G25.265: Statistical Mechanics Notes for Lecture 4 I. THE CLASSICAL VIRIAL THEOREM (MICROCANONICAL DERIVATION) Consider a system with Hamiltonian H(x). Let x i and x j be specic components of the phase space vector. The classical virial theorem states that i = kt ij where the average is taken with respect to a microcanonical ensemble. To prove the theorem, start with the denition of the average: i = C (E) dxx i (E? H(x)) where the fact that (x) = (?x) has been used. Also, the N and V dependence of the partition function have been suppressed. Note that the above average can be written as However, writing allows the average to be expressed as i = i = C (E) E = C (E) = C (E) E E dx x i (E? H(x)) dx x i dx x i (H? E) x i (H? E) = [x i (H? E)]? ij (H? E) C (E) E = C (E) E I H=E dx [x i (H? E)] + ij (E? H(x)) x i (H? E)dS j + ij dx (E? H(x)) The rst integral in the brackets is obtained by integrating the total derivative with respect to x j over the phase space variable x j. This leaves an integral that must be performed over all other variables at the boundary of phase space where H = E, as indicated by the surface element ds j. But the integrand involves the factor H? E, so this integral will vanish. This leaves: i = C ij (E) = C ij (E) E = ij (E) (E) dx H<E dx(e? H(x))
where (E) is the partition function of the uniform ensemble. Recalling that we have (E) = E (E) (E) i = ij x (E) j E = ij ln (E) E = k ij ~ S E = kt ij which proves the theorem. Example: x i = p i and i = j. The virial theorem says that hp i p i i = kt h p2 i m i i = kt h p2 i 2m i i = 2 kt Thus, at equilibrium, the kinetic energy of each particle must be kt=2. By summing both sides over all the particles, we obtain a well know result 3NX i= h p2 i 2m i i = 3NX i= h 2 m iv 2 i i = 3 2 NkT II. LEGENDRE TRANSFORMS The microcanonical ensemble involved the thermodynamic variables N, V and E as its variables. However, it is often convenient and desirable to work with other thermodynamic variables as the control variables. Legendre transforms provide a means by which one can determine how the energy functions for dierent sets of thermodynamic variables are related. The general theory is given below for functions of a single variable. Consider a function f(x) and its derivative y = f 0 (x) = df dx g(x) The equation y = g(x) denes a variable transformation from x to y. Is there a unique description of the function f(x) in terms of the variable y? That is, does there exist a function (y) that is equivalent to f(x)? Given a point x 0, can one determine the value of the function f(x 0 ) given only f 0 (x 0 )? No, for the reason that the function f(x 0 ) + c for any constant c will have the same value of f 0 (x 0 ) as shown in the gure below. 2
f(x) x 0 x b(x ) 0 FIG.. However, the value f(x 0 ) can be determined uniquely if we specify the slope of the line tangent to f at x 0, i.e., f 0 (x 0 ) and the y-intercept, b(x 0 ) of this line. Then, using the equation for the line, we have This relation must hold for any general x: f(x 0 ) = x 0 f 0 (x 0 ) + b(x 0 ) f(x) = xf 0 (x) + b(x) Note that f 0 (x) is the variable y, and x = g? (y), where g? is the functional inverse of g, i.e., g(g? (x)) = x. Solving for b(x) = b(g? (y)) gives b(g? (y)) = f(g? (y))? yg? (y) (y) where (y) is known as the Legendre transform of f(x). In shorthand notation, one writes (y) = f(x)? xy however, it must be kept in mind that x is a function of y. III. THE CANONICAL ENSEMBLE A. Basic Thermodynamics In the microcanonical ensemble, the entropy S is a natural function of N,V and E, i.e., S = S(N; V; E). This can be inverted to give the energy as a function of N,V, and S, i.e., E = E(N; V; S). Consider using Legendre transformation to change from S to T using the fact that 3
E T = S The Legendre transform ~ E of E(N; V; S) is ~E(N; V; T ) = E(N; V; S(T ))? S E S = E(N; V; S(T ))? T S The quantity ~ E(N; V; T ) is called the Hemlholtz free energy and is given the symbol A(N; V; T ). It is the fundamental energy in the canonical ensemble. The dierential of A is A da = T However, from A = E? T S, we have From the rst law, de is given by Thus, A dt + V N;T da = de? T ds? SdT A dv + N de = T ds? P dv + dn da =?P dv? SdT + dn Comparing the two expressions, we see that the thermodynamic relations are A S =? T A P =? V N;T A = N V;T T;V dn B. The partition function Consider two systems ( and 2) in thermal contact such that N 2 N E 2 E N = N + N 2 ; E = E + E 2 dim(x ) dim(x 2 ) and the total Hamiltonian is just H(x) = H (x ) + H 2 (x 2 ) Since system 2 is innitely large compared to system, it acts as an innite heat reservoir that keeps system at a constant temperature T without gaining or losing an appreciable amount of heat, itself. Thus, system is maintained at canonical conditions, N; V; T. The full partition function (N; V; E) for the combined system is the microcanonical partition function (N; V; E) = dx(h(x)? E) = dx dx 2 (H (x ) + H 2 (x 2 )? E) Now, we dene the distribution function, f(x ) of the phase space variables of system as 4
