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October 19

Falling into Jupiter

I was thinking one day. Imagine you are an astronaut in free fall to Jupiter. You are in a spacesuit with plenty of oxygen and food available so you are not dying from suffocation or starvation in your spacesuit. You have no way of escaping Jupiter's gravity. There are no other dangers than Jupiter itself. You will eventually enter Jupiter's atmosphere. At which point would you die? JIP | Talk 09:30, 19 October 2024 (UTC)[reply]

You would be moving so fast, that you would burn up in the upper atmosphere, turning into plasma temporarily. But when the space suit ruptured, by burning through, suffocation and depressurisation would be a terminal issue. Graeme Bartlett (talk) 10:53, 19 October 2024 (UTC)[reply]

The thing about vague hypotheticals is, they're vague, and hypothetical. The astronaut could bring along a bigass rocket, and once in a stable orbit around Jupiter fire it to cancel out their orbital momentum until they were at rest relative to Jupiter, then "let go" and just let gravity do its thing pulling them towards Jove's center of mass.

Most spacecraft don't do this b/c hauling reaction mass up a gravity well is a giant pain. The "easiest" way to slow down for landing, is to slam into the atmosphere and let that bleed off your velocity. If you can. If not, for ex the atmosphere is very thin, other methods are required: see the Mars probes, or lunar landings.

(The real non-hypothetical answer: they would be long-dead of acute radiation syndrome before anything else, unless their "spacesuit" was a massive, very dense and multilayered radiation shield chamber) Slowking Man (talk) 04:30, 20 October 2024 (UTC)[reply]

The onset of radiation syndrome is slow enough that the fall is over before the radiation really kills. Or the radiation is very high, then that should be said where and why. 176.0.161.3 (talk) 12:45, 20 October 2024 (UTC)[reply]

Since neutrinos (and dark matter) don't interact with light, so what should happen when light comes across them?

I can think about two options:

Option #1: The light keeps travelling "through" them, as if they don't exist. But if this is the case, then what does that mean, in terms of the neutrino's (and dark matter's) refractive index? Is it identical to the vacuum's refractive index?

Option #2: The light experiences absorption or reflection or scattering, in which case the neutrino's (and dark matter's) momentum must be influenced by that encounter with light, due to the conservation of momentum, so we do see them interact with light, in some sense...

So, what's the correct option? Is it #1 or #2 or another one?

HOTmag (talk) 16:29, 19 October 2024 (UTC)[reply]

Since neutrinos do not interact with light, the light never collides with them. Ruslik_Zero 20:42, 19 October 2024 (UTC)[reply]

By "collides" I meant "comes across" (due to your important comment I've just fixed that in the header) So, what should happen if a [a beam of] photon[s] and a [beam of] neutrino[s] travel towards each other, i.e. on the same route but in opposite directions? Similarly, what should happen when light comes across dark matter? HOTmag (talk) 21:04, 19 October 2024 (UTC)[reply]

protons and electrons interact with photons. What is their refractive index? Short answer: no one can tell, because the refractive index is not of a particle alone. It depends on the interaction. And neutrinos not only do not interact with photons, they don't interact with each other. So you have no interactions to base a refractive index on. 176.0.161.3 (talk) 22:25, 19 October 2024 (UTC)[reply]

The refractive index of a given medium is only relevant if [a beam of] photons can travel through that medium. In the case of protons-electrons you're talking about, nobody claims [a beam of] photons can travel through a proton or through an electron, so I can't see how any refractive index may be relevant in that case. But a refractive index may be relevant in option #1 I was talking about as you can see above in that option. HOTmag (talk) 02:16, 20 October 2024 (UTC)[reply]

Your fundamental problem is you keep trying to think about "quantum stuff" intuitively, in terms of the familiar everyday big world we all have direct experience of via our senses. You're asking what photons etc "really act like". Billiard balls? Pebbles? Ocean waves etc etc? The correct answer is, they act like none of those things. They act like photons. They don't "take up space" in any way we can visualize, or occupy a definite fixed position in space, or "move" by plodding around from point A to B in a fixed interval of time, or "pass through one another", anything like that.

A necessary precondition to really "grokking" "modern physics", is to throw out your preconceptions, and simply start with: what do our observations of things tell us. From those, we make predictions (hypotheses), and then we test them to see if they're right. That's how science is done. And if you think it's all made up, you're presumably reading this on some kind of electronic device, which simply wouldn't work if electrons were really tiny little balls, or photons were really little tiny beams or rays or water waves that "bounced off" other stuff when they "ran into it".

Richard Feynman: Things on a very small scale behave like nothing that you have any direct experience about. They do not behave like waves, they do not behave like particles, they do not behave like clouds, or billiard balls, or weights on springs, or like anything that you have ever seen. Slowking Man (talk) 05:02, 20 October 2024 (UTC)[reply]

What do you mean by keep trying? When did I try to do that for the first time?

