Calculation of No. of micro states (in equilibrium)

[Guffie]

Homework Statement

Hello,

I am required to determine the total number of micro states of a system in equilibrium within a certain value, 1/σ.

The number of micro states for this system is given by,

$\Omega =\frac{({{Q}_{A}}+{{N}_{A}}-1)!}{{{Q}_{A}}!({{N}_{A}}-1)!}\frac{({{Q}_{B}}+{{N}_{B}}-1)!}{{{Q}_{B}}!({{N}_{B}}-1)!}$

In the equilibrium state Q_a,e = 5000 and Q_b,e = 10000
Also Na = 5000 and Nb = 10000 and E = e(Qa+Qb)

So I need to find the total number of micro states in this system at equilibrium within 1/σ.
In this case σ=sqrt(5000).

Could anyone help me figure out how I would do this?

Homework Equations

The Attempt at a Solution

Is the correct way to do this by summing all the micro states together from

Q_a,e + 1/σ to Q_a,e + 1/σ,

if so how is that possible? the typical factorial function applies to integers.

1/σ is pretty small as well 0.014 so should i just ignore this?

The other part of the question asks to calculate hte total number of micro states outside of this interval.

So that means calculating every possible combination of Qa and Qb and adding them all together right?

Is there a way I can do this quickly in mathematica or something?

 

[DrClaude]

$\Omega =\frac{({{Q}_{A}}+{{N}_{A}}-1)!}{{{Q}_{A}}!({{N}_{A}}-1)!}\frac{({{Q}_{B}}+{{N}_{B}}-1)!}{{{Q}_{B}}!({{N}_{B}}-1)!}$

In the equilibrium state Q_a,e = 5000 and Q_b,e = 10000
Also Na = 5000 and Nb = 10000 and E = e(Qa+Qb)

These factorials will result in very large numbers. It will be easier to work with $\ln \Omega$, and take the exponential at the end to convert back to a multiplicity.

Can you convert write this $\ln \Omega$ as a function of $Q_A$? (Hint: You will need Stirling's approximation.)