Take a 600-cell in 4D with 120 vertices ±v_i, i=1…60
Define F(x)=𝚷_i v_i·x
Stereographically project F from a 4-sphere to 3D, then Fourier-transform from momentum space to position space.
Result: this hydrogen wave function, with energy level n=61.
https://gregegan.net/SCIENCE/SymmetricWaves/SymmetricWaves.html
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Reviews said “Prime Target” would be dumb fun, but it starts with a gratuitous scene of Iraqis dying in an explosion, and when it gets to the maths no amount of Ramanujan taxicab-number jokes and people feverishly scrawling Wilson’s Theorem can make up for the sheer inanity of every match for “prime” in the Cambridge library’s collection turning red and disappearing from the results page right before the hero’s eyes.
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We can find that angle with a simple formula from spherical trigonometry:
https://en.wikipedia.org/wiki/Spherical_law_of_cosines
cos C = –cos A cos B + sin A sin B cos c
where A, B, C are the angles at the corners of a spherical triangle, and c is the angle subtended at the centre of the sphere by the side of the triangle opposite the corner with angle C.
For our purposes:
c is the angle θ between the pole and the points we rotate around.
A is either π/N or π–π/N, depending on whether T and R^{–1} rotate in the same direction.
B is α/2.
C is an integer multiple of π/N, half the angle we want T R^{–1} to rotate by.
For the example we illustrated, with N=3 and θ=π/4, if we look for solutions where T rotates clockwise, we end up with the equation:
1 + cos(α/2) – √(3/2) sin(α/2) = 0
which is solved by:
cos(α/2) = 1/5
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There are a couple of ways we could do this. One would be to explicitly compute the matrix for
T R^{–1}
and find its trace, the sum of its diagonal elements. Any rotation matrix in 3D by an angle β has a trace of:
1 + 2 cos β
So if we solve:
trace(T R^{–1}) = 1 + 2 cos 2πp/N
for α, for any integer p, that value for α will do the job.
But if computing those matrices sounds like too much work, there’s an even simpler method. We can combine the two rotations, T and R^{–1}, using the “mirror trick”.
Any rotation by an angle β is equivalent to two successive reflections, by a pair of mirrors whose planes meet along the axis of rotation, and are separated by an angle of β/2. We are free to rotate the two mirrors around the axis in any way we like, so long as we keep the correct angle between them.
This means that if we want to combine two rotations around two different axes, we can choose the second mirror for the first rotation, and the first mirror for the second rotation, to lie in exactly the same plane: the plane that contains both axes of rotation.
But two successive reflections in the same plane cancel out. That means the result of performing the two rotations we’re combining is now equivalent to reflections in the first mirror of the first rotation and the second mirror of the second rotation, and the angle of rotation is just twice the angle between those mirrors!
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Pick N points on a sphere, equally spaced on the circle at an angle θ from the pole.
Rotate the sphere around each of the points in turn by the same angle, α.
What value should α be, in order for the net rotation to restore the sphere to its original orientation?
Although this problem is very symmetrical, it’s not so symmetrical as to be trivial! It might be tempting to think that the symmetry guarantees that the net rotation will always be around the pole, but that’s not true: because we rotate around each of the N points in a particular order, they are not all treated equally, which breaks the symmetry and means the overall axis of rotation need not be the pole.
But here’s a sketch of one way to find the solution in general.
Suppose we call a rotation around the first point T, and let R be a rotation by 2π/N around the pole.
Then the rotation around the (k+1)st point is:
R^k T R^{–k}
because R^{–k} moves the (k+1)st point to the first point, where we perform a rotation by the same angle in each case, and then R^k moves the first point back to the (k+1)st point.
This means the net result of performing all N rotations is:
R^{N–1} T R^{–(N–1)} R^{N–2} T R^{–(N–2)} ... R^2 T R^{–2} R T R^{–1} T
If we cancel positive and negative powers of R that are multiplied together, and note that
R^{N–1} = R^{–1}
this becomes:
R^{–1} (T R^{–1})^{N–1} T
Now, suppose this is equal to the identity matrix, I. We then have:
R^{–1} (T R^{–1})^{N–1} T = I
(T R^{–1})^{N–1} = R T^{–1} = (T R^{–1})^{–1}
(T R^{–1})^N = I
So we need to make the product of the rotation by α around the first point, and a rotation around the pole by –2π/N, into a rotation by an angle that is (any integer multiple of) 2π/N.
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If you advance one body-length for each fish currently at successive clockwise positions on the central quadrilateral, doing 4 such moves does not restore the colouring.
It takes 12 moves to restore everything, but the 6 in this loop gives a 180° turn, a symmetry of the tiling.
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If you let one row of fish follow their nose in Escher’s woodcut “Circle Limit III”, all the rest are forced to swing around wildly.
The fish here are strung out along “hypercycles”: curves in the hyperbolic plane that are equidistant from a given hyperbolic line, but no two hypercycles here share the same line.
Escher’s woodcut shows the “Poincaré disk model” of the hyperbolic plane. In the “hyperboloid model”, the hyperbolic plane is the set of points (x,y,t) in 3-D spacetime such that:
x^2 + y^2 – t^2 = –1
In this model, if you pick a spacelike vector V and slice the hyperboloid with a plane that is orthogonal to V, then:
• if the plane passes through the origin, you get a hyperbolic line;
• if the plane does NOT pass through the origin, you get a hypercycle equidistant from that line.
The translations of the hyperbolic plane shown in the movie, which preserve one particular hypercycle, correspond to boosts in the 3D spacetime that preserve the vector V.
If any other hypercycle in the woodcut came from a plane with the same normal, then it too would be unchanged by the boosts.
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I only saw “Inland Empire” once, and I found it frustrating, but it had the most compelling sense of sustained dream logic I’ve ever encountered. Having rewatched my absolute favourites many times (“Eraserhead”, “Blue Velvet”, “Mulholland Drive”), I should go back into the woods.
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Go home Safari, you’re drunk.
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Somewhat heartened to learn that “The Brutalist” has a 15-minute intermission, which really ought to be a binding legal requirement for any 215-minute film.
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Joan Quigley discussed her relationship with Nancy Reagan in a book, titled What Does Joan Say?
“Not since the days of the Roman emperors, and never in the history of the United States presidency, has an astrologer played such a significant role in the nation's affairs of State.”
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Every time I buy new earphones I hear new things in old tracks. Even the $10 ones I just bought from a supermarket because the cable in the last pair broke seems to reveal subtleties I never noticed before. Maybe it’s just a slight shift in EQ, not superior fidelity, but still …
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A beautifully clear analysis by @startswithabang of recent observations — in real time, across the electromagnetic spectrum — of a tidal disruption event, followed by the emission of relativistic jets of debris from around a supermassive black hole.
https://bigthink.com/starts-with-a-bang/supermassive-black-hole-caught-turning-on/
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They should add a footnote to the “Squid Game” subtitles giving the exchange rate from Korean won to the viewer’s currency. (It’s about $0.0011 Australian, so their billion is close to our million.)
Also … those votes to keep playing are so grimly believable it’s just crushing.
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Here is Escher’s “Circle Limit III” tiling applied to Dini’s surface, a shape built from helices that also has the same geometry as a portion of the hyperbolic plane.
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But they’re neither! The underlying symmetry can be seen in this tiling with triangles of angles π/2, π/8 & π/3. The “fundamental domain” of the Escher woodcut comes from 4 of these triangles.
The white curves are not hyperbolic lines. Rather, they are “hypercycles” or “equidistant curves”, which lie at a constant distance from true hyperbolic lines. The red curve is a true line, from which the white curve is equidistant.
https://en.wikipedia.org/wiki/Hypercycle_(geometry)
If that sounds strange, it’s because there’s nothing entirely analogous to it in Euclidean geometry: a curve equidistant from a straight line is just another straight line. And in spherical geometry, a curve equidistant from a great circle is a smaller circle.
A hypercycle is like a circle, in that it has a constant curvature along its length. But there is no point within the hyperbolic space itself from which it is equidistant.
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Escher learned about hyperbolic geometry from Coxeter, and Coxeter in turn wrote about “Circle Limit III”:
https://www.jstor.org/stable/1574078
That article made it much easier to repeat Escher’s construction, which I’ve done below.
At first glance, this image is very confusing, because it looks as if the white curves might be hyperbolic lines, and/or axes of symmetry.
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In Escher’s woodcut “Circle Limit III” the infinite hyperbolic plane (drawn as a finite disk) is tiled with interlocking fish. The tractroid is a surface in Euclidean space with the same geometry as a portion of the hyperbolic plane, so I thought it would be fun to show the fish.
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TIL that animals’ spots might play as much of a role in thermoregulation as camouflage.
The spots on a giraffe aren’t just superficial pigment; they mark regions of tissue vascularised by specific arteries, and blood flow to them can control heat loss.
https://www.abc.net.au/listen/programs/what-the-duck/summer-why-do-quolls-have-spots-/104462848
Reference: https://pubmed.ncbi.nlm.nih.gov/36790779/#full-view-affiliation-1
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I’m hugely thankful my mother’s phone is too old to be infested with “Apple Intelligence”, so the fight to spare her from the stress & danger of this gibbering inanity is (for now) confined to kicking Windows Copilot in the face whenever it raises its head.
https://www.crikey.com.au/2025/01/08/apple-new-artificial-intelligence-rewords-scam-messages-look-legitimate/
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