On Tue, Aug 18, 2015 at 09:27:11PM -0400, Jonathan Clark wrote:
> Or the nucleus of an atom. (I am not disagreeing with you here.) As atoms get heavier,
> the number of neutrons in a stable atomic nucleus seems to go up at a ratio of about 1.5
> neutrons per proton. The reasons for this seems to be built into the way that the universe
> is constructed (so it might as well be magic).

That part is pretty simple really: protons are charged, neutrons
aren't.  The strong force applies equally to neutrons and protons, but
ordinary electromagnetic repulsion means that for larger nuclei (that
is greater charge), fewer protons than neutrons is more stable.

> Here's one for any real fundamental particle physicists out there:
> if you had some neutronium which was gravity-stabilized (Pauli
> exclusion principle and all), and you removed the gravity field,
> which fundamental force would apply first, and what would the
> results be?

Neutronium is only stable under *extreme* pressure, so the first thing
that would happen is that its release would drive extremely rapid
expansion.  It would probably be similar in energy per nucleon as a
fusion reaction, though more of it would be in kinetic energy.

So that's not really any of the fundamental interaction forces, just
Fermi energy.  The strong force is too short ranged to have any
appreciable effects (except in some aspects of the potential function
that yields the Fermi energy), and weak interactions leading to decay
are very much slower.

The results would be an explosion of free neutrons at a small but
significant fraction of the speed of light, with energy probably on
the order of a kiloton of TNT per few grams of neutronium.  Decays of
free neutrons would be negligible; nearly all of them would be
captured by some nucleus after scattering from a whole lot of others
along the way.

High-energy neutrons are highly penetrating, so most of those
travelling upward would make it through kilometres of an Earth-like
atmosphere and many would reach space.  Those travelling more
horizontally would pass through kilometres of air and metres of most
solid materials.  Neutron scattering from ambient nuclei would produce
gamma rays within the area.

The net effect would be pretty similar to a nuclear explosion, with
the energy depending upon how much neutronium was present.  If there
was only a fraction of a milligram, then it might "only" irradiate the
hell out of everything in line of sight out to a few kilometres (and
some stuff behind relatively light shielding such as a few concrete
walls).  In any event, there are numerous materials that would become
themselves radioactive as a result of absorbing free neutrons.

> This is one reason I specified the neutronium blade as being
> sub-atomic radius. At least in my version of reality, these will end
> up pushing atoms aside. Yes, there is a physical force required to
> do so, but it will, I think, be less than that required to scrape
> off a layer or three of atoms, force them around the blade into the
> slot that you are cutting, and eject them (hope this is clear, I'm
> trying to describe in real-world terms what happens when you use a
> current-day saw blade on some material).

Both would require a great deal of energy per unit volume of blade
(hence force).  The difference is that in one case it's supplied
externally by the wielder, and in the other it's supplied by the
inbuilt power source.

Naturally the thinner the blade, the less energy required.  However
there are limits to how thin it can be due to buckling even with
various forms of unobtainium.  So at some scale or other, you have a
wedge pushing apart the material being cut to allow the blade to pass
through.  Vibration might reduce net friction across the wedge faces,
so it could still cut through a lot of softer materials almost
effortlessly.  Rocks or metals though would still pose problems.

- Tim