On Friday, April 12, 2019 at 5:48:23 AM UTC-6, [email protected] wrote:
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>
>
> On Thursday, April 11, 2019 at 10:56:08 PM UTC-6, Brent wrote:
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>> On 4/11/2019 9:33 PM, [email protected] wrote:
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>>
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>> 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
>>
>

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