# The Griffiths inequality

1. Sep 4, 2012

### matematikuvol

Zero field magnetisation like a function of temperature vanished in $T=T_c$ as $(T_c-T)^{\beta}$. Let $M_1$ be a magnetisation for temperature $T_1$. Since $\forall M<M_1$, $(\frac{\partial A}{\partial M})_T=H=0$ it follows that
$$A(T_1,M)=A(T_1,0)$$ for $M \leq M_1(T_1)$
Why only for $M \leq M_1(T_1)$?
Now define the function
$$A^{*}(T,M)=\{A(T,M)-A_c\}+(T-T_c)S_c$$
$$S^{*}(T,M)=S(T,M)-S_c$$
So $S^{*}(T,M)=-(\frac{\partial A^{*}}{\partial T})_M$
Now in Griffiths construction I need to drive a tangent in point $T=T_1$. Eq of tangent is
$$f(T)=A^{*}(T_1,M_1)+(T-T_1)(\frac{\partial A^{*}}{\partial T})_{T=T_1}$$
I can't visualise this.