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August 12

Moffat and ozone

Who was "Dr Moffat of Hawarden", who wrote papers about ozone, invented an ozonometer (see for example [1]), and after whom "Moffat's Ozone papers" ([2]) appear to be named? What more can we know about him? Was he Thomas Moffat MD FRAS ([3])?

How do the papers work? Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 14:44, 12 August 2024 (UTC)

Seems very likely it was Thomas Moffat, since the PMC2439974 article mentions ozone 13 times and was written in 1856, soon after ozone had been characterised for the first time by Christian Friedrich Schönbein in 1840. The Google books reference was also written in 1856, by David T. Ansted. Mike Turnbull (talk) 15:31, 12 August 2024 (UTC)

The full name is given as Thomas Barbour Moffat ([4], p.17), an obituary is here, and genealogical data here. --Wrongfilter (talk) 15:41, 12 August 2024 (UTC)

Thank you, both. I have compiled data from the above sources into Thomas Barbour Moffat (Q128923496). There is also an obituary in the BMJ, which describes him as "the author of many papers on geology, meteorology, sanitation etc."; though I can only find one other (On Medical Meteorology (Q58709971)).

Thomas Barbour Moffatt, of the distinguished Moffatt clan of Sundaywell, Dumfriesshire. What's the origin of the name "Sundaywell", sometimes written "Sunday Well"? 91.234.214.10 (talk) 17:41, 12 August 2024 (UTC)

According to the Dumfriesshire OS Name Books, 1848-1858 "Sundaywell - [Situation] At E. [East] end of Sundaywell Moor. A good Spring, the water of which is deepened by a Stone dam round it, seemingly very old, There is a tradition, that, at Some remote time, there were great numbers of people Baptised here, The farm takes the name from the well." and "Sundaywell - A large Farm House with extensive outbuildings and garden the property of trustees of the late Alexander Moffatt Occupied by John Edgar". Mikenorton (talk) 21:27, 12 August 2024 (UTC)

How did Carl Linnaeus know about the species he described?

I was reading about the Pied kingfisher, whose article states that it was first described by Linnaeus in 1758. I assumed he must have visited "Persia and Egypt" in order to catalogue the species, as that's where he says it lives—but I don't see reference to any such expedition on the catalogue's article. Were their earlier catalogues that Linnaeus drew from? How did he know about this bird? ꧁Zanahary꧂ 16:19, 12 August 2024 (UTC)

To quote from this webpage from Berkeley, "Linnaeus continued to revise his Systema Naturae, which grew from a slim pamphlet to a multivolume work, as his concepts were modified and as more and more plant and animal specimens were sent to him from every corner of the globe". Mikenorton (talk) 20:51, 12 August 2024 (UTC)

He was apparently quite happy to accept specimens, but very stingy in giving credit or sending some back (according to a German bio podcast I recently listened to). --Stephan Schulz (talk) 14:30, 14 August 2024 (UTC)

photons

Suppose the Sun were part of a binary: identical twins, separation several million miles. When the stars were 'side-by-side' in our line of sight, the radiation received on earth would be twice its present value. But what would happen when one star was eclipsing the other? Would photons from the distant star pass unscathed through the nearer one? If not what would the received intensity be? What spectral changes if any would be observed compared to the familiar G2 spectrum? Renshaw 2 (talk) 16:37, 12 August 2024 (UTC)

Subjecting our planet to double the usual solar radiation is likely to leave no one alive to make the interesting observation of whether a star is transparent. Philvoids (talk) 17:11, 12 August 2024 (UTC)

I heard that the average photon generated in the sun's core bounces around in there for hundreds of thousands of years before accidentally escaping to the surface. So I'm pretty certain that the sun isn't transparent. Abductive (reasoning) 23:10, 12 August 2024 (UTC)

"When this random walk process is applied to the interior of the sun, and an accurate model of the solar interior is used, most answers for the age of sunlight come out to be between 10,000 and 170,000 years."^[5]  --Lambiam 23:17, 12 August 2024 (UTC)

If the expected number of steps of a random walk process for getting a distance d away from the starting point equals n, the expected number of steps needed to traverse twice that distance, 2d, equals n². I don't know the average time between two hops of a photon, but I bet it is less than 3 seconds. This means that a randomly hopping photon takes more than 10⁷ hops in a year, and more than 10¹¹ hops in 10,000 years. To get all the way through the Sun instead of getting out from the core would then take more than 10²² hops or 10¹⁵ years, several orders of magnitude longer than the estimated age of the universe.. Of course, many incident photons would find their way out earlier, but in a very different direction than where they came from.  --Lambiam 23:42, 12 August 2024 (UTC)

Subjecting our planet to double the usual solar radiation is likely to leave no one alive

What about a hypothetical planetary environment with an unusually strong magnetosphere? Same result? Viriditas (talk) 23:42, 18 August 2024 (UTC)

Among the countless binary star systems there are probably such star systems elsewhere, two yellow dwarfs, each of approximately one solar mass. Photons reaching the surface of a hot star are almost immediately absorbed by the ions of it outer layer.  --Lambiam 23:12, 12 August 2024 (UTC)

Stars are pretty dense and full of free electrons, so they're very good at scattering light. The light from the other star will loose its identity in the photosphere (i.e., it will be scattered and frequency shifted to match the temperature of the gas).

