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I quote Sabine Hossenfelder:

Gravitation is a spin-2 interaction. It is straightforward to see that this means that like charges attract and unlike charges repel. The charge of gravity is the mass. This does not mean that negative gravitational mass repels everything. Negative gravitational mass repels positive mass but attracts negative mass. Reference

I also quote W. B. Bonnor:

Active gravitational mass occurs for the first time as a constant of integration in Schwarzschild's solution. If 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 (this) case, all bodies will be repelled. Reference

If gravity is mediated by a spin two field in which negative masses attract, how can the Schwarzchild solution with negative mass repel a negative mass test particle?

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  • $\begingroup$ Note that negative mass would violate the weak energy condition, which is something that you cannot do with ordinary matter. $\endgroup$
    – Andrew
    Commented Oct 8, 2023 at 23:11
  • $\begingroup$ More on negative mass in gravity. $\endgroup$
    – Qmechanic ♦
    Commented Oct 9, 2023 at 6:48

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In a spin-2 theory, you can prove that the direction of the force between like-charged particles is toward the other particle, but if the particles have negative inertial mass (which a particle with negative gravitational charge would have to have by the equivalence principle), then they accelerate in the opposite direction to the force.

Looking at the rest of Hossenfelder's blog post and some of the comments, it looks like she considers the standard approach to negative mass to be inconsistent because of the existence of self-accelerating matter configurations. (If you put a positive and negative mass next to each other, the positive mass attracts the negative mass and the negative mass repels the positive mass, so they both accelerate in the same direction indefinitely.) She linked a paper she wrote about a theory with two different metrics, the second metric, I guess, determining the geodesic motion of negative masses. The paper was published in Physical Review D, so at least a few people thought it had some merit, but it is definitely not standard. In standard GR, a negative mass has to repel everything (see Sten's answer), and the most likely resolution of the problem of self-accelerating systems is that negative masses just don't exist.

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If gravity could repel some masses while attracting others, then you could use this to locally measure the gravitational field, in violation of the equivalence principle. In a theory that satisfies the equivalence principle, a mass must either attract everything or repel everything.

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  • $\begingroup$ Would this measurement work by measuring how rapidly the attracted and repelled masses diverged from each other? $\endgroup$
    – Charles Hudgins
    Commented Oct 9, 2023 at 8:17
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    $\begingroup$ @CharlesHudgins Yeah, that's a possibility. For convenience, you could also fix them together with a spring and measure the displacement. $\endgroup$
    – Sten
    Commented Oct 9, 2023 at 13:41
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The source of gravity is energy-momentum tensor, which is related to mass-SQUARED, which means the sign of mass DOES NOT MATTER to gravity.

All the talk about negative mass is a wild goose chase.

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  • $\begingroup$ No, mass appears in the Schwarzschild solution as a constant of integration. It is a solution of the vacuum Einstein equations, therefore the stress-energy tensor is null everywhere the metric is defined. $\endgroup$
    – Independent Physics
    Commented Oct 9, 2023 at 20:16
  • $\begingroup$ @Manuel, the constant of integration is $\sqrt{m^2}$, not $m$. $\endgroup$
    – MadMax
    Commented Oct 9, 2023 at 21:15
  • $\begingroup$ I cite Hermann Bondi: "Active gravitational mass occurs for the first time as a constant of integration in Schwarzschild's solution. [...] 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." $\endgroup$
    – Independent Physics
    Commented Oct 9, 2023 at 21:50
  • $\begingroup$ @Manuel, please do a simple excise of calculating the energy-momentum tensor of for example the Klein–Gordon action and see how mass enters the picture, and you will understand what I am talking about. $\endgroup$
    – MadMax
    Commented Oct 10, 2023 at 13:44
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    $\begingroup$ Which is the energy momentum tensor in the Schwarzschild vacuum solution then? (or the one elsewhere) $\endgroup$
    – Independent Physics
    Commented Oct 10, 2023 at 22:58

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