Calculating the specific heat capacity for the 2D Ising model

  • #1

julian

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TL;DR Summary
I'm calculating the specific heat capacity. I nearly get the right answer. Is there a typo in the book?
So I'm looking at the book "Equilibrium Statistical physics" by Plischke and Bergersen. I'm doing the calculation of the specific heat of the 2D Ising model. I can't seen to quite get out the same expression as in the book - there are a coupe of minus signs that are different. I don't know if I have made a mistake or if the book has a typo. So here's the calculations that I did:

We have the identity

\begin{eqnarray}
\frac{d K_1 (q)}{d q} = \dfrac{E_1 (q)}{q (1 - q^2)} - \frac{K_1 (q)}{q} .
\nonumber
\end{eqnarray}

We now substitute in the explicit expression for q into this. The expression for q is

$$
q (K) = \dfrac{2 \sinh 2 K}{\cosh^2 2 K}
$$

implying

$$
q (1 - q^2) = \dfrac{2 \sinh 2 K}{\cosh^2 2 K} \left( 1 - \dfrac{4 \sinh^2 2 K}{\cosh^4 2 K} \right) = \dfrac{2 \sinh 2 K (1 - \sinh^2 2 K)^2}{\cosh^6 2 K}
$$

so

\begin{eqnarray}
\frac{d K_1 (q)}{d q} &=& \dfrac{E_1 (q)}{q (1 - q^2)} - \frac{K_1 (q)}{q}
\nonumber \\
&=& \dfrac{\cosh^6 2 K}{2 \sinh 2 K (1 - \sinh^2 2 K)^2} E_1 (q) - \dfrac{\cosh^2 2 K}{2 \sinh 2 K} K_1 (q)
\nonumber \\
\end{eqnarray}

We then have

\begin{eqnarray}
\frac{d K_1(q)}{d K} &=& \frac{d q}{d K} \frac{d K_1(q)}{d q}
\nonumber \\
&=& \dfrac{4 (1 - \sinh^2 2 K)}{\cosh^3 2 K} \left[ \dfrac{\cosh^6 2 K}{2 \sinh 2 K (1 - \sinh^2 2 K)^2} E_1 (q) - \dfrac{\cosh^2 2 K}{2 \sinh 2 K} K_1 (q) \right]
\nonumber \\
&=& \dfrac{2 \cosh^3 2 K}{\sinh 2 K (1 - \sinh^2 2 K)} E_1 (q) - \dfrac{2 (2 - \cosh^2 2 K)}{\cosh 2 K \sinh 2 K} K_1 (q)
\nonumber \\
&=& \dfrac{2 \coth 2 K \cosh^2 2 K}{1 - \sinh^2 2 K} E_1 (q) + 4 \coth 2 K (\tanh^2 2 K - 1) K_1 (q) + 2 \coth 2 K K_1 (q)
\nonumber \\
&=& \coth 2 K \left[ \dfrac{2 \cosh^2 2 K}{1 - \sinh^2 2 K} E_1 (q) + 2 (2 \tanh^2 2 K - 1) K_1 (q) \right]
\nonumber \\
\end{eqnarray}

The specific heat is given by

\begin{eqnarray}
\frac{1}{k_B} c (T) &=& \beta^2 \dfrac{\partial^2 \beta g(T)}{\partial \beta^2}
\nonumber \\
&=& \beta^2 \dfrac{\partial u(T)}{\partial \beta}
\nonumber
\end{eqnarray}

