TobyBartels

Probably the only anglophone Toby Bartels. They/them. Mathematics instructor at a community college in Nebraska. Tolstoyan ex-Catholic. 🏴🚩🏳️‍🌈🏳️‍⚧️

web: https://tobybartels.name/

Public Key

npub1x4dc9the5uen6unmwyn4whsuznj6jp2a004twuvjjcarkm745ems3z6lp6

NIP-05 Address

[email protected]

Profile Code

nprofile1qqsr2kuz4mu6wveawfahzf6htcwpfedfq4whh64hwxffvw3mdl26vacpz3mhxue69uhhyetvv9ujuerpd46hxtnfduqs6amnwvaz7tmwdaejumr0dss4w0gx

Publishing to

relay.damus.io

nos.lol

Author Public Key

npub1x4dc9the5uen6unmwyn4whsuznj6jp2a004twuvjjcarkm745ems3z6lp6

Show more details

Last Notes

2026-03-03 19:26:19 UTC

- reply

That's because bad cases don't necessarily make bad law; as you said before, it's *hard* cases that make bad law. And this case was easy!

2026-02-28 21:06:24 UTC

Q1: Which is heavier: a kilogram of gold or a kilogram of feathers?
Q2: Which is heavier: a pound of gold or a pound of feathers?
Q3: Which is heavier: an ounce of gold or an ounce of feathers?
(1/3, answers in next post)

2026-02-26 05:57:08 UTC

- reply

For a moment reading that, I thought that it might have gone a lot worse than it actually did.

2026-02-22 18:28:06 UTC

- reply

Even the shapes of the written letters show this effect, which is pretty wild when you consider all of the history that those letters went through.

2025-12-25 15:22:24 UTC

- reply

I was taught to call it the minor second, not the diminished second. The Wikipedia page that you linked agrees with me, calling the diminished second ‘enharmonically equivalent to a perfect unison’ in equal temperament.
But the question isn't (or at least shouldn't be) whether ‘major’ and ‘minor’ are good terms for these two modes once you've picked them out, but why the two terms should be used for *those* two modes. The truly minor mode is phrygian.

2025-12-24 17:00:55 UTC

- reply

We use those names in English as well, except with Ti instead of Si. (If you know your English-language musicals, then you may have heard them used in The Sound of Music.) But the difference between these and the letters is that the letter notes are at (or at least near) specific frequencies, while solfege (as we call it) is relative. That is, A is always about 440 Hz (modulo a power of 2), but La could be anywhere (with the others arranged at the same relative frequencies).

2025-12-24 16:48:43 UTC

- reply

In the picture, you say that Boethius used the letters A through O for the notes that a theoretical male singer could sing. Since I and J were not distinguished in his time, this is 14 letters, making two octaves. Since we didn't repeat letters, it seems strange to stop at the high G and not continue with an extra high A to finish off the scale, but perhaps he thought of scales differently (or just did not have faith in his male singers).

2025-12-10 09:52:47 UTC

- reply

Y'all need to read Curtain, the last Hercule Poirot story (not the last to be written by any means, but the one written to be the last, and published on Agatha Christie's death}. Poirot remarks that it's very suspicious if one is around a lot of murders (without an excuse like being a detective), but (without spoiling the ending) it goes a great deal further than that.

2025-11-28 13:26:54 UTC

- reply

Eagles > Seals

2025-11-28 04:08:12 UTC

- reply

Here's a takedown of the entire article (5 points, only #1 is about the graphic): https://pubpeer.com/publications/E1A0A9AC672336E92673154575091E

2025-09-30 13:13:14 UTC

- reply

I understand how the two valences of iron give rise to the two different colours (dull brown and bright red) of blood. But then you say that when the two are in close proximity, this can look blue, and I'm half expecting you to say that this is why veins look blue through the skin. But that seems like a little too much; it's just a coincidence, right?

2025-09-05 18:42:53 UTC

- reply

A minor point: it has been many years now since the minor-edit button was disabled for anonymous edits. While I would encourage anybody to create an account on Wikipedia, you definitely don't have to create an account to fix typos and punctuation; but it does mean that you can't mark these fixes as minor.

2025-08-26 13:43:05 UTC

- reply

This also works for d=1 (a line segment with F=2 endpoints).

2025-08-12 02:01:58 UTC

- reply

(Just kidding, I know what an Indigo printer is.) But seriously, I'm surprised to see you buying from HP. Maybe it's not enshittified on the high end?