Irradiating a star with light from a nearby star may raise the temperature a bit on the sides where they face each other. On the other hand, two stars close together will deform each other by tidal forces and the tidal bulges get a bit cooler than the lower parts. I don't know what effect will be dominant in the case of two G stars in a very tight orbit, but I expect the tidal deformation. Such an unequal temperature distribution can be seen in the spectrum. The absorption lines in the spectrum will deform as the binary spins and the hotter and colder parts alternate redshift and blueshift. PiusImpavidus (talk) 08:30, 13 August 2024 (UTC)

The OP's proposed reconfiguration of our solar system cannot be carried out as a practical test before reaching a political consensus on the work that it will involve. The project to split the Sun into a Binary star pair will lie beyond the resources of a space agency such as NASA unless significant new funding can be raised. In the present political climate this funding must wait while Environmental impact assessment statements are properly considered. The OP's proposal implies that the binary stars' orbit is in the Ecliptic plane of the Earth's present orbit so that solar-solar eclipses are periodically observed but a full calculation of the Orbital mechanics of the revised solar system will be needed. Shall we have an assessment of how Gravitational waves from a local binary star will affect life on Earth? An international body might be formed to collect, review and eventually override statements from affected stakeholders that will include publishers of tidal predictions, representatives of the Sunscreen manufacturers, weather forecasters whose computer models will need updating, Sun worshippers, astrologers and others. As an interim and less radical test of the Sun's opacity I suggest allocating more resources to Solar observation where one may try to detect prominent Sunspots as they rotate out of direct view to the far side of the Sun. Philvoids (talk) 20:45, 14 August 2024 (UTC)

Dear Pius,

Thanks for your reply to my query about photons (13 Aug.). Apologies for the belated reaction.

‘Fearless’ or ‘intrepid’ you may be, but that’s no bad thing in science is it. Breakthroughs such as relativity & quantum theory, among others, would never have happened if their pioneers hadn’t been brave enough to violate the constraints of ‘common sense’ and envisage things that were “clearly impossible”.

Though admittedly, even Einstein’s nerve failed him when it came to contemplating black holes, or gravitational waves, or ‘spooky action at a distance’, or a universe that might not be static in extent. (and he regretted that last one didn’t he; I think he called it “meine größte esel” (‘ass’ or ‘donkey’) a few years later, after Edwin Hubble’s findings.

Anyway, thanks for your useful comments. You possibly guessed that I was asking the question in the context of the prospects for a habitable zone in a binary star system. We have a tendency to immediately dismiss binaries as not worth investigation. But it seems to me that there’s considerable scope for them to incubate life.

One possibility is a closely coupled binary, if viewed in terms of the Solar System, say both stars within the orbit of Mercury (although Mercury itself probably wouldn’t survive!), and this is why I asked about the combined radiation from two stars that sometimes eclipsed. This would obviously be a very disruptive environment for emerging life. It seems to me that a planet at 1AU wouldn’t be viable, but that a habitable zone could exist at the distance of Mars or possibly the asteroid belt. (I’m not taking into account the various other properties such a planet would also need to have, merely surface temperature & tolerable variation.)

Another scenario would be a widely-spaced binary system (viewed in solar-system terms, the second Sun orbiting at about the distance of Neptune). The resulting environment might not be entirely to our liking but it should be quite habitable. A rough calculation suggests that the inner planets of each star would continue in stable –albeit slightly more elliptical– orbits around their own star & would not be at risk of being captured into an orbit round both stars. This seems a reasonably benign environment, and that such a binary would be certainly worth investigating.

Of course some configurations might not have a HZ at any distance. If the stars happened to be Sun-like and about 1AU apart, I’d guess there might be no stable planetary orbit – except maybe ones that were far too distant– and hence no HZ.

Thanks for your help. Renshaw 2 (talk) 11:56, 19 August 2024 (UTC)

"It is estimated that 50–60% of binary stars are capable of supporting habitable terrestrial planets within stable orbital ranges." from our article Habitability of binary star systems. Modocc (talk) 13:00, 19 August 2024 (UTC)

August 13

Dogs barking at some people

A friend of mine told me that on two recent occasions dogs started barking uncontrollably at a new acquaintance of her, a young man, in both cases much to the embarrassment of the dog owners, who could not silence their dogs and saw no other solution than to walk their dogs away. She did not see anything in his behaviour or demeanor that could have triggered this and asked me for an explanation. Now this is purely anecdotal, and I've heard similar anecdotes before. My question: is there evidence (beyond the anecdotal) for some people unwittingly tending to alarm dogs more than others, and if so, have possible science-based explanations been offered for the phenomenon?  --Lambiam 00:02, 13 August 2024 (UTC)

There seems to be evidence that dogs can smell biological changes in humans which indicate the presence of a disease (COVID, cancer, etc). Medical detection dogs exist, but they must be trained to detect VOCs (volatile organic compounds). Of course, canines may bark for multiple other reasons. --2001:871:6A:1B71:4CD8:804D:2688:FD83 (talk) 06:28, 13 August 2024 (UTC) Oops, --Cookatoo.ergo.ZooM (talk) 06:29, 13 August 2024 (UTC)

Who knows what it's like to be a dog. It's such a diverse group. I know a Siberian Husky street dog with an apparent racist hatred of white people. He's a good dog though, mostly. My dog is more aggressive towards male strangers but will change direction and try to follow attractive female strangers, which is a little awkward. He also barks aggressively at fallen leaves, beetles (no size dependency), mynah birds, skinks, many things...not seeing a pattern so far. They can smell adrenaline of course, which might explain some cases. Sean.hoyland (talk) 07:51, 13 August 2024 (UTC)

Did she ever see her new acquaintance in sunlight? —Tamfang (talk) 16:57, 16 August 2024 (UTC)

Lambiam, if you could provide the missing details, I believe we could solve this. Find out what type of breed the dogs are and what kind of work her acquaintance does. If he's into construction, landscpaing, or works in a mortuary, for example, he might have some very complex scents on him that are disturbing the dogs. I think it's also possible that it doesn't mean anything at all. Some of the dogs in my neighborhood will bark uncontrollably at strangers who talk to their owners. If they don't receive the voice command to be quiet, they will continue. Viriditas (talk) 23:18, 19 August 2024 (UTC)

Skin disorder

I have just got off the bus, where I was sitting behind a woman with globules all over her body. I have looked at List of skin conditions, (the longest article I have ever encountered), but am none the wiser. What is the name of it? 2A00:23D0:E1D:2D01:B1A1:2846:AA72:4934 (talk) 14:19, 13 August 2024 (UTC)

Lipoma...maybe. Sean.hoyland (talk) 14:56, 13 August 2024 (UTC)

Could also be neurofibromatosis. nmael^talk 15:08, 13 August 2024 (UTC)

Could be monkeypox. Graeme Bartlett (talk) 11:42, 17 August 2024 (UTC)

Inches and lines

I'm working on Phobaeticus annamallayanus, described by James Wood-Mason. As a taxon author he is quite unhelpful, since he left only a brief and vague description ([6], see the second entry, Phibalosoma annamallayanum). Worse, he gave his measurements in inches and lines, something which was apparently in vogue at the time but is a pain at the moment, since a "line" can be anywhere between a tenth and fortieth of an inch. I'm willing to make approximations like "between two and three inches", but that's less than ideal. Does anyone have any idea which meaning of line 19th century entomologists likely meant?

