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From the General Relativity class lectures I understood that this particular invariant, the Kretschmann scalar namely

$$R_{\mu\nu\lambda\rho} R^{\mu\nu\lambda\rho}$$

is really important because, through it, we can understand if a certain metric has removable singularities or essential singularities.

Why exactly does it work?

Am I wrong, or have I understood also that this scalar tells us if we are in presence of a gravitational field? Or was that a Riemann Tensor property?

Are there other important and useful scalar like the Kretschmann one?

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    $\begingroup$ Wouldn't it be easier to determine if a metric has a removable singularity by whether or not the volume element $\sqrt{det(g_{\mu \nu})} d^4 x$ makes sense? If it blows up or becomes zero at the singularity in question then it is not removable. $\endgroup$ – Jold Dec 23 '15 at 21:15
  • $\begingroup$ That would be easier, but our professor actually didn't mention it! At all.. He spoke only about that scalar quantity.. $\endgroup$ – Les Adieux Dec 23 '15 at 21:19
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    $\begingroup$ When you test whether or not a singularity is removable, you're basically just testing whether or not scalars make any sense there. The easiest way to do this is probably with the volume element. Or, more simply, the metric determinant. If det(g) is zero or infinite then volumes don't make sense, so the singularity is not removable. $\endgroup$ – Jold Dec 24 '15 at 18:21
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If you discover your metric has a singularity it can be very difficult to work out if this is a real singularity or just an artefact of the coordinate system you've used. If you've just run into a coordinate singularity then you can eliminate it by a coordinate transformation, but there is no easy way to work out what transformation is necessary. If you cannot find a coordinate transformation that eliminates the singularity this doesn't mean the singularity is real - it might be you just haven't stumbled across the required transformation yet.

But suppose you could find a quantity that didn't depend on whatever coordinate system you used i.e. it didn't change as you changed the coordinates. This would be a good way to look for singularities because if your quantity becomes infinite then it must be infinite in all coordinates, and that means the singularity is real.

General relativity has several such quantities, and we generically refer to them as scalars. The simplest is the Ricci scalar, but we are often interested in vacuum solutions and for a vacuum solution the Ricci scalar is zero everywhere so it isn't much use. For example this is the case for Schwarzschild and Kerr black holes.

The Kretschmann scalar is more complicated to compute, but unlike the Ricci scalar it (usually) isn't zero everywhere so it's far more useful. The Kretschmann scalar for a Schwarzschild black hole is given by:

$$ R_{\mu\nu\lambda\rho} R^{\mu\nu\lambda\rho} = \frac{48M^2}{r^6} $$

Since this isn't infinite at the event horizon $r = 2M$ we can tell immediately that the event horizon is a coordinate singularity not a real one. Likewise we can tell immediately that there is a real singularity at $r=0$.

There are a number of these scalars, and the Wikipedia article on curvature invariants lists a few of them.

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  • $\begingroup$ Suppose the Kretschman scalar is regular at some singularity. Is this necessary and sufficient for the singularity to be of the coordinate variant? $\endgroup$ – Ryan Unger Dec 25 '15 at 11:26
  • $\begingroup$ @0celo7: Yes. The curvature is undefined at a real singularity so the Kretschmann scalar cannot be calculated there. $\endgroup$ – John Rennie Dec 25 '15 at 11:29
  • $\begingroup$ That's necessity, not sufficiency. $\endgroup$ – Ryan Unger Dec 25 '15 at 11:33
  • $\begingroup$ @0celo7: to identify a coordinate singularity it is sufficient to show that (in some suitable coordinates) the Riemann tensor is defined there. If the Riemann tensor is defined there then so is the Kretschmann scalar and vice versa. So if the Kretschmann scalar can be calculated at a point that is sufficient proof that the geometry is not singular there. $\endgroup$ – John Rennie Dec 25 '15 at 11:56
  • $\begingroup$ I'm not entirely convinced of the "vice versa" part. Do you have a reference for the full proof? $\endgroup$ – Ryan Unger Dec 25 '15 at 11:58

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