- #1
roam
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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 :confused: :confused:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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.
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}##.
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.