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October 22

Although neutrinos can't interact with photons, can photons deliver momentum to neutrinos, via electrons as intermediaries which receive it from photons and deliver it to neutrinos?

Something that reminds of Newton's cradle (yet not exactly of course). HOTmag (talk) 16:06, 22 October 2024 (UTC)[reply]

This is the kind of thing that Feynam diagrams were invented for. You can have a neutrino and a photon going in, a neutrino and a photon going out, and all kinds of other particles running around in a loop in the middle. See for example [1] (paywalled unfortunately). --Amble (talk) 17:21, 22 October 2024 (UTC)[reply]

I gotcha covered on the link there

--Slowking Man (talk) 17:32, 22 October 2024 (UTC)[reply]

Why can't they interact? Both photons and neutrinos participate in the weak interaction. Difficulty: that cross-section is gonna be really small. Gamma ray cross section may be also of interest. But yes, the aforementioned besides, if you have a whole lot of photons and can make them go where you want and give them arbitrary energies, you can make the photons do all kinds of neat tricks with each other such as popping out other particles: see two-photon physics. The truly "high-energy physics" processes in our universe such as supernovas and neutron star collisions and gamma-ray bursts do plenty of this kind of stuff. (Another whopper is that we're now fairly sure type II supernovas are in essence "powered" mostly by the neutrino burst! The core collapse releases such a staggering amount of neutrinos that, if you somehow managed to get close enough and not have anything else kill you, you would be killed by the fatal neutrino radiation! As they say there, a phrase that just looks wrong if you know what it's talking about. Similarly: an "average supernova" releases about 10⁵⁷ neutrinos, 10 followed by a mere 57 zeroes. Kind of like: a "modest planetary collision" will only disrupt the crust and some of the mantle. (And one hypothesis is, the reason half of Mars is really different from the other half is that an even bigger impact happened to it!) --Slowking Man (talk) 17:29, 22 October 2024 (UTC)[reply]

Why can't they interact? Both photons and neutrinos participate in the weak interaction It seems that your question is addressed (also) to yourself, i.e. to what you wrote in another thread. HOTmag (talk) 18:17, 22 October 2024 (UTC)[reply]

Is it true the Pacific's where the Mars-sized moonmaker hit non-head on splashing off lava and boiled lava? I thought only continental crust can survive that long. Sagittarian Milky Way (talk) 22:21, 22 October 2024 (UTC)[reply]

If so it's not mentioned at our giant-impact hypothesis article, nor the Theia (planet) one. I would have been surprised if that were true; I assumed that the Earth's entire surface was liquid for a while after the impact so there shouldn't be any remaining localized remnant. But I could be completely wrong about that. --Trovatore (talk) 22:56, 22 October 2024 (UTC)[reply]

That's where the Kaiju come from right? But seeing as >4 bil ybp, there was no such thing as "the Pacific Ocean"... I dunno? Is this based on some specific thing from somewhere? A mere 220 mya there was just the one ocean with no major landmasses apart from "the one place where all the land is". If I recall right the hypothesized Theia impact is predicted to maybe have re-liquified Earth's entire crust, even vaporizing some of it to produce a temporary "rock vapor atmosphere", another one of those "phrases that just sound crazy". (Excellent band name up for grabs there btw) --Slowking Man (talk) 23:02, 22 October 2024 (UTC)[reply]

I heard that somewhere maybe whoever thought it was just ignorant? Sagittarian Milky Way (talk) 00:20, 23 October 2024 (UTC)[reply]

we can propose sending photons "this way" and neutrinos "that way", such that their worldlines at some point intersect, but we would expect to observe nothing (other than the extremely minute effects of their gravitational and weak interactions), (emphasis added). Okay, I was being a tad pedantic, but, being precise can matter for Science Stuff. You're the one proposing hypothetical scenarious here—you could always say "here, we're gonna fire enough photons such that their weak interactions w/ matter start adding up" (like in a supernova). Alternately if you just want to ask, "can photons and neutrinos interact with each other at all, directly or indirectly" just ask that and skip the trouble of crafting undergrad physics textbook study problems. (Keep in mind, people here are volunteering to devote some of their own time to responding to questioners' queries.) --Slowking Man (talk) 23:02, 22 October 2024 (UTC)[reply]

Oh sorry, I did read what you had written in parentheses "other than the extremely minute effects of their gravitational... interactions", but for some unknown reason I didn't notice the crucial words "and weak interactions". So you're right, sorry.

As for the two-photon physics you've mentioned: Well, AFAIK, two photons can turn into electron-positron (or muon-antimuon, tauon-anti-tauon), but never (directly) into neutrino-anti-neutrino. The same it true for the opposite direction: If a neutrino collides with its anti-particle, the direct result may be Z-boson, not photons (and with regard to what you wrote in your last parentheses: Of course, indeed you are always invited to remove my misunderstanding, but are never obligated to do that). HOTmag (talk) 00:24, 23 October 2024 (UTC)[reply]

That makes my (new) list of most exotic way to shuffle off this mortal coil. Clarityfiend (talk) 00:07, 23 October 2024 (UTC)[reply]

Victim [knocking on the Pearly gates]: Hello?

St. Peter: Sorry, guy. There's no room at the inn for you. Have you tried the Other Place?

Victim: But, but I'm high up on The List!

St. Peter: That cuts no ice up here.

Victim [in desperation]: Uh, I was struck down by neutrinos!

St. Peter: Way cool, dude! Okay, I can make an exception for you. Just don't tell anyone. Clarityfiend (talk) 03:07, 23 October 2024 (UTC)[reply]

October 24

Default multiplication tables in schools worldwide

According to multiplication table, in the English-speaking world schools use 1-12 multiplication tables, or 1-9. Can this maybe be distinguished between countries? In German-speaking Europe, schools use 1-10 by default. For sure in many countries they start with 0. What's the situation in different countries worldwide? --KnightMove (talk) 08:37, 24 October 2024 (UTC)[reply]

Wait, why would one start a multiplication table with zero? Remsense ‥ 论 08:43, 24 October 2024 (UTC)[reply]

I guess so that every possible product of two digits is included. Double sharp (talk) 09:36, 24 October 2024 (UTC)[reply]

Why would one start multiplication tables with 1? We started with 2. Shantavira|^{feed me} 09:00, 24 October 2024 (UTC)[reply]

Ok, and where did you go to school, please? --KnightMove (talk) 11:02, 24 October 2024 (UTC)[reply]

The curriculum in the UK is devolved.

The English curriculum includes multiplication tables up to 12 × 12 (no word on whether 0 or 1 are included as tables, but multiplying by 0 or 1 is included).

Wales (at least partly English speaking), only has up to 10x10

Scotland goes up to 12.

Northern Ireland doesn't explicitly have multiplication tables, but does cover "multiplication facts up to 10 x 10".

My understanding is that education is a state matter in the USA so maybe there are 50 different curricula?

