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Critical point exponents inequalities - The Rushbrooke inequality

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LagrangeEuler
#1
May1-12, 04:01 AM
P: 312
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{T-T_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)^{\beta-1}[/tex]

[tex](-\epsilon)^{-\alpha'} \geq \frac{(-\epsilon)^{2\beta-2}}{(-\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_H-C_M)=T\alpha_H^2[/tex]
But I don't know how?
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LagrangeEuler
#2
Jun4-12, 10:37 AM
P: 312
Any idea?

Problem is with

[tex]\chi_T(C_H-C_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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