.
The mass goes to zero, and the black hole evaporates after a time τ=
M03 5120G2 M ⊙3 = ≈ 2.2 × 1074 s, 3b ¯hc4
which is considerably longer than the current age of the universe (approximately ×1018 s). (f) What is the mass of a black hole that is in thermal equilibrium with the current cosmic background radiation at T = 2.7K? • The temperature and mass of a black hole are related by M = h ¯ c3 /(8πkB GT ). For a
black hole in thermal equilibrium with the current cosmic background radiation at T = 2.7◦ K, M≈
1.05 × 10−34 (J · s)(3 × 108 )3 (m/s)3 ≈ 4.5 × 1022 kg. 8π · 1.38 × 10−23 (J/K) · 6.7 × 10−11 (N · m2 /kg 2 ) · 2.7◦ K
(g) Consider a spherical volume of space of radius R. According to the recently formulated Holographic Principle there is a maximum to the amount of entropy that this volume of space can have, independent of its contents! What is this maximal entropy? • The mass inside the spherical volume of radius R must be less than the mass that would
make a black hole that fills this volume. Bring in additional mass (from infinity) inside
the volume, so as to make a volume-filling balck hole. Clearly the entropy of the system will increase in the process, and the final entropy, which is the entropy of the black hole is larger than the initial entropy in the volume, leading to the inequality S ≤ SBH =
kB c3 A, 4G¯h
where A = 4πR2 is the area enclosing the volume. The surprising observation is that the upper bound on the entropy is proportional to area, whereas for any system of particles we expect the entropy to be proportional to N . This should remain valid even at very high temperatures when interactions are unimportant. The ‘holographic principle’ is an allusion to the observation that it appears as if the degrees of freedom are living on the surface of the system, rather than its volume. It was formulated in the context of string theory which attempts to construct a consistent theory of quantum gravity, which replaces particles as degrees of freedom, with strings. 153
******** 3. Quantum harmonic oscillator: Consider a single harmonic oscillator with the Hamiltonian H=
p2 mω 2 q 2 + , 2m 2
with p =
¯ d h i dq
.
(a) Find the partition function Z, at a temperature T , and calculate the energy hHi. • The partition function Z, at a temperature T , is given by Z = tr ρ =
X
e−βEn .
n
As the energy levels for a harmonic oscillator are given by 0013 0012 1 , ǫn = h ¯ω n + 2 the partition function is Z=
X n
=
0014
0012 00130015 1 exp −β¯hω n + = e−β¯hω/2 + e−3β¯hω/2 + · · · 2
1 1 . = 2 sinh (β¯hω/2) eβ¯hω/2 − e−β¯hω/2
The expectation value of the energy is 0012 0013 0012 0013 ∂ ln Z 1 ¯hω cosh(β¯hω/2) ¯hω hHi = − = = . ∂β 2 sinh(β¯hω/2) 2 tanh(β¯hω/2) (b) Write down the formal expression for the canonical density matrix ρ in terms of the eigenstates ({|ni}), and energy levels ({ǫn }) of H.
• Using the formal representation of the energy eigenstates, the density matrix ρ is ! 0013 X 00130015 0014 0012 0012 1 β¯hω < n| . |n > exp −β¯hω n + ρ = 2 sinh 2 2 n In the coordinate representation, the eigenfunctions are in fact given by 0012 20013 0010 mω 00111/4 H (ξ) ξ n √ hn|qi = exp − , n π¯h 2 2 n! 154
where ξ≡ with
r
mω q, ¯h
0012
0013n
d dξ
exp(−ξ 2 ) Hn (ξ) = (−1) exp(ξ ) Z exp(ξ 2 ) ∞ (−2iu)n exp(−u2 + 2iξu)du. = π −∞ n
2
For example, H0 (ξ) = 1,
and
H1 (ξ) = − exp(ξ 2 )
d exp(−ξ 2 ) = 2ξ, dξ
result in the eigenstates h0|qi = and h1|qi =
0010 mω 00111/4 π¯h
0010 mω 0011 exp − q2 , 2¯ h
0010 mω 00111/4 r 2mω
0010 mω 0011 q · exp − q2 . ¯h 2¯ h
π¯ h
Using the above expressions, the matrix elements are obtained as
00010003 0002 1 · hq ′ |ni hn|qi exp −β¯ h ω n + 2 00010003 0002 hq ′ |ρ|qi = hq ′ |n′ i hn′ |ρ|ni hn|qi = n P 1 exp −β¯ h ω n + ′ n 2 n,n 0012 0013 X 0014 0012 00130015 β¯hω 1 = 2 sinh · exp −β¯hω n + · hq ′ |ni hn|qi . 2 2 n P
X
(c) Show that for a general operator A(x), ∂A ∂ exp [A(x)] 6= exp [A(x)] , ∂x ∂x while in all cases ∂ tr {exp [A(x)]} = tr ∂x • By definition
unless
001a
155
0015 ∂A = 0, A, ∂x
001b ∂A exp [A(x)] . ∂x
∞ X 1 n A , e = n! n=0 A
0014
and
But for a product of n operators,
The
∂A ∂x
∞ X ∂eA 1 ∂An = . ∂x n! ∂x n=0
∂A ∂A ∂A ∂ (A · A · · · A) = · A···A + A · ···A + ··· + A · A··· . ∂x ∂x ∂x ∂x 0002 0003 can be moved through the A′ s surrounding it only if A, ∂A ∂x = 0, in which case ∂A n−1 ∂A =n A , ∂x ∂x
and
∂eA ∂A A = e . ∂x ∂x
However, as we can always reorder operators inside a trace, i.e. tr(BC) = tr(CB), and
0013 0012 0013 0012 ∂A ∂A n−1 , tr A · · · A · · · · · · A = tr ·A ∂x ∂x
and the identity 0013 ∂A A , ·e ∂x 0002 0003 can always be satisfied, independent of any constraint on A, ∂A . ∂x 0001 ∂ tr eA = tr ∂x
0012
(d) Note that the partition function calculated in part (a) does not depend on the mass m, i.e. ∂Z/∂m = 0. Use this information, along with the result in part (c), to show that 001c 2001d 001c 001d p mω 2 q 2 = . 2m 2 • The expectation values of the kinetic and potential energy are given by 0012 2 0013 001d 0012 001c 0013 001c 2001d p mω 2 q 2 mω 2 q 2 p = tr = tr ρ , and ρ . 2m 2m 2 2 Noting that the expression for the partition function derived in 0011 part (a) is independent of 0010 mass, we know that ∂Z/∂m = 0. Starting with Z = tr e−β H , and differentiating 0015 0014 0011 0010 ∂Z ∂ ∂ −β H −β H = 0, = tr = tr e (−βH)e ∂m ∂m ∂m
where we have used the result in part (c). Differentiating the Hamiltonian, we find that 0015 0014 0015 0014 mω 2 q 2 −β H p2 −β H + tr −β = 0. e e tr β 2m2 2 156
Equivalently, 0015 0014 0015 p2 −β H mω 2 q 2 −β H tr = tr , e e 2m 2 0014
which shows that the expectation values of kinetic and potential energies are equal.
(e) Using the results in parts (d) and (a), or otherwise, calculate q 2 . How are the results in Problem #1 modified at low temperatures by inclusion of quantum mechanical effects. −1
• In part (a) it was found that hHi = (¯ hω/2) (tanh(β¯hω/2)) . Note that hHi =
2
p /2m + mω 2 q 2 /2 , and that in part (d) it was determined that the contribution from
the kinetic and potential energy terms are equal. Hence,
1 −1 hω/2) (tanh(β¯hω/2)) . mω 2 q 2 /2 = (¯ 2
Solving for q 2 ,
2 q =
¯ h ¯h −1 (tanh(β¯hω/2)) = coth(β¯hω/2). 2mω 2mω
While the classical result q 2 = kB T /mω 2 , vanishes as T → 0, the quantum result satu rates at T = 0 to a constant value of q 2 = h ¯ /(2mω). The amplitude of the displacement curves in Problem #1 are effected by exactly the same saturation factors.
