johncarlosbaez

johncarlosbaez, 7 days ago

@11011110 - I like these curiosities, but termites make that one a lot less charming.

johncarlosbaez, 7 days ago

@codrusofathens - A painter told us one of those now-closed windows was used to carry stuff into the building while it wa being built - the one that's now behind a 'press', a shallow closet common in these old buildings.

These windows face north, and keeping cool wasn't a big concern in Scotland - it tends to be chilly except for a short summer.

johncarlosbaez, 5 days ago

@dougmerritt - don't be sad that this doesn't fit into an infinite pattern. There are 'classical' algebraic structures, which fit into infinite patterns, and 'exceptional' algebraic structures, which don't. But most of the exceptional structures are related to each other! This means that there's a kind of pattern running through mathematics at right angles to our usual classifications. Nobody knows what this means, except for the obvious part:

"There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy."

johncarlosbaez, 5 days ago (edited 4 days ago)

@RefurioAnachro @dougmerritt -

A 'vector product algebra' is a finite-dimensional real vector space with an inner product I'll call the 'dot product' together with a bilinear operation I'll call the 'cross product', obeying three identities:

𝑢×𝑣 = −𝑣×𝑢

𝑢⋅(𝑣×𝑤) = 𝑣⋅(𝑤×𝑢)

(𝑢×𝑣)×𝑢 = (𝑢⋅𝑢)𝑣 − (𝑢⋅𝑣)𝑢

These imply a bunch more identities.

You can get a vector product algebra from a normed division algebra by taking the subspace of 'imaginary' elements, namely those orthogonal to 1. You can also reverse this process. Since there are only four normed division algebras, ℝ,ℂ,ℍ and 𝕆, there are only four vector product algebras! But you can also run this argument backwards, which is nice because there's a great string diagram proof that there are only four vector product algebras:

• Markus Rost, On vector product algebras, https://www.math.uni-bielefeld.de/~rost/data/vpg.pdf).

The four vector product algebras have dimensions 0, 1, 3, and 7. But only the last two are interesting, since in the first two the cross product is zero.

johncarlosbaez, 5 days ago

@dougmerritt - wow, I didn't know Rudvalis went into statistics. I don't know anything about him, except that there's a group called the Rudvalis group. I also don't know anything about the Rudvalis group! The sporadic finite simple groups are vastly stranger than the exceptional Lie groups (which fall out from the octonions in a simple yet mysterious way).

johncarlosbaez, 5 days ago

@dougmerritt - people are working on a simplified "second generation" classification, but it's easy to see why Rudvalis would want to try something new:

"The proof of the theorem, as it stood around 1985 or so, can be called first generation. Because of the extreme length of the first generation proof, much effort has been devoted to finding a simpler proof, called a second-generation classification proof. This effort, called "revisionism", was originally led by Daniel Gorenstein."

"As of 2023, ten volumes of the second generation proof have been published (Gorenstein, Lyons & Solomon 1994, 1996, 1998, 1999, 2002, 2005, 2018a, 2018b; & Capdeboscq, 2021, 2023). In 2012 Solomon estimated that the project would need another 5 volumes, but said that progress on them was slow. It is estimated that the new proof will eventually fill approximately 5,000 pages. (This length stems in part from the second generation proof being written in a more relaxed style.) However, with the publication of volume 9 of the GLS series, and including the Aschbacher–Smith contribution, this estimate was already reached, with several more volumes still in preparation (the rest of what was originally intended for volume 9, plus projected volumes 10 and 11). Aschbacher and Smith wrote their two volumes devoted to the quasithin case in such a way that those volumes can be part of the second generation proof."

johncarlosbaez, 6 days ago (edited 4 days ago)

@dougmerritt @mattmcirvin @idlestate @60sRefugee @pewnack -

Cool! To me the big mystery is not the chemistry of prometheum - since I'm not deep enough into lanthanide chemistry to be dying to collect the complete set of 14 and carefully compare them. It's the nuclear physics of prometheum! Why are all isotopes of this element radioactive?

I haven't checked lately, but it must be 'doubly odd' - an odd number of protons and an odd number of neutrons. That's a good rule of thumb but it leaves a lot of questions unanswered. (I'm sure there are people who know a lot more about these answers than I do.)

johncarlosbaez, 6 days ago

@mattmcirvin - makes sense; I should have guessed that since neutrons are a dime a dozen.

Maybe it's near some sort of 'anti-magic number'????