2025-08-12 02:00:14 UTC

- reply

I like indigo at least as much as the next person; in fact it's my favorite colour. But wouldn't a CMYK printer be more useful?

2025-08-08 10:15:45 UTC

- reply

There's also Fred Richman's approach, based not on the field of hyperreal numbers (most of which have no decimal representation as a standard ω-sequence of digits) but an algebraic system whose elements are precisely the infinite sequences of decimal digits (but which don't form a field or even a ring): https://web.archive.org/web/20230720165002/math.fau.edu/Richman/docs/999.pdf

2025-08-08 10:02:47 UTC

- reply

Anyone who thinks the calculator is always right, should not be teaching math.

2025-07-30 20:32:25 UTC

- reply

This is building on how the dangers of human-driven cars were justified, with the invention of ‘jaywalking’ a century ago.

2025-07-08 12:44:04 UTC

- reply

This makes sense (but is still misleading) if ‘Gemini Apps Activity’ and ‘Gemini for Phone, Messages, WhatsApp, and Utilities’ are separate settings.

2025-06-14 04:56:23 UTC

- reply

As long as you're an American citizen (or even a permanent resident), you can legally donate to candidates in a different state (or territory or federal district). But you can't donate to Bernie as a Canadian; you have to pay for a sticker instead.

2025-06-04 22:11:44 UTC

- reply

That's a quantum effect because of how USBs are spin-½ particles.

2025-05-13 11:02:57 UTC

- reply

1845 is too early for Doyle, he did publish his own medical research 3–4 decades later, before turning to fiction.

2025-05-08 07:39:33 UTC

- reply

So we want to know whether ((123123×5↑↑10)%999999)%1001 = 0, where a↑↑n is tetration and a%b is remainder modulo b. By Euler's theorem, a^n%b = a^(n%φ(b))%b if gcd(a,b)=1. Less famously, a^n%(ab) = a^(n%φ(b))%(ab) if gcd(a,b)=1 and n%φ(b)>0. Since a↑↑(n+1)=a^a↑↑n, we can find a↑↑(n+1)%b recursively as a^(a↑↑n%φ(b))%b, and a↑↑(n+1)%(ab) as a^(a↑↑n%φ(b))%(ab), if the relevant conditions are met, starting with a↑↑0 = 1.
Since the totient φ(999999) is 466560, we could be in trouble, but 466560/5 is 93312, so we'll be fine. Applying the totient function 9 more times, there are no more multiples of 5, and we find that φ^9(93312) is 16, so I'll start there.
5↑↑0%16 = 1%16 = 1.
5↑↑1%32 = 5^(5↑↑0%16)%32 = 5^1%32 = 5.
5↑↑2%64 = 5^(5↑↑1%32)%64 = 5^5%64 = 53.
5↑↑3%128 = 5^(5↑↑2%64)%128 = 5^53%128 = 117.
5↑↑4%384 = 5^(5↑↑3%128)%384 = 5^117%384 = 245.
5↑↑5%1152 = 5^(5↑↑4%384)%1152 = 5^245%1152 = 245.
5↑↑6%3456 = 5^(5↑↑5%1152)%3456 = 5^245%3456 = 245.
5↑↑7%10368 = 5^(5↑↑6%3456)%10368 = 5^245%10368 = 3701.
5↑↑8%31104 = 5^(5↑↑7%10368)%31104 = 5^3701%31104 = 3701.
5↑↑9%466560 = 5^(5↑↑8%31104)%466560 = 5^3701%466560 = 34805.
5↑↑10%999999 = 5^(5↑↑9%466560)%99999 = 5^34805%465560 = 345467.
(123123×5↑↑10)%999999 = (123123×345467)%999999 = 975975.
975975 % 1001 = 0.
So Yes!

2025-05-06 04:27:34 UTC

- reply

Not that the article explains it clearly, but I think we can extract answers to these questions (in the sense of finding his intended meaning without trying to justify his opinions).
• Algebra’s oldest problem: finding roots of polynomials.
• Previously closed book in mathematics history: finding roots of quintic and higher-order polynomials.
• Size of the universe: probably means the observable universe (I see these conflated a lot in the popular press), so maybe about 10⁸⁰ (approximate baryon number of the observable universe), possibly as high as about 10¹⁸⁰ (approximate volume of observable universe in planck units, if I didn't mess up).

