Possible to derive Boltzmann distribution using W, not lnW?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
Darren73
Messages
8
Reaction score
0
Hi all, in following the many available derivations of the Boltzmann distribution I was trying to do it by maximizing W, which is N choose n1,n2,...nt., instead of lnW, because it should give the same answer (since W is monotonically increasing with lnW, am I wrong?).

So given the two constraint equations of constant particle number and energy, [tex] g=\sum_{i}n_{i}=N, <br /> <br /> h=\sum_{i}n_{i}\epsilon_{i}=E[/tex]And the Stirling approximation of W, [tex] W=N^{N}n_{1}^{-n_{1}}n_{2}^{-n_{2}}...n_{t}^{-n_{t}} [/tex]

And maximizing W with the above constraints (using Lagrange multipliers) gives the following t equations, [tex] \frac{\partial W}{\partial n_{i}}-\alpha\frac{\partial g}{\partial n_{i}}-\beta\frac{\partial h}{\partial n_{i}}=0 [/tex]

Which gives,[tex] \frac{\partial W}{\partial n_{i}}-\alpha-\beta\epsilon_{i}=0 [/tex] [tex] \frac{\partial W}{\partial n_{i}}=C_{i}n_{i}^{n_{i}}\left(\ln n_{i}+1\right)[/tex] [tex] C_{i}n_{i}^{n_{i}}\left(\ln n_{i}+1\right)=\alpha+\beta\epsilon_{i}[/tex]

Where Ci is some constant of the other nj's and N. Proceeding from this point has proven fruitless for me to isolate ni and apply the constraints. Does anyone know if this can be done? Or do you have to use lnW? It would seem odd to me if this cannot be done by maximizing W directly. And they should give the same distribution, namely [itex]n_{i}=N\exp -\beta \epsilon_{i}[/itex], correct?
 
Physics news on Phys.org