2.10 A spacetime orbifold in two dimensions, 6

A First Course in String Theory

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(d) Consider the two curves \displaystyle{x^+ x^- = a^2} for some fixed \displaystyle{a}. The identification (2) makes each of these curves into a circle. Find the invariant circumference of this circle by integrating the appropriate root of \displaystyle{ds^2} between two neighboring identified points. Give your answers in terms of \displaystyle{a} and \displaystyle{\lambda}. Answer: \displaystyle{\sqrt{2} a \lambda}.

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Identification (2):

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

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\displaystyle{  \begin{aligned}  ds &= \sqrt{- ds^2} \\  \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds     &= \int_{x^+}^{e^{-\lambda} x^+} \sqrt{2 \frac{a^2}{(x^+)^2} (dx^+)^2}  \\     &= \sqrt{2} a \int_{x^+}^{e^{-\lambda} x^+} \sqrt{\frac{1}{(x^+)^2}} dx^+  \\   \end{aligned}}

Choose a segment on which \displaystyle{x^+ > 0} .

\displaystyle{  \begin{aligned}  \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds     &= \sqrt{2} a \int_{x^+}^{e^{-\lambda} x^+} \frac{1}{x^+} dx^+  \\     &= \sqrt{2} a \ln \left|\frac{e^{-\lambda} x^+}{x^+} \right| \\     &= - \sqrt{2} a \lambda \\   \end{aligned}}

The lower limit should be smaller than the upper limit.

\displaystyle{  \begin{aligned}  \left| \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds \right|    &= \sqrt{2} a \int_{e^{-\lambda} x^+}^{x^+} \frac{1}{x^+} dx^+  \\     &= \sqrt{2} a \lambda \\   \end{aligned}}

— Me@2022-01-05 02:40:45 PM

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2022.01.05 Wednesday (c) All rights reserved by ACHK