When I posted my question in the header, I had already been quite aware of the methodological idea you're describing quite well. Of course what you're depicting is the correct approach, methodologically speaking. But while you're portraying the correct methodological attitude one should take when thinking about modern physics, my question has nothing to do with methodology, because my question is only a practical one, empirically speaking (as follows), so the correct methodological approach you're quite well picturing has nothing to do with what I've practically asked about. To put it in a clear cut way: Just as you can practically ask, what is empirically expected to happen when one actualizes the photoelectric effect - although it heavily involves quantum physics that should be grokked by means of the methodological idea you're describing, so I can practically ask, what is empirically expected to happen when a [beam of] photon[s] and a [beam of] neutrino[s] travel towards each other, i.e. on the same route but in opposite directions, although both the photon and the neutrino are described in that quantum physics.

So, are you claiming that I can't suggest any experiment in which a [beam of] photon[s] and a [beam of] neutrino[s] travel towards each other, i.e. on the same route but in opposite directions? Similarly are you claiming, that once Science detects (somehow) any dark matter, still we won't be able to suggest any experiment in which we send light towards dark matter? Or what are you actually claiming, practically speaking? HOTmag (talk) 07:53, 20 October 2024 (UTC)[reply]

you can never design an experiment where a photon travels on a path. Whatever the path maybe. 😁 For example you can do a photon in a fiber travel from a source to a detector. You will never know that the photon really after the first atom leaves the fiber, travels to the black hole in the centre of the Andromeda galaxy, does half a round around the black hole, a quantum leap short of the event horizon, comes back at the last atom of the fiber and hitting the detector bearing the spectral attenuation of a sodium atom in the middle of the fiber it never passed on the way. Okay that is an extreme example for a very improbable but possible event in quantum mechanics. Now to your question. What if "never interacts with" is a code for active avoidance? That would mean a neutrino near another particle (photon, neutrinos...Whatever) goes out of the way and resumed its travel after the particle has passed. What would you do then? Even going out of time is possible. Think about the length of the way to Andromeda! 176.0.154.107 (talk) 13:13, 20 October 2024 (UTC)[reply]

According to your attitude, the very concept of "refractive index" of a given medium through which light travels, would have had no meaning. Additionally, please notice that my original post (i.e. the question in the header and under it), mentions no photons, but rather mentions light only, for example a beam of photons. Anyway, thanks to your comment, I've just added "a beam of" before every "photon" mentioned in my later responses (following my original post). To sum up: the main question in my original post still remains. HOTmag (talk) 13:36, 20 October 2024 (UTC)[reply]

Rewriting "photon" as "a beam of photons" changes the question. The exact path that a photon will follow cannot be predicted. Subsequent detection of a single photon is possible but allows only an estimate of the spread of likely refractive indices the photon has traversed. One increases the accuracy of a determination of refractive index by averaging measurements of many photons i.e. using "a beam of photons". However there will be practical limits to the focusing of both light sources and detectors. Philvoids (talk) 14:30, 20 October 2024 (UTC)[reply]

Rewriting "photon" as "a beam of photons" changes the question. Please re-read the second sentence (the one beginning with "Additionally") in my last response.

As for the rest of your response, I agree, but what's the answer to my original question summed up in the header? HOTmag (talk) 15:44, 20 October 2024 (UTC)[reply]

So, are you claiming that I can't suggest any experiment in which a [beam of] photon[s] and a [beam of] neutrino[s] travel towards each other, i.e. on the same route but in opposite directions? Similarly are you claiming, that once Science detects (somehow) any dark matter, still we won't be able to suggest any experiment in which we send light towards dark matter?

Essentially, yes, assuming we're right about dark matter not interacting with the electromagnetic field. Or, perhaps it could be put as: 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), because why would we?

In this vein: the neutrino fields don't interact with the EM field. The question "what if a beam of X and Y" travel towards each other" is still formulated in intuitive "classical" terms. Talking in strict QM terms, questions like "a beam of X and beam of Y" are ill-formed questions: to be meaningful (answerable) questions, rephrase them in terms of quantum operators, Dirac matrices, Hamiltonian mechanics etc.

Michio Kaku has some good advice for people "talking science": [1]:

Extended content

What to Do If You Have a Proposal for the Unified Field Theory?…and what not to do

