[S R (U 0 ɛ 1 ) S R (U 0 ɛ 2 ]. (0.1) k B

Transcription

1 Canonical ensemble (Two derivations) Determine the probability that a system S in contact with a reservoir 1 R to be in one particular microstate s with energy ɛ s. (If there is degeneracy we are picking one of them.) The reservoir has an energy U 0 ɛ s in this case and has g R (U 0 ɛ s ) accessible states; the entire isolated system has the same number of accessible states. (Why?) The relative probability of two states 1 and 2 is given by 2 P (ɛ 1 ) P (ɛ 2 ) = e 1 k B [S R (U 0 ɛ 1 ) S R (U 0 ɛ 2 ]. (0.1) Expand the entropy assuming that the reservoir is very large ( its degeneracy is fantastically large). Use the result we have for the argument of the exponential R /k B = 1 U=U0 k B T β (0.2) where T is the temperature of the reservoir, have P (ɛ 1 ) P (ɛ 2 ) = = e β(ɛ 1 ɛ 2 ) (0.3) This is Equation 9 of Chapter 3 and e βɛ is called the Boltzmann factor. If there are many states with the same energy then the probability of a state with that energy occurring is g(ɛ) e βɛ and this simple fact encapsulates a great deal of physics. Canonical ensemble: Use the Gibbsian approach. Suppose p j is the probability of obtaining the microstate j and we fix the average energy (there can be fluctuations in the total energy and so this is not microcanonical) and maximize the entropy. The mathematical setup is elementary and you should be able to write down the next equation without looking. We have to maximize S = p j ln(p j ) + λ p j 1 β p j ɛ j U. (0.4) k B j j j We obtain by differentiating with respect to p i 1 ln(p i ) + λ β ɛ i = 0. (0.5) Therefore, we obtain p i = e 1+λ e βɛ i (0.6) 1 What is a reservoir? 2 Employ the useful device and write g(u) = e S(U)/k B. 1

2 β = 1 k B T appears as the Lagrange multiplier. The constant λ is obtained by the normalization. We have e λ 1 = Therefore, p j = e βɛ j /Z. We also obtain 1 Z 1 j e βɛ j = 1 Z. (0.7) e βɛ i ɛ i = U. (0.8) i We define the partition function or Zustandsumme (sum over states) Z = s e βɛs.it is easy to verify that This is a useful formula. U = s P (ɛ s ) ɛ s = 1 Z s e βɛs ɛ s = β ln Z. What is the entropy in this formalism? Use Gibbs definition of entropy with p r = Z 1 e βɛr. We have S = k B r p r ln(p r ) = k B Z e βɛr (βɛ r + k B ln(z)) (0.9) r = U T + k B ln(z). (0.10) This is a useful place to record the (statistical mechanical 3 ) definition of the Helmholtz fee energy F = k B T ln Z. We have to connect this definition to the thermodynamically defined Helmholtz free energy and its implications as done in class and the book. As a first step we look at the expression for the entropy from the previous paragraph and the definition of F and see that F = U T S and this is precisely the thermodynamic definition. Two-state problem using the canonical ensemble. See the text page 62 for a clear description. I will just state the formulae assuming that the two states have energies 0 and ɛ. Z = 1 + e βɛ. (0.11) U = β ln Z = ɛ e βɛ 1 + e βɛ = ɛ. (0.12) 1 + eβɛ Often it is easier to compute this for two states 1 and 2 as 3 Why do we call this statistical mechanical? U = p 1 ɛ 1 + p 2 ɛ 2 = ɛ e βɛ 1 + e βɛ (0.13) 2

3 using the Boltzmann factor and avoiding differentiation. Look at high and low temperatures. What is high and low? ( ) Define C V = and plot it. See textbook and homework problems. T Z N 1 Ideal Gas Recapitulation: See page 72 in book. Recall that we wrote Z N (T, V ) = /N! where N! is the de-labeling factor introduced by Gibbs before he knew about distinguishable particles for internal consistency. Astounding! Going from a quantum mechanical sum to an integral we found Z 1 = V n Q = V λ 3 T = V ( ) 2πMkB T 3/2 (0.14) where λ T is the thermal de Broglie wavelength. What is the meaning of the thermal de Broglie wavelength? The characteristic magnitude of the momentum for an ideal gas at a temperature T is given by p2 2m = 3 2 k BT as we se below. Using λ = h p dependence apart from numerical factors. You should be able to recall the definition not the factors. h 2 we find the correct We can also do the problem classically (cheating appropriately 4 ) writing Z 1 = 1 h 3 d 3 x d 3 p2 β p e 2M. (0.15) V The coordinate integral yields V and the momentum integrals yield (do three identical Gaussian integrals in Cartesian coordinates) [ 2πMk B T ] 3. Z N This prescription along with = Z N 1 /N! yields the same answer. Please make sure you are clear about the h3 factor. Derive thermodynamic functions: [ F (T, V, N) = k B T ln Z N (T, V ) = Nk B T ln(v ) + 3 ( ) ] 2 ln 2πMkB T h 2 ln(n) + 1 (0.16) U = β ln Z N(t, V ) = 3 2 Nk BT. (0.17) ( ) C V = = 3 T 2 Nk B. (0.18) (How much energy does it take to heat 1 mole of nitrogen at room temperature by 1C? This is a trick question at this point. Why?) S = U F T [ = Nk B ln(n Q /n) + 5 ] 2 4 Can you identify the two places where we cheat? V ( ) V = Nk B [ln + 3 ( ) N 2 ln 2πMkB T h ] 2 (0.19) 3