AlmostReadytoFly (talk) 11:31, 24 October 2024 (UTC)[reply]

US states generally delegate to school districts, so each county or township has its own standards (and then there are tons of non-public schools). But there are a few standard curricula or textbooks that many of them use, and regardless of approach many follow the same or similar broad standards. And finally, it's sometimes just a semantic difference whether it's called part of the "table" or just a loose fact. Some relevant articles:

Mathematics education in the United States
Common Core (3.OA.C.7 specifies "By the end of Grade 3, know from memory all products of two one-digit numbers.")
Singapore math (popular in some parts of the US)

Some lead refs:

Olfos, Raimundo; Isoda, Masami (2021). "Teaching the Multiplication Table and Its Properties for Learning How to Learn". Teaching Multiplication with Lesson Study. pp. 133–154. doi:10.1007/978-3-030-28561-6_6. ISBN 978-3-030-28560-9.
Dotan, Dror; Zviran-Ginat, Sharon (2022). "Elementary math in elementary school: The effect of interference on learning the multiplication table". Cognitive Research: Principles and Implications. 7 (1): 101. doi:10.1186/s41235-022-00451-0. PMC 9716515. PMID 36459276.
Isoda, Masami; Olfos, Raimundo, eds. (2021). Teaching Multiplication with Lesson Study. doi:10.1007/978-3-030-28561-6. ISBN 978-3-030-28560-9.

DMacks (talk) 11:57, 24 October 2024 (UTC)[reply]

It's been a long time since I was at school but I dimly remember the multiplication tables in the booklet went up to 13x13 though we only had to learn up to 12x12. NadVolum (talk) 12:40, 24 October 2024 (UTC)[reply]

The 12 times table had a greater significance in my primary school days, since there were 12 pence in a shilling in those days. Alansplodge (talk) 14:10, 24 October 2024 (UTC)[reply]

Ditto. Learning the 13 and 17 times tables might be of some value; the others can be trivially derived mentally by doubling (e.g. 9x14 = (9x7)x2) and similar expedients, or for nx19 ((nx10)x2)-n. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 12:44, 27 October 2024 (UTC)[reply]

hm, Schoolhouse Rock (which has a fan wiki!) covered 2–9, 11, 12, plus zero (powers of ten). —Tamfang (talk) 21:40, 4 November 2024 (UTC)[reply]

Doesn't this thread belong more to the reference desk of Mathematics? HOTmag (talk) 07:37, 28 October 2024 (UTC)[reply]

October 26

Launch site identification

I am currently on a bridge on the Florida State Road 528 facing north, getting ready to view the Spacex launch[2] later today.

But I am not sure which one is the launch site that I should be looking at. I spotted 5 points of interest[3] (labelled A to E), and I can only positively identify A as the kennedy space center (which is not today's launch site).

I would really appreciate it if someone can tell me whether I should be looking at BCD or E. And also I'm interested to learn what BCDE each are. Epideurus (talk) 20:09, 26 October 2024 (UTC)[reply]

My exact location is 28.405476616667123, -80.65458604061091. Epideurus (talk) 20:31, 26 October 2024 (UTC)[reply]

At a guess, I'd say that I'd agree with your identification of "A" as the Vehicle Assembly Building at the JFK Space Center. My tentative identification of C and D are the Titan Solid Motor Assembly And Readiness Facility and the Titan Solid Motor Assembly building. E is possibly Cape Canaveral Space Launch Complex 37. PianoDan (talk) 17:47, 28 October 2024 (UTC)[reply]

October 27

Can sterile neutrinos (if exist) decay, if there don't really exist virtual particles?

The background of my question is the following two facts:

Our artricle sterile neutrino states: The production and decay of sterile neutrinos could happen through the mixing with virtual ("off mass shell") neutrinos.

While our article Virtual particles states: they are by no means a necessary feature of QFT, but rather are mathematical conveniences — as demonstrated by lattice field theory, which avoids using the concept altogether.

HOTmag (talk) 06:34, 27 October 2024 (UTC)[reply]

The statement that "x could happen if y" does not in itself exclude the possibilities that (absent y) x could also happen if z, or w, etc. Frankly, this is all so deep in the Jungles of Conjecture (that vast expanse beyond the Mountains of Hypotheses) that definitive answers probably don't yet exist, and Nobel prizes will probably be given for finding answers to such questions. Or so I think: perhaps some post-doctoral particle physicist will correct me. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 12:53, 27 October 2024 (UTC)[reply]

As for your first sentence: Of course. I didn't think otherwise. HOTmag (talk) 13:04, 27 October 2024 (UTC)[reply]

Does anything "really" exist? If we answer yes, what does it mean to "really" exist? We have models that make observable predictions. We say that atoms exist because the predictions of the atomic model were actually observed. One observable prediction of virtual particles is the Casimir effect, which was experimentally observed. They lack the stability of other particles, just like the waves that make the surf have no long-time stability. It may be possible to develop a form of hydrodynamic theory that accurately describes the observable effects of surf without introducing the concept of a wave. Does this then mean these waves do not "really" exist? --Lambiam 16:45, 27 October 2024 (UTC)[reply]

Are you responding to me, or (also?) to the quote I quoted from our article virtual particle? HOTmag (talk) 18:40, 27 October 2024 (UTC)[reply]

I am mainly responding to the question as formulated in the heading of this thread. The question is unanswerable if the meaning of "really exist" is not clear (which it isn't). --Lambiam 05:26, 28 October 2024 (UTC)[reply]

Let's put it this way: Do you agree to the content of the second quote under the header? IMO, it actually says that there don't necessarily exist virtual particles. This is what I meant by "don't really exist". But if you think it says something else, you're invited to tell what you think it says.

(Re. Casimir effect, its article in Wikipedia states: Although the Casimir effect can be expressed in terms of virtual particles interacting with the objects, it is best described and more easily calculated in terms of the zero-point energy of a quantized field in the intervening space between the objects).

HOTmag (talk) 06:32, 28 October 2024 (UTC)[reply]

I think it is theoretically possible to develop QCD and QFT to such an extent that it successfully describes macroscopically observable events without introducing the concept of particle. However, this theory would be extremely unwieldy and in practice mathematically untractable – and therefore pretty useless. Particles are a mathematical convenience, but a convenience we cannot do without. It may be possible that some practical version of QFT avoids the concept of virtual particle, but lattice field theory is not an alternative to QFT but a collection of mathematical approaches for obtaining QFT predictions by computer simulation. It is itself a mathematical convenience. Therefore, in my opinion, the sentence relegating virtual particles to a status of mere mathematical conveniences is not so much false as misleading. --Lambiam 19:28, 28 October 2024 (UTC)[reply]

Resolved

According to this video: "It's an easier transition from sterile neutrino to electron-neutrino".

So indeed, perhaps a sterile neutrino doesn't decay (as reqiured in the header), but still, it can (apparently) oscillate - becoming an electron neutrino. HOTmag (talk) 12:07, 28 October 2024 (UTC)[reply]

Global deforestation runoff

What is the global deforestation runoff in km3, which is part of the 40k km3 of global runoff* in general?

Trenberth KE, Smith L, Qian T, Dai A, Fasulo J (2007). Estimates of Global Water Budget and Its Annual Cycle Using Observational and Model Data. Journal of Hydrometeorolgy 8(4):758-769. DOI: org/10.1175/JHM600.1.

