Greg Egan

I am a science fiction writer and computer programmer.
Latest novel: MORPHOTROPHIC.
Latest collection: SLEEP AND THE SOUL.
Web site: gregegan.net

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Last Notes

2026-02-19 04:24:15 GMT

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@nprofile…hzzt Did you get it done in a public hospital? My GP referred me to a private cardiologist and I didn’t think to ask if I could go to a public hospital clinic instead. The private cardiologist charges about 3 times the Medicare rebate for everything.

2026-02-19 03:02:31 GMT

Echocardiograms are so expensive that I really think they should offer a 30 percent discount if they fail to locate an alien chest-burster.

2026-02-03 12:32:19 GMT

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It turns out there are 6 different digits between the barcodes, so it might have been some kind of bug in the checkout’s software, rather than the scanner misreading the code.

2026-02-03 11:35:19 GMT

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@nprofile…yqta
The barcodes on the two packs of flour that I scanned look completely identical to me. Not only is there no sign of tampering, there are no tears or stains etc. that might have confused the scanner.
Right now I have no idea what the frozen berries barcode number was, because I only noticed the error when I got home, but I should check next time I’m in the supermarket, to see how many digits were misread.

2026-02-03 11:15:53 GMT

Today in the supermarket I scanned 2 packs of flour … and one of them was listed on the receipt as frozen raspberries. So that must have been a genuine barcode misread, rather than just incorrect data in a product database. The barcodes looked fine, so maybe the reader glitched.

2026-02-01 14:19:20 GMT

Just rewatched “The Wages of Fear”, which suited the ambience on a 37C day.
I last saw it 40 years ago, and I’d forgotten everything but the overall premise, so it was every bit as jaw-dropping as if I was watching for the first time.
4.9/5 stars, with -0.1 for the very end.

2026-02-01 07:19:54 GMT

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To answer this, we can start with the fact that on a *sphere* of radius R, any two ants following neighbouring great circles would travel a distance of (π/2) R before their paths met. That behaviour comes from the intrinsic curvature of the sphere, which is measured by its Gaussian curvature, K = 1/R^2.
But what is the Gaussian curvature of our ellipsoid, at its equator? The equator itself has a curvature of 1, but the meridians are ellipses that are 3 times wider. What is the radius of curvature of an ellipse with semi-axis a and b, at the point where the minor semi-axis meets the ellipse?
The ellipse takes the form:
y(x) = b √[1-(x/a)^2]
For small x:
y(x) ≈ b (1-½(x/a)^2)
The second derivative is:
y''(x) = -b/a^2
If we set a = b = R this would be -1/R, so the curvature here is just the opposite of the second derivative.
The Gaussian curvature is the product of the curvatures along two perpendicular directions where it achieves it maximum and minimum values. For our ellipsoid, these curvatures are 1/b for the equator and b/a^2 for the meridian, so we have:
K = 1/a^2
In other words, the Gaussian curvature at the equator of the ellipsoid is completely independent of b (the radius of the equator) and is the same as that of a sphere of radius a (the polar radius)!
That in turn means that the ants following the red path will bump into each other after travelling a distance:
(π/2) a = 3π/2 ≈ 4.71239
So, perhaps a bit counterintuitively, the ants following the small red circle will travel *farther* before they converge than the ants following the larger blue ellipse.

2026-02-01 07:19:26 GMT

Two pairs of ants set off from the same starting point, walking side-by-side. One pair takes the red path, the other pair takes the blue path.
The red and blue paths are geodesics (the straightest possible paths, like great circles on a sphere), and each ant in a pair follows a geodesic that lies a small distance to either side of the marked one, initially pointing in the same direction.
If this ellipsoid has semi-axis of length 1 and 3, the ants taking the blue path will bump into each other at the “pole”, where all meridians converge, after travelling a distance of 3.34122.
How far will the ants following the red path travel before they bump into each other? More, less, the same?
https://media.mathstodon.xyz/media_attachments/files/115/994/151/014/718/903/original/194a2e7858959caa.png

2026-01-31 13:56:52 GMT

“Can a time capsule outlast geology?
A ridiculous but instructive thought experiment involving deep time, plate tectonics, erosion and the slow death of the sun”
This is hugely enjoyable! The subheading is exactly right: don’t take the premise seriously, just go along for the ride.
https://www.scientificamerican.com/article/can-a-buried-time-capsule-beat-earths-geology-and-deep-time/

2026-01-29 04:30:34 GMT

“Add personality to your work with new editable shapes!”
— Apple, fondly imagining that this prospect is so alluring that I will pay them A$19.99/month for “Creator Studio” when the current apps I have (Pages, Numbers and Keynote) work perfectly well.

2026-01-27 23:46:12 GMT

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@nprofile…gu9p @nprofile…g0zc OK. But given that I can think, observe and remember, I am not surprised at all to find that the universe I inhabit follows rules and contains patterns and regularities.

2026-01-27 23:12:34 GMT

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@nprofile…gu9p @nprofile…g0zc I’ve heard people suggest this a few times, but apart from anthropic-type arguments (could any organism exist in such a universe to have thoughts and observations about it, if *everything* was just random), I’m not sure why we’d even describe different subsets of these random events as belonging to the same “universe” if there was no structure whatsoever tying them all together.

2026-01-27 22:43:08 GMT

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@nprofile…g0zc @nprofile…gu9p Yeah, nothing surprises me less than the so-called “unreasonable effectiveness of mathematics” when mathematics was created, mostly, to capture empirical truths.

2026-01-21 10:22:16 GMT

For crying out loud, ABC news, never send a line graph to do a bar chart’s job.
Link: https://www.abc.net.au/news/2026-01-21/midnight-oil-rob-hirst-pancreatic-cancer-death/106252876
https://media.mathstodon.xyz/media_attachments/files/115/932/588/922/704/068/original/792a64d98ef3b792.png

2026-01-21 00:38:19 GMT

When the crackpot DIY Theory of Everything spammers forget to fill in the blanks in their email templates:
“I'm reaching out because your work in - particularly on - engages with fundamental questions that the U-Model addresses from a unique angle.”