Due to volume of e-mail I have received (several thousand at last count) I cannot answer all requests, especially those from individuals who have a new proposal for completing Einstein’s dream of a unified field theory, or a new theory of space and time. However, I would like to give some guidelines for people who have thoughtfully pondered the question of the meaning of space-time.
1) Try to summarize the main idea or theme in a single paragraph. As Einstein once said, unless a theory has a simple underlying picture that the layman can understand, the theory is probably worthless. I will try to answer those proposals which are short and succinct, but I simply do not have time for proposals where the main idea is spread over many pages.
2) If you have a serious proposal for a new physical theory, submit it to a physics journal, just as [sic] Physical Review D or Nuclear Physics B. There, it will get the referee and serious attention that it deserves.
3) Remember that your theory will receive more credibility if your theory builds on top of previous theories, rather than making claims like “Einstein was wrong! ” For example, our current understanding of the quantum theory and relativity, although incomplete, still gives us a framework for which we have not seen any experimental deviation.Even Newtonian gravity works quite well within its domain (e.g. small velocities). Relativity is useful in its domain of velocities near the speed of light. However, even relativity breaks down for atomic distances, or gravitational fields found in the center of a black hole or the Big Bang. Similarly, the quantum theory works quite well at atomic distances, but has problems with gravity. A crude combination of the quantum theory and relativity works quite well from sub-atomic distances (10^-15 cm.) to cosmological distances (10^10 km), so your theory must improve on this!
4) Try not to use vague expressions that cannot be formulated precisely or mathematically, such as “time is quantized, ” “energy is space, ” or “space is twisted, ” or “energy is a new dimension,” etc. Instead, try to use mathematics to express your ideas. Otherwise, it’s hard to understand what you are saying in a precise manner. Many referees will throw out papers which are just a collection of words, equating one mysterious concept (e.g. time) with another (e.g. light). The language of nature is mathematics (e.g. tensor calculus and Lie group theory). Try to formulate your ideas in mathematical form so that the referee has an idea of where you are coming from.
5) Once formulated mathematically, it’s then relatively easy for a theoretical physicist to determine the precise nature of the theory. At the very least, your theory must contain the tensor equations of Einstein and the quantum theory of the Standard Model. If they lack these two ingredients, then your theory probably cannot describe nature as we know it. The fundamental problem facing physicists is that General Relativity and the quantum theory, when combined into a single theory, is not “renormalizable, ” i.e. the theory blows up and becomes meaningless. Your proposal, therefore, has to give us a finite theory which combines these two formalisms. So far, only superstring theory can solve this key problem. Important: this means that, at the very minimum, your equations must contain the tensor equations of General Relativity and the Standard Model. If they do not include them, then your theory cannot qualify as a “theory of everything.”
6) Most important, try to formulate an experiment that can test your idea. All science is based on reproducible results. No matter how outlandish your idea is, it must be accepted if it holds up experimentally. So try to think up an experiment which will distinguish your result from others. But remember, your theory has to explain the experiments that have already been done, which vindicate General Relativity and the quantum theory.
Good luck!

--Slowking Man (talk) 16:28, 20 October 2024 (UTC)[reply]

In my view, presenting Michio Kaku's advice in this thread is redundant, as follows.

Introduction: the reason for which I mentioned the photoelectric effect in my last response to you, is because this effect can be formulated, not only in the language of Quantum chemistry, but also in the language of Classical electromagnetism - which indeed disagrees with this effect but can still tell us what it disagrees with.

The same is true for my original question in the header: It can also be formulated in the language of classical mechanics, as you have done yourself, stating in a classical language (a bit relativistic yet not the language of quantum mechanics): 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. To sum up, you agree to option #1 (in my original post), i.e.: We will see the beam of light keep travelling in the same 4D-route without any change, as if this route is not intersected by the 4D-route of the beam of neutrinos. Am I right? HOTmag (talk) 17:35, 20 October 2024 (UTC)[reply]

"Doing" special relativity, which ignores gravitation (assumes flat spacetime) and plotting it out on a Minkowski diagram, correct, b/c they only interact via the weak interaction and it's called that b/c seriously it's really weak. (Not as weak as gravity though!) Which is why gazillions of solar neutrinos are flooding through us and the entire Earth after plowing through half the Sun, like we're all not even there, constantly. (The neat fact being that what changes is simply the direction they come from: at night they're coming upwards from the ground having just zipped through the entire planet, after calling on Earth's day side!)

In gen rel, it would still be the same, b/c the only thing changing is adding in gravitation. The neutrino's mass is immensely tiny, thus its stress-energy-momentum tensor has accordingly miniscule effect on local spacetime geometry (which is what we call "gravity", in GR). Photon's mass is, well, zero, so it has even less effect. And gravitation is really weak.

...Unless, you can crank things up to truly mind-and-spacetime-warping energies, and channel and confine absolutely mind-boggling amounts of photons into a vanishingly-tiny volume of space. Somehow. Photons are bosons, which, unlike fermions such as quarks (or neutrinos), are "allowed" to have all identical quantum numbers if they feel like it. Meaning their wavefunctions can completely overlap and they can all "take up" an arbitrarily small volume. So if you figure how to do that out do tell the scientific journals, preferably before building your death ray and taking over the world.

Nota bene: when physicists these days are "talking shop" generally they only ever use "mass" to mean "the invariant or 'rest mass' according to GR": [2]. So I will try to do the same. --Slowking Man (talk) 19:38, 20 October 2024 (UTC)[reply]

Unless, you can crank things up to truly mind-and-spacetime-warping energies, and channel and confine absolutely mind-boggling amounts of photons into a vanishingly-tiny volume of space. Btw, some weeks ago I read a scientific article (I think in Nature or in Science) that discovered that a Kugelblitz was actually impossible, because it would've started to emit radiation before it became a black hole, so it would never become a black hole...

when physicists these days are "talking shop" generally they only ever use "mass" to mean "the invariant or 'rest mass' according to GR. I think you've noticed (as follows) that this fact is irrelevant, because the geometry of spcetrime is shaped by energy (and momentum) rather than by mass.