f(x ) = dx 2 (H (x ) + H 2 (x 2 )? E) Taking the natural log of both sides, we have ln f(x ) = ln dx 2 (H (x ) + H 2 (x 2 )? E) Since E 2 E, it follows that H 2 (x 2 ) H (x ), and we may expand the above expression about H = 0. To linear order, the expression becomes (x ) ln dx 2 (H (x ) + H 2 (x 2 )? E) ln f(x ) = ln dx 2 (H 2 (x 2 )? E) + H (x ) = ln dx 2 (H 2 (x 2 )? E)? H (x ) E ln dx 2 (H 2 (x 2 )? E) H (x)=0 where, in the last line, the dierentiation with respect to H is replaced by dierentiation with respect to E. Note that ln E ln dx 2 (H 2 ( 2 )? E) = S 2(E) k dx 2 (H 2 (x 2 )? E) = S 2 (E) E k = kt where T is the common temperature of the two systems. Using these two facts, we obtain ln f(x ) = S 2(E) k? H (x ) kt f(x ) = e S2(E)=k e?h(x)=kt Thus, the distribution function of the canonical ensemble is f(x) / e?h(x)=kt The prefactor exp(s 2 (E)=k) is an irrelevant constant that can be disregarded as it will not aect any physical properties. The normalization of the distribution function is the integral: dxe?h(x)=kt Q(N; V; T ) where Q(N; V; T ) is the canonical partition function. It is convenient to dene an inverse temperature = =kt. Q(N; V; T ) is the canonical partition function. As in the microcanonical case, we add in the ad hoc quantum corrections to the classical result to give Q(N; V; T ) = N!h 3N dxe?h(x) The thermodynamic relations are thus, Hemlholtz free energy: A(N; V; T ) =? ln Q(N; V; T ) To see that this must be the denition of A(N; V; T ), recall the denition of A: A = E? T S = hh(x)i? T S 5
But we saw that A S =? T Substituting this in gives A = hh(x)i? T A T or, noting that A T = A T =? A kt 2 it follows that A = hh(x)i + A This is a simple dierential equation that can be solved for A. We will show that the solution is A =? ln Q() Note that Substituting in gives, therefore so this form of A satises the dierential equation. Other thermodynamics follow: A = ln Q()? Q Q = A? hh(x)i A = hh(x)i + A? hh(x)i = A Average energy: E = hh(x)i = Q C N dxh(x)e?h(x) =? ln Q(N; V; T ) Pressure: A P =? V N;T = kt ln Q(N; V; T ) V N;T Entropy: S =? A T =? A T = A kt 2 = k 2? ln Q(N; V; T ) =?k ln Q + k ln Q = ke + k ln Q = k ln Q + E T 6
Heat capacity at constant volume: E C V = T = E T = k2 ln Q(N; V; T ) 2 C. Relation between canonical and microcanonical ensembles We saw that the E(N; V; S) and A(N; V; T ) could be related by a Legendre transformation. The partition functions (N; V; E) and Q(N; V; T ) can be related by a Laplace transform. Recall that the Laplace transform f() ~ of a function f(x) is given by ~f() = 0 dxe?x f(x) Let us compute the Laplace transform of (N; V; E) with respect to E: ~(N; V; ) = C N dee?e dx(h(x)? E) Using the -function to do the integral over E: ~(N; V; ) = C N 0 dxe?h(x) By identifying =, we see that the Laplace transform of the microcanonical partition function gives the canonical partition function Q(N; V; T ). D. Classical Virial Theorem (canonical ensemble derivation) Again, let x i and x j canonical average given by But Thus, be specic components of the phase space vector x = (p ; :::; p 3N ; q ; :::; q 3N ). Consider the x i i = Q C N = Q C N e?h(x) = i =? Q C N =? Q C N = i dxx i e?h(x) x j dxx i? e?h(x) x i e?h(x)? e?h(x) x i = x i e?h(x)? ij e?h(x) dx x i e?h(x) + Q ijc N dx 0 dx 0 x i e?h(x) dx j x i e?h(x) + kt ij x j =? + kt ij dxe?h(x) Several cases exist for the surface term x i exp(?h(x)): 7
. x i = p i a momentum variable. Then, since H p 2 i, exp(?h) evaluated at p i = clearly vanishes. 2. x i = q i and U! as q i!, thus representing a bound system. Then, exp(?h) also vanishes at q i =. 3. x i = q i and U! 0 as q i!, representing an unbound system. Then the exponential tends to both at q i =, hence the surface term vanishes. 4. x i = q i and the system is periodic, as in a solid. Then, the system will be represented by some supercell to which periodic boundary conditions can be applied, and the coordinates will take on the same value at the boundaries. Thus, H and exp(?h) will take on the same value at the boundaries and the surface term will vanish. 5. x i = q i, and the particles experience elastic collisions with the walls of the container. Then there is an innite potential at the walls so that U?! at the boundary and exp(?h)?! 0 at the bondary. Thus, we have the result i = kt ij The above cases cover many but not all situations, in particular, the case of a system conned within a volume V with reecting boundaries. Then, surface contributions actually give rise to an observable pressure (to be discussed in more detail in the next lecture). 8