(f) In a coordinate representation, calculate hq ′ |ρ|qi in the high temperature limit. One approach is to use the result 0002 0003 exp(βA) exp(βB) = exp β(A + B) + β 2 [A, B]/2 + O(β 3 ) . • Using the general operator identity 0002 0003 exp(βA) exp(βB) = exp β(A + B) + β 2 [A, B]/2 + O(β 3 ) , the Boltzmann operator can be decomposed in the high temperature limit into those for kinetic and potential energy; to the lowest order as 0013 0012 mω 2 q 2 p2 ≈ exp(−βp2 /2m) · exp(−βmω 2 q 2 /2). −β exp −β 2m 2 157
The first term is the Boltzmann operator for an ideal gas. The second term contains an operator diagonalized by |q >. The density matrix element < q ′ |ρ|q > =< q ′ | exp(−βp2 /2m) exp(−βmω 2 q 2 /2)|q > Z = dp′ < q ′ | exp(−βp2 /2m)|p′ >< p′ | exp(−βmω 2 q 2 /2)|q > Z = dp′ < q ′ |p′ >< p′ |q > exp(−βp′2 /2m) exp(−βq 2 mω 2 /2). Using the free particle basis < q ′ |p′ >=
h √ 1 e−iq·p/¯ , 2π¯ h
Z ′ ′ ′2 2 2 1 dp′ eip (q−q )/¯h e−βp /2m e−βq mω /2 < q |ρ|q >= 2π¯h  !2  r r 0013 0012 Z β 2m i 1 2m −βq 2 mω 2 /2 1 ′ ′ ′ ′ 2 =e dp exp − p + (q − q )  exp − (q − q ) , 2π¯h 2m 2¯ h β 4 β¯h2 ′
where we completed the square. Hence
0015 0014 1 −βq 2 mω2 /2 p mkB T ′ 2 2πmkB T exp − < q |ρ|q >= e (q − q ) . 2π¯h 2¯ h2 ′
The proper normalization in the high temperature limit is Z 2 2 2 Z = dq < q|e−βp /2m · e−βmω q /2 |q > Z Z 2 2 2 = dq dp′ < q|e−βp /2m |p′ >< p′ |e−βmω q /2 |q > Z Z ′2 2 2 kB T 2 = dq dp |< q|p >| e−βp /2m e−βmω q /2 = . ¯hω Hence the properly normalized matrix element in the high temperature limit is s 0013 0014 0015 0012 mω 2 mkB T mω 2 2 ′ ′ 2 < q |ρ|q >lim T →∞ = exp − q exp − (q − q ) . 2πkB T 2kB T 2¯ h2 (g) At low temperatures, ρ is dominated by low energy states. Use the ground state wave-function to evaluate the limiting behavior of hq ′ |ρ|qi as T → 0.
• In the low temperature limit, we retain only the first terms in the summation ρlim T →0 ≈
|0 > e−β¯hω/2 < 0| + |1 > e−3β¯hω/2 < 1| + · · · . e−β¯hω/2 + e−3β¯hω/2 158
Retaining only the term for the ground state in the numerator, but evaluating the geometric series in the denominator, 0011 0010 < q ′ |ρ|q >lim T →0 ≈< q ′ |0 >< 0|q > e−β¯hω/2 · eβ¯hω/2 − e−β¯hω/2 . Using the expression for < q|0 > given in part (b), ′
< q |ρ|q >lim T →0 ≈
r
h mω 0001i 0001 mω q 2 + q ′2 1 − e−β¯hω . exp − π¯h 2¯ h
(h) Calculate the exact expression for hq ′ |ρ|qi.
********
4. Relativistic Coulomb gas:
Consider a quantum system of N positive, and N negative
charged relativistic particles in box of volume V = L3 . The Hamiltonian is H=
2N X i=1
c|~ pi | +
2N X i −1, or d > s. Therefore, Bose-Einstein condensation occurs for d > s. For a two dimensional gas, d = s = 2, the integral diverges logarithmically, and hence Bose-Einstein condensation does not occur. ******** 3. Pauli paramagnetism: Calculate the contribution of electron spin to its magnetic susceptibility as follows. Consider non-interacting electrons, each subject to a Hamiltonian p~ 2 ~ , − µ0 ~σ · B 2m ~ are ±B. where µ0 = e¯h/2mc, and the eigenvalues of ~σ · B ~ has been ignored.) (The orbital effect, p~ → ~p − eA, H1 =
(a) Calculate the grand potential G − = −kB T ln Q− , at a chemical potential µ. • The energy of the electron gas is given by X − E≡ Ep (n+ p , np ), p
where n± p (= 0 or 1), denote the number of particles having ± spins and momentum p, and 0012 2 0013 0012 2 0013 p p + − + Ep (np , np ) ≡ − µ0 B np + + µ0 B n− p 2m 2m 2 − − p − (n+ = (n+ + n ) p − np )µ0 B. p p 2m The grand partition function of the system is P + − N= (n +n ) ∞ Xp p X 0002 0003 − exp −βEp (n+ Q= exp(−βµN ) p , np ) − {n+ p ,np }
N=0
X
=
− {n+ p ,np }
=
0002 0001 00010003 − + − exp βµ n+ p + np − βEp np , np
X
Y001a
00130015001b 001a 0014 0012 00130015001b 0014 0012 p2 p2 · 1 + exp β µ + µ0 B − 1 + exp β µ − µ0 B − 2m 2m
p {n+ ,n− } p p
=
001a 00140012 0013 0012 0013 0015001b p2 p2 + exp β µ − µ0 B − np + µ + µ0 B − n− p 2m 2m
Y p
= Q0 (µ + µ0 B) · Q0 (µ − µ0 B) , 176
where
0014 0012 00130015001b Y001a p2 Q0 (µ) ≡ 1 + exp β µ − . 2m p
Thus ln Q = ln Q0 (µ + µ0 B) + ln Q0 (µ − µ0 B) .
Each contribution is given by
0013 0012 Z 0002 p2 0001 V p2 0003 3 −β 2m ) = d p ln 1 + ze ln Q0 (µ) = ln 1 + exp β(µ − 2m (2π¯h)3 p r Z √ V 4πm 2m dx x ln(1 + ze−x ), where z ≡ eβµ , = 3 h β β X
and integrating by parts yields ln Q0 (µ) = V
0012
2πmkB T h2
00133/2
2 2 √ π3
Z
dx
x3/2 V − = f (z). z −1 ex + 1 λ3 5/2
The total grand free energy is obtained from ln Q(µ) = as
0001i 0001 V h − − −βµ0 B βµ0 B , ze ze + f f 5/2 λ3 5/2
G = −kB T ln Q(µ) = −kB T
0001i 0001 V h − − −βµ0 B βµ0 B . ze ze + f f 5/2 λ3 5/2
(b) Calculate the densities n+ = N+ /V , and n− = N− /V , of electrons pointing parallel and antiparallel to the field. • The number densities of electrons with up or down spins is given by
where we used
0001 N± ∂ V − ze±βµ0 B , =z ln Q± = 3 f3/2 V ∂z λ z
∂ − − f (z) = fn−1 (z). ∂z n
The total number of electrons is the sum of these, i.e. 0015 0014 0001 0001 V − − −βµ0 B βµ0 B . + f3/2 ze N = N+ + N− = 3 f3/2 ze λ 177
(c) Obtain the expression for the magnetization M = µ0 (N+ − N− ), and expand the result for small B.
• The magnetization is related to the difference between numbers of spin up and down
electrons as
0001 0001i V h − − βµ0 B −βµ0 B − f3/2 ze . M = µ0 (N+ − N− ) = µ0 3 f3/2 ze λ
Expanding the results for small B, gives
0001 ∂ − − − − (z) ± z · βµ0 B f3/2 [z (1 ± βµ0 B)] ≈ f3/2 ze±βµ0 B ≈ f3/2 f3/2 (z), ∂z
which results in
M = µ0
V 2µ20 V − − (z) = (z). (2βµ B) · f · B · f1/2 0 1/2 3 3 λ kB T λ
(d) Sketch the zero field susceptibility χ(T ) = ∂M/∂B|B=0 , and indicate its behavior at low and high temperatures. • The magnetic susceptibility is 2µ20 V ∂M · f − (z), = χ≡ ∂B B=0 kB T λ3 1/2
with z given by,
N =2
V − · f3/2 (z). 3 λ
In the low temperature limit, (ln z = βµ → ∞) Z ln(z) n 1 [ln(z)] n−1 dx x = , Γ(n) 0 nΓ(n) T →0 0 00132/3 0012 √ 3N π 3 V 4(ln z)3/2 √ , λ , =⇒ ln z = N =2 3 · λ 8V 3 π 00131/3 00131/3 00131/3 0012 0012 0012 √ 2µ2o V 4µ20 V 4πmµ20 V 3N 3N 3N π 3 = χ= √ = . λ · · · πkB T λ3 8V kB T λ2 πV h2 πV
fn− (z)
1 = Γ(n)
Z
∞
xn−1 dx 1 + ex−ln(z)
≈
Their ratio of the last two expressions gives − µ20 f1/2 3µ20 1 3µ20 1 χ 3µ20 = = = . = − N T →0 kB T f3/2 2kB T ln(z) 2kB T βεF 2kB TF 178
In the high temperature limit (z → 0), →
z fn (z) z→0 Γ(n)
Z
∞
dx xn−1 e−x = z,
0
and thus → 2V N · z, 3 β→0 λ
=⇒
N N z≈ · λ3 = · 2V 2V
0012
h2 2πmkB T
00133/2
→ 0,
which is consistent with β → 0. Using this result, χ≈ The result
2µ20 V N µ20 · z = . kB T λ3 kB T
0010χ0011 N
T →∞
=
µ20 , kB T
is known as the Curie susceptibility. χ/Nµο2
χ/Nµο2 ∼ 1/k BT
3/2k BTF
3/2k BTF
1/k BT
(e) Estimate the magnitude of χ/N for a typical metal at room temperature. • Since TRoom ≪ TF ≈ 104 K, we can take the low T limit for χ (see(d)), and 3µ20 3 × (9.3 × 10−24 )2 χ = ≈ ≈ 9.4 × 10−24 J/T2 , N 2kB TF 2 × 1.38 × 10−23 179
where we used µ0 =
eh ≃ 9.3 × 10−24 J/T. 2mc ********
4. Freezing of He3 :
At low temperatures He3 can be converted from liquid to solid by
application of pressure. A peculiar feature of its phase boundary is that (dP/dT )melting is negative at temperatures below 0.3 K [(dP/dT )m ≈ −30atm K−1 at T ≈ 0.1 K]. We will
use a simple model of liquid and solid phases of He3 to account for this feature.
