8.044 Lecture Notes Chapter 6: Statistical Mechanics at ...

1 Lecture NotesChapter 6: Statistical Mechanics at Fixed Temperature(Canonical Ensemble)Lecturer: Derivation of the Canonical Ensemble .. Monatomic, classical, ideal gas, at fixedT.. Two-level systems, re-revisited .. Classical harmonic oscillators and equipartition of energy .. Quantum harmonic oscillators .. Heat capacity of a diatomic ideal gas .. Paramagnetism .. by adiabatic demagnetization .. The Third Law of Thermodynamics .. 6-42 Reading:Greytak, Notes on the Canonical Ensemble;Baierlein, Chapters 5, 13, Reading for :Greytak, Notes on (Raman Spectroscopy of) Derivation of the Canonical EnsembleIn Chapter 4, we studied the Statistical Mechanics of anisolatedsystem.

2 This meant fixedE,V, some fundamental principles (really, postulates), we developed an algorithm for cal-culating (which turns out not to be so practical, as you ll have you thought aboutthe random 2-state systems on pset 6):1. Model the system2. Count microstates for givenE: (E,V,N).3. from , derive thermodynamics byS=kBln .4. from , derive microscopic information byp(X) = (X) .The fixed-energy constraint makes the counting difficult, in all but the simplest problems(the ones we ve done). Fixing the temperature happens to be easier to analyze in bath, TConsider a system1which is not isolated, but ratheris in thermal contact with a heat bath2, and thereforeheld at fixed temperatureTequal to that of the ensemble of such systems is called the canonical en-semble.

3 I don t know fact thatTis fixed meansEis not: energy can beexchanged between the system in question and the that1+2together are isolated, with fixed energyEtotal=E1+E2. Then wecan apply the microcanonical ensemble to1+2. Note that1could be itself macroscopic(it just has to have a much smallerCVthan2), in which case we can learn about itsthermodynamics. Alternatively,1could be microscopic, with just a few degrees of freedom(like one square of a grid of 2-state systems, in which case we can ask for the probabilitythat it is ON).6-2 Consider a specific microstate A of1with : What the equilibrium probability that system1is in state A? We can apply the methodof Chapter 4 to1+2:Prob(system1is in state A) =p({p1,q1} specific values for all vars of system 1 in state A) = 1+2(A) 1+2(Etotal)=# of microstates of system2with energyEtotal E1total # of microstates of1+2with energyEtotal=E1+E2, fixed= p({p1,q1}) = 2(Etotal E1) 1+2(Etotal)Take logs for smoother variation:kBlnp({p1,q1}) =S2(Etotal E1 E2) S1+2(Etotal)So far, what we ve done would be valid even if systems1and2were of the same now let s use the fact that2is a reservoir by recognizing thatE1 Etotal.

4 We shouldTaylor expand in smallE1:kBlnp({p1,q1})' S2(Etotal) E1 E2S2(E2)|E2=Etotal 1T2+O(E21) S1+2(Etotal) S2 E2=1T2=1 TwhereTis the temperature of the bath2. (Here we are using the fact that2is a reservoirin saying that its temperature remainsTeven if we ignore the contribution ofE1toEtotal.)kBlnp({p1,q1}) = E1T+S2(Etotal) S1+2(Etotal) independent of microstate of1+O(E21)= H1({p1,q1})T+..H1here is the energy of system1in the specified microstate ({p1,q1}) =e H1({p1,q1})kBT Boltzmann factoreS2(Etotal) Stotal(Etotal)kB C, indep of microstate of1(1)[End of Lecture 14.]1 His for Hamiltonian. It s not the : why can we ignore theO(E21) terms in the Taylor expansion?A: because they become small very rapidly as we make the reservoir larger, and we canmake the reservoir as large as we want.

5 In particular, the next term of the Taylor expansionofS2(Etotal E1) is:12E21 2S2 E2 What s this last term? 2S E2take one derivative= E(1T)chain rule= 1T2 T Edef ofCV= 1T21 CVThisCVis the heat capacity of the reservoir the defining property of the reservoir is thehugeness of its heat capacity. So the biggest term we are ignoring is of magnitudeE21T21CV(2)and we should compare this to the last term we keep, which isE1T. The term we are ignoringbecomes smaller and smaller as we add degrees of freedom to the reservoir ( would golike 1/Nif it were comprised ofNideal gas atoms), whereasE1/Tdoes factorIn particular,p({p1,q1}) e H1({p1,q1})T Energy scale is set bykBT. Recall that in the ensemble with fixed energy, we didn t ever compare microstates withdifferent energies.

