91Ó°ÊÓ

The simplified model of a magnet is defined as Ising model. For an Ising model of just two elementary dipoles, the energy is $-\mathrm{Î\mu }$when the dipoles are parallel and $+\mathrm{Î\mu }$when the dipoles are antiparallel.

The states of the system on their Boltzmann factors are as follows:

$↑↑:e^{\mathrm{Î\mu }/kT}$

$↑↓:e^{-\mathrm{Î\mu }/kT}$

$↓↑:e^{-\mathrm{Î\mu }/kT}$

$↓↓:e^{\mathrm{Î\mu }/kT}$

Here,

$k$=Boltzmann factor

$T=$temperature.

The above four equations represent Boltzmann's factors for the given four states.

For $\mathrm{Î\mu }$,

$Z_1=2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)$

For $-\mathrm{Î\mu }$,

$Z_2=2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)$

The partition function for the system with two elementary dipoles is as follows:

$Z=Z_1+Z_2$

Substituting the value of $2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)\text{=}{Z}_1\text{and}2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)\text{=}{Z}_2\text{.}$

$Z=2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)+2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)$

$=4\cos h\left(\frac{\mathrm{Î\mu }}{kT}\right)$

The partition function for the system with two elementary dipoles =$=4\cos h\left(\frac{\mathrm{Î\mu }}{kT}\right)$

${\mathrm{P}}_{\text{parallel}}=\frac{Z_2}{Z}$

$\text{Substitutingthevalueof}2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)\text{=}{Z}_2\text{and}2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)+2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)\text{=}Z$

${\mathrm{P}}_{\text{parallel}}=\frac{2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)}{2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)+2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)}$

$=\frac{1}{1+\exp \left(-\frac{2\mathrm{Î\mu }}{kT}\right)}$

Thus, the probability that the dipoles are parallels is=$=\frac{1}{1+\exp \left(-\frac{2\mathrm{Î\mu }}{kT}\right)}$

${\mathrm{P}}_{\text{antiparallel}}=\frac{Z_1}{Z}$

$\text{Substitutingthevalueof}2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)\text{=}{Z}_1\text{and}2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)+2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)\text{=}Z\text{.}$

${\mathrm{P}}_{\text{antiparallel}}=\frac{2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)}{2\exp \left(\frac{\mathrm{Î\mu }}{kT}\right)+2\exp \left(-\frac{\mathrm{Î\mu }}{kT}\right)}$

$=\frac{1}{1+\exp \left(\frac{2\mathrm{Î\mu }}{kT}\right)}$

Thus, the probability that the dipoles are anti-parallels =$=\frac{1}{1+\exp \left(\frac{2\mathrm{Î\mu }}{kT}\right)}$

Average energy of the system=

$U=⟨EâŸ\copyright$

$=-\frac{1}{Z}\frac{∂Z}{∂\mathrm{β}}$

Substituting the value of $4\cos h\left(\frac{\mathrm{Î\mu }}{kT}\right)\text{=}Z\text{.}$

$U=\frac{\mathrm{Î\mu }\left(4\sin h\right(\mathrm{Î\mu }/kT\left)\right)}{4\cos h(\mathrm{Î\mu }/kT)}$

$=-\mathrm{Î\mu }\tanh (\mathrm{Î\mu }/kT)$

$\frac{U}{\mathrm{Î\mu }}=-\tanh (\mathrm{Î\mu }/kT)$