CHM 6484 Lecture Notes
17 September 1996

No, we're not trying to present Statistical Mechanics as a Riddle of the Sphinx. Instead the focus is on the pyramids of Gizeh behind the Sphinx in this photo (from the WWW's World of Art Part 1). The pyramids rise from an imposing base to a trivial peak, mimicing the decline of state population with energy in the Boltmann Distribution.

The texts treat probability in a different way that we have in the past lecture. Instead of multiplying the probability of independent events to arrive at likelihood, they employ a count, W, of all the ways in which a given distribution of systems might have arisen.

For the case of molecules occupying the most likely 50:50 (left:right) distribution in the room, that number of ways, W* (the * implies maximal, thus most likely), is arrived at by counting all the options for placing N molecules in the room such that N/2 of them are on the left and N/2 are on the right.

It boils down to which molecules are where even though we agree that one can't tell one nitrogen from another! (We'll ignore the difference between oxygen and nitrogen for the moment and call them all "air" molecules.)

In any given distribution of air molecules, the first such molecule we come to might me any of N, but the second can only be any of N-1 (since we've already identified the first). Thus the number of ways of finding the first two molecules is N(N-1). By induction (thought you'd never use that again, eh?), the number of ways of finding all N molecules is N! or N factorial, N(N-1)(N-2)(N-3)...(3)(2)(1), the product of all integers from 1 through N.

Right off the bat, you know this number is enormous. Your calculator goes belly up trying to produce 70! since 69!=1.7x10⁹⁸, nearly a googol. 70! is over a googol, and hand calculators can't handle that. So it won't surprise you that Avogadro's Number Factorial is unthinkably large.

Although N! is a correct count of the ways to find molecules in the room (N=2.4x10²⁷), it is an amazingly large overcount of the ways of finding one particular half (neatly numbered) molecules in the left half of the room and the rest on the right. The error arises because we just don't care how many ways we could rearrange the molecules in either half of the room. So since N! was a multiplicative overcount, we must divide it by the ways of finding left (and right) half molecules.

Those smaller counts are accomplished exactly the same way, resulting in (N/2)! each. That makes the count we REALLY wanted as W*=(N!)/[(N/2)!(N/2)!]. But even this is impractically large to deal with. We'd be better off investigating the logarithm of this number both because it is considerably smaller and because a simple mathematical approximation exists which improves the larger the number becomes!

The approximation is Stirling's, and it is that ln(x) = xln(x) - x plus terms on the order of x^1/2 and smaller. If we're dealing with Avogadro's Number, its square root is almost 12 orders of magnitude smaller than the Number itself; trust me, we're willing to throw out errors in the 12th decimal place.

So we'd like to evaluate ln(W*) for our "coin toss" air molecule count. Since the logarithm of the quotient is the difference of the logs, and that of a product is the sum of the logs, the natural log of the best left-right half room distribution is:

ln(W*) = ln[N!] - 2ln[(N/2)!)

       = N ln[N] - N - 2(N/2) ln[N/2] + 2(N/2)  {Stirling}
                   /                    //////  {cancellations}

       = N ln[N] - 2(N/2) ln N + 2(N/2) ln[2]
         ///////   ///////////                  {cancellations}

       = N ln[2] = ln[2N]


or  W* = 2N

So since N=2.4x10²⁷, we get W*=2^(2.4x10²⁷)=10^(7.2x10²⁶) confirming our previous analysis that equal distribution of the air is W* times more likely than all of it in the left half.

We can, with only slightly more effort, confirm that the standard deviation, s, of molecules about this optimal value is about s=(1/2)N^1/2 by direct substitution into the W count for this non-optimal redistribution. What we hope to find is that the number of ways, W*, of being on the norm is e times larger than W being one standard deviation off the norm. This is the falloff we'd expect from a normal (Gaussian) distribution. Indeed, we'd expect to lose another factor of e (2.718...) for every s worth of maldistribution.

The derivation proceeds just as above:

     W = [N!] / [ (N/2 + s)! (N/2 - s)!]

 ln(W) = ln[N!] - ln[(N/2 + s)!) - ln[(N/2 - s)!]

