1. ME 083 The Statistical Interpretation Of Entropy Professor David M. Stepp Mechanical Engineering and Materials Science 189 Hudson Annex david.stepp@duke.edu 549-4329 or 660-5325 http://www.duke.edu/~dms1/stepp.htm 26 February 2003

2. From Last Time…. The Free Energy of a crystal can be written as –the Free Energy of the perfect crystal (G 0 ) –plus the free energy change necessary to create n defects (n*∆g), which is also internal energy change necessary to create these defects in the crystal (∆H) –minus the entropy increase which arises from the different possible ways in which the defects can be arranged (T∆S C ) ∆G = ∆H - T∆S C The Configurational Entropy of a crystal, ∆S C, is proportional to the number of ways in which the defects can be arranged (W) ∆S C = k B * ln(W)

3. Remember: ∆S C = k B * ln(W) Configurational Entropy is proportional to the number of ways in which defects can be arranged. Example: Vacancy defect (calculation of W) Imagine a crystal lattice with N sites:

4. Remember: ∆S C = k B * ln(W) Configurational Entropy is proportional to the number of ways in which defects can be arranged. Example: Vacancy defect (calculation of W) Imagine a crystal lattice with N sites:

5. Onepossiblecomplexion N sites and two states (full, vacant) n: Number of sites that are vacant n’: Number of sites that are full Then we have n + n’ = N Here, N = 21, n = 3, and n’ = 18 Note that n can assume a range of values between 0 and N

6. The situation where n lattice sites are vacant can be achieved in many alternate ways. Another arrangement (complexion) for our case is: N = 21 n = 3 n’ = 18 Keep in mind also that we can differentiate between full and empty sites (mathematically); however, all full and all empty sites are equivalent (physically)

7. In order to determine W (the number of ways in which defects can be arranged), we need to calculate all complexions that are distinguishable. N C n : The number of distinguishable (distinct) configurations of N lattice sites where any n are vacant and the remaining n’ are filled. = W

8. Example: N = 4 with n vacant sites █ and n’ full sites O 1234nn’ NCnNCnNCnNCn ████401 ███O314 ██O█31 █O██31 O███31 ██OO226 █OO█22 █O█O22 O█O█22 OO██22 O██O22 OOO█314 OO█O31 O█OO31 █OOO31 OOOO041

9. List all possible configurations for N = 4, n = 2 Process: Place █1 on one of the N lattice sites, and create configurations with █2 in each of the remaining (N-1) lattice sites. Now, label each distinct configuration. 1234 █1█2OOa █1O█2Ob █1OO█2c █2█1OOa O█1█2Od O█1O█2e █2O█1Ob O█2█1Od OO█1█2f █2OO█1c O█2O█1e OO█2█1f

10. Generally: The possible number J N(n) of distinct vacancy placements is obtained by multiplying the number of possible locations of each vacancy. In our last example with N = 4 and n = 2: J N(n) = (# possible █1 locations) * (# possible █2 locations) = (4) * (4-1) = 12 More generally: J N(n) = N * [N-1] * [N-2] * [N-3] *… [N-n+1]

11. Now, rewrite J N(n) in terms of factorials: N! = N * (N-1) * (N-2) * (N-3) * … (1) J N(n) = N * (N-1) * (N-2) * (N-3) *… (N-n+1) = N * (N-1) * (N-2) * (N-3) *… (N-n+1) * (N-n) *… (1) (N-n) *… (1) (N-n) *… (1) = N! (N-n)! But we cannot distinguish (physically) one vacancy from the next; therefore, J N(n) over estimates the number of distinguishable complexions.

12. Now consider the vacancies alone (i.e., for any set arrangement in a lattice): the possible number of distinct permutations (combinations) of n vacancies is n! In other words, the possible number of permutations of n vacancies (even though these permutations are indistinguishable physically) is n! Example: For n = 3, how many possible distinct permutations exist? 123213231 132312321 = 6 = 3!

13. Now, with: –The total possible number of distinct vacancy placements in our lattice –The total possible number of permutations of vacancies within a given placement n! The desired number, N C n, of distinguishable complexions is thus given by dividing J N(n) by n!

14. Verifying our earlier example: N = 4, n = 2 J N(n) = 12 n! = 2 N C n = 12/2 = 6 Recall all distinct configurations for N = 4, n = 2 (a+b+c+d+e+f = 6 distinct configurations)

15. Properties of N C n : –Symmetric under interchange of n and n’ N C n = N C n’ N C 0 = N C N = 1 The ratio of successive coefficients is initially large, but decreases monotonically with n. Staying larger than unity as long as n < ½ N, and becoming smaller than unity for n ≥ ½ N N C n has a maximum value near n = ½ N