2026-01-19 08:34:24 GMT

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@nprofile…tfwq I wonder if that and 7025869314 are the only ones that are pandigital in base 10. It’s not immediately obvious to me that there couldn’t be other 10-digit numbers that work with higher exponents.

2026-01-13 05:40:52 GMT

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Maybe the easiest mnemonic is: the spacetime curvature* due to the sun, at the distance where an orbit takes 1 year, is the same as the curvature on a sphere with a circumference of 1 light-year.
For other bodies and orbits, just scale the circumference with the orbital period.
* This is the spacetime curvature in either of the two directions at right angles to a radial vector, pointing towards the sun.
In the radial direction, the curvature is negative and twice as much (so that all three curvatures sum to zero, which is always true in a vacuum).

2026-01-11 10:53:03 GMT

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A measure of curvature itself is:
K = 1/R^2
= GM/(c^2 r^3)
≈ 1.707 × 10^{-23}
The sum of the angles of a triangle with area A in a space with curvature K is about π + K A, so if space at a fixed moment in time has a similar curvature to the spacetime curvature we’ve found, then an equilateral triangle with sides as big as the radius of the Earth would have an angular excess or deficit of about:
K A ≈ (1.707 × 10^{-23}) × (1.76 × 10^{13})
≈ 3 × 10^{-10} radians
That would not be easy to measure directly! But we can still quantify the curvature of the spacetime around the Earth using nothing but T, the period of a surface-grazing satellite, and c, the speed of light.

2026-01-11 10:52:42 GMT

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Now, consider two particles moving around a massive body, both in circular orbits of equal size, but in slightly different planes. The red and blue circles in the image show these orbits in space, while the red and blue helices represent the world lines of the particles in spacetime.
The world lines of free-falling particles are geodesics in spacetime. So the “radius of curvature” of that spacetime can be found from the period with which their separation repeats, which in this case will just be the period of the orbit. However, because an orbital period is a time, not a distance, we need to multiply it by the speed of light to convert it into a distance. For an orbit with radius r around a mass M, the period of the orbit is:
T = 2π √[r^3/(GM)]
where G is the gravitational constant. We convert this to a distance:
P = c T
and then the radius of curvature of spacetime is:
R = P / (2π) = c √[r^3/(GM)]
For M and r equal to the mass and radius of the Earth:
T = 5096 sec
and the radius of curvature of spacetime is:
R = 2.4 × 10^{11} metres
The spacetime curvature is actually positive in some directions and negative in others; for two particles that start at different distances from the mass, they move apart rather than together, and that corresponds to the diverging geodesics that we see under negative curvature, such as in the hyperbolic plane.
https://media.mathstodon.xyz/media_attachments/files/115/876/085/358/955/974/original/b621982432fbee19.png

2026-01-11 10:51:43 GMT

Gravity is the curvature of spacetime. So, how curved is the spacetime we are in right now?
Let’s start by looking at the effect of curvature in a very simple situation. Two meridians on a sphere, like the red and blue ones in the diagram, will come together and move apart as you follow them along their length. If their separation at the equator is s(0), after a distance d it will be:
s(d) ≈ s(0) cos(d/R)
where R is the radius of the sphere. The separation will repeat with a period of P = 2πR, so we can say that the radius of curvature of the space these geodesics are in is:
R = P / (2π)
https://media.mathstodon.xyz/media_attachments/files/115/876/079/233/590/239/original/e92dcd3a54fb6525.png

2026-01-07 13:33:54 GMT

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@nprofile…4hp5 It’s fun (though no doubt very silly) to imagine taking this to extremes. There are plenty of structures in modern mathematical physics that are as immutable as Euclidean space used to be as the stage on which Newtonian physics unfolded. What if every Hilbert space and Lie group was warped and distorted by the unruly behaviour of its constituents?

2026-01-06 08:41:34 GMT

Every Australian politician who has dragged their feet on emissions reductions is cordially invited to spend the upcoming 44°C day in Sydney’s western suburbs in a two-bedroom flat without air conditioning.

2026-01-05 12:32:43 GMT

Today in Rough Approximations For ArcSine:
The Taylor series for sin and cos start with:
sin θ ≈ θ - (1/6) θ^3
cos θ ≈ 1 - (1/2) θ^2
So:
(θ - sin θ) / (1 - cos θ) ≈ θ/3
Solving for θ:
θ ≈ 3 sin θ / (2 + cos θ)
or:
asin(x) ≈ 3 x / (2 + √[1-x^2])
This isn’t too bad for small x, and has the correct value and derivative at x = 0, but we can make it match those things at x = ±1 as well by tweaking the coefficients:
asin(x) ≈ π x / (2 + (π-2)√[1-x^2])
This has a maximum proportional error of 0.011.

2026-01-04 06:29:00 GMT

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The original post:
https://mathstodon.xyz/@[email protected]/115809815616652881

2026-01-04 06:28:15 GMT

A few days ago, I saw a nice approximation to the arcsine function by @nprofile…40t9
asin(x) ≈ a(x) = x + (π/2-x)(1-√[1-x^2])^2
I won’t bother plotting a(x) against asin(x), because the two curves are indistinguishable to the eye, but the proportional error is plotted below.
I spent some time trying to figure out if there’s an underlying *geometric* reason why this works so well ... in the same sense that there is for, say:
asin(x) ≈ chord(x) = √[2(1-√[1-x^2])]
the length of the chord that subtends the angle asin(x).
There are some obvious nice features built into a(x): it clearly must agree with asin at x=0 and 1, and less obviously it will match derivatives at those points as well.
But surely there had to be some special geometric relationship as well?
If so, I never did find it. Maybe someone else will (or already has). But I found another approximation, roughly as simple and roughly as good:
asin(x) ≈ b(x) = ½(π-4)x^2 + x + 1 - √[1-x^2]
which also matches values and derivatives with asin(x) at x=0 and 1, and whose proportional error is the gold curve in the plot below.
That partly cured me of my conviction that there had to be a nice geometrical account for any approximation this good. Maybe all that’s really needed is a low-degree polynomial and one function, √[1-x^2], with an infinite derivative at x=1 the same as asin(x).
https://media.mathstodon.xyz/media_attachments/files/115/835/253/029/996/791/original/ca74bc24e2a289df.png

