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March 15

Can't remember the name of this organic chemistry phenomenon where conjugation "extends" a functional group

I think I remember reading a Wikipedia article about this, but I can't find the name of this phenomenon at all. Pinging @Smokefoot: maybe you might know since you are an active editor of organic chemistry articles. Here I have drawn the concept:

This is the phenomenon in which a double bond "extends" a functional group by conjugation, so that it behaves similarly to the parent functional group. The bottom compound, an "extended" carboxylic acid, has similar chemistry as the top carboxylic acid, except the two parts of the functionality (C=O and -OH) are separated by an additional double bond. The reason for this similarity is because conjugation can transfer charges by resonance, e.g. the bottom compound is acidic like a carboxylic acid because the negative charge on the deprotonated -OH oxygen can move to the carbonyl oxygen by resonance. Michael7604 (talk) 05:50, 15 March 2024 (UTC)

See or search dicarbonyl or keto-enol equilibrium. @Michael D. Turnbull and DMacks:. --Smokefoot (talk) 14:37, 15 March 2024 (UTC)

@Michael7604 You already had the key concept in your title: see conjugated system and the articles linked from it. My favourite related topic is, of course, the Michael addition reaction. Mike Turnbull (talk) 16:28, 15 March 2024 (UTC)

A vinylogous group. DMacks (talk) 18:26, 15 March 2024 (UTC)

This is it, thank you Michael7604 (talk) 21:15, 15 March 2024 (UTC)

Fictitious force. Is the opposite phenomenon called a: "Real" force? "Physical" force? "External" force? "Natural" force?

I'm looking for the most useful term, intended to exclude fictitious forces. HOTmag (talk) 08:38, 15 March 2024 (UTC)

Each of the suggested terms is usable if you take away the ornamentation of Scare quotes. Philvoids (talk) 10:31, 15 March 2024 (UTC)

They should only be considered to be quotation marks. That's because I was looking for "the most useful term" (Btw now I'm quoting myself), i.e. a term that could be quoted from sources using useful terminology. HOTmag (talk) 10:45, 15 March 2024 (UTC)

A commonly used contrasting term is true force.^[1][2][3]  --Lambiam 17:25, 15 March 2024 (UTC)

Meh. Yes, true force has some popularity, sometimes even in quotes, but don't expect everybody to understand you when you mention true forces. And despite Newton's ideas on absolute space and preferred reference frames that only have constant velocity relative to it, in real life we deal with the equivalence principle, under which those fictitious forces are as real as gravity. Which is quite real in classical mechanics. In fact, they are gravity. In a free-falling lift, we just declare gravity zero. On a merry-go-round, we just declare the centrifugal force and Coriolis force part of gravity. Good luck describing orbital manoeuvring or accretion in compact binaries in an inertial frame. General Relativity (I think you'll love that) even makes it its basic principle. Turns out that the entire universe spinning around our stationary merry-go-round causes frame dragging, which exactly matches those fictitious forces. PiusImpavidus (talk) 19:57, 15 March 2024 (UTC)

That was Einstein's original view, but I think the matter has become somewhat controversial. We have an article on Mach's principle. --Trovatore (talk) 21:06, 15 March 2024 (UTC)

Just as a slight extension, in dynamics we invent a fictitious force m*a, called a D'Alembert force, and can then treat it as a statics problem. It isn't always useful. Greglocock (talk) 23:36, 15 March 2024 (UTC)

March 16

Weld; Syren

This 1878 report makes a passing reference to "Weld's Sound Experiment". What was that, and who was Weld?

It also mentiones "Syren and Galton's Whistle"; the latter is Francis Galton, but who was Syren? Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 21:28, 16 March 2024 (UTC)

My guess is a siren whistle. --Wrongfilter (talk) 21:49, 16 March 2024 (UTC)

Galton's whistle is similar to a boy scout's whistle . It differs in that it has a piston provided with a screw and a milled head... (it's an ultrasonic dog whistle).

From Oscillations and Waves p. 297. Alansplodge (talk) 21:56, 16 March 2024 (UTC)

The most promising Weld is Alfred Weld, who was a bit of a scientist, but I fail to find any indication that he may have dealt with sound. --Wrongfilter (talk) 22:02, 16 March 2024 (UTC)

And but nay. In duo together with Ernst Chladni, Alfred will make too much of a youngster, for those classicals. --Askedonty (talk) 01:33, 17 March 2024 (UTC)

What classicals? This is about the first annual meeting by the Midland Union of Natural History Societies. Also, they had no scruples looking at a toddler like Galton's whistle, which had been invented a mere two years before. But don't hesitate to suggest another Weld if you've got one. --Wrongfilter (talk) 07:29, 17 March 2024 (UTC)

Wiktionary:syren#Noun: "Obsolete form of siren". Alansplodge (talk) 22:03, 16 March 2024 (UTC)

For the meaning of "Galton's Whistle", see Dog whistle.  --Lambiam 00:53, 17 March 2024 (UTC)

Isaac Weld.^[4]  --Lambiam 10:24, 17 March 2024 (UTC)

Resolved

 – Thank you. all. Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 10:50, 17 March 2024 (UTC)

Length of a photon

A couple of questions about photons.

Firstly, am I right to feel slightly uncomfortable about the following passage? From the article photon:

However, experiments confirm that the photon is not a short pulse of electromagnetic radiation; a photon's Maxwell waves will diffract, but photon energy does not spread out as it propagates, nor does this energy divide when it encounters a beam splitter. Rather, the received photon acts like a point-like particle since it is absorbed or emitted as a whole by arbitrarily small systems, including systems much smaller than its wavelength, such as an atomic nucleus (≈10⁻¹⁵ m across) or even the point-like electron.

It's trying, I think, to make a point that a quantum-mechanical wavefunction is different from a classical wave, because it can suffer wavefunction collapse; and because it always represents a single whole particle (and also that a photon is a point particle -- which I understand to mean that its overall (extended) wavepacket can be evolved by integrating it forward via point to point propagators).

But am I right to feel uncomfortable with the argument about it not being distributed because it can interact with objects that have a very small scale? I seem to remember that in the classical treatment of radio waves and antennas, an antenna might be essentially just a single vertical pole; but may interact with a radio wave that may have a wavelength of hundreds of metres, if the antenna is part of a circuit that has the right resonant frequency. If I remember correctly, the radio wave induces a current in the circuit, which in turn can be thought of as exciting an induced radio wave of its own that has an opposite phase in the far field, thus effectively partially cancelling the original wave; and so, despite what may be its small physical size, the antenna can have an effective area ("aperture") that can be very considerably larger. (Did I remember that right? It's a long time ago since those lectures on EM ...)

Now I'm not saying that that is directly equivalent to quantum measurement, because measuring the location of a photon absorbs all of it, not just part of it (give or take whatever discussion we might have about decoherence...) But am I right that the antenna does maybe suggest that we should perhaps be just a little more careful before making a blanket statement that an object cannot be distributed if it can interact with a system much smaller than its own wavelength?

The remark that "the photon energy does not spread out as it propagates" also makes me feel a bit uncomfortable. If one thinks of gravitational lensing, surely it's reasonable to imagine a photon travelling both sides of the gravity well while still retaining self-coherence, with a corresponding energy flow associated with both sides?

Beyond that, the second thing I wanted to ask about (which was what led me to look at the photon article in the first place, to see if it anywhere touched on the subject) was this: I seem to remember, way back at school level, being told that a typical wavepacket of a photon of visible light (eg perhaps from a sodium vapour lamp) had a longitudinal length of the order of about a metre (maybe give or take an order of magnitude either way). Does that seem plausible / viable / about right ?

One constraint would seem to be the time <-> frequency uncertainty relation. The overall linewidth of the source would seem to set a limit to the maximum range of frequencies that could be associated with such a photon, and therefore a minimum to the length of time (and so therefore spatial length) that the wavepacket must persist. Presumably there is also a maximum length of time one pictures the emission event taking -- perhaps even measurable by trying to measure the coherence time of such a source using an interferometer at lower and lower light levels ? Is this sort of thinking at all on the right track?