(a) In the solid phase, the He3 atoms form a crystal lattice. Each atom has nuclear spin of 1/2. Ignoring the interaction between spins, what is the entropy per particle ss , due to the spin degrees of freedom? • Entropy of solid He3 comes from the nuclear spin degeneracies, and is given by Ss kB ln(2N ) ss = = = kB ln 2. N N
(b) Liquid He3 is modelled as an ideal Fermi gas, with a volume of 46˚ A3 per atom. What is its Fermi temperature TF , in degrees Kelvin? • The Fermi temperature for liquid 3 He may be obtained from its density as εF h2 TF = = kB 2mkB
0012
3N 8πV
00132/3
(6.7 × 10−34 )2 ≈ 2 · (6.8 × 10−27 )(1.38 × 10−23 )
0012
3 8π × 46 × 10−30
00132/3
≈ 9.2 K.
(c) How does the heat capacity of liquid He3 behave at low temperatures? Write down an expression for CV in terms of N, T, kB , TF , up to a numerical constant, that is valid for T ≪ TF .
• The heat capacity comes from the excited states at the fermi surface, and is given by CV = kB
π2 2 3N π2 T π2 kB T D(εF ) = kB T = N kB . 6 6 2kB TF 4 TF
180
(d) Using the result in (c), calculate the entropy per particle sℓ , in the liquid at low temperatures. For T ≪ TF , which phase (solid or liquid) has the higher entropy? • The entropy can be obtained from the heat capacity as T dS , CV = dT
⇒
1 sℓ = N
Z
T 0
CV dT π2 T = kB . T 4 TF
As T → 0, sℓ → 0, while ss remains finite. This is an unusual situation in which the solid
has more entropy than the liquid! (The finite entropy is due to treating the nuclear spins
as independent. There is actually a weak coupling between spins which causes magnetic ordering at a much lower temperature, removing the finite entropy.) (e) By equating chemical potentials, or by any other technique, prove the Clausius– Clapeyron equation (dP/dT )melting = (sℓ − ss )/(vℓ − vs ), where vℓ and vs are the volumes
per particle in the liquid and solid phases respectively.
• The Clausius-Clapeyron equation can be obtained by equating the chemical potentials at the phase boundary,
µℓ (T, P ) = µs (T, P ),
and µℓ (T + ∆T, P + ∆P ) = µs (T + ∆T, P + ∆P ).
Expanding the second equation, and using the thermodynamic identities 0012 0013 0013 0012 ∂µ S V ∂µ = − , and = , ∂T P N ∂P T N results in
0012
∂P ∂T
0013
=
melting
sℓ − ss . vℓ − vs
(f) It is found experimentally that vℓ − vs = 3˚ A3 per atom. Using this information, plus the results obtained in previous parts, estimate (dP/dT )melting at T ≪ TF .
• The negative slope of the phase boundary results from the solid having more entropy than the liquid, and can be calculated from the Clausius-Clapeyron relation 0010 0011 π2 T 0012 0013 − ln 2 4 TF ∂P sℓ − ss = ≈ kB . ∂T melting vℓ − vs vℓ − vs
Using the values, T = 0.1 K, TF = 9.2 J K, and vℓ − vs = 3 ˚ A3 , we estimate 0012 0013 ∂P ≈ −2.7 × 106 Pa ◦ K−1 , ∂T melting 181
in reasonable agreement with the observations. ******** 5. Non-interacting fermions:
Consider a grand canonical ensemble of non-interacting
fermions with chemical potential µ. The one–particle states are labelled by a wavevector ~k, and have energies E(~k). (a) What is the joint probability P ( n~k ), of finding a set of occupation numbers n~k , of the one–particle states?
• In the grand canonical ensemble with chemical potential µ, the joint probability of finding a set of occupation numbers n~k , for one–particle states of energies E(~k) is given by the Fermi distribution
i h ~ exp β(µ − E( k))n Y ~ k h i, P ( n~k ) = ~ 1 + exp β(µ − E(k)) ~ k
where
n~k = 0 or 1,
for each ~k.
(b) Express your answer to part (a) in terms of the average occupation numbers • The average occupation numbers are given by h i ~k)) exp β(µ − E(
h i, n~k − = ~ 1 + exp β(µ − E(k))
from which we obtain
h i exp β(µ − E(~k)) =
n o n~k − .
n~k −
. 1 − n~k −
This enables us to write the joint probability as 0015 Y 00140010 0011n~k 0010
00111−n~k . 1 − n~k − P ( n~k ) = n~k − ~ k
(c) A random variable has a set of ℓ discrete outcomes with probabilities pn , where n = 1, 2, · · · , ℓ. What is the entropy of this probability distribution? What is the maximum possible entropy?
• A random variable has a set of ℓ discrete outcomes with probabilities pn . The entropy of this probability distribution is calculated from S = −kB
ℓ X
pn ln pn
n=1
182
.
The maximum entropy is obtained if all probabilities are equal, pn = 1/ℓ, and given by Smax = kB ln ℓ. (d) Calculate the entropy of the probability distribution for fermion occupation numbers in part (b), and comment on its zero temperature limit. • Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by S = −kB
0010 X h
0011i
0011 0010 n~k − ln n~k − + 1 − n~k − ln 1 − n~k − . ~ k
In the zero temperature limit all occupation numbers are either 0 or 1. In either case the contribution to entropy is zero, and the fermi system at T = 0 has zero entropy.
(e) Calculate the variance of the total number of particles N 2 c , and comment on its zero temperature behavior.
• The total number of particles is given by N =
are independent
P
~ k
n~k . Since the occupation numbers
0013 X X 0012D E XD E
0011
0010
2
2 2 2 n~k − 1 − n~k − , − n~k − = n~k n~k = N c= ~ k
since
D
n~2k
vanishes.
E
−
c
~ k
−
~ k
= n~k − . Again, since at T = 0, n~k − = 0 or 1, the variance N 2 c
(f) The number fluctuations of a gas is related to its compressibility κT , and number density n = N/V , by
2 N c = N nkB T κT
.
Give a numerical estimate of the compressibility of the fermi gas in a metal at T = 0 in units of ˚ A3 eV −1 .
• To obtain the compressibility from N 2 c = N nkB T κT , we need to examine the behavior
at small but finite temperatures. At small but finite T , a small fraction of states around the fermi energy have occupation numbers around 1/2. The number of such states is roughly N kB T /εF , and hence we can estimate the variance as
2 1 N kB T . N c = N nkB T κT ≈ × 4 εF 183
The compressibility is then approximates as κT ≈
1 , 4nεF
where n = N/V is the density. For electrons in a typical metal n ≈ 1029 m−3 ≈ 0.1˚ A3 , and
εF ≈ 5eV ≈ 5 × 104 ◦ K, resulting in
κT ≈ 0.5˚ A3 eV −1 .
******** 6. Stoner ferromagnetism:
The conduction electrons in a metal can be treated as a
gas of fermions of spin 1/2 (with up/down degeneracy), and density n = N/V . The Coulomb repulsion favors wave functions which are antisymmetric in position coordinates, thus keeping the electrons apart. Because of the full (position and spin) antisymmetry of fermionic wave functions, this interaction may be approximated by an effective spin-spin coupling which favors states with parallel spins. In this simple approximation, the net effect is described by an interaction energy U =α
N+ N− , V
where N+ and N− = N − N+ are the numbers of electrons with up and down spins, and V is the volume. (The parameter α is related to the scattering length a by α = 4π¯h2 a/m.)
(a) The ground state has two fermi seas filled by the spin up and spin down electrons. Express the corresponding fermi wavevectors kF± in terms of the densities n± = N± /V . • In the ground state, all available wavevectors are filled up in a sphere. Using the appropriate density of states, the corresponding radii of kF± are calculated as N± = V
Z
k αc =
00012/3 h ¯ 2 −1/3 4 3π 2 n . 3 2m 185
(e) Explain qualitatively, and sketch the behavior of the spontaneous magnetization as a function of α. • For α > αc , the optimal value of δ is obtained by minimizing the energy density. Since the coefficient of the fourth order term is positive, and the optimal δ goes to zero continuously
as α → αc ; the minimum energy is obtained for a value of δ 2 ∝ (α−αc ). The magnetization √ is proportional to δ, and hence grows in the vicinity of αc as α − αc , as sketched below.