where

##
u (T) = - J \coth 2 K \left[ 1 + \dfrac{2}{\pi} (2 \tanh^2 2 K - 1) K_1 (q) \right]
##

Using this in the above expression for ##\frac{1}{k_B} c(T)##

\begin{eqnarray}
&\;& \frac{1}{k_B} c (T) = \beta^2 \frac{\partial}{\partial \beta} u (T)
\nonumber \\
&=& \beta^2 J \frac{d}{d K} \left( - J \coth 2 K \left[ 1 + \frac{2}{\pi} (2 \tanh^2 2 K - 1) K_1 (q) \right] \right)
\nonumber \\
&=& - \beta^2 J^2 \frac{d}{d K} \coth 2 K \left[ 1 + \frac{2}{\pi} (2 \tanh^2 2 K - 1) K_1 (q) \right]
\nonumber \\
&\;& - \beta^2 J^2 \coth 2 K \left[ \frac{8}{\pi} K_1 (q) \tanh 2 K \frac{d}{d K} \tanh 2 K \right]
- \beta^2 J^2 \coth 2 K \left[ \frac{2}{\pi} (2 \tanh^2 2 K - 1) \frac{d K_1 (q)}{d K} \right]
\nonumber \\
&=& - 2 \beta^2 J^2 (\coth^2 2 K - 1) \left[ 1 + \frac{2}{\pi} (2 \tanh^2 2 K - 1) K_1 (q) \right]
\nonumber \\
&\;& - \beta^2 J ^2 \coth 2 K \left[ \frac{16}{\pi} \tanh 2 K (1 - \tanh^2 2 K) K_1 (q) \right]
\nonumber \\
&\;& - \beta^2 J^2 \coth 2 K \left[ \frac{2}{\pi} (2 \tanh^2 2 K - 1) \coth 2 K \left( \dfrac{2 \cosh^2 2 K}{1 - \sinh^2 2 K} E_1 (q) + 2 (2 \tanh^2 2 K - 1) K_1 (q) \right) \right]
\nonumber \\
&=& - 2 \beta^2 J^2 (\coth^2 2 K - 1)
\nonumber \\
&\;& - \frac{4}{\pi} \beta^2 J^2 (\coth^2 2 K - 1) (2 \tanh^2 2 K - 1) K_1 (q)
\nonumber \\
&\;& - \frac{4}{\pi} (\beta J \coth 2 K)^2 (1 - \tanh^2 2 K) 4 \tanh^2 2 K K_1 (q)
- \frac{4}{\pi} (\beta J \coth 2 K)^2 (2 \tanh^2 2 K - 1)^2 K_1 (q)
\nonumber \\
&\;& - \frac{4}{\pi} (\beta J \coth 2 K)^2 \dfrac{(2 \tanh^2 2 K - 1) \cosh^2 2 K}{1 - \sinh^2 2 K} E_1 (q)
\nonumber \\
&=& - \frac{4}{\pi} (K \coth 2 K)^2 (1 - \tanh^2 2 K) \frac{\pi}{2}
\nonumber \\
&\;& - \frac{4}{\pi} (K \coth 2 K)^2 (1 - \tanh^2 2 K) (2 \tanh^2 2 K - 1) K_1 (q)
\nonumber \\
&\;& - \frac{4}{\pi} (K \cosh 2 K)^2 K_1 (q) + \frac{4}{\pi} (K \cosh 2 K)^2 E_1 (q)
\nonumber \\
&=& \frac{4}{\pi} (K \cosh^2 2 K)^2 \left\{ - K_1 (q) + E_1 (q) - (1 - \tanh^2 2 K) \left[ \frac{\pi}{2} + (2 \tanh^2 2 K - 1) K_1 (q) \right] \right\}
\nonumber
\end{eqnarray}

But the book says the answer is:

##
\frac{4}{\pi} (K \cosh^2 2 K)^2 \left\{ K_1 (q) - E_1 (q) - (1 - \tanh^2 2 K) \left[ \frac{\pi}{2} + (2 \tanh^2 2 K - 1) K_1 (q) \right] \right\}
##

Have I made a mistake or is there a typo in the book? (I think I have already found a couple of typos in the book).
 
