- #36
jonmtkisco
- 532
- 1
Hi smallphi, Garth & MeJennifer,
I understand that using different coordinates can cause different foliations of spacetime slices.
However, as I understand it, 4-dimensional "spacetime" is not a tangible, physical "geometry" per se, it is merely a particularly convenient abstraction for organizing coordinate systems as a function of the passage of time. In my mind, there has to be some sensible convention for separating spacelike and timelike slices which respects the geometric integrity of space at a chosen instant of time regardless of which particular clock is referred to.
Let's assume we're using the coordinate system of a distant observer whose inertial frame is approximately at rest with respect to the spherical central mass in the Schwarzschild metric. Assume that in this frame, the spherical mass is at rest for a long period of time. At any chosen single instant on the distant observer's clock (which particular instant is irrelevant), the observer measures (takes a "snapshot") of whether [tex]C = 2 \pi r[/tex] over a small set of transverse spatial coordinates very near to the central mass. If it does, then isn't the observer justified in concluding that the spatial curvature is zero?
It seems to me that even if this observer's coordinates are set in an inertial frame which is moving with respect to the rest frame of the mass, or in an accelerating frame (including but not limited to a frame in freefall with respect to the mass), at any arbitrary instant in time the observer will obtain a consistent spatial snapshot of whether [tex]C = 2 \pi r[/tex] over a small set of spatial coordinates near the central mass.
It seems reasonable for the observer to conclude that the spatial curvature is flat, while the spacetime curvature is curved. The curvature of spacetime constituting merely an abstraction of the idea that the timelike or null geodesic of a particle in freefall will be displaced away from a "straight" line as a function of the passage of time. If exactly zero time elapses on any given clock, then exactly zero time must elapse on every possible clock everywhere (even if those clocks do not register that instant in time to be "simultaneous" in their various time metrics), so the instantaneous spacetime curvature everywhere in the Schwarzschild gravitational field must be zero regardless of the coordinate system used. In that sense, I would consider "instantaneous spacetime curvature" to equate to "spatial curvature." Again, I'm assuming that the physical configuration of the central mass remains perfectly constant over a longer time period than is necessary for all possible observers to take their measurement snapshot.
Apparently all of the above is incorrect, so please explain why as explicitly as you can. Generic references to Riemann manifolds and Ricci tensors alone don't seem to provide an intuitive physical explanation.
Jon
I understand that using different coordinates can cause different foliations of spacetime slices.
However, as I understand it, 4-dimensional "spacetime" is not a tangible, physical "geometry" per se, it is merely a particularly convenient abstraction for organizing coordinate systems as a function of the passage of time. In my mind, there has to be some sensible convention for separating spacelike and timelike slices which respects the geometric integrity of space at a chosen instant of time regardless of which particular clock is referred to.
Let's assume we're using the coordinate system of a distant observer whose inertial frame is approximately at rest with respect to the spherical central mass in the Schwarzschild metric. Assume that in this frame, the spherical mass is at rest for a long period of time. At any chosen single instant on the distant observer's clock (which particular instant is irrelevant), the observer measures (takes a "snapshot") of whether [tex]C = 2 \pi r[/tex] over a small set of transverse spatial coordinates very near to the central mass. If it does, then isn't the observer justified in concluding that the spatial curvature is zero?
It seems to me that even if this observer's coordinates are set in an inertial frame which is moving with respect to the rest frame of the mass, or in an accelerating frame (including but not limited to a frame in freefall with respect to the mass), at any arbitrary instant in time the observer will obtain a consistent spatial snapshot of whether [tex]C = 2 \pi r[/tex] over a small set of spatial coordinates near the central mass.
It seems reasonable for the observer to conclude that the spatial curvature is flat, while the spacetime curvature is curved. The curvature of spacetime constituting merely an abstraction of the idea that the timelike or null geodesic of a particle in freefall will be displaced away from a "straight" line as a function of the passage of time. If exactly zero time elapses on any given clock, then exactly zero time must elapse on every possible clock everywhere (even if those clocks do not register that instant in time to be "simultaneous" in their various time metrics), so the instantaneous spacetime curvature everywhere in the Schwarzschild gravitational field must be zero regardless of the coordinate system used. In that sense, I would consider "instantaneous spacetime curvature" to equate to "spatial curvature." Again, I'm assuming that the physical configuration of the central mass remains perfectly constant over a longer time period than is necessary for all possible observers to take their measurement snapshot.
Apparently all of the above is incorrect, so please explain why as explicitly as you can. Generic references to Riemann manifolds and Ricci tensors alone don't seem to provide an intuitive physical explanation.
Jon
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