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I believe that equivalence theory only applies (in practice) to very small regions of space.The theory, itself, I think, states that a uniform gravitational field is equivalent to constant acceleration, which obviously doesn't apply to any actual gravitational field.
An observer on one of these earths may see theother approaching earth moving at a greater acceleration than a tennis ball falling from the same altitude.
Equivalence theory tells me I should not be able to distinguish between acceleration and gravity but when my tall space ship is sitting on the earth I can measure a difference in G whether my instrument is on the floor or near the ceiling.Presumably this would not happen when I was away from the Earth and accelerating in space.
The theory, itself, I think, states that a uniform gravitational field is equivalent to constant acceleration, which obviously doesn't apply to any actual gravitational field.
It does apply to fields which can exist in nature or in the presence of bodies which can be constructed. Consider the gravitational field of the Earth. We typically approximate it as a sphere of constant mass density. In that same approximation/idealization we can excavate a spherical cavity out of the Earth somewhere other than the center of the Earth. Perhaps 1,000 m at a depth of 10 km. In such a cavity the field is perfectly uniform. The strength of the field is proportional to the distance between the center of the cavity and the center of the Earth. For a derivation please see:...sorry, you cannot view external links. To see them, please
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Having read the derivation, I understand that the gravitational field in such a cavity is nearly uniform, not perfectly uniform, ..
..even if we assume the Earth to be perfectly spherical, of constant mass density, and not near any other gravitational bodies (tidal forces). In other words, I agree with the calculations of your "approximation/idealization," but would still argue that it doesn't hold for "any actual gravitational field."
For that matter, I will also argue that it doesn't hold for any actual accelerating frame of reference either (because there will be some finite, nonzero contribution from massive objects a finite distance away), but maybe I'm just being picky...
the gravitational field in such a cavity is nearly uniform, not perfectly uniform, ..