# Isothermal Magnetic Susceptibility

1. Apr 28, 2014

### roam

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.

2. Apr 28, 2014

### nasu

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.

3. Apr 29, 2014

### roam

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)?