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September 19

Metric tensor

This is a basic differential geometry question, though the context is that I just watched a video about general relativity. The lecturer describes spacetime as a curved surface. It's locally flat, so each point p has an associated matrix M_p. Then if you move away from p by some small vector ds, you can transform it by M_p, giving M_pds = (dx,dy,dz,dt). The distance then is sqrt(dx²+dy²+dx²-dt²) with the -dt² due to the Lorentzian signature.

Later though he calls M a metric tensor, and metric tensor says the tensor takes two vector inputs and produces a scalar output.

What is the deal here? Since the flatness is only local, to find the distance between arbitrary points p and q, first you have to find a geodesic and then integrate along it, amirite? Where does the tensor come in, that requires two vector inputs instead of just one?

Later he talks about contravariant and covariant vectors, which I think means column vectors and row vectors. I couldn't tell what that was about either, but I'll make another try at it later.

Thanks. 2601:644:8581:75B0:4C10:1C2B:E139:C42B (talk) 01:10, 19 September 2025 (UTC)[reply]

The metric tensor is essentially the inner product or dot product (not sure where those links will wind up). To use it to get the length of a vector, you take the dot product of the vector when itself (and then take the square root). --Trovatore (talk) 01:16, 19 September 2025 (UTC)[reply]

Thanks, that helps. I'll probably have to watch the whole video again, and it's a series, so it could be time consuming. Wow. 2601:644:8581:75B0:4C10:1C2B:E139:C42B (talk) 04:24, 19 September 2025 (UTC)[reply]

You might be talking about index raising and index lowering. I'm afraid the linked article (with the whimsical name "musical isomorphism", with which I was hitherto unfamiliar) is not going to be much use to you at the present time, but you can keep it in mind if you learn a whole bunch more about differential geometry. I (or someone else here) might be able to give you a gentler introduction to it, but I'd have to think about how. --Trovatore (talk) 04:28, 19 September 2025 (UTC)[reply]

Thanks. I have some familiarity with tensors from linear algebra, so maybe I can get through that article. I think the differential geometry used in the video I saw is not too bad, and I'm just confused by a single probably-minor point. Of course later videos in the series will be harder, but GR is looking less scary than it's reputed to be. 2601:644:8581:75B0:4C10:1C2B:E139:C42B (talk) 07:49, 19 September 2025 (UTC)[reply]

September 20

a variant of Wilson prime?

A prime p is Wilson if (p-1)!~-1(mod p^2). But as you know, (p-2)! is always congruent to 1 (mod p). So what type of prime is it if (p-2)!~1(mod p^2)? For example, 9! is 1(mod 121).Rich (talk) 02:44, 20 September 2025 (UTC)[reply]

I'm pretty sure these are "generalized Wilson primes" of order 2; see the corresponding section in the article. The section is not sourced though, so I have no idea how "official" that name is. --RDBury (talk) 15:07, 20 September 2025 (UTC)[reply]

September 28

Product of all positive integers

-1/12 is the sum of all positive integers, but what is the product of all positive integers? 61.229.98.9 (talk) 21:42, 28 September 2025 (UTC)[reply]

First of all, no, it's a divergent sum with no finite value. There are certain very generalized meanings of "sum" where the value turns out to be -1/12 (see Riemann zeta function#Specific values), but it's not the sum in the generally accepted sense. Similarly, the product diverges and has no finite value. There may be a way to do a similar generalized version of a product, but AFIAK there isn't a lot of interest in that kind of thing and there's no reason to expect there would be a meaningful value at the end of the process. --RDBury (talk) 00:30, 29 September 2025 (UTC)[reply]

The standard regularization of the product of the positive integers, analogous to the regularization of

\zeta (-1)=-1/12

is

e^{-\zeta '(0)}={\sqrt {2\pi }}

. I edited an article describing this, but can't find it now. You're welcome to search my contributions for it. (Of course the naive regularization, via the gamma function, has an essential singularity at infinity. ) Tito Omburo (talk) 00:54, 29 September 2025 (UTC)[reply]

September 29

Mercator projections with different poles

Mathematically speaking, does a Mercator projection have to be oriented so that the actual poles are at the top and bottom? Obviously it wouldn't be very useful, but would it be possible to produce a Mercator projection with random spots as poles and a random line on the surface (geometrically, a plane that's perpendicular to the line connecting the polar points) as an equator? I assume so, but I could be wrong, and I don't know what to call it; Mercator projection doesn't address "alternate" projections. Nyttend (talk) 03:42, 29 September 2025 (UTC)[reply]

