Given the distribution of #N# particles in #n# systems, #t = prod_N prod_i (n!)/(n_(N_i)!)#, and the following constraints, derive the grand canonical partition function and the distribution law for the grand ensemble?

1 Answer
Apr 7, 2017

Alright, so the first thing is to define the Lagrangian using our constraints:

#ℒ = lnt + alphasum_Nsum_i n_(N_i) - betasum_Nsum_i n_(N_i)E_(N_i) + ln lambdasum_Nsum_i n_(N_i)N#

For brevity and minimization of eyesores, I'll denote #sum_N sum_i# as #sum_(N,i)#. Hence, we have:

#ℒ = lnt + alphasum_(N,i) n_(N_i) - betasum_(N,i) n_(N_i)E_(N_i) + ln lambdasum_(N,i) n_(N_i)N#

The idea is that are finding the optimal distribution of systems, #n_(N_i)^"*"#, such that the Lagrangian is maximized. That means we have to perform the following derivative and set it equal to 0:

#((delℒ)/(deln_(N_i)))_(n_(N_(j ne i))) = 0#

Next, we would then need to convert the distribution expression into its #ln# form. The reason for this is that it is ridiculous to take the derivative of a factorial.

#ln(prod_N prod_i (n!)/(n_(N_i)!))#

#= sum_(N,i) ln((n!)/(n_(N_i)!))#

Since the sum is not over #n#, but the #n_(N_i)# systems and #N_i# particles, respectively, we can float the #n!# out.

#=> ln(n!) - sum_(N,i) ln(n_(N_i)!)#

Stirling's approximation, #ln(n!) ~~ nlnn - n#, is useful here. For statistical mechanics, the number of systems can get very large, so #n# can get very large. When that is the case, to a good approximation:

#lnt = nlnn - n - sum_(N,i) n_(N_i)ln(n_(N_i)) - n_(N_i)#

Now, we can take the derivative. Since the derivative is of a specific #n_(N_i)# (at a constant #n_(N_(j ne i))#), the sums all go away, and we focus on only the #n_(N_j)# where #j = i#.

#((delℒ)/(deln_(N_i)))_(n_(N_(j ne i))) = -(n_(N_i)*1/(n_(N_i)) + ln(n_(N_i)) - 1) + alpha (deln_(N_i))/(deln_(N_i)) - beta(deln_(N_i)E_(N_i))/(deln_(N_i)) + ln lambda (deln_(N_i)N)/(deln_(N_i))#

#= -ln(n_(N_i)) + alpha - betaE_(N_i) + Nln lambda = 0#

That was the hard part! Solving for #n_(N_i)^"*"#, we get:

#n_(N_i)^"*" = lambda^N e^(alpha)e^(-betaE_(N_i))#

Therefore, going back to the first constraint, we have:

#sum_(N,i) n_(N_i) = n = e^(alpha)sum_(N,i) lambda^N e^(-betaE_(N_i))#

We then define the grand canonical partition function as:

#color(blue)(Xi(lambda,beta,V) = sum_(N,i) lambda^N e^(-betaE_(N_i)))#

As a result, #e^(alpha) = n/Xi#, and if we want to find the distribution law, we multiply by #(n_(N_i))/(n_(N_i))# to get:

#e^alpha = n/Xi * (n_(N_i))/(n_(N_i))#

Therefore, the distribution law #p_(N_i) = n_(N_i)/n# is:

#color(blue)(p_(N_i)) = n_(N_i)/n = (n_(N_i))/(e^(alpha)Xi)#

#= (lambda^N cancel(e^(alpha))e^(-betaE_(N_i)))/(cancel(e^(alpha))sum_(N,i)lambda^N e^(-betaE_(N_i)))#

#= color(blue)((lambda^Ne^(-betaE_(N_i)))/(sum_(N,i)lambda^N e^(-betaE_(N_i))))#