ME 083ME 083

The Statistical InterpretationThe Statistical InterpretationOf EntropyOf Entropy

Professor David M. SteppProfessor David M. Stepp

Mechanical Engineering and Materials ScienceMechanical Engineering and Materials Science189 Hudson Annex189 Hudson Annex

david.stepp@duke.edudavid.stepp@duke.edu549-4329 or 660-5325549-4329 or 660-5325

http://www.duke.edu/~dms1/stepp.htmhttp://www.duke.edu/~dms1/stepp.htm

26 February 200326 February 2003

From Last Time….From Last Time….

The Free Energy of a crystal can be written asThe Free Energy of a crystal can be written as

– the Free Energy of the perfect crystal (Gthe Free Energy of the perfect crystal (G00) )

– plus the free energy change necessary to create n defects plus the free energy change necessary to create n defects (n*(n*∆g)∆g), which is also internal energy change necessary to , which is also internal energy change necessary to create these defects in the crystal (create these defects in the crystal (∆H)∆H)

– minus the entropy increase which arises from the different minus the entropy increase which arises from the different possible ways in which the defects can be arranged (T∆Spossible ways in which the defects can be arranged (T∆SCC))

∆∆G = ∆H - T∆SG = ∆H - T∆SCC

The Configurational Entropy of a crystal, The Configurational Entropy of a crystal, ∆S∆SCC, is proportional to , is proportional to

the number of ways in which the defects can be arranged (W)the number of ways in which the defects can be arranged (W)

∆∆SSCC = k = kBB* ln(W)* ln(W)

Remember: Remember: ∆S∆SCC = k = kBB * ln(W) * ln(W)

Configurational Entropy is proportional to the number Configurational Entropy is proportional to the number of ways in which defects can be arranged.of ways in which defects can be arranged.

Example: Vacancy defect (calculation of W)Example: Vacancy defect (calculation of W)

Imagine a crystal lattice with N sites:Imagine a crystal lattice with N sites:

Remember: Remember: ∆S∆SCC = k = kBB * ln(W) * ln(W)

Configurational Entropy is proportional to the number Configurational Entropy is proportional to the number of ways in which defects can be arranged.of ways in which defects can be arranged.

Example: Vacancy defect (calculation of W)Example: Vacancy defect (calculation of W)

Imagine a crystal lattice with N sites:Imagine a crystal lattice with N sites:

OneOne

possiblepossible

complexioncomplexion

N sites and two states (full, vacant)N sites and two states (full, vacant)

n: n: Number of sites that are vacantNumber of sites that are vacant

n’: n’: Number of sites that are fullNumber of sites that are full

Then we have n + n’ = NThen we have n + n’ = N

Here, N = 21, n = 3, and n’ = 18Here, N = 21, n = 3, and n’ = 18

Note thatNote that n can assume a range of values between 0 and n can assume a range of values between 0 and NN

The situation where n lattice sites are vacant can be The situation where n lattice sites are vacant can be achieved in many alternate ways.achieved in many alternate ways.

Another arrangement (complexion) for our case is:Another arrangement (complexion) for our case is:

N = 21N = 21

n = 3n = 3

n’ = 18n’ = 18

Keep in mind also that we can differentiate between Keep in mind also that we can differentiate between full and empty sites full and empty sites (mathematically)(mathematically); however, all ; however, all full and all empty sites are equivalent full and all empty sites are equivalent (physically)(physically)

In order to determine W (the number of ways in which In order to determine W (the number of ways in which defects can be arranged), we need to calculate defects can be arranged), we need to calculate allall complexions that are complexions that are distinguishabledistinguishable..

NNCCnn : : The number of distinguishable (distinct) The number of distinguishable (distinct)

configurations of N lattice sites where any n are configurations of N lattice sites where any n are vacant and the remaining n’ are filled.vacant and the remaining n’ are filled.

= W= W!)!(

!

nnN

NCnN

Example: N = 4 with n vacant sites Example: N = 4 with n vacant sites █ █ and n’ full sites Oand n’ full sites O11 22 33 44 nn n’n’ NNCCnn

█ █ █ █ 44 00 11

█ █ █ O 33 11 44

█ █ O █ 33 11

█ O █ █ 33 11

O █ █ █ 33 11

█ █ O O 22 22 66

█ O O █ 22 22

█ O █ O 22 22

O █ O █ 22 22

O O █ █ 22 22

O █ █ O 22 22

O O O █ 33 11 44

O O █ O 33 11

O █ O O 33 11

█ O O O 33 11

O O O O 00 44 11

List all possible configurations for N = 4, n = 2List all possible configurations for N = 4, n = 2

Process: Place Process: Place █1█1 on one of the N lattice sites, and create on one of the N lattice sites, and create configurations with configurations with █2█2 in each of the remaining (N-1) lattice in each of the remaining (N-1) lattice sites. sites. Now,Now, label each distinct configuration. label each distinct configuration.