@dougmerritt @idlestate @60sRefugee @pewnack

johncarlosbaez, 7 days ago

@TruthSandwich - a really low-grade crackpot, starting with 1×1=2.

johncarlosbaez, 4 days ago (edited 4 days ago)

@SvenGeier - there may be elements of 𝐴 that aren't in the sequence 𝑎₁,𝑎₂,𝑎₃,.... In fact if 𝐴 is uncountable it 𝑚𝑢𝑠𝑡 have lots of elements that aren't in this sequence!

johncarlosbaez, 4 days ago

@SvenGeier - and even in the countable case, there's no reason every element of 𝐴 needs to be on your list 𝑎₁,𝑎₂,𝑎₃,.... So you need to deal with ones that aren't. Luckily it's easy: leave them alone!

johncarlosbaez, 7 days ago

@CyberneticForests - it's good to get rid of Twitter. If you like posts about math, physics, and other bits of science try me.

johncarlosbaez, 9 days ago

@rbreich - there are such editorials, but the Republicans have gone full-on gangsta, and nobody expects them to stop now.

johncarlosbaez, 8 days ago

@franco_vazza - the pictures on my screen are just one inch tall. But in reality....

johncarlosbaez, 9 days ago

@pewnack - while the American Chemical Society includes both lanthanum and lutetium in a list of 15 lanthanoids, one of the alternatives I like even less than the two I presented, because there are just 14 orbitals in the f subshell.

johncarlosbaez, 9 days ago

@mattmcirvin - the entropy of ice is apparently largely due to the randomness of the positions of the hydrogens here.

From stuff I wrote:

The oxygen atoms form a bipartite lattice: in other words, they can be divided into two sets, with all the neighbors of an oxygen atom from one set lying in the other set. You can see this if you look.

Focus attention on the oxygen atoms in one set: there are N/2 of them. Each has 4 hydrogen bonds, with two hydrogens close to it and two far away. This means there are

(\binom{4}{2} = 6)

allowed configurations of hydrogens for this oxygen atom. Thus there are (6^{N/2}) configurations that satisfy these N/2 atoms.

But now consider the remaining N/2 oxygen atoms: in general they won’t be satisfied: they won’t have precisely two hydrogen atoms near them. For each of those, there are

2⁴=16

possible placements of the hydrogen atoms along their hydrogen bonds, of which 6 are allowed. So, naively, we would expect the total number of configurations to be

(6^{N/2} (6/16)^{N/2} = (3/2)^N)

Using Boltzmann’s ideas on entropy, we conclude that

S = Nk ln(3/2)

where k is Boltzmann’s constant. This gives an entropy of 3.37 joules per mole per kelvin, a value close to the measured value. But this estimate is ‘naive’ because it assumes the 6 out of 16 hydrogen configurations for oxygen atoms in the second set can be independently chosen, which is false. More complex methods can be employed to better approximate the exact number of possible configurations, and achieve results closer to measured values.

johncarlosbaez, 9 days ago (edited 9 days ago)

@mattmcirvin - It looks like Janet's table is implementing the 'Madelung rules' shown below - in fact reading some stuff I see some claim Janet discovered the Madelung rules.

As I guess you were trying to tell me, these rules have exceptions, the first being chromium, and then copper. Lanthanum is also an exception: instead of putting an electron in the 4f shell as it "should", it skips ahead and puts one in 5d subshell. Next cerium puts an electron in the 4f subshell as it should - but also keeps one in the 5d subshell, still violating the Madelung rules. Then praseodymium has 3 electrons in the 4f subshell, following the Madelung rules.

This is one of the various more erudite reasons for saying that lanthanum doesn't deserve to be a lanthanoid: it doesn't have any electrons in the 4f subshell as lanthanoids should.

johncarlosbaez, 9 days ago

@mattmcirvin - oh, I was looking at the wrong table just now! Here's Janet's table, for anyone curious yet lazy. So yeah, the noble gases are fucking weird in this table, and I don't think it nicely illustrates the Madelung rules.

johncarlosbaez, 9 days ago

@mattmcirvin - I consider scandium and yttrium transition metals simply because I insist that there be 10 metals, called transition metals, whose job it is to fill each d subshell - and scandium and yttrium are doing that job for the 3d and 4d subshells. If you look at Wikipedia's chart of electron configurations you'll see they're doing it just as you'd expect, with nothing anomalous about them:

https://en.wikipedia.org/wiki/Periodic_table_(electron_configurations)

With lathanum and actinium there's a stronger case to be made that they're doing something anomalous.

I am a mathematical physicist more than a chemist, so the beauty of the group representation theory that leads to the periodic table matters more to me than detailed chemical properties... when it comes to putting elements into a periodic table, I mean.

It's still very interesting that there's something "rare-earthy" about scandium and yttrium, though I'm not sure I know enough to say what it is!