Thanks in advance, Jheald (talk) 22:05, 16 March 2024 (UTC)

The following, unfortunately unsourced, passage is copied from Point particle § In quantum mechanics:

Nevertheless, there is good reason that an elementary particle is often called a point particle. Even if an elementary particle has a delocalized wavepacket, the wavepacket can be represented as a quantum superposition of quantum states wherein the particle is exactly localized. Moreover, the interactions of the particle can be represented as a superposition of interactions of individual states which are localized.

Does this help to ease your uncomfortability?  --Lambiam 00:48, 17 March 2024 (UTC)

Thanks @Lambiam:. No problem with what you've quoted, so far as it goes.

But I do think it can be useful (and meaningful) to try to give a sense of what the whole overall superposition - the whole wavefunction - may look like, when this is possible (for a typical environment or observational set-up).

So eg for an electron, it's a point particle, but it's also useful to be able to talk about the whole atomic or molecular orbital that may be a good representation of the whole wavefunction in particular circumstances.

And similarly, it seems to me, for a photon, it's useful to have an idea of what a whole photon wavefunction might be like in a particular circumstance; and in respect of an interaction, what shape the whole overall superposition of point interactions (as referred to in your quote) might take, to represent eg a whole molecular orbital interacting in typical way with the entirety of a photon wavepacket, and what basic picture that may give for a complete scattering (elastic or inelastic) or emission event. That's something I'd quite appreciate some input on, at least in respect of the second part of my question.

In respect of the first part of my question, while I feel comfortable with what you've just quoted from point particle, it doesn't seem to me to address the things I suggested gave me a bit of discomfort with the bit of text from photon. Do you think it should make me feel more comfortable with that text? Thanks, Jheald (talk) 14:00, 17 March 2024 (UTC)

I think you're quite right not to feel comfortable about its wavefunction. It is possible to get pictures that represent it. But thinking a particle was actually at any of the point when it has not been observed to be there - well ... have a look at Quantum Cheshire cat and marvel at its ... well whatever. NadVolum (talk) 15:51, 18 March 2024 (UTC)

• Just adding a holding message, to stop the thread being auto-archived while I continue to try to read into it. I do think it would be useful to try to establish where mainstream specialist thought is on this, as the variety of answers one can find on generalist sites like Quora, StackExchange, Reddit, PhysicsForums, etc and even preprints are all over the place. Even here on Wiki Reference Deska couple of years ago essentially the same question got shut down as not meaningful or not based on a proper understanding of the physics. (Pinging @Malypaet, PiusImpavidus, Jayron32, Nimur, and Rmhermen: from that discussion, if they'd like to come in).

Yet, despite the issues with setups like the Quantum Cheshire cat that User:NadVolum points to (and which I need to think some more about), or our article Wave packet bluntly making the statement that 'Physicists have concluded that "wave packets would not do as representations of subatomic particles"' without further explanation, it seems to me that a wave packet or evolving wavefunction is often a useful physical picture for a particle, at least until environmental interactions make considerations of decoherence or measurement unavoidable. They help us picture, for example, how single photons can exhibit interference effects from their own reflections, or between paths with different flight times in beam-splitter experiments. I need to read up more on Quantum Optical Coherence (eg [5]) as distinct from classical coherence, but it does seem to me that the coherence length of a source of incoherent photons probably can be reasonably pictured as the length of a photon, eg the 67mm for a low-pressure sodium vapour lamp given at Coherence_length#Other_light_sources, extending to six times this for such lamps cooled to liquid nitrogen temperatures.

Our article atomic electron transition seems once to focus on how short the time for atomic transitions can be (given very heavy environmental 'measurement' / decoherence) -- still a bit longer than zero, because (as I understand it) even then the electron wavefunction needs time to change without moving faster than the speed of light -- but little in that article that I could see as to how long in time the electron transition might be considered to take, in the absence of externally-induced decoherence, with measurement setups designed to maintain rather than destroy coherence. Perhaps there should be more of an investigation of this there ? Jheald (talk) 13:28, 24 March 2024 (UTC)

How long the transition can take instead of how short?, see Quantum Zeno effect ;-) NadVolum (talk) 16:45, 24 March 2024 (UTC)

@NadVolum: In fact pretty much the exact opposite of Quantum Zeno Effect. The Quantum Zeno Effect, as I understand it, arises if you are doing almost continuous measurement of the state of the test atom, destroying any superposition that may be beginning and forcing it into a state where either no transition has occurred or a very quick transition has occurred (and been completed) since the previous observation. The experiments to see how fast a transition could be observed were I believe developments of this set-up. The photons released from such a set-up I think have a very short coherence time (and coherence length), with a correspondingly very wide frequency line-width (like thermal broadening only very much more so), a much greater energy uncertainty made possible by the pronounced coupling of the system with the environment / measuring apparatus, that is used to achieve the decoherence.

In contrast I would like to see the articles make a bit more about how long the coherence times / coherence lengths of the emitted photons can be, corresponding to how long a time the transition superposition may have evolved over. Jheald (talk) 09:21, 25 March 2024 (UTC)

Ok @Jheald:, I'll give my opinion. Considering the orbital trajectory of an electron crossing a flow of electromagnetic radiation, how long will it receive enough radiation pressure energy added in this time to eject itself? If we consider the relation $E=h\nu$ as an energy over a second with $\nu$ not a frequency, but a number of cycles over this second, therefore with $h$ the energy of a cycle whatever the frequency, how many cycles of how many ray crossing this orbit will give their energy to eject the electron? I know, it's too simple to imagine with the analogy of an asteroid field crossing a satellite. It's not science fiction enough to enter the official domain of science. Maybe someday... Malypaet (talk) 22:40, 25 March 2024 (UTC)

@Malypaet: Interesting. But, to the extent that the energy density per metre of a photon can be conceptualised at all, I think I would have to come at it from the other direction, ie that the total energy of the photon is $E=h\nu$ (nb total energy, not energy per second), then the energy density per length would be this divided by the length of the photon -- which would depend, like the probability amplitude per metre, on the coherence length (or equivalently, the coherence time). A photon spread out over a long coherence length would have a lower energy density per metre (if such a concept is meaningful); a tightly bunched short coherence photon would have a higher energy density per metre.

But it would seem to me a secondary quantity, that might be derivable from the photon length; but with no apparent very ready way to estimate it in its own right seems of little use as a route in the other direction, to try to determine a photon length. Jheald (talk) 22:19, 28 March 2024 (UTC)

March 18

Special Relativity. Is it possible to calculate the velocity of a system, composed of two bodies, we given their different inertial masses and velocities?

An equivalent question: What's the system's (relativistic) inertial mass, we bearing in mind that the conservation of (relativistic) mass, as a general rule, does not hold in Relativity Theory.

2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 16:58, 18 March 2024 (UTC)

There is an article on the general problem Two-body problem in general relativity. Good luck with it! NadVolum (talk) 19:08, 18 March 2024 (UTC)

Oh, I forgot to point out that my question was only about Special Relativity, the given masses being the inertial ones (rather than the gravitational ones). Due to your comment, I've just added above, that I'm only interested in finding the system's velocity (or inertial mass) in Special Relativity. 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 19:34, 18 March 2024 (UTC)

Given the four-momenta of the two bodies, $p_{i}=(E_{i},\mathbf {p} _{i})=(\gamma _{i}m_{i},m_{i}\mathbf {v} _{i})$ ($i=1,2$; the last expression only for bodies with finite mass), the combined four-momentum is $p_{\mathrm {tot} }=p_{1}+p_{2}$. The mass of the two-body system is given by $m_{\mathrm {tot} }^{2}=p_{\mathrm {tot} }\cdot p_{\mathrm {tot} }=(E_{1}+E_{2})^{2}-(\mathbf {p} _{1}+\mathbf {p} _{2})^{2}$. The centre-of-mass velocity is given by $(\mathbf {p} _{1}+\mathbf {p} _{2})/m_{\mathrm {tot} }$. (I've set $c=1$; spatial (three-)momentum is denoted by $\mathbf {p}$). For photons use $E=|\mathbf {p} |=h\nu$. --Wrongfilter (talk) 09:58, 19 March 2024 (UTC)

For being more explicit, let's assume that the bodies: have masses m,n, and move at velocities u,v (respectively) on the same line. What's the system's velocity, using m,n,u,v only? 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 11:13, 19 March 2024 (UTC)

I told you everything you need to know, you just have to restrict to one-dimensional motion and use your notation. Why the anonymity today? --Wrongfilter (talk) 11:29, 19 March 2024 (UTC)

if you suggest that I study maths, then I accept the idea. For the time being, could you calculate it for me (using m,n,u,v only) ? I'm bad at mathematics. (If you also suggest that I register, then you are not the first one to suggest that, but I'd rather avoid registering.) 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 11:54, 19 March 2024 (UTC)

SOS. Despite my bad maths, I think I've found a contradiction in your method. Please help.