The article has sections on Transverse Mercator and Oblique Mercator, each with its own main article. Isn't that what you're thinking of? --Wrongfilter (talk) 05:20, 29 September 2025 (UTC)[reply]

Oops, yes it is. I skipped the "uses" section because it covers things like marine navigation; I didn't anticipate that types of the projection would be thrown in there too. Nyttend (talk) 09:20, 29 September 2025 (UTC)[reply]

For completeness, Transverse and Oblique Mercator require the poles to be antipodes, not just any two "random spots". At the extreme, if the spots coincide, one gets an azimuthal projection. cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 08:32, 30 September 2025 (UTC)[reply]

September 30

Sudoku ranking

Publishers typically sort their Sudoku puzzles into difficulties: easy, medium, hard, expert - or something like that. Is that process manual/subjective - or is there an algorithm that's able to sort the required moves into different types? For example, you might say that easy puzzles only require straightforward comparison checks, while medium puzzles might require some inferential checks or row/column based checks, while harder puzzles would require more complicated inferences (I assume there's proper jargon for this stuff, but hopefully I'm getting the gist across). The actual usage is of interest, but I'm more curious about the theoretical side, which is why I'm asking here rather than the Ent desk. Matt Deres (talk) 15:07, 30 September 2025 (UTC)[reply]

A couple of years ago I looked for material on this but came out empty-handed.

I thought of using an algorithm that imitates a plausible way in which humans solve these puzzles and counts the number of inferential steps but have not carried this through.

For one published series of puzzles, with difficulties ranging from 1 to 4 stars, I can report (from personal experience) that levels 1 and 2 are boringly easy, while there is substantial increase in difficulty from 2 to 3 (measured by the time needed to solve them, which goes up by a factor of 3, or so). But occasionally a 3-star puzzle is easier than the typical 2, or a 2-star puzzle proves more challenging than the typical 3. There is also a jump from 3 to 4, but not by that large a factor as from 2 to 3. Furthermore, there is no common difficulty measure like the Scoville scale for hotness; one publisher's "medium" may be easier than another publisher's "easy". ‑‑Lambiam 17:05, 30 September 2025 (UTC)[reply]

Yeah, that's broadly my experience as well and it was the "expert" ones that were perhaps only "hard" that prompted my question. In practice, I assume such stuff often comes down to happening to notice a particular item ahead of when it would more typically be found during a brute-force and it gives you an outsize advantage to complete the puzzle. Matt Deres (talk) 20:04, 30 September 2025 (UTC)[reply]

Spotting something ahead of when a systematic approach would have found it can help, but actively looking for such things may not pay off and slow one down. Also, in general you have to alternate between two modes, which I refer to as "digit seeks cell" and "cell seeks digit". I see no clear criterion when to choose which, although "cell seeks digit" is generally less fruitful in sparse puzzles. I have a rule of thumb to avoid unsuccessful "cell seeks digit" efforts, which often works but sometimes fails miserably in either direction. The choice is mostly a gamble, though, in which one can be lucky or unlucky. ‑‑Lambiam 23:43, 30 September 2025 (UTC)[reply]

Agreed. I'm not suggesting that going snipe hunting is the right thing to do; I'm just saying that sometimes you can stumble across a snipe and find that the expert puzzle is surprisingly easy. I think I understand what you mean by "digit seeks cell" and so on, though it raises the related question of whether the various processes players use have semi-formalized names. There are at least three very different methods I end up using for extreme puzzles (more than that, if we separate out rotations as discrete phases). If that terminology exists, it would be easier to search for whether puzzle-makers take them into account when setting their difficulty levels. Matt Deres (talk) 18:38, 1 October 2025 (UTC)[reply]

October 2

Lower tetration

Wikipedia's article Tetration says there is only one kind of tetration. But the talk page is talking about "lower tetration", which implies that there is more than one kind. Is this really true?? Georgia guy (talk) 16:05, 2 October 2025 (UTC)[reply]

I think it is a term invented by the poster. Note that, using left-associative grouping, we have

\left(\cdots {{{\left(\left({a_{0}}^{a_{1}}\right)^{a_{2}}\right)^{\cdot }}^{\cdot }}^{\cdot }}\right)^{a_{n}}=a_{0}^{a_{1}a_{2}\cdots a_{n}},

so this not very interesting. ‑‑Lambiam 17:59, 2 October 2025 (UTC)[reply]