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We can create plasmas quite easily, indeed you can buy a plasma cutter and generate it all day long for less than $500. Would it be possible to trap a plasma, say magnetically, and cool it so much that it became a solid/liquid/gas without allowing it to receive electrons? Or would the energy requirement exceed what is possible?

I'm mostly curious about how the nuclei would interact with each other in the absence of electrons as a solid or liquid. Is there some principle that makes such an endeavor impossible?

Ehryk
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    Soft-comment; would a neutron star fit the bill? – Argus Dec 18 '12 at 20:14
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    Meta-comment: What is a soft-comment? – gerrit Dec 18 '12 at 20:25
  • Hmm, a neutron star would be similar, but I was wondering if this was attainable on our planet or in our laboratories, which can't make quantum-degenerate matter as of yet (that I'm aware of). It's a good start though, thanks! – Ehryk Dec 18 '12 at 21:21
  • It's worth noting that very very many labs produce quantum-degenerate matter -- but with neutral atoms. – wsc Dec 18 '12 at 21:35
  • Is Metallic Hydrogen the only degenerate matter we've created, or is there anything else? http://en.wikipedia.org/wiki/Metallic_hydrogen – Ehryk Dec 19 '12 at 03:37
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    Bose-Einstein condensates and Fermi gases of neutral particles (atoms) are regularly created in labs, including mine. There are may labs that trap ions (chargred plasmas) in electrical (Paul) or magnetic (Penning) traps and cool them to incredibly low temperatures, but even at T=0 they are not degenerate because of the enormous Coulomb repulsion. Instead, they form ion crystals. I agree with Argus, neutron stars should fit the bill, but (to my knowledge) we are still far from forming a trapped degenerate neutron sample in a laboratory. – emarti Dec 19 '12 at 04:54
  • Even a neutron star is not electron-less. There is still a degenerate gas of electrons (and a corresponding gas of protons) in the body of the star, they are just numerically few compared to the neutrons. This is set by the energetics of the problem. – dmckee --- ex-moderator kitten Aug 09 '14 at 16:48

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Well you could replace the electrons with something else, such as muons. That is essentially what is proposed for this, although in extremely small quantities. Muons in this case replace some electrons, but very very few compared to the total:

http://en.wikipedia.org/wiki/Muon-catalyzed_fusion

Obviously, this is fraught with difficulties because there is a good reason we live in the world defined by electrons. You would have to scale up this idea beyond the fusion proposal dramatically, and even if you did, the fact that it can induce fusion is a problem... well, a big problem. Maybe you could replace the electrons with something other than muons, that isn't as heavy. But probably not.

If you're interested in how nuclei would interact with each other, you could consider a neutron star. That is rather like a big nucleus, but it's held together by gravity. It still doesn't have the electrons stripped out. As I understand, the electrons are still mixed in the mass.

Let's say you just took ordinary matter and removed the electrons. Let's look at a tennis ball sized thing, with a diameter of $6.7 cm$. The nuclei are now repelling each other. How much? Let's calculate the energy, with the electric analog to the gravitational binding energy.

$$ U = \frac{3GM^2}{5r} $$

here:

$$ U = \frac{ 3 k Q^2 }{5 r} $$

We can very simply find Q from the nuclei alone. If we assume the ball is the density of water and solid (unlike the tennis ball) then we find it weighs about $1.25 kg$. Let's assume it's Carbon-ish in terms of the formula mass. Divide by the formula mass, multiply by the electron charge (which is the same as the proton charge), and I find about 10.1 million Coulombs. Here is the total calculation string in Google, but i won't take the time to prettify it right now.

(8.987551e9 N*m^2/C^2)3((1.0 g/cm^3)*4/3*Pi*(6.7 cm)^3 *(electron charge) / (12 amu))^2/(5*6.7 cm)

This comes out to $8.258 \times 10^{24} J$. This is about 2.0 petatons of TNT equivalent. The Tsar Bomba was 100 megatons TNT equivalent. The KT extinction event was 100 million megatons of TNT, or 100 teratons of TNT.

A ball stripped of electrons would have so much energy it would possibly destroy life on Earth as we know it. That is how the nuclei would interact with each other.

mpv
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Alan Rominger
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  • Wow, point taken on the tennis ball sized matter calculation. Do we not have electric/magnetic fields that could hold together, say, a few hydrogen nuclei and repel any electrons from joining with them as they cooled? Or would any magnetic field be able to be that powerful / work in that manner? – Ehryk Dec 18 '12 at 21:24
  • @Ehryk We could hold together a few atoms, sure. In fact, I would say something very similar to this is done at the LHC. The task gets easier with scale according to about $r^5$. You were correct in thinking that such a configuration would be plasma-like. There is effectively no attractive force between ions, so they are only held together with external fields. Even an ordinary CRT has a region where it's mostly only electrons. But that's low density and a small total number, so the self-repulsion is very little. – Alan Rominger Dec 18 '12 at 21:29
  • So then, to finish the question off - would there ever be a temperature at which they would turn into a liquid or solid? Or conversely could they get all the way down to near absolute zero, and never be tempted (or be possible) to change their phase? – Ehryk Dec 19 '12 at 03:27
  • @Ehryk I'm not sure what substance you're talking about. If you're thinking about ions or electrons in particle accelerators, they're generally very hot, and have to be in motion to keep from touching the walls, so you couldn't freeze this. – Alan Rominger Dec 19 '12 at 13:08