On 4/11/2019 9:33 PM, agrayson2...@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 <javascript:> 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.  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.

Brent

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