       = Nln[N]-N-(N/2+s)ln[N/2+s]+(N/2+s)-(N/2-s)ln[N/2-s]+(N/2-s)
                /                   /// \                    /// \

       = Nln[N] - (N/2+s)ln{(N/2)[1+2s/N]} - (N/2-s)ln{(N/2)[1-2s/N]}

       = Nln[N] - (N/2+s)ln[1+2s/N] - (N/2-s)ln[1-2s/N]
                - (N/2+s)ln[N/2]    - (N/2-s)ln[N/2]
                       /                   /

       = Nln[N] - (N/2+s)ln[1+2s/N] - (N/2-s)ln[1-2s/N]
                - (N/2)ln[N] + (N/2)ln(2) - (N/2)ln[N] + (N/2)ln(2)
         //////   //////////                //////////

Since 2s/N <<< 1, and for very small x, ln(1+x) = x 
                                    and ln(1-x) = -x,

 ln(W) = Nln(2) - (N/2+s)(2s/N) - (N/2-s)(-2s/N)
                   ///             ///


       = Nln(2) - 4s2/N

and ln(W*/W) = ln(W*) - ln(W) = Nln(2) - { Nln(2) - 4s2/N }
                                //////     //////


             = 4s2/N = 1  if s = N1/2/2  as hoped.

thus  W*/W = e  as expected.

So equilibrium thermodynamic averages are really going to be 10 ^-12 narrow for Avogadro's Number of particles. Fluctuations off the optimal distribution will be virtually unmeasurable.

---

If there are more than two states to the system, W's denominator will involve each state's population factorial to correct the count by eliminating rearrangement of the systems in the same state as if they were distinguishable. The number of ways the system's optimal distribution over states might occur will then become:

where the capital pi is used to denote a product (as sigma would be a sum) over the many states i. Now i may start at zero (for the zero point energy, perhaps) but it doesn't stop at N; it just keeps going and going and going like the Energizer Bunny. We'll discover, for example, that there are enormously more translational states for a molecule than there ever are molecules. This has a (briefly) disturbing consequence. Many states will have NO population. What becomes of 0! in the denominator?

Not to worry. By definition, 0!=1; so all those unoccupied states add (or subtract) nothing from the count.

Were that to be the optimal W*, then the populations in the various states would have to be optimal n*, optimized for each i. Be "optimized," we mean the W* should be maximal; that identifies the most likely sequence of populations {n*}. So we really want to take the derivative of W with respect to every n and choose that sequence {n} such that dW=0. Not surprisingly, we'd rather deal with ln(W), given those factorials! And that's OK since d[ln(W)]=d[W]/W=0 must be true under the same circumstances (W not 0) that d[W]=0 is true. So d[ln(W)]=0 also identifies {n*}, thank Stirling!

We would dearly love to take the total differential, d[ln(W)], by the sum of partial differentials and then argue that since each of the d[n] can take one whatever value it chooses, the only way to zero the result is to zero each and every partial differential, d[ln(W)]/dn. But that is forbidden because the {n} cannot have just any values! The sum of the n's must be N, the total number of particles. Likewise the total energy is fixed; thus the sum of each nE product (one per state) must be exactly E_TOTAL. So the individual n's in {n} aren't independent. That means some other expression (involving those partials) must obtain.

Lagrange (he of Classical Action fame) produced a solution for this class of problems called the Method of Undetermined Multipliers. It involved scaling contributions from the total differential constraints (zero due to the fixed N and E) by unknown parameters; the values of the parameters are readily found once the form of the solution arises. Once the constraints have been so considered explicitly, there's no longer a need to consider them implicitly (as linking dependent n's), and the individual populations in the {n} sequence are thus imagined now to be truly independent.

No choice of the augmented partial differentials can yield the required d[ln(W)]=0 unless every one of them is zero individually...because the dn's can do anything. That finally gives us the form of the optimal population sequence {n} as Boltzmann's Distribution function.

Well, all but for those two undetermined multipliers. Fortunately, there's the two conditions they must satisfy!

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

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