Metric tensor at the earth surface

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SUMMARY

The discussion focuses on calculating the Ricci tensor and Ricci scalar for the space-time curvature at the Earth's surface using the Schwarzschild metric. It is established that the Ricci tensor and scalar are zero in vacuum solutions, such as outside the Earth, while they become non-zero inside the Earth. The discussion highlights the importance of the Weyl curvature tensor, which describes curvature in vacuum regions and is non-zero, contrasting with the Ricci curvature that vanishes in these areas. The Schwarzschild interior solution is suggested for modeling the Earth as a constant density, non-rotating sphere.

PREREQUISITES
  • Understanding of General Relativity concepts
  • Familiarity with the Schwarzschild metric
  • Knowledge of curvature tensors, specifically Ricci and Weyl tensors
  • Basic mathematical skills for tensor calculus
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  • Study the Schwarzschild interior solution for constant density spheres
  • Learn about the Weyl curvature tensor and its physical implications
  • Explore curvature invariants in General Relativity
  • Review geodesic congruences and their relationship with curvature quantities
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tm007
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I want to find the ricci tensor and ricci scalar for the space-time curvature at the Earth surface. Ignoring the moon and the sun. I have used the scwharzschilds metric, but then the ricci tensor and the scalar where equal to zero.
 
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The Schwarzschild metric is a vacuum solution so of course the Ricci tensor and Ricci scalar will vanish. The Schwarzschild metric is valid for the exterior of the Earth, ignoring the Earth's rotation and the presence of the other celestial bodies.
 
The Kretschmann scalar is non-zero M2/r6
 
Another thing you could look at is the Weyl curvature tensor. This specifically describes the type of curvature in vacuum regions in GR. Unfortunately, it is really hard to compute by hand, even for a metric as simple as the Scwharzschild.
 
The mass of the Earth should curve the space surounding the Earth. Why is then the ricci tensor equal to zero.
 
Outside the Earth, there is a vacuum, so Ricci is zero. Inside the Earth, Ricci is non-zero. A somewhat crude model takes the Earth as a constant density, non-rotating sphere. Then, Schwarzschild's interior solution can be used. When G=c=1,

<br /> ds^{2}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}-\left( 1-\frac{2Mr^{2}}{R^{3}}\right) ^{-1}dr^{2}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right),<br />

where R is the r coordinate at the surface of the Earth.

This metric is treated in many relativity texts, e.g., texts by Schutz, by Hobson, Efstathiou, Lasenby, and by Misner, Thorne, Wheeler.
 
tm007 said:
The mass of the Earth should curve the space surounding the Earth.

It does, but the curvature produced in the vacuum region around the Earth is Weyl curvature, not Ricci curvature.
 
tm007 said:
The mass of the Earth should curve the space surounding the Earth. Why is then the ricci tensor equal to zero.
Spacetime is curved, that's why several curvature invariants and Weyl curvature are non-zero, whereas Ricci-curvature is zero
 
tm007 said:
The mass of the Earth should curve the space surounding the Earth. Why is then the ricci tensor equal to zero.

Keep in mind that different curvature tensors measure different kinds of curvature; the Weyl curvature is the trace free part of the Riemann curvature and will not vanish in general for vacuum solutions whereas the Ricci tensor always vanishes for vacuum solutions. However you can always rely on the Riemann curvature tensor because this can vanish identically if and only if the manifold is locally isometric to euclidean space (which is of course flat).

Both the Ricci and Weyl curvatures can be made sense of physically in GR by looking at geodesic congruences and seeing which of the three kinematical quantities (expansion, shear, and twist) are dominated by which curvature quantity. See here for a start: http://en.wikipedia.org/wiki/Weyl_tensor
 

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