On 4/16/2019 11:41 AM, agrayson2...@gmail.com wrote:
On Monday, April 15, 2019 at 9:26:59 PM UTC-6, Brent wrote:
On 4/15/2019 7:14 PM, agrays...@gmail.com <javascript:> 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.
*IIUC, the tangent space is a vector space which has elements with
constant t. So its elements are linear combinations of t, x, y, and
z. How do you get accelerations from such sums (even if t is not
constant)? AG*
*
*
So you can transform to it, put in the accelerations, and
transform back.
*I see no way to put the accelerations into the tangent space at any
point in spacetime. AG*
The tangent space is just a patch of Minkowski space. d/t(dx/dt) =
acceleration.
Brent
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