pervect said:
I do believe there may be some difficulty with proving that a negative test mass follows a geodesic directly from Einstein's equation.
The geodesic equation in GR does not involve the mass of the test particle at all; it can be factored out in the process of deriving the equation. (As you note, this is just a manifestation of the fact that GR obeys the equivalence principle.) The only restriction is that the mass must be nonzero (more precisely, that deriving the geodesic equation for timelike geodesics involves a slightly different procedure than doing it for null geodesics). The sign of the mass plays no role at all.
That being the case, even to
define the "mass" of a test particle following a timelike geodesic involves some subtleties. By definition, the test particle does not affect the spacetime curvature at all, so there is no way to define a "mass" from its stress-energy since it has zero stress-energy by definition. The fact that its worldline is timelike doesn't help, because once you have made a choice of which direction is the "future" direction on the worldline, the sign of the particle's 4-momentum, which formally would depend on the sign of its mass, is AFAIK a convention (whether you want a 4-momentum pointing in the future direction to be positive or negative), not a physical quantity; therefore the sign of its mass is a convention as well. AFAIK no physical observable depends on the sign.
Another issue is the fact that in GR, unlike in Newtonian physics, there is no separate concept of "inertial" vs. "gravitational" mass (more precisely, "passive gravitational" mass, since as noted above, a test particle has zero "active" gravitational mass by definition since it is not a source of gravity). So the idea of switching the sign of one for a test particle while leaving the sign of the other the same isn't even well-defined in GR.
The concept of negative
active gravitational mass is well-defined in GR: the simplest example is Schwarzschild spacetime with ##M## negative in the line element. Geodesics in this spacetime "fall" away from the center instead of falling towards it, and it has no horizon and a timelike curvature singularity at ##r = 0##. But, as above, this is true regardless of the mass of test particles. Also, this solution is, AFAIK, not considered physically reasonable by any physicists working in the field.