Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of $-{\mu }_{\mathrm{B}}$B if it points up and $+{\mu }_{\mathrm{B}}$B if it points down (where${\mu }_{\mathrm{B}}$ is the dipole's magnetic moment). Analyse this system using the mean field approximation to find the analogue of equation 8.50. Study the solutions of the equation graphically, and discuss the magnetisation of this system as a function of both the external field strength and the temperature. Sketch the region in the T-B plane for which the equation has three solutions.

Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of $-{\mu }_{\mathrm{B}}$B if it points up and $+{\mu }_{\mathrm{B}}$B if it points down (where ${\mu }_{\mathrm{B}}$ 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:

$ϵ=\mu B$

Where,

$\mu$is the dipole moment and it has two allowed magnetic orientation, hence:

$ϵ=\pm \mu B$

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

$U=-ϵ\sum s_i{s}_j\pm \mu B$

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

$\uparrow \uparrow :-ϵ\downarrow \uparrow :ϵ\uparrow \downarrow :ϵ\downarrow \downarrow :-ϵ$

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:

$E_{\uparrow }=-ϵn\overline{s}-\mu B{E}_{\downarrow }=ϵn\overline{s}+\mu B$

The partition function is:

$$\begin{gathered} Z=\sum e^{-\beta E} \\ Z=e^{-\beta E_{\uparrow }}+e^{\beta E_{\downarrow }} \end{gathered}$$

Thus,

$$\begin{gathered} Z=e^{\beta (ϵn\overline{s}+\mu B)}+e^{-\beta (ϵn\overline{s}+\mu B)} \\ Z=2\cosh \left(\beta \right(ϵn\overline{s}+\mu B\left)\right)\left(1\right) \end{gathered}$$

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

$\overline{s}=\frac{1}{Z}\left[e^{\beta (ϵn\overline{s}+\mu B)}-e^{-\beta (ϵn\overline{s}+\mu B)}\right]$

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

$\overline{s}=\frac{2\sinh \left(\beta \right(ϵn\overline{s}+\mu B\left)\right)}{Z}$

Substitute from (1):

$$\begin{gathered} \overline{s}=\frac{2\sinh \left(\beta \right(ϵn\overline{s}+\mu B\left)\right)}{2\cosh \left(\beta \right(ϵn\overline{s}+\mu B\left)\right)} \\ \overline{s}=\tanh \left(\beta \right(ϵn\overline{s}+\mu B\left)\right) \end{gathered}$$

When magnetic field is zero, the equation will be:

$\overline{s}=\tanh \left(\beta ϵn\overline{s}\right)$

To plot this function, we substitute with: $\beta =1/k{T}_{c}andt=kT/nϵ$

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

we have two cases the first one when t = kT/n$\epsilon$> 1, in this case we have only one solution and when t = kT/n$\epsilon$<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: