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Isothermal Magnetic Susceptibility 
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#1
Apr2814, 05:30 AM

P: 916

My book says that "in the mean field approximation, the isothermal magnetic susceptibility just below the Curie temperature goes as ##(T_cT)^{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 M_{s}=0 we get: ##\chi_T = \frac{1}{k_B}\frac{1}{TT_c}## But how do we get ##T_c  T## in the denominator? We need ##\chi_T \propto (T_cT)^{1}## NOT ##(TT_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. 


#2
Apr2814, 08:06 AM

P: 1,970

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 1τ will be negative. So either kBT or the susceptibility should be negative in order to have that equation. 


#3
Apr2914, 08:20 AM

P: 916

Thank you for your response. Unfortunately that information is not provided.
So how else can we demonstrate that magnetic susceptibility is inversely proportional to (T_{c}T)? 


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