Can mass be negative in Schwarzschild metric? If we use $M<0$, will it still be a solution to EFE? If not, why?
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$\begingroup$ Related: physics.stackexchange.com/q/453939/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jun 18 at 7:59
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2 Answers
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Yes, the Schwarzschild solution with $M<0$ is still a vacuum solution to the Einstein equations. Bondi explored some ideas about negative mass in GR in this paper, in which he says
As long as relativity is considered purely as a theory of gravitation, the inertial and passive gravitational masses do not in fact appear. Active gravitational mass occurs for the first time as a constant of integration in Schwarzschild's solution. If this constant is taken to be positive, then test particles will, in the first approximation, describe the Newtonian orbits corresponding to an attractive body. If, however, the constant is taken to be negative then, in the first approximation, test particles will describe the orbits corresponding to the Newtonian case with repulsion. Note that in the first case all bodies will be attracted, in the second all bodies will be repelled.
With that being said, there are some issues with such a solution - for example, it possesses a naked singularity and therefore violates the cosmic censorship hypothesis. You might find the wiki article on negative mass to be useful.
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$\begingroup$ If we define Schwarzschild solution as a vacuum solution outside of spherical matter agglomeration then we face the question how that could be formed given the gravitational repulsion? $\endgroup$– JanGCommented Apr 9, 2023 at 11:21
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1$\begingroup$ @JanG The Schwarzchild solution is an idealized eternal, unchanging solution. I don't think we can apply it to such a scenario, which by definition would not be eternal or unchanging. $\endgroup$ Commented Apr 9, 2023 at 15:15
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$\begingroup$ @StephenG-HelpUkraine If we mean the eternal black hole spacetime without matter I would agree with you. I have thought more of vacuum metric outside a ball of matter like for example constant density star. $\endgroup$– JanGCommented Apr 9, 2023 at 18:36
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3$\begingroup$ @JanG The Schwarzschild metric is the unique spherically-symmetric vacuum solution to the EFE. If you are applying it to the exterior region of a spacetime which includes a spherically symmetric distribution of matter, it is agnostic as to the nature of the matter or how that distribution of matter came to be in the first place. There are plenty of reasons (see the linked wiki article) to suspect that negative mass (and aggregate collections thereof) is unphysical, but that's a separate issue to the hypothetical metric in the vacuum region around it. $\endgroup$ Commented Apr 9, 2023 at 19:38
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$\begingroup$ @J.Murray You are course right with your answer and comment. My remark about vacuum solutions touches on the question if spacetime without matter somewhere can be a physical solution of Einstein field equations at all. But this topic deserves a new question posted here. $\endgroup$– JanGCommented Apr 10, 2023 at 8:00
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No, actually you can't have negative energy/mass in classical general relativity, so $M<0$ is not a real solution of EFE (ie, it do not describe a possible geometry). There are a lot of proofs of this statement, see for example Witten's one.
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$\begingroup$ General relativity can cope with negative mass. It only is about additionally imposed conditions, like the positive energy theorem, that preclude negative mass. $\endgroup$– M.S.Commented Apr 10, 2023 at 13:38
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$\begingroup$ Yes, that's a theorem not a postulate. Is a necessary implication of GR. You can't have GR without the additional positive energy theorem $\endgroup$ Commented Apr 10, 2023 at 14:43