Since $I'(m)=\operatorname{artanh}m$, stationarity gives
$$
\operatorname{artanh}m_e
=\beta(h+g-2Jd\,m_o),
$$
$$
\operatorname{artanh}m_o
=\beta(h-g-2Jd\,m_e).
$$
Therefore
$$
m_e=\tanh[\beta A(m_o)],
\qquad
m_o=\tanh[\beta D(m_e)],
$$
where
$$
\boxed{A(m_o)=h+g-2Jd\,m_o,
\qquad D(m_e)=h-g-2Jd\,m_e.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 303 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-303/1/b/ii/solution