The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
Is there an estimate for the total global discharge rate of surface / ground water to the sea? It would be nice to state e.g. the Amazon as a percentage of the global total, just as we do for areas and populations of large countries. — kwami (talk) 06:23, 16 October 2024 (UTC)[reply]
I googled "amazon river total discharge rate" and it led me back to Amazon River, which says "The Amazon River has an average discharge of about 215,000–230,000 m3/s (7,600,000–8,100,000 cu ft/s)—approximately 6,591–7,570 km3 (1,581–1,816 cu mi) per year, greater than the next seven largest independent rivers combined." ←Baseball BugsWhat's up, Doc?carrots→ 07:25, 16 October 2024 (UTC)[reply]
Yes, rather amazing. We used to have a circle graph in that article that gave percentages, but the numbers were bullshit so I removed it. It would be nice to have an accurate graph, though: the full circle would be the global total, with pie slices for individual rivers. — kwami (talk) 07:55, 16 October 2024 (UTC)[reply]
The Caspian is an ocean, so certainly. As you say, I doubt the others would even be visible on a global scale. I'm not going to quibble with whatever I can find. — kwami (talk) 20:28, 17 October 2024 (UTC)[reply]
The place to search I would think would be in studies of the water cycle, and to look at the estimates that cut off that segment. Maybe check out some of the sources in that article to start (and their background sections to find sources for wider overviews, that might put down some hard estimates). SamuelRiv (talk) 15:50, 17 October 2024 (UTC)[reply]
"The estimated total from all rivers, large and small, measured and unmeasured, is about 9200 mi3 (38,300 km3) yearly (25 mi3 or 105 km3 daily)."[1] --Lambiam13:38, 19 October 2024 (UTC)[reply]
Rivers with an average discharge of 5,000 m3/s or greater, as a fraction of the estimated global total.
In Australia it's determined on a state-by-state basis. It's settled down now to a consistent set of dates, but Queensland and Western Australia both adopted and abandoned it more than once. For some decades now neither state has had DST, which makes it fun and games when working out times in the eastern states, and when travelling east-west or reverse, during the summer period. This is all because our Constitution makes no mention of time as a Commonwealth responsibility, which means it's automatically a state matter. -- Jack of Oz[pleasantries]17:07, 16 October 2024 (UTC)[reply]
Well, sort of. As I understand it, states have the option to observe DST or not, but they do not have the option to choose their own starting/ending dates. I'm not really convinced that this restriction is constitutional (I have a fairly narrow view of the Commerce Clause) but there doesn't seem to be any great advantage for a state to challenge it. --Trovatore (talk) 00:21, 17 October 2024 (UTC)[reply]
Yes, that's what I was trying to say. And presumably interstate commerce is the justification for the federal law. If I remember correctly, the need for standard time was driven by the railroads, in place of a myriad of local times. And also, if I recall correctly, it used to be that the railroads worked strictly within standard time, even during DST, as DST was only sporadically used until 1967. Once DST became standardized, the railroads could change to DST also. ←Baseball BugsWhat's up, Doc?carrots→ 01:15, 17 October 2024 (UTC)[reply]
While we have an article on Summer time in Europe, it is not very detailed. I found what appears at glance to be a reliable and well-researched article on the history of European time zones at "When Did DST Start in Europe?" via timeanddate, which suggests that Central European Summer TimeStandard Time as we know it (with its onset date) really began its continuity and spread from the Nazi conquests. (Other countries had been experimenting with daylight saving, but inconsistently, as the article explains.) The incongruity in clock-switch dates would be due to the haphazard nature of daylight saving being adopted in various countries, where countries tend to only finally decide to align their clocks with some treaty or conquest or absolute trade necessity. (Of course there are famous exceptions: the U.S. state of Indiana only adopted daylight saving in 2005, despite [EDIT: most of] the rest of the continental U.S., including the entire surrounding time zone, having uniform daylight saving dates.) SamuelRiv (talk) 01:56, 17 October 2024 (UTC)[reply]
Be careful — CEST does not stand for "Central European Standard Time" (easy mistake to make as a North American) but for "Central European Summer Time". It's the opposite of what you would expect from PST / PDT. CET is UTC+1; CEST is UTC+2. --Trovatore (talk) 06:08, 17 October 2024 (UTC)[reply]
As for Indiana, I assume that's because they're so far west in their time zone, making DST a double misery for night owls and farmers. They really should be on Central Time and then it would be much less of a problem. --Trovatore (talk) 06:18, 17 October 2024 (UTC)[reply]
Indeed. They did it to get the same time and business hours as the East (I've heard New York Stock Exchange is important somehow) and cause the enlatenization lobbies like businesses headquartered in EDT and golf are collectively stronger than the delayzation lobbies like CDT headquarters and nightclubs. Great for us New Yorkers, terrible for the Indianan colonists of the aggrandized UTC-4 Time Empire. So no, permanently stopping leap seconds won't cause 9 to 5 jobs to get later till the night owl torture is really bad like Urumqi. There's probably still 9 to 5 jobs in Urumqi, I wonder if they pay better than similar 12 to 8 jobs. Sagittarian Milky Way (talk) 21:11, 21 October 2024 (UTC)[reply]
Leap seconds should have been strangled in the cradle. Horrible idea from the very start. People will absolutely adjust nominal start and stop times to the Sun, no matter what the clock says. It just takes a while; doesn't usually happen in six months, which is why DST has any effect at all. --Trovatore (talk) 16:25, 23 October 2024 (UTC)[reply]
Oh, and not quite true about "the rest of the Continental US" -- Arizona, except for the Navajo Nation, does not observe DST. I expect that's because they're far enough south that the difference between summer and winter times doesn't justify it. Not sure why other southern states don't do the same thing. --Trovatore (talk) 06:21, 17 October 2024 (UTC)[reply]
It gets even more confusing. Within the Navajo Nation is the separate Hopi Reservation. Which does not observe DST. So, you go from Arizona in general (no DST), enter the Navajo Nation (has DST), and continue onward into the Hopi Reservation (no DST).--User:Khajidha (talk) (contributions) 15:08, 17 October 2024 (UTC)[reply]
Yes my mistakes, the article I linked specifies CEST = "Summer Time", not "Standard", so that's on me. Also I knew there were other U.S. states that didn't do daylight saving, but I had only remembered Hawaii, and Indiana has always felt like an odd one out for its area in so many ways (usually good ways, Hoosiers). SamuelRiv (talk) 15:57, 17 October 2024 (UTC)[reply]
My nerdy brain tells me that, logically, daylight saving should run for the same amount of time either side of the summer solstice. It doesn't in my state of Australia. My ageing memory tells me that the finishing date was extended further into autumn by a populist state premier who wanted more people to go to the Formula 1 Grand Prix. HiLo48 (talk) 02:16, 17 October 2024 (UTC)[reply]
Mid-sunlight isn't 12:00 at 150° East, it's 12:14 in February and 11:44 in early November and 12:00ish in April and December and early September and a few minutes early or late in May and July.
