Taking logarithms on both sides of equation 11 gives Log (W) = Log (N!) -∑log (n i!) (2.1) Using equation, which we derive in Appendix 3, in equation 2.1 gives, Stirling’s Log (W) = N Log (N) i log (n i )–n i ) (2.2) –N – (∑n Doing the summation the sum over the n i terms cancels with N and noting a subtraction for a log is the same as dividing we are left with Log (W) = - n i/N Log (n i/N) (2.3) ∑ The probability pi = n i/N (2.4) So Log (W) = - pi log (pi) (2.5) ∑ This is what we require for the second form of the Boltzmann equation. Appendix 3 Derivation of Stirlings Equation Stirlings equation is: Log (N!) = N log (N) (3.1) –N We can obtain this result easily starting from the definition N! = N (N-1) (N-2) = i (N-i) (3.2) ….. П Factorise this by taking N out of each bracket and writing xi = i /N we have (3.3) N! = NN (1-xi) (3.4) П for values of xi from 0 to 1/N.
Taking logarithms on both sides of equation 11 gives Log (W) = Log (N!) -∑log (n i!) (2.1) Using Stirling’s equation, which we derive in Appendix 3, in equation 2.1 gives, Log (W) = N Log (N) –N – (∑n i log (n i )–n i ) (2.2) Doing the summation the sum over the n i terms cancels with N and noting a subtraction for a log is the same as dividing we are left with Log (W) = - ∑ n i/N Log (n i/N) (2.3) The probability pi = n i/N (2.4) So Log (W) = - ∑ pi log (pi) (2.5) This is what we require for the second form of the Boltzmann equation. Appendix 3 Derivation of Stirlings Equation Stirlings equation is: Log (N!) = N log (N) –N (3.1) We can obtain this result easily starting from the definition N! = N (N-1) (N-2) ….. = П i (N-i) (3.2) Factorise this by taking N out of each bracket and writing xi = i /N we have (3.3) N! = NN П (1-xi) (3.4) for values of xi from 0 to 1/N.
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