Last edited:

Answers and Replies

  • #2
Have I written down the wrong formula for the specific heat? I think it should be:

##
\dfrac{1}{k_B} c (T) = \dfrac{1}{k_B} \dfrac{\partial }{\partial T} u (T) = \dfrac{1}{k_B} \dfrac{\partial \beta}{\partial T} \dfrac{\partial }{\partial \beta} u (T) = - \dfrac{1}{k_B} \dfrac{1}{k_B T^2} \dfrac{\partial }{\partial \beta} u (T) = - \beta^2 \dfrac{\partial }{\partial \beta} u (T) .
##

So that my expression becomes:

\begin{align*}
\dfrac{1}{k_B} c (T) = \frac{4}{\pi} (K \cosh^2 2 K)^2 \left\{ K_1 (q) - E_1 (q) + (1 - \tanh^2 2 K) \left[ \frac{\pi}{2} + (2 \tanh^2 2 K - 1) K_1 (q) \right] \right\}
\end{align*}

So there is still a different sign from the book's expression, but in front of the ##(1 - \tanh^2 2 K)## instead!

I think you need a positive sign in front of ##K_1 (q)## to get a positive value for the specific heat (calculated near the critical temperature, ##T_c##, defined by ##\sinh \dfrac{2J}{k_B T_c} = 1##) :

The Elliptic function of the first kind has the asymptotic behaviour:

##
K_1 (q) \sim - \frac{1}{2} \ln | 1 - q | \quad \text{as } q \rightarrow 1^-
##

Say ##C## is defined by ##\sinh C = 1##, then we have the expansions:

\begin{align*}
\sinh (C + x) &= \sinh C+ x \cosh C + \dfrac{x^2}{2!} \sinh C + \cdots = 1 + \sqrt{2} x + \dfrac{x^2}{2} + \cdots
\nonumber \\
\cosh (C + x) &= \cosh C + x \sinh C + \dfrac{x^2}{2!} \cosh C + \cdots = \sqrt{2} + x + \dfrac{1}{\sqrt{2}} x^2 + \cdots
\end{align*}

so that

\begin{align*}
q (C + x) &= \dfrac{2 \sinh (C+ x)}{\cosh^2 (C + x)}
\nonumber \\
&= \dfrac{2 (1 + \sqrt{2} x + x^2 + \cdots )}{(\sqrt{2} + x + \dfrac{1}{\sqrt{2}} x^2 + \cdots)^2} = \dfrac{1 + \sqrt{2} x + \cdots }{1 + \sqrt{2} x + \cdots}
\nonumber \\
&= (1 + \sqrt{2} x + \cdots ) (1 - \sqrt{2} x + \cdots )
\nonumber \\
&= 1 - 2 x^2 + \cdots
\end{align*}

Using this we approximate ##q (K)## for temperatures, ##T##, near to the critical temperature, ##T_c##. We have the definition ##2 K = \dfrac{2J}{k_B T}##, so that:

##
q (K) = q \left( \dfrac{2 J}{k_B T_c} \dfrac{1}{1 + \dfrac{T - T_c}{T_c}} \right) \approx q \left ( \dfrac{2 J}{k_B T_c} \left( 1 - \dfrac{T - T_c}{T_c} \right) \right) \approx 1 - 2 \left( \dfrac{2 J}{k_B T_c} \right)^2 \left( 1 - \dfrac{T}{T_c} \right)^2
##

where we assumed that ##T > T_c##.

##
q (K) = q \left( \dfrac{2 J}{k_B T_c} \dfrac{1}{1 - \dfrac{T_c - T}{T_c}} \right) \approx q \left ( \dfrac{2 J}{k_B T_c} \left( 1 + \dfrac{T_c - T}{T_c} \right) \right) \approx 1 - 2 \left( \dfrac{2 J}{k_B T_c} \right) \left( 1 - \dfrac{T}{T_c} \right)^2
##

where we assumed that ##T < T_c##. Using this in the asymptotic expression for ##K_1 (q)## we have

##
K_1 (q) \sim - \frac{1}{2} \ln | 1 - q | \sim - \ln \left| 1 - \dfrac{T}{T_c} \right|
##

and we have for the specific heat

##
\dfrac{1}{k_B} c (T) \approx - \dfrac{2}{\pi} \left( \dfrac{2J}{k_B T_C} \right)^2 \ln \left| 1 - \dfrac{T}{T_c} \right| + const
##

Which is the expression given in the book for the asymptotic behaviour of ##\dfrac{1}{k_B} c (T)##. Importantly it is positive.
 
Last edited:

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