10/27/09
Dear Dr. Laurendeau,
A couple of years ago I bought your book Statistical Thermodynamics. I have found it to be an excellent reference, particularly since you have provided detailed derivations of many topics that other texts have glossed over. However, I came across a derivation which I believe is incorrect, and thought my alternative might be of interest to you.
The derivation in question is described in section 3.4 The Most Probable Microstate (pp. 35-7). In the first paragraph you present your conclusion that “Surprisingly enough, we can actually demonstrate that a large majority of all possible microstates is affiliated with the most probable macrostate.” This bothered me, since it seemed to me that the probability of the macrostate adjacent to the most probable macrostate should have a similar probability. For instance, for an ideal gas composed of a million molecules, the most probable distribution of the molecules between two halves of a container would be 500,000 in each half. However, it seemed to me that having 500,001 on one side and 499,999 on the other should be almost as probable. This can be confirmed by comparing the probabilities of the two macrostates.
Using your notation, the number of microstates associated with the most probable macrostate for M = 2 is
The number of microstates in the nearest adjacent macrostate is
The ratio of the number of microstates in the most probable macrostate to the number of microstates in the nearest adjacent macrostate is therefore
which approaches 1 as N grows larger.
A formulation which is more directly related to your conclusion is to take the ratio R of Wmp to the total number of microstates W = MN,
The conventional way of evaluating this formula is to apply the truncated form of the Stirling approximation ln N! = N ln N – N, as given in section D.2 (p. 394). Applying this to the above equation results in ln R = 0, or R = 1, which is consistent with your conclusion. However, if one programs the exact formula for R into a spreadsheet, it appears that R approaches zero with increasing N. Unfortunately, one gets numeric overflow errors even for relatively low values of M and N, so this approach is not conclusive. There also remains the question of the inconsistency between the exact calculation and the Stirling approximation. This puzzled me for awhile, but was resolved by using the long form of the approximation (eq. D.12). This results in
which approaches -∞ as N increases, so that R approaches zero. Also, this approximation converges with the exact calculation for ln R as N increases, confirming that the exact formulation of R also approaches zero.
So it is not proven that “for macroscopic thermodynamic systems almost all microstates are associated with the most probable macrostate.” However, this has no practical significance, since eq. 3.6 (p. 36) still holds. As N increases, the distribution converges around the mean. However, the number of macrostates increases so rapidly that the probability of any particular macrostate, including the most probable macrostate, approaches zero. Therefore, it does not necessarily follow that “the most probable particle distribution must represent the equilibrium particle distribution!” (p. 37)
I have included an Excel file with the calculations discussed above for M = 2 and M = 4. These spreadsheets should be self-explanatory in the context of the above discussion. Note that column J of the spreadsheet labeled M=4 corroborates your eq. 3.5 (p. 36).
Best regards,
Bill Dreiss
Addendum 7/11/13
Larry Caretto’s 6/14/13 derivation for M = 2 was much easier to comprehend than the above logarithmic result, since he used the non-logarithmic version of Stirling’s approximation, N! = (N/e)N (2πN)1/2. This yields
where R is the probability of being in the most probable macrostate. For M = 2, this reduces to R = (2/πN)1/2, the same result obtained by Caretto, where R = Wmp / Wtotal.