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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nos.lol

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Last Notes

2026-02-01 07:19:54 UTC

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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 UTC

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

2026-01-27 23:46:12 UTC

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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 UTC

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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 UTC

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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 UTC

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

2025-12-23 11:21:21 UTC

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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.

2025-12-11 04:52:40 UTC

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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.

2025-12-05 10:52:03 UTC

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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 UTC

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 UTC

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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 03:17:22 UTC

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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/

2025-11-27 07:13:54 UTC

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.

2025-11-24 07:38:21 UTC

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

2025-11-20 12:34:06 UTC

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

2025-11-06 13:17:16 UTC

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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

2025-11-04 05:56:21 UTC

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!”

2025-10-22 12:38:30 UTC

In March 2016, someone conjectured on math stackexchange, in a 1-point question, that “Any entire function without zero is constant”. People quickly responded with the obvious counterexample of the exponential function, and explained why the OP’s argument failed.
But in October 2025, if you do a Google search for “non-zero complex function” its AI nonsense tells you, based on that obscure math stackexchange page, that:
“... a key theorem states that any entire (differentiable everywhere) function that is never zero must be constant, often written as f(z) = e^g(z) for some other entire function g(z), notes MathOverflow and Mathematics Stack Exchange.”
FFS Google, turn off this garbage generator and just link to entire pages people can read and assess for themselves, and which (if you put some effort into making your ranking algorithms work *against* the SEO spam and the AI slop) might have a decent chance of being accurate.
https://media.mathstodon.xyz/media_attachments/files/115/417/803/997/691/123/original/6d1093c6a6131650.png

2025-10-19 12:25:28 UTC

Somewhat disappointed to find that this is *not* a single word in German:
Die Genugtuung, festzustellen, dass das schwierigere Integral als Ableitung in Bezug auf einen Parameter des einfacheren Integrals gefunden werden kann.

2025-10-19 11:57:28 UTC

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https://media.mathstodon.xyz/media_attachments/files/115/400/703/445/278/617/original/889ef75256806667.png

2025-10-19 07:12:42 UTC

“About 2:00pm on Saturday, mine workers found the object burning near a remote access road and alerted emergency services.
Police said it may be "a composite-overwrapped pressure vessel or rocket tank", from a space vessel.”
https://www.abc.net.au/news/2025-10-19/wa-space-debris-reentry-investigation/105909612
https://media.mathstodon.xyz/media_attachments/files/115/399/580/103/846/671/original/9318047705006dca.png

2025-10-15 14:16:56 UTC

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@nprofile…9s72 Thanks! I see the prices there range from US$ 38.45 to US$ 1,479.30, which is insane.

2025-10-15 14:02:39 UTC

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@nprofile…9s72 Yes, I’ve partially read The Mathematical Theory of Black Holes. But least that one was not too expensive; this one is ludicrously overpriced! So I’m just going to have to get the things I need from other people’s accounts of the same calculations in various online sources.

2025-10-15 10:36:13 UTC

Yeah, sure Amazon, whatever you say.
https://media.mathstodon.xyz/media_attachments/files/115/377/736/690/073/383/original/431deed67eb314a7.jpg

2025-10-11 13:06:19 UTC

St. Peter: So ... you claim you studied Applied Mathematics at university?
Me: [Starting to sweat] Sure.
St. Peter: So, remind me ... which one is Neumann boundary conditions, and which one is Dirichlet?
Me: I am so screwed.

2025-10-09 13:04:57 UTC

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@nprofile…yajh There are a lot of very small singular values; the smallest is about 10^{-43}, and 70 of the 166 have absolute values less than 10^{-10}.

2025-10-09 12:41:26 UTC

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@nprofile…yajh I think you might be more or less correct!
The density eigenvectors with the smallest eigenvalues are roughly spherical clumps around the centre of the sphere.
I was a bit confused by the fact that I can pretty much randomise the sample point locations and *still* get the same kind of immensely small determinant … but that doesn’t really negate your argument for the case where the sample points lie on the surface of the sphere. It’s just that any argument for other configurations would be more complicated and less intuitive.

