Re: [math-fun] Diamonds for dopes
Henry Baker <hbaker1@pipeline.com> wrote:
Re Charles Stross & C12/C13 isotope lattices: Cool! Did Charles Stross have any ideas about how to read out these data?
Here are his original posts on the subject: Newsgroups: rec.arts.sf.composition From: Charlie Stross <charlie@antipope.org> Subject: Re: The Cafe on the Common Date: Thu, 17 Jul 2003 18:04:58 +0100 Stoned koala bears drooled eucalyptus spittle in awe as <oak@uniserve.com> declared:
(So is 'does not involve floppy disks in any way, shape, or form', these days. The good ones are old and the new ones you can get are junk.)
This has *always* been true. That is: there's a double-digit percentage defect rate on floppies. Reliable ones get reused indefinitely, defective ones get binned with a "the new ones you can get are junk" comment attached. Floppy disks -- better than saving to compact cassette, in much the same way that being subjected to the pilliwinks is 'better' than going to the gallows. I can't wait for diamond-phase storage to show up. (A zero is a carbon-12 nucleus, a one is a carbon-13 nucleus, you use nanotech assemblers to read/write the stuff, it stores 10^22 bits per gram with serious redundancy and error correction and has a half life measured in gigayears. Plus, you can wear it as jewellery.) -- Charlie Newsgroups: rec.arts.sf.composition From: Charlie Stross <charlie@antipope.org> Subject: Re: The Cafe on the Common Organization: foobar quux Date: Thu, 17 Jul 2003 19:14:48 +0100 Stoned koala bears drooled eucalyptus spittle in awe as <mra@pobox.com> declared:
It's a stone bitch to read, tho.
I'll grant you it *is* an archival medium, and Drexler-complete to boot (to the extent of requiring a mature nanotechnology base with general purpose assemblers to make it really practical). On the other hand, what's not to like about the idea of being able to carry the sum total of human recorded media (film, audio, video, text ...) around on a ring? Or a couple of thousand serialised and recorded human personalities, if you buy Moravec's estimate of the computational complexity of the human brain. -- Charlie
Re adjacent C12/C13 swapping: in a quantum universe, never say "never". The probability & half-life of such swapping should be calculable.
The atoms are locked into a rigid lattice. It's enormously more likely that the lattice will break down than that atoms will swap in an intact lattice. So it's the half-life of the lattice that you should calculate.
Perhaps it's 1000 years. Perhaps it's 1000 billion years.
I can't immediately find the answer online, but given how chemical and evaporative processes tend to scale with temperature, I'd bet money that it's enormously longer than 1000 billion years at room temperature. I really think cosmic rays will destroy it first. Or neutrinos (!) if you shield it from cosmic rays by placing it in the center of a large non-radioactive asteroid. Except that in the time it would take for neutrinos to do their damage, the background neutrino flux would probably drop off greatly due to the cosmological expansion of the universe. The same effect would also cool it to close to absolute zero. So might diamonds really be forever? No. Quantum tunneling will get them in the end if nothing else does, though that may take closer to a googolplex years than to a mere googol years. It's also possible that C12 and/or C13 is slightly radioactive. I don't think a half-life of much more than 10^24 years would have been noticed yet. Fred Lunnon <fred.lunnon@gmail.com> wrote:
Rather more than "all of Earth's carbon" I imagine, if literally "literally", since the recording process itself would be recorded;
That obviously depends on the resolution of the video recording. A resolution of one millimeter would more than suffice to determine who was doing what to whom. That's about 5.1E+20 pixels for the surface of the Earth. At 100 frames per second that's 1.6E+40 pixels for ten billion years of recording. At 30 bits per pixel, that's 4.8E+41 bits total, about 8E+17 moles of bits. One mole of bits requires 12.5 grams of carbon, so 6.4E+13 kilograms of carbon. Sources quoted in Wikipedia give crustal abundances of carbon from 200 to 1800 ppm. The mass of the Earth's crust is at least 2.4E22 kg, so at least 4.8E+18 kilograms of carbon. Only about one percent of that is C13, so 4.8E+16 of C13 plus another 4.8E+16 of C12 for your library. So the amount of carbon suffices to store 1500 copies of our video. And that's not taking into account video compression or the fact that the amount of C12 and C13 doesn't have to be equal. (What the most efficient data encoding is when one-bits cost more than zero-bits is a topic for another day.) Someone please double-check my arithmetic. Thanks. On the other hand, the Stross novella in question, "Palimpsest," depicts an Earth and sun which are re-engineered to last trillions of years, not mere billions. And not just one timeline is recorded, but a vast number of different ones, caused by time-traveler intervention. So maybe they'd better use compression and other techniques to maximize the use of that carbon. Of course they can also bring in more carbon from extraterrestrial sources, or from below Earth's crust.