My nerdy brain tells me that you can't save daylight hours; what you gain at one end of the day, is lost at the other. The intended result of shifting working hours relative to daylight hours can also be reached by shifting working hours, and there's nobody stopping us from doing that. I remember a ferry with the notice "Operating hours: 6:00–20:00 winter time, 7:00–21:00 summer time". Although it would be convenient to put zero o'clock, when the date changes, at a time when most people are asleep. Right, it's one o'clock now here, time to go to bed. PiusImpavidus (talk) 23:06, 17 October 2024 (UTC)[reply]
Here in Australia, one argument against daylight saving is the issue of milking cows. It's hard to tell cows with bursting udders to just hold on a bit longer. HiLo48 (talk) 23:14, 17 October 2024 (UTC)[reply]
You really expect America to do something that makes it easier to interact with the rest of the world? You know, like we've (the US) so readily done with the metric system? Or date format?--User:Khajidha (talk) (contributions) 20:29, 18 October 2024 (UTC)[reply]
Europe got tired of copying our extensions. Also Europe time's fucked up maybe they can't take more, 9 degrees West has the time of 30 degrees East in summer just cause they copied Hitler & seem to want all EU same time, (30+9)'s even worse than extreme west Indiana, or the "7.5 degrees plus or minus nearest country border" that it could be. Nov 7 Northwest Iberian sunrise would be late as hell at that latitude. Europe doesn't have many October 31st kids door-to-door afterschool either, Eurolatists can't say "won't someone think of the children!". Sagittarian Milky Way (talk) 21:47, 21 October 2024 (UTC)[reply]
A treaty would work, so would the governments of one side unilaterally wanting to shorten the "wrong offset time" from weeks to hours like they did till we extended for trick-or-treaters (and March for some reason) they didn't copy that one. The latism lobby is stronger (i.e. EUST creeped west from this) but they're not omnipotent. DST's 1:00 UT Eurowide (no earlier than 1 to 2/2 to 1 or later than 3 to 4/4 to 3 with intermediate values as disparate as 9.3 West & 31.6 East). If EU copied our dates a town of 4,736 would get Monday to Friday sunrises after 9:16, sunrises as late as ~9:19.0 11/6/88 at the cape (~43°10'N 9°13'W) seeing the Sun would take even longer cause hills/mountains instead of an idealized mirror-smooth globe. I don't know how late Spain starts but this is past where Americans would start saying "think of the children" outside in the dark & some would get miserable if it's too dark when they work, learn or wake. Everyone would hate it but Bizzaro World DST might be better for unconscientious night owls actually (our dates plus 6 months or centering the 34 weeks on Sunday nearest Dec solstice). At least kids wouldn't go weeks or months without full dark and mornings already suck if you can't sleep in may as well make them night so 16:30 isn't. Half of US time zones wouldn't even get that bad if centered on 75 90 105 120 like they should be only about 8:30 sunrise (nautical dawn would reach the threes though which would suck) Sagittarian Milky Way (talk) 21:42, 22 October 2024 (UTC)[reply]30 years after time zones no DST yetSagittarian Milky Way (talk) 21:42, 22 October 2024 (UTC)[reply]
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 | Talk09:30, 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 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?
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?
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 werereally 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.
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": [2]:
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!
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.
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]
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]
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]
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. --Lambiam15:31, 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
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]
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]
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]
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]
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]
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]
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,[6] this is not Earth-shattering news that deserves careful attention. --Lambiam06:10, 21 October 2024 (UTC)[reply]
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.
"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]
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]
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 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. --Lambiam06:00, 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]
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?
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 [7] (paywalled unfortunately). --Amble (talk) 17:21, 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]
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]
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]
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]
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
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:
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. ISBN978-3-030-28560-9.
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]
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]
I am currently on a bridge on the Florida State Road 528 facing north, getting ready to view the Spacex launch[8] 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[9] (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]
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]
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.
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]
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? --Lambiam16:45, 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). --Lambiam05:26, 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).
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. --Lambiam19:28, 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]
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.
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]
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 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]
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".[10]
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...
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 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]
(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. --Lambiam10:47, 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]
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]
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]
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?
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. --Lambiam12:32, 30 October 2024 (UTC)[reply]
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--Slowking Man (talk) 17:32, 22 October 2024 (UTC)
![[9]](https://i.imgur.com/kldMHwY.jpeg){kind=link}
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