Secondly, at the start of his measurements, he gives the total length as "8 in 9" lines. What does that mean? Surely not ⁸/₉.

Thank-you for any help you can provide. Cheers, Cremastra (talk) 16:20, 13 August 2024 (UTC)

In the last century, there was a useful booklet called Wightman's Arithmetical Tables. The first entry for Long Measure was "12 lines = 1 inch". This is also noted here [7]. These books have many alternative versions of the Thirty days hath September rhyme of which I was previously unaware. 91.234.214.10 (talk) 16:38, 13 August 2024 (UTC)

I feel that claiming it can mean anything from a tenth to a fortieth of an inch is an exaggeration. The different definitions were based on usage. It is not reasonable to think that he would be using the Russian ballistic definition of a line when referring to the size descriptions of an insect. The only real issue is that 1/12 of an inch was the standard for botany and entomology until the mid-1800's. Then, some (not all) areas of science (primarily the French) tried to "decimalize" it to 1/10 of an inch. So, the question is, did Wood-Mason use the newfangled definition of 1/10 of an inch or the long-standing definition of 1/12 of an inch. I personally see no reason he would update all of his research, which took place long before the 1/10 of an inch definition was introduced, to a new measurement. 12.116.29.106 (talk) 16:49, 13 August 2024 (UTC)

Agree that one twelfth is the most likely. See The Measures, Weights & Moneys of All Nations (p. 50) of 1890 which says: "A line is 1/12th part of an inch".

Also Great Britain's New Proposed Decimal Albert System of Weights, Measures and Coins (p.32) of 1869 which has "1 Line, or inch 1/12".

And an American textbook, Higher Arithmetic, Or, The Science and Application of Numbers (p. 150) of 1848 which says: "The inch is commonly divided into eigths or tenths; sometimes however, it is divided into twelfths which are called lines".

Alansplodge (talk) 17:40, 13 August 2024 (UTC)

The previous entry for Phibalosoma acanthopus (which should be the same as Phobaeticus serratipes, see [8]) gives us a simple math problem: "4 in. 6.5 lines + 12.5 lines = 5 in. 7 lines." That only works if one inch is twelve lines. The total length of "8 in. 9 lines" means 8 inches and 9 lines. --Amble (talk) 17:58, 13 August 2024 (UTC)

@Amble The total length of "8 in. 9 lines" means 8 inches and 9 lines Oh, I'm a moron. I didn't see the dot after "in".

Well, at least that wasn't too complicated.

Thanks for spotting the clue in the above entry; I was hoping it was 1/12, since that seemed standard, but I didn't want to guess. Thanks also to IPs and @Alansplodge for their research.

Cheers, Cremastra (talk) 18:58, 13 August 2024 (UTC)

In such cases I use BHL's "Search inside" button to search for "in. 10" and "in. 11". Indeed under "Phibalosoma acanthopus" is a "1 in. 11.75 lines". So clearly 12 lines per inch. JMCHutchinson (talk) 14:24, 15 August 2024 (UTC)

August 14

Indeed measuring the neutrino's restmass is regarded to be extremely problematic (or hard), but why can't the following simple experiment solve the problem?

In Cowan–Reines neutrino experiment, anti-neutrinos created in a nuclear reactor by beta decay, reacted with protons to produce neutrons and positrons, by the following reaction:

ν
_e +
p⁺
→
n⁰
+
e⁺

Now I suggest the following calculation:

1. Both the proton's restmass, the neutron's restmass, and the positron's restmass, are already known.

2. Hence, by the mass-energy equivalence, along with the kinetic energies in that experiment, both the proton's energy - the neutron's energy - and the positron's energy, are already known.

3. Hence, by the (quasi-)equation
ν
_e +
p⁺
→
n⁰
+
e⁺
, along with the conservation of energy, we receive the anti-neutrino's energy.

4. Hence, by the mass-energy equivalence, along with the anti-neutrino's kinetic energy in that experiment, we receive the anti-neutrino's restmass.

Where is the wrong stage? Maybe I'm not allowed to assume the conservation of energy? HOTmag (talk) 07:01, 14 August 2024 (UTC)

The neutrino speed is unknown.  --Lambiam 08:44, 14 August 2024 (UTC)

Unknown yet, but once we discover the anti-neutino's speed, the experiment I've suggested may simply solve the problem of measuring the anti-neutrino's restmass, right?

Additionally, even without measuring the anti-neutrino's velocity, still the quasi-equation
ν
_e +
p⁺
→
n⁰
+
e⁺
, along with the conservation of energy, and with the mass-energy equivalence, do determine the anti-neutrino's relativistic mass, right? HOTmag (talk) 09:39, 14 August 2024 (UTC)

The speed of a neutrino or anti-neutrino is not a physical constant. One may hope to measure it, but should then be aware that it varies with the frame of reference of the observer. In some frames of reference the (anti-)neutrino is at rest.  --Lambiam 22:42, 15 August 2024 (UTC)

I've never said this speed is constant. You said it was unknwon - you referring to the neutrino's varying speed, so I answered it was unknown yet - me referring to the neutrino's varying speed. Anyway, my previous comments still hold. HOTmag (talk) 08:07, 16 August 2024 (UTC)

You cannot measure its speed after it has disappeared in the interaction. If you succeed in measuring a neutrino's speed before the interaction with the positron has taken place, you thereby will have changed its speed. Moreover, how can you know that the particle whose speed you measured is the same particle that disappeared a few nanoseconds later? You can hardly tag it.  --Lambiam 01:30, 17 August 2024 (UTC)