Fred weiers (talk) 07:44, 27 October 2024 (UTC)[reply]

Please consult this article: "Deforestation-induced runoff changes dominated by forest-climate feedbacks". My layperson's summary: it's complicated. --Lambiam 16:20, 27 October 2024 (UTC)[reply]

Is Big Bang nucleosynthesis backwards?

The article says the universe started with electrons and protons then created neutrons through nucleosynthesis, but whataboutism those places in the current universe that are too dense for electrons to exist? Shouldn't the universe have started out with all sorts of exotic particles filling in every possible energy state that then phase changes to neutronium as soon as the density is low enough? The neutrons aren't bound together by the Strong Force and hence thermally scatter as the density drops to decay completely away over the next hour, with Lithium and Helium being the result of neutron capture, not fusion? Hcobb (talk) 14:01, 27 October 2024 (UTC)[reply]

In nucleosynthesis, including the Big Bang nucleosynthesis, neutrons and protons combine to form nuclei. In electron capture, a proton of a nucleus turns into a neutron. For example, a nickel nucleus with 28 protons may turn into a cobalt nucleus with 27 protons. In a sufficiently hot environment, nothing is stable; all reactions may go either way, as they certainly did in the first few seconds after the Big Bang. --Lambiam 15:56, 27 October 2024 (UTC)[reply]

1934–35 North American drought

The article on the 1934–35 North American drought appears to be erroneously dated. I only noticed this because I've been writing about Georgia O'Keeffe, who became widely known for collecting bones in the desert of New Mexico from late 1929 and into the 1930s. These bones are from wild horses and cows that died due to a drought. So when I discovered that this drought was described by Wikipedia as taking place in 1934, I could see something was wrong. Other sources indicate that the media of the time widely reported the drought in the general area as beginning in 1929, not 1934. Can anyone figure out why there is this discrepancy? I did make a comment on the talk page indicating one possible reason. Viriditas (talk) 21:21, 27 October 2024 (UTC)[reply]

The infamous Dust Bowl seems to cover a wider time period. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:54, 27 October 2024 (UTC)[reply]

Understood, but I'm wondering if our article on the 1934–35 North American drought should be widened in terms of the date range. Viriditas (talk) 21:59, 27 October 2024 (UTC)[reply]

The article is poorly written. If anything it should be just 1934 North American drought. The evidence is from tree rings which are incontrovertible, and the extent of the drought can be viewed in the North American Drought Atlas here. Abductive (reasoning) 10:42, 28 October 2024 (UTC)[reply]

It's very unlikely that this was the first and only drought in New Mexico – see Droughts in the United States#Events. In any case, cattle and horses in the wild die for other reasons than drought, and in a dry environment are slow to decay. The bones collected by O'Keefe could have been decades or even centuries old. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 12:49, 28 October 2024 (UTC)[reply]

It's important recognize that this drought was for nearly the whole of North America. Abductive (reasoning) 06:06, 29 October 2024 (UTC)[reply]

Yes, this 1934(–5?) drought was, but Viridas assumed that it was the same drought that was responsible for bones collected by Georgia O'Keefe in New Mexico from 1929 onwards, which is clearly unwarrented for the two reasons I have pointed out, over and above the date discrepancy. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 19:20, 29 October 2024 (UTC)[reply]

This is about Alberta, Canada, rather than New Mexico, but:

"Although the early years of the 19th century [presumably 20th century was intended] were wetter, drought returned [to the prairies] during the years of 1910, 1914, 1917, 1918 and 1919, and the drought between 1917-1926 was considered to be especially bad... The global stock market collapsed in 1929 and marked the beginning of the Great Depression. To make matters worse, a severe drought began in the prairies in 1929". [4]

Alansplodge (talk) 12:50, 28 October 2024 (UTC)[reply]

October 28

According to the common theory, is the mechanical equilibrium a necessary sufficient condition, for not emitting gravitational waves?

By mechanical equilibrium I mean, both equilibrium of forces and equilibrium of torques (moments). HOTmag (talk) 09:00, 28 October 2024 (UTC)[reply]

A sufficient condition is that quadrupole moment (and all higher moments) of an isolated system is constant. Ruslik_Zero 20:23, 28 October 2024 (UTC)[reply]

Is a mechanical equilibrium a necessary condition?
Do you think you can give a concrete example of a body which is in a mechanical equilibrium and which emits GWs? HOTmag (talk) 20:35, 28 October 2024 (UTC)[reply]
1: No. Take a cylindrically symmetric flywheel (i.e., solid, no spokes) and spin it on its axle. It won't emit gravitational waves, even when the rate of spin changes when you apply a torque. Its quadrupole moment is zero (or at least, the relevant components) and therefore doesn't change on rotation.

2: Yes. Take a flywheel with two masses attached to the rim, opposite to each other, and spin it. It will emit gravitational waves. Its quadrupole moment is non-zero and changes direction during rotation. The gravitational waves will carry away some angular momentum and thereby apply a torque, but with a small motor you can compensate for that and keep the flywheel in mechanical equilibrium. PiusImpavidus (talk) 09:03, 29 October 2024 (UTC)[reply]

Thx. Now, let's assume we don't add the small motor, so the flywheell won't be in mechanical equilibrium, because as you say: "The gravitational waves will carry away some angular momentum and thereby apply a torque". However, since the natural source of this torque can't be any "real force" (namely: the electromagnetic force, the strong force, and the weak force), so: is it really reasonable to conclude, that the natural source of this torque is the gravitational (fictitious) "force", even when the system is isolated, i.e. not close to any other mass? It sounds a bit strange to my ears... HOTmag (talk) 11:57, 29 October 2024 (UTC)[reply]

Yes, the source of that torque is gravity, even when spacetime were flat if the flywheel hadn't been there. It's the effect of the flywheel itself on the surrounding spacetime, not depending on any disruptions from nearby objects. Just as a rotating electric or magnetic dipole looses angular momentum to electromagnetic radiation, even without an externally applied electromagnetic field. PiusImpavidus (talk) 13:46, 29 October 2024 (UTC)[reply]

So your flywheel example - but without the motor (along with any two-body system satisfying the same principle), seems to be an extremely rare case (isn't it?), in which a system being "both isolated and neutral", i.e both - not close to any other gravitational mass - and not influeced by any external real force, is not in mechanical equilibrium...

When I was taught about mechanical equilibrium in school (not long ago), my teacher never mentioned this rare option... HOTmag (talk) 15:13, 29 October 2024 (UTC)[reply]

Well, what's isolated? The object isn't isolated from spacetime.

General relativity is hard. Most physics school teachers consider it hard for themselves (although they must have learned something about it) and don't want to go too deeply into it. Apart from some handwavy arguments, most students wouldn't understand it at all. PiusImpavidus (talk) 16:31, 30 October 2024 (UTC)[reply]

I've explained what I mean by "isolated": not close to any other gravitational mass.

I guess if Newton were asked about, what he thought about a body - not close to any other gravitational mass - and not influenced by any force other than the gravitational one, he would immediately determine: "The body is in mechanical equilibrium". This is a direct conclusion derived from the combination of his first two laws with his law of gravity.

So Relativity theory seems to contradict, not only the Galilean transformations and the like, but also the above combination.

