Physics 301 23-Sep-2002 5-1ReadingK&K, Chapter 3. Also, for a little culture, there is a handout which is a one pagearticle from Science, 1997, G. Bertsch, vol. 277, p. 1619. It describes melting in clustersconsisting of 139 atoms! So 1/√N ≈ 10%, quite a bit larger than the one part in a trillionwe’ve been talking about when there is a mole of particles! One might expect things to beless well defined, and sure enough, these clusters seem to melt over an appreciable rangein temperature (rather than a single T ). This may be a case where statistical mechanicsis just on, or just past the edge!The Boltzmann FactorLet’s imagine we have an object which is isolated from the rest of the world, so itsenergy, entropy, and so on are constant. Furthermore, it has come to equilibrium, so allparts of the object are at the same temperature, pressure, and so on. We imagine dividingthis object into two pieces: a small piece, called the system; and a large piece called theheat bath or heat reservoir (or just bath or reservoir). The system is supposed to besufficiently small that it’s useful to think of it as being in a single (quantum) state. Asystem might be a single atom or molecule, but it could be a larger entity if such an entitycan be described as being in a single state. The remainder of the object, the bath, issupposed to be very large and might consist of a very large number of similar atoms ormolecules. (In principle, we should let N →∞,whereN is the number of molecules inthe bath.) The system and the bath interact so that the system is in a particular stateonly with some probability. We are going to calculate this probability. We want to speakof the system as having energy E. This means that the interaction between the systemand the bath must be weak in order that we can ascribe a definite energy to the system.Consider two states of the system, 1 and 2, with energies E1and E2. When the systemis in state 1, the bath has energy U −E1,whereU is the total (fixed) energy of the bathplus system. The number of states that correspond to this energy in the system isg(U − E1) × 1=expσ(U − E1)× 1 .Where the first factor is the number of states in the bath and the second factor is thenumber of states (just 1) of the system. To have the factor 1 here is the reason that weinsistedthesystembeinasinglestate. Similarly,thenumberofstatesinthecasethatthe system has energy E2isg(U − E2) × 1=expσ(U − E2)× 1 .Our fundamental assumption is that each state (of the system plus bath together)that is compatible with the constraints is equally probable. So, the ratio of the probabilityCopyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 23-Sep-2002 5-2that the system is in state 2 to the probability that it is in state 1 isP (state 2)P (state 1)=g(U −E2)g(U −E1)= eσ(U − E2) − σ(U − E1).Now, E is an energy on the scale of an energy of a single molecule while U is an energytypical of a mole. So we can expand σ in a tailor series around E =0,σ(U − E)=σ(U) − E∂σ∂U+ ···= σ (U) −Eτ+ ··· .Note that The first term on the right is ∼ N times larger than the second term, and weexpect that the first omitted term will be another factor of ∼ 1/N smaller. Inserting thisexpansion into our expression for the probability ratio, we haveP (state 2)P (state 1)= eσ(U) − E2/τ − σ(U)+E1/τ= e−(E2− E1)/τ.The probability ratio is an exponential in the energy difference divided by the temperature.As noted when we discussed the model paramagnetic system, this is called the Boltzmannfactor, and we’ve just shown that this is a general result.It may seem we got something for nothing: we made a few definitions, did somemathematical hocus-pocus and voila, out pops the Boltzmann factor. A key ingredientis our postulate that all states which satisfy the constraints are equally probable. Thenumber of states goes up with increasing energy U. The rate of increase (in the logarithm)is measured by the temperature, τ . When the system is in a state with energy E,itnecessarily received that energy from the heat bath. The more energy the heat bath givesto the system, the fewer states it “has left.” This is what makes higher energy states ofthe system less probable.In principle, we should consider an ensemble of identically prepared heat baths andsystems. An ensemble in which the probability follows the Boltzmann form (∝ exp(−E/τ ))is called a canonical ensemble.Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 23-Sep-2002 5-3Systems with Several Forms of EnergyA single system in a canonical ensemble might have several ways to store energy. Forexample, if the system is a gas molecule, it has energy associated with translation of thecenter of mass, rotation about the center of mass, and various forms of internal energy suchas vibration or other electronic excitations and interactions among spins. If the energiescan be assigned independently, then the probabilities are independent. For example, if wewant to know the probability that a molecule is in a particular rotational state, with energyErot, we can think of the bath as including the translational and internal motions of themolecule under consideration as well as all the motions of all the other molecules. Similarly,if we consider the translational motion, with energy Etran, all other forms of energy in thismolecule and all other molecules may constitute the heat bath. The probabilities areP (Etran) ∝ e−Etran/τ,P (Erot) ∝ e−Erot/τ,··· .These probabilities are independent, so the probability that our molecule has a total energyEtot= Etran+ Erot+ ···is just the product,P (Etot) ∝ e−(Etran+ Erot+ ···)/τ= e−Etot/τ.Caveat 1: this is the probability for a single state of the system having the specifiedtotal energy. If the system has several different ways of distributing Etotamong its sub-systems, then each way has the same probability (our assumption that each state has thesame probability) and the probability of being in any state with that Etotis just the sum.Caveat 2: it may not always be true that the energies can be distributed independently.For example, the moment of inertia, and hence the rotational energy levels, may dependon the amplitude of vibration.In (older) thermo texts, you will often see expressions likeP (energy 2)P (energy 1)=g2g1e−(E2− E1)/τ,where g1and g2are called