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I'm a mathematical physicist who likes explaining stuff. Sometimes I work at the Topos Institute. Check out my blog! I'm also a member of the n-Category Café, a group blog on math with an emphasis on category theory. I also have a YouTube channel, full of talks about math, physics and the future.

This profile is from a federated server and may be incomplete. Browse more on the original instance.

johncarlosbaez, to random
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I quit Twitter and moved to Mastodon in October 2022 when Elon Musk took over. Many other academics also came to Mastodon, and I was very happy - see my pinned posts. But most of them later became inactive.

I stick around because I've decided that corporate-owned social media are dangerous - especially now, when authoritarians are spreading their reach. Turns out it's unrealistic to expect that most academics feel this way - yet. They may get the idea when it's too late.



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As for journalists, I've seen some recent articles saying they still feel a need to be on Twitter (or whatever it's called). This article says that's less true nowadays. But if the US goes full-on authoritarian, and our new rulers are big on social media, I bet journalists will feel the need to be there.



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@veronica - good! 👍 If we keep feeding the monster, it will keep growing bigger until it decides to eat us.

johncarlosbaez, (edited )
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@franco_vazza @albertcardona - it's too late to join when it becomes illegal, like it already is in China and maybe Russia. If the US becomes a dictatorship, we'll see how it goes.

johncarlosbaez, (edited ) to random
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If you could watch an individual water molecule, about once in 10 hours you'd see it do this!

As it bounces around, every so often it hits another water molecule hard enough enough for one to steal a hydrogen nucleus - that is, a proton - from the other!

The water molecule with the missing proton is called a hydroxide ion, OH⁻. The one with an extra proton is called a hydronium ion, H₃O⁺.

This process is called the 'autoionization' of water. Thanks to this, roughly one in ten million molecules in a glass of water are actually OH⁻ or H₃O⁺, not the H₂O you expect.

And this gives a cool way for protons to move through water. Let's watch it!


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How does electrical current move through water? Here's a little movie of it, made by Mark Petersen. A positively charged proton gets passed from one molecule to another!

This is called the 'Grotthuss mechanism' because Theodor Grotthuss proposed this theory in his paper “Theory of decomposition of liquids by electrical currents” back in 1806. It was quite revolutionary at the time, since ions were not well understood.

Something like this theory is true. But in fact all the pictures I've shown so far are oversimplified! A hydronium ion is too powerfully positive to remain a lone H₃O⁺. It usually attracts a bunch of other water molecules and creates a larger structure!


A chain of water molecules passing an extra proton along. https://commons.wikimedia.org/wiki/File:Proton_Zundel.gif This work has been released into the public domain by its author, Matt K. Petersen a.k.a. w:User:Pigwiggle. This applies worldwide.

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Very often a hydronium ion H₃O⁺ attaches itself to another water molecule, creating H₅O₂⁺.

This is called a 'Zundel cation', named after Georg Zundel, a German expert on hydrogen bonds.

In this picture, the H⁺ in the middle looks more tightly connected to the water at right than the water at left. But in fact it should be completely symmetrical. At least, that’s the theory of how a Zundel cation works!

However, the Zundel cation is not the end of the story. When you've got an extra proton around, it can really attract a lot of nearby water molecules. So read on....


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@RobertJackson58585858 - I'm trying to give lessons with all the fun stuff and none of the pain.

johncarlosbaez, (edited )
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Nice questions, @leemph!

  1. " Does this enables, somehow, "protonic" current flowing into suitable "water circuits"?"

Inside the Grotthuss mechanism I discussed, shown below, seems to be one of two important mechanisms for protons to move through water, the other being 'vehicular transport' where they move as part of hydronium ions H₃O⁺:


  1. "I believed that the electric charge of 1 electron balances perfectly the charge of 1 proton, as 1+(-1)=0."

That's true.

"So why the hydronium ion missing only 1 electron to be "neutral" attracts so many water molecules?"

In the van der Waals effect, a positively charged particle can polarize nearby neutral molecules, pulling their negatively charged bits towards it and repelling their positively charged bits. Then these neutral molecules are attracted to it!


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@leemph -

  1. "Such structures that are then formed in water... are they visible? Ideally how large they can get?"

I don't think they get so large as to be visible, but they get quite large! Here is a hypothesized "icosahedral water cluster". In reality they aren't so perfect, since they're always being buffeted around by other water molecules.

For more see this:


johncarlosbaez, (edited ) to random
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I'm having fun learning about 'crystal field theory'. It sounds so cool! It's actually a specialized subject where we look at a transition metal atom surrounded by nearby molecules called 'ligands'. Here you see 6 ligands at the vertices of an octahedron. Each repels the electrons of your transition metal atom in the middle.

A transition metal atom has up to 10 electrons in its 'd orbitals'. These are shown as blue blobs here. There are five d orbitals, and each can hold at most 2 electrons: one 'spin up' and one 'spin down'. That gives a total of 10.

Before you surround the atom with ligands, electrons in all five d orbitals have the same energy. But because the ligands repel these electrons, the energies of the d orbitals called B and C get pushed up compared to those called D, E, F - because B and C's blue blobs get closer to the ligands!