Indeed, I knew General relativity was a bit different from the Newtonian theory of gravitation, but I didn't expect such a basic controversy bewteen Einstein and Newton over the ~~necessary~~ sufficient condition for mechanical equilibrium. Newton could define this condition as: "not close to any other gravitational mass and not influenced by any force other than the gravitational one", but Einstein would disagree. This surprises me... HOTmag (talk) 20:02, 30 October 2024 (UTC)[reply]

Newton didn't know about gravitational waves, did he? In Newton's view, gravity is a force just like the pull on a rope, whilst the centrifugal force isn't real and only appears in invalid reference frames. In Newton's view, something is in mechanical equilibrium when all forces (including gravity) and all torques (including gravity) are balanced. In Newton's view, one can be isolated from gravity by being very far from the sun, as he wasn't aware of any object farther away than Saturn. And in Newton's view, bodies act on bodies at a distance.

In Einstein's view, gravity is as real as the centrifugal force and not really a force, but a deformation of spacetime. Gravity isn't directly considered when looking at mechanical equilibrium, but this is solved by having very complex coordinate transformations that can reintroduce the acceleration resulting from gravity. You cannot be isolated from gravity, as we are in a universe full of things, and the farther away, the more massive they get: we can have a star at one AU, but not an entire galaxy. We can have a galaxy at a megaparsec. And finally, bodies don't act on bodies at a distance, but act on local fields and are acted upon by such local fields. The fields provide for the propagation. PiusImpavidus (talk) 09:45, 31 October 2024 (UTC)[reply]

Yes, I'm aware of all these differences between both theories.

As for what I suggested as a condition for mechanical equilibrium, I was wrong when I described it as a "necessary" (and sufficient) condition, because as you say: "we are in a universe full of things", so I've just struck out the word "necessary" in my last response. Additionally: indeed, I defined an "isoloted" body as "not close to any other gravitational mass", but this definition can very easily be sharpened or idealized, by simply saying that an isolated body is a body in an ideal universe that only contains this body and not any other body. That said, Newton and Einstein wouldn't agree about the following intuitive sufficient condition for mechanical equilibrium: "being - both alone in an ideal (theoretical) universe - and also uninfluenced by any force other than the gravitational one". My point was, that Newton could agree to this sufficient condition, while Einstein would not, although this condition sounds very intuitive, if we consider both Newton's two first laws and his law of gravitation (which is of course different from Einstein's field equations). HOTmag (talk) 12:17, 31 October 2024 (UTC)[reply]

(edit conflict) A steadily rotating dumbbell in a zero-gravity environment consisting of a very thin and long bar connecting two extremely massive spheres will emit gravitational waves yet has constant linear and angular momentum. One can argue that this rotating system will actually loose angular momentum due to its rotational energy being transferred to energy dissipated by the gravitational waves. But this lack of rotational mechanical equilibrium is the result of the emission of the gravitational waves and not its cause. --Lambiam 10:47, 29 October 2024 (UTC)[reply]

Thx. Apparently, the natural source - of the torque applied on this system - can't be the gravitational (fictitious) "force", because you're referring to a "zero-gravity environment". Nor can the natural source of the torque be any other natural force (namely: the electromagnetic force, the strong force, and the weak force). Isn't this a bit bizzare? HOTmag (talk) 11:57, 29 October 2024 (UTC)[reply]

Have a look at Mach's principle and Frame dragging. Zero gravity just means no nearby masses. Mach's principle hasn't been verified but there's good reason to think it or something very like it holds so you get an inertial frame when there is 'zero gravity' set by the distant stars. By the way frame dragging happens round a rotating body even if there are no gravity waves. NadVolum (talk) 18:49, 31 October 2024 (UTC)[reply]

October 29

Which ones (if any) of the following seven bodies, have both a [mechanical] equilibrium and a varying quadrupole moment?

HOTmag (talk) 07:56, 29 October 2024 (UTC)[reply]

The four objects in the middle appear to be in equilibrium and have a constant quadrupole moment. The rod in the upper right has varying quadrupole moment, but is in equilibrium (apart from the torque resulting from the emission of gravitational waves). The rod on the lower right has a varying quadrupole moment and is not in mechanical equilibrium, as its centre of mass moves in a circle. I can't be sure about the teapot. It looks like its quadrupole moment varies, its centre of mass moves, so there's a net force, and it doesn't spin on a principal axis, so there's a net torque. PiusImpavidus (talk) 09:16, 29 October 2024 (UTC)[reply]

Thx. By equilibrium I meant mechanical equilibrium, i e. both of forces and of torques (Sorry for not adding this from the beginning). So is your answer still valid after adding this addition? HOTmag (talk) 10:34, 29 October 2024 (UTC)[reply]

Yes, I already assumed that was what you meant by mechanical equilibrium. PiusImpavidus (talk) 13:53, 29 October 2024 (UTC)[reply]

Thx. HOTmag (talk) 15:45, 29 October 2024 (UTC)[reply]

Aardvark cucumbers and Madagascar

in the article on aardvarks habitat and range it states that they are not found in Madagascar

the article on the aardvark cucumber states that it can be found in Madagascar

Yet. both articles make it clear that the aardvark is necessary for the cucumber to grow and thrive

So my question is, what animal assists the cucumber in madagascar where there are no aardvarks?140.147.160.225 (talk) 12:53, 29 October 2024 (UTC)[reply]

I find this interesting. I've found multiple sources for related information. The aardvark cucumber is only eaten by aardvarks (many sources state that, but surely SOMETHING will also eat it). Humans do not eat aardvark cucumbers, and therefore do not farm them. There are no aardvarks on Madagascar. A closely related animal was last noted in 1895. It is assumed extinct. The only source that claims there are aardvark cucumbers on Madagascar is Wikipedia. Every website I found that made that claim was simply a copy of the Wikipedia page. The claim on Wikipedia is unsourced, so it is most likely not true. Unless someone can find a reliable resource that is not a copy of the Wikipedia article, I suggest removing Madagascar from the aardvark cucumber article. 12.116.29.106 (talk) 13:51, 29 October 2024 (UTC)[reply]

Thanks 12.116! I have removed Madagascar from the cucumber article and left an edit summary explaining140.147.160.225 (talk) 16:46, 29 October 2024 (UTC)[reply]

Because resources are very hard to find, I sent emails to a few organizations to ask for help including some nature reserves on Madagascar, a few large botanical gardens with African collections, and a few botanist instructors that I know. If we are lucky, one of them will respond. 12.116.29.106 (talk) 16:51, 29 October 2024 (UTC)[reply]

Why aren't bone balls big enough that they can't leave the socket without cracking bone?