Your first postulate is simply false, per Amble below. Remsense ‥ 诉 09:47, 16 August 2024 (UTC)

Your current indent is simply false, because you're responding to my response to a user other than the one you indicated. By "my previous comments", I referred to my comments to the user I was responding to. HOTmag (talk) 11:01, 16 August 2024 (UTC)

It's false regardless of who you were replying to. Remsense ‥ 诉 11:05, 16 August 2024 (UTC)

What's false? My first postulate? So again, also your indent was false. HOTmag (talk) 11:07, 16 August 2024 (UTC)

The first statement is not "false". Those masses are known, just not with sufficient precision to make HOTmag's proposal a viable way for measuring neutrino masses.--Wrongfilter (talk) 11:33, 16 August 2024 (UTC)

They asked where the flaw in their reasoning was, and if one had to pick a spot, it's that the first point is not the case. Apologies for coming off more pedantic about it than it sounded in my head. Remsense ‥ 诉 11:49, 16 August 2024 (UTC)

Besides the difficulty in measuring the small difference between the neutrino's speed and the speed of light, there's the problem that the neutrino's mass is small compared to the uncertainty on the masses of the other particles. For example, the standard uncertainty on the proton mass is 0.29 eV [9] and the standard uncertainty on the neutron mass is 0.48 eV [10]. These are very small errors in fractional terms, but the neutrino mass is even smaller: the sum of the three neutrino masses is below 0.120 eV (Neutrino#cite_note-Mertens-2016-mν-1). On the whole, an endpoint experiment like KATRIN can achieve much better sensitivity to the neutrino mass. --Amble (talk) 16:04, 14 August 2024 (UTC)

To put it another way, it's not a logic puzzle where each item is either "known" or "unknown". We know all of these things to within some level of uncertainty. In absolute terms, we know the neutrino mass better than we know the neutron or proton mass, or the difference between the two. Therefore, your procedure wouldn't improve our knowledge of the neutrino mass, it would (if anything) improve our knowledge of the proton-neutron mass difference, using our current knowledge of the neutrino mass as a "known" input. --Amble (talk) 16:52, 14 August 2024 (UTC)

In short, OP has proposed a purely logical solution for what remains a purely empirical problem. Remsense ‥ 诉 23:07, 15 August 2024 (UTC)

Yes, in the same vein I could measure the mass of a table tennis ball by throwing it at a known speed at a large wrecking ball and measuring how far the wrecking ball is deflected. Other ways might work better. NadVolum (talk) 10:09, 17 August 2024 (UTC)

August 15

Outdated species names of bees in Nomia and Nomadopsis

I've found a list of bees that the bee fly Heterostylum robustum parasitizes in a book written in 1973, but a lot of them are outdated. The list is: "Nomia bakeri Cockerell, N. triangulifera Vachal; Nomadopsis anthidius Fowler, N. scutellaris Fowler, and Halictus rubicundus (Christ.)." I have suspicions but I can't confirm them and would like to be sure, but can't locate any authoritative resource.

• Nomia bakeri I know has been reclassified as Nomia nevadensis baker ([11]), and I think may be Dieunomia nevadensis bakeri.
• Nomia triangulifera I am guessing is probably Dieunomia triangulifera
• Nomadopsis anthidius I think is Calliopsis anthidia: Calliopsis is a synonym but the second part apparently changed?
• Nomia scutellaris is still valid
• Halictus rubicundus is still valid

Does anyone know how to confirm any of this, or how to source it? Mrfoogles (talk) 07:10, 15 August 2024 (UTC)

• Have you tried asking at WT:WikiProject Insects? Or look at the sources in the taxobox at the bottom of the Nomia (bee) article or the sources in List of Nomia species? Abductive (reasoning) 08:01, 15 August 2024 (UTC)

This page syas: "Calliopsis anthidia - Synonyms: Calliopsis anthidius". Alansplodge (talk) 10:54, 15 August 2024 (UTC)

Also this page: "Dieunomia triangulifera - Synonyms: Nomia triangulifera" using this as a reference. Alansplodge (talk) 10:59, 15 August 2024 (UTC)

Thanks, this worked with a little extra googling for the last sources. Pretty much resolved now. Odd WorldSpecies, sourced to GBIF, has synonym listings it does not. Mrfoogles (talk) 00:20, 16 August 2024 (UTC)

Ancient Greek ὄψις (ópsis) is feminine, so Calliopsis from kalli- + opsis should be considered feminine as well. This means that Calliopsis anthidius (Fowler, 1899) did not conform to the gender-agreement requirement of the International Rules of Zoological Nomenclature introduced in 1905, and was changed to Calliopsis anthidia to make the gender of the specific epithet agree. (Due to an error in the introduction of the genus name Nomadopsis (Ashmead, 1898), this genus name was adopted by some but not accepted universally, which has led to some confusion.^[12])  --Lambiam 11:01, 15 August 2024 (UTC)

I thought it might have something to do with gender -- makes sense. Mrfoogles (talk) 00:32, 16 August 2024 (UTC)

August 17

Transmutation

I'd like to see a video of an element transmuting into another element. I quote Richard Dawkins and Yan Wong in their wonderful book 'The Ancestor's Tale', "The half-life of carbon 15 is 2.4 seconds. After 2.4 seconds you'll be left with half of your original sample. After another 2.4 seconds you'll have only a quarter of your original sample..." I realize that carbon-15 is not something you are going to readily find in a laboratory. I'm guessing it would appear to be sublimating into a gas. Ifyoucrydon'tlose (talk) 11:21, 17 August 2024 (UTC)

Does he say what it transmutes into? ←Baseball Bugs ^{What's up, Doc?} carrots→ 11:24, 17 August 2024 (UTC)

No he doesn't. In the Wikipedia article 'Isotopes of Nitrogen', nitrogen-15 is said to come from carbon-15. Ifyoucrydon'tlose (talk) 11:33, 17 August 2024 (UTC)