I'm checking two cases, each of which involves two bodies, that have the same mass, and that move with the same speed in opposite directions, so according to your formula of the "combined four-momentum", we get:

First conclusion: $p_{\mathrm {tot} }=(\mathbf {p} _{1}+\mathbf {p} _{2})=0,$ right?

By combining, the first conclusion, along with your formula of "centre-of-mass velocity", we conclude:

Second conclusion: The velocity of the two-body system is $0$ as well, right?

By the second conclusion, which tells us that the two-body system is at rest, we conclude:

Third conclusion: The mass of the two-body system, $m_{\mathrm {tot} },$ is simply its rest mass, which is supposed to be invariant, right?

Fourth conclusion: In the first case checked out, the speed of both bodies is low, so $E_{i}$ is low for both bodies, so according to your formula of the "mass of the two-body system", along with the first conclusion, we conclude that $m_{\mathrm {tot} }$ (proportional now to the energy only) is small. While in the second case checked out, the speed of both bodies is high, so $E_{i}$ is high for both bodies, so according to your formula of the "mass of the two-body system", along with the first conclusion, we conclude that $m_{\mathrm {tot} }$ (again proportional now to the energy only) is big, right?

How can we avoid the contradiction between the third conclusion and the fourth one? Where is my mistake?

2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 14:27, 19 March 2024 (UTC)

The total mass of the 2-body system comprises the masses of the two components and their kinetic energies with respect to the common centre-of-mass rest frame. This is now to be considered as internal energy, i.e. energy in internal degrees of freedom. The total mass is invariant with respect to changes of reference frame. --Wrongfilter (talk) 14:45, 19 March 2024 (UTC)

Thank you. May I formulate it as follows: A rest mass is invariant in one-body systems only (e.g. an electron), rather than in multy-body systems, because of the internal energy that may change if several bodies are involved in the whole system? 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 14:52, 19 March 2024 (UTC)

In an isolated system the internal energy will not change. --Wrongfilter (talk) 14:54, 19 March 2024 (UTC)

Thank you. 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 13:28, 27 March 2024 (UTC)

Aliens: from cavemen to NASA

I have been reading many books about extraterrestrial life lately. Although there is no actual example to study, the usual approach is to study which things are required to happen or be there, which ones can be likely, which ones not so much, and the feasibility of achiving similar results in other ways (such as life being made of an element other than carbon, or with a solvent other than water). I have seen this approach used to see the chances of a given planet to turn into a Earth-like planet, and for life, once formed, to survive and evolve.

But there is a missing step: once we get full ecosystems with diverse animals, how likely would it be for some animal to evolve into a human-like intelligence? Which would be the requirements and obstacles? And if one actually does, how likely would it be to go all the way from rocks and sticks to space programs? I'm not asking to discuss that ourselves, but to point me some author that did. The article Extraterrestrial intelligence is almost devoid of content and does not help with this search. Cambalachero (talk) 19:44, 18 March 2024 (UTC)

Are you familiar with the Fermi paradox? 41.23.55.195 (talk) 05:39, 19 March 2024 (UTC)

There is a strong bias in the question itself. Which is more intelligent: A species of humans who invented war and terrorism or a species of dolphins that chat and play most of the day? What if a planet of "intelligent" creatures stopped at having fun and didn't feel the need to progress into colonialism and repression? It leads to two common points on the topic. The first is that we will meet technology from other space-exploring societies before we meet the society just as others will see our satellites and transmissions long before they meet a human. The second is that any species that travels through space to meet us isn't doing so to be nice. They see our planet as a source of profit in some way and will want to exploit it as much as possible. 12.116.29.106 (talk) 18:46, 19 March 2024 (UTC)

“For instance, on the planet Earth, man had always assumed that he was more intelligent than dolphins because he had achieved so much—the wheel, New York, wars and so on—whilst all the dolphins had ever done was muck about in the water having a good time. But conversely, the dolphins had always believed that they were far more intelligent than man—for precisely the same reasons.” ― Douglas Adams, The Hitchhiker’s Guide to the Galaxy I'm betting you were making a reference, but figured this was a good chance to introduce the unenlightened to Adams. --User:Khajidha (talk) (contributions) 13:19, 20 March 2024 (UTC)

Dolphins probably never advanced towards a civilization because, as an underwater species, they have no access to fire, the one element that allowed all human technology in the first place. And the lack of versatile limbs, such as our hands with opposing thumbs, probably does not help either. Those two are the kind of points I'm sure some author must have thought and organized in a related book. Cambalachero (talk) 14:04, 20 March 2024 (UTC)

Why would dolphins need fire? ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:56, 20 March 2024 (UTC)

To cook all the fish we gave them. --User:Khajidha (talk) (contributions) 17:32, 20 March 2024 (UTC)

Maybe they prefer sushi. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:26, 21 March 2024 (UTC)

Good luck getting to the copper age or further without fire. Cambalachero (talk) 17:25, 21 March 2024 (UTC)

Why would dolphins need metals? ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:10, 21 March 2024 (UTC)

How else could they become our tyrannical overlords, forcing us to frolic and have fun every single day? Clarityfiend (talk) 00:51, 22 March 2024 (UTC)

Evolution of human intelligence discusses how some apes got human-like intelligence. It includes links to cephalopod intelligence and cetacean intelligence. Also, possible extraterrestrial life has no obligation to develop intelligence, space pollution or Rubik cubes. To believe otherwise is probably a misleading use of teleology in biology. --Error (talk) 23:34, 19 March 2024 (UTC)

March 20

Removing air from water to reduce your water bill

There are a number of companies out there that purport to be able to reduce your water bill by reducing the volume of air in the pipe ahead of the meter so that you "just pay for water". (Example site). Leaving aside any issues with the water company, does the science of these things make sense? Matt Deres (talk) 19:32, 20 March 2024 (UTC)

The example site you provided is based on little more than advertising hyperbole! I have read it in its entirety and I saw nothing that made technical sense. The great majority of water systems connected to a water distribution network don’t contain air (unless the advertisement is referring to air dissolved in the water.)

If this hardware actually achieves something worthwhile this advertisement fails to disclose what it is or how it works. Dolphin (t) 21:11, 20 March 2024 (UTC)

In my area of the UK, meters are installed under the tarmac of the public pavement (sidewalk) outside the residence, accessed by a very small hatch (they are read remotely), so to treat the water before it reached a meter would likely require excavating the pavement – not a practical or legal proposition – and installing the device (whatever it is) on the water company's pipework, which would likely also be illegal. Moreover, even if it worked, it would likely be a violation of the water company's contractual terms. {The poster formerly known as 87.81.230.195} 51.241.39.117 (talk) 06:39, 21 March 2024 (UTC)

What a wonderful advice: "trusting advice about Smart Valves from anyone other than experts like us is ill-advised." Experts "like us" can easily be recognized, because they wouldn't recommend any other valves. So where the ad admonishes, "Don't fall victim to outdated first-generation valves", I fully agree. I'd even go further and say, "Don't fall victim to any valves, whether first, second or third generation."  --Lambiam 10:00, 21 March 2024 (UTC)

Water metering#Problems addresses the issue of air in the water, giving a reference to regulations, but the cite is to an 'International Recommendation' rather than any national law. -- Verbarson  ^talk_edits 12:23, 21 March 2024 (UTC)

...and the recommendation specifically refers to 'liquids other than water'. -- Verbarson  ^talk_edits 12:27, 21 March 2024 (UTC)

Faucets are equipped with aerators specifically to introduce air into the water at the point of use, reducing water use through more effective wetting in the dispersed water, and reducing splashing. Maybe these marketers are trying to make people think that it happens in the supply lines and they're getting cheated? Acroterion (talk) 12:33, 21 March 2024 (UTC)

It all sounds like a scam. And doesn't dissolved air in the water improve the taste? Otherwise you'd be drinking distilled water, which is pretty bland. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:37, 21 March 2024 (UTC)

