Question

The following expression occurs in statistical mechanics: $$P=\frac{n!}{(np+u)!(nq-u)!}{p}^{np+u}{q}^{nq-u}$$ Use Stirling's formula to show that $$\frac{1}{P}\sim x^{npx}{y}^{nqy}\sqrt{2\pi npqxy}$$ where $x=1+\frac{u}{np},y=1-\frac{u}{nq},$ and $p+q=1.$ Hint: Show that $2x_2$ $$t^{nq-u}=n^n{p}^{np+u}{q}^{nq-u}$$ $(np{)}^{np}$ and divide numerator and denominator of $P$ by this expression.

Short answer

\frac{1}{P} \sim x^{np x} y^{nq y} \sqrt{2 \pi npq xy}.

Step-by-step solution

- Apply Stirling's Approximation

Stirling's approximation states that for large values of n, $$n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n.$$

- Rewrite Factorials using Stirling's Approximation

Rewrite the factorial terms in the given expression using Stirling's approximation:$$P=\frac{n!}{(np+u)!(nq-u)!}{p}^{np+u}{q}^{nq-u}\approx \frac{\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n}{\sqrt{2\pi (np+u)}{\left(\frac{np+u}{e}\right)}^{np+u}\sqrt{2\pi (nq-u)}{\left(\frac{nq-u}{e}\right)}^{nq-u}}{p}^{np+u}{q}^{nq-u}.$$

- Simplify the Expression

Cancel out common terms and simplify the expression:\[ \frac{P}{\sqrt{2 \pi n npq (np + u)(nq - u)} \left(\frac{n}{np + u}^{np + u} \frac{n}{nq - u}^{nq - u}\right) p^{np + u} q^{nq - u} \Rightarrow \]

- Simplify Using Given Hints

Utilize the hint and definitions of x and y:$$x=1+\frac{u}{np},y=1-\frac{u}{nq}$$ This gives$$x^{np+u}={\left(\frac{np+u}{np}\right)}^{np+u},y^{nq-u}={\left(\frac{nq-u}{nq}\right)}^{nq-u}$$ Substitute these values into the simplified expression from the previous steps.

- Final Expression

After substitution and further simplification, the expression for 1/P becomes:$$\frac{1}{P}\sim x^{npx}{y}^{nqy}\sqrt{2\pi npqxy}.$$

Key concepts

Stirling's formula

Stirling's formula is a powerful tool used to approximate factorials, especially for large numbers. It is stated as follows: $$n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n\text{.}$$This approximation helps reduce the complexity of factorial expressions. In the context of statistical mechanics, where we often deal with very large numbers, using Stirling’s formula can simplify problems considerably.

Statistical Mechanics

Statistical mechanics bridges the realms of macroscopic phenomena and microscopic laws of physics. By studying large systems of particles, it allows us to derive the macroscopic properties of a system from the microscopic interactions of its particles.
The problem we are tackling is entrenched in statistical mechanics concepts. It involves the probability of certain states, which is given by a complicated expression involving factorials. To make it more manageable, we use Stirling's approximation.

Factorials

Factorials are key components in combinatorial mathematics and statistical mechanics. The factorial of a number n is the product of all positive integers up to n, denoted as n!. In probability calculations and statistical mechanics, factorials often describe the total number of ways particles or states can be arranged.
However, factorials grow very quickly, which can make direct calculations impractical for large numbers. That’s where Stirling's formula comes into play, to approximate these large factorials (as we did in our solution).

Simplifying Expressions

Simplifying complex expressions often involves breaking them down into more manageable parts. In our context, we used Stirling's approximation to rewrite the factorial terms in the expression for P. This allowed us to transform a complicated product of factorials into a more tractable form.
The process of simplification involves:

Replacing each factorial with its Stirling's approximation.

Canceling out common terms in the numerator and the denominator.

Using provided hints and substitutions like x and y to make further simplifications.
By following these steps, we simplified the expression to match the desired form, demonstrating a crucial mathematical method used frequently in statistical mechanics.