2025-05-06 04:07:14 UTC

- reply

In particular, there's a bunch of work on finitist ways of working with irrational algebraic numbers, going back to Kronecker. (For example, you can work with the number's minimal polynomial together with a rational (or gaussian) approximation that's precise enough to distinguish it from the other roots of that minimal polynomial.)

2025-05-01 04:21:16 UTC

- reply

you might want to make a note of this.

2025-04-25 00:06:46 UTC

- reply

Nice! Although to square the circle, you need a segment of length √π, not length π. (Both are impossible in finite time with straightedge and compass, since π is transcendental.)

2025-04-14 16:42:22 UTC

- reply

That's who I thought of, but I don't know.
It probably passed through the idea of differential operator. In local coordinates, a vector field (X¹,…,Xⁿ) defines a differential operator ΣᵢXⁱ∂ᵢ (where ∂ᵢ = ∂/∂xⁱ), which will get attention because it can be used in differential equations, and it doesn't take much to notice that the vector field can recovered from the operator (and even from its action only on infinitely differential functions if it occurs to anyone to consider that); but it could have been a while until someone worked out a coordinate-free description of which operators come from vector fields in this way.
This could be in Michael Spivak's Comprehensive Introduction to Differential Geometry, which has a lot of historical material.

2025-04-10 02:26:00 UTC

- reply

That's just straight-up corrupt, and yet compared to everything else, it barely deserves a mention.

2025-04-10 01:27:07 UTC

- reply

That last one at least has a reasonable explanation. The ε here is supposed to be elasticity of demand (the derivative of the logarithm of quantity demanded with respect to the logarithm of price, that is p/q⋅dq/dp), and since this is (as I have defined it) almost always negative, both this and its opposite are called ‘elasticity’ in practice. (In the applied calculus textbook that my school uses, they actually define elasticity as the absolute value, but I tell my students that this is a bad idea since elasticity of demand is not *always* negative (and elasticity of supply is not always positive); it's reasonable to swap the sign for demand so that it will usually be positive, but the sign still gives you information.) So while it's careless, it's not unheard of to say first that elasticity of demand is negative and later that it's 4 (when you really mean that it's −4).
What is unreasonable is to just make up a number out of whole cloth and pretend that it applies in every situation, just to balance out some other number that you made up out of whole cloth, all in a vain effort to retroactively justify your bullshit calculation.

2025-03-11 05:43:41 UTC

- reply

It's an easy place to make them!

2025-03-11 05:43:10 UTC

- reply

If it actually did what it said was going to do, it could have solved the problem. There is (up to equivalence) only one monoidal category V such that the category of truth values is equivalent to the category of V-enriched categories, and that's the answer. But of course it doesn't know how to reason through that.

2025-03-11 04:13:59 UTC

- reply

But its conclusion was wrong (this is all correct until the ‘solution’, which repeats the definition of a (−1)-category instead):
https://media.mathstodon.xyz/media_attachments/files/114/141/841/412/029/851/original/ce9dc5f0ade0e96d.jpg

2025-03-11 04:10:08 UTC

- reply

My attempt started off better (this is all correct):
https://media.mathstodon.xyz/media_attachments/files/114/141/830/692/849/527/original/935165f126b17a42.jpg

2025-03-11 04:04:10 UTC

- reply

Identity 2-morphisms?

2025-02-26 05:05:13 UTC

- reply

Not remotely interested, and yet I think that you took more care to make it believable and workable than they ever did. Of course it's always easier in fiction than in real life.

2025-02-26 04:42:06 UTC

- reply

They didn't do nearly as well as Stateless did in Distress!

2025-02-25 23:39:46 UTC

- reply

I think I would have answered something like ‘anything that you want to have immediate access to at any time’. It wouldn't have occurred to me to list what those things might be.

2025-02-20 17:11:08 UTC

- reply

OK, my use of excessive spaces confused it; without those it works. (Although ironically the \mathrm I put in to reproduce the roman subscripts that you had still breaks it, whereas without the \mathrm it produces roman Unicode since there are no italic subscripts in Unicode.)

2025-02-20 17:08:38 UTC

- reply

Except as soon as I start putting in the subscript, it takes away that option, so I still don't know how you did it.

2025-02-20 17:06:44 UTC

- reply

Oh I see, as you type it in there's a ‘Convert to Unicode’ option (which I normally never see because I'm not using the right app so when people post in LaTeX I just read the LaTeX).

2025-02-20 17:04:10 UTC

- reply

How do you get Mathstodon to do that? That's not the usual LaTeX support: \( T _ { \mathrm { i j } } \).

2025-02-20 16:26:05 UTC

- reply

For scripts, I like to copy paste the characters from the Wikipedia article on the Unicode characters https://en.wikipedia.org/wiki/Unicode_subscripts_and_superscripts . For Unicode characters more generally, if you know the name of the character then you can search for that and find web pages like https://www.compart.com/en/unicode/U+2084/.

2025-02-12 06:03:54 UTC

- reply

Apparently Google is doing that everywhere in the world except Mexico and the USA. Too insulting to do it in Mexico I guess, and full capitulation in the USA. https://blog.google/products/maps/united-states-geographic-name-change-feb-2025/

2025-02-09 22:35:15 UTC

- reply

: There's a mismatch that was acquired over time. But it's not English anymore; it's the Maori alphabet, and they (Maori speakers) wouldn't want to change it anymore than we (English speakers) would want to change the English alphabet because it no longer matches Latin.

2025-02-09 18:01:32 UTC

- reply

: It's because the sound was originally [ɸ] (the unvoiced bilabial fricative), which isn't used in English but lies between [f] (the unvoiced labiodental fricative, the usual sound of the phoneme /⁠f/ with the typical spelling ⟨f⟩) and [ʍ] (the unvoiced labiovelar fricative, the usual sound of the phoneme /⁠ʍ/ in conservative speech with the typical spelling ⟨wh⟩), which are. But [ʍ] is now largely obsolete, increasingly replaced with [w] (the voiced labiovelar approximant, the sound of the phoneme /⁠w/ with the typical spelling ⟨w⟩). So the first English speakers to transliterate Maori heard /⁠ɸ/ as /⁠ʍ/ rather than /⁠f/ and so wrote it as ⟨wh⟩, even though English speakers today almost always hear it as /⁠f/. And to make the choice even stranger, many Maori speakers today say [f], thanks to the influence of English. So the spelling only makes any sense if you compare conservative Maori pronunciation and conservative English pronunciation.

2025-01-26 02:46:19 UTC

- reply

: Re: ‘Protect the decent and hardworking members of your communities who are undocumented or whose parents are undocumented’
Protect people who are undocumented or have undocumented parents regardless of whether they're decent or hard-working. We don't need to waste time or weaken precedent deciding which are ‘the good ones’.

2025-01-24 01:09:25 UTC

- reply

: Since I didn't know that verb, I gave your sentence to Google Translate, and it even translated the noun that way. (‘The pukes make me puke’, which is certainly a thing that can happen.)

2025-01-01 01:02:48 UTC

- reply

: I think that I fixed the echo, and also completed a sentence (the last one before the last parenthesis) that didn't quite contain all that I'd intended.

2024-12-31 23:19:50 UTC

- reply

: Seeing Planck's constant here reminds me that this was originally a course-graining parameter in statistical physics that was meant to go to zero, but Planck discovered that it doesn't quite go to zero (and classical statistical physics would lead to infinities if it did). Quantum gravity renormalization might have a similar cutoff. (Naively, thi@johncarlosbaez : Seeing Planck's constant here reminds me that this was originally a course-graining parameter in statistical physics that was meant to go to zero, but Planck discovered that it doesn't quite go to zero (and continuum statistical physics would lead to infinities if it did). Quantum gravity renormalization might have a similar cutoff. (Naively, this suggests a discrete space-time at the Planck scale, but of course that's hard to make Lorentz-invariant, as people have been discussing elsewhere in this thread.)s suggests a discrete space-time at the Planck scale, but of course that's hard to make Lorentz-invariant, as people have been discussing elsewhere in this thread.)

2024-12-28 20:48:37 UTC

- reply

: Good, it's the same (except that my n is Leo's k, and Leo's n is my 2n).

2024-12-28 19:53:05 UTC

- reply

: During one of the periodic SE crises, @npub1d60…mhwm tried to plan by asking on Meta what should trigger MO to leave. And when I answered that we should leave right away, and the only question is how to do it, my answer got the most upvotes (both gross and net). But I got no comments about how to do it (or anything else), and of course it hasn't been done. I have to conclude that it's more hassle than it's worth to anyone who would be doing the actual work. https://meta.mathoverflow.net/questions/5669/at-what-point-should-mo-recognise-it-needs-to-leave-stack-exchange-inc/

2024-12-28 19:23:53 UTC

- reply

: The circle is \( r = \cos \theta \), not \( r^2 = \cos \theta \). This suggests that the generalization is \( r^n = \cos(n\theta) \), but I haven't checked if this is the same as Leo's generalization. (Which for those who haven't clicked through is based on \( \arcsin x = \int_0 (1 - x^2)^{-1/2} \) and \( \operatorname{arcsl} x = \int_0 (1 - x^4)^{-1/2} \), so \( \int_0 (1-x^{2n})^{-1/2} \) in general.)

