91Ó°ÊÓ

Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of $-{\mathrm{μ}}_{\mathrm{B}}$B if it points up and $+{\mathrm{μ}}_{\mathrm{B}}$B if it points down (where ${\mathrm{μ}}_{\mathrm{B}}$ is the dipole's magnetic moment).

Assume we have dipoles on an external magnetic field, and the energy of the interaction between the dipole and the external magnetic field is:

$\mathrm{Ï\mu }=\mathrm{μ}B$

Where,

$\mathrm{μ}$is the dipole moment and it has two allowed magnetic orientation, hence:

$\mathrm{Ï\mu }=Â\pm \mathrm{μ}B$

The total energy of the system resulting from all interactions with its nearest neighbours is:

$U=-\mathrm{Ï\mu }∑s_i{s}_jÂ\pm \mathrm{μ}B$

Consider a lsing model with two dipoles. This system has four alignments, which are as follows:

$↑↑:-\mathrm{Ï\mu }↓↑:\mathrm{Ï\mu }↑↓:\mathrm{Ï\mu }↓↓:-\mathrm{Ï\mu }$

where s_i =-1 when the dipole pointing down and s_i= l when the dipole pointing up, for two dipoles in an external magnetic field, we have two energies, that are:

$E_{↑}=-\mathrm{Ï\mu }n\stackrel{¯}{s}-\mathrm{μ}B{E}_{↓}=\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B$

The partition function is:

$$\begin{gathered} Z=∑e^{-\mathrm{β}E} \\ Z=e^{-\mathrm{β}{E}_{↑}}+e^{\mathrm{β}{E}_{↓}} \end{gathered}$$

Thus,

$$\begin{gathered} Z=e^{\mathrm{β}(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B)}+e^{-\mathrm{β}(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B)} \\ Z=2\cosh \left(\mathrm{β}\right(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B\left)\right)\left(1\right) \end{gathered}$$

The average expected value for the spin alignment is given by (from equation 8.49):

$\stackrel{¯}{s}=\frac{1}{Z}\left[e^{\mathrm{β}(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B)}-e^{-\mathrm{β}(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B)}\right]$

Using $\sinh \left(x\right)=\left(e^x-e^{-x}\right)/2$

$\stackrel{¯}{s}=\frac{2\sinh \left(\mathrm{β}\right(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B\left)\right)}{Z}$

Substitute from (1):

$$\begin{gathered} \stackrel{¯}{s}=\frac{2\sinh \left(\mathrm{β}\right(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B\left)\right)}{2\cosh \left(\mathrm{β}\right(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B\left)\right)} \\ \stackrel{¯}{s}=\tanh \left(\mathrm{β}\right(\mathrm{Ï\mu }n\stackrel{¯}{s}+\mathrm{μ}B\left)\right) \end{gathered}$$

When magnetic field is zero, the equation will be:

$\stackrel{¯}{s}=\tanh \left(\mathrm{β}\mathrm{Ï\mu }n\stackrel{¯}{s}\right)$

To plot this function, we substitute with: $\mathrm{β}=1/k{T}_{c}andt=kT/n\mathrm{Ï\mu }$

$\stackrel{¯}{s}=\tanh \left(\frac{\stackrel{¯}{s}}{t}\right)$

we have two cases the first one when t = kT/n$\mathrm{Î\mu }$> 1, in this case we have only one solution and when t = kT/n$\mathrm{Î\mu }$<l we have three solutions, two of them are stable and the third one is unstable, as shown in the following figure (the first figure is when t>1 and the second one is when t < 1)

The following codes were used to plot the graph:

When we have a non zero magnetic field, then we modify the code as follows to have the following graph: