Notices by Greg Egan (gregegansf@mathstodon.xyz)

*Edit: Relativistic momentum isn’t proportional to velocity, so reducing the momentum by 4/5 actually changes the velocity to 41.9% of lightspeed.

The exact calculation is:

p = m v/√(1-v^2) in units where c=1.

v = 1/2 gives p = m/√3
p = 4m/(5√3) gives v = 4/√(91) ≈ 0.4193

All of this is pretty obvious in the flat geometry of the plane, but it continues to hold true in the curved geometry of an expanding spacetime, with the following provisos:

• Instead of a straight line AB, we have a geodesic AB, the closest thing to a straight path through spacetime, which is the worldline of a body in free fall.

• Instead of the w vectors showing how a point would move if we rotated the plane, we consider a symmetry of the whole of spacetime that shifts everything in the direction in space that the spacecraft travelled. Because the overall size of spacetime scales with a(t), so must the vectors w, in order to slide everything along rigidly, without changing the geometry.

The spacecraft’s relativistic energy-momentum vector P is a vector of a fixed length that is tangent to its worldline, and its momentum vector p is just the projection of P into the direction we are treating as “space”.

But that is also the direction of the w vectors. So a(t) p(t) is the length of the projection of the w vectors onto a unit vector that points along the spacecraft’s worldline. And just as in the plane, it remains the same all along the geodesic!

An intergalactic spacecraft cruises past a galaxy at 50% the speed of light, and without firing its engines, travels for so long before it passes another galaxy that the length scale of the universe has increased by 25%.

At what speed does it pass the second galaxy?

In an expanding universe energy and momentum aren’t conserved, but to the extent that the universe is homogeneous and isotropic on large scales, we can use *that* to answer our question. It turns out that the spatial symmetry of the universe lets us determine that the quantity:

a(t) p(t)

*is* conserved, where a(t) is the length scale of the universe, and p(t) is the momentum of a body in free fall, measured relative to the background of the overall expansion.

So if a(t) has increased by a factor of 5/4, p(t) must have decreased by a factor of 4/5, and the spacecraft will pass the second galaxy at 40% of lightspeed.

Why is a(t) p(t) conserved? This follows from a remarkably simple geometric fact, known as Killing’s Theorem.

If we start with the geometry of the plane, suppose we draw two vectors, w_A and w_B, based at points on the straight line segment AB, which show the rate at which the points A and B would move if we rotated everything in the plane around some point.

Then the difference, w_B-w_A, will always be at right angles to AB. Why? Because rotating everything in the plane won’t change the length of AB, so the motion can only turn it, not bring its endpoints closer together or further apart.

This in turn means that the length of the projection of the two w vectors onto a unit-length vector tangent to the line segment at each point will be the same.

Breaking: DOGE team has barricaded the Sistine Chapel, refusing to release the cardinals until they agree to replace the PopeMobile with a CyberTruck.

[Serious documentary voice]: “Observers will be paying close attention to the colour of the smoke emerging from the Sistine Chapel next week, as the conclave follows a time-honoured tradition: blue smoke means the new Pope will be a boy, while pink smoke indicates a girl.”

Curious to see what ads Netflix shows me now that they know I was born in 1900.

Extreme UV lithography for microchips is wild.

“Getting to EUV light with a wavelength of 13.5 nm requires a source ... around 200,000 °C.”

This turns out to involve laser-zapping tin droplets into exploding bursts of plasma, creating shock waves similar to those of supernovae.

https://spectrum.ieee.org/euv-light-source

This is a fun list of unsolved mathematical problems, described by professional mathematicians who are working on them, ranging from the easily understood (are there any odd perfect numbers?) to some arcane questions in number theory and topology.

[One peculiar typo, at least at the time I’m posting this: y^2 = x^5 – x + 1 is obviously *not* solved by x=0, y=0. I guess they meant x=0, ±1, y=±1.]

https://www.scientificamerican.com/article/9-unsolved-mysteries-in-mathematics/

This is ingenious: astronomers have identified 138 new asteroids just a few tens of metres across by re-analysing JWST images taken for other purposes, shifting and recombining the images to mimic tracking the asteroids and synthesise an effective longer-term exposure.

https://badastronomy.beehiiv.com/p/hundreds-of-small-asteroids-found-in-the-main-belt-using-jwst

Apropos of nothing:

∅

U+2205

https://en.wikipedia.org/wiki/Empty_set

A rigid ring or sphere surrounding a single massive body will not be stable, but in the “restricted 3-body problem” where the ring/sphere is infinitesimally light, and one massive body is orbiting another, there can be some stable configurations.