2025-10-09 12:05:44 UTC

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@nprofile…hsm0
The determinant is algebraically non-zero. It is just truly very, very small.
No, the vector sum of the fields at the sample points is not zero.
The 83 sample points are arranged perfectly symmetrically around the x-axis, but that really has nothing to do with it. Changing the number of grid points (where the rings intersect the plane) and sample points, and moving them into either less symmetrical *or* more symmetrical configurations leaves det(T) of the same order of magnitude.

2025-10-09 11:29:48 UTC

From the department of very small numbers that are not zero.
Suppose you want to approximate the gravitational field of a sphere whose density is rotationally symmetrical around some axis, but is otherwise unconstrained. You could discretise the mass within the sphere as a number of rings, centred on that axis, each of them individually having a different uniform linear density.
The image shows a cross-section through the sphere, with each black dot one of 166 rings. The 83 red dots are points on a meridian of the sphere where we sample the gravitational field, which we can treat as a 2-dimensional vector because it will lie precisely in the plane of the meridian.
For simplicity, assume a gravitational constant of 1, and a unit sphere. There is a linear map from the 166 densities of the rings to the 2 × 83 = 166 components of the gravitational field vectors at the 83 sample points:
T (densities vector) = gravitational field vector
What would you guess the order of magnitude of det(T) to be, if I tell you that -26.9302 ≤ 𝑇ᵢⱼ ≤ 23.9687?
Of course if T is degenerate, det(T) will be precisely zero. But why would it be degenerate?
The answer is:
det(T) ≈ -6.28 × 10⁻¹⁷⁶⁷
When I saw that, I thought: “Oh, T must be degenerate, that’s a numerical error for a true value of 0.”
But it’s not an error! It’s robust even when recomputed at very high precision.
So T is *not* degenerate.
https://media.mathstodon.xyz/media_attachments/files/115/343/974/246/979/451/original/5f983d7c2d017e2f.png

2025-10-07 08:06:28 UTC

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@nprofile…q8c7 Is that analogous to an Erdős number? What does “of the second type” mean?

2025-10-07 04:52:49 UTC

Just received author’s copies of this translation of “The Clockwork Rocket” [Book One of the Orthogonal Trilogy] from Explorer Books, a Russian-language publisher in the Netherlands.
Link to Explorer Books: https://explorerbooks.org/product/clockwork-rocket/
All about the Orthogonal Trilogy:
https://www.gregegan.net/ORTHOGONAL/ORTHOGONAL.html
https://media.mathstodon.xyz/media_attachments/files/115/331/084/833/903/227/original/dfba5ccadeb8c154.jpg

2025-10-07 04:25:47 UTC

An astonishing result, beautifully described in @QuantaMagazine: the mathematics linking scattering amplitudes in particle physics to a geometric object called the amplituhedron can be better understood by also mapping the amplituhedron to the mathematics of origami.
https://www.quantamagazine.org/origami-patterns-solve-a-major-physics-riddle-20251006/

2025-10-06 10:24:18 UTC

Cringing listening to Avi Loeb on Australian Radio National. The journalist tried valiantly to push back as Loeb compared himself to Galileo and dismissed every expert in comets and planetary science as deluded, but I wish they’d had someone qualified to debate him in real time.

2025-10-05 13:35:59 UTC

“Flow” was eerily beautiful, charming, sad, strange and hopeful.
I must admit a part of my brain uncanny-valleyed a bit at the disparity between the lush, near-photorealistic landscapes and architecture versus the blocky palette in which the animals were rendered, but in the end I managed to shut that out and go with the story.

2025-10-05 11:21:38 UTC

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@nprofile…4rgp
This was all based on Schwarzschild”s second metric, which does indeed respect the shell theorem.
https://www.gregegan.net/SCIENCE/Borehole/Borehole.html#GRRO

2025-10-05 10:22:48 UTC

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How does the proper time measured by clocks on the orbiting body and the radially oscillating body compare?
The radially oscillating clock always measures slightly less proper time than the orbiting clock.
https://media.mathstodon.xyz/media_attachments/files/115/321/059/994/060/313/original/fd6484638bf1acd6.png

2025-10-05 10:21:32 UTC

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There is an orbital radius that matches the period of the orbit to the period of the “vacuum+borehole” oscillation for a uniform ball of matter of any allowed radius (but R_ball/M must be greater than 9/4 for the pressure at the centre of the ball to be finite). The corresponding R_orbit/M is never less than 6, so that is always a physically possible, stable orbit.
https://media.mathstodon.xyz/media_attachments/files/115/321/050/457/920/736/original/958a3014891a52e0.png

2025-10-05 10:19:19 UTC

In Newtonian gravity, if you drop an object down a borehole in a ball of uniform density it will oscillate at the same rate as an orbiting body grazing the surface.
In General Relativity that perfect match is lost … unless you drop the object from higher up and raise the orbit.
https://media.mathstodon.xyz/media_attachments/files/115/321/047/061/939/550/original/c4f689649f39d32b.mp4

2025-10-05 10:06:47 UTC

Whew, some good news for a change!
Just spoke to an Afghani friend (in Australia) who said the internet is back across Afghanistan since Wednesday, and incoming flights resumed just in time for some medication he’d sent to a relative to reach her before the last of the previous batch ran out.
It was looking terrifying for a while —for three days, *all* mobile calls were cut, domestic and international, not just data over mobile.

2025-10-05 00:17:30 UTC

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@nprofile…fa4k That sounds right. Though I haven’t done any calculations for light emitted by a falling lantern, the borehole calculations that I plotted in this thread show that if you drop a material object down a borehole that goes all the way through to the other side, or shine a light from the surface of the ball down into the borehole, for R=9M/4 neither the material object nor the light will emerge at the other end of the borehole in any finite time as measured at the surface of the ball.

2025-10-04 23:54:04 UTC

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@nprofile…fa4k I guess this is fine-tuned because the matter is magically incompressible, or at least magically possessing uniform density in spite of the pressure. With any real matter, the density profile would be different, so there would have to be something else fine-tuned to avoid forming a horizon before the singularity appears.

2025-10-04 23:21:06 UTC

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@nprofile…fa4k
By “the pressure is objectively unbounded” I meant the pressure in any coordinate system where the material is at rest.
But in any case, the Ricci curvature scalar is unbounded at r=0. For M=1, R=9/4, it is:
$ \frac{512}{243}-\frac{2 \left(\sqrt{729-128 r^2}+27\right)}{9 r^2} $

2025-10-04 23:02:22 UTC

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@nprofile…fa4k The metric certainly stays differentiable. The blowups in the Riemann tensor (in some mixes of upper/lower indices, but not others) at r = 0 is due to 𝑔ₜₜ going to zero. Of course 𝑔ₜₜ goes to zero at the event horizon in the first Schwarzschild metric without anything physically pathological happening there, but in this case the pressure at the centre is objectively unbounded as 𝑟−>0 and switching to different coordinates can’t change that.

2025-10-04 06:18:57 UTC

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What if we use mirrors to bounce a pulse of light back and forth between the ends of the borehole?
For large values of R/M, the light will cycle much faster than an object in a grazing orbit. The two cycles become identical only at R/M=3, when the orbital velocity itself equals the speed of light.
For smaller values of R/M, the time measured at the top of the borehole for the light to cycle increases without bound, but by then no grazing orbit is possible.
https://media.mathstodon.xyz/media_attachments/files/115/314/440/109/650/901/original/bfefc9268f8946f1.png

2025-10-03 12:42:26 UTC

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But the proper time, measured by a clock that you dropped down the borehole, would not stretch out like that.
There are many beautiful aspects to Schwarzschild's less-famous second solution!
You can read more about these calculations here:
https://www.gregegan.net/SCIENCE/Borehole/Borehole.html
https://media.mathstodon.xyz/media_attachments/files/115/310/285/488/960/385/original/1c9035eb02920776.png

2025-10-03 12:41:36 UTC

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If we compute the rate of oscillation in the borehole versus a circular orbit, in Schwarzschild coordinate time, the upshot is that the borehole oscillations are *slower* than the orbits.
This is a bit like the time dilation effects near a black hole, where objects can take a long time to approach the black hole and return to someone waiting at a distance.
https://media.mathstodon.xyz/media_attachments/files/115/310/281/233/243/615/original/759ae3489058669d.png

2025-10-03 12:40:06 UTC

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But that’s spatial curvature. What about spacetime?
We can embed a slice that includes time in a flat spacetime with two time dimensions and one space, and see the world line of an object oscillating in the borehole as a geodesic on that slice (the red curve here).
https://media.mathstodon.xyz/media_attachments/files/115/310/276/222/490/636/original/9ff2cbb44128b6e3.png

