On 04/02/2010 08:28 AM, Roarty, Francis X wrote: > Stephan, > > This isn’t meant > to be philosophical but if these fields meet in the cavity and there > is no mass there for them to fight over, will the fields even sum or > just pass through each other?
That question can only be answered in the context of a model of gravity. The only models I am somewhat familiar with are Newtonian gravity and GR geometric gravity. In Newtonian gravity the fields superpose linearly, just like electric fields. So, to find the aggregate field, you just add up the fields contributed by each bit of mass. If the vector sum is zero, then there's no "field" present at that point (the "field" in Newtonian gravity is defined only by its effect on matter). In GR geometric gravity, there isn't any gravitational "field" at all. A "field", in GR, is a tensor field on the spacetime manifold, and there is no tensor field associated with gravitation (as opposed to, say, electromagnetism). All there is, is the bare spacetime manifold with a metric tensor, with some choice of coordinates. If the coordinates have been chosen such that the partial derivatives of the metric tensor are nonzero, you'll measure a gravitational acceleration relative to that coordinate system. If the coordinates have been chosen such that the diagonal components in the metric tensor are not +/- 1 or the off diagonal components aren't zero, you'll measure some distortion of space or time in that coordinate system. Where there are tidal forces, you'll find that the curvature tensor (which is derived from the second partial derivatives of the metric tensor) is nonzero. You can always choose coordinates such that the local acceleration is nil and there isn't any distortion of either space or time. That's the coordinate system which would be used by someone in free-fall at that point in spacetime. However, you can't always choose coordinates such that the curvature vanishes; tidal forces are present even for someone in freefall. This is sometimes taken to mean that the "fingerprint" of a gravitational field (versus acceleration) is curvature, but we can arrange to have gravity present with arbitrarily small curvature, so it's a pretty flimsy fingerprint. For gravitatational fields which are not outrageously strong and for regions which are reasonably well contained, the Newtonian and GR models produce essentially identical results, and both include conservation of energy. Consequently, we can use the Newtonian gravity model with COE to predict gravitational time dilation in simple cases. When you start getting into black holes, small energy transfers over celestial distances, and other "fringe" areas, the two models diverge and it's a lot harder to picture what's going on in the GR cases.
