Answer to Question #104027 in Molecular Physics | Thermodynamics for Ajay

Question

For a Bose-Einstein system, the expression for the thermodynamic probability is:
W=π{(gi+Ni–1)!/(Ni !(gi–1)!)}
Derive an expression for the Bose-Einstein distribution function.

Answer 1

As per the Bose-Einstein system,

The particles are indistinguishable.

Let total number of particle in the system is $n_i$ ​ , energy $E_i$

​ and let the statistical weight factor(degeneracy) is $g_i$

​ .Now, the way of the distribution of $n_i$ particle into $g_i$ space,

So, the distribution $\dfrac{(n_i-g_i-1)!}{n_i(g_i-1)!}$

for $n_o$ particle,

we know that thermodynamic probability $w=\dfrac{(n_o-g_o-1)!}{n_o(g_o-1)!}.\dfrac{(n_1-g_1-1)!}{n_1(g_1-1)!}.........\dfrac{(n_i-g_i-1)!}{n_i(g_i-1)!}$

So, we can write it as $w=\Pi\dfrac{(n_i-g_i-1)!}{n_i(g_i-1)!}$

Hence $\ln w=\sum {\ln((n_i-g_i-1)!)- \ln n_i-\ln(g_i-1)!}$