Critical exponents of Weiss mean field theory for the Ising model

We can also try to understand something more about our system, for example computing its critical exponents.

Exponent beta

Let us begin with the exponent $\beta$$\beta$, which (see Critical exponents and universality) is the one relative to the order parameter, namely the magnetization $m$$m$. In order to find it we must see how the magnetization varies as a function of $T_{c}-T$$T_{c}-T$; from the expansion of $f$$f$ we have just seen, imposing the extremal condition we get:

$$0={\frac {\partial f}{\partial m}}_{|{\overline {m}}}\sim Jz(1-\beta Jz){\overline {m}}+{\frac {\beta ^{3}}{3}}J^{4}z^{4}{\overline {m}}^{3}={\overline {m}}\left[{\frac {Jz}{T}}\left(T-T_{c}\right)+{\frac {\beta ^{3}}{3}}J^{4}z^{4}{\overline {m}}^{2}\right]$$

$$0={\frac {\partial f}{\partial m}}_{|{\overline {m}}}\sim Jz(1-\beta Jz){\overline {m}}+{\frac {\beta ^{3}}{3}}J^{4}z^{4}{\overline {m}}^{3}={\overline {m}}\left[{\frac {Jz}{T}}\left(T-T_{c}\right)+{\frac {\beta ^{3}}{3}}J^{4}z^{4}{\overline {m}}^{2}\right]$$

where we have used the definition of critical temperature $T_{c}=Jz/k_{B}$$T_{c}=Jz/k_{B}$. Therefore (excluding of course the case ${\overline {m}}=0$${\overline {m}}=0$):

$${\overline {m}}^{2}\sim {\text{const.}}\cdot (T_{c}-T)\quad \Rightarrow \quad {\overline {m}}\sim (T_{c}-T)^{1/2}$$

$${\overline {m}}^{2}\sim {\text{const.}}\cdot (T_{c}-T)\quad \Rightarrow \quad {\overline {m}}\sim (T_{c}-T)^{1/2}$$

We therefore have:

$$\beta ={\frac {1}{2}}$$

$$\beta ={\frac {1}{2}}$$

Exponent delta

The exponent $\delta$$\delta$ is the one that describes the behaviour of $H$$H$ as a function of $m$$m$ when $T=T_{c}$$T=T_{c}$. To compute that let us begin from the self-consistency equation for $m$$m$ in the case $H\neq 0$$H\neq 0$, and invert it:

$$m=\tanh(\beta (Jzm+H))\quad \Rightarrow \quad \beta (Jzm+H)=\tanh ^{-1}m$$

$$m=\tanh(\beta (Jzm+H))\quad \Rightarrow \quad \beta (Jzm+H)=\tanh ^{-1}m$$

Now expanding for small $m$$m$, since $\tanh ^{-1}x\sim x+x^{3}/3+x^{5}/5$$\tanh ^{-1}x\sim x+x^{3}/3+x^{5}/5$ we have:

$$Jzm+H\sim k_{B}Tm+{\frac {k_{B}T}{3}}m^{3}\quad \Rightarrow \quad H\sim k_{B}m(T-T_{c})+{\frac {k_{B}T}{3}}m^{3}$$

$$Jzm+H\sim k_{B}Tm+{\frac {k_{B}T}{3}}m^{3}\quad \Rightarrow \quad H\sim k_{B}m(T-T_{c})+{\frac {k_{B}T}{3}}m^{3}$$

and if we set $T=T_{c}$$T=T_{c}$:

$$H\sim m^{3}$$

$$H\sim m^{3}$$

Therefore:

$$\delta =3$$

$$\delta =3$$

Exponent alpha

This exponent is the one that describes the behaviour of the specific heat $C_{H}=-T\partial ^{2}f/\partial T_{|H}^{2}$$C_{H}=-T\partial ^{2}f/\partial T_{|H}^{2}$ when $H=0$$H=0$. If $T>T_{c}$$T>T_{c}$ then ${\overline {m}}=0$${\overline {m}}=0$, and since $f\sim -k_{B}T\ln 2$$f\sim -k_{B}T\ln 2$ we have $C_{H}=0$$C_{H}=0$. Therefore:

$$\alpha =0$$

$$\alpha =0$$

For $T<T_{c}$$T<T_{c}$ the specific heat has a different dependence on the temperature, but in the end it turns out that $C_{H}$$C_{H}$ has a jump discontinuity at $T=T_{c}$$T=T_{c}$, so from the definition of critical exponent (see Critical exponents and universality) we indeed have $\alpha =0$$\alpha =0$ (see Landau theory for the Ising model for a slightly more detailed computation).

Exponent gamma

This is the exponent that regulates the isothermal susceptibility:

$$\chi _{T}={\frac {\partial m}{\partial H}}={\frac {1}{\partial H/\partial m}}$$

$$\chi _{T}={\frac {\partial m}{\partial H}}={\frac {1}{\partial H/\partial m}}$$

From the computation of the exponent $\delta$$\delta$ we have seen that:

$$H\sim k_{B}m(T-T_{c})+{\frac {k_{B}T}{3}}m^{3}$$

$$H\sim k_{B}m(T-T_{c})+{\frac {k_{B}T}{3}}m^{3}$$

so:

$${\frac {\partial H}{\partial m}}\sim k_{B}(T-T_{c})+k_{B}Tm^{2}$$

$${\frac {\partial H}{\partial m}}\sim k_{B}(T-T_{c})+k_{B}Tm^{2}$$

Therefore for small $m$$m$, neglecting the quadratic term:

$$\chi _{T}\sim {\frac {1}{k_{B}}}(T-T_{c})^{-1}$$

$$\chi _{T}\sim {\frac {1}{k_{B}}}(T-T_{c})^{-1}$$

and thus:

$$\gamma =1$$

$$\gamma =1$$

To recap, the Weiss mean field theory for magnetic systems predicts the following values for the critical exponents:

$$\alpha =0\quad \qquad \beta ={\frac {1}{2}}\quad \qquad \gamma =1\quad \qquad \delta =3$$

$$\alpha =0\quad \qquad \beta ={\frac {1}{2}}\quad \qquad \gamma =1\quad \qquad \delta =3$$

We can immediately note that these exponents are different from those found by Onsager for the Ising model in two dimensions, so the mean field theory is giving us wrong predictions. As we have already stated (but this will be treated in much more detail later on) this is because mean field theories are good approximations only if the system has a high enough dimensionality (and $d=2$$d=2$ is still too low for the Ising model, see Coarse graining procedure for the Ising model).