******** 7. Boson magnetism: Consider a gas of non-interacting spin 1 bosons, each subject to a Hamiltonian H1 (~ p, sz ) =
p~ 2 − µ0 sz B 2m
,
where µ0 = e¯h/mc, and sz takes three possible values of (-1, 0, +1). (The orbital effect, ~ has been ignored.) p~ → p~ − eA, (a) In a grand canonical ensemble of chemical potential µ, what are the average occupation n o numbers hn+ (~k)i, hn0 (~k)i, hn−(~k)i , of one-particle states of wavenumber ~k = p~/¯h?
• Average occupation numbers of the one-particle states in the grand canonical ensemble of chemical potential µ, are given by the Bose-Einstein distribution ns (~k) = =
1 eβ [H(s)−µ] − 1
,
(for s = −1, 0, 1)
1 h 0010 2 0011 i p exp β 2m − µ0 sB − βµ − 1
(b) Calculate the average total numbers {N+ , N0 , N− }, of bosons with the three possible
+ values of sz in terms of the functions fm (z).
186
• The total numbers of particles with spin s are given by Z X 1 V 3 h 0010 0011 i . d k Ns = ns (~k), =⇒ Ns = p2 (2π)3 exp β − µ sB − βµ − 1 0 ~ 2m {k} After a change of variables, k ≡ x1/2
√
Ns = where + fm (z)
1 ≡ Γ(m)
Z
0
∞
2mkB T /h, we get
0001 V + βµ0 sB f ze , λ3 3/2
dx xm−1 , z −1 ex − 1
λ≡ √
h , 2πmkB T
z ≡ eβµ .
(c) Write down the expression for the magnetization M (T, µ) = µ0 (N+ − N− ), and by expanding the result for small B find the zero field susceptibility χ(T, µ) = ∂M/∂B|B=0 . • The magnetization is obtained from M (T, µ) = µ0 (N+ − N− ) 0001i 0001 V h + + −βµ0 sB βµ0 B . − f3/2 ze = µ0 3 f3/2 ze λ
Expanding the result for small B gives
0001 ∂ + + + + (z) ± z · βµ0 B f3/2 (z[1 ± βµ0 B]) ≈ f3/2 ze±βµ0 B ≈ f3/2 f3/2 (z). ∂z
+ + Using zdfm (z)/dz = fm−1 (z), we obtain
M = µ0 and
V 2µ20 V + + (z) = (z), (2βµ B) · f · B · f1/2 0 1/2 λ3 kB T λ3 2µ20 V ∂M = · f + (z). χ≡ ∂B B=0 kB T λ3 1/2
To find the behavior of χ(T, n), where n = N/V is the total density, proceed as follows: (d) For B = 0, find the high temperature expansion for z(β, n) = eβµ , correct to second order in n. Hence obtain the first correction from quantum statistics to χ(T, n) at high temperatures. 187
+ • In the high temperature limit, z is small. Use the Taylor expansion for fm (z) to write
the total density n(B = 0), as
3 + N+ + N0 + N− (z) = 3 f3/2 n(B = 0) = V λ B=0 0013 0012 z3 3 z2 ≈ 3 z + 3/2 + 3/2 + · · · . λ 2 3 Inverting the above equation gives z=
0012
nλ3 3
0013
0012
1
−
23/2
nλ3 3
00132
+ ···.
The susceptibility is then calculated as 2µ20 V · f + (z), kB T λ3 1/2 0013 0012 2µ20 1 z2 χ/N = z + 1/2 + · · · kB T nλ3 2 00130012 30013 0015 0014 0012 2 0001 1 nλ 2µ0 1 2 . 1 + − 3/2 + 1/2 +O n = 3kB T 3 2 2 χ=
(e) Find the temperature Tc (n, B = 0), of Bose–Einstein condensation. What happens to χ(T, n) on approaching Tc (n) from the high temperature side? • Bose-Einstein condensation occurs when z = 1, at a density n=
3 + f (1), λ3 3/2
or a temperature h2 Tc (n) = 2πmkB
0012
n 3 ζ 3/2
00132/3
,
+ + (z) = ∞, the susceptibility χ(T, n) diverges (1) ≈ 2.61. Since limz→1 f1/2 where ζ 3/2 ≡ f3/2
on approaching Tc (n) from the high temperature side.
(f) What is the chemical potential µ for T < Tc (n), at a small but finite value of B? Which one-particle state has a macroscopic occupation number? 0002 0003−1 ~ • Since ns (~k, B) = z −1 eβ E s (k,B) −1 is a positive number for all ~k and sz , µ is bounded
above by the minimum possible energy, i.e. for
T < Tc ,
and
B finite,
zeβµ0 B = 1, 188
=⇒
µ = −µ0 B.
Hence the macroscopically occupied one particle state has ~k = 0, and sz = +1. (g) Using the result in (f), find the spontaneous magnetization, M (T, n) = lim M (T, n, B). B→0
• Contribution of the excited states to the magnetization vanishes as B → 0. Therefore the
total magnetization for T < Tc is due to the macroscopic occupation of the (k = 0, sz = +1) state, and M (T, n) = µ0 V n+ (k = 0) = µ0 V n − nexcited
0001
0013 0012 3V = µ0 N − 3 ζ 3/2 . λ
******** 8. Dirac fermions are non-interacting particles of spin 1/2. The one-particle states come in pairs of positive and negative energies, p E ± (~k) = ± m2 c4 + h ¯ 2 k 2 c2
,
independent of spin. (a) For any fermionic system of chemical potential µ, show that the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. (δ is any constant energy.)
• According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] =
eβ(µ−E )n , 1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ, p [n(µ + δ)] =
eβδn , 1 + eβδ
=⇒
p [n(µ + δ) = 1] =
eβδ 1 = . 1 + eβδ 1 + e−βδ
Similarly, for a state of energy µ − δ, p [n(µ − δ)] =
e−βδn , 1 + e−βδ
=⇒
p [n(µ − δ) = 0] = 189
1 = p [n(µ + δ) = 1] , 1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. (b) At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. Using the result in (a) find the chemical potential at finite temperatures T . • The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for an tem-
perature; any particle leaving an occupied negative energy state goes to the corresponding
unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enfrces µ(T ) = 0. (c) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) − E(0) = 4V
Z
d3~k E + (~k) 0010 0011 (2π)3 exp βE (~k) + 1
.
+
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated
as
E(T ) − E(0) =
X
k,sz
=2
[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]
X k
2 hn+ (k)i E + (k) = 4V
Z
d3~k E + (~k) 0010 0011 . (2π)3 exp βE (~k) + 1 +
(d) Evaluate the integral in part (c) for massless Dirac particles (i.e. for m = 0). • For m = 0, E + (k) = h ¯ c|k|, and ∞
4πk 2 dk ¯hck E(T ) − E(0) = 4V = (set β¯hck = x) 8π 3 eβ¯hck + 1 0 00133 Z ∞ 0012 2V x3 kB T = 2 kB T dx x π ¯hc e +1 0 0013 0012 3 kB T 7π 2 V kB T . = 60 ¯hc Z
For the final expression, we have noted that the needed integral is 3!f4− (1), and used the given value of f4− (1) = 7π 4 /720. (e) Calculate the heat capacity, CV , of such massless Dirac particles. 190
• The heat capacity can now be evaluated as 00133 0012 ∂E 7π 2 kB T CV = = . V kB ∂T V 15 ¯hc (f) Describe the qualitative dependence of the heat capacity at low temperature if the particles are massive. • When m 6= 0, there is an energy gap between occupied and empty states, and we thus
expect an exponentially activated energy, and hence heat capacity. For the low energy excitations, E + (k) ≈ mc2 + and thus
¯ 2 k2 h + ···, 2m
√ Z ∞ 2V 2 −βmc2 4π π dxx2 e−x E(T ) − E(0) ≈ 2 mc e π λ3 0 2 48 V = √ 3 mc2 e−βmc . πλ
The corresponding heat capacity, to leading order thus behaves as C(T ) ∝ kB
0001 V 2 2 −βmc2 βmc e . λ3 ********
9. Numerical estimates: The following table provides typical values for the Fermi energy and Fermi temperature for (i) Electrons in a typical metal; (ii) Nucleons in a heavy nucleus; ˚3 per atom). and (iii) He3 atoms in liquid He3 (atomic volume = 46.2A n(1/m3 )
m(Kg)
εF (eV)
TF (K)
electron
1029
4.4
nucleons
1044
9 × 10−31
5 × 104
liquid He3
2.6 × 1028
1.6 × 10−27
4.6 × 10−27
1.0 × 108 ×10−3
1.1 × 1012 101
(a) Estimate the ratio of the electron and phonon heat capacities at room temperature for a typical metal. • For an electron gas, TF ≈ 5 × 104 K, TF ≫ Troom ,
=⇒
Celectron π2 T ≈ · ≈ 0.025. N kB 2 TF 191
For the phonon gas in iron, the Debye temperature is TD ≈ 470K, and hence # ' 0012 00132 T 1 Cphonon + . . . ≈ 3, ≈3 1− N kB 20 TD resulting in
Celectron ≈ 8 × 10−3 . Cphonon
(b) Compare the thermal wavelength of a neutron at room temperature to the minimum wavelength of a phonon in a typical crystal. • Thermal wavelengths are given by λ≡ √
h . 2πmkB T
For a neutron at room temperature, using the values m = 1.67 × 10−27 kg,
T = 300 K,
kB = 1.38 × 10−23 JK−1 ,
h = 6.67 × 10−34 Js,
we obtain λ = 1˚ A. The typical wavelength of a phonon in a solid is λ = 0.01 m, which is much longer than the neutron wavelength. The minimum wavelength is, however, of the order of atomic spacing (3 − 5 ˚ A), which is comparable to the neutron thermal wavelength. (c) Estimate the degeneracy discriminant, nλ3 , for hydrogen, helium, and oxygen gases
at room temperature and pressure. At what temperatures do quantum mechanical effects become important for these gases? • Quantum mechanical effects become important if nλ3 ≥ 1. In the high temperature
limit the ideal gas law is valid, and the degeneracy criterion can be reexpressed in terms of pressure P = nkB T , as nλ3 =
nh3 h3 P = ≪ 1. (2πmkB T )3/2 (kB T )5/2 (2πm)3/2
It is convenient to express the answers starting with an imaginary gas of ‘protons’ at room temperature and pressure, for which mp = 1.7 × 10−34 Kg, and
(nλ3 )proton =
P = 1 atm. = 105 Nm−2 ,
(6.7 × 10−34 )3 10−5 = 2 × 10−5 . −21 5/2 −27 3/2 (4.1 × 10 ) (2π · 1.7 × 10 ) 192
The quantum effects appear below T = TQ , at which nλ3 becomes order of unity. Using 3
3
nλ = (nλ )proton
0010 m 00113/2 p
m
and TQ = Troom (nλ3 )3/2 ,
,
we obtain the following table: (d) Experiments on He4 indicate that at temperatures below 1K, the heat capacity is given by CV = 20.4T 3 JKg −1 K−1 . Find the low energy excitation spectrum, E(k), of He4 . (Hint: There is only one non-degenerate branch of such excitations.) • A spectrum of low energy excitations scaling as E(k) ∝ k s , in d-dimensional space, leads to a low temperature heat capacity that vanishes as C ∝ T d/s . Therefore, from CV = 20.4 T 3 JKg−1 K−1 in d = 3, we can conclude s = 1, i.e. a spectrum of the form E(k) = h ¯ cs |~k|, corresponding to sound waves of speed cs . Inserting all the numerical factors, we have 12π 4 N kB CV = 5
0012
T Θ
00133
,
where
¯ cs h Θ= kB
0012
6π 2 N V
00131/3
.
