Solution

The mean extension in an ensemble of

A

molecules is

L = \frac{1}{A} \sum_{i} a_{i} l_{i},

with constraints

\sum_{i} a_{i} = A

and

\sum_{i} a_{i} l_{i} = A L

. For occupancies

{a_{i}}

, the number of assignments of the

A

distinguishable ensemble members to the states is the multinomial coefficient

W = \frac{A !}{\prod _{i} a _{i} !} .

Thus maximizing the probability at fixed

A

and

L

is equivalent to maximizing the stated Lagrange multiplier expression.

Because

A ≫ 1

lo g (a_{i}!) ≃ a_{i} lo g a_{i} - a_{i}

. Differentiating with respect to each

a_{i}

gives

- lo g a_{i} + τ l_{i} - α = 0, a_{i} = e^{- α} e^{τ l_{i}} .

Normalization by

\sum_{i} a_{i} = A

p_{i} = \frac{a _{i}}{A} = \frac{e ^{τ l_{i}}}{Z}, Z = i \sum e^{τ l_{i}} .

Here

Z

is the fixed-tension partition function that normalizes the probabilities.

Solved by gpt-5.6-sol high.

 Ancestors (11)