Chapter 11: Q11.9P (page 554)

The following expression occurs in statistical mechanics:
$P = \frac{n!}{(n p + u)! (n q - u)!} p^{n p + u} q^{n q - u} .$
Use Stirling’s formula to show that
$\frac{1}{P} □ x^{n p x} y^{n q y} \sqrt{2 πnpqxy}, w h e r e x = 1 + \frac{u}{n p}, y = 1 - \frac{u}{n q}, p + q = 1 .$
Hint: Show that ${(n p)}^{n p + u} {(n q)}^{n q - u} = n^{n} p^{n p + u} q^{n q - u}$ .

Short Answer

Expert verified

The statement has been proved.

Step by step solution

Given Information

The expression is $P = \frac{n!}{(n p + u)! (n q - u)!} p^{n p + u} q^{n q - u}$ .

Definition of the sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial. $n! ~ n^{n} e^{- n} \sqrt{2 πn}$ .

Prove the statement.

The expression is $P = \frac{n!}{(n p + u)! (n q - u)!} p^{n p + u} q^{n q - u}$ .

The sterling’s formula is $n! ~ n^{n} e^{- n} \sqrt{2 πn}$ .

Use Sterling’s formula in the expression.

The expression becomes as follows.

$P = \frac{n!}{(n p + u)! (n q - u)!} p^{n p + u} q^{n q - u} P = \frac{n^{n} e^{- n} \sqrt{2 πn} p^{n p + u} q^{n q - u}}{{(n p + u)}^{n p + u} e^{- (n p + u)} \sqrt{2 π (np + u)} {(nq - u)}^{(nq - u)} e^{- (nq - u)} \sqrt{2 π (np - u)}} p^{n p + u} q^{n q - u}$

The value (np)^np+u(nq)^nq-u becomes as follows.

${(n p)}^{n p + u} {(n q)}^{n q - u} = n^{n p + u} p^{n p + u} n^{n q - u} q^{n q - u} {(n p)}^{n p + u} {(n q)}^{n q - u} = n^{n (p + q)} p^{n p + u} q^{n q - u} {(n p)}^{n p + u} {(n q)}^{n q - u} = n^{n} p^{n p + u} q^{n q - u}$

Substitute the above value in the equation mentioned below.

$P = \frac{n^{n} e^{- n} \sqrt{2 πn} p^{n p + u} q^{n q - u}}{{(n p)}^{n p + u} x^{n p x} e^{- (n p + u)} \sqrt{2 πnpx} {(n q)}^{(n q - u)} y^{n q y} e^{- (n q - u)} \sqrt{2 π (nqy)}} P = \frac{e^{- n} \sqrt{2 πn}}{x^{n p x} y^{n q y} 2 πn \sqrt{pxqy} e^{- n}} P = \frac{1}{x^{n p x} y^{n q y} \sqrt{2 πnpxqy}}$

Hence, the statement has been proven.

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Short Answer

Step by step solution

Given Information

Definition of the sterling’s formula.

Prove the statement.

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The following expression occurs in statistical mechanics:P=n!(np+u)!(nq-u)!pnp+uqnq-u.Use Stirling’s formula to show that1P□xnpxynqy2πnpqxy,wherex=1+unp,y=1-unq,p+q=1.Hint: Show that(np)np+u(nq)nq-u=nnpnp+uqnq-u.

Short Answer

Step by step solution

Given Information

Definition of the sterling’s formula.

Prove the statement.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Thermodynamics

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Linear Momentum

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Study anywhere. Anytime. Across all devices.