The neutrino's mass is immensely tiny...Photon's mass is, well, zero, so it has even less effect. Correct, less effect but not zero effect, because the geometry of spcetrime is shaped by energy/momentum rather than by mass. Anyway, thank you for noticing the (immensely tiny) generally-relativistic effect being done to the geometry of spacetime by each beam, thus influencing the other beam, so actually they do have some impact on each other after all, yet not via any force other than the fictitious force of gravitation. Well, your noticing this generally-relativistic effect was an important reservation. Anyway, what I'm taking from your answer to my original question is as follows: Both beams don't interact with each other, as far as gravity is ignored. HOTmag (talk) 22:35, 20 October 2024 (UTC)[reply]

I think you've noticed (as follows) that this fact is irrelevant, because the geometry of spcetrime is shaped by energy (and momentum) rather than by mass.

See invariant mass: a bunch of photons with the same momentum vector have zero effect on spacetime geometry, because a single photon (having no invariant mass) has no center-of-momentum frame where it is at rest, no matter what Lorentz transformations are applied to it. This is why a beam of light always travels at c in vacuum. If you have two+ photons with different vectors relative to each other (they have a scattering angle) then you can calc a center-of-momentum frame for the multiple-photon system, and then the photons have an effect on the spacetime metric. Hence the kugelblitz idea: if hypothetically you could stuff tons (heh) of photons into a tiny volume of space, they can't be all at rest relative to each other, so the whole system would have a CoM frame and the photons would affect the local metric. In general, only changes in energy have physical significance. --Slowking Man (talk) 15:10, 22 October 2024 (UTC)[reply]

when I wrote that the geometry of spacetime was shaped by energy and momentum rather than by mass, I tried to be brief. actually I meant what I'd already written in my last response of this old thread: that the geometry of spacetime was shaped by the "density and flux of momentum and of energy".

Anyway, my main point in my last response, was not about energy (or about momentum), but rather about mass, that is: as far as gravity is concerned, mass is irrelevant. HOTmag (talk) 15:57, 22 October 2024 (UTC)[reply]

Interactions are predicted. See [3]. Modocc (talk) 22:44, 20 October 2024 (UTC)[reply]

Thank you for this source. HOTmag (talk) 23:02, 20 October 2024 (UTC)[reply]

October 20

Soap bubbles and flatulence

Let's say that someone ate a whole can of beans for lunch and had a piece of steak, and some milk to wash it all down. A couple of hours later, they're suffering from farting problems, as they have to fart a lot, and the gas doesn't smell good at all. They have filled a bathtub full of water and added a generous amount of soap, and in they go the bathtub. They fart in the there and those bubbles of soap caused by the release of gas travel up onto the surface. If they had a lighter nearby (for whatever reason) and tried to ignite those bubbles, would the bubbles catch on fire? Kurnahusa (talk) 05:16, 20 October 2024 (UTC)[reply]

I don't think the human digestive system works as fast as that, but leaving that aside, it's well known that human flatulance is inflammable – lighting one's farts is a widespread activity; I recall a story that one squad of British Army recruits managed to burn down their barracks hut while indulging in it; I would have liked to have been in the Colonel's office the following morning. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 05:40, 20 October 2024 (UTC)[reply]

I guess? Why wouldn't they? Is there a reason you were expecting them not to? Fun bio facts time the flammable stuff in human flatus is mostly hydrogen gas, made by your little belly buddies fermenting complex carbohydrates that your digestive system can't tackle. And they actually share some of the products with your cells and they're probably good for gut health. (Non-human primates consume buttloads of fiber, as did all humans pre-agriculture.)

The rest of it is mostly swallowed air which makes its way down there, along with small amounts of volatile sulfur compounds also produced by your flora, thiols, which your smeller is extremely sensitive to. That's the smelly stuff. Slowking Man (talk) 19:52, 20 October 2024 (UTC)[reply]

Haha, no reason why I wouldn't think the bubbles were not flammable; after all, the gases are basically trapped inside, but very interesting as well as informative - thank you! A while back I sprayed some gas from a nearly empty alcohol rub bottle into water and ignited it, and so I thought, if it was possible, the same stuff applies to farts. Kurnahusa (talk) 23:16, 20 October 2024 (UTC)[reply]

A bubble of flammable gas in water is an interesting apparatus where you can see inside the bubble before and during it popping. With an electronic igniter, could be fun to try to ignite the bubble itself to demonstate the effect of UEL. Are H2 and methane actually "flammable" when pure? DMacks (talk) 02:11, 21 October 2024 (UTC)[reply]

"Flame"/"fire" is a redox reaction, between at least 2 reactants, a fuel and an oxidizer. H2 and CH4 fill the "fuel" role while most commonly O2 takes the "oxidizer" one. If by "pure" you mean, "a volume of gas which is 100% eg H2 and no other substance", then no, no "flame" can occur without mixing w/ an oxidizer first. H2 is routinely used as a coolant in environments where lots of "sparky"/"arc-y" stuff is liable to happen; other stuff is simply kept out of said environment (already necessary anyway for things to work right). H2 is hard to beat in terms of drag coefficient!

