How Does Weiss Molecular Field Theory Model Magnetization Symmetry?

[MathematicalPhysicist]

This question is more about the maths than the physics.

So I am reading the textbook by Bergersen and Plischke, and they get the following:

$$m= \tanh [ \beta (qJm+h)]$$

where $m$ is the magnetization, $q$ is the number of nearest neighbours of site $0$, $J$ and $h$are the coefficients in the Hamiltonian: $H = -J\sum_{<ij>} \sigma_i \sigma_j -h \sum_i \sigma_i$;

For the question, they write that $m(h,T)$ satisfies: $m(h,T) = -m(-h,T)$;
but I tried to showed this and I didn't succeed.

Here's my attempt:

$$-m(-h,T) = -\tanh [ \beta(-qJm(-h,T)-h) ] = m(-h,T)$$

I used the fact that $\tanh(-x) = -\tanh(x)$.

Am I wrong?
Did they mean something else here?

Thanks.

 

[DeathbyGreen]

In your original expression for $m$ is seems it is a function of $q,J, \beta ,h$ and there is also another $m$ inside this expression. Why are there two m's? And I also don't see the T dependence from your statement $m \rightarrow m(h,T)$, either in your m expression or in the Hamiltonian, unless it's somehow derived from J.

 

[MathematicalPhysicist]

In your original expression for $m$ is seems it is a function of $q,J, \beta ,h$ and there is also another $m$ inside this expression. Why are there two m's? And I also don't see the T dependence from your statement $m \rightarrow m(h,T)$, either in your m expression or in the Hamiltonian, unless it's somehow derived from J.

The $m$ in the LHS and the one in the RHS are the same or at least that's what I think is the case.

As for the $T$ dependence, as is well known in statistical mechanics textbooks we denote by $\beta := \frac{1}{k_B T}$ where $k_B$ is Boltzmann constant and $T$ is the temperature.

 

[DeathbyGreen]

Its confusing to think about if it is the same m on both sides. But maybe they mean:
$-m(-h,T) = -\tanh(\beta[qJ(-m(-h,T))-h]) = -\tanh(-\beta[qJm(-h,T)+h])=\tanh(\beta[qJm(-h,T)+h])=m(h,T,m(-h,T))$

In other words, the equality is $-m(-h,T,m(-h,T))=m(h,T,m(-h,T))$

 

[MathematicalPhysicist]

Its confusing to think about if it is the same m on both sides. But maybe they mean:
$-m(-h,T) = -\tanh(\beta[qJ(-m(-h,T))-h]) = -\tanh(-\beta[qJm(-h,T)+h])=\tanh(\beta[qJm(-h,T)+h])=m(h,T,m(-h,T))$

In other words, the equality is $-m(-h,T,m(-h,T))=m(h,T,m(-h,T))$

Seems so, thanks.

Quite recursive here.

 

[DrClaude]

The choice of the orientation of the axis along which spin is measured is arbitrary. Therefore, the solution should be invariant to that choice of orientation. Changing $h$ to $-h$ corresponds to inverting which spin state is the ground state and which is the excited state, so the magnetization should change sign, but not amplitude.

If $m$ is a solution of
$$
m= \tanh [ \beta (qJm+h)]
$$
then let $\bar{m}$ be the result when both $m$ and $h$ change sign:
$$
\begin{align*}
\bar{m} &= \tanh [ \beta (qJ(-m)-h)] \\
&= \tanh [ -\beta (qJm+h)] \\
&= -m
\end{align*}
$$
which is consistent with the assumption that $-m$ is a solution to the equation when the substitution $h \rightarrow -h$ is made, so $-m(-h,T) = m(h,T)$.