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avirab
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A (3-d or higher) metric which is flat except for one non-trivial metric function of a different coordinate - eg changing dx2 to f(y)dx2 in Euclidean or Minkowski metric [but not f(x)dx2] - is curved if f(y) has a non-zero second derivative; there is no way to make the f(y) 'disappear', ie to make the metric flat, via a coordinate transformation.
Einstein's 'entwurf' metric was of this type.
Would it be true to say that the metric of this type can't be made flat because there are no other metric functions to 'cancel' with this one during the coordinate transformations? And thus the presence of only one such non-trivial metric function guarantees curvature?
The classical limit (Newtonian regime) of the r-geodesic equation for the Schwarzschild (and entwurf) metric has one dominant connection coefficient, all the others are 'small' (or 'much smaller'), which is why it reduces to the Newtonian gravity acceleration equation. Would it be true to say that no coordinate transformation could make all the connection coefficients 'small', so that the presence of only one connection coefficient - or one very dominant one - in a particular coordinate system guarantees curvature?
Einstein's 'entwurf' metric was of this type.
Would it be true to say that the metric of this type can't be made flat because there are no other metric functions to 'cancel' with this one during the coordinate transformations? And thus the presence of only one such non-trivial metric function guarantees curvature?
The classical limit (Newtonian regime) of the r-geodesic equation for the Schwarzschild (and entwurf) metric has one dominant connection coefficient, all the others are 'small' (or 'much smaller'), which is why it reduces to the Newtonian gravity acceleration equation. Would it be true to say that no coordinate transformation could make all the connection coefficients 'small', so that the presence of only one connection coefficient - or one very dominant one - in a particular coordinate system guarantees curvature?
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