2025-12-31 13:24:09 GMT

Of course They Might Be Giants have the perfect song to prepare you for 2026:
https://www.youtube.com/watch?v=rBCVGshPGBk

2025-12-27 10:31:04 GMT

Three movies I’m looking forward to in 2026. (The dates here are when they open in my home town, at Perth’s Luna Cinemas.)
“No Other Choice” Jan 15
“The Secret Agent” Jan 22
“It Was Just An Accident” Jan 29

2025-12-26 21:58:16 GMT

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@nprofile…nt94
No, just a different directory.

2025-12-26 08:25:33 GMT

This is the weirdest internet glitch I have encountered.
When I access any arxiv dot org abstract page via my home ISP, the page takes about 5 minutes to load. But the corresponding PDF for the same paper takes a second or two, so I’m not experiencing slow access to arxiv dot org in general.
And when I access the abstract pages via my mobile provider, there is no delay. So the problem isn’t anything on the arxiv servers.
Maybe this is Cloudflare, or some other content distribution network, interacting differently with my home and mobile ISPs. But it’s baffling that the (very small) abstract pages ever take so long to load.

2025-12-24 12:20:43 GMT

Chekhov would be ... pleased?
#Pluribus

2025-12-24 10:19:00 GMT

My Christmas was just saved from being fairly dire by D. C. Stillson, inventor of the pipe wrench. My place is so old you can’t buy the square-plug tools for opening the drain access hatches anymore, but a pipe wrench did the trick, and now the kitchen sink ... empties again.

2025-12-23 11:21:21 GMT

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@nprofile…8eua This is very nice, but I hope it won’t end up attracting the awful kind of pop-sci coverage that makes it sound like the Halting Problem throws some shocking new spanner into the works of classical physics ... as if it had somehow been the case, before Turing and Gödel, that physicists had expected to be able to predict the behaviour of arbitrary systems at arbitrary future times with magical short-cuts that did not require them to calculate arbitrary numbers of intermediate states.

2025-12-20 21:57:52 GMT

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@nprofile…tctg @nprofile…8eua @nprofile…pwph
Larson is amusing, but maybe I should have said “Gold Age of animation” to make it clear what kind of work I meant.
https://en.wikipedia.org/wiki/Magical_Maestro

2025-12-20 11:01:06 GMT

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@nprofile…8eua @nprofile…pwph This sounds like something that would have appeared in the Golden Age of cartoons: a farmyard full of animals frenetically plucking vegetables from the fields, carving them into instruments, and playing a symphony as hundreds of slapstick subplots unfolded around them.

2025-12-19 09:58:34 GMT

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@nprofile…8eua
Wikipedia’s article on sundogs claims:
“It is possible to predict the forms of sun dogs as would be seen on other planets and moons. Mars might have sun dogs formed by both water-ice and CO2-ice. On the giant planets—Jupiter, Saturn, Uranus, and Neptune—other crystals form clouds of ammonia, methane, and other substances that can produce halos with four or more sun dogs.[9]”
Ref [9] was to a fairly old web site, https://www.atoptics.org.uk/halo/oworld.htm, which made several predictions, including:
“Halos from octahedral ammonia crystals as might exist in the cold high level clouds of Jupiter and Saturn. The 42° circular halo has four associated sundogs. The inner halos are produced by rays reflected within the crystal.”
That web site has no actual Martian haloes. However, we have seen a halo on Mars:
“Graphical Abstract
For the first time, a halo has been photographed from the surface of another planet. Its angular distance from the Sun indicates that the halo-generating Mars cloud consisted of water–ice crystals, not CO2 crystals.”
https://rmets.onlinelibrary.wiley.com/doi/10.1002/wea.4197 [Paywalled, alas]
https://media.mathstodon.xyz/media_attachments/files/115/745/607/691/657/334/original/af46af24e5a029a1.png

2025-12-18 02:00:45 GMT

I just got my annual COVID booster. The vaccine updated for LP.8.1 recently became available in Australia, so if you were waiting for that, now’s the time to get a pre-Christmas boost.

2025-12-16 12:14:43 GMT

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@nprofile…9s72 Time to try transporting a steamship over the Andes.

2025-12-16 11:19:38 GMT

Until today I would have been hard-pressed to name a surface arising naturally in physics that’s described by a degree-6 polynomial, but there’s a family of them hiding in plain sight:
x^2 – k^2 (x^2 + y^2 + z^2)^3 = 0
These are the surfaces of equal electrostatic potential for a dipole pointing along the x-axis! The contours in the image come from a 2D slice through the axis.
But these surfaces also have two nice properties in Newtonian gravity.
[1] A point mass of m located *anywhere* on a k-surface has the same x-component for its gravitational attraction at the origin:
g_x = G k m
[2] If you fill the interior of a k-surface with matter of uniform density ρ, the gravitational attraction at the origin is *the maximum possible* for that density and total mass.
g(k) = 4πGρ/(5√k)
The volume of the interior of a k-surface is:
vol(k) = 4π/(15√[k^3])
In terms of total mass M and density ρ:
g_max(M,ρ) = G ((48/25)π^2 M ρ^2)^{1/3}
The attraction at the surface of a sphere is:
g_sph(M,ρ) = G ((16/9)π^2 M ρ^2)^{1/3}
So g_max/g_sph = (27/25)^{1/3} ≈ 1.026
The polar gravity of the “best” oblate spheroid can improve on a sphere *almost* as much:
g_oblate(ecc = 0.694464)/g_sphere ≈ 1.022
The max gravity result is proved in:
https://arxiv.org/abs/1412.5541
H/T @nprofile…ks48
https://media.mathstodon.xyz/media_attachments/files/115/728/927/828/744/715/original/5eab6d56d4e793e5.png

2025-12-15 23:55:27 GMT

Orwell’s widow in 1974: “No, you may *not* make a musical adaptation of 1984!” But “Diamond Dogs” was probably ten times better for needing to add and change enough to stand on its own.
Orwell’s estate in 2025: “Knock yourself out!”
Serkis: “Thank you! Here is my awful remake of ‘Chicken Run’”.

