The Onsager solution for specific heat is

$$C\approx -Nk\frac{2}{\pi}\bigg(\frac{2J}{kT_c}\bigg)^2\ln\Big|1-\frac{T}{T_c} \Big|\qquad (T \textrm{ near } T_c)$$

Critical exponent $\alpha\neq 0$. Neither it follows a power law. But when we talk about scaling relation or the result we get for mean-field calculation, we say $\alpha = 0$.

Even when we plot any graph

Specific heat

(On x-axis temperature in units T$\equiv Tk_B/J$)

I cannot say $$C = \begin{cases} C_1 & \textrm{for $T<T_c$} \\C_2 & \textrm{for $T>T_c$} \end{cases}$$

where $C_1,C_2$ are some constants.


1 Answer 1


Let $\tau=(T-T_{\rm c})/T_{\rm c}$ and consider the behavior of some quantity $X$ as $T$ approaches $T_{\rm c}$ (that is, $\tau$ approaches $0$). Informally, if the quantity $X$ has a behavior of the type $$ X \sim \tau^a $$ as, say, $\tau\to 0^+$ (convergence to $T_{\rm c}$ form above), then $a$ is the corresponding critical exponent. Now, what would be a precise definition of the exponent $a$? The standard way to define it is as $$ a = \lim_{\tau\to 0^+} \frac{\log \lvert X\rvert}{\log \tau}. \tag{$\star$} $$ Check that, if $X \sim \tau^a$, then the right-hand side gives you indeed the correct exponent. (Note that often the exponent is negative; in this case, one usually considers its absolute value.)

Now, in the case of the specific heat of the 2d Ising model, $C \sim \log\lvert\tau\rvert$. Plugging this into ($\star$) indeed gives you the critical exponent $\alpha = \lim_{\tau\to 0^+} \log\lvert\log\tau\rvert/\log\tau = 0$. (The same is obviously also true when $\tau\to 0^-$.)


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