For example, if the lighter of the two bodies is less than 1/9 the mass of the heavier one, a rigid sphere enclosing the lighter body can be stable.

H/T Shubhendu Trivedi

https://academic.oup.com/mnras/article/537/2/1249/7989465

Samples from asteroid Bennu contain A, G, C, T & U nucleotide bases, and 14 of 20 amino acids used by life — but while we use only left-handed versions of these molecules, Bennu has a roughly equal L/R mixture, puncturing theories that the bias on Earth came from an initial cosmic seeding.

https://www.nature.com/articles/d41586-025-00264-3

This analysis of the highly non-uniform distribution of 4-digit PINs has no big surprises, but it’s still sobering to see it spelled out in detail.

https://www.abc.net.au/news/2025-01-28/almost-one-in-ten-people-use-the-same-four-digit-pin/103946842

@Infoseepage

I saw “Killers of the Flower Moon” at the cinema, which probably helps; it’s easier to commit the time, and focus, if there are no distractions [assuming fellow audience members aren’t arseholes, which is the gamble there].

It was pretty satisfying, though I might have enjoyed it just as much as a limited series on streaming, when it could have been twice as long!

Me making plans: “Over the New Year break, I *will* find 3½ hours when I’m rested and focused enough to watch Scorsese’s ‘The Irishman’ in one sitting.”

Me in reality: “Ooh, a new ‘Wallace and Gromit’ movie from Nick Park! With ‘Cape Fear’ [and ‘Aliens’] references! Cracking!”

What counts as a domestic “robot” right now:

“Hashme told me that he expects the first release of Alfie to handle only about 20% of tasks on his own. The rest will be assisted by a Prosper team of “remote assistants,” at least some of them based in the Philippines, who will have the ability to remotely control Alfie’s movements.”

Also:

“We don’t want you to have to place as much trust in the company or the people the company hires. We’d rather you place trust in the device, and the device is the robot, and the robot is making sure the company doesn’t do something they’re not supposed to do.”

No thanks.

https://www.technologyreview.com/2024/12/23/1108466/general-purpose-robots-humanoids-ai-remote-assistants/

TIL that in 2018 Angélique Kidjo recorded an entire album covering every track of Talking Heads’ “Remain In Light”!

I haven’t heard the whole thing, but the samples here sound good.

https://www.abc.net.au/listen/programs/musicshow/vikingur-olafsson-bach-goldberg-variations-maurizio-pollini/104462508

@mpesce Not mirror life, but an alternative chemistry for the DNA bases. (People really have synthesised those alternative bases, but as with mirror DNA, they’re a long, long way from being able to build a whole organism that uses them.)

https://en.wikipedia.org/wiki/Xeno_nucleic_acid

Headlines for news articles too long for you to read in your busy day?

Why not have an LLM probabilistically “summarise” them into something even shorter that’s easier to take in at a glance, and … might or might not say the same thing as the original.

https://www.bbc.com/news/articles/cd0elzk24dno

In some previous posts, I’ve been chipping away at the question of how two objects falling into a black hole appear from each other’s vantage, and now I’ve put all the pieces together to answer a question @johncarlosbaez posed on his blog:

If a fleet of spaceships all fall into a black hole from different starting points along the same radial line, how would the other ships appear to one in the middle of the fleet?

https://johncarlosbaez.wordpress.com/2024/11/30/black-hole-puzzle/

The plot here shows the “apparent distance” of the other ships (as determined by the angle each ship subtends, if you know its size, or by its parallax if you view it from slightly different positions). It also shows where the ships were located, when the light now being received from them was emitted.

The details are quite different depending on whether we’re seeing a ship that is closer to the black hole than us, by means of “outgoing light” — light moving away from the hole — versus a ship further from the hole that we see by “incoming light”.

Because the event horizon itself is formed by the paths through spacetime traced by outgoing light rays, we see all the ships closer to the hole than us cross the event horizon just as we cross it ourself.

But the incoming light from ships that are further from the hole than us reaches us later, for any given r coordinate where it was emitted, and we hit the singularity at the centre of the hole before we see any light emitted from those ships when they crossed the horizon.