2025-12-12 07:44:25 GMT

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@nprofile…468s I’m pretty sure you’re right!
In fact, if my numerical integrals can be trusted, for a cylinder of height 2 and radius 1.01, the gravitational attraction at the “equator” is 1.00455 times the gravitational attraction at the “poles”, despite the distance from the centre being greater at the equator. (This also kills off a potential alternative conjecture, where the maximum was achieved by the normal component of the gravity rather than the magnitude.)

2025-12-12 05:52:22 GMT

Given a homogeneous, centrally symmetric convex body, has it been proved that its gravitational acceleration on its surface attains its greatest magnitude at the closest point to the centre of mass, and its smallest magnitude at the farthest point?
Alternatively, are there known counterexamples?
This is certainly true for ellipsoids, and Newton more or less proved it for an oblate spheroid ( https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html#NPG ) but I can only get the ellipsoid result from an explicit calculation.

2025-12-11 12:06:47 GMT

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Just as a=b makes a spheroid special among ellipsoids, there is a condition:
a^2 b^2 (B – A) / (a^2 – b^2) – C c^2 = 0
that makes an ellipsoid special, where A, B, C measure how fast the gravitational potential increases as you move away from the centre along the x, y, z axes.
Those coefficients A, B, C themselves depend on the semi-axes a, b, c, and are given by the values of certain integrals.
But the upshot is that we can find a curve in the space of eccentricities for two meridians of the ellipsoid, say e_xz and e_yz, in two orthogonal planes.
For any tri-axial ellipsoid with two meridians whose eccentricities lie on this curve, it is possible for *the same value of spin* to bring the corresponding meridians of the equipotential ellipsoids into alignment with the meridians of the body itself!
So, a shape like this is a possible shape for “sea level” in a spinning body of fluid.
Note that one of the meridians *must* have an eccentricity greater than a critical value, e_crit ≈ 0.812670. This means that the angular momentum must also exceed a threshold.
The second issue is whether the Jacobi ellipsoid has less total energy than the Maclaurin spheroid, for the same angular momentum.
The potential energy, U, of the Maclaurin spheroid is actually slightly *lower* than that of the Jacobi ellipsoid!
But the Jacobi ellipsoid, with a greater moment of inertia I, reaches the same angular momentum, L, with a smaller angular velocity ω … and ends up with a kinetic energy, K, sufficiently smaller to make its total energy, E, the lower of the two.
More at https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html
https://media.mathstodon.xyz/media_attachments/files/115/700/840/750/847/526/original/680d2ce3261b4048.png

2025-12-11 12:04:48 GMT

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But how can this possibly work for a tri-axial ellipsoid? We only have one parameter to tweak, the rate of spin, so how can we match two *differently shaped* meridians?
The answer is that you *can’t* pull off this trick for a generic tri-axial ellipsoid, with any old a, b and c.
https://media.mathstodon.xyz/media_attachments/files/115/700/835/792/255/659/original/70e1e13a3ceae2af.mp4

2025-12-11 12:03:44 GMT

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Given any oblate spheroid (a = b > c), made of rock, say, so we can fix the shape, the surfaces of constant potential will also be spheroids — but if it’s not spinning they will not match the shape of the body.
However, by adding just enough spin the shapes can be made the same.
All the “equipotential ellipsoids” are just scaled versions of a single shape. Once you add enough spin to make one meridian of these ellipsoids the same shape as a meridian of the body itself, the problem is solved, since the rotational symmetry makes all meridians the same.
https://media.mathstodon.xyz/media_attachments/files/115/700/830/951/812/877/original/20785d80e0bd74e4.mp4

2025-12-11 12:02:26 GMT

If a mass of incompressible fluid is floating in space, subject only to its own gravity and centrifugal force, what shape will it adopt as its angular momentum is increased?
Initially, it will form an oblate spheroid, with two equal semi-axes shorter than its axis of rotation.
However, past a certain value for the angular momentum, L, the fluid can have a lower total energy by breaking that symmetry and forming a “tri-axial” ellipsoid, with the semi-axes perpendicular to the axis, a and b, taking on different lengths!
The first configuration is known as a “Maclaurin spheroid”, after Colin Maclaurin, who studied it in 1742.
The second is called a “Jacobi ellipsoid”, after Carl Gustav Jacob Jacobi, who in 1834 realised that a fluid could be in equilibrium in this less symmetrical state.
There are two separate issues here.
One is the question of when the “level surfaces” of constant gravitational-plus-centrifugal potential for an ellipsoid can match the shape of the ellipsoid itself. What shapes can “sea level” take for a spinning mass of fluid?
https://media.mathstodon.xyz/media_attachments/files/115/700/825/786/077/375/original/373dedb145afde74.mp4

2025-12-11 04:52:40 GMT

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@nprofile…stct Once a year, I need to print my French tax forms (in triplicate for each publisher), so I buy a new ink cartridge and use the brief window of opportunity to print anything else I might conceivably want for the next 12 months, because the cartridges will dry up and become hopelessly clogged during any period of prolonged disuse.

