91影视

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

$系=渭B$

Where,

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

$系=卤渭B$

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

$U=-系鈭�s_i{s}_j卤渭B$

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

$鈫�鈫�:-系鈫�鈫�:系鈫�鈫�:系鈫�鈫�:-系$

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:

$E_{鈫�}=-系n\stackrel{炉}{s}-渭B{E}_{鈫�}=系n\stackrel{炉}{s}+渭B$

The partition function is:

$$\begin{gathered} Z=鈭�e^{-尾E} \\ Z=e^{-尾E_{鈫�}}+e^{尾E_{鈫�}} \end{gathered}$$

Thus,

$$\begin{gathered} Z=e^{尾(系n\stackrel{炉}{s}+渭B)}+e^{-尾(系n\stackrel{炉}{s}+渭B)} \\ Z=2\cosh \left(尾\right(系n\stackrel{炉}{s}+渭B\left)\right)\left(1\right) \end{gathered}$$

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

$\stackrel{炉}{s}=\frac{1}{Z}\left[e^{尾(系n\stackrel{炉}{s}+渭B)}-e^{-尾(系n\stackrel{炉}{s}+渭B)}\right]$

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

$\stackrel{炉}{s}=\frac{2\sinh \left(尾\right(系n\stackrel{炉}{s}+渭B\left)\right)}{Z}$

Substitute from (1):

$$\begin{gathered} \stackrel{炉}{s}=\frac{2\sinh \left(尾\right(系n\stackrel{炉}{s}+渭B\left)\right)}{2\cosh \left(尾\right(系n\stackrel{炉}{s}+渭B\left)\right)} \\ \stackrel{炉}{s}=\tanh \left(尾\right(系n\stackrel{炉}{s}+渭B\left)\right) \end{gathered}$$

When magnetic field is zero, the equation will be:

$\stackrel{炉}{s}=\tanh \left(尾系n\stackrel{炉}{s}\right)$

To plot this function, we substitute with: $尾=1/k{T}_{c}andt=kT/n系$

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

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