Chapter 6: Q 6.16 (page 231)

Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy is
$\bar{E} = - \frac{1}{Z} \frac{\partial Z}{\partial β} = - \frac{\partial}{\partial β} \ln Z$
where $β = 1 / k T$ . These formulas can be extremely useful when you have an explicit formula for the partition function.

Short Answer

Expert verified

Hence proved,

$- \frac{1}{Z} \frac{\partial Z}{\partial β} = \bar{E}$

Step by step solution

Given information

For any system in equilibrium with a reservoir at temperature T, the average value of the energy is

$\bar{E} = - \frac{1}{Z} \frac{\partial Z}{\partial β} = - \frac{\partial}{\partial β} \ln Z$

Explanation

The partition function is given by:

$Z = \sum_{s} e^{- β E (s)}$

Where $β$ is Boltzmann's constant and is given by $β = \frac{1}{k T}$

By partial derivative of partition equation with respect to $β$

$\frac{\partial Z}{\partial β} = \sum_{s} - E (s) e^{- β E (s)}$

Multiplying by a factor of $- 1 / Z$

$- \frac{1}{Z} \frac{\partial Z}{\partial β} = \sum_{s} E (s) \frac{e^{- β E (s)}}{Z}$

The probability is given as:

$P = \frac{1}{Z} e^{- β E (s)}$

Equation (1) becomes,

role="math" localid="1647371210348" $- \frac{1}{Z} \frac{\partial Z}{\partial β} = \sum_{s} E (s) P (s)$

The average energy is:

role="math" localid="1647371238971" $\bar{E} = \sum_{s} E (s) P (s)$

Therefore,

$- \frac{1}{Z} \frac{\partial Z}{\partial β} = \bar{E}$

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Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy is E¯=-1Z∂Z∂β=-∂∂βlnZwhere β=1/kT. These formulas can be extremely useful when you have an explicit formula for the partition function.

Short Answer

Step by step solution

Given information

Explanation

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Most popular questions from this chapter

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