Hence, we obtain E =h ¯ cs k = kB
0012
2π 2 kB V T 3 5 CV
00131/3
k = (2 × 10−32 Jm) k,
corresponding to a sound speed of cs ≈ 2 × 102 ms−1 . ********
10. Solar interior: According to astrophysical data, the plasma at the center of the sun has the following properties: Temperature: Hydrogen density: Helium density:
T = 1.6 × 107 K
ρH = 6 × 104 kg m−3
ρHe = 1 × 105 kg m−3 .
(a) Obtain the thermal wavelengths for electrons, protons, and α-particles (nuclei of He). 193
• The thermal wavelengths of electrons, protons, and α-particles in the sun are obtained
from
λ= √
h , 2πmkB T
and T = 1.6 × 107 K, as λelectron ≈ p λproton ≈ p
2π × (9.1 ×
10−31
2π × (1.7 ×
10−27
6.7 × 10−34 J/s
Kg) · (1.4 ×
10−23
Kg) · (1.4 ×
10−23
6.7 × 10−34 J/s
and
J/K) · (1.6 ×
107 107
K)
≈ 1.9 × 10−11 m, ≈ 4.3 × 10−13 m,
J/K) · (1.6 × K) 1 λα−particle = λproton ≈ 2.2 × 10−13 m. 2
(b) Assuming that the gas is ideal, determine whether the electron, proton, or α-particle gases are degenerate in the quantum mechanical sense. • The corresponding number densities are given by 3
nH ≈ 3.5 × 1031 m−3 , ρHe 3 ≈ 1.5 × 1031 m−3 , ρHe ≈ 1.0 × 105 Kg/m =⇒ nHe = 4mH ne = 2nHe + nH ≈ 8.5 × 1031 m3 . ρH ≈ 6 × 104 kg/m
=⇒
The criterion for degeneracy is nλ3 ≥ 1, and nH · λ3H ≈2.8 × 10−6 ≪ 1,
nHe · λ3He ≈1.6 × 10−7 ≪ 1, ne · λ3e ≈0.58 ∼ 1. Thus the electrons are weakly degenerate, and the nuclei are not. (c) Estimate the total gas pressure due to these gas particles near the center of the sun. • Since the nuclei are non-degenerate, and even the electrons are only weakly degenerate,
their contributions to the overall pressure can be approximately calculated using the ideal gas law, as P ≈ (nH + nhe + ne ) · kB T ≈ 13.5 × 1031 (m−3 ) · 1.38 × 10−23 (J/K) · 1.6 × 107 (K) 2
≈ 3.0 × 1016 N/m .
194
(d) Estimate the total radiation pressure close to the center of the sun. Is it matter, or radiation pressure, that prevents the gravitational collapse of the sun? • The Radiation pressure at the center of the sun can be calculated using the black body formulas,
P = as P =
1U , 3V
1U π 2 k4 4 c= T = σT 4 , 4V 60¯ h3 c3
and
4 4 · 5.7 × 10−8 W/(m2 K4 ) · (1.6 × 107 K)4 2 σT 4 = ≈ 1.7 × 1013 N/m . 8 3c 3 · 3.0 × 10 m/s
Thus at the pressure in the solar interior is dominated by the particles. ******** 11. Bose condensation in d–dimensions:
Consider a gas of non-interacting (spinless)
bosons with an energy spectrum ǫ = p2 /2m, contained in a box of “volume” V = Ld in d dimensions. (a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical + potential µ. Express your answers in terms of d and fm (z), where z = eβµ , and
1 = Γ (m)
+ fm (z)
Z
∞ 0
xm−1 dx. z −1 ex − 1
(Hint: Use integration by parts on the expression for ln Q.)
• We have
Q= =
∞ X
P
eNβµ
N=0
n =N
i X
i
{ni }
YX
exp −β
eβ(µ−ǫi )ni =
i {ni }
Y i
X i
ni ǫi
!
,
1
1 − eβ(µ−ǫi )
0001 P P whence ln Q = − i ln 1 − eβ(µ−ǫi ) . Replacing the summation i with a d dimensional h i R d d R d−1 integration V dd k/ (2π) = V Sd / (2π) k dk, where Sd = 2π d/2 / (d/2 − 1)!, leads
to
ln Q = −
V Sd
d
(2π)
Z
0011 0010 2 2 k d−1 dk ln 1 − ze−β¯h k /2m .
The change of variable x = β¯h2 k 2 /2m (⇒ k = results in V Sd 1 ln Q = − d (2π) 2
0012
2m ¯h2 β
0013d/2 Z 195
p p 2mx/β/¯h and dk = dx 2m/βx/2¯ h)
0001 xd/2−1 dx ln 1 − ze−x .
Finally, integration by parts yields V Sd 1 ln Q = d (2π) d
0012
2m ¯h2 β
0013d/2 Z
x
d/2
ze−x Sd dx = V 1 − ze−x d
i.e. Sd G = −kB T ln Q = −V d
0012
2m h2 β
0013d/2
kB T Γ
0012
0012
2m h2 β
0013d/2 Z
dx
0013 d (z) , + 1 f+ d 2 +1 2
which can be simplified, using the property Γ (x + 1) = xΓ (x), to G=−
V (z) . kB T f + d 2 +1 λd
The average number of particles is calculated as 0012 0013d/2 Z ze−x Sd 2m ∂ d/2−1 x dx ln Q = V N= ∂ (βµ) d h2 β 1 − ze−x , 0012 0013d/2 0012 0013 Sd 2m V + d + =V f d (z) = d f d (z) Γ 2 2 h2 β 2 λ 2 i.e. n=
1 + f d (z) . λd 2
(b) Calculate the ratio P V /E, and compare it to the classical value. • We have P V = −G, while E=−
d ln Q d ∂ ln Q = + = − G. ∂β 2 β 2
Thus P V /E = 2/d, identical to the classical value. (c) Find the critical temperature, Tc (n), for Bose-Einstein condensation. • The critical temperature Tc (n) is given by n=
1 + 1 f d (1) = d ζ d , d λ 2 λ 2
for d > 2, i.e. h2 Tc = 2mkB
n ζd 2
!2/d
(d) Calculate the heat capacity C (T ) for T < Tc (n). 196
.
xd/2 , z −1 ex − 1
• At T < Tc , z = 1 and 0012 0012 0013 0013 ∂E G d d V d ∂G d d C (T ) = +1 = +1 kB ζ d +1 . =− =− 2 ∂T z=1 2 ∂T z=1 2 2 T 2 2 λd (e) Sketch the heat capacity at all temperatures. •
.