This is why if you block an ICE's air intake, it's not going to be running much longer, and why jet engines don't work in space, and why rockets generally need two things, fuel and oxidizer. (The latter being often O2, sometimes even nastier stuff.) As do most chemical explosives—a rocket is in essence just a bomb that explodes more slowly, if everything goes right. (And when it doesn't...) --Slowking Man (talk) 22:28, 22 October 2024 (UTC)[reply]

Conventional jet engines may not work in space, but see Bussard ramjet. --Lambiam 16:55, 27 October 2024 (UTC)[reply]

For pure hydrogen or methane, not by itself, no. The combustion of the gases inside the bubbles are using the air around it, which doesn't contain too much oxygen. Fun thing is if you managed to inject some oxygen into those fart bubbles, the mixture would be potentially explosive, and when ignited can create a small bang. People do, however, like to weld with fuel and oxidizer mixtures (see oxy-fuel welding); the flame of oxygen and acetylene can go up to 3,500 °C (6,330 °F), hot enough to melt a variety of metals. Kurnahusa (talk) 00:56, 23 October 2024 (UTC)[reply]

For the record, I was proposing experiments, not being ignorant of UEL that I had mentioned:) DMacks (talk) 12:01, 24 October 2024 (UTC)[reply]

Must every moving [accelerating] body lose energy, namely the energy of the gravitational waves emitted by that body while moving [accelerating]?

HOTmag (talk) 10:08, 20 October 2024 (UTC)[reply]

Can you accept as a counter example the body envisaged in Newton's first law of motion that remains at rest, or in motion at a constant speed in a straight line, except insofar as it is acted upon by a force? Philvoids (talk) 13:36, 20 October 2024 (UTC)[reply]

Given a body moving at a constant speed, there is a reference frame in which the body is at rest, from which it follows that it does not emit gravitational waves. Only a change in the gravitational interaction between massive bodies can stir up the gravitational field. --Lambiam 15:31, 20 October 2024 (UTC)[reply]

OP's apology: Sorry for replacing the correct word "accelerating" by the wrong word "moving". I've just fixed that in the header [by brackets]. HOTmag (talk) 16:00, 20 October 2024 (UTC)[reply]

Responders can lose interest in freely helping a questioner who keeps changing their question. Can you accept that energy must always be added (or subtracted) to accelerate (or decelerate) a body? Reference: Kinetic energy. Philvoids (talk) 19:45, 20 October 2024 (UTC)[reply]

Responders can lose interest in freely helping a questioner who keeps changing their question. What do you mean by keeps changing my question? When did I change my question, excluding this single time (for which I have already apologized)?

Can you accept that energy must always be added (or subtracted) to accelerate (or decelerate) a body? Yes, I can, and I do accept. Still, I'm asking if, besides the energy your're talking about, one should also take into account another amount of energy that should actually be subtracted because of the gravitational waves which are always emitted by every accelerating body. HOTmag (talk) 22:59, 20 October 2024 (UTC)[reply]

The term Gravitational wave rather than gravity wave is used in article space.

Energy (luminosity) carried away by gravitational waves is purportedly given by Einstein's Quadrupole formula

{\frac {dE}{dt}}=\sum _{ij}{\frac {G}{5c^{5}}}\left({\frac {d^{3}I_{ij}^{T}}{dt^{3}}}\right)^{2}

As yet this has been only partially confirmed by an observation of a binary star combination of a neutron star and a pulsar (earning the 1993 Nobel Prize in Physics). Research continues and I think we are far from the kind of laboratory demonstration that is needed to cement this theory to the same extent as, for example, the refined measurements of Gravitational constant G initiated (effectively but not deliberately) by Cavendish. Philvoids (talk) 02:11, 21 October 2024 (UTC)[reply]

The term Gravitational wave rather than gravity wave is used in article space. Of course. Has anyone ever used the term "gravity wave", in this thread?

Energy (luminosity) carried away by gravitational waves is purportedly given by Einstein's Quadrupole formula. Now let's assume that Einstein was correct. So, regardless of the other kind of energy mentioned in your previous response, must every accelerating body lose the kind of energy you're mentioning in your last response, namely the energy of the gravitational waves emitted by that body while accelerating? HOTmag (talk) 09:42, 21 October 2024 (UTC)[reply]

According to the theory, a spherically symmetric pulsing body wouldn't emit gravitational waves. NadVolum (talk) 10:46, 21 October 2024 (UTC)[reply]

Does it mean that spherically asymmetric pulsing bodies would? HOTmag (talk) 14:37, 21 October 2024 (UTC)[reply]

The opposite of "spherically symmetric" is "not spherically symmetric". The pulsation of body needs to respect certain symmetries in order to keep its centre of mass at rest. As the formula says, the quadrupole of the mass distribution needs to change. Think of a wobbling drop of water. A rotating barbell emits gravitational waves, as does (in a similar way) a pair of orbiting black holes. All the detected gravitational wave events are of that type, occasionally with a neutron star in place of a black hole. --Wrongfilter (talk) 16:12, 21 October 2024 (UTC)[reply]

I am content that the single counter-example provided by NadVolume answers the OP's question. Philvoids (talk) 16:41, 21 October 2024 (UTC)[reply]

Well, if you're content, I guess I'll just keep my mouth shut in the future. --Wrongfilter (talk) 16:45, 21 October 2024 (UTC)[reply]

Thank you for these examples. So, regardless of the kinetic energy added to an accelerating body, do the bodies in your examples also lose radiant energy - due to the emission of gravitational waves? HOTmag (talk) 19:30, 21 October 2024 (UTC)[reply]

