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Talk:Essay:Quantifying Openmindedness (view source)
Revision as of 03:53, October 2, 2009
, 03:53, October 2, 2009→Just my answers, for fun!
* Gravitational fields are divergence-free away from a source. This physically means that there is no contribution to the gravitational field of a mass from any point that's not part of the mass.
* Gravitational fields are divergence-free away from a source. This physically means that there is no contribution to the gravitational field of a mass from any point that's not part of the mass.
::This is basically a consequence of the [[divergence theorem]] (if I'm thinking straight). These are both basic tenets of the Newtonian picture. It's really not possible to do Newtonian mechanics where gravity is an inverse-2.00001 law: it would look to bodies as if gravitational force was being exerted by a body outside the mass in question! Similarly the fact that Coulomb's law is precisely an inverse-square law is a consequence of Gauss's law, which also follows from very basic assumptions. On the other hand, Newtonian physics isn't exact in the first place. I think I have read that gravity seems to be weaker than an inverse square law over extremely large distances (on the astronomical scale). This is because of some non-Newtonian, and perhaps not fully understood, physics. But there's no way to describe gravity in GR by something as simple as an inverse-square law anyway! --[[User:MarkGall|MarkGall]] 23:31, 1 October 2009 (EDT)
::This is basically a consequence of the [[divergence theorem]] (if I'm thinking straight). These are both basic tenets of the Newtonian picture. It's really not possible to do Newtonian mechanics where gravity is an inverse-2.00001 law: it would look to bodies as if gravitational force was being exerted by a body outside the mass in question! Similarly the fact that Coulomb's law is precisely an inverse-square law is a consequence of Gauss's law, which also follows from very basic assumptions. On the other hand, Newtonian physics isn't exact in the first place. I think I have read that gravity seems to be weaker than an inverse square law over extremely large distances (on the astronomical scale). This is because of some non-Newtonian, and perhaps not fully understood, physics. But there's no way to describe gravity in GR by something as simple as an inverse-square law anyway! --[[User:MarkGall|MarkGall]] 23:31, 1 October 2009 (EDT)
That's a great analysis, Mark. Even before the invention of calculus, Newton figured this with basic geometry, on the assumption that the force of gravity caused by a point mass would be the same in any direction, that is, at any other point at some distance from the mass. This logically means the force would be spread out equally over a sphere at a given radius r. Then the strength of the force at a point at radius r would be some "inherent" strength, divided by the area of the sphere. The surface area of a sphere is 4*pi*r^2. That is where the exponent 2 comes from.
That formula for the area is geometric; the exponent couldn't possibly be anything other than 2.
ON THE OTHER HAND, this is all in Newtonian gravity. In GR, a derivation of the exponents works similarly, but the possibility of them being slightly off 2 is equally remote, that is, 0, but even then, GR is not an ultimate formulation of gravity. Who knows how the world really works. [[User:JacobB|JacobB]] 23:53, 1 October 2009 (EDT)