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The universe so far:

If charged lepton fields are eliminated from the universe, charged pions become stable (having no decay path that preserves charge), replacing electrons to form bound "atomic" states with protons, as do free neutrons. Protium, however, is not stable, as it is energetically favorable for a proton and pion to combine, releasing a gamma ray to produce a neutron. Thus, the largest component of the material universe is not hydrogen gas, but free neutron gas. Stellar fusion is, therefore, much easier (exactly how much easier depends on the stability of dineutronium in this universe, which I am not sure about, but that's not super relevant here), so stars tend to be smaller. Neutrons and protons bound in nuclei are stable against much larger nuclear shell energy differences, since conversion between neutrons and protons needs to release enough energy to produce a massive pion, rather than a comparatively light electron.

As a result, we get pionic "atoms" that are relatively rich in neutrons, with the positive nuclear charge being balanced by a cloud of negative pions. What with being bosons, the pion cloud does not contribute to any particularly interesting chemistry like electron clouds do in our universe. Instead, neutrons take on the "chemical" role vacated by electrons. The neutron content of pionic atoms is not dictated by the nuclear charge like the electron shell structure of a neutral electronic atom is, but a maximum neutron count is set by the point at which filling ever higher neutron shells results in a sufficiently large energy difference between the next proton orbital and the next neutron orbital that adding a neutron will result in pion-decay to produce another proton (thus producing a much closer relationship between "chemical" reactions and nuclear reactions in this universe than exists in our universe). The excess of neutrons, resulting in larger nuclei, and the small size of the massive pion cloud in a pionic atom compared to the electron cloud around atoms in our universe makes it possible for nuclei to approach close enough to each other for neutron-chemistry to occur, forming covalent bonds in which neutrons are shared between multiple proton cores.

As far as I can tell, there is no obvious equivalent of polar or ionic bonding based on neutrons.

It is still conceivable that polyatomic neutron shells may end up forming a "conduction band" in which neutrons can flow freely over long distances... but this obviously doesn't result in the net transport of electric charge.

Now, on to the question: can this universe support electricity as we know it, based either on the flow of negative pions or positive ions? With a bosonic pion cloud, can arrangements of different neutron-bonded protonic nuclei still result in net charge separation, giving rise to polar molecules, ions, and static electric effects? If not, is there some other way to begin inducing charge separation and current flows that can form the basis of electromagnetic technology?

Maybe it's as simple as relying on permanent magnets, which should still exist based on nuclear spin alignment...

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  • $\begingroup$ What is stellar fusion process in your universe is going to be? $\endgroup$ – Alexander Aug 1 at 18:20
  • $\begingroup$ @Alexander Depends on the stability of dineutronium. If that doesn't exist, it'll be a triple-neutron process (analogous to our universe's triple-alpha process) involving neutron decay to produce tritium, followed by rapid single-neutron accretion, and the occasional heavy-nucleus merger. $\endgroup$ – Logan R. Kearsley Aug 1 at 18:34
  • $\begingroup$ It's unclear to me why you consider invidual charge carrier (proton or pion or whatever) maybe not possible to have. Can you explain what is your concern about it a bit better? I mean, you don't have hard-science tag in your question, so it's all just wild speculation (and hard-science would probably be full of QED and kinda beyond 99.9% of the readers here, me included). $\endgroup$ – hyde Aug 1 at 18:34
  • $\begingroup$ @hyde Because bosonic pions do not replicate the shell structure of electrons, so it's not obvious that there would be any energetic advantage to be had in jumping from a (not-higher) orbital in one atom to a (not-lower) orbital in a different atom, which is what induces charge separation in our universe's chemistry. $\endgroup$ – Logan R. Kearsley Aug 1 at 18:42
  • $\begingroup$ @Logan R. Kearsley existence of dineutronium (or trineutronium) is a big if here. $\endgroup$ – Alexander Aug 1 at 18:49
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Short version

Everything should work the same but at much higher energies.

Long version

Given the relatively high mass and close proximity of the pion cloud to the atomic nucleus compared to the electron cloud of a standard atom all chemistry is going to require higher reaction energies. At the same time the atoms have far more embodied energy as the pions in the cloud have far greater kinetic energy and momentum than their electron counterparts, being more massive and in tighter faster orbits around a much heavier nucleus. The creation of ions as well as the formation of covalent bonds is made harder by how tightly atoms will hold their external charge carriers but both forms of chemistry should still be possible.

Given the possibility of ion formation it follows that ionic chemistry will still be practical and will probably be unchanged in effect and proportional energy discharge though the absolute energy states of the reactants will be far higher. It also follows that sufficiently large voltages will act to cause ionic migration and sufficient charge will cause neutralisation so processes like electroplating and electrolysis will still work but only at far higher voltages than we're used to.

Given covalent bonds exist temporary dipole formation is to be expected but I'm not exactly sure how electronegativity and permanent molecular dipoles will be effected by this scenario. I expect a linear increased such that relative electronegativity remains fairly consistent and thus so does molecular polarity and hydrogen bonding behaviour etc...

Static electricity is effectively mechanical, as opposed to chemical, ion creation, it will still be possible but the charge accumulated for a given input of mechanical action will be lower. So it will be much harder to generate large static charge voltages but they should still occur, the voltages necessary to create static discharges like lightening may well be lower due to how tightly nuclei hold their pions.

Current in wires depends on the special "sea of electrons" particular to metallic bonding that allows electrons to move independently through the atomic structure to which they are associated. I can argue both for and against this phenomenon occurring in the scenario presented but on balance it would appear to be unaltered with the proviso that higher voltages will likely be required to give impetus to the much heavier pions involved in the current flowing through the material.

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  • $\begingroup$ This all makes sense in a scenario where electrons are significantly heavier--but does ionic and covalent bonding based on pions still exist given that pions are bosons? $\endgroup$ – Logan R. Kearsley Aug 7 at 15:30
  • $\begingroup$ @LoganR.Kearsley Assuming that their behaviour is substantially similar to that of heavy electrons, which appears to be the case in the scenario presented, then their particle class shouldn't make any difference. $\endgroup$ – Ash Aug 7 at 15:39
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    $\begingroup$ Except, the behavior of bosons is not substantially similar to that of fermions. Bosons won't form multiple distinct energy shells like fermionic electrons do--in the ground state, every pion in a large atom shares the same state, equivalent to an electron s-shell. $\endgroup$ – Logan R. Kearsley Aug 7 at 15:42
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    $\begingroup$ @LoganR.Kearsley Their behaviour is substantially similar in that they each carry a discrete amount of charge and if you remove one you'll get a +1 ionisation state. The fact that all of them have an s-shell energy state makes them harder to move but it doesn't change what happens when they do. $\endgroup$ – Ash Aug 7 at 15:55
  • $\begingroup$ would you mind looking at these related physics stackexchange questions? physics.stackexchange.com/questions/491951/… & physics.stackexchange.com/questions/491980/… $\endgroup$ – Logan R. Kearsley Aug 12 at 19:33

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