Hi Keith: Although I don't know enough quantum physics to do the calculation, my gut still tells me that the probability of exchanging an adjacent C12 for a C13 nucleus is non-negligible. The problem with crystal lattices is that the constituent atoms are usually the same -- e.g., diamond or graphene. In the case of crystals like NaCl, there is no hope of exchanging a Na for a Cl, and if a Na<->Na happened or a Cl<->Cl happened, we'd not notice. In the case of a metal, the electrons become smeared all over the crystal, so they can't be localized at all. In the case of a diamond crystal, the location of "a" carbon nucleus is localized to within the distance to its 4 (?) nearest neighbors. https://en.wikipedia.org/wiki/Diamond_cubic Delocalizing a carbon nucleus would make diamond look a lot more like a liquid. Diamond won't melt at 1 atmosphere, as 1 atmosphere is below its triple point at 107 atmospheres and 7820 degrees F. So I would guess that the probability of an adjacent C12<->C13 exchange would be dramatically improved with 100x atmospheric pressure, and floating on top of molten tungsten at 6500-7000 degrees F. However, if we heat up diamond at 1 atmosphere, past about 3500F it first turns to graphite before finally melting at ~7600F. So somewhere in the vicinity of 3500F and 1 atmosphere, the localization of the carbon atoms must start to break down, and the exchange probability must rise sharply. So the real question is: what is the falloff in exchange probability as we lower the temperature from 3500F to 100F ? https://en.wikipedia.org/wiki/File:Carbon_basic_phase_diagram.png At 02:13 PM 6/5/2016, Keith F. Lynch wrote:
Henry Baker <hbaker1@pipeline.com> wrote:
Re adjacent C12/C13 swapping: in a quantum universe, never say "never". The probability & half-life of such swapping should be calculable.
The atoms are locked into a rigid lattice. It's enormously more likely that the lattice will break down than that atoms will swap in an intact lattice. So it's the half-life of the lattice that you should calculate.
An old geometry problem discussed here asks for a finite set of points in the plane such that the perpendicular bisector of any 2 of the points contains exactly 2 of the points. At last check there was just one known solution (up to rotation and scaling). In case anyone feels like looking for a solution, I won't mention the answer unless asked to. * * * OK, what about a solution in R^3 or R^n ? We would of course want it to be more than just the planar solution, so we'll require that "an n-dimensional solution" to this question must not lie in any affine Euclidean space R^k in R^n. I know a little, but not much, about the answers. —Dan
On Jun 6, 2016, at 7:38 AM, Dan Asimov <asimov@msri.org> wrote:
An old geometry problem discussed here asks for a finite set of points in the plane such that the perpendicular bisector of any 2 of the points contains exactly 2 of the points.
At last check there was just one known solution (up to rotation and scaling). In case anyone feels like looking for a solution, I won't mention the answer unless asked to.
Are there entries in OEIS that have just one integer, as in “number of points in known solutions to Dan’s problem”? You would have to add terms such as “bisector” to search such items. -Veit
There are a few A058445 2236081408416666 A058446 5000060065066660656065066555556 A072288 316912650057057350374175801344000001 A076337 509203 A115453 1414213562373095048801688724209698078569671875376948073 A118329 9159655941772190150546035149323841107741493742816721 A122036 351351 A144134 62527434837271029229 A230528 375494703 A245206 1019 but generally we require 3+ terms. Charles Greathouse Case Western Reserve University On Mon, Jun 6, 2016 at 9:20 AM, Veit Elser <ve10@cornell.edu> wrote:
On Jun 6, 2016, at 7:38 AM, Dan Asimov <asimov@msri.org> wrote:
An old geometry problem discussed here asks for a finite set of points in the plane such that the perpendicular bisector of any 2 of the points contains exactly 2 of the points.
At last check there was just one known solution (up to rotation and scaling). In case anyone feels like looking for a solution, I won't mention the answer unless asked to.
Are there entries in OEIS that have just one integer, as in “number of points in known solutions to Dan’s problem”? You would have to add terms such as “bisector” to search such items.
-Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
But something like A115453 = "Primes formed from the initial digits of sqrt(2)" is only sort of a single term: We know what the next several are, they are just too large to fit on one line. So any sufficiently-quickly-growing sequence will only have one term in OEIS by that definition. --Michael On Tue, Jun 7, 2016 at 12:29 PM, Charles Greathouse < charles.greathouse@case.edu> wrote:
There are a few
A058445 2236081408416666 A058446 5000060065066660656065066555556 A072288 316912650057057350374175801344000001 A076337 509203 A115453 1414213562373095048801688724209698078569671875376948073 A118329 9159655941772190150546035149323841107741493742816721 A122036 351351 A144134 62527434837271029229 A230528 375494703 A245206 1019
but generally we require 3+ terms.