Is this kind of ball-and-socket joint a local fitness maximum ? Sagittarian Milky Way (talk) 19:25, 29 October 2024 (UTC)[reply]

Isn't the question its own answer? Abductive (reasoning) 01:20, 30 October 2024 (UTC)[reply]

Yeah, you're probably better off with a dislocated limb that might pop back in place, than a nasty fracture to a precise structure. In fact there's probably a name for this concept in engineering, something like redundancy (engineering). Except not that. Fault tolerance? Fail-safe? That general conceptual area. Apple's MagSafe system comes to mind. Card Zero (talk) 06:07, 30 October 2024 (UTC)[reply]

There is also the concept of ~~graceful degredation^[5] (Wot? No article??)~~ graceful degradation, which seems to have been engineered by evolution into the functioning of the brain. --Lambiam 12:10, 30 October 2024 (UTC)[reply]

There is a long-standing redirect from graceful degradation [sic] to fault tolerance. Mike Turnbull (talk) 16:22, 30 October 2024 (UTC)[reply]

Wouldn't the possibility of "overshooting the mark" and having the head become to big to move within the cavity also be relevant? --User:Khajidha (talk) (contributions) 12:20, 30 October 2024 (UTC)[reply]

It's been on Wikipedia over 241 months just not a similar-sounding typo of it. Sagittarian Milky Way (talk) 15:47, 30 October 2024 (UTC)[reply]

A generic engineering term is fail-safe: a feature of a design whose purpose is to reduce harm in the event of a failure of the design. An example are pressure relief valves; gradual release of possibly noxious gases is better than a catastrophic blowup. --Lambiam 12:32, 30 October 2024 (UTC)[reply]

The incidental Cam out feature of Phillips screwdrivers similarly reduces damage when stress exceeds a normal range. Philvoids (talk) 14:09, 30 October 2024 (UTC)[reply]

It would give you either a thin connection between the ball and the rest of the bone it belongs to, or a limited range of motion. PiusImpavidus (talk) 16:07, 30 October 2024 (UTC)[reply]

Is the shoulder socket even over a full hemisphere? Sagittarian Milky Way (talk) 17:50, 30 October 2024 (UTC)[reply]

Nowhere near – see the illustrations in Scapula, particularly Figure 1, and in Glenoid fossa, which is the actual 'socket'. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 23:09, 30 October 2024 (UTC)[reply]

Why'd they draw a minimalist mechanical ball joint next to an anatomical one and compare them to mechanical ball joints when there's so many more accurate possible names like ball-and-wok joint or ball-and-hollow joint or ball-and-bowl joint or ball-and-depression joint or ball-and-crater joint? Sagittarian Milky Way (talk) 15:06, 31 October 2024 (UTC)[reply]

Anatomists have traditionally used the 'ball and socket' terminology for a long time, perhaps because the term is well established and familiar to most people even outside of anatomy, whereas the others you mention are not in general use (I assume you just made them up). There is no implication that the socket has to be at least a (concave) hemisphere – see for example Ground glass joint#Ball-and-socket joints.

Note that the linked article mentions the condition you specified in your query title: "An enarthrosis is a special kind of spheroidal joint in which the socket covers the sphere beyond its equator", and the linked reference notes that the (human) hip joint is an example of this. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 15:32, 31 October 2024 (UTC)[reply]

How can hips sometimes pop out without any bone fractures? Is it mostly cause everything's at least slightly compressible even bone and cartilage or is it mostly abrasion from being shoved through a slightly too small hole or is it that the ball would only need help to fit if it filled the socket to the exclusion of the liquid joint lubricant? Sagittarian Milky Way (talk) 18:19, 31 October 2024 (UTC)[reply]

Because, as you suggest, not just bone is involved. There are (in a healthy joint) layers of cartilage and other tissues between the bone of the ball and the wall and rim of the socket, which can be somewhat (painfully) compressed, and bone can be slightly flexible. Also, the socket is only just more than a hemisphere (and individuals will differ a little). {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 19:28, 31 October 2024 (UTC)[reply]

That the ball normally maintains contact with the socket is due to the strength of the somewhat elastic joint capsule keeping them together. But the application of too large a force can distend the membranes of this joint capsule enough for the ball to dislocate (jump out of its usual seat). --Lambiam 20:19, 31 October 2024 (UTC)[reply]

October 31

Does an accelerated body at (instantaneous) rest, emit gravitational waves?

According to LIGO, "every physical object that accelerates produces gravitational waves". But how can the GWs, emitted from the accelerated body, carry away momentum from the body being at (instantaneous) rest? HOTmag (talk) 14:30, 31 October 2024 (UTC)[reply]

Perhaps because "instantaneous rest" is a mathematical abstraction, not a real-world condition that applies to a real-world accelerating body? Others more expert can doubtless address this concept better.

My impression is that you are just making up puzzles using random concepts, as you have previously been doing under this and your previous User name HOOTmag for more than ten years. I'm not intending to play anymore. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 15:39, 31 October 2024 (UTC)[reply]

Sorry, but my question is serious (like all of my questions here). I really don't know how to anwser it correctly (I don't remember I ever made up puzzles using random concepts). HOTmag (talk) 16:11, 31 October 2024 (UTC)[reply]

Zeno of Elea c. 490 – c. 430 BC tried "seriously" to divide time into instants. In his arrow paradox, Zeno states that for motion to occur, an object must change the position which it occupies. He gives an example of an arrow in flight. He states that at any one (durationless) instant of time, the arrow is neither moving to where it is, nor to where it is not. It cannot move to where it is not, because no time elapses for it to move there; it cannot move to where it is, because it is already there. In other words, at every instant of time there is no motion occurring. If everything is motionless at every instant, and time is entirely composed of instants, then motion is impossible. Philvoids (talk) 16:18, 31 October 2024 (UTC)[reply]

Zeno's paradoxes are well known, but the mistake hidden in them has already been discovered, actually after Calculus was discovered. The way to remove Zeno's paradox is achieved by the concept of mathematical limit.

But my question has nothing to do with those old paradoxes, because as opposed to them, I can phrase my question without using any instantaneous velocity, but rather with the rigorous concept of mathermatical limit. I've only used the concept "instantaneous rest" for letting you grasp my question intuitively. If I had used the concept of mathermatical limit, I would have phrased my question otherwise, but the question would have still remained. For example I could ask: What's happening to the GWs emitted by the body, when the body's velocity approaches to zero? Does the GW emission approach to a zero radiation emitted by the body? HOTmag (talk) 16:48, 31 October 2024 (UTC)[reply]

The statement is simplified and not strictly true. A linearly accelerating body without rotation (or with rotational symmetry around the axis of rotation) does not emit gravitational waves — as discussed in another thread the quadrupole moment of the mass distribution needs to change. This is true for almost any real-life body or mass distribution, and this argument could be used to justify making that statement in a non-technical web page for the lay public. The instantaneous rest thing is fine, by the way and in a frame-dependent way, but not particularly relevant here. --Wrongfilter (talk) 16:50, 31 October 2024 (UTC)[reply]

Thank you for this important clarification. I'm quite amazed. Previous threads mentioned the quadrupole moment of the mass distribution, but none of them mentioned what you're caliming now, that "A linearly accelerating body without rotation...does not emit gravitational waves". On the contrary, some users claimed that an acceleration was sufficient for emitting GWs, and nobody disagreed, so I thought they were correct. Now you're surprising me.

Anyway, according to your clarification, I wonder now why our article Gravitational wave claims "An isolated non-spinning solid object moving at a constant velocity will not radiate". Aren't the words "moving at a constant velocity" redundant? HOTmag (talk) 17:02, 31 October 2024 (UTC)[reply]

"A linearly accelerating body without rotation (or with rotational symmetry around the axis of rotation) does not emit gravitational waves ... This is true for almost any real-life body or mass distribution." (emphasis mine).