The "bodies" as a whole (barbell, binary BH) are not accelerated (although parts of them are, e.g. the individual BHs), and no kinetic energy is added to them. Yet their rotation causes them to emit gravitational waves and they lose energy through them. The barbell's rotation slows down, the BHs approach each other and finally coalesce. Incidentally, I mentioned the barbell because Bernard Schutz uses that example (in Class. Quantum Grav. 16, A131 (1999)) to illustrate the strength of the emitted gravitational waves, which is tiny except for the most extreme cases. --Wrongfilter (talk) 05:52, 22 October 2024 (UTC)[reply]

If the rotating barbell is in vacuum, it's not supposed to slow down, is it? If it doesn't slow down, and it doesn't receive any kinetic energy, then this barbell will keep losing energy "for ever", or rather until it eventually "evaporates" (like a BH), am I right? HOTmag (talk) 12:49, 22 October 2024 (UTC)[reply]

Emitting gravitational waves slows its rotation. If the reason it emits gravitational waves is that it's rotating, it should cease emitting them (and thus cease losing energy) when it's lost enough energy to stop rotating. Its mass won't evaporate. -- Avocado (talk) 22:40, 22 October 2024 (UTC)[reply]

Why losing energy may make the body slow down or stop rotating? What about the conservation of angular momentum? HOTmag (talk) 23:32, 22 October 2024 (UTC)[reply]

You are not right, there is no reason for the barbell to evaporate. The rotation slows down because the GWs carry off energy (and angular momentum). If the thing is not in vacuum the GW effect is overwhelmed by friction. But this is a highly idealised example meant to illustrate the momentary emission of GWs. It does not occur as such in nature, and it is not worth thinking it through to the end. --Wrongfilter (talk) 13:02, 22 October 2024 (UTC)[reply]

I still think, Bernard schultz's example you've mentioned, is interesting, and worth thinking.

I've asked about vacuum, in which no friction exists.

Apparently, according to the conservation of angular momentum, this momentum is not supposed to change. So, I still wonder, what may prevent the rotating barbell from continuing to rotate "for ever". As long as it rotates, it emits GWs, thus loses energy, without changing the angular momentum, until the barbell loses all of its energy, i.e. until it "evaporates". I wonder where the mistake lies.

Is it possible, that I'm wrong with the conservation of angular momentum, so that what I was taught in school about this conservation in vacuum was not that accurate? HOTmag (talk) 14:10, 22 October 2024 (UTC)[reply]

Gravitational_wave#Energy,_momentum,_and_angular_momentum. --Wrongfilter (talk) 16:42, 22 October 2024 (UTC)[reply]

Thank you for this clarification. Consequently, all of the basic laws of conservation (i,.e. conservation of energy, of linear momentum and of angular momentum), are only true for spherical symmetric bodies, or bodies that don't rotate. All other bodies, emit GWs, so they can't satisfy those laws of conservation any more, at least according to the theory. Do you think we should mention this fact (an important one IMO) in the respective articles about those laws? HOTmag (talk) 17:36, 22 October 2024 (UTC)[reply]

No!!!!! Do you not understand how conservation laws work???? --Wrongfilter (talk) 18:09, 22 October 2024 (UTC)[reply]

Our article Angular momentum indicates "The symmetry associated with conservation of angular momentum is rotational invariance.". Does this reservation exclude every non-spherical symmetric body? If it does then all is fine. But if it doesn't then:

AFAIK, if no external forces act, then the angular momentum must be conserved. Correct? However, non-spherical symmetric bodies that rotate, emit GWs, so the angular momentum of those bodies is not conserved. Correct?

If I'm correct, then the conservation of angular momentum does not hold in all cases where no external forces act. What's wrong? HOTmag (talk) 18:26, 22 October 2024 (UTC)[reply]

The single counter-example provided by NadVolume, only answers the question in the header, but I was waiting for an answer to my follow-up question addressed to NadVolume. It seems that Wrongfilter gives a positive answer, by two theoretical examples: the wobbling drop of water, and the rotating barbell (besides the empirical example of a pair of orbiting black holes). HOTmag (talk) 19:21, 21 October 2024 (UTC)[reply]

How is the center of mass of the rotating barbell not at rest? The distribution of mass in the volume it rotates within changes, but if it's rotating around the center of mass, isn't the center of mass stationary? -- Avocado (talk) 22:28, 21 October 2024 (UTC)[reply]

There may be a misunderstanding here about what a gravitational wave is like. It does not push and pull in the direction from which it comes - it is polarized and squishes and stretches at right angles to its path. If what the observer sees is symmetric then they won't see gravitational waves. NadVolum (talk) 23:47, 21 October 2024 (UTC)[reply]

I did not say (or at least tried not to say) that the CM of the barbell is not at rest, quite the opposite. The rotating barbell is not spherically symmetric but still has some symmetry. My comment was triggered by the word "asymmetric", which I think is not entirely appropriate, and then tried to illustrate systems with a varying quadrupole moment. --Wrongfilter (talk) 05:52, 22 October 2024 (UTC)[reply]

Got it - thanks for the clarification, and sorry for misreading! -- Avocado (talk) 22:34, 22 October 2024 (UTC)[reply]

it is polarized and squishes and stretches at right angles to its path. Are the GWs emitted from both sides at right angles to the body's path? If they are then they have opposite directions, don't they, so the total momenta carried by them is zero, isn't it, so they don't have to carry momentum away from the body emitting them, do they? But our article about gravitational waves claims they do, so I'm probably missing something here... HOTmag (talk) 09:26, 1 November 2024 (UTC)[reply]

