On 4/15/2019 8:08 PM, [email protected] wrote:
On Monday, April 15, 2019 at 8:14:35 PM UTC-6, [email protected] wrote:
On Friday, April 12, 2019 at 5:48:23 AM UTC-6, [email protected]
wrote:
On Thursday, April 11, 2019 at 10:56:08 PM UTC-6, Brent wrote:
On 4/11/2019 9:33 PM, [email protected] wrote:
On Thursday, April 11, 2019 at 7:12:17 PM UTC-6, Brent
wrote:
On 4/11/2019 4:53 PM, [email protected] wrote:
On Thursday, April 11, 2019 at 4:37:39 PM UTC-6,
Brent wrote:
On 4/11/2019 1:58 PM, [email protected] wrote:
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.
Yeah, that's what he's doing. He's assuming a
given coordinate system and some arbitrary
point in a non-empty spacetime. So spacetime
has a non zero curvature and the derivative of
the metric tensor is generally non-zero at that
arbitrary point, however small we assume the
region around that point. But applying the EEP,
we can transform to the tangent space at that
point to diagonalize the metric tensor and have
its derivative as zero at that point. Does THIS
make sense? AG
Yep. That's pretty much the defining
characteristic of a Riemannian space.
Brent
But isn't it weird that changing labels on spacetime
points by transforming coordinates has the result of
putting the test particle in local free fall, when
it wasn't prior to the transformation? AG
It doesn't put it in free-fall. If the particle has
EM forces on it, it will deviate from the geodesic in
the tangent space coordinates. The transformation is
just adapting the coordinates to the local free-fall
which removes gravity as a force...but not other forces.
Brent
In both cases, with and without non-gravitational forces
acting on test particle, I assume the trajectory appears
identical to an external observer, before and after
coordinate transformation to the tangent plane at some
point; all that's changed are the labels of spacetime
points. If this is true, it's still hard to see why
changing labels can remove the gravitational forces. And
what does this buy us? AG
You're looking at it the wrong way around. There never
were any gravitational forces, just your choice of
coordinate system made fictitious forces appear; just like
when you use a merry-go-round as your reference frame you
get coriolis forces.
If gravity is a fictitious force produced by the choice of
coordinate system, in its absence (due to a change in
coordinate system) how does GR explain motion? Test particles
move on geodesics in the absence of non-gravitational forces,
but why do they move at all? AG
Maybe GR assumes motion but doesn't explain it. AG
Another problem is the inconsistency of the fictitious
gravitational force, and how the other forces function; EM,
Strong, and Weak, which apparently can't be removed by changes
in coordinates systems. AG
It's said that consistency is the hobgoblin of small minds. I am
merely pointing out the inconsistency of the gravitational force
with the other forces. Maybe gravity is just different. AG
What is gets you is it enforces and explains the
equivalence principle. And of course Einstein's theory
also correctly predicted the bending of light,
gravitational waves, time dilation and the precession of
the perhelion of Mercury.
I was referring earlier just to the transformation to the
tangent space; what specifically does it buy us; why would we
want to execute this particular transformation? AG
Brent
*I could be mistaken, I usually am, but ISTM that labeling all points
in spacetime as (t, x, y, z) makes no sense since there is no
universal clock in GR. Each observer has his own clock in GR. No
"Bird's Eye" observer GR. So what could the same t for all spatial
points mean, or increasing t's as time evolves? AG*
The "t" in the coordinate point is just a label. The difference of of
t-values at two events is not in general the elapsed time between them.
The elapsed time between them depends on the path taken (twin paradox)
and has to be calculated by integrating the metric along the path. The
metric is given as a function of (t,x,y,z).
Brent
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