It has been widely said that if $\alpha$ were to increase by only a few percent, life would be impossible because of the lack of carbon production in stars--and if it is increased too much, electrostatic repulsion would make every nucleus larger than hydrogen unstable.

But, in the vein of the Weakless Universe model, could those problems be solved by tweaking other parameters to compensate? E.g., could the strength of the strong nuclear force be increased, or the masses of quarks (and thus pions) decreased to increase the residual force range, so as to maintain the stability of heavier isotopes while increasing $\alpha$?


While a weakless universe prevents beta decays in a nucleus, it may still be unstable due to the possibility of emitting protons or neutrons. Because the strong force exhibits a faster than inverse-square decay, it cannot compensate for electromagnetic repulsion between protons unless (i) its $\alpha$ is greater than that of electromagnetism, or (ii) there are plenty of neutrons to serve as nuclear glue. Since nucleons can't change species in a weakless universe - in particular, the neutron would be stable, rather than having a half-life of $15$ minutes - primordial neutrons would persist.

In our universe, helium formation occurs in the stages $2\,{}^1\text{H}^+\to{}^2\text{He}^{2+}\to{}^2\text{H}^++\text{e}^++\nu_\text{e}$ (which would be problematic with $\alpha_\text{EM}\ge\alpha_\text{S}$ because (i) helium-2 shouldn't have enough binding energy to form even as a transient species and (ii) we've assumed beta-decays are off-limits so we can't produce deuterium this way), $\text{e}^++\text{e}^-+2\gamma$ (no problem there), ${}^1\text{H}^++{}^2\text{H}^+\to{}^3\text{He}^{2+}+\gamma$ (but I doubt in this scenario a single neutron is enough to stabilize helium) and $2\,{}^3\text{He}^{2+}\to{}^4\text{He}^{2+}+2\,{}^1\text{H}^+$ (but the briefer helium-3 exists, the harder it is for two nuclei of it to collide). Higher elements form by joining two alpha particles to make beryllium-8, and adding successive alphas, all the way up to nickel-$56$. Further decay processes can make other species before nickel.

Most of the above objections don't care whether the weak interaction exists. But let's consider what would happen in a weakless $\alpha_\text{EM}\ge\alpha_\text{S}$ universe. We would probably see multiple neutrons fusing together into neutronium isotopes. It would take a detailed analysis of this hypothetical nuclear physics to see which numbers of neutrons would be magic, e.g. because of how they form orbitals, but let's make the simplifying assumption that quite large nuclei are achievable, often exothermically. (That nuclei of multiple neutrons could fuse makes it easier to achieve many different values of $N$.) As these nuclei grow, they gradually increase the number of protons they can stably maintain. Reactions that added protons to neutronium to produce hydrogen onwards would be endothermic, but not impossible. Stars in our universe also contain endothermic fusions, buoyed by some exothermic processes listed above.

So if helium etc. as far as you like are achievable in this scenario, they'll have far more neutrons than in our universe. Famously $\frac{N}{A-N}\approx 1+kA^{2/3}$ determines the optimal $N$ for a given $A$, with $k\propto\tfrac{\alpha_\text{EM}}{\alpha_\text{S}}$. Your plan is to multiply $k$ by as much as $137$ or so, or maybe only $55$ or so if we achieve $\alpha_\text{S}\approx1$ instead of $\alpha_\text{EM}\approx0.4$. Clearly, this makes a big difference to neutron counts. Unfortunately, there would only be so many primordial neutrons per primordial proton, so the quantities of post-hydrogen elements would be very low. Most protons would persist as hydrogen, as in our universe (albeit with probably more deuterium since hydrogen won't care too much how many neutrons it has), while the second most abundant element would probably be neutronium. Chemistry, of course, would be left to elements other than neutronium (unless you argue electride ions (electrons) are of neutronium-$0$).


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