On 4/15/2019 7:14 PM, agrayson2...@gmail.com wrote:
On Friday, April 12, 2019 at 5:48:23 AM UTC-6, agrays...@gmail.com wrote:
On Thursday, April 11, 2019 at 10:56:08 PM UTC-6, Brent wrote:
On 4/11/2019 9:33 PM, agrays...@gmail.com wrote:
On Thursday, April 11, 2019 at 7:12:17 PM UTC-6, Brent wrote:
On 4/11/2019 4:53 PM, agrays...@gmail.com wrote:
On Thursday, April 11, 2019 at 4:37:39 PM UTC-6, Brent
wrote:
On 4/11/2019 1:58 PM, agrays...@gmail.com 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
The sciences do not try to explain, they hardly even try to interpret,
they mainly make models. By a model is meant a mathematical construct
which, with the addition of certain verbal interpretations, describes
observed phenomena. The justification of such a mathematical construct
is solely and precisely that it is expected to work.
--—John von Neumann
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
That's one possibility, e.g entropic gravity.
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
For one thing, you know the acceleration due to non-gravitational forces
in this frame. So you can transform to it, put in the accelerations,
and transform back. So all the "gravitation" is in the transform.
Brent
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