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I am trying to understand how the Euclidean "cigar" is built. I understand how and why the time is periodic, as for the radius of the cigar I am confused, it should be constant far from the horizon (cylinder shape). If I use the paper of Gibbons-Hawking "Action integrals and partition functions in quantum gravity", Eq.(2.7)

$$\xi^2+y^2=(r/2M-1)*e^{r/2M}$$

it looks like the radius diverge..

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Gary's coordinate ``$r$'' is parameterising the distance from the event horizon, and so can be as large as you want. It is not the radius of the cylinder.

Take the metric to be of Schwarzschild form $$ d\sigma^2 = f(r) d\tau^2 +\frac 1 {f(r)} dr^2. $$ (I ignore the angular coordinates as they do nothing). The function $f(r)$ is zero at the the horizon $r_H$.
Introduce a new radial co-ordinate $$ \rho = \int_{r_H}^r \frac{dr'}{\sqrt{f(r')}}\approx \frac{2}{\sqrt{f'(r_H)}}\sqrt{r-r_H}, $$ where the approximation holds for $r$ just above $r_H$. Then, in this same region, $$ d\sigma^2 =f(r)d\tau^2+\frac 1{f(r)}dr^2\nonumber\\ = f(r) d\tau^2 +d\rho^2,\nonumber\\ \approx f'(r_H) (r-r_H)d\tau^2+d\rho^2\nonumber\\ =\kappa^2 \rho^2d\tau^2+d\rho ^2. $$ Here $\kappa$ is the surface gravity defined by $$ \kappa =f'(r_H)/2 $$ Comparison with the metric of plane polar coordinates now shows that if there is to be no conical singularity at $r_H$ we must identify $ \kappa \tau $ with the polar angle $\theta$. Thus the smooth euclidean manifold described looks like the skin of a salami sausage in which the circumferential coordinate $\theta$ is identified with $\theta+2\pi$, or equivalently $\tau$ is identified with $\tau+\beta$ where $\beta= 2\pi/\kappa$. As the sausage circumference is $\beta$, its radius is $\beta/2\pi= \kappa^{-1}$.

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  • $\begingroup$ So how the radius of the cylinder is defined? $\endgroup$
    – TTT
    Jun 4, 2023 at 12:00
  • $\begingroup$ I have extended my answer, $\endgroup$
    – mike stone
    Jun 4, 2023 at 12:20
  • $\begingroup$ thank you. According to your explanation, the geometry is a cylinder near the horizon, but as far as I know near the horizon it looks like a paraboloid, and then far enough like a cylinder. $\endgroup$
    – TTT
    Jun 4, 2023 at 12:58
  • $\begingroup$ Why do you think that? I have shown, from the short distance geometry of the end-cap, that $\kappa \tau=\theta$. At large distance $f\to 1$ so that $d\sigma^2\to \kappa^{-2} d\theta^2+dr^2$ which is the metric of a cylinder of radius $\kappa^{-1}$. $\endgroup$
    – mike stone
    Jun 4, 2023 at 13:42
  • $\begingroup$ You can find more details of the coordinate transformations in arXiv:1804.08668. $\endgroup$
    – mike stone
    Jun 4, 2023 at 13:48

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