On Monday, April 15, 2019 at 9:26:59 PM UTC-6, Brent wrote: > > > > On 4/15/2019 7:14 PM, [email protected] <javascript:> 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* > So all the "gravitation" is in the transform. > > Brent > -- 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.
