Now I want to write a novel where aliens kill us by stripping away all the entanglement that causes our density matrices to have tr(Ο) < 1.
11.12.2025 13:35 β π 5 π 0 π¬ 1 π 1@gregegansf.bsky.social
SF writer / computer programmer Latest novel: MORPHOTROPHIC Latest collection: SLEEP AND THE SOUL Web site: http://gregegan.net Also: @gregeganSF@mathstodon.xyz
Now I want to write a novel where aliens kill us by stripping away all the entanglement that causes our density matrices to have tr(Ο) < 1.
11.12.2025 13:35 β π 5 π 0 π¬ 1 π 1βInitially, it will form an oblate spheroid, with two equal semi-axes shorter* than its axis of rotation.β
This should be βlongerβ of course.
But the Jacobi ellipsoid, with a greater moment of inertia I, reaches the same angular momentum, L, with a smaller angular velocity Ο β¦ and ends up with a kinetic energy, K, sufficiently smaller to make its total energy, E, the lower of the two.
More at www.gregegan.net/SCIENCE/Elli...
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!
So, a shape like this is a possible shape for βsea levelβ in a spinning body of fluid.
Note that one of the meridians *must* have an eccentricity greater than a critical value, e_crit β 0.812670. This means that the angular momentum must also exceed a threshold.
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!
11.12.2025 12:21 β π 2 π 0 π¬ 1 π 0A plot of e_xz versus e_yz, which is like one quadrant of a smoothed square. The curve is symmetrical in the diagonal e_xz = e_yz, and wherever one of the two eccentricities is greater than the critical value of 0.812, where the curve crosses the diagonal, the other eccentricity is less.
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.
Just as a=b makes a spheroid special among ellipsoids, there is a condition:
a^2 b^2 (B β A) / (a^2 β b^2) β 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.
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 *canβt* pull off this trick for a generic tri-axial ellipsoid, with any old a, b and c.
All the βequipotential ellipsoidsβ 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.
11.12.2025 12:19 β π 1 π 0 π¬ 1 π 0Given 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 β but if itβs not spinning they will not match the shape of the body.
However, by adding just enough spin the shapes can be made the same.
There are two separate issues here.
One is the question of when the βlevel surfacesβ of constant gravitational-plus-centrifugal potential for an ellipsoid can match the shape of the ellipsoid itself. What shapes can βsea levelβ take for a spinning mass of fluid?
The first configuration is known as a βMaclaurin spheroidβ, after Colin Maclaurin, who studied it in 1742.
The second is called a βJacobi ellipsoidβ, after Carl Gustav Jacob Jacobi, who in 1834 realised that a fluid could be in equilibrium in this less symmetrical state.
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 βtri-axialβ ellipsoid, with the semi-axes perpendicular to the axis, a and b, taking on different lengths!
11.12.2025 12:17 β π 6 π 0 π¬ 1 π 0If 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.
The US social media vetting for visas will be devastating for scientific and journalistic conferences, fellowships etc. No global organisation can seriously consider holding an international conference in the US while this policy exists.
11.12.2025 05:35 β π 57 π 19 π¬ 2 π 1I see your βJournal of Artificial Intelligence and Consciousnessβ and raise you my βJournal of Earth and Flatnessβ
10.12.2025 20:56 β π 53 π 8 π¬ 4 π 0My dad got shingles two years ago and is still recovering. Sometimes just sitting activates the nerve pain. Get your shingles vaccine!
09.12.2025 05:48 β π 75 π 14 π¬ 12 π 4βI brought this horrid creature into the world, and now I must take him out!β Winfrey said
09.12.2025 00:30 β π 14882 π 2737 π¬ 405 π 237This entire story is justβ¦ π€―
#giftarticle
www.nytimes.com/2025/12/07/u...
New article inspired by the recent hoopla surrounding Karen Hao's book (re: water use). I argue that "no one should look to the EA community as exemplifying good habits of epistemic and moral conduct." Here's why: www.realtimetechpocalypse.com/p/how-effect...
28.11.2025 18:56 β π 156 π 53 π¬ 4 π 8We've been regularly receiving variations on this scam for weeks now. Sometimes 2-3 in a day. Today is special though. It's only 1:15 PM and I already have 18, each for a different issue of Clarkesworld. It's likely that there are more in my spam folder.
05.12.2025 18:14 β π 100 π 22 π¬ 4 π 0Great writeup from @startswithabang.bsky.social 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!
01.12.2025 08:33 β π 41 π 5 π¬ 0 π 1Some percentage of the population consistently reports that they are reincarnations of Jesus, that the CIA are monitoring us with subcutaneously injected spy drones, and that the moon landing is a hoax. Is it time to consider whether theyβre onto something?
30.11.2025 22:38 β π 201 π 27 π¬ 15 π 2On the Factor Fexcectorn and autism bicycle AI slop study: I got an answer from Springer Nature this morning that this scientific paper will be retracted! π§ͺ
Full story: nobreakthroughs.substack.com/p/riding-the...
Figure 1 from a paper published in Nature (linked in post). The figure purports to be an infographic depiction of the "Overall working of the framework" for 'explainably' diagnosing autism spectrum disorder. The infographic is plainly nonsense, containing many spelling errors, nonsense words, meaningless images and graphs. Clearly created using generative AI and (crucially) NEVER CHECKED AT ANY STEP OF THE WRITING OR PUBLICATION PROCESS.
Hey @nature.com, have you got an explanation for how the hell THIS happened? & especially why you accepted a paper with such a bizarre piece of genAI slop in it?!
& more to the point, why we should take you seriously at all going forward?
www.nature.com/articles/s41...
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.
FFS, Carol, if the Hive Mind really will do anything you ask, tell it to get the internet working again.
#Pluribus
For an account of how Newton reached this result (and also a different way to obtain the same formula, using quadrupole moments):
www.gregegan.net/SCIENCE/Elli...
St Peter: You studied Newtonian gravity, right?
Me: Umm, a bit.
SP: Explain to me, like Iβm 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, didnβt you?