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nutgeb
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My understanding is that SR time dilation and Lorentz contraction are flip sides of the same coin. The two always occur together, never separately. That is implicit in spacetime diagrams. I think this duality requirement is necessary so that light always travels at the speed of c in every local inertial frame.
So a question occurs to me. In a scenario where a Lorentz spatial contraction is exactly offset by positive spatial curvature caused by gravity, is SR time dilation implicitly eliminated?
Consider first an empty Milne model cosmology in Minkowski space. Test particles all depart the origin at t = 0 and move away from it in uniform quantities at all speeds up to but not including c. At any point in time the distribution of test particles (as observed from the origin) is inhomogeneous, because the farther a particle is from the origin, the faster it is moving, and therefore the more Lorentz contracted the radial distance is between the origin and the particle. Clocks attached to the test particles are time dilated (relative to the origin clock) in the same proportion as the Lorentz contraction of the particle's distance to the origin.
Now consider a Schwarzschild idealized point mass in empty space. The Schwarzschild metric says that the point mass causes a positive spatial curvature which manifests itself physically as a radial expansion of distances from the point mass. This radial expansion is a function of radial proximity to the point mass -- the closer a test particle is, the greater the proportional radial expansion of distance due to positive spatial curvature. If a test particle is moving radially away from, or toward, the point mass at exactly the escape velocity of the point mass, then this radial expansion of distance mathematically exactly offsets the radial contraction of distance (between the point mass and the test particle) caused by the SR Lorentz contraction resulting from the test particle's velocity. The result is that, in this scenario, the test particle locally measures the space between itself and the point mass to be flat -- zero spatial curvature. (I'm not referring to spacetime curvature, which obviously isn't zero here.)
I submit that because the locally positive spatial curvature in this scenario has counteracted the Lorentz contraction of the distance between the point mass and the test particle, it also prevents any SR time dilation from occurring, as between the test particle and the point mass. Returning to first principles, SR time dilation cannot occur in the absence of Lorentz contraction. Otherwise the locally measured speed of light would depart from c.
In this scenario I'm not referring to gravitational time dilation. The Schwarzschild metric says that gravitational time dilation will occur here as between the point mass and the test particle. Specifically, a clock at the point mass will run slower than the test particle's clock, because the point mass is deeper in its own gravity well than the test particle is.
So a question occurs to me. In a scenario where a Lorentz spatial contraction is exactly offset by positive spatial curvature caused by gravity, is SR time dilation implicitly eliminated?
Consider first an empty Milne model cosmology in Minkowski space. Test particles all depart the origin at t = 0 and move away from it in uniform quantities at all speeds up to but not including c. At any point in time the distribution of test particles (as observed from the origin) is inhomogeneous, because the farther a particle is from the origin, the faster it is moving, and therefore the more Lorentz contracted the radial distance is between the origin and the particle. Clocks attached to the test particles are time dilated (relative to the origin clock) in the same proportion as the Lorentz contraction of the particle's distance to the origin.
Now consider a Schwarzschild idealized point mass in empty space. The Schwarzschild metric says that the point mass causes a positive spatial curvature which manifests itself physically as a radial expansion of distances from the point mass. This radial expansion is a function of radial proximity to the point mass -- the closer a test particle is, the greater the proportional radial expansion of distance due to positive spatial curvature. If a test particle is moving radially away from, or toward, the point mass at exactly the escape velocity of the point mass, then this radial expansion of distance mathematically exactly offsets the radial contraction of distance (between the point mass and the test particle) caused by the SR Lorentz contraction resulting from the test particle's velocity. The result is that, in this scenario, the test particle locally measures the space between itself and the point mass to be flat -- zero spatial curvature. (I'm not referring to spacetime curvature, which obviously isn't zero here.)
I submit that because the locally positive spatial curvature in this scenario has counteracted the Lorentz contraction of the distance between the point mass and the test particle, it also prevents any SR time dilation from occurring, as between the test particle and the point mass. Returning to first principles, SR time dilation cannot occur in the absence of Lorentz contraction. Otherwise the locally measured speed of light would depart from c.
In this scenario I'm not referring to gravitational time dilation. The Schwarzschild metric says that gravitational time dilation will occur here as between the point mass and the test particle. Specifically, a clock at the point mass will run slower than the test particle's clock, because the point mass is deeper in its own gravity well than the test particle is.
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