2025-10-03 12:39:13 UTC

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The geometry of a slice through a combined spacetime containing a solid ball and the surrounding vacuum looks like this, with the part of the slice within the ball curved just like a piece of a sphere, and the rest like part of a parabola lying on its side rotated around an axis.
https://media.mathstodon.xyz/media_attachments/files/115/310/272/933/368/758/original/9aa3c421d307d606.png

2025-10-03 12:38:26 UTC

It’s a famous result in Newtonian gravity that if you drop an object down a borehole in a ball of uniform density it will oscillate at the same rate as an orbiting body grazing the surface.
But what happens in General Relativity?
The answer lies in the 2nd Schwarzschild metric!
In 1916, shortly before his death, Schwarzschild published *two* exact solutions to Einstein’s equations for gravity.
The first was for a point mass, and it’s now famous for describing a black hole.
The second was for a solid ball of uniform density.
You can find these two metrics by first writing down what an unchanging, spherically symmetric metric must look like, with two unknown functions.
You then compute the corresponding Einstein tensor, which describes aspects of the space-time curvature that need to be matched by the density and the pressure (in the radial direction, and the directions perpendicular to it) of the matter responsible for that curvature.
Setting density and pressure to zero gives the first Schwarzschild metric.
Constant non-zero density, and equal pressures in the radial and non-radial directions, give the second.
https://media.mathstodon.xyz/media_attachments/files/115/310/264/378/642/568/original/2aa99672c1fb398e.mp4

2025-09-30 15:52:27 UTC

I haven’t had to actually think about PC video connections for 20 years, so when my mother’s monitor started complaining there was “no VGA cable”, I spent 15 minutes checking, plugging and unplugging the (DVI) cable before realising the VGA input had been selected by mistake.

2025-09-30 05:08:36 UTC

2025: Hoorah, this *looks* perfect! The chatbots will do all the calculus for us!
2030: Wait ... why did that bridge fall down?
https://media.mathstodon.xyz/media_attachments/files/115/291/501/568/022/832/original/95eb7f9ddc8f1eec.png

2025-09-12 11:53:55 UTC

Doctor: I need to test you for proBNP.
Me: What? I promise you, I don't support British fascists.

2025-09-10 11:01:50 UTC

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[2/2] What's more, they are all invertible; given a pair of points in that half of Opp, we can always locate the correct red corner, reconstruct the parallelogram, and find the two black dots in Adj. This might seem ambiguous at first; there is another potential corner to use, the blue dot in each panel. How do we know which one is correct? But depending on whether the line we get from our Opp points slopes up or down, we can compare the two candidate lines that join each corner to one of the points, and pick the correct one by its position at the start or end of the counterclockwise arc running between their directions.
So, these maps tells us that:
Vol(Adj) = ½ Vol(Opp)
because by symmetry we know that the two halves of Opp have equal volume.
But together, the two sets comprise the whole unit hypercube of all pairs of points in the square:
Vol(Adj) + Vol(Opp) = 1
It follows that:
Vol(Adj) = 1/3
Vol(Opp) = 2/3

2025-09-10 11:01:18 UTC

Suppose we have a unit square, and we want to find the volumes of two 4-dimensional sets — call them “Opp” and “Adj” — consisting of pairs of points inside the square, such that the line containing both points intersects either two opposite sides of the square (Opp), or two adjacent sides (Adj).
I’ve posted previously about ways to compute these volumes by integration, but it turns out there’s another easy way: we can just map various subsets of these sets into each other.
Suppose we have two points such that the line through them cuts across one of the corners of the square, like the two black dots in each panel in the image. These pairs of black dots belong to Adj, and these four cases cover all possible pairs of adjacent sides of the square.
If we place a red dot at the horizontally opposite corner of the square to the corner we cut across, there is a unique parallelogram (if we require it to be inside the square) whose vertices are the two black dots, the red corner dot, and a second red dot that completes the parallelogram.
The second red dot and the black dot with which it shares an edge of the parallelogram now comprise a new pair of points, such that the line containing them ALWAYS intersects the two opposite vertical sides of the square. So this new pair belongs to a subset of Opp: the half of it where the opposite sides are vertical rather than horizontal.
The four maps we have described, from four disjoint subsets of Adj to that half of Opp, all preserve volume: they take a 4-dimensional region in their part of Adj to another region of the same volume in Opp.
https://media.mathstodon.xyz/media_attachments/files/115/179/654/098/055/376/original/30d4c70c9978d284.png

