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In a NASA script on negative masses, there was the following statement:

enter image description here

How is it understandable that, there, charge "acts negative on mass"? In general relativity, all kinds of energy density (in the stress-energy-momentum tensor) lead to curvature of spacetime and curvature of spacetime leads to an attractive force. This seems to be contradictory to "something acting negative to mass". Could someone please clarify this contradiction?

What I have thought so far on that, there are three possibilities:

  1. Reissner-Nordström metric is calculated wrong there

  2. Curvature exists also as "opposite curvature" (which is unfortunately not mainstream physics) or

  3. I got something really wrong.

The Link to the presentation (kindly provided by @JanGogolin): https://ntrs.nasa.gov/api/citations/20200000366/downloads/20200000366.pdf

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Irreducible mass is like the "rest mass for a black hole". Just like for a normal particle if you remove the kinetic energy all that remains is the rest mass, for a Kerr–Newman metric after removing electromagnetic and rotational energy what remains is $${\displaystyle M_{\rm {irr}}={\frac {1}{2}}{\sqrt {2M^{2}-r_{Q}^{2}c^{4}/G^{2}+2M{\sqrt {M^{2}-(r_{Q}^{2}+a^{2})c^{4}/G^{2}}}}}}$$

Now let us take a Schwarzschild metric with mass $M$. If you somehow add charge to it without adding energy (like think of adding massless particles with charge* that have very very low frequency). Now your total energy is the same $M$. But the $M_{\rm {irr}}$ is (I neglected rotation i.e. $a=0$) $${\displaystyle M_{\rm {irr}}={\frac {1}{2}}{\sqrt {2M^{2}-r_{Q}^{2}c^{4}/G^{2}+2M{\sqrt {M^{2}-(r_{Q}^{2})c^{4}/G^{2}}}}}}$$ here ${\displaystyle r_{Q}^{2}={\frac {Q^{2}G}{4\pi \varepsilon _{0}c^{4}}}}$. Now you can see that the $M_{\rm {irr}}$ has decreased from initial value which was $M$.

This they might be interpreting as adding negative energy. But I don't think it is the proper way to state that. Strictly speaking the field generated by a point charge has positive energy.

* Such particles will have naked singularities which violate causality. So these are assumed to not exist.

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  • $\begingroup$ And in our universe, mass and charge come always together, therefore, nowhere really exists something what you can call negative mass from charge? It's only that charged mass is a little smaller than mass without charge? And that's the way the contradiction is solved? $\endgroup$ Commented Apr 4, 2022 at 13:39
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    $\begingroup$ @BarrierRemoval yes in our universe mass and charge always come together. If a massless particle has charge then it will have a naked singularity and it will violate causality. $\endgroup$ Commented Apr 4, 2022 at 14:07
  • $\begingroup$ Why it will violate causality? Sounds like time runs backwards $\endgroup$ Commented Apr 4, 2022 at 14:45
  • $\begingroup$ @BarrierRemoval I have never studied how they violate causality but only read that they violate. You can search in arXiv to find the exact mechanism through which they violate causality. $\endgroup$ Commented Apr 4, 2022 at 16:45

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