Statistical Mechanics Phys504 Fall 2006 Lecture #3 Anthony J. Leggett Department of Physics, UIUC Thermodynamic equilibrium Let s consider a situation where the Universe, i.e. system plus its environment E, possesses a total energy 1 E 0 and has come to complete thermodynamic equilibrium. 2 What now will be the energy ε of the system? Consider a small range ε surrounding ε. The weight of the range ε, ε + ε in the total density of states available to the universe will be proportional to the numbers of states of in this range the number of states of E. The former is by definition W s (ε) ε, and because of energy conservation the latter is W Env (E o ε) (E o ε) W env (E 0 ε) ε. Thus the total number of states available to +E with the energy of in the specified range is N(ε) = W s (ε)w env (E o ε)( ε) 2 This function is very sharply peaked as a f(ε) around its maximum, so we can say that the equilibrium (or overwhelmingly probable) value of ε is very close to the maximum, i.e. ε is determined (since ε is fixed) by W s (ε)w env (E o ε) = max. Let us take the logarithm and differentiate with respect to ε; d dε ln W s(ε) = d dε ln W env(e o ε) + d ln W (ε env ) dε env (ε env E o ε). But apart from the constant k B, ln W (ε) is just S(ε), so the condition for equilibrium is ds dε = ds env dε env But in thermodynamics, the condition for two bodies which are in thermal contact (i.e. can exchange energy freely) to be in thermal equilibrium is that their temperatures are equal, thus we define T (ds/dε) 1 (not T = ds/dε, cf. below) 1 Or an energy in a small range [E o, E o + E]; this does not affect the outcome. See problem. 2 For the moment we suppose and E can exchange energy but not particles or volume. 1
This immediately explains the so-called zeroth low of thermodynamics : two bodies which are each in thermal equilibrium with a third are automatically, if placed in contact, in thermal eqm. with one another. Also, let s consider the case that initially and E are not in equilibrium, so that the quantity N(ε) is not a maximum. According to the considerations of l.2, it will tend to a maximum, which is equivalent to the statement that the total entropy, k B (ln W s (ε) + ln W env (E o ε)) will increase: d dt (S(ε) + S env(ε env )) > 0 Writing ds/dt = ds/dε dε/dt and dε env /dt = dε/dt and using the above definition of temperature, this gives ( dε 1 dt T 1 ) > 0 i.e. heat flows into (out of) the system if it is colder (hotter) than the environment. (Note that had we defined T = ds/dε rather than (ds/dε) 1, we would have got the opposite result). Suppose the system is just slightly out of equilibrium with the environment: (T = + δt, δt 0). This means we are close to the maximum of total entropy; the deviation of T / ε from its eqm. value env / ε is (by definition!) prop. to δt, so the deviation of the entropy of the Universe from its maximum value is of order (δt ) 2. Thus to order δt S = S env i.e. the process is reversible by changing the sign of δt. Under these conditions we have E = T S dq ( dqis by definition energy input at constant N, V ) which is a special case of the first law of thermodynamics (see below). What can we say about the case of finite disequilibrium? It is convenient to redefine ε as the difference between the value of the system energy and its value in thermal equilibrium with the environment at temperature 3. Consider first the entropy of the Universe, 3 Note that because the environment is by construction very large, its temperature is negligibly changed when energy is transferred to or from the system (cf. below). 2
which is a sum of the entropies of the system and the environment, and expand it in ε: ( ) ( ) S univ = S eqm univ + env ε ε + S (2) + S env (2) ε eqm ε env = S eqm univ + S(2) + S (2) env where S (2) denotes all the higher-order terms in the expansion. Now, since the environment is very large, S (2) env is vanishingly small (e.g. the second-order term is 2 S env / ε 2 env ( / ε env ) (1/ ) = 1/T 2 envc env, and c env ). Thus, S univ = S eqn univ + S(2) and since we know that S univ is a maximum at equilibrium, it follows that S (2) < 0. Now expand S itself in ε: S(ε) = S eqn + ( ) ε eqm + S (2) (ε) S eqm + ε ε T + S 2(ε) Suppose the system goes from a nonzero value of ε to equilibrium (ε = 0). The change in energy E is then ε, and we know that S 2 (ε) < 0, so S S eqm S(ε) = E S 2 (ε) E dq i.e. the increase in entropy is always the heat input/. From the above argument it also follows that S univ is of second order in the deviation from eqm., δt. The Helmholtz free energy In general, when a system is in thermal isolation from any environment, it will tend to maximize its entropy at constant energy. On the other hand, if it is in thermal contact with an environment (at constant V and N) then as we have seen it is the total entropy of the Universe, ( +E), which is maximized, and this leads to the equality of temperature: T (/ ε) 1 =. In fact, we already saw that for such a system S E/ the equality being reached when T =. Consequently, the quantity F ( ) E S (*) 3
