# 2.11 Extra dimension and statistical mechanics

A First Course in String Theory

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(a) Explicitly calculate $\displaystyle{Z(a, R)}$ in the very high temperature limit $\displaystyle{\left( \beta = \frac{1}{kT} \to 0\right)}$.

~~~ \displaystyle{ \begin{aligned} Z &= Z(a) \tilde{Z}(R) \\ \end{aligned}} \displaystyle{ \begin{aligned} Z(a) &= \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k \pi}{a} \right)^2 \right] \\ \tilde Z (R) &= 1 + 2 Z(R \pi) \\ \end{aligned}}

. \displaystyle{ \begin{aligned} \Delta k &= 1 \\ Z(a) &= \frac{1}{\beta} \sum_{\beta k = \beta, 2\beta, ...} \exp\left[- \frac{1}{\beta} \left( \frac{\hbar^2}{2m} \right) \left(\frac{\beta k \pi}{a} \right)^2 \right] \Delta (\beta k) \\ \end{aligned}} \displaystyle{ \begin{aligned} \int_{0}^{\infty } e^{-ax^{2}}\,dx &= \frac{1}{2} {\sqrt {\frac {\pi }{a}}} \\ \end{aligned}}

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When $\displaystyle{ \beta \to 0 }$, \displaystyle{ \begin{aligned} Z(a) &\approx \frac{1}{\beta} \int_{x = 0}^\infty e^{\left[- \frac{1}{\beta} \left( \frac{\hbar^2}{2m} \right) \left(\frac{x \pi}{a} \right)^2 \right]} dx = \sqrt{\frac{m a^2}{2 \beta \pi \hbar^2}} \\ \end{aligned}} \displaystyle{ \begin{aligned} \tilde Z (R) &\approx 1 + \sqrt{\frac{2 m R^2 \pi}{\beta \hbar^2}} \approx \sqrt{\frac{2 m R^2 \pi}{\beta \hbar^2}} = Z(2 \pi R) \\\\ \end{aligned}}

. \displaystyle{ \begin{aligned} Z &\approx Z(a) Z(2 \pi R) \\ \end{aligned}}

— Me@2022-03-04 07:12:49 PM

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