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Metric Tensor of Earth

  1. Dec 11, 2008 #1
    I would like to know the Metric Tensor of the Earth in the form of g = [g11,g21,g31;g12,g22,g32;g13,g23,g33].
  2. jcsd
  3. Dec 11, 2008 #2


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    There is no such thing as a "metric tensor" of an object. Do you mean the metric tensor for space around the earth, ignoring the sun, moon, and other planets?
  4. Dec 11, 2008 #3


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    Outside of the earth, as a good approximation (assuming the earth is spherical) you could use the Schwarzschild metric.
  5. Dec 11, 2008 #4


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    Last edited: Dec 11, 2008
  6. Dec 11, 2008 #5
    I'm guessing he might mean the metric for the surface of the earth, as in given two lats and lons, find distance between them. In that case, it's a bumpy sphere.

    [tex] ds^2 = R^{2}(d\theta^2 + sin^{2}\theta d\phi^2)[/tex]

    is the metric on a sphere. The mountains/valleys and all the rest that make the earth interesting are going to be impossible to get into a metric.
  7. Dec 11, 2008 #6
    Will the Schwarzschild metric work for the atmosphere, the edge of the atmosphere? Are there any hints on formulating the Schwarzschild metric into the Metric Tensor.
  8. Dec 11, 2008 #7
  9. Dec 12, 2008 #8
    Is the Metric Tensor derived directly from Schwarzchild Metric or is it derived from the spherical metric of a sphere?
    Last edited: Dec 12, 2008
  10. Dec 14, 2008 #9
    The schwarzschild metric is the metric for the spacetime surrounding a sphere, as derived from einstein's equations in general relativity.

    The metric for a sphere is a more basic geometric idea, and not derived from einstein's equations. It has little to do with general relativity.

    You can call either of these a metric tensor.
  11. Dec 14, 2008 #10
    Hi Jack. There is no unique metric tensor for any given point in spacetime. But given that, the Schwarzschild metic is one metric, in spherical coordinates, applicable to the near-space points lying outside the surface of a spherically symmtrical mass such as the Earth, as long as the Earth can be approximated as spherical, and having zero charge and angular momentum. For a mass as small as the Earth, the R variable in the Schwarzschild metric can be taken as the radius of the Earth.

    You may not be aware of this, but AstroRoyale gave you a metric at the surface of a spherical mass:

    [tex] ds^2 = R^{2}(d\theta^2 + sin^{2}\theta d\phi^2)[/tex]

    [tex]\ ds^2[/tex] is the metric. I don't know how to put it in the form you want. The metric elements range from 0 to 3, where 0 indexes time, and 1-3 index spatial coordinates. In the case of the full Schwarzschild metric, the spatial coordinates are spherical coordinates of an inertial frame centered on a mass at rest, asymptotic at infinity.
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