On 4/11/2019 4:53 PM, [email protected] wrote:
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.On Thursday, April 11, 2019 at 4:37:39 PM UTC-6, Brent wrote: On 4/11/2019 1:58 PM, [email protected] <javascript:> 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? AGYep. That's pretty much the defining characteristic of a Riemannian space. BrentBut 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
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