Metric tensor at the earth surface


by tm007
Tags: earth, metric, surface, tensor
tm007
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#1
Jul31-13, 06:13 PM
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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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WannabeNewton
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Jul31-13, 06:17 PM
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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.
tom.stoer
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Jul31-13, 07:09 PM
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The Kretschmann scalar is non-zero M2/r6

PAllen
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Jul31-13, 08:47 PM
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Metric tensor at the earth surface


Another thing you could look at is the Weyl curvature tensor. This specifically describes the type of curvature in vaccuum regions in GR. Unfortunately, it is really hard to compute by hand, even for a metric as simple as the Scwharzschild.
tm007
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Aug1-13, 09:35 AM
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The mass of the Earth should curve the space surounding the Earth. Why is then the ricci tensor equal to zero.
George Jones
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Aug1-13, 09:46 AM
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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 [itex]G=c=1[/itex],

[tex]
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),
[/tex]

where [itex]R[/itex] is the [itex]r[/itex] 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.
PeterDonis
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Aug1-13, 09:55 AM
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Quote Quote by tm007 View Post
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.
tom.stoer
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Aug1-13, 02:32 PM
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Quote Quote by tm007 View Post
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
WannabeNewton
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Aug1-13, 03:42 PM
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Quote Quote by tm007 View Post
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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