2025-09-07 08:52:57 UTC

How did I miss last year’s most gonzo biotech story?
Arthur “Jack” Schubarth, 81, illegally imported body parts of Marco Polo argali sheep, the world's largest, from Kyrgyzstan, and sent the genetic material to a lab to create cloned embryos, court documents say.
https://www.bbc.com/news/articles/cdenrg7wy5ro

2025-09-05 23:11:16 UTC

In a dangerous escalation in nominative determinism, several nations have acted to rename the bureaucracies governing their militaries “the Department of Victory”, “the Department of Unassailable Triumphs”, and “the Department of World Conquest Preordained by Manifest Destiny”.

2025-09-04 05:53:52 UTC

The Bertrand “paradox” is the famous observation that asking for the probability that a “random chord” on a circle spans more than 120° is meaningless until you specify exactly how you choose the chord. The Wikipedia article describes three methods that give P = 1/3, 1/2 and 1/4.
So … having played around with a lot of calculations for what happens when you pick two points inside a set and draw a line through them, I couldn’t resist asking what happens if you choose a “random chord” that way.
The answer is:
P(θ > 120°) = 1/3 + (3√3)/(4π) ≈ 0.74683
https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)

2025-09-04 01:02:03 UTC

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@nprofile…radz Ha! I just sold a story to Asimov's about Alpha-Gal Syndrome.

2025-09-04 00:58:24 UTC

Breaking: Vladimir Putin announces that from 2026, all livestock raised in Russian farms will be genetically engineered to match his histocompatibility profile.

2025-09-03 21:46:04 UTC

Sergei Drozdov has published his nice proof using hyperbolic simplexes of the necessary and sufficient condition on the radii of spheres that sit inside and outside a Euclidean simplex in any dimension.
https://www.sciencedirect.com/science/article/pii/S0166218X25004809?dgcid=author

2025-08-26 13:15:07 UTC

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But of course it wouldn’t say this without some very good reasons. Like, starting out with some perfectly sane observations, then deciding that convex hulls in four dimensions are magical corridors that defy all the ordinary rules of, umm, convex hulls in any dimension.
https://media.mathstodon.xyz/media_attachments/files/115/095/244/881/154/282/original/30197fedbdfd8c76.png

2025-08-26 12:57:46 UTC

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OK, this is genuinely hilarious.
https://media.mathstodon.xyz/media_attachments/files/115/095/178/105/249/020/original/413728ff3fc83026.png

2025-08-26 03:34:07 UTC

A great story from @nprofile…l42a about the 50-year effort to understand the physics and mathematics of the Hofstadter Butterfly.
https://www.quantamagazine.org/ten-martini-proof-uses-number-theory-to-explain-quantum-fractals-20250825/

2025-08-26 03:28:14 UTC

One formula to rule them all!
P(d, F) = 4d / ((d+1) F)
In d-dimensional space, d≥2, given a regular polytope with F faces that come in opposite pairs, if you pick two points at random from the interior of the polytope and draw the line that contains them, the probability that the line will intersect both faces of any of the F/2 opposite pairs is P(d, F).
For d=2, this applies to all polygons with an even number of edges.
For d=3, the four Platonic solids apart from the tetrahedron.
For d=4, the hypercube, 16-cell, 24-cell, 120-cell and 600-cell.
For d>4, the hypercubes and the cross-polytopes.

2025-08-25 23:26:05 UTC

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@nprofile…fa4k It turns out to be pretty straightforward to prove this in general, without worrying about the specific geometry of the (d-1)-dimensional faces of each regular polytope in R^d.
The Jacobian that gives the 2d-volume of the space of pairs of points on every line segment joining two parallel, congruent (d-1)-dimensional faces factors into three pieces, that when integrated give a product that contains the square of the (d-1)-volume of the faces, the square of the orthogonal distance between them, and some factors that only depend on d. But the total volume of the polytope, squared — which is the total 2d-volume in the denominator when computing the probability — also contains the first two factors, along with a factor of F^2.
When we multiply by F/2 to account for all the different pairs of opposite faces, we are left with k_d / F, where k_d depends only on the dimension.

