Wikipedia:Reference desk/Science

Welcome to the science section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

April 22

Detecting bird "pregnancy"

Recently I was talking to someone who was feeding gulls, and he pointed out a gull which he claimed was a "pregnant" female, i.e. one that would be laying eggs soon, because it was slightly fatter than the other gulls. (I suggested the neologism "eggnant".) Is it actually possible to tell by looking at a female bird if she's "eggnant," as with pregnant mammals? 169.228.151.215 (talk) 03:27, 22 April 2018 (UTC)

Sometimes. --jpgordon^{𝄢𝄆 𝄐𝄇} 04:31, 22 April 2018 (UTC)

Interesting. The gull I mentioned didn't appear to have a bulge on its abdomen, though, it just looked wider across the torso than most. 169.228.151.215 (talk) 04:52, 22 April 2018 (UTC)

Hens become "eggbound" when they've a bun in the oven too long. They don't look so well but don't look fatter. Gulls are probably similar. InedibleHulk (talk) 06:01, April 22, 2018 (UTC)

Just to note, "probably similar" there is a link to egg binding (unpiped to help readers realize WHAAOE). DMacks (talk) 20:34, 22 April 2018 (UTC)

They are. InedibleHulk (talk) 06:07, April 22, 2018 (UTC)

The word you're looking for is "gravid". Abductive (reasoning) 06:14, 22 April 2018 (UTC)

TIL gravid means "carrying eggs" as well as "pregnant". Thanks, Abductive! -Nunh-huh 04:52, 25 April 2018 (UTC)

What ever became of the user named something like Kurt Shaped Box who lived for questions like this?73.211.241.4 (talk) 02:31, 23 April 2018 (UTC)

User:Kurt Shaped Box silently stopped contributing in September 2016. DMacks (talk) 02:54, 23 April 2018 (UTC)

Shark bite

Can a shark bite through a knight's armor? 2601:646:8E01:7E0B:0:0:0:9A39 (talk) 06:23, 22 April 2018 (UTC)

That would depend on the shark, and the armor. According to Shark suit, chainmail seems to be preferred, and that was something that knights wore. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:11, 22 April 2018 (UTC)

Sharks are known for their numerous teeth and their powerful shearing effect. But what's the simple bite force like, compared to a large feline or hyena? - animals which are known particularly for their crushing ability, rather than sharpness. Andy Dingley (talk) 08:41, 22 April 2018 (UTC)

"A 2008 computer model estimated that a 21-foot (6.5-meter) great white shark would produce nearly 4,000 psi (17,790 newtons) of bite force, that figure hasn't been directly measured". [1]

Tests by Robert Hardy using a 710 Newton English longbow found that it could penetrate medieval armour at very close range; see English longbow#Armour penetration.

So the answer seems to be "yes", Alansplodge (talk) 12:12, 22 April 2018 (UTC)

Yes, but this is just a computer model, and this is the peak force at the back of the bite. Add to this that sharks would not apply all their force to all bites.

In reality sharks sharks have saw-like teeth and they shake the prey once bitten to cause a sawing effect. It's about the sharpness more than about the force. --Hofhof (talk) 13:01, 22 April 2018 (UTC)

• Further, I would have to consider the source Alan cites as unreliable, if it can't even get the units right. The PSI is a pressure unit—pounds per square inch, where "pound" means pound force— but the newton is a force unit. 17,790 newtons is 4,000 pounds force, not psi. Or if it meant to talk about pressures, 4,000 psi is 27.6 megapascals, i.e. meganewtons per square meter. --69.159.62.113 (talk) 21:17, 22 April 2018 (UTC)

• The original Journal of Zoology paper seems to be here [2]. It uses Newton except in the abstract. The relevant part seems to be "Assuming isometry, scaling the 35° gape angle data for a 6.4 m, 3324 kg white shark yields maximum anterior and posterior bite forces of 9320 and 18 216 N". While doubts over pop science reports are quite resonable, it's often helpful to at least check out where this claim is likely coming from, particularly if it's peer reviewed. Nil Einne (talk) 07:15, 23 April 2018 (UTC)

