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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Outdated species names of bees in Nomia and Nomadopsis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a list of bees that the bee fly Heterostylum robustum parasitizes in a book written in 1973, but a lot of them are outdated.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 ([1]), 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.^[2])  --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)

I don't know about most continental European countries, but a quick browse to the recipe of the day on chefkoch.de shows some ingredients, including chopped parsley, measured in spoons (EL (Esslöffel) and TL (Teelöffel) (sizes discussed in German Wikipedia here)).

Likewise, a tabbouleh recipe on French site marmiton.org has parsley measured by the soupspoon. AlmostReadytoFly (talk) 10:56, 27 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)

The heaviest isotope of tin known is ¹⁴⁰
₅₀Sn
, which lives less 50 ms and already drips neutrons. The existence of an isotope as heavy as ¹⁷⁶
₅₀Sn
seems unlikely. Magic number itself does not mean that the nucleus exists in any meaningful way. You can look at ¹⁰
₂He
. Ruslik_Zero 20:48, 26 August 2024 (UTC)

That's beta-delayed neutron emission, so the drip line hasn't been reached yet, as expected.

It seems then that the best we could find at the moment are the papers Nucleus hydro elemon found at first, which suggest that ¹⁷⁶Sn should be more or less on the border between being bound and being unbound. Those two papers suggest N = 126 is still magic (because the two-neutron separation energy has a big jump going from ¹⁷⁶Sn to ¹⁷⁸Sn), but this one makes it less clear. Since this is so far from what's currently known, it's probably not possible to do better at the moment. I'd guess, therefore, that the best possible answer to my question at the moment would have to be "no one really knows; could be either way". Double sharp (talk) 06:27, 27 August 2024 (UTC)

I found a reference by Fang et al. about beta decay of ¹⁷⁶Sn. Its β⁻ decay energy is around 22 MeV (comparable with ²⁹F 21.7 MeV) and has a half-life of <1 ms. ¹⁷⁶Sn should undergo β⁻n instead of only β⁻.

The calculated mass excess of ¹⁷⁶Sn is 217.59 MeV, as predicted by KTUY. Mass excess of ¹⁷⁶Sb is predicted to be 195.49 MeV, thus the β⁻ decay energy will be 22.10 MeV, not far away from Fang et al. Somehow the decay process β⁻,23n to ¹⁵³Sb+23n is actually possible with decay energy 2.47 MeV.

KTUY predicts S_2n of ^174,176,178Sn are −0.42,−0.52,−2.69 MeV, implies all of them can possibly undergo 2n emission. The big jump from ¹⁷⁶Sn to ¹⁷⁸Sn suggest N = 126 is still magic.

I think ¹⁷⁶Sn wouldn't get seriously affected by 2n emission, due to some trends related to atomic number. There is no heavy 2n emitters (the heaviest is ^26,28O with Z = 8), so I just show it with 2p emitters. Despite ¹²O (S_2p = −1.737 MeV, Z = 8) has a lower decay energy than ⁶⁷Kr (S_2p = −2.89 MeV, Z = 36), it decays much faster (8.9×10⁻²¹ s vs 7.4 ms). If the trend follows, then 2n emission of ¹⁷⁶Sn is just not important compared to beta decay.

So, I think there is nothing that forbids the existence of ¹⁷⁶Sn. Nucleus hydro elemon (talk) 14:50, 27 August 2024 (UTC)

@Nucleus hydro elemon: Thanks, very cool!

I think I'll upgrade my personal hunch to bet on ¹⁷⁶Sn being doubly magic, but I'll be interested as new predictions come. :) Double sharp (talk) 04:47, 28 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)

Now I'm curious: Do we know, then, what stars it's going to actually pass close by? -- Avocado (talk) 21:15, 26 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)

[3] says that the gain of Arecibo antenna at 2.38 GHz was 77 dBi, only 600 mdB short of Count Iblis's estimate from the physical diameter (an aperture efficiency × antenna efficiency of 87% if true). A receiver temperature of 20°C is a little pessimistic; usually the receiver would be cooled (it is not necessary to cool the antenna, assuming that it is low-loss). catslash (talk) 00:09, 28 August 2024 (UTC)

Thanks for the informative replies. Bubba73 ^{You talkin' to me?} 04:47, 28 August 2024 (UTC)

Regarding the integration time: when there are two possible symbols (0 and 1) represented by two orthogonal signals of equal energy $E$ ( = received power × time) then the bit error rate is something like

$P_{\mathrm {err} }={\frac {1}{2}}\mathrm {erfc} \left({\sqrt {\frac {E}{2kT}}}\right)$

where $\mathrm {erfc} (x)$ is the complementary error function. This assumes (1) that it is known exactly what the two signals are - there is no random change in the phase between symbols, and (2) that the prior probability of each symbol is equal.

Without error-correcting codes it is impossible to reduce the error rate to zero, so it is necessary to decide what rate is acceptable before building the receiving antenna. catslash (talk)