2025-08-25 14:25:44 UTC

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@nprofile…fa4k Yes! I just completed the cross-polytope calculations, and they agree with the prediction.
That is, for a cross-polytope in dimension d, the probability of the line containing two uniformly random points from the interior of the polytope intersecting any of the pairs of opposite faces is:
P(cross-polytope, d) = k_d / 2^d
where k_d = 4d/(d+1)
and we also have:
P(hypercube, d) = k_d / (2d)

2025-08-25 13:15:25 UTC

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@nprofile…fa4k I did the general d-dimensional hypercube, which gives 2/(d+1).
So if the pattern k_d/n holds for d=4, we’d have:
k_4/8 = 2/5
k_4 = 16/5

2025-08-25 12:56:21 UTC

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@nprofile…fa4k
“For even F” should be “For F greater than 4, i.e. excluding the tetrahedron, which doesn’t have pairs of opposite faces.”
I did have to do it case by case! When I get a chance I’ll look more closely at the calculations and see if I can isolate the underlying reason.

2025-08-25 11:56:55 UTC

Pick two points uniformly at random inside a regular n-gon. For even n≥4, the probability that the line containing these points will pass through any pair of opposite edges of the polygon is:
8/(3n)
Pick two points uniformly at random inside a regular polyhedron with F faces. For even F, the probability that the line containing these points will pass through any pair of opposite faces of the polyhedron is:
3/F

2025-08-25 08:51:42 UTC

∑ᵢ₌₀ⁿ⁻²∑ⱼ₌ᵢ₊₁ⁿ⁻¹𝑠𝑖𝑛((𝑗−𝑖)π/𝑛)⁴=3𝑛²/16
Well, obviously.

2025-08-25 04:42:02 UTC

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@nprofile…3ek5 @nprofile…h3jy
Proof (in fact this holds even for a parallelogram):
https://mathstodon.xyz/@gregeganSF/115087563535440300

2025-08-24 14:07:47 UTC

@nprofile…h3jy In a rectangle the opposite vs adjacent probabilities are the same as for a square.

2025-08-24 13:55:10 UTC

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Pick two points uniformly at random inside a regular n-gon.
If you specify any two sides by their order, say side 0 and side j, then the probability that the line containing the two random points passes through those two sides is:
16 sin(j π/n)^4 /(3n^2)

2025-08-24 12:30:06 UTC

Pick two points uniformly at random inside a regular n-gon.
For even n≥4, the probability that the line containing them will pass through two opposite sides of the n-gon is:
8/(3n)
For odd or even n≥3, the probability that the line will pass through two adjacent sides is:
16 sin(π/n)^4 /(3n)

2025-08-24 08:32:20 UTC

Robert Musil, “A Man Without Qualities”
https://media.mathstodon.xyz/media_attachments/files/115/082/806/326/718/346/original/26bc0ba7909f3ed3.jpg

2025-08-23 21:12:58 UTC

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@nprofile…zsze Yes, I read the Python code carefully before paying any attention to the output. I don’t need Gemini to teach me anything; I only use LLMs to laugh at their mistakes.

2025-08-23 21:09:54 UTC

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@nprofile…pxhq No, “crate” isn’t meant to imply open. Crates usually come with lids that turn them into complete cubes.

2025-08-23 13:04:50 UTC

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@nprofile…dkn0 No.

2025-08-23 12:10:07 UTC

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@nprofile…jt59
OK, sorry for accusing you of using a chatbot!
Your reasoning is false for n=3, because it is not equally likely that a line through two random points inside the cube that intersects one face will be equally likely to intersect any of the other five. The other five faces are all possibilities, but the opposite face and the other four faces do not have identical probabilities.

2025-08-23 11:54:03 UTC

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@nprofile…vst6 Take it as shorthand for the local version of cubic, i.e. d-dimensional in a d-dimensional universe. But I promise d > 3.

2025-08-23 11:37:40 UTC

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@nprofile…rshq @nprofile…args
Sorry … but you really have no reason to be reasonably sure of those inequalities.