will decrease until it reaches its minimum value in the equilibrium state. Thus, for a system in thermal equilibrium, at constant V and N, with an environment at temperature T, we have the general principle F (T ) E T S = min. Evidently for a system whose environment is at zero temperature, this leads to E = min. as we expect. We will see below that (*) is a special case of a class of minimum principles. Adiabatic processes So far, we have supposed that all the parameters which control he Hamiltonian of the system (volume, particle number, etc.) are fixed. (so that the only interactions of with E is by energy exchange at the microlevel). Let us now for the moment thermally isolate the system, and suppose that the Hamiltonian depends on some parameter λ which we allow to vary slowly in time. (An example might be a slow change in one of the dimensions of the system). The energy of the system will then evidently also be a function of time. What about the entropy? It is actually easiest to consider this question in a QM picture. Suppose that we have a system subject to a Hamiltonian which for t < 0 is constant and for > 0 is a function of some continuously varying control parameter λ(t): Ĥ = Ĥo + Ĥ[λ(t)] Ĥ = 0, t < 0 Suppose that at time 0 the system is known to be in a particular eigenstate ψ n of Ĥ o with eigenvalue E n. For t > 0 both the eigenfunctions and the eigenvalues of Ĥ will be time-dependent. However, provided that λ(t) is indeed continuously varying, we can keep track of each of the eigenvalues and eigenfunctions E m (t), ψ m (t). The quantum adiabatic theorem then states that provided λ(t) is sufficiently slowly varying, the system will, with probability approaching 1, stay in its original state n, i.e. in the state ψ n (t) with energy E n (t). The criterion for λ to be sufficiently slowly varying is, crudely speaking, that for all states m n ( ) ( t Ĥ(λ(t)) E m (t) mn ) 2 E n (t) ( ) At first sight this criterion looks very difficult to fulfill in a many-body system since as we have seen the energy levels of such a system are densely spaced. However, it often turns 4
out that for such pairs of levels the LHS of ( ) is also very small, so that the condition is nevertheless fulfilled. A process taking place under the condition ( ) is called adiabatic (in the quantum context). Suppose now we start with our thermodynamic system in a (pseudo-) microcanonical ensemble, so that formally the DM is ˆρ = ˆ1/D Θ (E o, E, E o + E) i.e. the system is with equal probability in any of the energy eigenstates lying in E. The entropy is just k B lnd. It is clear that an evolution which is adiabatic in the quantum sense cannot change those probabilities, and since the system is thermally isolated there are no other effects. Consequently, if we define adiabatic process (SM) adiabatic (QM) and thermally isolated then in an adiabatic (SM) process the entropy of the system cannot change. Thus, the dependence of any thermodynamic quantity on λ in an adiabatic process must be just the derivative at constant S, and in particular ( ) ( ) E E = λ λ ad Note that according to the above definition every adiabatic process must be reversible (since the entropy of the system and hence of the Universe is unchanged) but not every reversible process is adiabatic (e.g. isothermal heat transfer is not). C.f. LL 11 (p. 37 in 1958 edn.). Also note that in an adiabatic process the temperature of the system in general changes. Let s consider the quantity ( E/ λ) s. We know that ds = 0 in an adiabatic process, so ( ) ( ) ( ) dλ + de dλ + T 1 de = 0 λ E E λ λ E so ( E/ λ) s = T ( ) λ E. S We can define X λ ( E/ λ) s as the generalized force exerted on the body by the source of λ. Consider a system in contact with an environment, and suppose that the total λ is conserved: λ P + λ E = const. (Example: λ = volume, particle number, magnetization 5
