Critical Exponents in the 1D Ising Model

  • Thread starter MisterX
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  • #1
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Homework Statement


Obtain the critical exponents for specific heat, susceptibility, and the order parameter (magnetization).

Homework Equations


$$A = -k_B T N \ln \left[e^{\beta J} \cosh (\beta h) +\sqrt{ e^{2\beta J}\sinh^2 \beta h + e^{-2\beta J} }\right]$$
$$\left<m \right> \propto \frac{\partial A}{\partial h} \to \frac{\sinh (\beta h)}{\sqrt{\sinh ^2(\beta h)+e^{-4 \beta J}}}$$
$$\chi_T = \left(\frac {\partial \left<m \right>}{\partial h} \right)_T = \frac{\beta \cosh (\beta h)}{\sqrt{\sinh ^2(\beta h)+e^{-4 \beta J}}}-\frac{\beta \sinh ^2(\beta h) \cosh (\beta h)}{\left(\sinh ^2(\beta h)+e^{-4 \beta J}\right)^{3/2}} $$

The Attempt at a Solution


I can understand how to derive these various expressions above but I don't know how to determine the critical exponent using them. The "base" for this exponent shold be ##e^{-\beta J}##.

So I need to show for example ## \chi \sim \left( e^{-\beta J}\right)^p ## as ##\beta \to 0##. I have tried using the formal definition of asymptotic and found exponent 0 for magnetization, but I'm not sure this is correct. Also it appears for the susceptibility there is no value of p for which the susceptibility will be asymptotic. So maybe they critical exponents come from a looser definition then asymptoticity?
 

Answers and Replies

  • #2
760
69
Sorry that should be ##\beta\to \infty##
 

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