2.10 A spacetime orbifold in two dimensions, 2

A First Course in String Theory

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(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.

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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}}

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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

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2021.05.17 Monday (c) All rights reserved by ACHK