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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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 Spring '08
 Bowers

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