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Critical point exponents inequalities  The Rushbrooke inequality 
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#1
May112, 04:01 AM

P: 316

The Rushbrooke inequality: [tex]H=0, T\rightarrow T_c^[/tex]
[tex]C_H \geq \frac{T\{(\frac{\partial M}{\partial T})_H\}^2}{\chi_T}[/tex] [tex]\epsilon=\frac{TT_c}{T_c}[/tex] [tex]C_H \sim (\epsilon)^{\alpha'}[/tex] [tex]\chi_T \sim (\epsilon)^{\gamma'}[/tex] [tex]M \sim (\epsilon)^{\beta}[/tex] [tex](\frac{\partial M}{\partial T})_H \sim (\epsilon)^{\beta1}[/tex] [tex](\epsilon)^{\alpha'} \geq \frac{(\epsilon)^{2\beta2}}{(\epsilon)^{\gamma'}}[/tex] and we get Rushbrooke inequality [tex]\alpha'+2\beta+\gamma' \geq 2[/tex] My only problem here is first step [tex]C_H \geq \frac{T\{(\frac{\partial M}{\partial T})_H\}^2}{\chi_T}[/tex] we get this from identity [tex]\chi_T(C_HC_M)=T\alpha_H^2[/tex] But I don't know how? 


#2
Jun412, 10:37 AM

P: 316

Any idea?
Problem is with [tex]\chi_T(C_HC_M)=T\alpha^2_H[/tex] from that relation we obtain [tex]C_H=\frac{T\alpha^2_H}{\chi_T}+C_M[/tex] So if [tex]C_M>0[/tex] than [tex]C_H>\frac{T\alpha^2_H}{\chi_T}[/tex] Equality is in the game if and only if [tex]C_M=0[/tex]. Or when [tex](\frac{\partial^2 F}{\partial T^2})_M=0[/tex]. Is that possible? F is Helmholtz free energy. Can you tell me something more about that physically? 


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