(f) Find the ratio, Cmax /C (T → ∞), of the maximum heat capacity to its classical limit,
and evaluate it in d = 3.
• As the maximum of the heat capacity occurs at the transition, Cmax
d = C (Tc ) = 2
Thus
0012
d +1 2
0013
V 0010
ζ d /n 2
0011 kB f + d 2 +1
Cmax = C (T → ∞)
0012
d +1 2
d (1) = N kB 2
0013
ζ d +1 2
ζd
0012
d +1 2
0013
ζ d +1 2
ζd
.
2
,
2
which evaluates to 1.283 in d = 3. (g) How does the above calculated ratio behave as d → 2? In what dimensions are your
results valid? Explain.
+ • The maximum heat capacity, as it stands above, vanishes as d → 2! Since fm (x → 1) →
∞ if m ≤ 2, the fugacuty z is always smaller than 1. Hence, there is no macroscopic
occupation of the ground state, even at the lowest temperatures, i.e. no Bose-Einstein condensation in d ≤ 2. The above results are thus only valid for d ≥ 2. 197
******** 12. Exciton dissociation in a semiconductor:
Shining an intense laser beam on a semi-
conductor can create a metastable collection of electrons (charge −e, and effective mass
me ) and holes (charge +e, and effective mass mh ) in the bulk. The oppositely charged particles may pair up (as in a hydrogen atom) to form a gas of excitons, or they may dissociate into a plasma. We shall examine a much simplified model of this process. (a) Calculate the free energy of a gas composed of Ne electrons and Nh holes, at temperature T , treating them as classical non-interacting particles of masses me and mh .
• The canonical partition function of gas of non-interacting electrons and holes is the product of contributions from the electron gas, and from the hole gas, as Ze−h
1 = Ze Zh = Ne !
0012
V λ3e
0013Ne
1 · Nh !
0012
V λ3h
0013Nh
,
√ where λα = h/ 2πmα kB T (α =e, h). Evaluating the factorials in Stirling’s approximation, we obtain the free energy Fe−h = −kB T ln Ze−h = Ne kB T ln
0012
Ne 3 λ eV e
0013
+ Nh kB T ln
0012
0013 Nh 3 λ . eV h
(b) By pairing into an excition, the electron hole pair lowers its energy by ε. [The binding energy of a hydrogen-like exciton is ε ≈ me4 /(2¯ h2 ǫ2 ), where ǫ is the dielectric constant,
−1 and m−1 = m−1 e + mh .] Calculate the free energy of a gas of Np excitons, treating them
as classical non-interacting particles of mass m = me + mh . • Similarly, the partition function of the exciton gas is calculated as 1 Zp = Np !
0012
V λ3p
0013Np
e−β(−Np ǫ) ,
leading to the free energy Fp = Np kB T ln where λp = h/
p
0012
Np 3 λ eV p
0013
− Np ǫ,
2π (me + mh ) kB T .
(c) Calculate the chemical potentials µe , µh , and µp of the electron, hole, and exciton states, respectively. 198
• The chemical potentials are derived from the free energies, through 0001 ∂Fe−h 3 = k T ln n λ µe = , B e e ∂Ne T,V 0001 ∂Fe−h = kB T ln nh λ3h , µh = ∂Nh T,V 0001 ∂Fp = kB T ln np λ3p − ǫ, µp = ∂Np T,V
where nα = Nα /V (α =e, h, p).
(d) Express the equilibrium condition between excitons and electron/holes in terms of their chemical potentials. • The equilibrium condition is obtained by equating the chemical potentials of the electron
and hole gas with that of the exciton gas, since the exciton results from the pairing of an electron and a hole, electron + hole ⇀ ↽ exciton. Thus, at equilibrium µe (ne , T ) + µh (nh , T ) = µp (np , T ) , which is equivalent, after exponentiation, to ne λ3e · nh λ3h = np λ3p e−βǫ .
(e) At a high temperature T , find the density np of excitons, as a function of the total density of excitations n ≈ ne + nh .
• The equilibrium condition yields np = ne nh
λ3e λ3h βǫ e . λ3p
At high temperature, np ≪ ne = nh ≈ n/2, and 0010 n 00112 h3 λ3 λ3 np = ne nh e 3 h eβǫ = 3/2 λp 2 (2πkB T ) ******** 199
0012
me + mh me mh
00133/2
eβǫ .
13. Freezing of He4 :
At low temperatures He4 can be converted from liquid to solid by
application of pressure. An interesting feature of the phase boundary is that the melting pressure is reduced slightly from its T = 0K value, by approximately 20Nm−2 at its minimum at T = 0.8K. We will use a simple model of liquid and solid phases of 4 He to account for this feature. (a) The important excitations in liquid 4 He at T < 1K are phonons of velocity c. Calculate the contribution of these modes to the heat capacity per particle CVℓ /N , of the liquid. • The dominant excitations in liquid 4 He at T < 1K are phonons of velocity c. The corresponding dispersion relation is ε(k) = h ¯ ck. From the average number of phonons in D E −1 mode ~k, given by n(~k) = [exp(β¯hck) − 1] , we obtain the net excitation energy as ¯ ck h exp(β¯hck) − 1 ~ k Z 4πk 2 dk ¯hck =V × (change variables to x = β¯hck) 3 (2π) exp(β¯hck) − 1 00134 Z ∞ 00134 0012 0012 x3 6 kB T π2 kB T V dx x , ¯hc = V ¯hc = 2π 2 ¯hc 3! 0 e −1 30 ¯hc
Ephonons =
X
where we have used 1 ζ4 ≡ 3!
Z
0
∞
dx
x3 π4 = . ex − 1 90
The corresponding heat capacity is now obtained as dE 2π 2 CV = = V kB dT 15
0012
kB T ¯hc
00133
,
resulting in a heat capacity per particle for the liquid of 2π 2 CVℓ = kB vℓ N 15
0012
kB T ¯hc
00133
.
(b) Calculate the low temperature heat capacity per particle CVs /N , of solid 4 He in terms of longitudinal and transverse sound velocities cL , and cT . • The elementary excitations of the solid are also phonons, but there are now two transverse sound modes of velocity cT , and one longitudinal sound mode of velocity cL . The 200
contributions of these modes are additive, each similar inform to the liquid result calculated above, resulting in the final expression for solid heat capacity of 2π 2 CVs = kB vs N 15
0012
kB T ¯h
00133
0012
×
2 1 + 3 3 cT cL
0013
.
(c) Using the above results calculate the entropy difference (sℓ − ss ), assuming a single
sound velocity c ≈ cL ≈ cT , and approximately equal volumes per particle vℓ ≈ vs ≈ v. Which phase (solid or liquid) has the higher entropy?
• The entropies can be calculated from the heat capacities as sℓ (T ) = ss (T ) =
Z
Z
T 0 T 0
CVℓ (T ′ )dT ′ 2π 2 = kB vℓ T′ 45 2π 2 CVs (T ′ )dT ′ = kB vs T′ 45
0012
kB T ¯hc
0012
kB T ¯h
00133
,
00133
×
0012
2 1 + 3 3 cT cL
0013
.
Assuming approximately equal sound speeds c ≈ cL ≈ cT ≈ 300ms−1 , and specific volumes vℓ ≈ vs ≈ v = 46˚ A3 , we obtain the entropy difference 4π 2 kB v sℓ − ss ≈ − 45
0012
kB T ¯hc
00133
.
The solid phase has more entropy than the liquid because it has two more phonon excitation bands. (d) Assuming a small (temperature independent) volume difference δv = vℓ − vs , calculate the form of the melting curve. To explain the anomaly described at the beginning, which phase (solid or liquid) must have the higher density? • Using the Clausius-Clapeyron equation, and the above calculation of the entropy difference, we get
0012
∂P ∂T
0013
melting
sℓ − ss 4π 2 v = =− kB vℓ − vs 45 δv
0012
kB T ¯hc
00133
.
Integrating the above equation gives the melting curve π2 v Pmelt (T ) = P (0) − kB 45 δv
0012
kB T ¯hc
00133
T.
To explain the reduction in pressure, we need δv = vℓ − vs > 0, i.e. the solid phase has
the higher density, which is normal.
201
******** 14. Neutron star core:
Professor Rajagopal’s group at MIT has proposed that a new
phase of QCD matter may exist in the core of neutron stars. This phase can be viewed as a condensate of quarks in which the low energy excitations are approximately E(~k)± =± h ¯2
0010 00112 ~ | k | − kF 2M
.
The excitations are fermionic, with a degeneracy of g = 2 from spin. (a) At zero temperature all negative energy states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. By relating occupation numbers of states of energies µ + δ and µ − δ, or otherwise, find the chemical potential at finite temperatures T .
