On Tuesday, April 9, 2019 at 6:41:52 PM UTC-6, Brent wrote: > > > > On 4/9/2019 5:20 PM, agrays...@gmail.com <javascript:> wrote: > > > > On Tuesday, April 9, 2019 at 2:40:16 PM UTC-6, Brent wrote: >> >> >> >> On 4/9/2019 12:47 PM, agrays...@gmail.com wrote: >> >> >> >> On Tuesday, April 9, 2019 at 1:35:34 PM UTC-6, Brent wrote: >>> >>> >>> >>> On 4/9/2019 11:55 AM, agrays...@gmail.com 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 >> >> >> Like "geodesic" being "locally flat" is a physical characteristic of the >> spacetime. It's just part of being a Riemannian space that there is a >> sufficiently small region around any point that is "flat". This is the >> mathematical correlate of Einstein's equivalence principle. So it is not >> the coordinate system or any transformation that "puts the particle in a >> flat region". It's just a property of the space being smooth and >> differentiable so that even a curved spacetime at every point has a flat >> tangent space. >> >> Brent >> > > What you're saying is pretty easy to understand. So I wonder why the > "expert" I was discussing this with, claimed something about a > transformation existing from one coordinate system to another, to make the > particle to be locally in an inertial condition, when that's always the > case? Do you have any idea what he was referring to? AG > > > Well, it's not always the case. There are other forces that can act on a > particle besides gravity. So the the fact that you can always transform to > a free-falling local reference frame and eliminate "gravitational force" > doesn't mean that a particle may not be accelerated by EM or other forces. > > Brent >
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? I am not sure what initial conditions he assumed for the test particle, whether or not it was under the influence of non gravitational forces. 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.
