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

  • Thread starter Thread starter Bishal Banjara
  • Start date Start date
  • Tags Tags
    Curvature Riemann
Click For Summary
The discussion centers on calculating the Riemann curvature tensor at the coordinate singularity r=2GM in the Schwarzschild solution, emphasizing that this point does not represent a physical singularity. Participants seek authentic references and clarification on the calculations presented in the "Gravitation" textbook by Misner, Thorne, and Wheeler, particularly regarding the Riemann and Ricci tensors. There is confusion about the expected values of the Ricci tensor components, which should be zero, and the number of non-zero components of the Riemann tensor, with references to various sources for further understanding. The conversation highlights the importance of correctly applying tensor symmetries and index manipulations in calculations. Overall, the thread reflects a collaborative effort to clarify complex concepts in general relativity.
  • #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.
 
  • Like
Likes Bishal Banjara
Physics news on Phys.org
  • #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.
 
  • Like
Likes Bishal Banjara

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
4K
Replies
1
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
7K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
7K