Metric Tensor of Earth

1. Dec 11, 2008

Jack3145

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. Dec 11, 2008

HallsofIvy

Staff Emeritus
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?

3. Dec 11, 2008

vanesch

Staff Emeritus
Outside of the earth, as a good approximation (assuming the earth is spherical) you could use the Schwarzschild metric.

4. Dec 11, 2008

atyy

Last edited: Dec 11, 2008
5. Dec 11, 2008

AstroRoyale

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.

$$ds^2 = R^{2}(d\theta^2 + sin^{2}\theta d\phi^2)$$

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.

6. Dec 11, 2008

Jack3145

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.

7. Dec 11, 2008

AstroRoyale

8. Dec 12, 2008

Jack3145

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
9. Dec 14, 2008

akelleh

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.

10. Dec 14, 2008

Phrak

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:

$$ds^2 = R^{2}(d\theta^2 + sin^{2}\theta d\phi^2)$$

$$\ ds^2$$ 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.