How to calculate the Riemann curvature at r=2GM?

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SUMMARY

The discussion focuses on calculating the Riemann curvature tensor at the coordinate singularity r=2GM in the context of the Schwarzschild solution. Participants clarify that the Riemann curvature tensor remains smooth despite the singular behavior of the metric. Key references include the "Gravitation" textbook by Misner, Thorne, and Wheeler (MTW), specifically pages 821 and equations 14.50 and 14.51. The conversation emphasizes the importance of correctly substituting values and understanding the relationship between the Ricci tensor and the Riemann tensor, with participants noting that individual components of the Ricci tensor should be zero in this scenario.

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  • Understanding of General Relativity concepts, particularly the Schwarzschild solution.
  • Familiarity with the Riemann curvature tensor and Ricci tensor.
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  • Knowledge of differential geometry and its application in physics.
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  • Study the derivation of the Riemann curvature tensor components from the Schwarzschild metric.
  • Review the equations 14.50 and 14.51 in the "Gravitation" textbook by Misner, Thorne, and Wheeler.
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  • #61
Actually, your proposed metric (with the change of signature) will be a solution to the EFE's in vacuum. In the typical derivation of the Schwarzschild metric you end up with a differential equation of the form
$$
\partial_r ( r f(r)) = 1,
$$
where ##f(r)## is the coefficient of ##- dt^2## in the line element. The solution to this is obviously
$$
r f(r) = r + C,
$$
where ##C## is an integration constant. This tells you nothing about the sign of ##C## and generally you have
$$
f(r) = 1 + \frac{C}{r}.
$$
Asking for the correct behaviour in the weak field limit then identifies ##C = - 2MG##, but there is nothing preventing you from putting ##C## positive apart from the fact that we have not observed anything that actually behaves like that (essentially the weak field limit becomes a limit where a point "mass" repels rather than attracts). Celestial objects would move on approximate hyperbolae rather than approximate ellipses, etc. The case ##C > 0## is therefore just not physically interesting.
 
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  • #62
The case C>o is therefore just not physically interesting. It is not the matter of interest, its the matter of reality. Then such teste-parameters would get varied?
 
  • #63
Bishal Banjara said:
Then such teste-parameters would get varied?

This question has already been answered:

PeterDonis said:
The values of observable quantities related to the spacetime geometry (which I think is what you mean by "Tests") will obviously be different for different spacetime geometries.
There is no point in continuing to repeat the same answer. Thread closed.
 
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