2.10 A spacetime orbifold in two dimensions, 2

A First Course in String Theory

(a) Use the result of Problem 2.2, part (a), to recast (1) as

$\displaystyle{(x^+, x^-) \sim \left( e^{-\lambda} x^+, e^{\lambda} x^- \right)}$ , where $\displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}$ .

What is the range of $\lambda$ ? What is the orbifold fixed point? Assume now that $\beta > 0$ , and thus $\lambda > 0$ .

~~~

Range of $\displaystyle{\lambda}$ :

$\displaystyle{ \begin{aligned} 0 &< \beta < \infty \\ 1 &< \frac{1 + \beta}{1 - \beta} < \infty \\ 0 &< \ln \frac{1 + \beta}{1 - \beta} < \infty \\ 0 &< \frac{1}{2} \ln \frac{1 + \beta}{1 - \beta} < \infty \\ 0 &< \lambda < \infty \\ \end{aligned}}$

Fixed points:

$\displaystyle{ \begin{aligned} \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix} &= \begin{bmatrix} e^{- \lambda} x^+ \\ e^{\lambda} x^- \\ \end{bmatrix} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} (x^+, x^-) &= (0, 0) \\ \end{aligned}}$

— Me@2021-05-16 06:31:12 PM

Physics Town

continues

2.10 A spacetime orbifold in two dimensions, 2

Related