On Tuesday, April 9, 2019 at 1:35:34 PM UTC-6, Brent wrote: > > > > On 4/9/2019 11:55 AM, [email protected] <javascript:> wrote: > > > > On Tuesday, April 9, 2019 at 12:05:11 PM UTC-6, Brent wrote: >> >> >> >> On 4/9/2019 7:52 AM, [email protected] wrote: >> >> >> >> On Monday, April 8, 2019 at 11:16:25 PM UTC-6, [email protected] >> 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 [email protected] <javascript:>. > To post to this group, send email to [email protected] > <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 [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.
