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The Griffiths inequality

  1. Sep 4, 2012 #1
    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
    [tex]A(T_1,M)=A(T_1,0)[/tex] for ##M \leq M_1(T_1)##
    Why only for ##M \leq M_1(T_1)##?
    Now define the function
    [tex]A^{*}(T,M)=\{A(T,M)-A_c\}+(T-T_c)S_c[/tex]
    [tex]S^{*}(T,M)=S(T,M)-S_c[/tex]
    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
    [tex]f(T)=A^{*}(T_1,M_1)+(T-T_1)(\frac{\partial A^{*}}{\partial T})_{T=T_1}[/tex]
    I can't visualise this.
     
  2. jcsd
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