The Schwarzschild-deSitter spacetime in Euclidean signature is given by:

$$ ds_{E}^{2}=\left(1-\frac{2M}{r}-\frac{H^{2}r^{2}}{3}\right)d\tau^{2}+\frac{dr^{2}}{\left(1-\frac{2M}{r}-\frac{H^{2}r^{2}}{3}\right)}+r^{2}d\Omega_{2}^{2} $$ where $\tau$ is the Euclidean time, r is the radial coordinate, M is the mass of the black hole and $\Lambda=H^{2}$ is the cosmological constant.

If we set $H=0$ we retrieve the familiar Schwarzschild metric which is asymptotically flat. The latter has only one horizon given by $r=2GM/c^{2}$. Since the horizon singularity is a coordinate artifact, when going to Euclidean time and near the Rindler wedge we can define a periodicity of $\tau$ such that the horizon is regular.

However, when $H\neq0$ the spacetime has two horizons, the black hole and the de Sitter. In this case Hawking and Bousso showed that the metric cannot be made regular in Euclidean signature since one cannot match the periodicity of $\tau$ near both horizons, except for the degenerate case in which they coincide.

The question is: Will an observer crossing either of the two horizon experience a singularity? I am confused about this because the curvature tensor is not singular at any of the horizons.

  • $\begingroup$ Will an observer crossing either of the two horizon experience a singularity?” - Can you clarify what exactly you mean by “singularity” and what specific “experience” of the observer you are referring to? $\endgroup$
    – safesphere
    Commented May 5, 2022 at 18:03
  • $\begingroup$ All I am trying to say is this: If the metric singularity at the horizons was due to this choice of coordinates then I would expect it to be remedied in Euclidean time by choosing appropriate periodicity. This is not the case, however. On the other hand the Riemann tensor is not singular on the horizons. So is this a coordinate singularity or not? $\endgroup$ Commented May 5, 2022 at 18:57
  • 1
    $\begingroup$ When you look at a sheet of paper, it looks “singular” if you look at its edge. Your point of view is “a coordinate singularity” (although induced by the shape of the paper). Imagine the sheet is bent along a line. Now you have two viewpoints showing one half of the sheet as “singular” while the other as “normal”. Obviously, you have two coordinate singularities, but you cannot “remove” both by a simple coordinate transformation, such as changing your angle of view. This is because coordinate singularities are induces by the shape of the manifold, but not merely by the choice of coordinates. $\endgroup$
    – safesphere
    Commented May 5, 2022 at 20:24
  • $\begingroup$ “… induced by the shape of the manifold…” (sorry for the typo). $\endgroup$
    – safesphere
    Commented May 6, 2022 at 2:13
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    $\begingroup$ Thank you for the example with the sheet of paper! It really helps visualize the issue. So if I wrap the paper to a cylinder I will get rid of one of the singularities, but then I cannot do the same with the other edges. Thanks! $\endgroup$ Commented May 16, 2022 at 3:05

1 Answer 1


George Fanaras wrote: "Λ=H² is the cosmological constant."

In my books $ \rm{\Lambda = 3 H^2} $ , if you have the line element in terms of the Hubble parameter $\rm H$ like you do then you don't need the $3$ in the denominator.

George Fanaras asked: "The question is: Will an observer crossing either of the two horizon experience a singularity?"

The answer is no, neither the black hole horizon nor the Hubble radius or cosmic horizon are real singularities. You can cross them without any trouble, since the Kretschmann scalar $ \rm K $ for the SSdS metric

$$ \rm K = \frac{48 M^2}{r^6}+24 H^4 $$

which tells you if a singularity is real or just a coordinate artifact is only singular in the center of the black hole where $\rm r=0 $. Since the Kretschmann scalar is an invariant, if it is finite in one coordinate system it is finite in all other coordinate systems as well, and as you can clearly see, it is only infinite at the center of the black hole.

  • $\begingroup$ To nitpick, is not that simple to check for absence of singularities: physics.stackexchange.com/questions/249115/… $\endgroup$
    – Rexcirus
    Commented May 6, 2022 at 8:29
  • $\begingroup$ The safest way is to show a coordinate change which makes the metric regular at that point. $\endgroup$
    – Rexcirus
    Commented May 6, 2022 at 8:35
  • $\begingroup$ In this case we have here it might be easier, since we know that far from the black hole the Schwarzschild part vanishes and the de Sitter part remains, we have "The de Sitter space-time universe is geodesically complete, [...]. Hence, it is free from singularity (Barata et al. 2017)" - mpra.ub.uni-muenchen.de/83187/1/MPRA_paper_83187.pdf#page=3 but we can also switch to comoving coordinates to have the spacetime regular everywhere above r=0. $\endgroup$
    – Yukterez
    Commented May 6, 2022 at 17:48
  • $\begingroup$ Thank you for the comments! $\endgroup$ Commented May 16, 2022 at 3:06

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