# Calculating the specific heat capacity for the 2D Ising model

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• julian
In summary: J}{k_B T} = 1 + \dfrac{2}{k_B T} \left( \sqrt{2} x + \cdots + \left( \dfrac{x}{\sqrt{2}} \right)^{2n} + \cdots \right)##and##\sqrt{1 - q} = \dfrac{1}{\sqrt{k_BT}} \left( \sqrt{2} -... - \dfrac{1}{2
julian
Gold Member
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:
Delta2
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:

## 1. What is the 2D Ising model and why is it important in physics?

The 2D Ising model is a mathematical model used to study the behavior of magnetic materials. It consists of a lattice of spins (magnetic moments) that can either be in an "up" or "down" state. This model is important because it helps us understand the properties of real materials, such as phase transitions and critical phenomena.

## 2. How is the specific heat capacity calculated for the 2D Ising model?

The specific heat capacity for the 2D Ising model is calculated by taking the derivative of the average energy with respect to temperature. This can be done analytically or numerically using computer simulations.

## 3. What factors affect the specific heat capacity of the 2D Ising model?

The specific heat capacity of the 2D Ising model is affected by the strength of the interactions between spins, the size of the lattice, and the temperature. It also depends on the boundary conditions and the type of spin arrangement (e.g. ordered or disordered).

## 4. How does the specific heat capacity of the 2D Ising model change during a phase transition?

During a phase transition, the specific heat capacity of the 2D Ising model shows a sharp peak or discontinuity at the critical temperature. This is because the system undergoes a sudden change in its energy and behavior, which is reflected in the specific heat capacity.

## 5. What are some real-world applications of studying the specific heat capacity for the 2D Ising model?

Studying the specific heat capacity for the 2D Ising model can help us understand the behavior of real materials, such as magnetic materials and alloys. This knowledge can be applied in fields such as materials science, engineering, and physics to develop new materials with desired properties.

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