This made me curious. There are exceptions to this rule? Would you be willing to give some examples, please? (Asking as a member of said lay public.)

Thanks! -- Avocado (talk) 12:51, 1 November 2024 (UTC)[reply]

It's a word of caution. If I had written "all real-life bodies" somebody would have blasted me for that. If you wish you can read it as almost every. --Wrongfilter (talk) 13:06, 1 November 2024 (UTC)[reply]

Yes, this is also what I wondered about, but I finally didn't ask you about that, because I guessed you had only wanted to use a word of caution, as you say now. So it seems you don't rule out NadVolum's reservation "a constant linear acceleration doesn't generate gravitational waves", do you? HOTmag (talk) 13:18, 1 November 2024 (UTC)[reply]

Just a little extra on that - a constant linear acceleration doesn't generate gravitational waves. NadVolum (talk) 18:40, 31 October 2024 (UTC)[reply]

Now you add: "constant". But if the acceleration is not constant, then my question in the header comes back... HOTmag (talk) 18:45, 31 October 2024 (UTC)[reply]

A single, accelerating body doesn't even exist: conservation of momentum says that there must be at least second body, accelerating in the opposite direction. And although a single body has no quadrupole moment, the pair of two bodies has. So the issue is avoided. PiusImpavidus (talk) 20:24, 31 October 2024 (UTC)[reply]

This thread hasn't mentioned a "single" body, and I can't see how a universe containing more than one object avoids the issue. My question is actually: how can the GWs, emitted from a given body accelerated by a jerk (i.e. by a non constant acceleration), carry momentum away from the body being at (instantaneous) rest? HOTmag (talk) 08:52, 1 November 2024 (UTC)[reply]

The question is poorly phrased and possibly based on a misunderstanding.

"But how can the GWs, emitted from the accelerated body," The GWs are produced by the body, but the body doesn't do that on its own. The GWs are emitted by the space surrounding the body and its reaction mass. The waves are, as usual, a far-field approximation. The near-field is a bit more complex. "carry away momentum from the body" Who said that? In the discussion a few topics up on spinning rods I mentioned angular momentum. "being at (instantaneous) rest" Here you make the same error as Zeno (good he was mentioned). The accelerated object is at rest for a time interval of zero, so it must emit zero waves during that time interval, as waves are a continuous phenomenon. You have to consider the emission of waves over a time interval equal to the inverse of the wave's frequency. That's rather basic.

BTW, you won't find a constant acceleration in the universe. Also, a speed of zero is physically irrelevant. You can always make the speed zero by coordinate transformation, which cannot change the physics.

Now I'm wondering, you ask questions on general relativity, which I consider academic master's level of physics, yet make such basic errors, third year secondary school. I can't squeeze seven years of physics education in an answer here; that's a pile of physics books. If you aren't making fun of us, then you started reading that pile from the wrong end. I like to assume good faith and love a good physics question, but that's why I don't always respond to your questions. PiusImpavidus (talk) 17:12, 1 November 2024 (UTC)[reply]

I like to assume good faith. Thank you, and please keep assuming good faith. Yes, I graduated secondry school not long ago, so I may make mistakes sometimes. Anyway, when I state any statement, I only rely on articles in Wikipedia. If you think my wording is wrong, don't hesitate and please tell where my mistake is, and I will thank you from the bottom of my heart.

The GWs are produced by the body, but the body doesn't do that on its own. When I wrote "GWs, emitted from the accelerated body", I used a wording used in our article Gravitational wave: "This gives the star a quadrupole moment that changes with time, and it will emit gravitational waves". Anyway, if you're trying to claim that the wording in Wikipedia is wrong and that a single body cannot emit GWs, then please don't hesitate to say that (I'm still not sure if that's the case because you haven't said this yet), and I will thank you from the bottom of my heart for removing this error - not only from Wikipedia - but mainly from me, because I've always thought that also a single body can emit GWs (provided that its quadrupole moment varies).

"carry away momentum from the body" Who said that? Again, I'm only relying on Wikipedia. Please see our article Gravitational wave: "Water waves, sound waves, and electromagnetic waves are able to carry energy, momentum, and angular momentum and by doing so they carry those away from the source. Gravitational waves perform the same function. Thus, for example, a binary system loses angular momentum as the two orbiting objects spiral towards each other – the angular momentum is radiated away by gravitational waves".

The accelerated object is at rest for a time interval of zero, so it must emit zero waves during that time interval, as waves are a continuous phenomenon. You have to consider the emission of waves over a time interval equal to the inverse of the wave's frequency. That's rather basic. AFAIK, you can always use the well known mathematical operation called integration, for "collecting (uncountably) infinitely many infinitesimals" of instants at which the accelerated system was at rest, and then you receive a non zero quantity of energy of GWs produced by the accelerated system at those instants of rest. The question is, where was the energy/momentum of those GWs carried away from? But maybe this integration is actually impossible, not only physically - because (as you say): "waves are a continuous phenomenon", but also mathematically - because the set of those infinitely many instants (at which the system producing the GWs is at rest), is always a countable set only. Am I right?

BTW, you won't find a constant acceleration in the universe. Yes, but AFAIK you can always conduct an experiment which can artificially create a constant acceleration. Additionally, AFAIK a constant acceleration is important in theoretical physics, so you can regard my question as a theoretical one.

Also, a speed of zero is physically irrelevant. You can always make the speed zero by coordinate transformation, which cannot change the physics. AFAIK, physics does consider a speed of zero, for many purposes, e.g for deciding whether there is some momentum, and whether there is some kinetic energy, and the like. Additioanlly the speed of zero is important for establishing many relativistic concpets, e.g. proper frame, proper reference frame, proper length - being the longest length a given body can have, proper time - being the shortest life-time a given body can have, rest mass - being the smallest mass a given body can have (for those physicists who make a distinction between a rest mass and a relativistic mass), and likewise. HOTmag (talk) 19:03, 2 November 2024 (UTC)[reply]

The multiverse of science

Superhero fiction loves the concept of a "multiverse", that is, infinite universes that exist somewhere and that completely similar to the main universe, except for some details. It can be something big (as in, all heroes are villains instead, something turned the world into a dystopia or a post-apocalyptic wasteland) or something minor (as in, the radiactive spider does not bite Peter Parker but someone else), but in the grand scheme of things the history of the Solar System, Earth, life on Earth and human history are all basically completely the same. Needless to say, that's just a narrative device, one that has led to some awesome stories (and other so-so ones), but no more than that.