Even more interesting would be a video of a solid transmuting into another solid. But when I search for videos on the subject, all I find are chalk talks. Ifyoucrydon'tlose (talk) 11:43, 17 August 2024 (UTC)

Our article Isotopes of carbon agrees with the half-life. It transforms into nitrogen-15. But it would not be possible to make a visible solid piece of carbon-15 as it decays so fast, and in decaying releases so much energy. I could imagine a solid lump would just explode into a hot carbon plasma fireball in a fraction of a second. An example of a video could be of a supernova, where nickel-56 decays into iron-56 over about a month. Graeme Bartlett (talk) 11:38, 17 August 2024 (UTC)

You're right. These isotopes are unstable and are involved in long chains of reactions. Ifyoucrydon'tlose (talk) 11:46, 17 August 2024 (UTC)

I'd like to see a video though of the transmutation of solid lead into solid gold.  --Lambiam 21:27, 17 August 2024 (UTC)

It could be done with some isotope with a half-life of a few decades, then make a time-lapse video. But you have to put the highly radioactive sample in front of a camera (or automatically position is in front of the camera for a few seconds every week), instead of safely locked away in some nuclear waste storage facility. A risky move for a time-lapse video that will only be seen by the next generation. PiusImpavidus (talk) 08:35, 18 August 2024 (UTC)

See also long-term experiment. DMacks (talk) 17:07, 18 August 2024 (UTC)

Any time you look at something that is radioactive, or at a nuclear explosion, you're seeing it. I think the interesting examples would be when the substance changes appearance. Watching one metal (or metal-oxide) become another metal...are you actually seeing a change? Something like a solid to a gas, or evidence of releasing a gas from a solid (alpha particles) would be nice. But it is indeed an annoying trade-off between how long you want to spend making a time-lapse video vs how willing and able to are to handle extremely hot isotopes. Sevearl isotopes of radium have a nice balance...easily handled by those who do that sort of thing, a half-life of a few days, and radium metal becomes radon and helium gasses. "All you need" is a large enough sample to see it and not mind wasting it. Iodine has an available isotope with a half-life of hours and another with a half-life of days, which both decay into tellurium, so "purple gas or condensed phase becomes metal", but I would assume it would be microscopic particles rather than a noticeable solid aggregate. DMacks (talk) 09:22, 18 August 2024 (UTC)

What happens when kids don't get enough sugar?

What happens when kids don't get enough sugar? 2A00:23C5:1C00:C001:8DB5:7D3A:7C98:F54A (talk) 11:24, 17 August 2024 (UTC)

see Protein poisoning if there is no carbohydrate in the diet. Carbohydrate does not have to be in the form of "sugar" but could be starch. Graeme Bartlett (talk) 11:41, 17 August 2024 (UTC)

The name of the article is misleading. This is not poisoning due to proteins. Also, reading the article, it seems it's more the lacking of fats rather than carbohydrates. 2A0D:6FC0:982:9000:411B:FF20:C5F4:B172 (talk) 18:21, 17 August 2024 (UTC)

The health problems, metabolic disturbances such as hyperammonemia and hyperinsulinemia, are caused by the excess digestion of proteins due to a lack of metabolically less problematic energy sources. So it might be more accurate to speak of poisoning by protein metabolites, but for many poisons it is not the compound as ingested that impacts a patient's health, but metabolites of the ingested substance.  --Lambiam 21:18, 17 August 2024 (UTC)

Basically, nothing. Especially if sugar means simple sugars or disaccharides, where the result would likely be better health and teeth. But unlike the other two macronutrients, proteins and fats, carbohydrates are not essential to human diet, and communities have lived for millennia, and millions of people today with close to zero carbohydrate consumption. This refers to those beyond infancy, Breast milk of course has substantial sugar/carbohydrate. Removing sugars from an infant's diet might lower insulin levels and inhibit growth, just as modern high sugar consumption (together with ample fat and protein) is thought to promote growth, sometimes deleteriously.John Z (talk) 00:44, 19 August 2024 (UTC)

I'm a bit confused about the formal definition of the work done by a constant force: Is that work defind, as a product, or as an integral?

Regarding the work W done by a constant force ${\vec {F}}$ along a displacement ${\vec {s}}$ in a straight line in the direction of the force, our article work (physics) claims, that if the force is constant then the work done by that force is $W={\vec {F}}\cdot {\vec {s}},$ whereas if the force is variable then the work done by that force is $W=\int {\vec {F}}\cdot d{\vec {s}}.$

For simplifying my question, let's assume the force is constant, so that it exerts a constant acceleration $a$ on a given mass $m$ being constant (in classical mechanics).

1. Using the first formula $W={\vec {F}}\cdot {\vec {s}},$ we receive the potential energy only: $W={\vec {F}}\cdot {\vec {s}}=(m{\vec {a}})\cdot {\vec {s}}.$

2. Using the second formula $W=\int {\vec {F}}\cdot d{\vec {s}},$ we receive the change in the kinetic energy only: $W=\int {\vec {F}}\cdot d{\vec {s}}=\int ma\cdot d{\vec {s}}=\int m{\frac {d{\vec {v}}}{dt}}\cdot d{\vec {s}}=\int m{\frac {d{\vec {s}}}{dt}}\cdot d{\vec {v}}=\int m{\vec {v}}\cdot d{\vec {v}}=\Delta ({\frac {1}{2}}mv^{2}).$

3. To sum up: It seems there are apparently two kinds of work (assuming that the force doing the work is constant): The work equivalent to the potential energy should only be defined according to the first formula $W={\vec {F}}\cdot {\vec {s}},$ whereas the work equivalent to the change in the kinetic energy should only be defined according to the second formula $W=\int {\vec {F}}\cdot d{\vec {s}}.$

I wonder where my mistake is. HOTmag (talk) 20:43, 17 August 2024 (UTC)

One issue is that you are using the notation of indefinite integrals, although the integrals here should definitely be definite, giving the change in energy. By the conservation of energy, for an object in free fall, $|\Delta E_{\text{pot}}|=|\Delta E_{\text{kin}}|.$ Apart from the sign, they are the same. If the object is not in free fall, you may not replace $g$ by ${\frac {d{\vec {v}}}{dt}}.$  --Lambiam 21:03, 17 August 2024 (UTC)

As for the notation, you are right, but I followed the notation in our article work (physics), and I thought it was clear what this notation meant.

