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I am investigating the metric

$ds^2 = -dt^2 + (1+C)dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2, $

where $C$ is a constant. This intuitively seems like flat space but actually has a non-zero Kretschmann scalar, which goes as $K=4C/r^4$ and therefore diverges at the origin. Has this metric been studied before?

N.B. By transforming into Cartesian coordinates, we can write this metric as

$-dt^2 + dx^2 + dy^2 + dz^2 + C \frac{\left(x dx + y dy + zdz\right)^2}{x^2 + y^2 + z^2},$

which is clearly not flat space.

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  • $\begingroup$ $C$ depends on what? $\endgroup$
    – magma
    Apr 3, 2017 at 22:32
  • $\begingroup$ If you reduce that to 2D+1 the spatial part is just the metric for the surface of a cone. $\endgroup$ Apr 4, 2017 at 12:40
  • $\begingroup$ Aha, my friendly differential geometers tell me the spatial part is a cone on a 2-sphere i.e. the cone whose cross sections are 2-spheres. $\endgroup$ Apr 4, 2017 at 13:01

1 Answer 1

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As has been said in the comments, it's a conical geometry with the 'tip' at $r=0$.

Let's assume that $1+C>0$, because otherwise you'll have a signature $(--++)$ metric, which is weird. Apart rom that, the time direction does not play a rôle here, so I'll ignore it for now.

By a rescaling of the $r$ coordinate, you can get the spatial matric $$\text{d} s^2 = \text{d} r^2+ \frac{r^2}{1+C} \text{d} \Omega^2 \,.$$ Here, $\text{d} \Omega^2$ is the usual spherical volume element. Hence, a sphere around the origin with radius $r=R=\text{const.}$ will have an area of $$\frac{4\pi R^2}{1+C}\,,$$ that is, it will be smaller (for $C>0$, larger otherwise) than in flat space.

A two-dimensional analogy that is simpler to visualise is an actual cone: Take a 2d plane, cut out a wedge with tip at the origin and opening angle ('deficit angle') $\phi=2\pi \frac{C}{1+C}$ and glue the edges. Then you end end up with a conical singularity in otherwise flat space, which you can notice by looking at circles surrounding the origin.

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