On Thursday, April 11, 2019 at 10:56:08 PM UTC-6, Brent wrote: > > > > On 4/11/2019 9:33 PM, [email protected] <javascript:> 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 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 > 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 > > 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.
