What motivates your first factor? 0.782343 MeV is the free neutron beta decay; where in the solar system are the free neutrons minutes after they are magically teleported to terrestrial ground zero as a something like a (degenerate, possibly ultra-relativistic) Fermi gas?

I think most attempts to arrive at an answer will end up somewhere between half and virtually all of them being "not very close" (~ light-minutes) away, and that's assuming one corrects for the differences in escape velocities. (The equatorial escape velocity of a spinning neutron star is in tenths of the speed of light, thus the sobriquet "relativistic star"). Without this correction, it is likely the bulk of the expanding drop of Fermi gas just exits the atmosphere in milliseconds (timed by terrestrial stopwatches), with time dilation extending the mean lifetime of the free neutrons in the drop comparably to the extended lifetime of atmospheric muons from cosmic rays. The bulk of the beta decays happen at a distance from terrestrial ground zero best measured in astronomical units.

If we play Star Trek transporter games such that the neutrons arrive at ground zero at rest in local East-North-Up coordinates, you'd want to know the internal kinetic energy (KE) density of the (pure-)neutron star, which will be in the range of 20-40 for x in 10^{x} J m^-3. The 10^25ish or even 10^30ish joules of KE will be released from our several cm^3 spoonful practically all at once and practically omnidirectionally from ground zero (so again, most free neutron decays happen at ~ AU distances from ground zero because they'll zip right through the atmosphere). The expansion of the suddenly unpressurized gas of neutrons will make a mess, particularly the fraction that slams into and through the ground. Part of the mess is neutron scattering physics, and I have no expertise there, but I would guess there wouldn't be any free neutrons near ground zero (and probably not within the solid Earth) in ~minutes.

Additionally, one might compare the R-process <https://en.wikipedia.org/wiki/R-process> for kilonovas in which a binary neutron star collision ejects high-neutron-density matter which decompresses pretty spectacularly, forming lots of heavy elements.

To summarize, I think the free neutron decay timescale (mean lifetime ~ 15 minutes, multiply by ln 2 if you prefer half-life) is simply too long after the neutron star material is teleported to Earth: any free neutrons that haven't been absorbed into heavy nuclei likely will be millions of kilometres away from ground zero when they decay.

> I think most attempts to arrive at an answer will end up somewhere between half and virtually all of them being "not very close" (~ light-minutes) away

Mean free path of free neutrons moving past normal matter is only in the order of centimetres, exactly how many centimetres depends on the neutron energy and the specific nuclei it's interacting with, but still order of centimetres.

Given the relative masses, I can assume the air above will be exploded out of the way; but the half going down will have all of the earth as a moderator… and also serve as a neutron-absorbing backstop that will probably increase the actual yield.

I'm also ignoring any binding energy between the neutrons. I'm basically treating them as disconnected from the first moment, which may be a terrible idea, but AFAIK nobody actually knows how long a macroscopic combination of this scale would remain stable for.

I don't know enough about neutron physics to comment usefully on your mean free path logic, but I do know that solar eruptive activity can launch relativistic neutrons at Earth which can be detected even at sea level using scintillators, and that mountaintop detection has been around since the early 1980s. Shibata 1994, Propagation of Solar Neutrons <https://sci-hub.se/https://doi.org/10.1029/93JA03175>, §4.2.1 (Fig 3) higher energy neutrons get further into the atmosphere, so I don't think the atmosphere is much of a barrier for the comparable (MeV-GeV) teleported neutron-star neutrons.

We seem to agree that free neutrons don't stay free neutrons when they slam into the solid earth.

I too wanted to think about neutrons as a non-self-interacting gas, but that just doesn't work: Meyer 1994, https://ned.ipac.caltech.edu/level5/Sept01/Meyer/Meyer3.html (Paragraph beginning with, "Only the strong gravity of the neutron star keeps such matter from exploding apart." Cold in this context is partly explained in the preceding paragraph; in inner regions the matter is a degenerate gas meaning the particle kinetic energy becomes dependent on the density or equivalently pressure becomes independent of temperature; even at enormous pressures, degenerate gases don't hold much thermal energy -- that was practically all radiated away when the NS was young. Our teleporting (of inner region matter) therefore engages a very low-entropy r-process.

Outer regions are just too complicated and varied for a HN comment. The crust is thin -- a few to a few hundred metres or so compared to an NS radius of ~ 10km. It's also much less dense, so is a small fraction of the NS mass, and thus maybe not a target for our teleportation. Here's a 180-page open access review: https://link.springer.com/article/10.12942/lrr-2008-10 Pesky electrons and protons complicating things.

Thanks for all three links; it's getting late here, so this is only going to touch on the first part of your message.

> Shibata 1994, Propagation of Solar Neutrons <https://sci-hub.se/https://doi.org/10.1029/93JA03175>

If I'm reading that figure right, at sea level the attenuation is at least a factor of 2000 for all energies they're graphing. That sounds about right to me.

I realise now that I may have been unclear in intent previously: if you look at figure 2, and then consider a typical solid or liquid's cross sectional mass density, hopefully that explains why I was speaking of neutron mean free path of centimetres — 100g/cm^2 is 1m of water.

However this is just the initial condition, and I don't think this scenario is one where the atmospheric density can be accurately approximated as constant over time.

I wouldn't sweat it, and I don't know enough about the nuclear physics to keep up (and we haven't even been talking about the neutrino energy in beta- decays, the gamma spectrum, or what becomes of the electrons; resonances go way over my head). This isn't really a gravitational problem (but...footnote [1]), so I'm not so useful here.

So, more for the original questioner than for us:

What's inside a neutron star stays inside a neutron star. Unless of course the NS is destroyed via e.g. collision, tidal disruption, or infall pushing it over a mass limit like Tolman-Oppenheimer-Volkoff. Sci-Fi teleporters don't exist, and there's no basis to think they ever will.

The closest neutron stars are between hundreds and a thousand light-years away and IIRC all the close ones are isolated (in the sense of no stellar multiplicity; they have no binary partner(s)).

Consequently what ben_w and I have been yakking about is inaccessible to experiment (we can't generate the relevant pressures, and artificial neutron sources are not very bright yet (pardon the BrightnESS pun, <https://europeanspallationsource.se/about>)).

It's not accessible to astronomical observation either. The closest physical phenomenon I can think of is an NS mass ejection (for which there is an ample and active academic literature), and that's far from a close match. At least in some parts of the spectrum we can see a large NS mass ejection -- large meaning somewhere around 10% of the mass of the sun -- but there's practically no hope to see just a spoonful, and not hurled into a close-by planet's atmosphere or even that of a noncompact companion star.

So the answer to the question ultimately is -- if we imagined the magical arrival of a small ball of NS matter on Earth at rest on the Earth's surface -- "complex nuclear physics" is in the details of the practically-instantaneous kaboom, and a lot of that complexity is because the Earth is not the practical vacuum around a neutron-star/neutron-star collision that ejects a lot more than a spoonful of material.

- --

footnote [1]: I mean, one can think of it in terms of Raychaudhuri's equation (and that's where I started, in fact): the initial radial divergence of the acceleration vector from the sudden release of pressure dominates, causing the bits to tend to fly away beyond the hope of recollapse. But the solid earth (and as the thread involved, considerations of nuclear interactions even in the atmosphere) generates enormous shear via contact forces, so some of the energy-density of the NS matter will stick around, and in due course what wasn't ejected "to infinity" settles back to a basically round Earth (hydrostatic equilibirium returns). From this perspective comparing the NS matter with an asteroid impact makes sense to me, but probably undersells the nuclear fallout.