That would be minerals, at least up to a point. Air, unless it's carbonation, doesn't affect taste significantly. Acroterion (talk) 02:56, 22 March 2024 (UTC)

Our colleagues over at WikiHow (and many plumbing sites) have addressed this - "Water occasionally comes out of the tap with a cloudy or milky appearance. In most cases, cloudy water is caused by air bubbles in the water, and these will dissipate on their own if you let the water sit for a few minutes." HiLo48 (talk) 03:55, 22 March 2024 (UTC)

Here's another site offering the same kind of thing: [6], but searching for water valves that remove air to reduce water bills will give you no end of similar items. This one at least has a demonstration video of how it's supposed to work. Many, like the Watergater, tout that they are 'certified' and/or 'approved' which always strike me as weasel-words in this kind of context. Like, they were maybe approved in the sense that they wouldn't explode or poison the water or something, but meanwhile implying that an impartial researcher has confirmed their claims. But just because it sounds fishy doesn't mean that it is, so I guess my questions are: 1) does municipal water typically have significant air in it? 2) does that air actually impact the reading of a standard water meter and 3) is there any chance this doohickey could affect that in a meaningful way? Matt Deres (talk) 14:05, 22 March 2024 (UTC)

Regarding (1), the charts here indicate that at normal temperatures (40-80 F) and typical municipal water pressures (60 psi), the ratio of the volume of air to the volume of water is roughly 8% to 13%. This is more than I would have guessed. Whether a valve can change that in any significant way, without changing the pressure, seems doubtful to me but I don't know for sure. CodeTalker (talk) 02:58, 23 March 2024 (UTC)

There is a difference between an air and water mixture, and water containing dissolved air (properly, dissolved nitrogen and oxygen, being the significant components of air). To my understanding, dissolving gases in water does not greatly increase its volume (if at all), so it's important to be sure what is being measured: this reference discusses air dissolved in water. {The poster formerly known as 87.81.230.195} 51.241.39.117 (talk) 09:26, 23 March 2024 (UTC)

March 22

2 related questions about thought

I've read novels where a character says thinking too hard about relativity or some other difficult subject "made his head hurt." I've also had heard people say that, jokingly or not. Q1). Shoud I take it literally? Q2). Could there be microbes in the brain that cause inflamation and make hard thinking painful and exhausting? Rich (talk) 03:48, 22 March 2024 (UTC)

I think it is just idiom, like saying that something makes one's blood boil. Tension headache can result from being literally tense, for which stress can be a factor. So in a desperate last-minute study of a difficult subject for an impending exam, finding it inscrutable, a student may clench their teeth and at some time get a headache. But then it is not caused directly by the brain activity itself. The blood–brain barrier should defend the brain against invading microbes. If pathogenic microbes nevertheless manage to get in, as sometimes happens, a patient will likely develop encephalitis, which can be debilitating and life-threatening. Early symptoms include headache and confusion.  --Lambiam 10:38, 22 March 2024 (UTC)

I should add that the brain itself does not contain nociceptors. Brain surgeons poking around in the brain of a conscious and unanesthetized patient may cause all kinds of subjective sensations, but this is not by itself painful.  --Lambiam 14:44, 22 March 2024 (UTC)

Lapse rate between sub-adiabatic and inversion

Simplified graph of atmospheric lapse rate near sea level

I sketched this graph but am unsure what to label the teal vertical line. Does it have a special name? Thanks, cmɢʟee⎆τaʟκ 08:13, 22 March 2024 (UTC)

"Isothermal lapse rate" is used e.g. in the abstract of this article. --Wrongfilter (talk) 09:58, 22 March 2024 (UTC)

@Wrongfilter: Thank you very much. Updated... cmɢʟee⎆τaʟκ 13:13, 22 March 2024 (UTC)

The area between dry adiabatic lapse rate and wet adiabatic lapse rate is marked as neutral. It could be stable or unstable, depending on the humidity. Neutral is a single line, that must be in the area marked as neutral. PiusImpavidus (talk) 09:36, 23 March 2024 (UTC)

Fair point, thanks. It was too wordy to label ("region with neutral"?) so I just omitted the label. cmɢʟee⎆τaʟκ 12:29, 23 March 2024 (UTC)

Chemical X real or not?

Blocked user. Matt Deres (talk) 16:53, 22 March 2024 (UTC)
The following discussion has been closed. Please do not modify it.
From cartoon references this chemical they can real or not? 2001:44C8:40E2:8DFE:54A:6AF:F2CA:27EA (talk) 08:18, 22 March 2024 (UTC) Chemical X is clearly fictional. Shantavira|feed me 09:44, 22 March 2024 (UTC) It is a plot device. Graeme Bartlett (talk) 11:09, 22 March 2024 (UTC) As with Roger Ramjet and his proton pills, and Mighty Mouse with his super cheese. ←Baseball Bugs What's up, Doc? carrots→ 11:20, 22 March 2024 (UTC)

Personality of science deniers

Have any studies been done whether, say, climate change deniers and flat earthers typically share specific personality traits?  --Lambiam 14:32, 22 March 2024 (UTC)

Somewhat related: The article Anti-vaccine activism may contain relevant links / references. --Cookatoo.ergo.ZooM (talk) 18:42, 22 March 2024 (UTC)

The article contains 180 wikilinks and 172 references with 183 external links, but I did not spot anything suggesting any studies of personality traits of anti-vaccine activists.  --Lambiam 20:19, 22 March 2024 (UTC)

Not sure that there's much on personality traits but this paper may be of interest. Mikenorton (talk) 20:31, 22 March 2024 (UTC)

Denialism is the Wikipedia article that should cover that, but it doesn't say anything much about it. Some of the external references though could be of interest to you. Quite a few of them are in it for the money from their audiences, but there's lots of perfectly reasonable people out there who seem to just get this urge to spend time learning more just so they can defend their denial and spread it to the world. NadVolum (talk) 20:57, 22 March 2024 (UTC)

The article Sectarianism may contain some interesting links, eg Collective Narcissim et al. It seems that S Freud (and, by implications, others) were speculating on these internecine dynamics. --Cookatoo.ergo.ZooM (talk) 07:17, 23 March 2024 (UTC)

PS: Using Google Scholar with suitable query terms seems to bring up useful references on psychological parameters. --Cookatoo.ergo.ZooM (talk) 14:05, 23 March 2024 (UTC)

March 23

Why do Blattella germanica cerci point up then out?

Not adult: they're in a vertical plane, adult: in horizontal plane. Sagittarian Milky Way (talk) 05:43, 23 March 2024 (UTC) Okay that's not very accurate but it sure does look that way from above. They have different orientations before and after sexual maturity. Sagittarian Milky Way (talk) 06:16, 23 March 2024 (UTC)

Courtesy links: Blattella germanica, cerci.-Gadfium (talk) 05:46, 23 March 2024 (UTC)

Hours of daylight vs latitude vs day of year

Hours of daylight vs latitude vs day of year

@SebastianHelm: modified a vectorisation I made of File:Hours_of_daylight_vs_latitude_vs_day_of_year.png. Yesterday, @Episcophagus: posted some questions which I couldn't answer:

1. Why are the borders to constant night and constant day labelled with ‘1 hour’ and ‘23 hours’, respectively?
2. Why isn't ‘equal day length’-latitude south of the equator, as the rate of change in the equation of time affects the daylength (the steepness of the ecliptic at the equinoxes make the sun to move slower in rectascension and thus the daylength is shorter than at the solstices)?

Can anyone help them, please? Thanks, cmɢʟee⎆τaʟκ 12:11, 23 March 2024 (UTC)

1. I don't know but it increases asymptotically "exponentially" cause geometry, the 1 and 0 hour lines would be very close at this scale. I don't know if it's an exponential function but the layman meaning of exponential. Half of 12 hours is most of the way to the polar circle 1 hour would be over 2 more halvings, at the polar circle refraction and Sun radius add at least 50 nautical miles of midnight Sun above simple geometric anyway. Very close it could rise after set while still going down or vice versa (at extremely glancing angle) just from refraction fluctuations. And with a Titanic-like extremely calm sea horizon you could float in the water and climb 5 feet when it set and it rises, float in the water again and it set repeat (eyeballs 5 feet above ocean lowers the horizon about 0.1X Sun radii with ever diminishing returns with height. 10 times more horizon dip than 5 feet takes hundreds of feet, 1,000 times more dip takes millions of feet and 2,000 times more dip takes over infinity light years cause the horizon can't dip more than 90 degrees). At medium latitude this period only lasts seconds.
2. At the equator the Sonnar goes up about 3 and one third minutes before the center would be risen without refraction. Daytime would be about 12 hours 6 minutes and 40 seconds at the equator if there was no equation of time.