2025-12-11 03:44:44 GMT

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@nprofile…mmff
For Dichronauts:
Books-A-Million:
http://www.booksamillion.com/p/9781597806053
Apple Books:
https://books.apple.com/us/book/dichronauts/id1171077019
Barnes and Noble:
https://www.barnesandnoble.com/w/?ean=9781597806053
Bookshop dot org
https://bookshop.org/a/1688/9781597806053
Kobo:
https://www.kobo.com/us/en/ebook/dichronauts?sId=8ac3153c-656f-4c4b-bba1-685c935ecd1b&ssId=3OYA7vbjgc7ujdoXQhfBA
(Those books where I have the US epub rights can be bought DRM-free from Smashwords and various other sites, but Dichronauts is one where the US publisher still has the US rights, so I can only sell it outside the US. All the books I self-publish are listed here:
https://www.gregegan.net/BIBLIOGRAPHY/Ebooks.html )

2025-12-10 04:40:17 GMT

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@nprofile…67su Euler?

2025-12-06 00:25:31 GMT

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@nprofile…zlur Very nice!

2025-12-05 13:20:35 GMT

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@nprofile…zlur
Is there an elegant proof of this? I think I can kludge together a proof by induction that relies on looking at how many carries can occur each time you add 999 to the previous multiple, with some tedious book-keeping relating zeroes in the last 3 digits and 9's in the digits to the left of the last 3. But is there something nicer?

2025-12-05 11:15:14 GMT

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@nprofile…y0ah @nprofile…8eua I think SO(3) can be embedded in R^5, but even that’s not particularly useful or interesting for most purposes.

2025-12-05 10:52:03 GMT

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@nprofile…8eua That’s so frustrating, because SO(3) is one of the easiest Lie groups by far to visualise without having to worry about higher dimensions!
And once you have that model of it — as a sphere with antipodal points identified — you can easily understand cool things like the Dirac belt trick, and why an electron needs to be rotated twice to return to its original state:
https://www.gregegan.net/APPLETS/21/21.html

2025-12-04 13:57:15 GMT

A grim, but fascinating, unsolved mystery. A mostly peaceful and prosperous farming culture known as LBK spread across Europe from Anatolia, from 11,000 to 7,000 years ago. When it eventually came to an end, in many places the transition appears to have been unremarkable, but in some sites there are vast mass graves, and other signs of strange tensions arising in various settlements.
https://www.science.org/content/article/headless-bodies-hint-why-europe-s-first-farmers-vanished

2025-12-04 13:26:01 GMT

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@nprofile…8eua @nprofile…y0ah @nprofile…l4xy @nprofile…4ndu
At the linked post, it says:
“range of y coordinates is about +/-5537521000”

2025-12-01 08:34:40 GMT

Great writeup from @nprofile…26f8 of an extraordinary hypothesis: the reason the two sides of the moon are so different might come down to tidal locking when the moon was still forming, and chemical gradients in the proto-lunar material due to radiant heat from the Earth!
https://bigthink.com/starts-with-a-bang/moon-two-faces-different/

2025-12-01 03:17:22 GMT

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@nprofile…9s72 You might find this interesting. There probably isn’t a consensus, but the study discussed here suggests that some very large gas clouds might have directly collapsed into black holes at a very early time.
https://bigthink.com/starts-with-a-bang/first-supermassive-black-holes/

2025-11-28 06:22:36 GMT

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@nprofile…w3j3 Sadly it’s not a joke, but if it’s any small consolation, “Scientific Reports”, though it comes under the Nature umbrella, is not “Nature” as such.

2025-11-28 04:01:40 GMT

Is there any word-producing thing in the universe that understands even less about autism than RFK Jr and Andrew Wakefield?
Amazingly, yes: whatever 90% chatbot / 10% human chimera cranked out this article about, err, AIs diagnosing autism.
H/T feralrobots on BlueSky.
https://www.nature.com/articles/s41598-025-24662-9
https://media.mathstodon.xyz/media_attachments/files/115/625/318/221/683/648/original/51f39be9eba650d7.webp

2025-11-27 08:01:16 GMT

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@nprofile…5gem No, they’re not that magical; you can only split them into measurable parts.

2025-11-27 07:13:54 GMT

Black Friday Sale! Our 1,000-dimensional magic golden spheres[*] are discounted by a whopping 50%!
* Discounted 1,000-dimensional magic golden spheres are 99% the radius of normally priced spheres, which may result in a small decrease in magicness and total gold volume.

2025-11-26 22:56:43 GMT

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@nprofile…8eua One thing that confuses me is that (unless I’m mistaken) the standard model gives accurate predictions of when and how leptons in the heavier generations decay into lighter versions, and yet the symmetry group underlying the generation structure remains unknown.
I thought the coupling that enables a decay like this tells you what gauge boson is involved, and if you know what gauge boson is involved, you know the symmetry group. Obviously I’m wrong about something major here! Can you tell me what I’m fundamentally wrong about?

2025-11-26 22:23:08 GMT

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@nprofile…ylz3 By making the internet work again, I mean restore access to all the static resources that would still have been useful without any ongoing edits, e.g. the last pre-apocalypse version of Wikipedia, and various web sites covering technical issues that she would benefit greatly from learning something about. Potentially, she could also maintain direct contact with those other survivors she hadn’t completely alienated.

2025-11-26 13:11:14 GMT

FFS, Carol, if the Hive Mind really will do anything you ask, tell it to get the internet working again.
#Pluribus

2025-11-26 11:35:06 GMT

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For an account of how Newton reached this result (and also a different way to obtain the same formula, using quadrupole moments):
https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html#NPG

2025-11-25 13:11:37 GMT

St Peter: You studied Newtonian gravity, right?
Me: Umm, a bit.
SP: Explain to me, like I’m 5, why the ratio of polar gravity to equatorial for a uniform oblate spheroid is approximately 1 + 1/5 the ellipticity of the meridians.
Me: You let Isaac set the questions, didn’t you?

2025-11-24 14:26:16 GMT

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@nprofile…eqr8 I’ve switched off all the explicitly AI nonsense in MacOS, but Preview still has some over-excitable “data detectors” that think any sequence of digits must be a phone number that needs to be added to your contacts.