October 21

datetime for cub birth

[4] when did this happen. date? time? story out oct-21 but it doesn't say time of event. .... it's for news. -- Gryllida (talk, e-mail) 04:41, 21 October 2024 (UTC)[reply]

Such news articles are based on press releases put out by the organizations featured in the news, in this case Cotswold Wildlife Park. Large parts of it are taken from a "Park News" item on Cotswold's website. The latter also does not mention when the young was born. The release date of this news item was likely inspired by World Lemur Day being celebrated on the last Friday of October, this year 25 October, and a reasonable guess it was released just before the article in The Guardian was published, which has publication date 20 October. The "Park News" item features a photo whose caption reads, "The Greater Bamboo Lemur Baby bred at Cotswold Wildlife Park – aged 5 weeks", so the baby probably arrived near mid-September. Since the park has successfully bred more than 70 lemurs,^[5] this is not Earth-shattering news that deserves careful attention. --Lambiam 06:10, 21 October 2024 (UTC)[reply]

Thanks for checking. The 5 weeks age helps. Gryllida (talk, e-mail) 06:47, 21 October 2024 (UTC)[reply]

"successfully bred more than 70 lemurs" Different species. As the linked article says, "Only 36 greater bamboo lemurs are in captivity globally". Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 13:25, 22 October 2024 (UTC)[reply]

Why do we use 12 o'clock to represent midnight and noon?

It occurred to me recently that the way we number and label hours is rather odd. We divide the day into two twelve-hour sections, starting at midnight and noon, but we number the hours starting an hour after that. This leads to various oddities: 11am is followed by 12pm, not 12am; likewise 11pm is followed by 12am (something that people often get confused about). 11:59pm and 12:01am are different days, despite the numbering logically implying that they are part of the same day. If the 12-hour clock was invented now, I suspect we would define midnight and noon as zero hours, but the concept of 12-hour semi-days predates the concept of zero. But given the way things were typically numbered in the absence of zero, and the way we still number dates, it occurred to me that it would be more sensible, and more expected, to use 1 o'clock to mark the start of the day, and the start of the afternoon. (That would give us a morning running from 1:00am to 12:59am, and an afternoon running from 1:00pm to 12:59pm. No weird flipping between am and pm at 12, all consecutive numbers are in the same semi-day). So I'm wondering: why was the modern notation adopted? I've looked at 12-hour clock and Hour but they don't explain why this system was adopted, only that it started to become common in the 14th century, displacing the earlier system of using twelve (seasonally-varying) hours for the period of sunrise to sunset. Iapetus (talk) 15:12, 21 October 2024 (UTC)[reply]

This doesn't answer your question, but in Japanese usage 午前12時 ("12am") means noon and 午後12時 ("12pm") means midnight. Also possible are 午前0時 ("0am") for midnight and 午後0時 ("0pm") for noon. :) Double sharp (talk) 15:17, 21 October 2024 (UTC)[reply]

Japanese time is high IQ! Not only does it do XX:XX to 24:00 like much of the world instead of XX:XX to 0:00 or 00:00 it also does things like this bar's open 16:00 to 28:00 or "trains run till 25:00". Mechanical clocks once had only hour hands and had to have their drift fixed every day with a glance at a sundial, it took a long time for people to stop thinking in Roman numerals and "this is the first hour" instead of "it's 12:27". If all civilizations had 0 clocks would probably not illogically have a 1 at the top instead of 0. Also am and pm mean ante and post meridian, they CAN'T change at 1:00. After the noon meridian not midnight cause the Sun's midnight meridian crossing is invisible unless midnight is in the day. Sagittarian Milky Way (talk) 16:15, 21 October 2024 (UTC)[reply]

Time is measured continuously: it's now 8 hours plus 32 minutes plus 7 seconds past midnight and this is only the case for a single moment. Days are counted discretely: it's now the 22nd day of the 10th month of the 2024th year since the epoch and this is the case for the entire day. That's why time starts at 0 and dates at 1. It's also why time is in big endian order and date in little endian in most European languages. PiusImpavidus (talk) 08:32, 22 October 2024 (UTC)[reply]

11am is followed by 12pm It is not, it is followed by noon. And then by 12:01pm.

11pm is followed by 12am It is not, it is followed by midnight. And then by 12:01am.

something that people often get confused about Well, quite.

There are no such things as 12am or 12pm, by definition.

It may help to clear your confusion if you consider what "am" and "pm" mean. "Before meridian" and "after meridian". In other words, before/after noon. Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 09:49, 22 October 2024 (UTC)[reply]

"There are no such things as 12am or 12pm, by definition" - an exception being in the datetime libraires of programming languages where they are defined. Sean.hoyland (talk) 10:55, 22 October 2024 (UTC)[reply]

My dictionary defines various gods. They don't exist either. Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 13:01, 22 October 2024 (UTC)[reply]

There are probably some people out there that approach the whole 12am/12pm issue from a mathematical Platonist perspective. Sean.hoyland (talk) 14:51, 22 October 2024 (UTC)[reply]

"Time is an illusion; lunchtime, doubly so."

