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Positronium is an atom with one electron and one positron. It's mass is 1.022MeV which is almost twice the electron mass: The ground-state (1S orbital) binding energy of -6.8eV reduces the total mass ever so slightly.

The ratio of mass to binding energy is very close to $8\alpha^2$ where $\alpha$ is the fine structure constant of about 1/137. This is not a coincidence: $\alpha^2$ determines how strong relativistic effects are in QED.

So just make $\alpha > \sqrt {8}$ and get a negative ground-state mass? My reasoning is no. QED is pertubative: you get a Taylor series in $\alpha$ and neglect high order terms. If $\alpha$ was large this process wouldn't work.

My understanding is that negative mass in a vacuum is nonsensical and the mass of positronium would stay positive no matter how big $\alpha$ became. Is this reasoning correct?

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    $\begingroup$ The question is how $mc^2$ depends on $\alpha$. $\endgroup$
    – my2cts
    Commented Feb 7 at 10:59
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    $\begingroup$ Ground state binding energy is $-6.8 eV$, not $6.8$,- you forgot negative sign. So if you'll subtract negative binding,- you'll get positive increased total energy. Otherwise, you will not get 1.022 MeV. $\endgroup$
    – Agnius Vasiliauskas
    Commented Feb 7 at 12:56
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    $\begingroup$ @AgniusVasiliauskas You are wrong on both counts. The mass of positronium is lower than the mass of two electrons by the binding energy. And whether or not the 6.8eV reduces the mass, the mass still rounds to 1.022MeV. Two free electrons have a mass of 1.0219979MeV. The binding energy is 6.8eV=0.0000068MeV for a total of 1.0218811MeV. Whether or not the "binding energy" is negative depends on your convention - if you think binding energy should be added to mass energy, then it is negative. If you think it should be subtracted, it is positive. But it is definitely true that the mass is reduced. $\endgroup$
    – AXensen
    Commented Feb 8 at 18:04
  • $\begingroup$ Yes, you are right. I understood that I've made an error in analysis too lately. I've removed my answer for similar reasons, but forgot to delete the comment. Everybody makes mistakes. The only thing I may not fully agree with you is ground energy sign. As I've understood it, it's accepted to let energy be negative in bounded system, and positive in unbounded ones. But maybe you are right here too in respect that sign can just indicate energy difference with respect to some reference frame, similarly like for gravitational potential energy $\endgroup$
    – Agnius Vasiliauskas
    Commented Feb 8 at 18:56

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If $\alpha$ is too large, the electron and positron will annihilate before positronium can form.

As you note, this problem becomes non-perturbative, and I am not aware of any precise calculations. There is, however, a significant literature on the closely related "Z>137 Catastrophe" that happens if you have a large enough point charge. This is important in trying to understand if there is an end to the periodic table.

The hand-waving explanation is that as the electron and positron approach each other, at some distance it becomes energetically possible for an $e^+e^-$ pair to pop out of the vacuum between initial electron and positron. The $e^-$ from the pair will annihilate with the initial positron and the $e^+$ will annihilate with the initial electron producing photons. This will happen before any positronium state can form.

A theorist might be able to provide more details. There is a large literature on how electron-positron pairs are created in strong electric fields, and this paper on the breakdown of perturbative QED in strong electromagnetic fields is recommended in a comment.

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  • $\begingroup$ Might adduce papers. $\endgroup$
    – Cosmas Zachos
    Commented Feb 8 at 18:05
  • $\begingroup$ Thanks @Cosmas-Zachos. I have added your suggested paper and another link to help people find the relevant literature. $\endgroup$
    – David Bailey
    Commented Feb 9 at 17:40

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