But lately I have noticed in actual science publications people who talk about the multiverse, in the real world. And we do have an article about that, Multiverse. Before breaking my head trying to understand the fine details, just a quick question: would the real multiverse be, at least in principle, similar to the multiverse as seen in fiction, or is it a completely different idea that got distorted? And if there are parallel universes, where would they physically be? I dismiss such a question with fiction because of the willing suspension of disbelief, but in science you can't get away that easily... Cambalachero (talk) 18:40, 31 October 2024 (UTC)[reply]

Try and wrap your mind round things like Elitzur–Vaidman bomb tester and if you really can explain it all well I'll be glad to listen! :-) NadVolum (talk) 19:06, 31 October 2024 (UTC)[reply]

In Multiverse § Types you can see different – sometimes very different – meanings of this term. Authors of pop-science articles who bandy the term accordingly do not all use the term with the same meaning. All of it is purely speculative and in most versions has the problem that the theory is unfalsifiable because it makes no testable predictions that differ from current theory, and is therefore generally deemed to fall outside the scope of science proper. --Lambiam 20:04, 31 October 2024 (UTC)[reply]

It is among the Interpretations of quantum mechanics#Influential_interpretations. Card Zero (talk) 23:33, 31 October 2024 (UTC)[reply]

If there was only one universe, where would it physically be? For multiple universes, we can use the same answer. If you own a car, where do you keep it? It's a deep question, but it's not an obstacle to the concept of a person owning two or three cars. Card Zero (talk) 23:47, 31 October 2024 (UTC)[reply]

So the multiverse would be like a multi-car garage? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:56, 31 October 2024 (UTC)[reply]

OK, if you have more than one parking spot it raises the question "how does one parking spot relate to the next, in physical space?", and that's a reasonable question if you know the first parking spot's location in physical space. But when the parking spots are universes, the answer might be "they don't", because nobody ever said the first parking spot was located anywhere anyway. Card Zero (talk) 00:08, 1 November 2024 (UTC)[reply]

Yes. It would be in some other dimension beyond mere physicality. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:23, 1 November 2024 (UTC)[reply]

They are all in your mind. --Lambiam 07:06, 1 November 2024 (UTC)[reply]

I recall one professor contradicting the famous "Cogito ergo sum / I think, therefore I am" as "Maybe he only thinks that he thinks." ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:04, 1 November 2024 (UTC)[reply]

As a few people have said, "The problem with thinking about the universe is that there's nothing to compare it to". That's based on the assumption that universe = "everything that can possibly exist, anywhere". And yet, the human mind can comprehend the concept of an infinite number of different universes, each containing everything that can possibly exist, anywhere. It's no more difficult to work with such an idea than to make great use of the square root of -1. But is it actually true? I'm glad you asked ... -- Jack of Oz ^{[pleasantries]} 17:29, 1 November 2024 (UTC)[reply]

Now imagine a (strongly) inaccessible cardinality of universes. --Lambiam 20:46, 1 November 2024 (UTC)[reply]

I reckon the multiverse concept was inspired by the failure of the universe that we know to contain everything that can possibly exist. This assumes it is bounded and doesn't contain, for instance, versions of itself at all other ages, and versions of itself where the laws of physics are different such that life is impossible. Then the other universes are the other possibilities. They're often synonymous with moments of time, in which case we are constantly moving through universes (or perhaps "featuring in a causally related series of moments" rather than moving - same difference). The parallel universes are moments of time that we don't go to, or in many cases couldn't possibly go to. On the other hand, especially in fiction, the term more often means a causally related series of moments, a timeline or "environment", where subsequent moments are constantly branching and diverging but share a common history. Card Zero (talk) 21:44, 1 November 2024 (UTC)[reply]

Is there a common term indicating that the current value of every derivative (of any order) of the position over time is zero?

That is, the current velocity iz zero, and so is the currect acceleartion, and so is the current jerk, and so forth...

I thought about "static" as a sufficient condition, but I'm not sure, so I'm also asking: Should every "static" body be considered to satisfy the property mentioned in the header? HOTmag (talk) 21:56, 31 October 2024 (UTC)[reply]

Something that doesn't move at all is described as a fixed point or fixed object. This usually means that the object won't move even if a force is applied to it. --Amble (talk) 23:34, 31 October 2024 (UTC)[reply]

Having all time derivatives zero at a particular instant is not the same as having constant position. The standard example is

e^{-{\frac {1}{t^{2}}}}

, but there are lots of other possibilities.

This is important to know when formulating differential topology, as it enables finding bump functions and partitions of unity in the smooth (

C^{\infty }

) category. --Trovatore (talk) 05:31, 1 November 2024 (UTC)[reply]

A term used by mathematicians for the function giving such a position as a function of time is "flat function". --Lambiam 07:03, 1 November 2024 (UTC)[reply]

So my original question can be phrased as follows: Is there a common adjective, describing an object, and indicating that the object's location with respect to time is a flat function? Additionally, can the body's adjective "static" be a sufficient condition, for the above location to be a flat function? HOTmag (talk) 08:28, 1 November 2024 (UTC)[reply]

Extended content

1. immovable 2. immoveable 3. fixed 4. immotile 5. unmovable 6. fast 7. nonmotile 8. stiff 9. firm 10. stabile 11. amovable 12. unmoveable 13. static 14. immoble 15. moveless 16. irremovable 17. rooted 18. stationary 19. nonmobile 20. standing 21. dead 22. nonmoving 23. rigid 24. motionless 25. unbudgeable 26. inamovable 27. unimmobilized 28. unshiftable 29. nonimmobilized 30. unmoving 31. staid 32. immoved 33. nonmutable 34. nonchangeable 35. undeposable 36. untransmutable 37. unrelocatable 38. nonremovable 39. inflexible 40. untranslocatable 41. unfluid 42. nonchanging 43. nonrotatable 44. non-mobile 45. unmigratable 46. unmobilized 47. unstationary 48. non-stationary 49. nonflexible 50. incommutable 51. nonmodifiable 52. nonrelocatable 53. unresizable 54. unfixed 55. unmodifiable 56. unchangeable 57. untransformable 58. untransportable 59. unadjustable 60. intransmutable 61. unflexible 62. nontransportable 63. sedentary 64. invariable 65. nonmigratable 66. nonvariable 67. non-animate 68. noncommutable 69. nondisplacable 70. nondisplaceable 71. immalleable 72. unvariable 73. unmechanizable 74. inanimate 75. unlocomotive 76. confined 77. unmanipulatable 78. nonadjustable 79. undisplaceable 80. uninclinable 81. nondetachable 82. unalterable 83. undislodgeable 84. intransformable 85. unmutable 86. inanimated 87. restagnant 88. nonstationary 89. torpid 90. semistationary 91. unfluidizable 92. unfixable 93. unmaneuverable 94. nonportable 95. unbending 96. nonrotative

Philvoids (talk) 17:16, 1 November 2024 (UTC)[reply]

You need a preferential reference frame to get zero velocity, so the property is not intrinsic but observer-dependent. Have objects with this property been the subject of studies in theoretical physics? If not (and I can't think of a reason why they should be of interest to physicists), it is very unlikely that there is a term of art for the property. --Lambiam 20:00, 1 November 2024 (UTC)[reply]

Time crystal? 176.2.70.177 (talk) 18:03, 2 November 2024 (UTC)[reply]

November 1

Midnight sun in Norway

Directives for military officers and military commanders in the event of an armed attack on Norway has a completely uncited section discussing an alleged event from 1968 on the Russo-Norwegian border:

On the evening of 7 June, the garrison heard the noise of powerful engines coming from the manoeuvres...Actual observations were not possible over the border in the dark...At daybreak the impressive numbers of the Soviet forces staged along the entire border became visible.