2024-12-28 19:07:49 UTC

- reply

: StackExchange is like Twitter and Bluesky (and Zulip as a host): owned by a company that could be bought out and could do (and has done) evil regardless. Technically, MathOverflow is independent, but in practice, it's too tied in to StackExchange to leave.

2024-12-28 16:57:57 UTC

- reply

: But notice that \( (x^2 + y^2)^2 = x^2 - y^2 \) is not analogous to \( x^2 + y^2 = 1 \) but rather to the rightward-shifted half-unit circle \( x^2 + y^2 = x \). Whereas on \( x^2 + y^2 = 1 \), the coordinates of a point are the sine and cosine of a related arclength; on \( x^2 + y^2 = x \), some related arclengths are the sine and cosine of the distance of a point from the origin. And on \( (x^2 + y^2)^2 = x^2 - y^2 \), the analogous arclengths are the lemniscate sine and lemniscate cosine of the distance of a point from the origin. See https://en.wikipedia.org/wiki/Lemniscate_elliptic_functions#Relation_to_geometric_shapes for diagrams.

2024-12-28 15:55:37 UTC

- reply

: If I had a kind of shape named after me, then I'd seriously consider it for anything I could as well! (Joke since presumably he didn't call them that. He probably just called them ‘ovals’.)

2024-11-30 22:04:19 UTC

- reply

: So this explanation doesn't explain why Kheitin's constant is what it is (for that we should be able to just pick a random irrational and see what that one gives us), but it explains why it's a constant.
With, of course, some exceptions (such as the quadratic numbers, which are the ones that are eventually cyclic under this map).

2024-11-30 02:56:52 UTC

- reply

: That is wild! Considering that any sequence of positive integers whatsoever is the continued-fraction expansion of some real number (giving a bijection between such sequences and the positive irrational reals, indeed a homeomorphism under the usual topologies), you wouldn't expect anything like this.

2024-11-25 16:00:50 UTC

- reply

(3/3):
Anyway, as Bob falls in, he knows that Alice hit the event horizon before he did, since she fell in ahead of him. But he doesn't actually see that until the light reaches him. And the light reaches him right when he reaches the event horizon; the event horizon is travelling at the same speed as the light, after all. So he sees Alice reach the horizon at exactly the same moment as he reaches it, even though she reached it at an earlier time, when the horizon was in a different place from his perspective (although this is the same place from the perspective of some distant and/or accelerating observers, like Charlie on the ship or us when we were observing from lightyears away).

2024-11-25 15:58:40 UTC

- reply

(2/3):
As we approach the black hole, we see our astronauts' spaceship, and at first (except when Alice and Bob jump in) everything appears motionless. But as we approach and get captured by the black hole ourselves, we realize that the ship is rushing towards us, accelerating even (which explains its rocket fire). And the event horizon seems to be accelerating towards us too. As we fall closer, the apparent speed of both the ship and the event horizon increases, but only as we pass them do we observe their true proper acceleration. The ship is accelerating just as much as it would be if the black hole were a planet like Earth, with the ship sitting on it, being pushed by the planet's surface.
As for the event horizon, as we pass it, we finally observe its true motion, which is at the speed of light. So it's actually not accelerating at all, but it's also moving faster than we were able to observe from outside. If we regarded the event horizon as at rest, then we would think that we ourselves were moving at lightspeed! So the event horizon cannot be at rest relative to any local inertial observer; anyone who sees it thus must be some combination of distant and accelerating, since they can't be travelling at lightspeed themself.
(2/3)