Sagittarian Milky Way (talk) 15:57, 23 March 2024 (UTC)

Thanks, @Sagittarian Milky Way: cmɢʟee⎆τaʟκ 14:29, 24 March 2024 (UTC)

British physician who debunked Lourdes miracles

Hi, I've read in a book by Piero Angela (Italian science writer) an interview with William A. Nolen who said a British physician once wrote a book debunking alleged Lourdes miracles. Does anyone know who he was? Thanks.-- Carnby (talk) 12:39, 23 March 2024 (UTC)

Probably Donald J. West in Eleven Lourdes Miracles (1957). See this URL Mike Turnbull (talk) 12:47, 23 March 2024 (UTC)

Thank you!-- Carnby (talk) 12:57, 23 March 2024 (UTC)

The definition of relativistic momentum

As opposed to Newtonian Mechanics, which defines momentum as $P=mv,$ Relativity theory defines momentum as $P=\gamma mv,$ i.e. as $P=m(dx/d\tau ),$ but I wonder why $d\tau ,$ rather than $dt$ - thus sticking to the original Newtonian definition of momentum...

If the reason for preferring the "rest" time $\tau$ is because also the mass $m$ being referred to is the "rest" mass, then also the velocity referred to should be the "rest" velocity, i.e. $v=0,$ shouldn't it?

On the other hand, since the velocity being referred is not the rest velocity $v=0$ but rather the real veloctity, so isn't it usually defined elsewhere as $dx/dt,$ rather than as $dx/d\tau ?$

On the third hand, If the reason for defining the relativistic momentum as $P=m(dx/d\tau )$ is our desire to let Newton's second law $F=dp/dt$ remain invariant under the Lorentz transformations, then why do we replace the expected definition $P=m(dx/dt),$ by the less intuitive definition $P=m(dx/d\tau ),$ rather than by any other definition (e.g. $P=2m(dx/d\tau )$ or whatever), that lets Newton's second law remain invariant under the Lorentz transformations? 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 22:21, 23 March 2024 (UTC)

τ represents proper time, as distinct from coordinate time represented by t. See those two articles for more details, although to be honest, neither article does a great job of explaining the difference. CodeTalker (talk) 01:26, 24 March 2024 (UTC)

Yes, I know that, but how does that answer my question? isn't the proper velocity usually defined elsewhere as $dx/dt,$ rather than as $dx/d\tau ?$ 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 01:29, 24 March 2024 (UTC)

Four-momentum ${\vec {p}}=m{\vec {U}}$ and four-velocity ${\vec {U}}={\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }}$ have to be covariant, i.e. frame-independent, vectors; this is only possible by taking the derivative with respect to a frame-independent quantity, $\tau$, but not with respect to a frame-dependent coordinate time $t$. Proper time $\tau$ parameterises the space-time trajectory ${\vec {x}}(\tau )$ of the particle in question, and ${\vec {U}}(\tau )$ is the tangent vector to that curve. In the non-relativistic limit these definitions reduce to the standard Newtonian momentum and velocity, so there is no contradiction. --Wrongfilter (talk) 09:09, 24 March 2024 (UTC)

If the only reason for defining, the four-momentum as ${\vec {p}}=m{\vec {U}},$ and the four-velocity as ${\vec {U}}={\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }},$ is our desire to let the four-momentum and the four-velocity be covariant, then why do we replace the apparently-more-intuitive definition $P=m(dx/dt),$ by the less intuitive definition $P=m{\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }},$ rather than by any other less intuitive definition (e.g. $P=2m{\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }}$ or whatever), that lets the four-momentum and the four-velocity be covariant? I'm pretty sure there are infinitely many options for the four-momentum and four-velocity to be covariant, aren't there? 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 12:59, 24 March 2024 (UTC)

Most people would want the relativistic momentum to coincide with the Newtonian momentum in the non-relativistic limit. That puts a lot of restrictions on the relativistic form (not sure whether that makes it unique). Your factor 2, for instance, just wouldn't go away. --Wrongfilter (talk) 13:14, 24 March 2024 (UTC)

Well, your remark in the parentheses is the most important one so far, in my eyes. I'm really curious to know (even though you are not sure), if the common relativistic form of momentum is unique, for it to be frame independent and to coincide with the Newtonian momentum in the non-relativistic limit. If the relativistic form of momentum is really unique, then there must be a rigorous proof that derives this relativistic form somehow, so this proof is probably mentioned in textbooks, but it's not, AFAIK. On the other hand, if the relativistic form of momentum is not unique, then I wonder what other forms the relativistic momentum could have, while still being frame independent and coinciding with the Newtonian momentum in the non-relativistic limit. 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 13:39, 24 March 2024 (UTC)

This is the unique definition that (1) reduces to the Newtonian momentum for small speeds, and (2) makes conservation of momentum frame-independent. See [7]. --Amble (talk) 20:59, 26 March 2024 (UTC)

Thank you. However, I noticed that (3.6) and (3.7) in the chapter, are definitions rather than conclusions.

Anyway, also the continuity of the function should be assumed, unless it's considered to be included in your first assumption (1). 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 13:30, 27 March 2024 (UTC)

Continuity follows from (1) and the Lorentz transformations, since a discontinuity anywhere will be a discontinuity at v=0 in some frame of reference. Eq. 3.6 is a definition, but it just attaches a convenient label to a particular quantity. Equation 3.7 defines relativistic momentum and shows that this definition satisfies the relevant criteria, while p=m(dx/dt) does not -- which I believe was your original question. Quoting from [8]: "Indeed, the underlying philosophy is that energy and momentum are nothing else than functions of mass and velocity that, under suitable conditions, happen to be conserved. This is why we treat in a special way those functions, rather than others." --Amble (talk) 04:53, 28 March 2024 (UTC)

March 24

Smallest electric current intensity

Can we consider that the smallest electrical intensity corresponds to one electron per second? Or is there another value? Malypaet (talk) 23:55, 24 March 2024 (UTC)

Surely one electron per hour is a smaller intensity than one electron per second. And one electron per day is smaller than that. I don't see that there's any meaningful definition of the "smallest electrical intensity". CodeTalker (talk) 01:53, 25 March 2024 (UTC)

Thanks, I hadn't seen things from that angle. The electron is a discrete quantity, but indeed the intensity also depends on time which is continuous. However, if we know the start date of this current, we still know the minimum intensity at a given date: an electron between these two dates. Malypaet (talk) 22:06, 25 March 2024 (UTC)

Imagine an infinitesimally thin plane the electron is passing through. Since it is a wave in the electromagnetic field, it is not possible to define when it passes through. But using the probability associated with the wave, we can consider the probability $p(t)$ that, at time $t$, it has already passed through. The intensity of the current would then be $e{\tfrac {{\text{d}}p(t)}{{\text{d}}t}},$ in which $e$ denotes the elementary charge. By slowing down the electron, this can be made (mathematically) arbitrarily small. Perhaps the wave packet started to move through when the first electrons came into being and has not made substantial progress since, the prognosis being that it will not even be halfway through at the end of time, $10^{10^{10^{.^{.^{.}}}}}$ eons into the future.  --Lambiam 12:00, 26 March 2024 (UTC)

My question was about physics, not math. An electron being a discrete quantity and matter also, in physics, an infinitely thin plane does not exist. Unless they are electrons in a vacuum. But since the measurement of intensity is done by material instruments, we return to discrete quantities. Physics is an experiment. Malypaet (talk) 21:55, 26 March 2024 (UTC)

Can you define the notion of the magnitude of the electric current represented by a single moving electron without appeal to mathematical concepts? If a regular stream of electrons moving at a certain velocity represents a certain current, an equally spaced stream at half that velocity will represent half that current. However you choose to define the magnitude of the current of a single electron, the same should hold for this single electron: halve its speed, and you halve its current. So now we should wonder, what is the smallest speed?  --Lambiam 09:22, 27 March 2024 (UTC)

We can make that speed arbitrarily small, but Heisenberg tells us that we may have to increase the spacing of the stream. Which makes the current even lower. PiusImpavidus (talk) 17:06, 27 March 2024 (UTC)

So, the smallest strictly positive current you can measure depends on the time period over which you average it. Yes, an electron can be halfway through, in the sense that it is on both sides at the same time, but when we take a measurement, the wave will collapse and it will be on one side or the other. And taking such a measurement will affect the current. PiusImpavidus (talk) 17:01, 27 March 2024 (UTC)

March 26

If Neptune was named Janus, what will be the name of element 93?