6 Microstates with high/low energy are less/more probable. This last statement is NOT the same as higher energy is less probable : Supposethere is some set of microstates of1with the same energyE1. Then:p(1is in some state with energyE1) e E1kBT degeneracy # of microstates with energyE1 This last factor, called the density of states can contain a lot of physics. It is thenumber of microstates of system1with energyE1, also known as 1(E1) =eS1(E1) that it depends functionMissing: the normalization of the probability distribution:1!= all microstates of system1{dp1dq1}p({p1,q1})= {dp1dq1}e H1({p1,q1})/(kBT) C the thing above, determined by this equation= p({p1,q1}) =e H1({p1,q1})/(kBT)Z=e H1({p1,q1})/(kBT) {dp1dq1}e H1({p1,q1})/(kBT).

7 This quantity which enters our consciousness as a normalization factor,Z {dp1dq1}e H1({p1,q1})/(kBT)partition functionis called thepartition function, and it is the central object in the canonical ensemble. ( Z isforZustandssumme, German for state sum .)To recap, our answer for the equilibrium probability distribution at fixed temperature is:p({p1,q1}) =1Ze H1({p1,q1})/(kBT)Boltzmann distributionThis is the probability that system1is in the microstate labelled by{p1,q1}when it isin contact with a heat bath at temperatureT(and in equilibrium). We derived this byapplying the microcanonical ensemble to system1plusthe heat :(microcanonical, Chapter 4)Constrained integral that counts microstates with = 1 Limits of integration: :(canonical, Chapter 6)Suitably weighted integral overallmicrostates ise H, not 1 (not just counting).

8 Limits of integration: straightforward. Integrate over everything (including some very un-likely states).Note: for a system with discrete states labelledi= 1, :p(system1is in a specific microstate i) =1Ze EikBTwithZ i,all states of the systeme EikBT6-5 Thermodynamics from the partition functionSo we already have the microscopic info: we ve foundp(microstate) at fixedT. We haven tyet assumed that1is thermodynamically large. Next:suppose1is also macroscopic,and let s learn to extract its thermodynamics from the partition expressions for1 Zfrom (1):1Z=e(S2(Etotal) S1+2(Etotal))/kBDoes the RHS really depend on system2?S1+2(Etotal) =S1( E1 meanE1in equilibrium) +S2( E2 meanE2in equilibrium)S2(Etotal)'S2 Etotal E1 E2 + E1 S2 E2|E2= E2 +.

9 =S2( E2 ) + E1 T= 1Z=e1kB( S1( E1 )+ E1 T)Everything having to do with2has disappeared, and we ve written the answer in terms ofthermodynamic variables of1. So we have two true expressions forZ:Z= microstates of1e H1/(kBT)andZ=e 1kB( S1( E1 )+ E1 T)the left one involving only microscopic information about1and the right one involvingonly between thermodynamics and canonical ensembleWe can now drop the notational baggage we ve been carrying around (the subscripts andthe .. s), since everything refers to system1in thermal equilibrium. We ve writtenZinterms of the thermodynamic variables of the system of interest (1):Z=e 1kBT(E TS)=e FkBTwhere I remind you thatF(T,V,N) =E TSis the Helmholtz free energy,dF= SdT PdV .(2)This is the bridge between microscopic stuff (Z) and thermodynamics (F) in the canonicalensemble:F= kBTlnZ(T,V,N)All of thermodynamics follows from (ch 4): count, find , findS=kBln , take (ch 6): computeZ, findF= kBTlnZ, then from (2), we have:S= ( F T)V, P= ( F V)T.

10 E=F+TS,H=E+PV,G=H simpler way to getE:E=F T( F T)V= T2 T(FT)V= T2 T|V( kBlnZ) = lnZ|Vwith concludes the derivation of the canonical ensemble. The canonical ensemble is theprimary tool of the practicing Statistical mechanic. What to remember from Chapter 4, most important application of the microcanonical ensemble: how to derive the : a warning about a common misconception, then an important special case. Thenmany examples the rest of for a fixed microstatevsprobability for a fixed energyFor a system in equilibrium at fixed temperatureT, we have:p(system is in microstate A) =Z 1e EA/(kBT)However, the probability distribution for the energy of the system,p(E) is NOT proportional toe E/(kBT).Rather, the dependence of this quantity on the energy must also include a degeneracy factor:p(E) =Z 1e E/(kBT)(degeneracy).