—Douglas Adams, The Hitchhiker's Guide to the Galaxy

--Slowking Man (talk) 15:50, 22 October 2024 (UTC)[reply]

Assistance is available at 12-hour clock#Confusion at noon and midnight. In some applications of the 24-hour clock, midnight is denoted as 0000 or 00:00, rather than 1200 or 12:00. There is a lot of logic in that! Dolphin (t) 10:49, 22 October 2024 (UTC)[reply]

In some applications of the 24-hour clock. Isn't this the case for all applications of the 24-hour clock? I've never seen one that doesn't use 00:00, and if there was an exception, I would expect it to be using 24:00. Iapetus (talk) 12:25, 22 October 2024 (UTC)[reply]

Your typical digital clock will display non as 12:00PM, not 12:00. There is an infinitesimally short period of time at exact noon when it is neither AM nor PM. But for your typical digital clock (with minute-precision) there will be a whole minute (minus that infinitesimal) where it is showing 12:00 post meridian, so use of that PM is probably not unreasonable. Iapetus (talk) 12:23, 22 October 2024 (UTC)[reply]

I think the South Koreans used to say people were in their first year when they were born but the west has always said they were no years old until theiy were one year old. Clocks follow that western rule. NadVolum (talk) 11:52, 22 October 2024 (UTC)[reply]

Yup, see East Asian age reckoning. Double sharp (talk) 13:07, 22 October 2024 (UTC)[reply]

That system is the one used worldwide for horses, irrespective of whether they were foaled in the northern or southern hemisphere. All share a common birthday - 1 January. Lots of previous discussion Wikipedia:Reference desk/Archives/Humanities/2011 December 8#Afghanistan time and Wikipedia:Reference desk/Archives/Humanities/2012 September 18#Either the Calendar is wrong or the Clock is. It's that simple. 2A02:C7B:21A:700:AD1D:15C5:CE76:21CE (talk) 16:17, 22 October 2024 (UTC)[reply]

Timekeeping systems are all socially constructed human systems, defined by humans—though (usually) linked to one or more "real-world" physical referent(s). Which makes this more of a Humanities desk question. This is the thing that people are to some degree going in circles about here. The only stuff that's physically "real" as in, the consequences of underlying invariant physical laws existing outside of humans, time-wise, are spacetime and the things embedded in it, which we humans model and interpret using tools like Minkowski diagrams and four-vectors. (In Earth orbit time passes more quickly than stuck down here, b/c time is relative. So don't take a trip up to orbit if you really wanna "make every minute last"!)

A "day" if defined as, one full spin of Earth about its major axis, never takes exactly 24 "hours" to complete; we just pretend it does for convenience (humans like nice simple round numbers) and occasionally arbitrarily adjust our major timekeeping systems to compensate for the accumulating measurement error. More starkly sometimes various places decide arbitrarily boom now it's a different time because we want it to be. In the past various human communities decided "okay now it's instantly like 2 weeks later. Japan used to use a calendar based on just, once in a while we change the current "era" and now it's a new one, and still does ceremonially though they cut back on the frequent "time-skips". Etc etc. --Slowking Man (talk) 16:45, 22 October 2024 (UTC)[reply]

A full spin of Earth about its major axis is a sidereal day, which is closer to 23 hours, 56 minutes and 4 seconds. Leap second adjustments are not made arbitrarily, but to keep our clocks in sync with solar time. --Lambiam 06:00, 23 October 2024 (UTC)[reply]

Japan also had a year count that's approaching 2700 now but the Western year count's been more popular for awhile. The era system can cause cool names like calling a skilled sportsman Monster of the Reiwa Era. But also causes 1926-89 to be named for a semi-figurehead who didn't try to reduce evil till he sped up surrender when he was 44. Sagittarian Milky Way (talk) 22:15, 22 October 2024 (UTC)[reply]

Apparent misunderstanding here. All clocks other than atomic clocks (including radio-controlled ones) tick mean solar time. That's because they are not capable of doing anything else. Your sundial, naturally enough, cannot show anything other than apparent solar time, but over the long term that's the same as mean solar time (that's why it's called mean solar time). Coordinated Universal Time is Atomic Time plus an offset (regulated by means of the leap second) which keeps it so close to mean solar time that nobody can tell the difference. This is why all countries (bar a handful that don't) use mean solar time. It avoids argument:

Traffic warden: You are allowed to park for one hour. You overstayed by one second.

Motorist: No I didn't. I parked at 12 midnight and left at 1 AM.

Warden: Yes you did. There are sixty minutes in an hour. You parked at 12:00:00 and your time expired at 12:59:60. You left a second later.

As little as once or twice a month your radio-controlled clock is adjusted by means of a radio signal to show Coordinated Universal Time. Twice a year the signal adjusts it to show (or stop showing) what is in effect "Coordinated Universal Summer Time" (although nobody calls it that). 2A02:C7B:21A:700:6CF1:15C9:BDB3:F61A (talk) 11:19, 23 October 2024 (UTC)[reply]

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 [6] (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]

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[7] 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[8] (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". [9]

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 Newfton 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^[10] (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]

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]

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]