As for the "change" in the kinetic energy: Thanks to your comment, I've just added the word "change" (See above).

As for your last remark about objects in free fall: Please notice I've only replaced the acceleration $a$ by ${\frac {d{\vec {v}}}{dt}}$ in #2, that discussed the kinetic energy (in free fall), so what was wrong there?

Let me ask you that more explicitly: What is the formal definition of the work $W$ done by a constant force ${\vec {F}}$ along a displacement ${\vec {s}}$ in a straight line in the direction of the force? Is it $W={\vec {F}}\cdot {\vec {s}},$ or $W=\int {\vec {F}}\cdot d{\vec {s}}?$ HOTmag (talk) 21:15, 17 August 2024 (UTC)

If ${\vec {F}}$ is constant, you can take it out of the integral:

$W=\int _{start}^{end}{\vec {F}}\cdot d{\vec {s}}={\vec {F}}\cdot \int _{start}^{end}d{\vec {s}}={\vec {F}}\cdot {\vec {s}}$

so both definitions are equivalent. If ${\vec {F}}$ is not constant, only the version with the integral works. So the version with the integral is the proper way to do it, the version without is a simplification possible with a constant force. PiusImpavidus (talk) 23:13, 17 August 2024 (UTC)

On the other hand, if the force is not constant, then from the proper definition (in that case): $W=\int {\vec {F}}\cdot d{\vec {s}},$ we receive the change in the kinetic energy only: $W=\int {\vec {F}}\cdot d{\vec {s}}=\int ma\cdot d{\vec {s}}=\int m{\frac {d{\vec {v}}}{dt}}\cdot d{\vec {s}}=\int m{\frac {d{\vec {s}}}{dt}}\cdot d{\vec {v}}=\int m{\vec {v}}\cdot d{\vec {v}}=\Delta ({\frac {1}{2}}mv^{2}),$ yet we cannot receive the change in the potential energy - from that definition, unless we assume the conservation of the sum of both energies, am I right? If I am, then the kinetic energy has apparently a "technical advantage" over the potential energy, mathematically speaking, right? HOTmag (talk) 08:04, 18 August 2024 (UTC)

August 18

Colour of francium

Is francium predicted to have a colour other than silverish, just like caesium above? Nucleus hydro elemon (talk) 13:17, 18 August 2024 (UTC)

Rubidium and caesium crystals – francium should have a colour intermediate between the two, I think. Double sharp (talk) 09:21, 19 August 2024 (UTC)

Cs is a bit yellow because its plasma wavelength is in the very near ultraviolet: here it is calculated as 351 nm. In francium we cite Arblaster for the lattice parameter and crystal structure for Fr; he extrapolates them from the trend, and that would suggest a higher wavelength for Fr, and consequently a more yellow colour than Cs.

However, per Covalent radius#Radius for multiple bonds, relativistic effects may make Fr atoms slightly smaller than Cs atoms, not bigger. Since such a relativistic prediction has been experimentally validated for the ionisation energy of Fr, I am inclined to think that this should be correct, and that extrapolation for Fr will give the wrong answer. If that's so, the plasma wavelength of Fr should be between those of Cs and Rb, and as a result, it could at most be very pale yellow – closer to silvery than Cs. At least, Droog Andrey (a computational chemist) also answered this way when I asked him back in 2018.

There's of course an obligatory disclaimer here that any macroscopic sample of Fr would vapourise itself immediately, leaving you with bigger problems than worrying about its colour. But it's a fun question indeed! :D Double sharp (talk) 09:21, 19 August 2024 (UTC)

Thanks, the properties of Po~Ac are really interesting! It's pity to see them die in extreme radioactivity. By the way, both Po²⁺ (pink) and Po⁴⁺ (yellow) are coloured. Ignoring that large amounts of At will immediately boil the solution, will similar things happen in astatine cations At⁺ and AtO⁺, making them coloured? Nucleus hydro elemon (talk) 11:01, 19 August 2024 (UTC)

I think it's not possible to answer this because the structures of the At aqueous ions are not really known (and AFAIK there are not yet predictions). Famously, unhydrolysed [Fe(H₂O)₆]³⁺ has a quite different colour from its hydrolysis products.

I also wish Po~Ac were more stable: they'd be superbly interesting to study. In particular, I'd like to know how the chemistry of Rn compares with that of Xe, and if it might be possible for Fr to break into the 6p shell. But if we go that far, then why stop there? Copernicium would be amazing to study. (Though I guess its compounds would be pretty toxic. It must be a very soft cation indeed.) :) Double sharp (talk) 14:03, 19 August 2024 (UTC)

I just start to think an alternative universe where these elements (possibly Tc and Pm too) are more stable. That universe might found more interesting properties, which we can't due to their extreme radioactivity. :) Nucleus hydro elemon (talk) 15:01, 19 August 2024 (UTC)

Dextrin and wetting

On the Joseph Lister article it now mentions this oiled green silk used as a wound dressing that has been painted one side with dextrin to ensure a thorough wetting with carbolic acid to steralize it. What is actually happening here? Why would it be painted on one side with dextrin. There does seems to a lot of references when you search dextrin and wetting, even in the modern context, but what does it mean exactly? It is an oiled silk dressing from 1865, sold under the brand "Oiled Green Silk Protective". There is several images of it. scope_creep^Talk 20:27, 18 August 2024 (UTC)

Silk is hydrophobic (not rabies) - water on silk beads up rather than soaking in. The powder (dextrin would be used because it is fine and easily available) would absorb the acid and allow it to permeate the silk. Or so my natural dyeing wife tells me. Greglocock (talk) 04:23, 19 August 2024 (UTC)

@Greglocock: Right. That is really cool. That makes sense, finally. Yipeeee. I will make a note to that effect. I now need to find a reference to support it but should be easier now there is an explanation. I didn't think that was going to answered. That is excellent. Give yourself and your wife a gold star. I'm happy. The simple things in life.:) scope_creep^Talk 11:17, 19 August 2024 (UTC)

August 19

How did people find out which mushrooms are safe to eat?