Google Maps says that the southernmost point of the border is about 69°N, and based on Midnight sun, it looks like anywhere north of 67°13'N experiences midnight sun by the end of May. Consequently, this means that all points on the Russo-Norwegian border experience midnight sun on 7-8 June, so the whole scenario is impossible. Am I understanding rightly, or have I missed something? This isn't some recent vandalism; it's present in the first version of the page history, apparently translated from the corresponding article in the Norwegian Bokmal Wikipedia. Nyttend (talk) 18:59, 1 November 2024 (UTC)[reply]

The article on the Norwegian bokmål Wikipedia ascribes the difficulty in observing the cause of the hubbub to dårlig vær, bad weather. In the original version on the Norwegian bokmål Wikipedia, the difficulty is said to have been, specifically fog. BTW, in this original bokmål version the alleged incident took place on 7 June 1967. Half a year later, "1967" was changed to "1968" by a user whose only contribution was this change. --Lambiam 20:33, 1 November 2024 (UTC)[reply]

User:Lambiam, there's also a no:Sovjets demonstrasjon av militær styrke ved den norsk-russiske grensen i 1968, with several sources. Do the sources confirm the year, or is 1968 an error? Maybe Theohein was just fixing a typo. Nyttend (talk) 21:25, 1 November 2024 (UTC)[reply]

The sources confirm June 1968 but appear to name 6 June 1968 as the date when Norwegian soldiers stationed along the border with the Soviet Union became alarmed by a sudden advance of Soviet tanks and heavily armed soldiers, stopping only within metres of the border. There is no mention of any difficulties in observing this. Apparently, the information has been kept classified for 40 years. --Lambiam 07:00, 2 November 2024 (UTC)[reply]

November 2

Is there a name for 0.001 miles?

I need to work on a software that works in units of one one-thousandth of a mile internally. Is there a name for such a unit? --193.83.24.42 (talk) 00:35, 2 November 2024 (UTC)[reply]

American silliness? HiLo48 (talk) 00:45, 2 November 2024 (UTC)[reply]

As opposed to French silliness, such as the met-ray. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:09, 2 November 2024 (UTC)[reply]

The Roman mile was by definition a thousand paces (milia passuum). catslash (talk) 01:24, 2 November 2024 (UTC)[reply]

The obvious millimile gets a little use. (The author there can't redefine span (unit) so easily though.) I also found it in a more modern book about chemistry, where it seems to be part of a quiz designed to test the reader's understanding of units: Which length is longer, a millimile or a decameter? but archive.org has stopped showing snippet views of in-copyright books. Millimole tends to pollute search results, which is perhaps why a chemist would be inspired to invoke millimiles. Card Zero (talk) 01:46, 2 November 2024 (UTC)[reply]

See pace (unit). When I worked as a surveyor, we often used this informal unit and with practice it became 99% accurate, good enough for most purposes. Shantavira|^{feed me} 09:21, 2 November 2024 (UTC)[reply]

There is an existing more accurate unit, used by the railways, known as the "link". Your unit is 8 links. The link is divided decimally and contains 7.92 inches. Your unit is therefore (7.92 x 8) = 63.36 inches. Before metrication, Ordnance Survey maps were scaled at 1 inch to a mile. The scale was therefore 1/63 360. 2A00:23D0:FFC:3901:90DF:CC65:72B0:11FF (talk) 14:51, 2 November 2024 (UTC)[reply]

Can I determine distance from size of a photographed object

If a photograph contains an object of known real size, can I use that to determine how far it was from the camera or other device that took the picture? Also, would this question fit better here or on the math reference desk? Primal Groudon (talk) 16:18, 2 November 2024 (UTC)[reply]

For one thing, you would have to know the type of lens. A wide-angle or fisheye lens makes things look farther away than they actually are. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:55, 2 November 2024 (UTC)[reply]

Even with a pinhole camera you couldn't, unless you know the distance inside the camera between the pinhole and the photographic plate or film. Define four variables:

$h$ _ext is the real size of the object;
$d$ _ext is the distance between the pinhole and the object;
$h$ _int is the size of the object's image;
$d$ _int is the distance between the pinhole and the photographic plate or film.

Then

h

_ext :

d

_ext =

h

_int :

d

_int.

If you have the values of three of these variables, you can determine that of the fourth. With simple fixed lenses, this also gives a reasonable estimate.

To determine the value of

d

_int, you don't have to look inside the camera. Just take a picture of an object of known size at a known distance and measure

h

_int. You can now calculate

d

_int. --Lambiam 20:16, 2 November 2024 (UTC)[reply]

Empirical calibration is surely the best approach, but if your camera has a zoom, then you will need to ensure the same level of zoom is used for the measurement and the calibration. You can do this by checking the 35 mm-equivalent focal length in the exif meta-data in the jpeg files. catslash (talk) 23:49, 2 November 2024 (UTC)[reply]

Here is a scenario, if you have an uncropped picture of the White House and you know the Camera Model and the Fixed Len used. You can buy the Camera and Len and take the same Picture at vairous distance of the White House until the your picture matches up to the target Picture. Then you can estimate the distance. 2001:8003:429D:4100:5CCA:727C:C447:3877 (talk) 22:07, 3 November 2024 (UTC)[reply]

There are two factors that change the size of an object in a photo when you fix the lens and zoom. One is the distance to the object. The other is the scaling of the photo itself. If I print a photo as 4x6 it will be much smaller than the same photo printed on 8x10. To measure distance, you really need another photo with known distance from the same camera with same lens and same zoom so you can measure the width of an object of known size on the size of the photo you printed. Then, you know the scaling factor to work out the distance to an object in another photo. Another option is to have two (or more) objects of known size at different distances where you know the distances between them. When in the same photo, you can use the sizes of the objects and the known distance between them to estimate the distance to the camera. 12.116.29.106 (talk) 18:26, 5 November 2024 (UTC)[reply]

Projective Geometry can help if you know the length of something in a picture and the shape of something, e.g. a square on the ground. NadVolum (talk) 18:47, 5 November 2024 (UTC)[reply]

Magenta vs. purple

How come mixing blue and red with equal gets magenta in modern color theory but a bluer purple (more violet-like) in RYB color theory?? Georgia guy (talk) 16:36, 2 November 2024 (UTC)[reply]

No colour theory is perfect, but, more importantly, it is ill-defined what it means to "mix" colours. Different physical procedures will also give different outcomes. --Lambiam 20:20, 2 November 2024 (UTC)[reply]

it depends! Red in RYB is something "not (subtracting) blue" and blue is somewhat "not (subtracting) red". So the red takes the role of the blue in the other system and vice versa. If you constructed a special "blue pigment" with some green in it and a special red pigment with some orange in it, you would mix either the pigments and have a tone between magenta and purple or you would mix the reflected light from both pigments, and have a more bright version of exactly the same tone. I don't know if you can make the brightness the same too. So it doesn't have to be a different colour, but that only works with purple because of the complementarity. Generally the colours in both systems are very differently mixed. 176.2.70.177 (talk) 20:48, 2 November 2024 (UTC)[reply]

November 5