Greg Egan @gregegansf.bsky.social · 04/10/2026If you pick 3 points at random in the unit disk, the probability that they lie on a perfectly straight line is zero. However, the probability that the curvature of the circle they lie on is less than ε is, in the limit of small ε, equal to: 1024 ε / (525 π²) ≈ 0.2 ε 2270
Greg Egan @gregegansf.bsky.social · 04/10/2026… is the same knot, but which cannot be deformed into each other when embedded in either 3 or 4 dimensions. 0132
Greg Egan @gregegansf.bsky.social · 04/10/2026I don’t know what cheesy LLM these scammers are using, but the glitches can be hilarious. [Though I have written a novel called “Morphotrophic”, “Morphological Regency” sounds like a token midway between that and the average of all titles on Smashwords.] 2142
Greg Egan @gregegansf.bsky.social · 29/09/2026L’adaptation en bande dessinée de la nouvelle Le Coffre-Fort de Greg Egan, sublimée par le graphisme singulier de Clément C. Fabre. Une histoire fascinante autour de l’identité, de la mémoire et des vies que l’on traverse… 1184
Greg Egan @gregegansf.bsky.social · 14/09/2026At first I thought it was some oddly ungrammatical reference to electric fields, but no ... en.wikipedia.org/wiki/Ionosph... 160
Greg Egan @gregegansf.bsky.social · 06/09/2026In the free group generated by four letters, if you remove any two letters (and their inverse) from this product of 16 elements: 𝑎𝑏𝑎⁻¹𝑏⁻¹𝑐𝑑𝑏𝑎𝑑⁻¹𝑐⁻¹𝑑𝑐𝑏⁻¹𝑎⁻¹𝑐⁻¹𝑑⁻¹ it simplifies to 1, but dropping any single letter does not give 1. 2488
Greg Egan @gregegansf.bsky.social · 02/09/2026I’ve walked past this door (in Sir Charles Gairdner Hospital) a few times, but today I finally gave in to the urge to take a picture, because ... physics! [I once worked in the Medical Physics dept of a different hospital, so I guess this is the same kind of dept, though I’m not actually sure.] 2100
Greg Egan @gregegansf.bsky.social · 19/08/2026along with the location of the centre of the circle and the positions of the 3 points around the circle. Details at gregegan.net/SCIENCE/Thre... BTW, if you want a laugh, here is one of the results offered up by the clown car that used to be Google Search. The MathOverflow post… 150
Greg Egan @gregegansf.bsky.social · 19/08/2026If you pick 3 points at random on a unit sphere, the average radius, R, of the circle that contains them is 7/6. For 3 points on a flat disk of radius ρ, mean(R) diverges. However, the curvature, κ = 1/R, has a finite mean which can be computed exactly: 16 (3 – 16/π^2) / (15ρ) 2110
Greg Egan @gregegansf.bsky.social · 18/08/2026TIL about “cancellous bone”. I’d previously misunderstood what the solid sample taken in a bone marrow biopsy was composed of; I thought it was part of the outer, “cortical” bone. But today the doctor doing the biopsy showed me the sample and explained exactly what it was. 1140
Greg Egan @gregegansf.bsky.social · 06/08/2026The image shows two cases, where the plane through 3 points intersects the hyperboloid either in a closed curve or an open one, with the latter corresponding to a hypercycle. 150
Greg Egan @gregegansf.bsky.social · 28/07/2026After recent observations with the VLT: “While there was evidence for the existence of this companion star, including a possible direct detection with the Gemini North Telescope in Hawaiʻi, USA, this is the strongest evidence for, and clearest image yet of, Betelgeuse B.” www.eso.org/public/news/... 191
Greg Egan @gregegansf.bsky.social · 27/07/2026The Burau representation of 𝐵ₙ associates a matrix with each element of the group, in such a way that multiplying these matrices is the same as joining braids together to make new ones. 1633
Greg Egan @gregegansf.bsky.social · 27/07/2026An element of the Braid group 𝐵ₙ is a way of joining two sets of n points with strings, where two elements are the same if the strings can be moved around and/or stretched to follow the same paths in 3-dimensional space while keeping their endpoints fixed. 2792
Greg Egan @gregegansf.bsky.social · 14/06/2026... “a 3D version of cryo-EM called cryo–electron tomography” that will reveal all kinds of cellular structures in situ with molecular resolution. The picture here shows two cryo–electron microscope images of the protein apoferritin, with and without dual “laser phase plates”. 161
Greg Egan @gregegansf.bsky.social · 14/06/2026How does this work? Light passing through the sample is both scattered slightly off-course, and shifted in phase by about 90 degrees. If the sample is illuminated by an annular light source, it is possible to use the angular separation of light scattered by the sample ... 130
Greg Egan @gregegansf.bsky.social · 14/06/2026Phase contrast microscopes were invented in the 1930s by the Dutch physicist Frits Zernike, and developed and manufactured by physicist and entrepreneur Caroline “Lili” Bleeker. Zernike won the Nobel Prize for this in 1953 ... 1112
Greg Egan @gregegansf.bsky.social · 31/05/2026These are the tallies of cones for each dimension from 2 to 10. The first number in each pair is the number of vectors that generate the cone, the second is the count of such cones. 110
Greg Egan @gregegansf.bsky.social · 31/05/2026… the set of linear combinations of these vectors with positive coefficients. Then these 2^n-1 cones will all fit together to form a half-space! This is what we get for triangles. 150
Greg Egan @gregegansf.bsky.social · 31/05/2026How can we generalise the fact that the angles at the vertices of a triangle add up to 180°, to apply to tetrahedra and higher-dimensional n-simplexes? Here’s a result discovered by Benoît Bertrand and Lucía López de Medrano that I learned from Omar Antolín. 2275
Greg Egan @gregegansf.bsky.social · 27/05/2026Most likely, no human typed the word “permission” here. Rather, they misspelled “persimmon” in such a way that autocorrect considered “permission” the closest valid word. Pure spellchecking that *flagged* the typo would have alerted the user, who would have made the right choice. 1289
Greg Egan @gregegansf.bsky.social · 17/05/2026By reflecting the solid triangle in the side AB, we can see that its height from BC to A is Q'A = QA = 1. Equating area from ½bh with area from vertex coords ½det(AB,AC) gives: ½√(x²+y²) = ½(x+y-xy) Solving this gives us coords for C', which then yields: AC·AC' = 0 So ∠CAC' = 90° Red ∠CAB = 45° 110
Greg Egan @gregegansf.bsky.social · 27/04/2026A new knot invariant that “uniquely identifies more than 97% of the knots with 18 crossings. By comparison, the Jones polynomial, one of the most widely used invariants for cataloging knots, identifies about 42%, and the Alexander polynomial only about 11%.” 1346
Greg Egan @gregegansf.bsky.social · 01/04/2026A more objective answer is that the lightlike geodesics remain “equally spaced” … but the spacing itself has no observer-independent value! What does that mean? The cyan curves in this image join neighbouring geodesics with equal-length segments … 140
Greg Egan @gregegansf.bsky.social · 01/04/2026Here lightlike geodesics grow further apart, as measured along the grey circles. This is how the cosmological red shift works: wavefronts in an expanding universe move apart. But that’s a measurement by *particular observers*, at rest in galaxies that are themselves moving apart. 110
Greg Egan @gregegansf.bsky.social · 01/04/2026But the timelike geodesics behave like lines in the hyperbolic plane: if they start out parallel, they accelerate *away* from each other. The grey curves here, perpendicular to the geodesics and equally spaced along them, have a length that grows with the cosh of the distance. 110
Greg Egan @gregegansf.bsky.social · 01/04/2026The cyan curves here, perpendicular to the blue geodesics and equally spaced along them, vary in length like the circles of latitude on a sphere crossing the great circles of meridians, with the cosine of the latitude. 110
Greg Egan @gregegansf.bsky.social · 01/04/2026In 3D Euclidean space, the surface equidistant from a point is a sphere, with constant curvature. In (2+1)-dimensional Minkowski spacetime, the equidistant surface is a (1+1)-dimensional de Sitter spacetime, which also has constant curvature. 1243
Greg Egan @gregegansf.bsky.social · 25/03/2026Those staff layoffs and vibe coding seem to be really working out for Amazon ... 2318
Greg Egan @gregegansf.bsky.social · 24/03/2026This image (from www.gregegan.net/SCIENCE/Elli...) sums up for me (roughly, intuitively) why there is SO(4) symmetry for orbits: the two equal-sized vectors A and B can be rotated *independently* while preserving the semi-major axis, and hence the energy, of a Kepler orbit. And so(3)+so(3) = so(4). 1224
Greg Egan @gregegansf.bsky.social · 07/03/2026A new collection of my stories in Korean 잠과 영혼, East Asia Publishing, Seoul, 2026. Translated by Kim Sang-hoon. ISBN13 979-1193078808 Contents: “Crystal Nights” / “크리스털의 밤” “This Is Not the Way Home” / “고향으로 돌아가는 길” “You and Whose Army?” / “너 혼자서?” “Dream Factory” / “꿈 공장” 12310
Greg Egan @gregegansf.bsky.social · 05/02/2026In special relativity, observers in relative motion will disagree about various measurements. But if Alice draws any planar figure, and fires a simultaneous pulse of light from each point on it, perpendicular to the plane, any observer will agree on the figure’s shape and size. 1150
Greg Egan @gregegansf.bsky.social · 02/02/2026These are the paths of the individual ants, from the starting point until they bump into each other. 030
Greg Egan @gregegansf.bsky.social · 01/02/2026Two 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) … 1140
Greg Egan @gregegansf.bsky.social · 21/01/2026"We're trying to compete with the big boys, and part of that is you've got to keep your content refreshed and new all of the time," Mr Hennessy said. Refreshed, new and hallucinated: tourists book trips to nonexistent hot springs. www.abc.net.au/news/2026-01... 1141
Greg Egan @gregegansf.bsky.social · 21/01/2026For crying out loud, ABC news, never send a line graph to do a bar chart’s job. Link: www.abc.net.au/news/2026-01... 3274
Greg Egan @gregegansf.bsky.social · 11/01/2026Now, 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. 120
Greg Egan @gregegansf.bsky.social · 11/01/2026Gravity 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 … 2322
Greg Egan @gregegansf.bsky.social · 16/12/2025What surface arising naturally in physics is described by a degree-6 polynomial? There’s a whole 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! 1163
Greg Egan @gregegansf.bsky.social · 11/12/2025Those 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. 110
Greg Egan @gregegansf.bsky.social · 11/12/2025But 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. 120
Greg Egan @gregegansf.bsky.social · 11/12/2025Given 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. 110
Greg Egan @gregegansf.bsky.social · 11/12/2025If 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. 3151
Greg Egan @gregegansf.bsky.social · 24/11/2025One mouse click away from butt-dialling the Big Bang. 0214
Greg Egan @gregegansf.bsky.social · 06/11/2025Every 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. 130
Greg Egan @gregegansf.bsky.social · 06/11/2025Every 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. 3351
Greg Egan @gregegansf.bsky.social · 01/11/2025The 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. 120
Greg Egan @gregegansf.bsky.social · 01/11/2025Here is a version where the point on the ellipse is held still while the point on the hyperbola is swept along it. 110