Edible mushroom says: "To ensure safety, wild mushrooms must be correctly identified before their edibility can be assumed. Deadly poisonous mushrooms that are frequently confused with edible mushrooms include several species of the genus Amanita, particularly A. phalloides, the death cap. Some mushrooms that are edible for most people can cause allergic reactions in others; old or improperly stored specimens can go rancid and cause food poisoning. Additionally, mushrooms can absorb chemicals within polluted locations, accumulating pollutants and heavy metals including arsenic and iron—sometimes in lethal concentrations."

So, if some primitive society of humans had no knowledge about whether mushrooms are edible at all, how would they end up obtaining the detailed knowledge about which mushrooms are edible and which not without that leading to many deaths? Because once people start to die then you would think that they would no longer be interested in this and just classify most mushrooms as toxic. Count Iblis (talk) 15:11, 19 August 2024 (UTC)

People would rub the item on their arm, taste and spit it out, eat a small amount and wait for any discomfort, and (probably) try to feed it to dogs or other tamed animals. Abductive (reasoning) 16:54, 19 August 2024 (UTC)

The obvious answer is trial and error. Early hominins were mainly hunter-gatherers it seems, relying heavily on nuts, seeds, fruits and fungi, all of which may be toxic. If the more adventurous members of a group of such people tried out a few new things every generation and passed on the results in their oral history e.g. "this one made me very sick and killed my brother, but this one tastes really good", I think that over longer periods this would build up their "repertoire". Periods where normal food sources became scarce would provide more than enough incentive to give something a try. I've tried to find sources for this, but struggled so far. One thing I did find is the suggestion that the early hominins would check out what other animals were eating. Mikenorton (talk) 21:04, 19 August 2024 (UTC)

Also keep in mind that there are edible fungi that don't resemble poisonous ones; that not everything grows everywhere; and that traditional knowledge includes knowing where to find particular foods, not just how to identify them out of context. --Amble (talk) 21:25, 19 August 2024 (UTC)

That is why, in every language, humans built up little sayings to remember things. We don't need them now because we avoid nature. So, we've mostly forgotten the exact words and come up with silly jibberish like "red on black makes a wasp attack by pee turns yellow is a happy fellow." Therefore, it is difficult to imagine a society where these sayings were well known and taught important lessons. I just wonder about Australia. Wouldn't the sayings simply have been: "If it moves, it will kill you. If it doesn't move, it will likely still kill you." 12.116.29.106 (talk) 12:57, 20 August 2024 (UTC)

We still use those sorts of sayings. Things like "red touches black, ok Jack; red touches yellow, turns your blood to Jello" for colorations of snakes that aren't vs are venomous; "touch the white, you'll be alright; touch the black, you won't be back" for the neutral vs hot conductors in US electrical wiring. DMacks (talk) 14:06, 20 August 2024 (UTC)

There are old mushroom hunters. There are bold mushroom hunters. There are no old, bold mushroom hunters. 41.23.55.195 (talk) 04:47, 21 August 2024 (UTC)

Thanks everyone for your answers! Count Iblis (talk) 18:00, 25 August 2024 (UTC)

Too big

I was watching an episode of Star Trek: Discovery, and the people is having a problem with a black hole that is way too big. As in, five light-years in size. So I checked the article Supermassive black hole, to check how big can they get, and it says "with its mass being on the order of hundreds of thousands, or millions to billions, of times the mass of the Sun". It's hard to keep perspective with numbers, sizes and distances so high, so just to be clear... 5 light-years in size would be obcenely big even for a supermassive black hole, right? Cambalachero (talk) 18:57, 19 August 2024 (UTC)

Sounds like it was this item: Dark Matter Anomaly. The Schwarzschild radius of a black hole is directly proportional to its mass. If you take the Schwarzschild radius to be (5 light-years) / 2 = 2.5 light years and plug it into the formula, you will get a mass of 8 x 10^12 solar masses. That is several times larger than the mass of the Milky Way galaxy, or around 1000 times as large as the mass of the very large supermassive black hole M87*. So yes, that is way too big. —Amble (talk) 20:02, 19 August 2024 (UTC)

August 21

Two types of units

1. Are cup, metric teaspoon and tablespoon used in recipes in most continental European countries? Are non-liquid things ever measures in these units?
2. Why screen sizes for smartphones, tablets, computers and TVs are usually measured in inches, even in mostly metric countries?

--40bus (talk) 18:29, 21 August 2024 (UTC)

Re 1. In the UK, it was common to measure liquids in teaspoons, tablespoons and cups (all standard volumes), and to measure granular solids such as sugar in both 'level' and 'heaped' tea- and tablespoons, and flour (for example) in cups (usually not heaped). Fluid ounces and (where applicable) pints and other fractions of pints (a UK pint being 20 fl. oz.) were also used: it depended on the preferences of the recipe writer. {The poster formerly known as 87.81.230.195} 94.1.209.45 (talk) 00:28, 22 August 2024 (UTC)

Re 2. Marketing of consumer electronics products is influenced by the major market of the USA where inches is a customary unit. Philvoids (talk) 11:09, 23 August 2024 (UTC)

August 23

Planck's law 1901 article and reverse function

At the end of the chapter "...Wien's displacement...", after equation (8), Max Planck gives the formula:
$U=\nu \cdot f_{1}({\frac {\theta }{\nu }})$
Then a new formula:
$\theta =\nu \cdot f_{2}({\frac {U}{\nu }})$
Ok, but the second formula that follows from it is incomprehensible to me:
${\frac {1}{\theta }}={\frac {1}{\nu }}\cdot f_{3}({\frac {U}{\nu }})$
One has:
$f_{3}({\frac {U}{\nu }})={\frac {1}{f_{2}({\frac {U}{\nu }})}}???$
Any idea ?
Malypaet (talk) 12:58, 23 August 2024 (UTC)

