Q 8.24 In this problem you will use the... [FREE SOLUTION]

91影视

An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 8: Q 8.24 (page 346)

In this problem you will use the mean field approximation to analyse the behaviour of the Ising model near the critical point.
(a) Prove that, when $x 鈮� 1, \tanh x 鈮� x - \frac{1}{3} x^{3}$
(b) Use the result of part (a) to find an expression for the magnetisation of the Ising model, in the mean field approximation, when T is very close to the critical temperature. You should find $M 鈭� {(T_{c} - T)}^{\overset{炉}{尾}}, where 尾$ (not to be confused with 1/kT) is a critical exponent, analogous to the f defined for a fluid in Problem 5.55. Onsager's exact solution shows that $尾 = 1 / 8$ in two dimensions, while experiments and more sophisticated approximations show that $尾鈮� 1 / 3$ in three dimensions. The mean field approximation, however, predicts a larger value.
(c) The magnetic susceptibility $蠂$ is defined as $蠂鈮� (鈭� M / 鈭� B)_{T}$ . The behaviour of this quantity near the critical point is conventionally written as $蠂鈭� {(T - T_{c})}^{- 纬}$ , where y is another critical exponent. Find the value of in the mean field approximation, and show that it does not depend on whether T is slightly above or slightly below Te. (The exact value of y in two dimensions turns out to be 7/4, while in three dimensions $纬鈮� 1.24$ .)

Short Answer

Expert verified

Hence proved that when $x 鈮� 1, \tanh x 鈮� x - \frac{1}{3} x^{3}$

$b) 尾 = 0.5 c) 纬 = 1$

Step by step solution

Given information

In this problem we will use the mean field approximation to analyse the behaviour of the Ising model near the critical point.

Explanation

a) The function tanh x has the following definition:

$\tanh x = \frac{e^{2 x} - 1}{e^{2 x} + 1} e^{x} 鈮� 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥� \tanh x 鈮� \frac{1 + 2 x + \frac{(2 x)^{2}}{2} + \frac{(2 x)^{3}}{6} - 1}{1 + 2 x + 1} 鈮� \frac{2 x + \frac{(2 x)^{2}}{2} + \frac{(2 x)^{3}}{6}}{2 + 2 x} 鈮� \frac{x + \frac{2 x^{2}}{2} + \frac{4 x^{3}}{6}}{1 + x}; (1 + x)^{- 1} 鈮� 1 - x 鈮� (x + x^{2} + \frac{2 x^{3}}{3}) (1 - x) = x - x^{2} + x^{2} - x^{3} + \frac{2 x^{3}}{3} = x - \frac{x^{3}}{3}$

(b)This expression will now be used to expand relation 8.50:

role="math" $\overset{炉}{s} = \tanh (尾系 n \overset{炉}{s}) T_{c} = \frac{n 系}{k}, 尾 = \frac{1}{k T}, 尾 n 系 = \frac{T_{c}}{T} 鈮� \frac{T_{c}}{T} 路 \overset{炉}{s} - \frac{{(\overset{炉}{s} T_{c} / T)}^{3}}{3} \overset{炉}{s} = \frac{T_{c}}{T} 路 \overset{炉}{s} [1 - \frac{{(T_{c} / T)}^{2}}{3} 路 {\overset{炉}{s}}^{2}] \frac{T}{T_{c}} = 1 - \frac{{(T_{c} / T)}^{2}}{3} 路 {\overset{炉}{s}}^{2} \frac{T_{c}^{2}}{3 T^{2}} 路 {\overset{炉}{s}}^{2} = 1 - \frac{T}{T_{c}} \frac{T_{c}^{2}}{3 T^{2}} 路 {\overset{炉}{s}}^{2} = 1 - \frac{T}{T_{c}} / 路 \frac{3 T^{2}}{T_{c}^{2}} {\overset{炉}{s}}^{2} = 3 路 \frac{T^{2}}{T_{c}^{2}} 路 (1 - \frac{T}{T_{c}}) \overset{炉}{s} = \sqrt{3 路 \frac{T^{2}}{T_{c}^{2}} (1 - \frac{T}{T_{c}})}, T 鈫� T_{c}^{-} = \sqrt{3 路 \frac{T^{2}}{T_{c}^{2}} \frac{T_{c} - T}{T_{c}}}, t = \frac{T - T_{c}}{T_{c}} = \sqrt{- 3 t (1 + t)^{2}}, | t | < < 1, (1 + t)^{2} 鈮� 1 + 2 r 鈮� \sqrt{- 3 t (1 + 2 t)} 鈮� \sqrt{- 3 t} = \sqrt{\frac{- 3 (T - T_{c})}{T_{c}}} \overset{炉}{s} ~ \sqrt{T_{c} - T} 尾 = 0.5$

We may conclude that the same critical exponent applies to magnetisation because it is proportional to $s$ .

Explanation

(c) To calculate susceptibility, we must differentiate the non-approximate formula for $s$ in relation to B, which we shall manually insert:

$\overset{炉}{s} = \tanh (尾 (系 n \overset{炉}{s} + B)) / \frac{d}{d B} 蠂 = \frac{尾 (1 + 系 n 蠂}{\cosh^{2} [尾 (系 n \overset{炉}{s} + B)]}$

We can get $蠂$ by setting B = 0 and rearranging terms:

$蠂 = \frac{尾 (1 + 系 n 蠂)}{\cosh^{2} [尾系 n \overset{炉}{s}]} 蠂 = \frac{尾}{\cosh^{2} [尾系 n \overset{炉}{s}] - 尾系 n} 尾系 n = \frac{T_{c}}{T} \overset{炉}{s} 鈮� (3 | t |)^{0.5} for T 鈫� T_{c}^{-} \cosh (x) 鈮� 1 + \frac{x^{2}}{2} \cosh^{2} [\frac{T_{c}}{T} \overset{炉}{s}] 鈮� {(1 + \frac{{(T_{c} / T)}^{2} 路 3 | t |}{2})}^{2} 鈮� 1 + 3 \frac{T_{c}^{2}}{T^{2}} | t | 蠂鈮� \frac{尾}{1 + 3 \frac{T_{c}^{2}}{T^{2}} | t | - \frac{T_{c}}{T}} 1 + 3 \frac{T_{c}^{2}}{T^{2}} | t | - \frac{T_{c}}{T} = 3 \frac{T_{c}^{2}}{T^{2}} | t | + 1 - \frac{T_{c}}{T} = 3 \frac{T_{c}^{2}}{T^{2}} | t | - | t | = | t | (3 \frac{T_{c}^{2}}{T^{2}} - 1) = | t | 路 \frac{3 T_{c}^{2} - T^{2}}{T^{2}} 蠂鈮� \frac{尾}{| t | 路 \frac{3 T_{c}^{2} - T^{2}}{T^{2}}} 鈮� \frac{1}{k_{B} T} 路 \frac{T^{2}}{| t | 路 (3 T_{c}^{2} - T^{2})} 鈮� \frac{1}{k_{B}} 路 \frac{T}{| t | 路 (3 T_{c}^{2} - T^{2})} T 鈮� T_{c} 鈮� \frac{1}{k_{B}} 路 \frac{T_{c}}{| t | 路 2 T_{c}^{2}} 蠂鈮� \frac{1}{2 k_{B}} 路 \frac{1}{| t | T_{c}} 蠂鈮� {(T - T_{c})}^{- 1} 纬 = 1$