• According to Fermi statistics, the probability of occupation of a state of of energy E is eβ(µ−E )n p [n(E)] = , 1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ, eβδn , p [n(µ + δ)] = 1 + eβδ
=⇒
eβδ 1 p [n(µ + δ) = 1] = = . βδ 1+e 1 + e−βδ
Similarly, for a state of energy µ − δ, p [n(µ − δ)] =
e−βδn , 1 + e−βδ
=⇒
p [n(µ − δ) = 0] =
1 = p [n(µ + δ) = 1] , 1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. This implies that for µ = 0, hn(E)i + hn(−E)i is unchanged for an temperature; for every particle leaving an occupied negative energy
state a particle goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Assuming a constant density of states near k = kF , i.e. setting d3 k ≈ 4πkF2 dq with q = |~k | − kF , show that the mean excitation energy of this system at finite temperature is k2 E(T ) − E(0) ≈ 2gV F2 π
Z
∞
0
202
dq
E + (q) exp (βE + (q)) + 1
.
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated as
E(T ) − E(0) =
X k,s
=g
[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]
X k
2 hn+ (k)i E + (k) = 2gV
Z
E + (~k) d3~k 0010 0011 . (2π)3 exp βE (~k) + 1 +
The largest contribution to the integral comes for |~k | ≈ kF . and setting q = (|~k | − kF ) and using d3 k ≈ 4πkF2 dq, we obtain
4πkF2 E(T ) − E(0) ≈ 2gV 2 8π 3
Z
∞
0
E + (q) k2 dq = 2gV F2 exp (βE + (q)) + 1 π
Z
0
∞
dq
E + (q) exp (βE + (q)) + 1
.
(c) Give a closed form answer for the excitation energy by evaluating the above integral. • For E + (q) = h ¯ 2 q 2 /(2M ), we have ∞
¯ 2 q 2 /2M h = (set β¯h2 q 2 /2M = x) β¯ h2 q 2 /2M + 1 e 0 00131/2 Z ∞ 0012 2 gV kF x1/2 2M kB T = dx k T B π2 ex + 1 ¯h2 0 0013 0012 0013 00131/2 √ 0012 0012 ζ3/2 V kF2 1 gV kF2 π 1 2M kB T √ √ ζ = 1 − = 2 kB T 1 − kB T. 3/2 π 2 π λ ¯h2 2 2
k2 E(T ) − E(0) = 2gV F2 π
Z
dq
− For the final expression, we have used the value of fm (1), and introduced the thermal √ wavelength λ = h/ 2πM kB T .
(d) Calculate the heat capacity, CV , of this system, and comment on its behavior at low temperature. • Since E ∝ T 3/2 , 0012 0013 √ 3ζ3/2 1 3E V kF2 ∂E √ 1 − = k ∝ = CV = T. B ∂T V 2T 2π λ 2
This is similar to the behavior of a one dimensional system of bosons (since the density of states is constant in q as in d = 1). Of course, for any fermionic system the density of states close to the Fermi surface has this character. The difference with the usual Fermi systems is the quadratic nature of the excitations above the Fermi surface. ******** 203
15.
Non-interacting bosons:
Consider a grand canonical ensemble of non-interacting
bosons with chemical potential µ. The one–particle states are labelled by a wavevector ~ q, and have energies E(~q). (a) What is the joint probability P ({nq~ }), of finding a set of occupation numbers {nq~}, of
the one–particle states, in terms of the fugacities zq~ ≡ exp [β(µ − E(~q))]?
• In the grand canonical ensemble with chemical potential µ, the joint probability of finding a set of occupation numbers {nq~}, for one–particle states of energies E(~q) is given by the normalized bose distribution P ({nq~ }) = =
Y q ~
Y q ~
{1 − exp [β(µ − E(~q))]} exp [β(µ − E(~q))nq~] n
(1 − zq~ ) zq~ q~ ,
with nq~ = 0, 1, 2, · · · ,
for each ~q.
(b) For a particular ~q, calculate the characteristic function hexp [iknq~ ]i. 0001n • Summing the geometric series with terms growing as zq~ eik q~ , gives hexp [iknq~ ]i =
1 − zq~ 1 − exp [β(µ − E(~q))] = . 1 − exp [β(µ − E(~q)) + ik] 1 − zq~ eik
(c) Using the result of part (b), or otherwise, give expressions for the mean and variance of nq~ . occupation number hnq~ i.
• Cumulnats can be generated by expanding the logarithm of the characteristic function in powers of k. Using the expansion formula for ln(1 + x), we obtain
0002 00010003 ln hexp [iknq~ ]i = ln (1 − zq~ ) − ln 1 − zq~ 1 + ik − k 2 /2 + · · · 0014 0015 zq~ k 2 zq~ = − ln 1 − ik + +··· 1 − zq~ 2 1 − zq~ ' 0012 00132 # zq~ zq~ zq~ k2 = ik − + +··· 1 − zq~ 2 1 − zq~ 1 − zq~ = ik
zq~ zq~ k2 − + ···. 1 − zq~ 2 (1 − zq~ )2
From the coefficients in the expansion, we can read off the mean and variance hnq~ i =
zq~ , 1 − zq~
and 204
2 nq~ c =
zq~
2.
(1 − zq~ )
(d) Express the variance in part (c) in terms of the mean occupation number hnq~ i. • Inverting the relation relating nq~ to zq~ , we obtain zq~ =
hnq~ i . 1 + hnq~i
Substituting this value in the expression for the variance gives
2 nq~ c =
zq~ (1 − zq~ )
2
= hnq~ i (1 + hnq~ i) .
(e) Express your answer to part (a) in terms of the occupation numbers {hnq~i}.
• Using the relation between zq~ and nq~ , the joint probability can be reexpressed as i Yh nq~ −1−nq~ . P ({nq~}) = (hnq~ i) (1 + hnq~ i) q ~
(f) Calculate the entropy of the probability distribution for bosons, in terms of {hnq~ i}, and comment on its zero temperature limit.
• Quite generally, the entropy of a probability distribution P is given by S = −kB hln P i. Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by S = −kB
X q ~
[hnq~ i ln hnq~ i − (1 + hnq~ i) ln (1 + hnq~i)] .
In the zero temperature limit all occupation numbers are either 0 (for excited states) or infinity (for the ground states). In either case the contribution to entropy is zero, and the system at T = 0 has zero entropy. ******** 16. Relativistic Bose gas in d dimensions:
Consider a gas of non-interacting (spinless)
bosons with energy ǫ = c |~ p |, contained in a box of “volume” V = Ld in d dimensions. (a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical
+ (z), where z = eβµ , and potential µ. Express your answers in terms of d and fm + (z) fm
1 = (m − 1)!
Z
0
205
∞
xm−1 dx. z −1 ex − 1
(Hint: Use integration by parts on the expression for ln Q.)
• We have
Q= =
∞ X
P
eNβµ
N=0
n =N
i X
i
exp −β
{ni }
YX
eβ(µ−ǫi )ni =
i {ni }
Y i
X
ni ǫi
i
!
,
1
1−
eβ(µ−ǫi )
0001 P P i) whence ln Q = − i ln 1 − eβ(µ−ǫ . Replacing the summation i with a d dimensional iR h R∞ ∞ d d k d−1 dk, where Sd = 2π d/2 / (d/2 − 1)!, integration 0 V dd k/ (2π) = V Sd / (2π) 0
leads to
ln Q = −
V Sd
d
(2π)
Z
∞
0
0001 k d−1 dk ln 1 − ze−β¯hck .
The change of variable x = β¯hck results in ln Q = −
V Sd d
(2π)
0012
kB T ¯hc
0013d Z
∞ 0
0001 xd−1 dx ln 1 − ze−x .
Finally, integration by parts yields V Sd 1 ln Q = d (2π) d
0012
kB T ¯hc
0013d Z
∞ 0
ze−x Sd x dx = V 1 − ze−x d d
0012
kB T hc
0013d Z
∞
dx
0
xd , z −1 ex − 1
leading to Sd G = −kB T ln Q = −V d
0012
kB T hc
0013d
+ kB T d!fd+1 (z) ,
which can be somewhat simplified to G = −kB T
V π d/2 d! + f (z) , λdc (d/2)! d+1
where λc ≡ hc/(kB T ). The average number of particles is calculated as N =−
∂G ∂G V π d/2 d! + = −βz = d f (z) , ∂µ ∂z λc (d/2)! d
where we have used z∂fd+1 (z)/∂z = fd (z). Dividing by volume, the density is obtained as n=
1 π d/2 d! + f (z) . λdc (d/2)! d
206
(b) Calculate the gas pressure P , its energy E, and compare the ratio E/(P V ) to the classical value. • We have P V = −G, while ∂ ln Q ln Q E=− = +d = −dG. ∂β z β
Thus E/(P V ) = d, identical to the classical value for a relativistic gas. (c) Find the critical temperature, Tc (n), for Bose-Einstein condensation, indicating the dimensions where there is a transition. • The critical temperature Tc (n) is given by n=
1 π d/2 d! 1 π d/2 d! + f (z = 1) = ζd . λdc (d/2)! d λdc (d/2)!
This leads to hc Tc = kB
0012
n(d/2)! π d/2 d!ζd
00131/d
.
However, ζd is finite only for d > 1, and thus a transition exists for all d > 1. (d) What is the temperature dependence of the heat capacity C (T ) for T < Tc (n)? • At T < Tc , z = 1 and E = −dG ∝ T d+1 , resulting in π d/2 d! G V E ∂E k = −d(d + 1) = d(d + 1) ζd+1 ∝ T d . = (d + 1) C (T ) = B ∂T z=1 T T λdc (d/2)! (e) Evaluate the dimensionless heat capacity C(T )/(N kB ) at the critical temperature T = Tc , and compare its value to the classical (high temperature) limit. • We can divide the above formula of C(T ≤ T c), and the one obtained earlier for N (T ≥ T c), and evaluate the result at T = Tc (z = 1) to obtain d(d + 1)ζd+1 C(Tc ) = . N kB ζd In the absence of quantum effects, the heat capacity of a relativistic gas is C/(N kB ) = d; this is the limiting value for the quantum gas at infinite temperature. ******** 207
17. Graphene is a single sheet of carbon atoms bonded into a two dimensional hexagonal lattice. It can be obtained by exfoliation (repeated peeling) of graphite. The band structure of graphene is such that the single particles excitations behave as relativistic Dirac fermions, with a spectrum that at low energies can be approximated by E ± (~k) = ±¯hv ~k .
There is spin degeneracy of g = 2, and v ≈ 106 ms−1 . Recent experiments on unusual
transport properties of graphene were reported in Nature 438, 197 (2005). In this problem, you shall calculate the heat capacity of this material. (a) If at zero temperature all negative energy states are occupied and all positive energy ones are empty, find the chemical potential µ(T ). • According to Fermi statistics, the probability of occupation of a state of of energy E is eβ(µ−E )n p [n(E)] = , 1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ, eβδn , p [n(µ + δ)] = 1 + eβδ
eβδ 1 p [n(µ + δ) = 1] = = . βδ 1+e 1 + e−βδ
=⇒
Similarly, for a state of energy µ − δ, p [n(µ − δ)] =
e−βδn , 1 + e−βδ
=⇒
p [n(µ − δ) = 0] =
1 = p [n(µ + δ) = 1] , 1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for all temperatures; any particle leaving an occupied
negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at
zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) − E(0) = 4A
Z
d2~k E + (~k) 0010 0011 (2π)2 exp βE (~k) + 1 +
208
.
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated as E(T ) − E(0) =
X
k,sz
=2
[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]
X k
2 hn+ (k)i E + (k) = 4A
Z
E + (~k) d2~k 0010 0011 . (2π)2 exp βE (~k) + 1 +
(c) Give a closed form answer for the excitation energy by evaluating the above integral. • For E + (k) = h ¯ v|k|, and ∞
2πkdk ¯hvk = (set β¯hck = x) 4π 2 eβ¯hvk + 1 0 00132 Z ∞ 0012 2A x2 kB T = dx x kB T π ¯hv e +1 0 00132 0012 kB T 3ζ3 . AkB T = π ¯hv
E(T ) − E(0) = 4A
Z
For the final expression, we have noted that the needed integral is 2!f3− (1), and used f3− (1) = 3ζ3 /4. E(T ) − E(0) = A
Z
E + (~k) d2~k 0010 0011 (2π)2 exp βE (~k) − 1
.
+
(d) Calculate the heat capacity, CV , of such massless Dirac particles. • The heat capacity can now be evaluated as 00132 0012 9ζ3 ∂E kB T = CV = . AkB ∂T V π ¯hv (e) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capacity of graphene. The typical sound velocity in graphite is of the order of 2×104 ms−1 . Is the low temperature heat capacity of graphene controlled by phonon or electron contributions? • The single particle excitations for phonons also have a linear spectrum, with E p = h ¯ vp |k| and correspond to µ = 0. Thus qualitatively they give the same type of contribution to 209
energy and heat capacity. The difference is only in numerical pre-factors. The precise contribution from a single phonon branch is given by Z ∞ 2πkdk ¯hvp k Ep (T ) − Ep (0) = A = (set β¯hck = x) 4π 2 eβ¯hvp k − 1 0 00132 Z ∞ 0012 A x2 kB T = dx x kB T 2π ¯hvp e −1 0 00132 00132 0012 0012 ζ3 3ζ3 kB T kB T = AkB T , CV,p = . AkB π ¯hvp π ¯hvp We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since the speed of Dirac fermions is considerably larger than that of phonons, their contribution to heat capacity of graphene is negligible. ******** 18. Graphene bilayer:
The layers of graphite can be peeled apart through different
exfoliation processes. Many such processes generate single sheets of carbon atoms, as well as bilayers in which the two sheets are weakly coupled. The hexagonal lattice of the single layer graphene, leads to a band structure that at low energies can be approximated by E 1 layer (~k) = ±tk (ak), as in relativistic Dirac fermions. (Here k = ~k , a is a lattice ±
spacing, and tk is a typical in-plane hopping energy.) A weak hopping energy t⊥ between the two sheets of the bilayer modifies the low energy excitations drastically, to E bilayer (~k) ±
=±
t2k
2t⊥
(ka)2
,
i.e. resembling massive Dirac fermions. In addition to the spin degeneracy, there are two branches of such excitations per unit cell, for an overall degeneracy of g = 4. (a) For the undoped material with one electron per site, at zero temperature all negative energy states are occupied and all positive energy ones are empty. Find the chemical potential µ(T ). • According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] =
eβ(µ−E )n , 1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ, p [n(µ + δ)] =
eβδn , 1 + eβδ
=⇒
p [n(µ + δ) = 1] = 210
eβδ 1 = . βδ 1+e 1 + e−βδ
Similarly, for a state of energy µ − δ, e−βδn , p [n(µ − δ)] = 1 + e−βδ
p [n(µ − δ) = 0] =
=⇒
1 = p [n(µ + δ) = 1] , 1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for all temperatures; any particle leaving an occupied
negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at
zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) − E(0) = 2gA
Z
d2~k E + (~k) 0010 0011 (2π)2 exp βE (~k) + 1
.
+
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated
as
E(T ) − E(0) =
X
k,sz ,α
=g
X k
[hn+ (k)i E + (k) + (1 − hn− (k)i) E − (k)] 2 hn+ (k)i E + (k) = 2gA
Z
d2~k E + (~k) 0010 0011 . (2π)2 exp βE (~k) + 1 +
(c) Give a closed form answer for the excitation energy of the bilayer by evaluating the above integral. • Let E + (k) = αk 2 , with α = (tk a)2 /(2t⊥ ), to get ∞
2πkdk αk 2 = (set βαk 2 = x) 2 eβαk2 + 1 4π 0 0013Z ∞ 0012 x kB T gA dx x kB T = 2π α e +1 0 0013 0012 0012 00132 gπ π A kB T kB T = = AkB T t⊥ . 24 α 3 a2 tk
E(T ) − E(0) = 2gA
Z
For the final expression, we have noted that the needed integral is f2− (1), and used f2− (1) = ζ2 /2 = π 2 /12. 211
(d) Calculate the heat capacity, CA , of such massive Dirac particles. • The heat capacity can now be evaluated as 2π A ∂E = kB CA = ∂T A 3 a2
k B T t⊥ t2k
!
.
(e) A sample contains an equal proportion of single and bilayers. Estimate (in terms of the hopping energies) the temperature below which the electronic heat capacity is dominated by the bilayers. • As stated earlier, the monolayer excitations for phonons have a linear spectrum, with
E1
layer
= ±tk (ka). Their contribution to energy and heat capacity can be calculated as
before. Including the various prefactors (which are not required for the solution), we have Z ∞ 2πkdk tk ak 1 layer E (T ) = 2gA = (set βtk ak = x) 4π 2 eβtk ak + 1 0 00132 Z ∞ 0012 x2 gA kB T dx x = kB T π tk a e +1 0 0012 0012 00132 00132 kB T kB T A A 1 layer ∝ 2 kB , CA . ∝ 2 kB T a tk a tk The bilayer heat capacity, which is proportional to T is more important at lower temper-
atures. By comparing the two expressions, it is apparent that the electronic heat capacity per particle is larger in the bilayer for temperatures smaller than T ∗ ≈ t⊥ /kB . (f) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capacity of graphene. The typical sound velocity in graphite is of the order of 2 × 104 ms−1 .
Is the low temperature heat capacity of (monolayer) graphene controlled by phonon or electron contributions? • The single particle excitations for phonons also have a linear spectrum, with E p = h ¯ vp |k| and correspond to µ = 0. Thus qualitatively they give the same type of contribution to
energy and heat capacity. The difference is only in numerical pre-factors. The precise contribution from a single phonon branch is given by Z ∞ 2πkdk ¯hvp k = (set β¯hck = x) Ep (T ) − Ep (0) = A 4π 2 eβ¯hvp k − 1 0 00132 Z ∞ 0012 A x2 kB T = dx x kB T 2π ¯hvp e −1 0 00132 00132 0012 0012 3ζ3 kB T kB T ζ3 , CV,p = . AkB = AkB T π ¯hvp π ¯hvp 212
We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since the speed of Dirac fermions is considerably larger than that of phonons, their contribution to heat capacity of graphene is negligible. ********
213