- #1

julian

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- 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).

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: