time-of-flight

I don't think it's egregious at all. We use terms like "gnocchi phase", "lasagna phase", and neutron-star mountains regularly in the nuclear-matter literature, so these aren't indicative of any pop-sci "massaging" by the author.

Just a clarification: A here doesn't mean Angstroms. It means "number of nucleons", as in carbon-12 is an A=12 system.

So they're saying A>300 means "a quark system with >900 quarks"; normal protons and neutrons have three quarks a piece.

If I misinterpreted your comment and the above was already clear to you, please forgive my mistake (I didn't want anyone to confuse A for angstroms here).

The udQM they're talking about is in a very different regime than the quark matter produced at RHIC (Relativistic Heavy Ion Collider) and other colliders. In those, heavy ions are collided ultrarelativistically (speed just under C) and for a very brief instant, quarks and gluons are liberated from their confinement inside proton and neutrons to become quark-gluon plasma, or QGC. There are up and down quarks just like in normal protons/neutrons, but also other quarks can be created (strange being the most common). At high enough energies and with particular kinematics, you can make all six quark types: see BaBar experiment, LHC, Stanford Linear Accelerator (SLAC) experiments.

In the paper, they're talking about udQM being more stable at zero temperature and pressure, basically "competing" with normal elements. The wiki diagram is useful for reference: https://en.wikipedia.org/wiki/Quark%E2%80%93gluon_plasma#/me...

Yes, one could phrase it that way... but keep in mind that all these theoretical models suggesting an "end to the table" make huge assumptions that are not well-grounded in experimental results. We don't even know where the neutron dripline is for light systems (Z=12), much less the high end of the table (Z=100+).

The periodic table is only a useful model (and it is very useful) for nuclear systems acting as a zero-temperature Fermi gas (e.g., nuclei here on Earth). At extremes of temperature and pressure (neutron star merger, say), the idea of a discrete periodic table loses meaning.

No, they mean that cosmic rays would be udQM (presumably produced somewhere in the universe and accelerated to us somehow). The premise of the paper is that you need a large number of quarks (for 300+ nucleon system, that's 900+ quarks) to start to enter this udQM regime of stability, and as you say cosmic rays typically have A<40, which is only 120 quarks...

Neutrons are a pain to control as a beam because they have no charge, so methods employing electric or magnetic optics are not applicable. Typical pocket sources use the alpha(9Be,12C)n reaction to produce neutrons; if you have a reactor handy, you can moderate (slow) the neutrons produced and use them... this would be by far the most economical option.

As a nuclear chemistry PhD, I am highly skeptical of the practicality of this as an energy source.

Check the Nature paper (and I think this is one of the biggest take-aways from AlphaGo Zero):

"Finally, it uses a simpler tree search that relies upon this single neural network to evaluate positions and sample moves, without performing any Monte Carlo rollouts."

In this new version, MCTS is not even used to evaluate a position! Speaking as a Go player, the ability for the neural network to accurately evaluate a position without "reading" ahead is phenomenal (again, read the Nature paper last page for details).

I chose a coincidentally easy problem (whoops).

Here's another one, less accessible to attack: Find e^7. Or 25000^(1/8).

The trick to using the musical scale is that it's logarithmic in frequency but linear in key (ABCDEFG). The frequency step between adjacent notes is 2^(1/12); the frequency jump over an octave is 2; the frequency jump over 40 notes is 10. There are other handy ratios to remember, like that going up a whole step is a frequency increase of 9/8, and the more you know the more powerful this method is.

To solve 25000^(1/8) = x, you're looking for an interval x on the keyboard such that if you were to go up 8 of those intervals, the frequency would jump by 25000. Well, 25000 is just 10^5 / 2^2, or in terms of keyboard intervals, "Go up 40 notes (repeat four more times), then down two octaves". This means to represent a frequency increase by a factor of 25000, I need to go up 105-212 = 176 half steps on a keyboard.

Divide by 8 to find x = 22 steps, which is two octaves minus a whole step on the keyboard. The frequency change over this 22-step interval is thus (2*2) [up two octaves] / (9/8) [down a whole step] ~ 3.5. So that's the final answer: 3.5

The actual value of 25000^(1/8) is 3.546, so this approximation is good to within a couple of percent.

Back in 2010 I ran across Sanjoy's method for using the musical scale to perform more complicated arithmetic (mentally calculate pi^6 +/- 5% within 30 seconds).

Can't wait to read the entire work.

I'm using CAEN (Italian company) digitizers in nuclear research, and I've expended at least 6 months sussing out firmware problems for our simple experiments.

[x++] italians