2025-08-23 11:32:10 UTC

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@nprofile…rshq @nprofile…args
See: https://mathstodon.xyz/@gregeganSF/115076414261569820

2025-08-23 11:27:39 UTC

Here’s some Gemini gaslighting:
• giving a wrong answer to a puzzle
• giving Python code that could test its claim
• asserting it obtained results from the code supporting its claim
• but when I ran the code myself, it showed the claim was false.
https://g.co/gemini/share/2188cebf58cc

2025-08-23 11:27:20 UTC

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@nprofile…jt59 No. (Maybe you asked a Chatbot, which is always a bad idea.)

2025-08-23 11:08:27 UTC

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@nprofile…nj54 No.

2025-08-23 11:02:36 UTC

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@nprofile…args The hypercubes are all definitely convex. But the line that contains the two points (extended indefinitely beyond them, not the line segment that lies only between them) might intersect two sides of the hypercube that are not an opposite pair, just as the line through two points inside a square might intersect either two opposite edges, or two adjacent edges.

2025-08-23 10:32:09 UTC

You wake up in a universe with a different number of spatial dimensions.
Your senses are confused, and you’re too embarrassed to ask someone on the street “Hey, how many dimensions does space have?”
But then you chance upon someone who likes mathematical puzzles, and ask them, “If I picked two points at random inside a hypercubic crate, what would the probability be that the line joining them intersected two opposite sides of the crate?”
They scribble on some paper for a while, then reply “One ninth.”
How many dimensions does space have here?

2025-08-23 05:25:31 UTC

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Here is a more geometrically intuitive solution.
We assume the square has both x and y coordinates ranging from 0 to 1.
Imagine sweeping a line across the square (red in the animation) so that it always joins the top and bottom edges, and one endpoint or the other is always in one of the corners of the square.
If we call the two points (x₁,y₁) and (x₂,y₂), suppose we hold y₁ and y₂ fixed, and look at the values of x₁ and x₂ so that both points lie on the red line.
In the (x₁, x₂) plane, their values sweep out a parallelogram with vertices at (0,0), (y₁,y₂), (1,1) and (1-y₁,1-y₂). The area of this parallelogram is:
|y₁-y₂|
So for any fixed y₁ and y₂, this is the probability that the line between the two points with uniformly random x₁ and x₂ joins the top and bottom edges of the square.
What is the average of |y₁-y₂| across all values of y₁ and y₂? This is just the volume of two pyramids with equal-sized triangular bases, one with y₁ ≤ y₂ and one with y₁ ≥ y₂. The height of both pyramids is 1, since that is the maximum that |y₁-y₂| reaches in both triangles. So the volume of each pyramid is 1/6, for a total of 1/3.
We multiply by two to account for the other case, where the pair of opposite edges are the left and right edges.
So P(opposite) = 2/3.
https://media.mathstodon.xyz/media_attachments/files/115/076/409/735/105/362/original/80992a3c723b9c2c.mp4

2025-08-22 13:13:09 UTC

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SOLUTION:
Take the unit square [0,1] × [0,1], and consider the two horizontal edges.
Parameterise the endpoints of a line joining those two edges as (α,0) and (β,1), then parameterise two points along that line as:
(1-λ) (α,0) + λ (β,1)
(1-μ) (α,0) + μ (β,1)
where α, β, λ, μ all range from 0 to 1.
So we have a map from the parameter space [0,1]^4 to the geometric space of pairs of points [0,1]^4:
F(α,β,λ,μ) = (β λ + α (1-λ), λ, β μ + α (1-μ), μ)
The absolute value of the Jacobian determinant of F is just:
|J(F)| = |μ-y|
The integral of this over [0,1]^4 in the parameter space gives the 4-volume in the geometric space:
∫_[0,1]^4 |μ-y| = 1/3
We multiply by two to account for the other case, where the pair of opposite edges are the two vertical edges.
So P(opposite) = 2/3.
I think this is a simpler approach than trying to specify the required subset of the geometric space in terms of a collection of inequalities in the Cartesian coordinates. My original intuition was that this subset ought to consist of a union of convex polytopes in R^4 ... but it doesn't, the boundaries are not hyperplanes!