2025-11-24 07:38:21 GMT

One mouse click away from butt-dialling the Big Bang.
https://media.mathstodon.xyz/media_attachments/files/115/603/525/026/534/969/original/f2d3765109aab32f.png

2025-11-24 04:28:43 GMT

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@nprofile…u6hy I’m still a bit curious as to why you mentioned credit cards! I guess it can’t be a coincidence that they have exactly the same aspect ratio as a sheet of foolscap, and in the imperial measurements the specified sizes are exactly 4:1.
I’m used to all the cool relationships between the sizes of paper in the ISO 216 standard, but I never knew about the credit card / foolscap link.

2025-11-21 09:00:22 GMT

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I learned about “mirror spiders” from here (despite living in Australia I’d never heard of them before), and here’s an open access paper in Nature that dives into the mechanism:
“Animals precisely control the morphology and assembly of guanine crystals to produce diverse optical phenomena in coloration and vision.”
https://www.nature.com/articles/s41467-023-35894-6

2025-11-21 03:59:04 GMT

A huge, beautiful library of tessellated materials in biology!
https://tessellated-materials.mpikg.mpg.de/collection
And an interview in Science with the creators of this library:
https://www.science.org/content/article/why-does-biology-keep-building-things-out-tiles

2025-11-20 22:33:52 GMT

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Some embarrassing Grok responses like this are vanishing from X right now. As of this moment, this one is still online, but it will probably be deleted eventually.
FWIW, here’s someone who independently verified the existence of the screenshotted Grok response earlier, in case you’re wondering if it was faked.
https://mathstodon.xyz/@[email protected]/115582077236839502

2025-11-20 12:34:06 GMT

Who could have guessed that the end product of a fearless, intellectually rigorous research program whose sole aim was to produce an LLM with the most reliable, objective and trustworthy responses possible would sound so much like a sycophantic courtier flattering a demented, narcissistic monarch?
https://media.mathstodon.xyz/media_attachments/files/115/582/015/043/225/242/original/b177cba3d28214d5.png

2025-11-20 12:08:40 GMT

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@nprofile…e4ry I lived without a washing machine for about 20 years, soaking/handwashing everything. That started when I was living in tiny rented flats with no laundry, and I was too lazy and/or poor to go to a laundromat, but once I was used to it, it was just part of the daily routine.

2025-11-18 04:10:58 GMT

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@nprofile…nzfz @nprofile…8eua I also found that puzzling. For some kinds of neuropharmaceutical testing in organoids, you would definitely want human neurons, but for biocomputing per se it seems irrelevant.

2025-11-17 13:03:57 GMT

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@nprofile…8eua
Maybe he’s decided that the future is human neurons in dishes.
“Biocomputing advocates claim that these systems could one day rival the capability of artificial intelligence and the potential of quantum computers.”
[There’s some genuinely interesting research here, but also a lot of ridiculous hype, like this quote.]
https://www.nature.com/articles/d41586-025-03633-0

2025-11-16 12:36:25 GMT

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@nprofile…gfh6 This looks more or less identical to waving a magnet in front of a 1980s CRT television tuned to a station that is currently off the air.

2025-11-15 14:28:40 GMT

Google’s “AI overview”, pointlessly appended to their usual weather forecast for my suburb:
“The weather in XXX is currently sunny with a temperature of about 17° C ... ”
The forecast already states the current temperature, but no, it is not “currently sunny” as it’s 10:23 pm.
The forecast by itself is perfectly understandable. It does not need a summary, an overview, an interpretation, a Reader’s Digest Condensed Version.
FFS, Google, is this what you’re building new data centres and buying billions of dollars worth of GPUs to do?

2025-11-12 23:33:50 GMT

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@nprofile…xprf
You can see the relationship of the focal conics to the surface in this construction:
https://mathstodon.xyz/@gregeganSF/115468415319176584

2025-11-12 23:30:59 GMT

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@nprofile…xprf
The simplest definition is that two ellipsoids with the same centre, and with their axes aligned, are confocal if the squares of their semi-axes all differ by the same amount.
For an ellipse with semi-axes a and b, the distance f from the centre to the foci is given by:
f^2 = a^2 - b^2
so adding the same quantity to both a^2 and b^2 preserves the foci.
So algebraically this generalises that relationship: we add the same quantity to a^2, b^2, c^2 for the ellipsoid.
Geometrically, you can also identify “focal conics” of an ellipsoid that remain the same under this transformation.
https://en.wikipedia.org/wiki/Focal_conics

2025-11-12 12:40:11 GMT

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[3] The gravitational field at a given point outside an ellipsoid will be identical for any two ellipsoids that have the same mass, and that are “confocal” with each other.
Why? It turns out that if you map points between two confocal ellipsoids with a map T that rescales the three coordinates in proportion to the ratios between the ellipsoids’ axes, then for any points P and Q on the same ellipsoid you have:
dist(P, T(Q)) = dist(T(P), Q)
If we take two confocal ellipsoids with the same density and dissect them into rectangular rods, like AB and T(A)T(B) in the diagram, the component of the gravitational force those rods exert, in the direction of their own axes, at the points T(P) and P will depend only on the cross-sections of the rods and the distances of the endpoints of the rods from the point where the force is measured.
But the distances from the endpoints are the same! So we have a simple relationship between the force that the ellipsoid E_1 exerts on an exterior point T(P), and the force that the ellipsoid E_2 exerts on the interior point P.
Combined with the linearity of T and the linear force law in the interior, we end up with a force at the exterior point that is independent of the particular ellipsoid confocal with E_1 that we choose, and simply scales with its mass.
More details at: https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html
https://media.mathstodon.xyz/media_attachments/files/115/536/769/652/855/421/original/3f3c229fc5d69a0a.png

