2.11 Extra dimension and statistical mechanics, 3

Posted on by

A First Course in String Theory

.

(b) Assume that \displaystyle{R \ll a} in such a way that …

…

~~~

[guess]

\displaystyle{\frac{\hbar^2}{ma^2} \ll k T \ll \frac{\hbar^2}{mR^2}}

\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{\beta = \frac{1}{kT} \ll \frac{ma^2}{\hbar^2}}

\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{\beta \gg \frac{mR^2}{\hbar^2}}

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

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

[guess]

— Me@2022-03-10 10:56:10 AM

.

.

2022.03.10 Thursday (c) All rights reserved by ACHK