On Tuesday, April 9, 2019 at 6:41:52 PM UTC-6, Brent wrote: > > > > On 4/9/2019 5:20 PM, [email protected] <javascript:> wrote: > > > > On Tuesday, April 9, 2019 at 2:40:16 PM UTC-6, Brent wrote: >> >> >> >> On 4/9/2019 12:47 PM, [email protected] wrote: >> >> >> >> On Tuesday, April 9, 2019 at 1:35:34 PM UTC-6, Brent wrote: >>> >>> >>> >>> On 4/9/2019 11:55 AM, [email protected] 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 >> >> >> 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 [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.