Now for the math. A 'd orbital' corresponds to a homogeneous quadratic polynomial in the 𝑥,𝑦,𝑧 coordinates that has Laplacian equal to zero. Every d orbital is a linear combination of these five:

B: 𝑥²−𝑦²
C: 2𝑧²-𝑥²-𝑦²
D: 𝑥𝑦
E: 𝑥𝑧
F: 𝑦𝑧

Note the Laplacian of each function is zero! Also note that 𝑧² in the picture is short for 2𝑧²-𝑥²-𝑦².

It's purely arbitrary that the two weirdo orbitals, B and C, treat the z axis differently than the x and y axis. We could choose other d orbitals that treat the x or y axes differently.

If you rotate a d orbital you get another d orbital, so we say d orbitals form a 'representation' of the rotation group. But the ligands break the rotational symmetry! With the ligands around, the symmetry group is reduced to the symmetry group of an octahedron.


johncarlosbaez, (edited )
@johncarlosbaez@mathstodon.xyz avatar

What's the point of this math? It's used to explain the colors of transition metal compounds. Yes, the symmetries of polynomials are important for understanding the colorful world of pigments!

More of the math:

Rotation and reflection symmetries of an octahedron permute the x, y, and z coordinates and also introduce minus signs. These transformations send the d orbitals

D: 𝑥𝑦
E: 𝑥𝑧
F: 𝑦𝑧

to themselves, or minus themselves, so these d orbitals are basis vectors for a 3d representation of the octahedral symmetry group.

Less obviously, these transformations send the d orbitals

B: 𝑥²−𝑦²
C: 2𝑧²-𝑥²-𝑦²

to linear combinations of themselves, so these d orbitals are basis vectors for a 2d representation of the octahedral group!

Why is that last bit true? Say we do a 90 degree rotation in the xz plane. Then 𝑥²−𝑦² becomes 𝑧²−𝑦², but this is still a linear combination of functions B and C, since

B + C = (𝑥²−𝑦²) + (2𝑧²-𝑥²-𝑦²) = 2𝑧² - 2𝑦²

Similarly 2𝑧²-𝑥²-𝑦² becomes 2𝑥²-𝑧²-𝑦², which, you can check, is -B/2 + 3C/2. So even though the weirdo orbitals B and C aren't preserved by octahedral symmetries, linear combinations of them are!

You can learn the chemistry here:

The Transition Elements and Their Coordination Compounds, https://www.slideserve.com/mai/chapter-23

and that's where I got the pictures!


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@adityakhanna - I simplified slightly, since actually the d orbitals are described by these quadratic polynomials in x,y,z times the function exp(-r/3a), where a is the Bohr radius. We need that extra exp(-r/3a) to make the wavefunctions go to zero as r → ∞. But this function exp(-r/3a) is spherically symmetric so it doesn't affect how the orbitals transform under rotations!

johncarlosbaez, (edited ) to random
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One string theorist wonders what will happen when "the founding members of modern string theory, who have been so influential thus far, will gradually retire and/or go to their next stage of existence". And he suggests that we

"Train an LLM with the very best papers written by the founding members, so that it can continue to set the trend of the community."

I think this is a GREAT idea.

Then we should get LLMs to write papers on string theory.

And then we should train LLMs to referee papers on string theory.

And then we should train LLMs to write grant proposals on string theory.

And then we should train LLMs to referee these grant proposals!

For more, see Peter Woit's blog:


johncarlosbaez, (edited )
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@dougmerritt - String theory has had a wonderfully transformative impact on many branches of algebra, geometry, topology, number theory, etc. It's really a godsend to mathematicians - and while I can't keep up with it, I do spend time studying it. Alas, string theory has never made a verified prediction in particle physics, gravity or cosmology... and there have always been good reasons for why this is the case.

The string theorist Nima Arkani-Hamed recently put it this way:

“String theory is spectacular. Many string theorists are wonderful. But the track record for qualitatively correct statements about the universe is really garbage.”

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@dougmerritt - the situation is actually much worse than that!

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@dougmerritt - the problem is not just that too many things fit in string theory, it's that people don't know how to fit what we actually see in string theory. (This is a very brief summary, and string theorists would of course argue against it, and then I'd have to argue with them, etc.)

johncarlosbaez, (edited )
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@MartinEscardo wrote: "As you said, the mathematics is sound."

I didn't exactly say that. Some mathematics generated from string theory is sound, and some of that is downright revolutionary. Mathematicians generally focus on that. But the closer string theory gets toward actually making predictions about real-world physics, the more shaky the math becomes. Some of it is the usual lack of rigor in physics - e.g., nobody knows that the Standard Model of particle physics is mathematically consistent, but few physicists care because they can use it to make predictions that match experiment and they have ways of avoiding the known problems. But I'm not talking about that - I'm talking about worse problems.

"But exactly how is the physics considered to be unsound, other than "it doesn't have observable content"?"

