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WannabeNewton

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WannabeNewton

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WannabeNewton

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WannabeNewton

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This is no different from classical EM. When ##\nabla \times E = 0## and ##\nabla \cdot E = 0## we get Laplace's equation ##\nabla^2 \varphi = 0## for the electric potential in a charge free region of space. This doesn't imply ##\varphi = 0## necessarily obviously because things like the Coloumb field satisfy Laplace's equation in the charge free regions.

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thanks

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The Weyl Tensor governs tidal forces and gravitational radiation in free space devoid of matter and is not in Einsteins Field Equation.

Yes, but the EFE is not the equation that governs how particles move in a particular spacetime (i.e., a particular spacetime geometry). Solutions of the EFE describe particular spacetimes; but once you have a solution, you have to use its particular properties (such as the Weyl tensor) to find the equations of motion for particles in that solution.

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WannabeNewton

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To draw analogy again with classical EM, we have four governing equations ##\nabla \cdot E = \rho, \nabla \cdot B = 0## and ##\nabla \times E = -\partial_t B, \nabla \times B = j + \partial_t E## which determine the evolution of the electric and magnetic fields ##E,B##. We use them to solve for the electric and magnetic fields generated by ##\rho,j## and then use Maxwell's equations to determine their evolution explicitly in accordance with ##\rho,j##. To determine the motion of test charges interacting with ##E,B## we use the Lorentz force law ##a = \frac{q}{m}(E + v\times B)##.

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PAllen

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You can't use special relativity. For practical purposes, you use the geodesic law of motion.

It is not invalid to use the EFE, just hard and rarely necessary. An example of where it would be necessary is if you ask what happens if a neutron star approaches a stellar mass black hole. Here, the geodesic law breaks down (the neutron star is not approximately a test body), and you must numerically solve the EFE for two body initial conditions.

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so if you establish a particular spacetime geometry, it's invalid to use the EFE

That's not what I said. I said the EFE is what you solve to figure out what the spacetime geometry is. Once you've figured that out, the EFE doesn't tell you anything else. That doesn't mean the EFE is invalid.

as it is no longer background independent?

Solving the EFE doesn't change anything about background independence.

Can you then use Special Relativity on this particular geometry?

Only if the particular solution of the EFE that describes the geometry is the flat solution, i.e., Minkowski spacetime.

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Schwarzschild geodesics: http://en.wikipedia.org/wiki/Geodesics_of_the_Schwarzschild_vacuum

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