On Tuesday, April 16, 2019 at 6:39:11 PM UTC-6, [email protected] wrote: > > > > On Tuesday, April 16, 2019 at 6:10:16 PM UTC-6, Brent wrote: >> >> >> >> On 4/16/2019 11:41 AM, [email protected] wrote: >> >> >> >> On Monday, April 15, 2019 at 9:26:59 PM UTC-6, Brent wrote: >>> >>> >>> >>> On 4/15/2019 7:14 PM, [email protected] wrote: >>> >>> >>> >>> On Friday, April 12, 2019 at 5:48:23 AM UTC-6, [email protected] >>> wrote: >>>> >>>> >>>> >>>> On Thursday, April 11, 2019 at 10:56:08 PM UTC-6, Brent wrote: >>>>> >>>>> >>>>> >>>>> On 4/11/2019 9:33 PM, [email protected] wrote: >>>>> >>>>> >>>>> >>>>> On Thursday, April 11, 2019 at 7:12:17 PM UTC-6, Brent wrote: >>>>>> >>>>>> >>>>>> >>>>>> On 4/11/2019 4:53 PM, [email protected] wrote: >>>>>> >>>>>> >>>>>> >>>>>> On Thursday, April 11, 2019 at 4:37:39 PM UTC-6, Brent wrote: >>>>>>> >>>>>>> >>>>>>> >>>>>>> On 4/11/2019 1:58 PM, [email protected] 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? AG >>>>>>> >>>>>>> >>>>>>> Yep. 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? AG >>>>>> >>>>>> It 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? AG >>>>> >>>>> >>>>> You'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. AG >>> >>> >>> The 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 >>> >>> >>>> Another problem is the inconsistency of the fictitious gravitational >>>> force, and how the other forces function; EM, Strong, and Weak, which >>>> apparently can't be removed by changes in coordinates systems. AG >>>> >>> >>> It's said that consistency is the hobgoblin of small minds. I am merely >>> pointing out the inconsistency of the gravitational force with the other >>> forces. Maybe gravity is just different. AG >>> >>> >>> That's one possibility, e.g entropic gravity. >>> >>> >>>> >>>> >>>>> What is gets you is it enforces and explains the equivalence >>>>> principle. And of course Einstein's theory also correctly predicted the >>>>> bending of light, gravitational waves, time dilation and the precession >>>>> of >>>>> the perhelion of Mercury. >>>>> >>>> >>>> I was referring earlier just to the transformation to the tangent >>>> space; what specifically does it buy us; why would we want to execute this >>>> particular transformation? AG >>>> >>> >>> For one thing, you know the acceleration due to non-gravitational forces >>> in this frame. >>> >> >> *IIUC, the tangent space is a vector space which has elements with >> constant t. So its elements are linear combinations of t, x, y, and z. How >> do you get accelerations from such sums (even if t is not constant)? AG* >> >> So you can transform to it, put in the accelerations, and transform back. >>> >> >> *I see no way to put the accelerations into the tangent space at any >> point in spacetime. AG* >> >> >> The tangent space is just a patch of Minkowski space. d/t(dx/dt) = >> acceleration. >> >> Brent >> > > *Sorry; I was thinking about QM, where the state of the system is a linear > combination of component states of the vector space representing it. In GR, > since there is an infinite uncountable set of tangent spaces, how can we be > sure that our test particle is in one of those subspaces, called tangent > states? That would be the case, I surmise, if the tangent spaces spanned > the manifold. I think they do so since there's a tangent space at every > point in the manifold. AG * >
*The presumed test particle has a history, and each tangent space is a proper subset of the manifold. So is there a guarantee that an arbitrary test particle will have a history contained in a particular tangent space? AG * -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
