Isothermal Magnetic Susceptibility

  1. My book says that "in the mean field approximation, the isothermal magnetic susceptibility just below the Curie temperature goes as ##(T_c-T)^{-1}##". I need some help understanding how to get this proportionality. My book does not contain any derivation or further explanations.

    According to my notes the isothermal magnetic susceptibility ##\chi_T## diverges near ##T_c##:

    ##\chi_T = \frac{\partial M}{\partial H} |_T##

    Differentiating the equation of state we get:

    ##\frac{1}{k_B T} = \chi_T (1- \tau) +3M_s^2 \chi_T \left( \tau - \tau^2 + \frac{\tau^3}{3} \right)##

    Where ##\tau=T_c/T##. If Ms=0 we get:

    ##\chi_T = \frac{1}{k_B}\frac{1}{T-T_c}##

    But how do we get ##T_c - T## in the denominator? We need ##\chi_T \propto (T_c-T)^{-1}## NOT ##(T-T_c)^{-1}##. :confused:

    Also are we justified to set magnetization to 0 for ##T<T_c##? I did this because the books says "just below the Curie temperature", so I assumed it's almost 0 just as it would be for ##T>T_c##.

    Any explanation is greatly appreciated.
     
  2. jcsd
  3. Can you show the equation of state?
    Something is not right. If τ=Tc/T, at T<Tc this will be larger than 1 so 1-τ will be negative.
    So either kBT or the susceptibility should be negative in order to have that equation.
     
  4. Thank you for your response. Unfortunately that information is not provided.

    So how else can we demonstrate that magnetic susceptibility is inversely proportional to (Tc-T)? :confused:
     
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