# 2.10 A spacetime orbifold in two dimensions, 6

A First Course in String Theory

.

(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: $\displaystyle{\sqrt{2} a \lambda}$.

~~~

Identification (2): $\displaystyle{(x^+, x^-) \sim \left( e^{-\lambda} x^+, e^{\lambda} x^- \right)}$, where $\displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}$.

. \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

.

.