This question is based on page 71 of Thomas Hartman's notes on Quantum Gravity and Black Holes.

The Euclidean Schwarzschild black hole

$$ds^{2} = \left(1-\frac{2M}{r}\right)d\tau^{2} + \frac{dr^{2}}{1-\frac{2M}{r}} + r^{2}d\Omega_{2}^{2}$$

is obtained from the Lorentzian Schwarzschild black hole via Wick rotation $t \to -i\tau$.

Why does the fact that the coordinates must be regular at the origin imply that the angular coordinate must be identified as

$$\tau \sim \tau + 8\pi M?$$


1 Answer 1


This is so that there is no "conical singularity" at the horizon (not the origin). Let me elaborate.

In general, how do you know if a spacetime has a curvature singularity? Well if any metric components diverge or vanish at any point, then you know you might be in trouble. However you still need to determine if you have a curvature singularity (i.e. a true singularity) or a coordinate singularity (i.e. it was just a bad coordinate system you chose). The failsafe way to determine if it's merely a coordinate singularity is to find a coordinate system where the metric is regular at the point of interest.

For the Schwarzschild metric, we know the singularity at $r=2M$ is a coordinate singularity because there exists a coordinate transformation where the metric is smooth at that point (Kruskal coordinates or Eddington-Finkelstein coordinates).

Similarly, if you were faced with a metric $$ ds^2 = dr^2 + r^2 d\theta^2$$ you might be concerned that the space is singular at $r=0$. However you know that this is merely a coordinate singularity because if you do the coordinate transformation $x = r \, \cos\theta$ and $y=r\, sin\theta$ the metric is now $$ds^2 = dx^2 +dy^2$$ and is regular everywhere.

However there is a big catch! In order for that coordinate transformation to be well defined, $\theta$ had to be periodic with period $2\pi$. If not, then you would have a space where you took a sheet of paper, cut out a slice from it, and glued it back together, forming a cone. Therefore there is a singularity right at the tip, and we call this a "conical singularity."

Now going to the Euclidean black hole case, if you do a coordinate transformation $\rho^2 = 8M(r-2M)$ and expand near $\rho =0$ you get something like $$ds^2 = d\rho^2 + \rho^2 \beta^2 d\tau^2+ \cdots$$ Therefore, in order for $\rho=0$ to remain merely a coordinate singularity the coordinate $\beta \tau$ must have periodicity $2\pi$ which means $\tau$ must have periodicity $2 \pi/\beta$. As a fun exercise determine $\beta$!

  • $\begingroup$ Would it be possible to explain how you found that the coordinate transformation is $\rho^{2} = 8M (r-2M)$ and why you have to expand near $\rho =0$? Also, why can't the polar coordinates extend beyond the horizon? $\endgroup$ Jun 16, 2017 at 2:26
  • $\begingroup$ The idea is to look at the metric near the horizon, so define a coordinate $\rho$ that is zero at the horizon. Further in order to investigate the singularity I want the $g_{\rho\rho}$ component of the metric to be one. You can verify this transformation is the one that satisfies both these conditions. Additionally the Euclidean spacetime pinches off like a cigar at the horizon, so there is no behind the horizon region. You can see this happens because the size of the time circle vanishes at the horizon. $\endgroup$
    – jswien
    Jun 16, 2017 at 6:46

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