Question

Starting from the partition function, calculate the average energy of the one-dimensional Ising model, to verify equation 8.44. Sketch the average energy as a function of temperature.

Short answer

The average energy =$-N\epsilon \tanh \beta \epsilon$

Sketch for the average energy as function of temperature

Step-by-step solution

Step 1. Given information

The partition function of the system:

$Z=\sum _s{e}^{-{\epsilon }_i/r}$

$Z=\sum _s{e}^{-{\epsilon }_i/r}$

Step 2. To find the expression for average energy 

We substitute$\beta \text{for}1/\tau$

$\mathit{Z}\mathbf{=}\mathbf{\sum }_{\mathbf{s}}{\mathit{e}}^{\mathbf{-}{\mathbf{\epsilon }}_{\mathbf{n}}\mathbf{\beta }}$

The average energy of the system is,

$U=⟨\epsilon ⟩$

$=\frac{1}{Z}\sum _s{\epsilon }_s{e}^{-\beta c_s}$

$=-\frac{1}{Z}{\epsilon }_s\sum _s{e}^{-\beta s_s}$

Substituting the value of $\frac{dZ}{d\beta }\text{=}{\epsilon }_s\sum _s{e}^{-\beta c_s}$

$U=-\frac{1}{Z}\frac{dZ}{d\beta }$

$=-\frac{\partial }{\partial \beta }\left(\ln Z\right)$

The expression for partition function for final sum of $N$dipole is,

$Z=\left(2\cosh \beta \epsilon \right)$

$\ln Z=\ln \left(2\cosh \beta \epsilon \right)$

$U=-\frac{1}{Z}\frac{dZ}{d\beta }$

$=-\frac{\partial }{\partial \beta }\left(\ln \right(2\cosh \beta \epsilon \left)\right)$

$=\frac{1}{\left(2\cosh \beta \epsilon \right)}\left(\right(2\sinh \beta \epsilon \left)\right)\alpha$

$=-N\epsilon \tanh \beta \epsilon$

The average energy =$=-N\epsilon \tanh \beta \epsilon$

Step 3. Sketch for the following function