2024-11-25 15:56:54 UTC

- reply

:
>A free falling observer is in an accelerating frame,
This is the big difference between how Newtonian mechanics treats gravity and how General Relativity treats it: a free-falling observer is *not* in an accelerated frame. In fact, that's what makes the fall free: there are no forces on the observer (since gravity is not a force in GR), so Alice and Bob are (after they jump) in inertial frames. Only Charlie, who stayed back on the ship, is accelerating.
Standing still on Earth as I am, I might seem to be at rest, and that's how I would think of it while doing a high-school physics experiment. But even according to Newton, I'm really in a revolving frame with a cycle of a day (actually one a bit more complicated than that). Similarly, Einstein treats me as inertial in his thought experiments with trains describing Special Relativity, but he knows (since he didn't write that stuff until after he came up with GR) that in reality I am (in addition to revolving) being constantly pushed away from Earth by the floor, as if in a great space elevator (which was another of his thought experiments).
A distant observer, watching me from space, would see me revolving, but they wouldn't otherwise see me accelerating (although they could infer it from their knowledge of gravity). However, if they get closer to Earth, certainly if they're close enough to get captured by Earth's gravity themself, then they'll see me (along with the rest of Earth) rushing towards them (if they're in free fall themself), and even accelerating. So accelerating objects can appear motionless from a distance, but up close you can observe the truth.
(1/3)

2024-11-24 05:53:59 UTC

- reply

: It's also on the Wayback Machine (just barely, with the first archive only 3 months before it was removed).

2024-11-24 05:15:35 UTC

- reply

: It sounds like The Onion, but if it's in The Guardian, then it must be true!

2024-11-24 03:17:18 UTC

- reply

: You never see your hand (or anything) the way it looks right now; you see it as it appeared in the past, when the light left it. As you fall in, hand first, your hand passes the event horizon (entering the black hole) while you're receiving the light from before it entered. Then when your head enters, you receive the light from when your hand entered. Later, you'll see the light that left your hand at the time when your head entered.

2024-11-22 08:27:21 UTC

- reply

: I stand corrected. And delighted but not surprised that you have this all worked out for us already.

2024-11-22 08:10:12 UTC

- reply

: The sky facing the black hole wouldn't be black, no matter how much (or how little) matter is falling in.
Above, when I compared falling into a black hole to falling into the future, that wasn't a vague analogy; the future timelike direction inside a black hole *is* towards the centre. We don't get any light from the future, but there's nowhere in our field of vision where we expect such light and get blackness instead; the same goes for the absence of light from a black hole.
If we're a little ways outside a black hole, facing it, then the light we see will be light left over from before its formation, light that almost fell in but didn't quite, light that's been moving away ever so slowly (relative to a distant observer) as it fights the gravitational pull of the collapsed star behind it, until we finally come along and see it. There's no such thing as the direction from the black hole, only the direction from a little before the black hole, and that's what we're looking at.
You can still tell that the black hole is there, but from the distortion, not from any absence.

2024-11-22 03:15:01 UTC

- reply

: And if we're falling into a black hole just at our threshold of being able to tell, how long do we have until tidal forces destroy us, and how does that compare to other end-of-the-world scenarios like the Sun engulfing the Earth in 5 billion years?
At some point, it doesn't matter if we're falling into a black hole, any more than it matters that we're falling into the future.

2024-11-21 19:46:33 UTC

- reply

: Neat! Above, when I described the Moon as ‘smaller but much closer’ than the Sun, by ‘smaller’ I meant ‹less massive›. But I could have just written ‘denser’, or maybe ‘denser (and closer, but this is almost perfectly cancelled by its smaller size)’ for more detail. (Looking up these densities on Google, it looks like their ratio is about 2.4.)

2024-11-21 19:31:23 UTC

- reply

: So when a supervillain threatens to destroy the Moon, ending the tides and disrupting the ecology of bays, it's not that serious?

2024-11-21 17:59:33 UTC

- reply

:
>water on the sunny side of the Earth is pulled toward the Sun more than water on the night side
While the Sun does cause a tidal force as you describe, the tides that we see are mainly due to the Moon (which is smaller but much closer). So water on the moony side (if you will) is pulled more, while water on the opposite side is pulled less, and the land (which moves more as a solid chunk) is pulled some amount in the middle. So we get high tide on both the moony side (where up is towards the moon) and on the opposite side (where up is away from the moon).
You probably know this, but it seems weird to focus on the relatively minor tidal effect of the Sun, so I thought that I would clarify for other readers.

2024-10-29 16:24:24 UTC

- reply

: If the fine was, say, a billion rubles per week, then Google's lawyers might be worried that somewhere down the line, when relations between countries improve, they might have to pay some of it. But with this, you have nothing to worry about; nobody's ever going to expect you to pay that.

2024-10-17 20:50:16 UTC

- reply

: OK, now I want to read this.

2024-10-05 02:22:33 UTC

- reply

: One heart will break into two, and then each half will combine with another, separate heart; so that, due (in part) to a break, three hearts will become two.