On Tuesday, April 16, 2019 at 9:15:40 PM UTC-6, Brent wrote: > > > > On 4/16/2019 6:14 PM, agrays...@gmail.com <javascript:> wrote: > > > > On Tuesday, April 16, 2019 at 6:39:11 PM UTC-6, agrays...@gmail.com > wrote: >> >> >> >> On Tuesday, April 16, 2019 at 6:10:16 PM UTC-6, Brent wrote: >>> >>> >>> >>> On 4/16/2019 11:41 AM, agrays...@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 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? 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* > > > No. It's guaranteed that at every point on the particles world line there > is a tangent space. > > Brent >
*On a different issue, if you agree with Stenger that time is what is read on a clock, how do you justify labeling all spacetime points with a t component, which is called "time", and overwhelmingly will never be read on any clock? 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 everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