It's easy to use string theory to make predictions that are obviously wrong. To get it to make predictions that might be right, one needs to add elaborate ad hoc mechanisms "by hand" to the basic theory. These ad hoc mechanisms are so flexible that every time a prediction of the resulting elaborate theory is falsified by experiment, string theorists can to tweak the elaborate setup to change the predictions and dodge the problem.

These elaborate mechanisms are so removed from the basic core of string theory that the people who work on them, and try to invent theories that might match experiment, are sometimes not considered string theorists. They are called "phenomenologists".


johncarlosbaez, (edited )
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@leemph - I can't really answer a question like "is it really that bad?" - it's too open-ended. I will just quote the string theorist Nima Arkani-Hamed:

“String theory is spectacular. Many string theorists are wonderful. But the track record for qualitatively correct statements about the universe is really garbage."

Notice: even qualitatively, there are big problems!

johncarlosbaez, (edited )
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@_spiralsource - no, those are not ad hoc: we don't put those into the theory, we get them out. So we get some very beautiful supersymmetric theories in 10 and 11 dimensions. The ad hocness begins when we try to make those theories match our non-supersymmetric 4-dimensional universe and its particular assortment of particles and forces. Then it's a matter of endless tweaks, each of which causes new problems, never quite getting what we want.

johncarlosbaez, (edited ) to random
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Q: Why is an Ab-enriched star-category like the drummer in a Beatles tribute band?

A: Because it's a star-ringoid.

To make up for that pun:

'Ringoids' are actually quite important - but like 'groupoid', this term means two things. A groupoid is a category where all morphisms are isomorphisms - or morally speaking, a 'many-object group'. These are important all over mathematics. But in older literature, 'groupoid' sometimes means a set with a binary operation. Bourbaki renamed these 'magmas', and I prefer that term.

Similarly, a 'ringoid' is a 'many-object ring': that is, a category where you can add morphisms f,g: x → y, making each homset into an abelian group, and composition of morphisms gets along with addition, as follows:

f ∘ (g + h) = f ∘ g + f ∘ h
(f + g) ∘ h = f ∘ g + f ∘ h
f ∘ 0 = 0
0 ∘ f = 0

For example, the category of abelian groups is a ringoid, and so is the category of representations of any group. The slick way to summarize all the axioms here is that a ringoid is a category enriched over abelian groups, or 'Ab-category' for short.

But 'ringoid' has also been used to mean a set with binary operations + and × obeying only

f × (g + h) = f × g + f × h
(f + g) × h = f × h + g × h

These are not so important in my experience.

A 'star-category' is a category where for each morphism f: x → y we have a morphism going back, f*: y → x, obeying

f** = f
(f ∘ g)* = g* ∘ f*

So it's a category where you can turn around morphisms.

If you know all this stuff, the concept 'star-ringoid' is self-explanatory, and you might never even notice that it sounds quite silly.

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@ProfKinyon - just so you know, I have no idea who that is.

@johncarlosbaez@mathstodon.xyz avatar

@ProfKinyon - oh, so Ringo was the second best!

@johncarlosbaez@mathstodon.xyz avatar

@dougmerritt @ProfKinyon - all the music YouTubers I watch agree that Ringo was a great drummer, and anyone who disagrees has got to listen to Tomorrow Never Knows or the first 10 seconds of Come Together.

gregeganSF, (edited ) to random
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Every solid object, such as the rods in the image below, has an inertia tensor. We can write the inertia tensor for the black rod, which keeps changing its orientation, as a sum of various multiples of the inertia tensors for the 6 fixed rods. The plot at the bottom shows how these 6 components change as the black rod points in different directions.

But why do we need 6 components? In 3D space, shouldn’t 3 be enough?

To see why we really need 6, let’s look at the inertia tensor. This is a matrix, I, that we use to multiply the angular velocity vector of an object, ω, to get the angular momentum, L.

The velocity of a point mass whose angular velocity is ω is given by:

v = ω × r

where r=(x,y,z) gives the coordinates of the point mass.

The linear momentum is:

p = m v = m ω × r

and the angular momentum is:

L = r × p = m r × (ω × r)

To get the same thing using matrices, note that we can write the cross product of r with any vector like ω as:

r × ω = C(r) ω

where C(r) is the matrix:

0 –z y
z 0 –x
–y x 0

This lets us write:

L = m r × (ω × r)
= –m r × (r × ω)
= –m C(r)^2 ω

Minus the square of the matrix C(r) is:

y^2+z^2 –xy –xz
–xy x^2+z^2 –yz
–xz –yz x^2+y^2

So the inertia tensor for a point mass is:

I = –m C(r)^2

This is the matrix above, multiplied by m.

In general, we add up (or integrate) over all the point masses in the body, to get the inertia tensor for the whole thing.

The inertia tensor we obtain this way will always be a symmetric matrix, so it can be specified with 6 numbers: 3 on the diagonal, and 3 above. So the inertia tensors belong to a 6-dimensional space.

johncarlosbaez, (edited )
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@gregeganSF - yes, I just happen to be into transition metals these days, so if you say"5d irrep of SO(3)" I think "d orbitals!".

The blue stuff in those pictures of orbials would be negative density regions in an object of total mass 0 in your approach.

(I added some text correcting that picture a bit.)

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