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I asked this question on the astronomy stack exchange: https://astronomy.stackexchange.com/q/30477 and it was recommended to me that I instead ask it here.

My question is related to the following question: Conversion of matter into antimatter

In the top answer the following is stated: "Because there are no long-range forces that couple to baryon number or lepton number, a black hole is under no obligation to conserve them, and thus should radiate equal amounts of matter and antimatter, regardless of what fell into it."

Is it therefor possible to set up a "matter-antimatter converter"? That is, a way of turning matter into antimatter or vice versa? In particular consider the following process:

Acquire some matter and force it together so as to form a black hole then wait for it to decay and then acquire the energy and matter-antimatter pairs that leave it. Turn any energy into matter-antimatter pairs. After doing so there should be an equal amount of matter and anti matter. Now repeat this process recursively with the remaining matter, that is, turning the matter into a black hole and letting it decay into matter-antimatter pairs. In the limit all the matter should turn to antimatter.

Is there a reason this process can't be performed? I am ignoring things such as the amount of time needed to do this and other similar constraints.

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    $\begingroup$ Ben has answered your question, and you said you want to ignore practical constraints. But (presuming Hawking radiation exists), this sounds like a really impractical way to make antimatter. You need really tiny BHs for the HR to emit anything apart from photons. If you've got tools that can achieve that insane energy density, you probably have much easier ways to churn out antimatter. $\endgroup$
    – PM 2Ring
    Commented Apr 16, 2019 at 8:02
  • $\begingroup$ A better logic would be as follows. Since matter cannot be converted to anti-matter, there is no Hawking radiation. $\endgroup$
    – safesphere
    Commented Apr 17, 2019 at 4:33

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The answer is yes. There is, however, an error in the quoted material:

Because there are no long-range forces that couple to baryon number or lepton number, a black hole is under no obligation to conserve them, and thus should radiate equal amounts of matter and antimatter, regardless of what fell into it.

This is a non sequitur. The reason that we expect black hole evaporation not to conserve baryon or lepton number is that no-hair theorems tell us that a black hole doesn't "remember" quantum numbers such as baryon or lepton number.

And keep in mind that we don't really know whether black holes evaporate. Semiclassical gravity has never made a correct, testable prediction, it has foundational problems, and there is no reason to trust it.

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  • $\begingroup$ Of course black holes evaporate. There were several flashes that were not able to be answered. It also merges general relativity and quantum mechanics. Of course hawking radiation could work a little differently and the decay products might be different but of course hawking radiation happens. $\endgroup$
    – Roghan Arun
    Commented May 18, 2020 at 15:00

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