11 22 33 44

██11 ██22 OO OO aa

██11 OO ██22 OO bb

██11 OO OO ██22 cc

██22 ██11 OO OO aa

OO ██11 ██22 OO dd

OO ██11 OO ██22 ee

██22 OO ██11 OO bb

OO ██22 ██11 OO dd

OO OO ██11 ██22 ff

██22 OO OO ██11 cc

OO ██22 OO ██11 ee

OO OO ██22 ██11 ff

Generally: The possible number JGenerally: The possible number JN(n)N(n) of of distinctdistinct

vacancy placements is obtained by multiplying the vacancy placements is obtained by multiplying the number of number of possiblepossible locations of each vacancy. locations of each vacancy.

In our last example with N = 4 and n = 2:In our last example with N = 4 and n = 2:JJN(n)N(n) = (# possible = (# possible █1█1 locations) * (# possible locations) * (# possible █2 locations)█2 locations)

= (4) * (4-1)= (4) * (4-1)

= 12= 12

More generally:More generally:

JJN(n)N(n) = N * [N-1] * [N-2] * [N-3] *… [N-n+1] = N * [N-1] * [N-2] * [N-3] *… [N-n+1]

Now, rewrite JNow, rewrite JN(n)N(n) in terms of factorials: in terms of factorials:

N! = N * (N-1) * (N-2) * (N-3) * … (1)N! = N * (N-1) * (N-2) * (N-3) * … (1)

JJN(n)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-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) *… (1)

= N!= N!

(N-n)!(N-n)!

But we cannot distinguish But we cannot distinguish (physically)(physically) one vacancy one vacancy from the next; therefore, Jfrom the next; therefore, JN(n)N(n) over estimatesover estimates the the

number of distinguishable complexions.number of distinguishable complexions.

Now consider the vacancies alone (i.e., for any set Now consider the vacancies alone (i.e., for any set arrangement in a lattice): the possible number of arrangement in a lattice): the possible number of distinct permutations (combinations) of n vacancies distinct permutations (combinations) of n vacancies is n!is n!

In other words, the possible number of permutations In other words, the possible number of permutations of n vacancies of n vacancies (even though these permutations are (even though these permutations are indistinguishable physically)indistinguishable physically) is n! is n!

Example: For n = 3, how many possible distinct Example: For n = 3, how many possible distinct permutations exist?permutations exist?

123123 213213 231231

132132 312312 321321 = 6 = 3!= 6 = 3!

Now, with:Now, with:– The total possible number of distinct vacancy The total possible number of distinct vacancy

placements in our latticeplacements in our lattice

– The total possible number of permutations of vacancies The total possible number of permutations of vacancies within a given placementwithin a given placement

n!n!

The desired number, The desired number, NNCCnn, of distinguishable , of distinguishable

complexions is thus given by dividing Jcomplexions is thus given by dividing JN(n)N(n) by n! by n!

)!(

!)( nN

NJ nN

!!

!

!)(

nnN

N

n

JC nN

nN

Verifying our earlier example: N = 4, n = 2Verifying our earlier example: N = 4, n = 2

JJN(n)N(n) = 12 = 12

n! = 2n! = 2

NNCCnn = 12/2 = 6 = 12/2 = 6

Recall Recall all distinct configurations for N = 4, n = 2all distinct configurations for N = 4, n = 2

(a+b+c+d+e+f = 6 distinct configurations)(a+b+c+d+e+f = 6 distinct configurations)

!!

!

!)(

nnN

N

n

JC nN

nN

Properties of Properties of NNCCnn::– Symmetric under interchange of n and n’Symmetric under interchange of n and n’

NNCCnn = = NNCCn’n’

NNCC00 = = NNCCNN = 1 = 1

The ratio of successive coefficients is initially large, but decreases The ratio of successive coefficients is initially large, but decreases monotonically with n. Staying larger than unity as long as n < monotonically with n. Staying larger than unity as long as n < ½ N, and becoming smaller than unity for n ½ N, and becoming smaller than unity for n ≥ ½ N≥ ½ N

NNCCnn has a maximum value near n = ½ N has a maximum value near n = ½ N

1)!1()!1(

!)!(1

n

nN

nnN

nnN

C

C

nN

nN