# Critical point exponents inequalities - The Rushbrooke inequality

1. May 1, 2012

### 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?

2. Jun 4, 2012

### 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?