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
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
> 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
>
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