0
$\begingroup$

Elaborating on antimatter, where atoms consist of antiprotons, antineutrons, and positrons

Is it possible to have something in between? with antiprotons, positrons, and regular neutrons? Can such an atom exist in a stable mode without the anti-upquarks annihilating the upquarks and likewise for the antidowns and downs?

Taking, for simplicity, an example analogous to deuterium, an atom consisting of an antiproton, a neutron and a positron. Is then the reaction$$\bar{u}\bar{u}\bar{d}+udd\rightarrow\bar{u}d+Energy$$inevitable? Or can these quarks/antiquarks exist safely confined within their respective particle in the nucleus to make a stable atom?

$\endgroup$
2
  • 2
    $\begingroup$ @CosmasZachos No, antiprotons (negatively charged) would attract positions (positively charged), resulting in the same closed orbits seen in atomic physics. What matters is whether neutrons are problematic. $\endgroup$
    – J.G.
    Sep 14, 2021 at 13:41
  • $\begingroup$ @J.G. sorry, you are right... I was turned around... $\endgroup$ Sep 14, 2021 at 14:09

2 Answers 2

2
$\begingroup$

I don't think it's annihilation you have to worry about. It's my understanding that at typical energies adjacent (anti)baryons in a cluster confine their respective valence (anti)quarks too tightly for them to annihilate those in another (anti)baryon.

The real stability issue for a neutron/antiproton nucleus is that beta-decay of one or more neutrons to protons will reduce the electrostatic potential. Having said that, it's possible some specific numbers of neutrons & antiprotons are especially stable; you'd just need to deduce their equivalent of the whole of neutron-proton physics: magic numbers, drip lines etc. That's well beyond the scope of this question.

(A thought on terminology: if a cluster of nucleons leads to nuclear physics, we may as well call the antinucleon case antinuclear physics, while your mixed case would be sesquinuclear physics.)

$\endgroup$
2
  • $\begingroup$ +1 for sesquinuclear! But the sesquideuteron never stood a chance, given intercommunication of its constituents via mesons... $\endgroup$ Sep 14, 2021 at 14:16
  • $\begingroup$ You may have a point about tightness of individual nucleons; the charge radius of a deuteron is 2.1 fm, vs 0.9 fm of the proton... $\endgroup$ Sep 14, 2021 at 14:39
0
$\begingroup$

Antinucleons in our present standard model can exist and be stable in anti atoms. The Alice experiment at CERN experiments with antinuclei.Antimatter annihilates if it interacts with matter, otherwise it is stable enough to experiment with.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.