The first suggestion of Neptune is Janus (by Galle), if Neptune was named Janus instead of Neptune, what will be the name (and the symbol) of the element neptunium (element 93), which was named after Neptune? 125.230.237.23 (talk) 04:05, 26 March 2024 (UTC)

We don't answer requests for opinions, predictions or debate. 41.23.55.195 (talk) 05:07, 26 March 2024 (UTC)

It's safe to say the element would have a different name. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:24, 26 March 2024 (UTC)

The elements named after astronomical objects named after mythological figures whose name in Latin ends in the second-declension suffix -us (Neptunus, Uranus) replace the suffix by -ium. This is also the most common suffix for other element names. An author writing a novel set in an alternate universe in which a planet came to be named Janus could believably use the name janium for a later discovered or synthesized element. In the novel's universe so many things may diverge from our universe (in fact, should, to make it a compelling read) that there is little reason for this thus-named element to be specifically that with the atomic number 93.  --Lambiam 11:27, 26 March 2024 (UTC)

Copper IUD end of life

The Copper IUD article mentions that they have a limited life time, but does not mention why the life time is limited.

I am imagining two possible ways in which this happens:

1. the copper slowly erodes. eventually the amount of copper left is too small to be effective

2. the amount of copper left is always sufficient, but a coating slowly develops over time. Eventually the amount of coating is so great that there is insufficient copper released.

Which is more correct in this case? Or is it a combination of both? Or is it some other mechanism? OptoFidelty (talk) 05:25, 26 March 2024 (UTC)

"Unlike an ordinary IUD which is left in the uterus indefinitely, the copper IUD may need to be replaced when the copper is exhausted, usually after about two years, in order to retain contraceptive efficacy."^[9]
    However, "The higher pregnancy rate among the women who continued to use their copper 7 device for a third year suggests that although the copper is still there, it is not available for contraceptive action. ... Examination under the microscope showed hard deposits of material on the surface of these copper 7 IUDs, which Gosden et a1 have shown is a layer of calcium, which would seriously interfere with the contraceptive action of the available copper. This may explain the highly significant difference in the pregnancy rates seen in our two groups-12 pregnancies in the continuation group during the third year of use and no pregnancies in the replacement group. Even allowing for possible differences in the fertility rates of these two groups, which were similar in mean age and parity, this striking difference is remarkable and seems to confirm that although the copper may not be exhausted by three years of use, it may not be available for contraceptive action."^[10]  --Lambiam 10:44, 26 March 2024 (UTC)

March 27

No-signaling and the infinite hat game

(posting in RDS instead of RDMath because the math is pretty clear; it's the alleged "physics" that is not making sense to me)

Recently someone added a note to our article on the axiom of choice a reference to an interesting paper that describes an allegedly counterintuitive consequence of AC. The first bit is fun but not super-surprising to a set theorist (which I am by academic training though not by current employment). The game involves a countably infinite collection of men (just to make the pronouns easy) lined up in the order of the natural numbers. They are given hats, either red or white. No one can see his own hat, but each can see the hats of every man in front of him. Each is supposed to guess the color of his own hat. They are allowed to agree on a strategy beforehand.

It turns out that if they choose and agree on a transversal of the equivalence relation of finite differences on sequences of red and white (that is, two sequences are equivalent if they differ in only finitely many places), then no matter how the hats are placed, all but finitely many of them can guess correctly. The proof of this is essentially trivial once you've understood the definitions. AC guarantees that such a transversal exists.

Now, the counterintuitive thing here is that no man can get any information whatsoever about the color of his own hat just from the colors of the hats in front of him, but nevertheless the strategy works perfectly after some point. I guess if you want you can take this as a point against AC, though from my perspective it doesn't really compete with the clear and simple intuition in favor of AC.

But where I think the paper might go off into the weeds a bit, or maybe I just haven't been clever enough to get what it's talking about, is when it tries to connect this with forms of the no-signaling principle. They identify a "probabilistic" and "functional" version of the NSP, and claim to show that, allegedly counterintuitively, the "functional" version is a stronger "resource".

This I cannot make heads or tails of. First, I don't see how either version of the NSP is a "resource", unless the word is being used in some way I'm not familiar with. As far as I can tell they are constraints, not resources. So my first clear question is, is there some physics-specific sense of the word "resource" that would make this make sense?

Then, in the "functional" version, they allow an oracle for a transversal of the equivalence relation. I don't think they allow that in the "probabilistic" version, though I haven't combed over every line of it, because if they did, you could just use it and ignore the probabilistic part. So isn't it the oracle that's the extra "resource" here? And then, second clear question, why should it be counterintuitive that you can do more with the oracle than without it? --Trovatore (talk) 05:19, 27 March 2024 (UTC) Pinging GiordanoB, who added the note. --Trovatore (talk) 06:35, 27 March 2024 (UTC)

Not relevant to your question, but I find this statement in the paper questionable: "After all, the whole of the theoretical apparatus of physics relies heavily on the use of set theory, thus on the AC." I doubt the theoretical apparatus of physics needs more than naive set theory; if formalized, I expect it can be accommodated equally comfortably in ZF+AD.  --Lambiam 07:46, 27 March 2024 (UTC)

Well, physics does use functional analysis, which is most comfortably formulated in a context where AC is true. For example the Hahn–Banach theorem is inconsistent with AD. --Trovatore (talk) 16:16, 27 March 2024 (UTC)

The idea that the infinite hat game can tell us something about physics is IMO absurd. It should be completely obvious that the lined-up players cannot do better than one would expect from guessing in a purely Newtonian universe, so this has nothing to do with space-like separation. At any time, a player can only have information about a finite number of other hats, so good luck in forming the equivalence class. The equivalence class exists in a Platonic sense, but the player has no access to it. One might as well let player $k$ apply AC to the set $\{\{c_{k}\}\},$ in which $c_{i}$ is the hat colour of the $i$-th player. While the player cannot know this set of sets, it "exists", so in the logic of the puzzle one may apply AC. So if the player now does apply AC, we also apply some smoke, and boom!, paradox.  --Lambiam 08:05, 27 March 2024 (UTC)

The authors understand that, I believe. That's not exactly the point of my question. I still can't follow why they think it's especially counterintuitive that they get more with a "functional" no-signaling than an ordinary one, or even just what they mean by that. As far as I can tell, probability is a fifth wheel in the whole argument. --Trovatore (talk) 16:19, 27 March 2024 (UTC)

Here, in a book titled Bell Nonlocality, the author refers to "a hypothetical no-signaling resource called a PR-box (Popescu and Rohrlich, 1994)". The reference is to: Popescu, Sandu and Rohrlich, Daniel (1994). Quantum nonlocality as an axiom. Foundations of Physics, 24, 379–385.  --Lambiam 08:45, 27 March 2024 (UTC)

Hmm, that does sound like somewhere I might figure out what they mean by "resource"; thanks. --Trovatore (talk) 16:21, 27 March 2024 (UTC)

Gah. And if you replace the Axiom of Choice with Axiom of determinacy which is supposed to make things simpler I think you'd probably still end up with this problem. NadVolum (talk) 10:07, 27 March 2024 (UTC)

NadVolum It's not totally clear what you mean by "this problem", but I do think that AD should imply that the hat wearers almost surely can't do better (or indeed worse) than an asymptotic density of 1/2 correct, when the hats are placed with random colors. That's because AD implies that all the sets involved are measurable, and since each wearer has no information about the color of his own hat, his guess can be no better (and also no worse) than a coin flip, regardless of what (pure or mixed) strategy he plays. I guess that might be what the authors are calling out as "unintuitive", given that AC implies the wearers have a pure strategy that gets them to asymptotic density 1 of correct guesses. They seem to think that mixed strategies ought to be more powerful than pure strategies and they find it surprising that this is not true; I am not sure why they find that surprising.

