On 4/16/2019 7:56 AM, agrayson2...@gmail.com wrote:
On Monday, April 15, 2019 at 9:26:59 PM UTC-6, Brent wrote: On 4/15/2019 7:14 PM, agrays...@gmail.com <javascript:> wrote:On Friday, April 12, 2019 at 5:48:23 AM UTC-6, agrays...@gmail.com wrote: On Thursday, April 11, 2019 at 10:56:08 PM UTC-6, Brent wrote: On 4/11/2019 9:33 PM, agrays...@gmail.com wrote:On Thursday, April 11, 2019 at 7:12:17 PM UTC-6, Brent wrote: On 4/11/2019 4:53 PM, agrays...@gmail.com wrote:On Thursday, April 11, 2019 at 4:37:39 PM UTC-6, Brent wrote: On 4/11/2019 1:58 PM, agrays...@gmail.com wrote:He might have been referring to a transformation to a tangent space where the metric tensor is diagonalized and its derivative at that point in spacetime is zero. Does this make any sense?Sort of. Yeah, that's what he's doing. He's assuming a given coordinate system and some arbitrary point in a non-empty spacetime. So spacetime has a non zero curvature and the derivative of the metric tensor is generally non-zero at that arbitrary point, however small we assume the region around that point. But applying the EEP, we can transform to the tangent space at that point to diagonalize the metric tensor and have its derivative as zero at that point. Does THIS make sense? AGYep. That's pretty much the defining characteristic of a Riemannian space. Brent But isn't it weird that changing labels on spacetime points by transforming coordinates has the result of putting the test particle in local free fall, when it wasn't prior to the transformation? AGIt doesn't put it in free-fall. If the particle has EM forces on it, it will deviate from the geodesic in the tangent space coordinates. The transformation is just adapting the coordinates to the local free-fall which removes gravity as a force...but not other forces. Brent In both cases, with and without non-gravitational forces acting on test particle, I assume the trajectory appears identical to an external observer, before and after coordinate transformation to the tangent plane at some point; all that's changed are the labels of spacetime points. If this is true, it's still hard to see why changing labels can remove the gravitational forces. And what does this buy us? AGYou're looking at it the wrong way around. There never were any gravitational forces, just your choice of coordinate system made fictitious forces appear; just like when you use a merry-go-round as your reference frame you get coriolis forces. If gravity is a fictitious force produced by the choice of coordinate system, in its absence (due to a change in coordinate system) how does GR explain motion? Test particles move on geodesics in the absence of non-gravitational forces, but why do they move at all? AG Maybe GR assumes motion but doesn't explain it. AGThe sciences do not try to explain, they hardly even try to interpret, they mainly make models. By a model is meant a mathematical construct which, with the addition of certain verbal interpretations, describes observed phenomena. The justification of such a mathematical construct is solely and precisely that it is expected to work. --—John von Neumann*This is straight out of the "shut up and calculate" school, and I don't completely buy it. E.g., the Principle of Relativity and Least Action Principle give strong indications of not only how the universe works, but why. That is, they're somewhat explanatory in nature. AG*
Fine, then take them as explanations. But to ask that they be explained is to misunderstand their status. It's possible that they could be explained; but only by finding a more fundamental theory that includes them as consequences or special cases. Whatever theory is fundamental cannot have an explanation in the sense you want because then it would not be fundamental.
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