On Tuesday, April 9, 2019 at 1:35:34 PM UTC-6, Brent wrote: > > > > On 4/9/2019 11:55 AM, agrays...@gmail.com <javascript:> wrote: > > > > On Tuesday, April 9, 2019 at 12:05:11 PM UTC-6, Brent wrote: >> >> >> >> On 4/9/2019 7:52 AM, agrays...@gmail.com wrote: >> >> >> >> On Monday, April 8, 2019 at 11:16:25 PM UTC-6, agrays...@gmail.com >> wrote: >>> >>> In GR, is there a distinction between coordinate systems and frames of >>> reference? AG?? >>> >> >> Here's the problem; there's a GR expert known to some members of this >> list, who claims GR does NOT distinguish coordinate systems from frames of >> reference. He also claims that given an arbitrary coordinate system on a >> manifold, and any given point in space-time, it's possible to find a >> transformation from the given coordinate system (and using Einstein's >> Equivalence Principle), to another coordinate system which is locally flat >> at the arbitrarily given point in space-time. This implies that a test >> particle is in free fall at that point in space-time. But how can changing >> labels on space-time points, change the physical properties of a test >> particle at some arbitrarily chosen point in space-time? I believe that >> such a transformation implies a DIFFERENT frame of reference, in motion, >> possibly accelerated, from the original frame or coordinate system. Am I >> correct? TIA, AG >> >> >> You're right that a coordinate system is just a function for labeling >> points and, while is may make the equations messy or simple, it doesn't >> change the physics.?? If you have two different coordinate systems the >> transformation between them may be arbitrarily complicated.?? But your last >> sentence referring to motion as distinguishing a coordinate transform from >> a reference frame seems to have slipped into a 3D picture.?? In a 4D >> spacetime, block universe there's no difference between an accelerated >> reference frame and one defined by coordinates that are not geodesic. >> >> Brent >> > > Suppose the test particle is on a geodesic path in one coordinate system, > but in another it's on an approximately flat 4D surface at some point in > the transformed coordinate system. > > > A geodesic is a physically defined path, one of extremal length. It's > independent of coordinate systems and reference frames. If a geodesic is > not a geodesic in your transformed coordinate system, then you've done > something wrong in transforming the metric. > > Brent >
It would clarify the situation if you would state the acceptable before and after states of a coordinate transformation that puts the test particle in a locally flat region for some chosen point in the transformed coordinate system. AG > > Doesn't this represent a change in the physics via a change in labeling > the space-time points? How is this possible without a change in the frame > of reference, and if so, how would that be described if not by > acceleration? 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 everyth...@googlegroups.com <javascript:>. > To post to this group, send email to everyth...@googlegroups.com > <javascript:>. > Visit this group at https://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/d/optout. > > > -- 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.