On a side note, AD does not always make things "simpler". It does imply that sets of reals have nice regularity properties. But it has some odd implications elsewhere (for example that the cofinality of $\aleph _{3}$ is $\aleph _{2}$). --Trovatore (talk) 16:00, 28 March 2024 (UTC)

I believe you're right okay, and have shown why the AC case isn't a problem either. For any particular finite number a random ordering of the men would have a zero chance of having got to the last difference between their's own sequence and the representative of the equivalence class by then - even if the probability is one at infinity. NadVolum (talk) 17:02, 28 March 2024 (UTC)

A red photon and a blue photon are approaching each other in opposite directions. Is the whole system at rest?

On the one hand, it seems the whole system is apparently at rest, because they are approaching each other in opposite directions, by the same speed, while carrying identical masses (being zero). On the other hand, both photons have different momenta, so the whole system's momentum is not zero, so apparently the whole system can't be at rest. That's why I wonder, what's the correct answer to the question: Is the whole system at rest?

I guess it's not at rest, yet I'm not sure about what's the mistake in my first consideration, by which I've concluded the whole system is apparently at rest. 147.235.215.72 (talk) 14:03, 27 March 2024 (UTC)

At rest relative to what? There's no priviliged reference frame to automatically define "at rest" against. If the two photons are approaching each other, you can CHOSE a reference frame where the net momentum of the system is zero. PianoDan (talk) 14:23, 27 March 2024 (UTC)

Here is my answer to your question (in your first sentence): Relative to whoever sees the first photon as red and the second photon as blue. 147.235.215.72 (talk) 15:29, 27 March 2024 (UTC)

The mass $m_{\lambda }$ of a photon in empty space is given by $m_{\lambda }=E_{\lambda }/c^{2}=(hc/\lambda )/c^{2}=(h/\lambda )/c.$ The magnitude of its momentum, $p_{\lambda },$ is given by $p_{\lambda }=h/\lambda .$ We see that the two are related by $m_{\lambda }=cp_{\lambda }.$ So if their momenta are not equal in magnitude, neither are their masses.  --Lambiam 14:31, 27 March 2024 (UTC)

Actually, there is a dispute over whether light has mass. My question assumes light has no mass, yet I know my question could have a clear answer if we assumed light had mass. 147.235.215.72 (talk) 15:28, 27 March 2024 (UTC)

There is no serious dispute over whether light has (rest) mass. People have come up with modifications to Maxwell's equations that would apply if it did, but there's no evidence that it does, there are very tiny experimental bounds on what it can be, and there would be (as I understand it, though it's not my area) some thorny theoretical problems if it did. Basically it can be taken as established (subject to revision, as science always is) that the rest mass of a photon is exactly zero.

Whether light should be considered as having nonzero relativistic mass is a different sort of "dispute", not really about facts but about terminology and bookkeeping. Most writers today consider it better to avoid the concept of relativistic mass. --Trovatore (talk) 16:12, 27 March 2024 (UTC)

Actually the sort of mass I referred to when I responded to Lambiam, was exactly the sort of mass they had referred to, i.e. relativistic mass. Wasn't it clear from the very beginning? Anyway, I agree with you that this dispute is about terminology, but my previous response still holds. 147.235.215.72 (talk) 16:35, 27 March 2024 (UTC) 16:35, 27 March 2024 (UTC)

Suppose that in our frame of reference $F$ the two photons approach each other head-on. Now in every frame of reference $F'$ moving relative to $F$ at a velocity parallel to that of the photons, the photons approach each other head-on with the speed of light too (although Doppler-shifted). So if that's sufficient to consider $F$ the rest frame of the system of two photons, then $F'$ must be one too, but as $F'$ moves relative to $F$, that doesn't work so well.

No, the rest frame of a system of photons would be the one where their total momentum is zero. Such a frame exists provided there are at least two photons not on parallel trajectories. PiusImpavidus (talk) 17:28, 27 March 2024 (UTC)

Regarding your last sentence:

(a) Please notice the two photons I'm asking about are moving in parallel trajectories (in opposite directions).

(b) Additionally, I can't understand why you had to add this stipulation about the non-parallel trajectories: In my view, it's quite easy to find an appropriate reference frame in which both photons have idetical (yet opposite) momenta, as follows: Since the original reference frame I'm asking about sees the first photon as red and the second photon as blue, so the new reference frame is supposed to see both photons as having the same color, indeed Doppler-shifted, but still the same one, probably yellow, because this is the medium color on the spectrum between red and blue. Don't you think it's quite easy to find such a new reference frame?

Regarding your answer: What you wrote, only supports my second consideration (in my original post), by which I concluded the whole system in the original reference frame was not at rest. Actually, your conclusion is exactly the conclusion in my second consideration, ibid. However, please notice I didn't ask for additional considerations (as your ones) that support my second consideration ibid., but rather for an explanation about what's wrong in my first consideration ibid., by which I concluded the whole system in the original reference frame was at rest. Do you think you can point at the exact stage (in my first consideration ibid.) that contains a mistake? 147.235.215.72 (talk) 18:42, 27 March 2024 (UTC)

The system is considered at rest if its total momentum is zero. In your situation the momentum is not zero. Therefore the system is not at rest. Ruslik_Zero 20:09, 27 March 2024 (UTC)

Sources?

This is new to me. Given a multi-particle system, I've always thought that if the particles have identical masses, then the system's velocity should be calculated as the average of the velocities of the particles in the system. At least, this is how we could calculate the system's velocity if the particles having identical masses were massive ones. 147.235.215.72 (talk) 20:22, 27 March 2024 (UTC)

You obviously thought wrong. Photons have no mass and $p=\gamma mv$ simply does not apply. Do yourself (and ourselves) a favour, accept that and move on. --Wrongfilter (talk) 20:26, 27 March 2024 (UTC)

Sorry, but this is exactly what I've said to Lambiam: "My question assumes light has no mass". In other words, photons have zero mass. Hence they have "identical masses (being zero)", as I wrote in my original post.

But my question was not answered yet. Is the system at rest? 147.235.215.72 (talk) 20:43, 27 March 2024 (UTC)

Your question has been answered many times. The system is not at rest because the total momentum is not zero. --Wrongfilter (talk) 20:53, 27 March 2024 (UTC)

This explanation had only been given by Ruslik_Zero (before you repeated their answer), but the other editors gave me other explantions, e.g. that light has mass, while I didn't understand their explanations, because I assume photons have no mass, i.e. they have identical masses being zero.

As for Ruslik_Zero's explanation, which is now also your one, did you read my response to them? If you did, then what's the answer to what I asked them? 147.235.215.72 (talk) 21:09, 27 March 2024 (UTC)

What question? "Sources?"? Any decent textbook on special relativity. Listen, HOTmag, in a thread from a couple of days ago you said yourself that relativity defines momentum as $p=\gamma mv$ (and that would be the formal basis for your handwaving argument). Now plug in what you know about photons: $m=0$, $v=c$. This gives $\gamma =\infty$ and $p=\infty \cdot 0$, which is of course undefined. You cannot compute the momentum of a photon that way. The momentum of a photon is given by the de Broglie relation, $p=hc/\lambda$. --Wrongfilter (talk) 21:23, 27 March 2024 (UTC)

I remember I mentioned the formula of relativistic momentum (By the way I don't remember HOTMAG has responded to my current question, yet I do remember they responsed to me on other occasions in this reference desk), but my question did not assume this relativistic formula. My question has only assumed that if the particles have identical masses (whether a zero mass or a non-zero mass), then the system's velocity can be calculated as the average of the velocities of the particles in the system. I didn't connect this calculation with any momentum. 147.235.215.72 (talk) 21:56, 27 March 2024 (UTC)

The average velocity equals ${\frac {m_{1}\mathbf {v} _{1}+m_{2}\mathbf {v} _{2}}{m_{1}+m_{2}}}.$ If $m_{1}=m_{2}=0,$ this comes out as ${\frac {\mathbf {0} }{0}}.$  --Lambiam 23:03, 27 March 2024 (UTC)

Again, you've used the concept of momentum. This is also the concept I used when I concluded in my second consideraion (of my first post) that the whole system was not at rest, yet I didn't have to devide by zero (as you had), because I didn't use the formula you've used, but rather the formula of the sum of momenta (when a given photon's momentum is received by e.g. the photon's wavelength) for getting the whole system's momentum.

