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The densest elements on Earth are metals. Osmium is the most dense. The densest matter in the Universe is Neutronium which is not an element but degenerate matter. If we put Osmium and Neutronium on the same scale what happens in between? I would like to know how we get from Osmium to the next densest material and all the way to Neutronium. I realize that it may be more complicated than a simple range because elements might behave differently under different conditions but could anyone give this a bit of a go or give me a few pointers?

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    $\begingroup$ What does "If we put Osmium and Neutronium on the same scale" mean? FWIW, the transuranic elements meitnerium & hassium are predicted to be much denser than osmium, with densities of 37.4 & 41 g/cm³, respectively. $\endgroup$ – PM 2Ring Jan 23 at 21:00
  • $\begingroup$ @PM2Ring - I meant a scale as in an ordered sequence. $\endgroup$ – Clock Jan 24 at 21:05
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"Neutronium" is a science fiction term. Pure neutron matter does not exist, since neutrons decay. But something like it exists in neutron star interiors in the form of a n, p, e fluid (mainly n, but a small fraction of p, e must be present - see https://physics.stackexchange.com/a/149656/43351) . This fluid is stable at densities from about $3\times 10^{16}$ kg/m$^3$ to around $10^{18}$ kg/m$^3$.

At lower densities, the next phase is "nuclear pasta" - very neutron-rich nuclear matter bound into planar or spaghetti-like forms surrounded by a (degenerate) free neutron plus electron fluid.

Below a few $10^{15}$ kg/m$^3$, more "normal" nuclei form with pseudo-spherical shapes. These nuclei are still very neutron-rich (n/p ratios of 3 or more and masses of several hundred amu) locked into some sort of solid lattice by Coulomb forces and bathed in a fluid of neutrons and relativistic degenerate electrons.

At densities below $4\times 10^{14}$ kg/m$^3$ the lowest energy configuration sees the free neutrons absorbed into the neutron-rich nuclei. The equilibrium nuclei are still very heavy, but not so heavy as at higher densities.

At lower densities, we gradually get back to material which has the n/p ratios of the stable elements we are familiar with.

Thus there aren't really too many abrupt phase changes. The surfaces of neutron stars are made of familiar elements, probably even a bit of osmium, and are already at densities of $\sim 10^9$ kg/m$^3$, compressed by the enormous gravity.

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    $\begingroup$ Might want to mention the matter in a white dwarf star as intermediate to what you describe at 1e14 and lower $\endgroup$ – dllahr Jan 24 at 3:07
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    $\begingroup$ @dllahr The electron degenerate matter at white dwarf densities is similar to the outer crust of a neutron star in terms of a lattice of nuclei with degenerate electrons. But the white dwarf matter is made of lighter nuclei that have not managed to reach the "ground state" of the iron-peak of the binding energy curve at "low" densities, because of Coulomb repulsion. TBH this was just a stab at answering a question that I don't really understand. $\endgroup$ – Rob Jeffries Jan 24 at 7:09
  • $\begingroup$ @RobJeffries - May I ask - I thought the "nuclear pasta" would be the densest form as the p,e, would be released as neutrinos. How come they they are part of the fluid at the greatest density as you mention? $\endgroup$ – Clock Jan 24 at 21:25
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    $\begingroup$ @OldHagYouHaveKilledMe "p,e released as neutrinos"? How does that conserve baryon number? If you are asking why protons and electrons MUST be present in a neutron fluid: physics.stackexchange.com/a/149656/43351 $\endgroup$ – Rob Jeffries Jan 24 at 22:36
  • $\begingroup$ @RobJeffries - I was under the impression that the neutrinos basically caused a supernova and would mostly be expelled leaving only neutrons. Thank you for clearing that up. I read the link you added and your answer there makes me think you might be the man to answer this question: physics.stackexchange.com/q/525607/251767 - it is very intriguing and I can't answer it - good luck $\endgroup$ – Clock Jan 25 at 20:31

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