Lecture 6 - Stat. Mech. Postulate: I f you can calculatea m...

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Unformatted text preview: Stat. Mech. Postulate: I f you can calculatea m chanical prope Xi consiste with e rty nt them acroscopic param te the <Xi> =m e rs, n, acroscopic the odynam X rm ic Now that we know the probability of finding the quantum state -E with Ej at a given N,V e j Pj = -E e j j e j -Ej = Q(N,V, ) PARTITION FUNCTION P1 a1 = e- E = P2 a2 for states 1 and 2 separated by ( E1 - E2 ) =E, We can calculate other mechanical thermodynamic properties Mj M = j Pj M j For the internal energy: E j (N,V , ) e E = E (N,V , ) = j - E j ( N ,V ) e j - E j ( N ,V ) E j (N,V , ) e = j - E j ( N ,V ) Q(N,V , ) Q(N,V , ) considering that =- ,V N Q(N,V , ) ,V N E =- Q(N,V , ) E j (N,V , ) e j - E j ( N ,V ) ln Q(N,V , ) E = - ,V N Going back to the calculation of mechanical properties... For the pressure: the work done ON the system to induce dV in a canonical ensemble dE j = - p j d V - p = j U Ej - = pj V N E j - E j ( N ,V ) V e N - E ( N ,V ) e j j With these tools in hand , we can now combine mechanical properties with thermodynamics and extract information about nonmechanical functions (like S or T) Consider 2 ensembles A and B that become in thermal contact, with no change in volume o dV = 0 ; dA = 0 dB = 0 A B A B A B A B A B heat bath(T) Initially W({a,b}) = W({a*}) x W({b*}) Ensemble A o A Ensemble B o B What are the values at the new equilibrium? W ( { a,b} ) will evolve until it reaches maximum dW ( { a,b} ) d ( { a} ) W ( { b} ) 0 0 W exchange of E is associated with W W ( { a,b} ) >0 =0 ({ a ,b }) has reached maximum; a small change in E will not change W ({ a ,b }) dW ( { a,b} ) = W ( { b} ) ( { a} ) + W ( { a} ) ( { b} ) dW dW by W ( { a,b} ) = = W ( { a} ) + W ( { b} ) 0 dW ( { a,b} ) W ( { a,b} ) dW ( { a} ) W ( { a} ) dW ( { b} ) W ( { b} ) d lnW ( { a,b} ) = d lnW ( { a} ) + d lnW ( { b} ) 0 in this case, the E j will not change, the only change is on the distributions { a} { b} to reach a ,b { } now consider, d ln W { a} = d ln N - N - ak ln ak + ak N k = -d ak ln ak + ak - ak d ak = ln k k ak a0 =e - Ek ln ak = ln a0 - Ek for a , we know that j d ln W { a} = - d ak = 0, k k ( ln a 0 - Ek ) d ak Ek d ak since d ln W { a} = k The sum of the changes in energy in each system yields the infinitesimal change in the total energy d Ei = k ki + ki k E i dP P dE i i for dV = 0 Pki dEk = 0 (changes in Ei d E = i are produced by heat transport, not work) d ln W { a} = d E E i k d ak A and therefore, for d ln W ( { a,b} ) = d ln W ( { a} ) + d ln W ( { b} ) 0 A d E A + B d EB 0 since the ensembles are isolated, E = E A + EB; ak EkA + bk EkB = E d E = 0 d E A + d E B = 0 d E A = -d E B A Ad E Bd E A A B Approaching equilibrium, there is an increase in the value of EA, heat is transferred from B to A. A> B At equilibrium A = B any change in W({a}) will be canceled by a similar change in W({b}) 1st law: equilibrium means TA =TB, and approaching equilibrium heat will be transferred from B to A when TA<TB 1 T = 1 k BoltzT heat exchange E j dPj + j j work Pj d E j w h e re w e u s e d 1 - ( ln P 1 d Erev = = - j + ln Q ) = - ln Q 1 e - E j Q E j 1 = - Pj + ln Q ) dPj + j dV (ln P j j V N and since d P ln P P ln P + ln P = l n P dP = d dP j j j j j j j j j j j j and ln dP Q j j Ej = Ej = ln Q j = 0 dP j , E E j V = -pj E N we obtain 1 d E = - d Pj ln Pj p dV - j j From thermodynamics, we already know that dE = T - p + dS dV j j j dN 1 - d Pj ln Pj TdS = j Gibbs Entropy j j thus for a system with dN j = 0, or S(N,V ,T ) = -k j E E - - kT kT e e -k = ln ( N ,V , T ) Q( N ,V , T ) Q j j j Pj ln Pj S = -k j E - kT E - kT e ln e - ln Q( N ,V ,T ) ( N ,V , T ) Q Ej = -k - Q(N,V ,T ) kT j e kT - Ej e kT + ln k Q(N,V ,T ) (N,V ,T ) Q - Ej ln Q(N,V ,T ) S = kT + k ln Q(N,V ,T ) ET ,V N Using = 1 kBT and considering that N,V,T are the independent variables in th e canonical ensemble,we look at A(N,V,T) as the characteristic function. dA = -SdT - pdV + dN A(N,V ,T ) = -kT ln Q(N,V ,T ) where Q(N,V,T) = canonical partition function W e can write again all the realationships: ln Q A p = kT = -V N V ,T ,T N ln Q A S = kT + k ln Q = - T N T ,V ,V N A ln Q E = kT 2 = -T 2 T T T ,V N and e ven lnQ = -k T N N ,V ,T ,N A = N ,V ,T ,N N ,V N So far we worked with energy states, but we can easily change to levels, by counting how many times each "state" is repeated o degeneracy= ( N,V) Q(N,V ,T ) = j e - E j ( N ,V ) kT = E (N,V ) e - E ( N ,V ) kT in the limit of To0 Q(N,V ,T ) e - E1 kT E -E E -E - 2 1 - 3 1 kT kT + 3e + K an d since 1 = 1 1 + 2e E -E E -E - E1 - 2 1 - 3 1 1 + 2e kT + 3e kT + K e kT Q(N,V ,T ) e - E1 kT in the limit of To o Q(N,V,T) o constant = number of states ...
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