So $\theta =\nu \cdot f_{2}({\frac {U}{\nu }})$ and ${\frac {1}{\theta }}={\frac {1}{\nu }}\cdot {\frac {1}{f_{2}({\frac {U}{\nu }})}}$, it's just taking the reciprocal of both sides. But I don't do physics so I'm probably missing the point.  Card Zero  (talk) 13:38, 23 August 2024 (UTC)

My question is:
on what logic can we write:
$f_{3}({\frac {U}{\nu }})={\frac {1}{f_{2}({\frac {U}{\nu }})}}???$
Malypaet (talk) 14:14, 23 August 2024 (UTC)

Still unsure if I'm really helping, but so long as I don't have to know anything about black-body radiation or whatever,

If $a=b\cdot c$ then ${\frac {1}{a}}={\frac {1}{b}}\cdot {\frac {1}{c}}$, so

If $a=b\cdot {f_{2}(x)}$ then ${\frac {1}{a}}={\frac {1}{b}}\cdot {\frac {1}{f_{2}(x)}}$, and

If ${\frac {1}{a}}={\frac {1}{b}}\cdot {f_{3}(x)}$ then

${f_{3}(x)}={\frac {1}{f_{2}(x)}}.$ But I'm just filling space until somebody comes along who knows what you were getting at.  Card Zero  (talk) 15:08, 23 August 2024 (UTC)

Nobody knows what Malypaet is trying to get at... The answer here, I guess, is simply that $f_{3}$ is a new name for $1/f_{2}$, nothing more, nothing less. Planck doesn't know what $f_{2}$ looks like (all he knows is that its argument is $U/\nu$), and he doesn't know what $f_{3}$ looks like (all he knows is that, because $f_{3}=1/f_{2}$, it is also a function of $U/\nu$). --Wrongfilter (talk) 15:46, 23 August 2024 (UTC)

Written like that, we can admit it. In his combinatorial demonstration we find this analogy of functions between logarithms and exponentials. But he does not write it.
Thank you. Malypaet (talk) 18:59, 23 August 2024 (UTC)

August 24

The magic number N = 126

This question is mostly theoretical, because there's no feasible way to create such heavy neutron-rich isotopes at present. But: what predictions are there on the N = 126 shell closure at low proton numbers? In particular, is ¹⁷⁶Sn (Z = 50, N = 126) expected to be doubly magic, or will this shell closure disappear that far from the valley of stability (like N = 20 does)?

(I got some links about this at User talk:ComplexRational#fluorine-30: thanks, Nucleus hydro elemon! But I thought it'd be worth asking for more answers.) Double sharp (talk) 17:41, 24 August 2024 (UTC)

Even if it exists it will be extremally unstable relative beta decay. Ruslik_Zero 19:57, 24 August 2024 (UTC)

Yes of course, since ⁷⁸Ni is also quite unstable to beta decay. What I'm curious about is (1) whether ¹⁷⁶Sn should exist and (2) whether it does close the neutron shell, or if the energy gaps are expected to change in this extremely neutron-rich region. Double sharp (talk) 04:33, 25 August 2024 (UTC)

August 25

Arecibo message distance

The Arecibo message was broadcast towards M13, 25,000 light years away. Is there any chance that the message could be received that far away, or would it be long lost in noise? Bubba73 ^{You talkin' to me?} 04:00, 25 August 2024 (UTC)

Not only would the signal-to-noise ratio be minuscule, but due to its orbit around the galactic center the signal would have to be aimed at where the Messier 13 cluster will be 25k years from now. 136.54.237.174 (talk) 18:05, 25 August 2024 (UTC)

From the data given in the article, I arrive at a minimum diameter of the receiving antenna of 2 kilometers. The diameter of the Arecibo dish is $D=305$ meters, the wavelength of the signal is $\lambda =12.6$ cm. The beam divergence angle is then $\theta =1.22{\frac {\lambda }{D}}=5\times 10^{-4}$ radians. Because $\theta$ is very small, the solid angle is to a good approximation $\Omega =\pi \theta ^{2}$, the exact formula is $\Omega =2\pi \left[1-\cos(\theta )\right]$. The area of the beam after traveling a distance of r is then $\Omega r^{2}$. Then with the power of the beam of 405 kW, at a distance of $22.2\times 10^{3}$ lightyears, the flux of the signal will be $F=1.3\times 10^{-26}$ Watt/m^2 at M13. This signal can then be detected using one or multiple antennas. If the total area of the antennas is A, then the received power is F A. If we assume that the temperature of the antennas and receivers are T = 20 C = 293.15 K, then the noise power will be $P=kT\Delta f$ where $\Delta f$ is the bandwith, that in this case must be 10 Hz or larger, as this is the frequency shift used to modulate the signal. The signal power must be larger than the noise power. If we then equate F A to P and solve for A and then assume a single antenna is used, and put $A=\pi r^{2}$ then the diameter of the receiving dish is 2 r and if I didn't make any mistakes, this yields a minimum diameter of approximately 2 kilometers. Count Iblis (talk) 19:25, 25 August 2024 (UTC)

A 2 km dish is feasible, but will the signal get lost in the noise at that distance? Bubba73 ^{You talkin' to me?} 04:57, 26 August 2024 (UTC)

There's also the matter of integration time. Noise adds incoherently, signal adds (hopefully) coherently, so with a longer integration time, the signal may rise above the noise. In this case, the integration time is limited to no more than 100 ms by the 10 Hz bitrate. The difference between the 0 bit and the 1 bit was only one wave, so a longer integration time doesn't help to decode the signal, but it may still help to detect the carrier wave.

Beam size matters too. The wider the beam, the more noise from other sources like stars; the narrower the beam, the less likely those aliens pointed it well enough at Earth. PiusImpavidus (talk) 10:31, 26 August 2024 (UTC)