In the presence of a gravitational field, the spacetime metric, $$ds^2=\eta_{ab}dx^a dx^b,$$ should be changed to, $$ds^2=g_{ab}(x)dx^adx^b.$$ What are the convincing physical arguments that motivate this point?
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asked Apr 12 at 17:07
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$\begingroup$ en.wikipedia.org/wiki/General_relativity $\endgroup$– hftCommented Apr 12 at 17:17
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4$\begingroup$ This is a standard definition in geometry. There is nothing to motivate. The whole point is that in GR, spacetime isn't necessarily flat. So $g$ is no longer necessarily equal to $\eta$. $\endgroup$– Vincent ThackerCommented Apr 12 at 17:19
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1$\begingroup$ "The whole point is that in GR, spacetime isn't necessarily flat." Do you take that as a postulate out of the blue? $\endgroup$– SolidificationCommented Apr 12 at 17:35
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1$\begingroup$ @VincentThacker, using a postulated theory without the physical justifications that gave a physical sense of the theory is bad physics, in my opinion. General Relativity was founded on simple ideas and principles (the equivalence principle, that started the whole theory). These ideas may become "useless" after the whole theory have been formulated, but we still need its physical justifications to motivate its formulation. $\endgroup$– ChamCommented Apr 12 at 17:41
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1$\begingroup$ @Cham I agree. But I originally interpreted the question as asking why the expression of the interval takes the form it takes, rather than why spacetime is described by a manifold. $\endgroup$– Vincent ThackerCommented Apr 12 at 18:16
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1 Answer
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It's a consequence of the equivalence principle, for which a gravitational field is locally indiscernible from an acceleration effect, for any moving small body. And the corollary: a free fall in a gravitational field is locally indiscernible from an inertial motion in Minkowski spacetime. Since the observer could have any motion and may use any coordinates system, the Minkowski metric should be promoted to a general spacetime metric.
This is so simple and beautiful that I have a very hard time in thinking how this could be false! Unless we discover one day that different particles could have different accelerations in a given gravitational field.
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$\begingroup$ "Since the observer could have any motion and may use any coordinates system, the Minkowski metric should be promoted to a general spacetime metric." But the observer can use any coordinate system in in flat spacetime. $\endgroup$ Commented Apr 12 at 17:39
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$\begingroup$ @Solidification, the Minkowski metric may be formulated for any coordinates system and any observer motion, yes, but it's still the Minkowski metric: flat geometry (without curvature) and without any real gravitational field. The equivalence principle is a local principle only. So to add a gravitational field, we need to use 1) a scalar field in Minkowski spacetime (aka Nordstrom theory or its variants), or 2) use a more general metric (with curvature) that extends the Minkowski metric. It appears that the simple scalar theory is too simple and doesn't fit with observations. $\endgroup$– ChamCommented Apr 12 at 17:49