On 4/9/2019 6:52 PM, agrayson2...@gmail.com wrote:
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?
Sort of. Where ever the particle is in spacetime there is a tangent
space at that point which is locally Minkowski. This is just another
way of describing Einstein's equivalence principle. If you take this
tangent space as your reference frame then the particle is stationary in
free fall. This is a useful technique if, for example, the "particle"
is a rocket ship. In the ship's instantaneous reference frame it will
follow a geodesic unless it fires it's rockets. If it fires it's
rockets then it experiences a proper acceleration relative to that
free-fall frame that is just F/m. Having found that, then you can
transform the result to some other reference frame you're interested,
e.g. the Earth.
Brent
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
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