Charles Greathouse Case Western Reserve University
On Mon, Jun 6, 2016 at 9:20 AM, Veit Elser <ve10@cornell.edu> wrote:
On Jun 6, 2016, at 7:38 AM, Dan Asimov <asimov@msri.org> wrote:
An old geometry problem discussed here asks for a finite set of points in the plane such that the perpendicular bisector of any 2 of the points contains exactly 2 of the points.
At last check there was just one known solution (up to rotation and scaling). In case anyone feels like looking for a solution, I won't mention the answer unless asked to.
Are there entries in OEIS that have just one integer, as in “number of points in known solutions to Dan’s problem”? You would have to add terms such as “bisector” to search such items.
-Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Forewarned is worth an octopus in the bush.
Not sure what’s driving this discussion, or the fascination with carbon isotopes, but here are some things to consider: 1) Beta-brass is a 50-50 mixture of Cu and Zn (atomic numbers 29 & 30) on a bcc lattice. Above about 1200K the two types of atoms are equally likely to be on either of the two cubic sub-lattices (this doesn’t preclude some short-range order). Because the phase transition to the ordered state, of Cu and Zn perfectly segregated on the two sub-lattices, is second order the segregation is small just below 1200K. By quenching to low temperature, the mostly random occupation of sites by Cu and Zn is preserved — in other words, the phase transition has negligible effect on the information content. The rate of a Cu-Zn pair swapping positions is given by the product of an atomic oscillation frequency, several terahertz for brass, and an exponential factor exp(-c T_D / T), where T = temperature, T_D = 340K is the Debye temperature, and c is a numerical constant of order unity. Most atom pairs, in a chunk of beta-brass in equilibrium with the balmy 3K microwave background of the cosmos, will not have changed places over the lifetime of the universe. 2) A different, temperature-independent, exponential suppression factor takes over at low temperatures. The exponent scales as the square root of the product of the atomic mass and the potential energy barrier. The system most studied in this regime is He, which forms a crystal under pressure. Atom-swapping in both He3-He4 mixture crystals and pure He3 can be studied because He3 has a nuclear spin-1/2 that serves as a label. The dominant quantum tunneling mechanisms in these crystals are small permutation cycles (I recall 3- and 4-cycles occur at higher rates than pair-exchange). These quantum effects actually order the crystal, now at temperatures below mK. Missing atoms in the crystal (“vacancies”) greatly enhance the rate of position scrambling, much like the Sam Loyd 15-puzzle. 3) We asked Feynman in Physics-X what he thought about the net effect of human civilization on the entropy balance of the universe. He told us he had compared illuminated manuscript production (middle ages) with a bottle of gas and that it wasn’t even close. -Veit
On Jun 6, 2016, at 1:21 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Hi Keith:
Although I don't know enough quantum physics to do the calculation, my gut still tells me that the probability of exchanging an adjacent C12 for a C13 nucleus is non-negligible.
The problem with crystal lattices is that the constituent atoms are usually the same -- e.g., diamond or graphene. In the case of crystals like NaCl, there is no hope of exchanging a Na for a Cl, and if a Na<->Na happened or a Cl<->Cl happened, we'd not notice.
In the case of a metal, the electrons become smeared all over the crystal, so they can't be localized at all.
In the case of a diamond crystal, the location of "a" carbon nucleus is localized to within the distance to its 4 (?) nearest neighbors.
https://en.wikipedia.org/wiki/Diamond_cubic
Delocalizing a carbon nucleus would make diamond look a lot more like a liquid.
Diamond won't melt at 1 atmosphere, as 1 atmosphere is below its triple point at 107 atmospheres and 7820 degrees F. So I would guess that the probability of an adjacent C12<->C13 exchange would be dramatically improved with 100x atmospheric pressure, and floating on top of molten tungsten at 6500-7000 degrees F.
However, if we heat up diamond at 1 atmosphere, past about 3500F it first turns to graphite before finally melting at ~7600F.
So somewhere in the vicinity of 3500F and 1 atmosphere, the localization of the carbon atoms must start to break down, and the exchange probability must rise sharply.
So the real question is: what is the falloff in exchange probability as we lower the temperature from 3500F to 100F ?
https://en.wikipedia.org/wiki/File:Carbon_basic_phase_diagram.png
At 02:13 PM 6/5/2016, Keith F. Lynch wrote:
Henry Baker <hbaker1@pipeline.com> wrote:
Re adjacent C12/C13 swapping: in a quantum universe, never say "never". The probability & half-life of such swapping should be calculable.
The atoms are locked into a rigid lattice. It's enormously more likely that the lattice will break down than that atoms will swap in an intact lattice. So it's the half-life of the lattice that you should calculate.
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participants (6)
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Charles Greathouse -
Dan Asimov -
Henry Baker -
Keith F. Lynch -
Michael Kleber -
Veit Elser