Critical point exponents inequalities - The Rushbrooke inequality

[LagrangeEuler]

The Rushbrooke inequality: H=0, T\rightarrow T_c^-

C_H \geq \frac{T\{(\frac{\partial M}{\partial T})_H\}^2}{\chi_T}
\epsilon=\frac{T-T_c}{T_c}

C_H \sim (-\epsilon)^{-\alpha'}
\chi_T \sim (-\epsilon)^{-\gamma'}
M \sim (-\epsilon)^{\beta}
(\frac{\partial M}{\partial T})_H \sim (-\epsilon)^{\beta-1}

(-\epsilon)^{-\alpha'} \geq \frac{(-\epsilon)^{2\beta-2}}{(-\epsilon)^{-\gamma'}}

and we get Rushbrooke inequality
\alpha'+2\beta+\gamma' \geq 2
My only problem here is first step

C_H \geq \frac{T\{(\frac{\partial M}{\partial T})_H\}^2}{\chi_T}
we get this from identity
\chi_T(C_H-C_M)=T\alpha_H^2
But I don't know how?

 

[LagrangeEuler]

Any idea?

Problem is with

\chi_T(C_H-C_M)=T\alpha^2_H

from that relation we obtain

C_H=\frac{T\alpha^2_H}{\chi_T}+C_M

So if C_M>0 than

C_H>\frac{T\alpha^2_H}{\chi_T}

Equality is in the game if and only if C_M=0.

Or when (\frac{\partial^2 F}{\partial T^2})_M=0. Is that possible?

F is Helmholtz free energy. Can you tell me something more about that physically?