2025-11-12 12:39:10 GMT

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[2] Inside a solid ellipsoid, the components of the gravitational force along each of the three axes is proportional to the distance from the centre along the corresponding axis ... but unlike a spherical ball, the constants of proportionality will differ for the three axes:
g = –(A x, B y, C z)
We can prove this by looking at two double cones through a point P = (x, y, z). As mentioned upthread, the gravitational force from the material in a cone will be proportional to the length of the cone, and so the x-component of the force will be proportional to the *extent* of each cone along the x-axis, Δx. If the two double cones differ by flipping the y- and z-components of the direction of their axes, the sum of all four values for Δx will be the sum of both roots of two quadratic equations, and the flip in y and z will cause the final answer to simply be proportional to x, with no other dependence on the coordinates of P.
So when we cover all the material in the ellipsoid by integrating over all such cones, the total remains proportional to x, and we can reach the same kind of result for the y- and z-axes.
https://media.mathstodon.xyz/media_attachments/files/115/536/764/746/169/630/original/067b84d25b60aca1.png

2025-11-12 12:37:43 GMT

Three fun things about the gravity of ellipsoids.
[1] There is no gravitational force inside an ellipsoidal shell of uniform density ... but only if it’s the *right kind* of shell, known as a homoeoid. The inner and outer boundaries of such a shell must be concentric, axis-aligned, scaled copies of each other, rather than the shell having uniform thickness.
If you take a spherical shell whose boundaries are concentric spheres, it’s clear by symmetry that any straight line through a point P in the interior will intersect the shell along two segments, AB and CD, of equal length. But we can turn such a shell into a homoeoid with a linear transformation, and a linear transformation rescales all parallel line segments by the same factor, so AB and CD will also be equal for any homoeoid.
Now, consider a double cone with its apex at P. The amount of material in slices of each cone at a distance r grows with r^2, exactly cancelling the 1/r^2 diminution of gravity, so the attraction at P from the material will (in the limit of an infinitesimal cone) be proportional to the lengths AB and CD. And since AB and CD are equal, there will be equal and opposite forces at P, cancelling each other out.
https://media.mathstodon.xyz/media_attachments/files/115/536/755/350/776/636/original/98f0604b128bfef2.png

2025-11-06 13:17:16 GMT

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Every parallelepiped that you place around an ellipsoid whose faces are tangent to the ellipsoid at their centres has the same volume for a given ellipsoid: 8 a b c, where a, b and c are the semi-axes of the ellipsoid.
It’s not hard to prove that this generalises to n dimensions.
Suppose we have an n-ellipsoid given by the equation:
∑ᵢ₌₁ⁿ𝑥ᵢ²/𝑎ᵢ² = 1
where the 𝑥ᵢ are coordinates (x,y,z,...) and the 𝑎ᵢ are the semi-axes in each direction.
A vector normal to the surface can be found from the gradient of the left-hand side of this equation; since that function is constant on the surface, the gradient is orthogonal to the surface.
norm(𝑥) = (𝑥ᵢ/𝑎ᵢ²)
Suppose we choose n points on the surface of the ellipsoid, 𝑥ˢ, s=1,...n, such that:
𝑥ᵗ·norm(𝑥ˢ) = 0 when t≠s
We will also have:
𝑥ˢ·norm(𝑥ˢ) = ∑ᵢ₌₁ⁿ𝑥ˢᵢ²/𝑎ᵢ² = 1
If we define the matrix 𝑋 to have the vectors 𝑥ˢ as its rows, and the matrix 𝐴 to have diagonal elements 1/𝑎ᵢ² and zeroes elsewhere, then the dot products above correspond to:
𝑋 𝐴 𝑋ᵀ = 𝐼
If we take the determinants of both sides of this equation, we have:
det(𝑋)² det(𝐴) = 1
So det(𝑋) will always be the same, the product of all the semi-axes, and the volume of the parallelotope that encloses the ellipsoid will be 2ⁿ det(𝑋), with each of its faces formed by taking ±𝑥ˢ as the centre and adding ±𝑥ᵗ, for all the choices of t≠s. Since those 𝑥ᵗ will be orthogonal to norm(𝑥ˢ), each face will be tangent to the ellipsoid at its centre.
https://media.mathstodon.xyz/media_attachments/files/115/502/938/114/861/700/original/6f8b8446923a5d3c.mp4

2025-11-06 13:15:32 GMT

Every parallelogram that you draw around an ellipse whose sides are tangent to the ellipse at their midpoints has the same area for a given ellipse: 4 a b, where a and b are the semi-axes of the ellipse.
https://media.mathstodon.xyz/media_attachments/files/115/502/934/309/988/377/original/51124b70418cdf9b.mp4

2025-11-04 12:15:59 GMT

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@nprofile…9s72 I’m finding it mostly pretty silly so far, but Emma Thompson plays an interesting character and I’ll probably stick with it and see if there’s enough entertainment value to compensate for the implausibility. (It’s based on a book by Mick Herron, who wrote the books on which the series “Slow Horses” is based, so if you enjoyed that you might like this as well ... though the intelligence agency portrayed in “Cemetery Road” seems to be ten times more cartoonish than the version of MI5 in “Slow Horses”.)

2025-11-04 05:56:21 GMT

It’s old news that most screenwriters are clueless about scientists and mathematicians, but you’d hope they'd be a *little* more familiar with what makes a good humanities student.
“Down Cemetery Road”: “She was brilliant! She memorised The Wasteland, including the footnotes!”

2025-11-03 12:41:20 GMT

It’s nice that an expert on aerosols has won the 2025 Prime Minister's Prize for Science … but the govt still can’t be bothered acting on her expertise.
"The reality is that there are no regulations for indoor air quality," she said.
https://www.abc.net.au/news/science/2025-11-03/lidia-morawska-prime-ministers-science-prize-air-quality/105959260

2025-11-03 08:00:30 GMT

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@nprofile…esx5 I liked it a lot! I was a bit afraid at first that we’d be stuck with Julia Garner just spiralling into alcoholism as the town shunned her ever more cruelly, but once they unleashed ... no spoilers, so I’ll just say “the thing that comes next” ... I really started to enjoy it.