2025-08-22 13:12:02 UTC

Here’s a fun puzzle I heard from @nprofile…3uyp
Pick two points uniformly at random in a square. What is the probability that the line that contains both points intersects two opposite edges of the square, rather than two adjacent edges?

2025-08-22 13:10:55 UTC

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@nprofile…yajh No, it’s really nothing so sinister. There is an existing Catholic college run by the Servite order, and this is just a small community school for adult education in the arts, being built on some adjoining land, which they are naming after Montosorli because he was a Servite friar as well as a sculptor.

2025-08-22 07:37:21 UTC

A letter from my local council announcing plans for a “Montorsoli school” made me wonder if it was a typo for “Montessori” … but on closer scrutiny, a religious order really is planning to name this facility after the 16th century Florentine sculptor & friar Giovanni Montorsoli.

2025-08-19 07:11:31 UTC

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@nprofile…yg3m I did it as an integral from first principles, and also using Mathematica’s boolean operations on its Region structures, and I got the same answer by both methods.
You can probably cajole Wolfram Alpha into doing boolean operations on regions as well, but I haven’t tried; I guess it takes verbatim Mathematica code, though sometimes it gets sulky because it wants you to flatter it by pretending it can interpret natural language.

2025-08-19 06:16:38 UTC

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@nprofile…yg3m Unless I made an error, it’s about 1.42364.

2025-08-19 02:33:27 UTC

Let R be a rectangle with side lengths 1 and 2, and let S be the subset of R consisting of all points x in R such that there exists a point y in R that is a distance of 3/2 from x. What is the area of S?
This is a high school-level calculus problem, but it takes a bit of care and for most people probably a sketch or two to get it right. I don’t doubt that some current LLMs can solve it correctly, but it’s still instructive to see Gemini 2.5 Pro (“Reasoning, math and code”) stitch together some relevant arguments and some correct computations and still get it wrong.
Why does it get it wrong? Because it’s stitching pieces of correct reasoning from its training set together, but has no overarching model of what the actual question here refers to. Getting it right or wrong is down to sheer luck.
https://g.co/gemini/share/83d10d221066

2025-08-16 14:02:48 UTC

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@nprofile…mq03 That’s reassuring.

2025-08-16 13:33:59 UTC

I read some reviews praising “Platonic”, and tried watching a few episodes, but it’s not my thing. But I got as far as episode 5, where Seth Rogen’s craft-beer-blah character who knows as much about physics as I do about craft beer is inexplicably wearing a T-shirt of the cover of Peskin and Schroeder’s “An Introduction to Quantum Field Theory”.
Is that really a thing, or was the costume designer just playing a strange joke? No such T-shirt showed up when I googled. Do people who actually study QFT wear Peskin and Schroeder T-shirts?

2025-08-16 12:02:27 UTC

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@nprofile…n9dm @nprofile…fa4k
What a great project, I hadn’t heard about that!
(I had a momentary freakout while contemplating their formula:
Δθ = 4 G M_sun / (R_sun c^2) 1/R
How can the amount of lensing possibly depend on the radius of the sun? If the sun contracted to half its current radius, the deflection angle shouldn’t change in the slightest.
But then I saw that they were measuring R, the distance of closest approach of the light ray to the centre of the sun, in units of solar radii. So really the formula is just 4 G M_sun / (c^2) 1/R, if you measure R in physical units.)

2025-08-16 11:23:52 UTC

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@nprofile…fa4k I love the fact that 106 years ago Eddington struggled to detect a minuscule amount of gravitational lensing by the sun during an eclipse, but these days every tenth Hubble image of a massive galaxy needs a footnote saying which parts of the picture are the actually galaxy and which are the hugely magnified image of another, far more distant galaxy that just happens to lie behind it.
When I travel back in time to visit Einstein on his death bed and console him about his unified theory not working out, I hope to cheer him up with a collection of Einstein rings, along with a compendium of gravitational wave recordings.

2025-08-15 13:38:57 UTC

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@nprofile…xy93 I think there’s plenty of blame to go around. The AI hucksters are lying and hyping in the media, but their products all come with very visible disclaimers about their limitations — disclaimers that lawyers, of all people, have no excuse for ignoring. And these cases of false citations have been widely publicised for years now. Nobody in a legal practice who hasn't had their head in the sand can be unaware that this has happened on literally dozens of occasions.