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The geodesic equation used in general relativity is the following: $$ {\mathrm d^2 x^\mu \over \mathrm ds^2} =- \Gamma^\mu {}_{\alpha \beta}{\mathrm d x^\alpha \over\mathrm ds}{\mathrm d x^\beta \over\mathrm ds}. $$ It states that the acceleration of the test particle is a function of the metric (Chistoffel symbol) and the derivative of coordinates with respect to "a scalar parameter of motion s ex.: proper time".

Also, the Wikipedia page on the Schwarzschild metric states the following: "[...] [Schwarzschild metric] is the solution to the Einstein field equations that describes the gravitational field outside a spherical mass, on the assumption that the electric charge of the mass, angular momentum of the mass, and universal cosmological constant are all zero." and the metric is the following: $${c^2 \mathrm d\tau^2 =}\left({1-{r_s \over r}}\right)c^2\mathrm dt^2-\left(1-{r_s \over r}\right)^{-1}\mathrm dr^2-r^2\mathrm d\Omega^2$$

Assuming all these conditions are true, does the Schwarzschild metric apply to the context of a particle in the vicinity of Earth's gravitational field? If so, can you give an example?

If, for some reason, the metric in question does not apply to the context of Earth, why not?

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closed as unclear what you're asking by DilithiumMatrix, AccidentalFourierTransform, ACuriousMind, Asher, CuriousOne May 25 '16 at 21:55

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is a good question, but should probably be restructured somewhat as it is borderline off-topic because of the homework-like question policy. $\endgroup$ – DilithiumMatrix May 24 '16 at 2:35
  • $\begingroup$ I'm not sure what the question is - you present all the assumptions that are needed for Schwarzschild, and then ask "assuming all these conditions are true, does the Schwarzschild metric apply" - and of course it applies, because you just assumed it! Are you asking whether the assumptions are a good approximation for the conditions around Earth? $\endgroup$ – ACuriousMind May 24 '16 at 15:55
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Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?.

An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS satellites. Strictly speaking, since the Earth is rotating, the spacetime around it is described by the Kerr metric rather than the Schwarzschild metric, though the difference is so small as to be barely detectable. An example of this is the measurement of the Lens-Thirring effect by the Gravity Probe B satellite, though I think the jury is out on whether GPB actually managed to measure the Lens-Thirring effect or not.

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  • $\begingroup$ If I remember correctly (and did the numbers correctly...) Earth can only very, very asymptotically be described by Kerr. It's easy to check: If Jc/(GM^2) is less than 1, everything's fine. But for Earth's values you end up with a supercritical Kerr solution. The gravito-magnetism, i.e. weak field approximation works, but thanks to the lack of Birkhoff's theorem, things are a lot messier than "Kerr describes Earth better than Schwarzschild"...:( $\endgroup$ – Wraith of Seth May 24 '16 at 6:02
  • $\begingroup$ The radius of the earth is still much larger than its gravitational radius, so the asymptotical description with Kerr works just fine. $\endgroup$ – Yukterez Oct 30 '18 at 2:29

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