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Isothermal Magnetic Susceptibility

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roam
#1
Apr28-14, 05:30 AM
P: 895
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}##.

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.
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nasu
#2
Apr28-14, 08:06 AM
P: 1,969
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.
roam
#3
Apr29-14, 08:20 AM
P: 895
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)?


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