Answer to Question #169633 in Molecular Physics | Thermodynamics for karan saxena

Question

For a Bose-Einstein system, the expression for the thermodynamic probability is + − =  − ( 1)! !( 1)! i i i i g N W N g Derive an expression for the Bose-Einstein distribution function

Answer 1

Suppose we have a number of energy levels, labeled by index ${\displaystyle \displaystyle i}$, each level having energy ${\displaystyle \displaystyle \varepsilon _{i}}$ and containing a total of ${\displaystyle \displaystyle n_{i}}$ particles. Suppose each level contains ${\displaystyle \displaystyle g_{i}}$ distinct sublevels, all of which have the same energy, and which are distinguishable. For example, two particles may have different momenta, in which case they are distinguishable from each other, yet they can still have the same energy. The value of ${\displaystyle \displaystyle g_{i}}$ associated with level ${\displaystyle \displaystyle i}$ is called the "degeneracy" of that energy level. Any number of bosons can occupy the same sublevel.

Let ${\displaystyle \displaystyle w(n,g)}$ be the number of ways of distributing ${\displaystyle \displaystyle n}$ particles among the ${\displaystyle \displaystyle g}$ sublevels of an energy level. There is only one way of distributing ${\displaystyle \displaystyle n}$ particles with one sublevel, therefore ${\displaystyle \displaystyle w(n,1)=1}$. It is easy to see that there are ${\displaystyle \displaystyle (n+1)}$ ways of distributing ${\displaystyle \displaystyle n}$ particles in two sublevels which will be writen as:

${{\dfrac {(n+1)!}{n!1!}}}$

so that,

${\displaystyle w(n,3)=\sum _{k=0}^{n}w(n-k,2)=\sum _{k=0}^{n}{\frac {(n-k+1)!}{(n-k)!1!}}={\frac {(n+2)!}{n!2!}}}$

Using the ${\displaystyle n_{i}\gg 1}$ approximation and using Stirling's approximation for the factorials

${\displaystyle \left(x!\approx x^{x}\,e^{-x}\,{\sqrt {2\pi x}}\right)}$ gives;

${\displaystyle f(n_{i})=\sum _{i}(n_{i}+g_{i})\ln(n_{i}+g_{i})-n_{i}\ln(n_{i})+\alpha \left(N-\sum n_{i}\right)+\beta \left(E-\sum n_{i}\varepsilon _{i}\right)+K.}$

Where K is the sum of a number of terms which are not functions of the ${\displaystyle n_{i}}$. Taking the derivative with respect to ${\displaystyle \displaystyle n_{i}}$, and setting the result to zero and solving for ${\displaystyle \displaystyle n_{i}}$, yields the Bose–Einstein population numbers:

$\displaystyle n_{i}={\dfrac {g_{i}}{e^{\alpha +\beta \varepsilon _{i}}-1}}$