- #1
highflyyer
- 28
- 1
Consider the transformation from Poincare-AdS##_3## geometry to global AdS##_3## geometry:
$$ds^{2} = \frac{dr^{2}}{r^{2}} + r^{2}g_{\alpha\beta}dx^{\alpha}dx^{\beta}, \qquad \text{Poincare-AdS$_3$}$$
$$ds^{2} = \frac{dr^{2}}{r^{2}} + r^{2}\left(-dt^{2}+r^{2}d\phi^{2}\right), \qquad \text{Poincare-AdS$_3$}$$
$$ds^{2} = - r^{2}dt^{2} + \frac{dr^{2}}{r^{2}} + r^{4}d\phi^{2}, \qquad \text{Poincare-AdS$_3$}$$
$$ds^{2} = -\cosh^{2}\rho\ d\tau^{2} + d\rho^{2} + \sinh^{2}\rho\ d\varphi^{2}, \qquad \text{global AdS$_3$}$$
where the transformation of coordinates is as follows:
$$\rho = \ln r, \qquad \tau = \left(\frac{2e^{\rho}}{e^{\rho}+e^{-\rho}}\right)t, \qquad \varphi = \left(\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}\right)\phi.$$
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The transformation ##\rho = \ln r## simply rescales the radial distance ##r## by the logarithmic function.
The transformation with ##\displaystyle{\tau = \left(\frac{2e^{\rho}}{e^{\rho}+e^{-\rho}}\right)t}## rescales the time ##t## by the factor ##\displaystyle{\frac{2e^{\rho}}{e^{\rho}+e^{-\rho}}}##. For example, at ##\rho = 0##, we have ##\tau = t##, and at ##\rho = \infty##, we have ##\tau = 2t##.
The transformation with ##\displaystyle{\varphi = \left(\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}\right)\phi}## rescales the angle ##\phi## by the factor ##\displaystyle{\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}}##. For example, at ##\rho = 0##, we have ##\varphi = \infty##, and at ##\rho = \infty##, we have ##\varphi = \infty##.
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Have I made a mistake in my interpretation of the transformation ##\displaystyle{\varphi = \left(\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}\right)\phi}##?
$$ds^{2} = \frac{dr^{2}}{r^{2}} + r^{2}g_{\alpha\beta}dx^{\alpha}dx^{\beta}, \qquad \text{Poincare-AdS$_3$}$$
$$ds^{2} = \frac{dr^{2}}{r^{2}} + r^{2}\left(-dt^{2}+r^{2}d\phi^{2}\right), \qquad \text{Poincare-AdS$_3$}$$
$$ds^{2} = - r^{2}dt^{2} + \frac{dr^{2}}{r^{2}} + r^{4}d\phi^{2}, \qquad \text{Poincare-AdS$_3$}$$
$$ds^{2} = -\cosh^{2}\rho\ d\tau^{2} + d\rho^{2} + \sinh^{2}\rho\ d\varphi^{2}, \qquad \text{global AdS$_3$}$$
where the transformation of coordinates is as follows:
$$\rho = \ln r, \qquad \tau = \left(\frac{2e^{\rho}}{e^{\rho}+e^{-\rho}}\right)t, \qquad \varphi = \left(\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}\right)\phi.$$
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The transformation ##\rho = \ln r## simply rescales the radial distance ##r## by the logarithmic function.
The transformation with ##\displaystyle{\tau = \left(\frac{2e^{\rho}}{e^{\rho}+e^{-\rho}}\right)t}## rescales the time ##t## by the factor ##\displaystyle{\frac{2e^{\rho}}{e^{\rho}+e^{-\rho}}}##. For example, at ##\rho = 0##, we have ##\tau = t##, and at ##\rho = \infty##, we have ##\tau = 2t##.
The transformation with ##\displaystyle{\varphi = \left(\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}\right)\phi}## rescales the angle ##\phi## by the factor ##\displaystyle{\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}}##. For example, at ##\rho = 0##, we have ##\varphi = \infty##, and at ##\rho = \infty##, we have ##\varphi = \infty##.
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Have I made a mistake in my interpretation of the transformation ##\displaystyle{\varphi = \left(\frac{2e^{2\rho}}{e^{\rho}-e^{-\rho}}\right)\phi}##?