However, in my first consideration (ibid.), as well as in my previous response, I bypassed the very concept of momentum, by using a totally different argument for concluding that the whole system was at rest...

ALL IN ALL, my question is not about what else we can claim (as you're claiming now) for proving that the whole system is not at rest, but rather about what's wrong in my first consideration (of my original post) or in my argument of my previous response, wherein I bypassed the very concept of momentum and concluded that the whole system was at rest. 147.235.215.72 (talk) 01:01, 28 March 2024 (UTC)

My reply used no other concepts than mass and velocity, taking the average velocity to be the weighted average, being the velocity of the centre of mass.  --Lambiam 12:43, 28 March 2024 (UTC)

Indeed, you didn't utter the explicit word "momentum", yet you did use products of mass and velocity, which is actually the same as using momentum, even without saying explicitly "momentum". This is how I describe your method of calculation, as far as you are concerned. However as far as I'm concerned, my first consideration (of my original post) hasn't used momenta, nor products of velocities multiplied by anything else whatever it is. My first consideration (ibid.) has only used velocities, being actually the velocities of particles that have the same mass (while assuming light has no mass). Why doesn't my method work with photons (having identical masses), whereas it does work with massive bodies (having identical masses), while in both cases I bypass the concept of momentum, as well as any product of a velocity multiplied by anything else? 147.235.215.72 (talk) 16:41, 28 March 2024 (UTC)

I could have written it as $\sum _{i}w_{i}\mathbf {v} _{i},$ were $w_{k}={\frac {m_{k}}{\sum _{i}m_{i}}}.$ No velocities were harmed in this formula by multiplying them by velocities.  --Lambiam 22:57, 28 March 2024 (UTC)

Rule A: In a system of particles with equal mass, the rest frame is the frame where the average velocity of the particles is zero.

Rule B: In a system of particles, the rest frame is the frame where the total momentum is zero.

Rule A follows from rule B, assuming that the mass isn't zero (and it isn't), but we have to be careful how to define that mass that must be equal. Rule A only follows if the particles have identical relativistic masses, $\gamma m_{0}$ for massive particles or $h/(\lambda c)$ for photons, the thing you get when dividing momentum by velocity. You, in your first consideration, used the rest mass of the particles. That works for non-relativistic particles, but not for photons. PiusImpavidus (talk) 11:30, 28 March 2024 (UTC)

If they move in opposite directions, I call them antiparallel. PiusImpavidus (talk) 09:55, 28 March 2024 (UTC)

In the first paragraph of my first answer, I told what was wrong with your first consideration: if the first consideration were true, it follows that there is a second rest frame, moving relative to the first. But two frames moving relative to each other cannot both be rest frames for the same system of particles, so the premise must be wrong. Therefore, the first consideration must be false. PiusImpavidus (talk) 10:12, 28 March 2024 (UTC)

The contradiction is rooted in relativity. Classically, the antiparallel photons' speeds should entail the velocity additions of the classical Galilean transformations, but according to relativity light speed is invariant (instead of invariant distances, time intervals and simultaneity). Had the OP been talking simply about two sound waves (red-like and blue-like) having a medium's speed of sound converging from opposite directions then their first consideration that they are at rest is accurate and it follows that the waves' inertial mass and momentum is moving with respect to them in the same direction as the red wave moves. Modocc (talk) 15:16, 28 March 2024 (UTC)

@PiusImpavidus: You claim my first consideration (of my original post) leads to a contradiction. But I have been aware of that, from the very beginning, in my oringinal post, where I presented also the second consideration which contradicts the first one, so you didn't have to remind me of what I've been aware of. i.e. that the first consideration leads to a contradiction.

When we encounter two apparently correct arguments contradicting each other, while we don't know yet which one is the wrong one, then for finding the wrong argument, we must point at the wrong stage of the argument, by declaring "This stage is unjustified because it's not based on any well established rule". I claim, that my first consideration (ibid.) is apparently well established, beacsue it uses the same method we could have used if the photons had been massive bodies. Just think, what would occur if two massive bodies, having the same masses, were approaching each other in opposite directions, but at the same speed: Wouldn't we agree, that the whole system were at rest, we bypassing the very concept of momentum? We would, so isn't this fact sufficient for using the same method for photons, we bypassing the very concept of momentum? The question is, why do we have to discriminate a pair of particles having zero mass, while we actually manage to bypass the very concept of momentum, just as we would've been allowed to bypass it - had we used the same method for massive particles? 147.235.215.72 (talk) 16:41, 28 March 2024 (UTC)

"...my first consideration (ibid.) is apparently well established,.." What do you mean by it is well established? If you mean photon's lightspeed is c, yes that has been. Neither the observer(s) or the net momentum of the photons are necessarily at rest except as you have referenced them. Modocc (talk) 17:15, 28 March 2024 (UTC)

I'm well aware that your first consideration and your second consideration contradict each other. That was clear from the beginning. What I'm telling you (for the third time now) is that the first consideration also contradicts itself, therefore can never be true.

If you want more details (which I already gave above): your argument depends on all particles having the same mass. But what mass is that? It's the relativistic mass, but you used rest mass. PiusImpavidus (talk) 00:16, 29 March 2024 (UTC)

Photon wavelengths are not fixed invariably at their emission. Neither an indelibly red nor an indelibly blue photon exists in timespace because the wavelength of any received photon depends on the relative motions of source and receiver. The scenario of "a red photon and a blue photon...approaching each other in opposite directions" may be merely the perception of a viewer who is moving fast enough along the line joining the two sources in the direction of the latter photon's source. Philvoids (talk) 18:28, 28 March 2024 (UTC)

March 28

Magnet affecting resonance.

My teen grandson has a home-craft steel anvil, about 15lbs in weight. When he taps it with a hammer it resonates with a high note for about 3 seconds. However he discovered that when he attached a small but strong magnet to the anvil the resonance is shortened to about 1 second - audibly obvious.
Can someone kindly explain why that should be. Thank you. Richard Avery (talk) 07:27, 28 March 2024 (UTC)

Surely the same thing would happen if any mass were added to the anvil. Strike tone is somewhat relevant here. I doubt the fact that it is a magnet is relevant. Shantavira|^{feed me} 09:34, 28 March 2024 (UTC)

An interesting reference, but the magnet is small, about 1 inch long and ovoid, with a minute surface contact with the anvil. But I take your point and perhaps some further experimentation is needed. Thank you. Richard Avery (talk) 10:55, 28 March 2024 (UTC)

(edit conflict) An effect easily demonstrated by a teaspoon and coffee cup or mug. Taking the handle as 000°, gently tap at 090°, 135° and 180° and you'll hear three different notes. The reason is the mass of the handle lowers the note when it is at an anti-node, but of course has no effect at a node. Martin of Sheffield (talk) 11:14, 28 March 2024 (UTC)

The magnet turns a high note lasting 3 seconds into an unspecified note lasting one second. It appears that adding the magnet increased the damping of the vibration. Where the magnet touches the anvil, there can be sliding action, leading to dissipation of energy. The same happens if there's a crack in the anvil. In solid metal, there's only elastic deformation, which dissipates no energy. The magnet may also cause dissipation through eddy currents. PiusImpavidus (talk) 11:54, 28 March 2024 (UTC)

Thank you for the responses, magnetism has little to do with it. A damping effect is at play. Richard Avery (talk) 15:43, 28 March 2024 (UTC)

March 29

Looking for Map with Eclipse Path and Actual Weather Conditions

I'm looking for an interactive map showing the path of totality of the Solar eclipse of April 8, 2024 and current weather conditions. I live near Chicago so I'm primarily interested in the weather for Illinois, Indiana and Ohio. I've found plenty of maps showing the path of totality and plenty of maps showing the probability of clouds based on historic data, but I have not yet found a map showing the path of totality and current weather conditions. To put it another way, if today was the eclipse, where can I find a map showing the path of totality and today's weather forecast? A Quest For Knowledge (talk) 14:04, 29 March 2024 (UTC)