Is Space-Time Curvature Zero Inside a Spherical Cavity at Earth's Center?

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SUMMARY

Inside a spherical cavity centered at the Earth's center, both Newtonian gravity and General Relativity (GR) predict that the gravitational field is zero. However, GR indicates that the curvature, as measured by the curvature tensors G_{ab} and R_{abcd}, is non-zero at the center due to the non-zero stress-energy tensor T_{ab}. This discrepancy arises from the ambiguity in the term "gravitational field" and how it translates into popular language. The Riemann curvature tensor R_{abcd} specifically measures Newtonian tidal gravity, providing more detailed information than G_{ab}.

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  • Knowledge of curvature tensors, specifically G_{ab} and R_{abcd}
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  • #31
You need to be more precise

mendocino said:
Let's forget about gravitational field for a moment and talk about measured quantity? For example, will the clock runs slowly at the center and will someone experienced a force inside the cavity?

Hi, mendocino, this question actually is a very good illustration of why I keep insisting that saying "the clock runs slow" somewhere or even that "time slows" somewhere is Not A Good Idea. Rather, the world lines of radially outgoing time signals are modeled by radial null geodesics and these tend to spread apart in a Schwarzschild vacuum. IOW, an ideal clock can only run slow compared to another ideal clock, and the way in which this comparison is carried out is crucially important.

To keep things manageable, we typically seek a simple operationally significant way of effecting the comparison; one of the simplest is to assume that both clocks A,B lie on the same radius and to look at radial null geodesics from A to B and vice versa. In this case, we have the added complication that the homogeneous isotropic material of which our spherical object with concentric cavity is made, is presumably not transparent to light, and light (or radio waves) presumably travels more slowly inside this material than in vacuo.

Be this as it may, you more or less said what A you have in mind (unless you don't mean, as I guess, the static observer at the very center of the cavity), but not what B, so you need to clarify that before we can compute the answer you seek. You also need to clarify whether you are willing to accept a weak-field approximation or not. And if the answer is "not", you should wait until I get around to generalizing the Newtonian theory in "What is the Theory of Elasticity?" Or else look up previous posts by myself from many years ago in sci.physics.relativity in which I worked this out for the case of a simple model of a perfect fluid ball (no cavity), the Schwarzschild fluid.

However, it should be clear just from looking at the Newtonian potential (see my plot in "What is the Theory of Elasticity?") that this scenario involves the behavior of geodesics "in the large". That is, we should expect a "time dilation" effect (the magnitude depends upon the details you haven't specified), even though the cavity is locally isomorphic to Minkowski spacetime, i.e. locally flat.

(Re what pervect said, this is a completely different concept from the concept of "locally geodesic" or sometimes "locally flat" charts, among which Riemann normal charts are particularly important. The terminology in this subject can sometimes be confusing, and is sometimes incompletely standardarized! If you stay alert to the possibility of confusion, you can probably avoid trouble. In particular, note that "locally flat spacetime" uses "local" in the standard sense of manifold theory, meaning "local neighborhood", whereas "locally flat charts" are defined at the level of the metric tensor and its derivatives at a single event. The latter usage is deprecated because it conflicts with the mathematical terminology required for much modern "geometric" physics, e.g. for fiber bundles. For more about normal charts, see Poisson, A Relativist's Toolkit.)

As for "force", if you were asking whether a stationary observer at the very center of the cavity in a Newtonian model experiences a force, the answer is self-evident from symmetry. (In what direction would the force vector point?) In gtr, there is no notion of "gravitational force", but you can ask whether the stationary observer at the center of the cavity is an inertial observer or not. If not, he must use a rocket engine (or taut cables) to maintain his position, and thus must feel a nongravitational force, corresponding geometrically to the path curvature of his world line. (In what direction would the force vector point?)
 
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