4 where n N V. The above is the Sackur-Tetrode formula and was verified experimentally confirming the h 3 factor. Of course, the same factor arises when we do the particle in a box and take the limit. The molar entropy of Argon measured at p = 1 atm is (J/K) mol 1. What does the above equation yield? Statistical mechanics: Start from U = r p r ɛ r and write du = r ɛ r dp r + r p r dɛ r and consider the first term. We have p r = e βɛr Z ɛ r = k B T (ln Z + ln p r ). Using this relation for ɛ r we obtain for the first term du 1 = r ɛ r dp r = k B T r [dp r ln(p r ) k B T ln(z) r dp r (0.20) Thus we obtain The other term yields pdv. We write du 2 = r ( ) du 1 = k B T d p r ln(p r ) = T ds. (0.21) p r dɛ r = r r p r ɛ r V dv = dv ɛr V = p dv. (0.22) We have shown that du = T ds pdv. The derivation is for completeness and can be forgotten (if there was any danger that you would remember it). The second term is the work done on the system. If V decreases then dw is negative and positive work is done on the system. Thermodynamics: Define F U T S. This is a Legendre transformation and implies F = F (T, V, N). Why? We know T ds = du + pdv keeping N a constant from the fundamental thermodynamic relation. df = du T ds SdT = du du + pdv SdT = SdT pdv. Thus F/ T = S and F/ V = p. This is very important. Derive Maxwell s relation from above: / V = p/ T. (Equation 51) Therefore we have F k B T ln ZS is thermodynamically defined Helmholtz free energy. They obey the same differential relation derived above. Another important result is that F is an extremum in equilibrium at constant V and T. What exactly does this mean? 4

5 Legendre Transformation We consider U(S, V, N) where S and V are used as independent variables (In statistical mechanics we imagine we are in the microcanonical ensemble and we have inverted S(U, V, N) to obtain U(S, V, N).) We recall the fundamental relation du = T ds pdv (0.23) ( ) ( ) ( ) = ds + dv + (0.24) V N Clearly, this implies that T = ( ). We would like to trade S for the partial derivative with respect to S, i.e., T, so that T and V can be used as independent variables. S,N S,V Note that the mathematical problem to be solved can be abstracted and posed as follows: Consider the curve y = f(x). Define the slope z by z = y x where we have used a partial derivative to emphasize that all other dependences of f are kept fixed. We will not be so pedantic in the rest of this section and use simple derivatives. We wish to use the slope as the independent variable. Can one describe the curve in terms of z, the slope? (Convince yourself that this is the same problem as the one raised above in the context of thermodynamics. Instead of U(S) we wish to describe the system in terms of (/) the derivative.) First attempt: Let y = 4x 3. Hence, z = dy/dx = 12x 2. The latter can be inverted to obtain x = z/12. Substituting for x in terms of z we obtain y = 4(z/12) 3/2. So we have expressed y in terms of the slope z, we are done, right? Not! This is obvious geometrically because the curve is not uniquely specified by the slope: Translates of the curve are not distinguished. We can ascertain this analytically by noting that when y is expressed as a function of (dy/dx), a first-order differential equation has to be solved there is a constant of integration to be determined. We can solve the problem geometrically: Let (x, y) denote points on the curve y = f(x) shown below. We can represent the curve by (ψ, z) where z is the slope at a given point on the curve and ψ is the intercept along the y axis of the tangent to the curve as shown below. Draw a figure. The required curve is obtained as the envelope of the family of the tangent lines. Using the figure we see that z = y ψ x, and therefore, we have ψ = y xz. More explicitly, ψ = y x dy dx. (0.25) Thus we have the definition of the Legendre transform and its geometrical interpretation. In the thermodynamic problem we have U = U(S, V, N). To trade S for the derivative 5

6 with respect to S, T = ( ), we use F = ( ) U S V (0.26) = U T S. (0.27) This is the definition of the Helmholtz Free Energy. Check that the definitions of the enthalpy and the Gibbs Free Energy can be written down analogously. It is easiest to think of G(T, p, N) = F (T, V, N) + pv and remember that p = ( F/ V ) T,N. Please pay attention to signs. Important Aside: In classical mechanics the Lagrangian L is a function of the coordinates {q i } and { q i }. We define the momenta by p i = L q i and we would like trade { q i } for the momenta that are the derivatives of L with respect to { q i }. This is again done by a Legendre transformation defining the Hamiltonian by convention as H = L i p i q i. (0.28) 6