2025-11-01 01:53:25 GMT

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The portions of the string where it departs from the hyperbola or the ellipse both lie on cones whose axis is a tangent to the curve at that point, and which make an angle with the tangent that is the same as the adjacent segment of the string.
This is a geometrical way of saying that putting tension on the string won’t make it slide along these curves, because the sum of the distances of the adjoining segments is a local minimum, so it can’t offer up slack by moving the point.
https://media.mathstodon.xyz/media_attachments/files/115/471/939/104/536/505/original/c0187bc3591e62ff.png

2025-11-01 01:52:31 GMT

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Here is a version where the point on the ellipse is held still while the point on the hyperbola is swept along it.
https://media.mathstodon.xyz/media_attachments/files/115/471/937/273/526/761/original/57919493d79dec75.mp4

2025-10-31 10:56:32 GMT

Most people know how to draw an ellipse by pinning two ends of a string to a board and sweeping a pencil around inside the string, keeping it taut.
But what about the 3D equivalent?
Start with an ellipse (blue) and a hyperbola (green) in orthogonal planes, with each curve’s vertices (extreme points) being the other’s foci.
Pin one end of a string (red and black) to one focus of the hyperbola, and the other to the opposite focus of the ellipse, and then pull the string taut, while also requiring it to pass through whatever point on each curve minimises the distance. The point the string reaches in between will trace out an ellipsoid!
This construction was found in the late 1800s by the German mathematician Staude (Wikipedia says the famous physicist Maxwell also made a start on the problem a bit earlier). I haven’t read Staude’s proof, but you can read a modern treatment here:
https://arxiv.org/abs/1707.01233v1
https://media.mathstodon.xyz/media_attachments/files/115/468/412/937/841/071/original/12af8952546697aa.mp4

2025-10-30 23:56:25 GMT

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@nprofile…9s72 Yes, but more of what?
The trouble with this kind of second-hand, context-free quote is that the speaker is not around to stop people twisting their words. I’m not accusing you of doing that, but honestly I hate the whole cottage industry of mining famous people’s utterances and dredging up fragments of speech or writing that they never intended to be treated as epigrams as if they are some kind of distillation of wisdom.

2025-10-28 22:15:21 GMT

Since the Harvard Astronomer is getting excited about the Wow! signal now, this is worth boosting:
“[...] We report the detection of narrowband signals (10 kHz) near the hydrogen line similar to the Wow! Signal, although two-orders of magnitude less intense and in multiple locations. Despite the similarities, these signals are easily identifiable as small interstellar clouds of cold hydrogen (HI) in the galaxy. We hypothesize that the Wow! Signal was caused by a sudden brightening of the hydrogen line in these clouds triggered by a strong transient radiation source, such as a magnetar flare or a soft gamma repeater (SGR). A maser flare or superradiance mechanisms can produce stimulated emission consistent with the Wow! Signal. Our hypothesis explains all observed properties of the Wow! Signal, proposes a new source of false positives in technosignature searches, and suggests that the Wow! Signal could be the first recorded event of an astronomical maser-like flare in the hydrogen line.”
https://arxiv.org/abs/2408.08513

2025-10-28 13:50:42 GMT

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@nprofile…02fp @nprofile…8eua
https://authorsguild.org/advocacy/artificial-intelligence/anthropic-settlement-faq/
“Does the settlement mean that Anthropic can now continue to use pirated books to train AI?
No. The settlement does not give Anthropic—or any AI company—permission to use pirated books going forward. It only resolves Anthropic’s liability for past use of books. In fact, the agreement requires Anthropic to destroy all copies in its possession.”

2025-10-28 11:34:16 GMT

“The grand mental panorama that was thus created was so intoxicating that those who followed Poincaré were not to recover from its pursuit.”
— S. Chandrasekhar, in “Ellipsoidal Figures of Equilibrium”, on the “fission theory” of double stars that turned out to be misconceived.

2025-10-28 06:28:24 GMT

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@nprofile…ssrx I do still get those now and then, but it’s only about 3 or 4 a year.

2025-10-28 03:18:37 GMT

The NO JUNK MAIL sign on my letter box is mostly respected, so I don’t get much paper advertising these days, but I was amused when a parcel from an American 2nd-hand bookseller arrived with my purchases … plus a Doordash coupon and a flyer from a U.S. burial insurance company.

2025-10-27 23:03:55 GMT

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@nprofile…4k4l @nprofile…8eua
I wrote “For b>3√3 light can’t enter the black hole ... ”
That is true for light that enters from r > 3M, but as r approaches 2M the ceiling rises again. The maximum value of b at any r > 2M is:
(r/M) / √(1 - 2M/r)
which has a minimum of 3√3 at r = 3M, but in the limit of light emitted at the horizon, there is no upper bound on b.
So my calculation of the change in φ for b = 3√3 isn’t really relevant for the scenario where people who have just fallen through the horizon emit light rays.
The relevant calculation is for b = ∞, which gives Δφ = π as you go from the horizon to the singularity, as Phil said.

2025-10-27 15:03:00 GMT

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@nprofile…4k4l @nprofile…8eua
The differential equation for φ as a function of u=M/r for a null geodesic with impact parameter b M is:
dφ/du = ±1/√[1/b^2 + (2u–1)u^2]
This is derived as eqn (21) on this page:
https://www.gregegan.net/PLANCK/Tour/TourNotes.html
but it is equivalent to Misner, Thorne and Wheeler’s eqn (25.27).
For b>3√3 light can’t enter the black hole, but for any b<3√3 this describes a possible null geodesic.
I haven’t been able to find an analytic expression for Δφ as u goes from 1/2 (i.e. the event horizon) to ∞ (the singularity) for the limit as b approaches 3√3 from below, but numerical integration seems to put it at around 2.63 radians.

2025-10-26 10:37:48 GMT

Oh, just a caterpillar that keeps all the husks it shed from its head when it was smaller as a kind of elaborate hat … because why wouldn’t you?
Congratulations to Georgina Steytler, who just won a wildlife photography award for this extraordinary image!
https://www.abc.net.au/news/2025-10-26/five-headed-caterpillar-wins-photo-award/105926608
https://media.mathstodon.xyz/media_attachments/files/115/440/025/034/871/675/original/8b949bc85a561f62.png