(under appropriate circumstances)). Suppose that λ can be freely exchanged between the system and its environment, whereas energy may or may not be exchangeable. What is the condition for equilibrium? Suppose we first assume that the system is thermally isolated, so that no heat can be transferred between and E. Then by an argument exactly similar to the one at the beginning of this lecture, we must adjust the λ s so as to maximize the total entropy of the Universe, and this leads to in other words ( ) = λ X λ T ( env λ env = X(env) λ In general, therefore, the generalized forces are not equal. However, if in addition we now allow transfer of energy between the two systems, then we have in eqm. (independently of λ) T =, and so X λ = X (env) λ ) as above The best-known cases of λ are the volume and the particle number. Recalling that X λ is ( E/ λ) S, we see that in the first case the generalized force is ( E/ V ) S,N P, the pressure, and in the second it is ( ( E/ N) S,V ) µ, the negative of the chemical potential. If then we exclude for the moment other λ s (corresponding to magnetization) we can write for any infinitesimal reversible change ( ) ( ) E E de = ds + V V,N S,N dv + T ds P dv + µ dn ( ) E dn N S,V which is just the first law of thermodynamics. ( thermodynamic identity ) Thermodynamic potentials Let us consider a system which possibly starts out of equilibrium with its environment, but eventually comes into equilibrium with it under various possible conditions (isothermal/adiabatic, isochoric/isobaric, etc.). For any given conditions of contact between and E, we can ask (a) what thermodynamic property of the system has to be minimized in 6
equilibrium? (b) if we start from a nonequilibrium state, how much work can we get out of the system during the transition to equilibrium? The answer to both questions is given by the relevant (free) energy. At first sight one would think that there should be 8 different free energies, corresponding to the 8 different conditions of equilibrium between and E, but we shall see that in effect there are only 4. In the following T o and P o represent the temperature and pressure of the environment, which can be regarded as constant during any process occurring in the system. Consider the energy balance of the system, for the moment treating N as constant. We have for any process, reversible or irreversible, the relation E = Q P o V env + W = T o S env P o V env W. where Q is the heat input from the environment and W is the (non-p dv ) work done, on a third system distinct from the environment, e.g. by magnetization, or... (we do not need to be explicit). Now evidently V = V env, and from the fact that the entropy of the Universe can only increase, S S env. Hence W ( E T o S + P o V ) W max W max is the maximum work obtainable from the system as it comes into equilibrium with the environment. Note that this maximum is obtained when the inequality S S env is an equality, i.e. when the process is reversible. Since W cannot be negative (equilibration is a spontaneous process!) it follows that the quantity E T o S + P o V is 0, i.e. if we define a quantity K such that K E T o S + P o V then (a) in the process of attaining equilibrium, dk/dt 0, and (b) K is the maximum work obtainable. Let s consider some special cases Case I. S = V = 0 (adiabatic isochoric process). Clearly in this case K = E E(S, V ) i.e. the relevant energy is just the total internal energy of the body. This is natural since no heat is absorbed and no external P dv work is done. 7
Case II. S = 0 but V 0: rather, P = P o (adiabatic, isobaric process). Now K = E + P V, or since P is constant, (E + P V ). Hence in this case K = E + P V H(S, P ) the so-called enthalpy or heat function. Case III. V = 0 but S 0, rather T = T o (isothermal isochoric process). Now V = E T S, or since T is constant, (E T S). Thus, K = E T S F (T, V ) the Helmholtz free energy. Finally, Case IV. Neither S or V are zero, but T = T o and P = P o (isothermal isobaric process): Now V = E T S + P V, so K = E T S + P V G(T, P ) the Gibbs free energy. The reason for expressing the four free energies as functions of four different pairs of the variables (S, V : T, P ) (since only 2 are independent, we could of course in principle write G as a f(t, V )!) is that their differentials then come out in simple form: e.g. dg = de T ds SdT + P dv + V dp but using the 1st law, de T ds + P dv = 0, so dg = SdT + V dp (etc.) From these differentials we can obtain a large number of Maxwell relations, e.g. from the above and ( ) ( ) ( ) ( ) 2 G 2 G V = P T P T T P T P Finally, we need to return to the role of total particle number. When we include this, the thermodynamic identity is generalized to (above) de = T ds P dv + µdn 8
where µ is the chemical potential. So in analogy to the way we have treated the entropy and volume, for the case of osmotic contact with the environment, let us say also isothermal and isobonic, it would be natural to introduce a quantity Φ E T S + P V µn G µn, µ ( E/ N) S,V ( G/ N) P,T which means µ ( G/ N) P,T = f(p T ) G/N. Thus, Φ = 0 identically! It further follows that, for example, the quantity which we would want to introduce for a process at constant T, V and µ, i.e. F µn, is just P V and does not contain any explicit reference to E. This quantity, and the other two (corresponding to S, V, µ and S, P, µ) are occasionally introduced but do not seem to have standard names. (cf. KIUH 1.14). 9