• BTW, the PLOS One paper mentioned in the National Geographic source is here [3] and well it's PLOS One so open access. It uses MPa with a psi conversion for the pressures and Newton with a lbs (appears to accurately be pound force) for the forces. The relevant parts seem to be "taxon representative molariform bite forces ranging from 900 to 8,983 N (202 to 2,019 lbs)" and "taxon representative caniniform tooth-pressure values ranged from 195 to 1,344 MPa (28,282 to 194,931 psi)" and "Taxon representative molariform tooth-pressure values ranged from 203 to 1,388 MPa (29,443 to 201,312 psi)". As sort of mentioned in the National Geographic source, the bite forces were measured. (The bite pressures were inferred.) Yes, these are crocodiles, but there's another model mentioned in the Zoology paper "applying a similar FE methodology, computation of the maximal anterior bite force in a 267 kg African lion yields a figure of c. 3300N". I mention these because if Alansplodge suggestion that 710 Newton is enough to penetrate armour, from a pure force standpoint, it doesn't even seem to be even close to being in doubt unless shark (and maybe lion) bite forces are very unusual so the models are way, way, way, way off. I make no comment on other factors. Nil Einne (talk) 07:36, 23 April 2018 (UTC)

• Oh I also came across [4], it seems at least one company does make wetsuits from either stainless steel or titanium mesh [5]. I don't know how these compare to older armour, they're likely designed much more for flexibility and lightness but also likely have a fair amount more tech. According to the first source, the founder doesn't recommend them for great white shark bites although as noted, even without the armour being penetrated there is still risk of significant injury. Another thing mentioned, also in some non RS discussions I came across, is the fact it can doesn't mean it will. Sharks have a tendency to let go if they come across something hard. Nil Einne (talk) 07:55, 23 April 2018 (UTC)

• This video, while not even close to being an RS [6] also mentions the likely ineffectiveness of chainmail wetsuits against a dedicated great white shark. This on cage divers mentions the general reluctance of sharks to bite cages since they don't look like food although I'm not sure how much this applies to a human in armour. (But my understanding is generally humans aren't interesting to sharks, and I wouldn't be surprised if sharks are even less interested in humans in chain mail.) Of course being rare doesn't preclude it happening if you try hard enough, as the few shark attacks and incidents like these [7] [8] show, crap can happen. Nil Einne (talk) 09:09, 23 April 2018 (UTC)

As some have noted, you don't necessarily have to penetrate armor to harm whatever's inside. Humans are squishy bags of mostly water. If you get hit, the energy still has to go somewhere. Armor simply distributes the force throughout the body, which protects against sharp object penetration but doesn't prevent blunt force injury. This is why blunt weapons like maces were useful, to attack armored combatants by causing blunt trauma. --47.146.63.87 (talk) 09:31, 23 April 2018 (UTC)

Tsar Bomba's effect on the Earth's Position in space

I was wondering about the effect that a massive explosion like the Tsar Bomba would have on the earth's position in space. I did some back of the envelope calculations as follows: we know that the total yield of Tsar Bomba was 50MT. That's about 210PJ. Most of that energy would have gone into heating the atmosphere, deforming/destroying all sorts of things, creating its crater, etc. But some of the energy would have been a force against the earth's surface. The bomb produced a shock wave which propagated omnidirectionally. Some of it propagated down to the earth's surface (where it in fact, reflected off it). Some of it also propagated upwards. Upon reaching the top of the atmosphere, it would be expected to blow some of the air molecules off into space. It also radiated a large amount of particles upward into space. This mass being displaced away from the earth should have imparted an acceleration on the earth. Even if the mass push upwards was slower than the escape velocity of earth and was simply forced into orbit, the earth would still experience acceleration, and the mass would likely never return to earth as it would be blown off by solar wind. In addition, the large flux of particle and electromagnetic radiation that went downwards towards the earth should have imparted a radiation pressure upon the earth.

I guessed that the yield energy converted into a net kinetic energy change of the earth would be say 10%. I don't really know if 10% is fair or not. I suspect it's possible that almost all of the energy going into motion of the constituent parts of the earth's structure would have been fully dissipated in the deformation of the same, making this an entirely inelastic process. In any case, the 10% figure gives 21PJ. Given that the earth is 5.972e24kg, we therefore have a change in the earth's velocity of 8.39e-5m/s. That's a tiny change, but since 1782259200 seconds have expired since the detonation, we'd expect by now that the earth is around 150km away from where it would otherwise have been. By the year 3061, we'd expect the earth to be shifted by 3000km, and after 1,000,000 years the shift would be 3,000,000km. But when I looked for confirmation of my reasoning I found these links [9] and [10]. They both seem to think that ejection of mass was zero (which I think is a totally incorrect assumption) and therefore conclude that no acceleration was imparted (also incorrect since it ignores the radiation pressure). 103.228.155.110 (talk) 09:52, 22 April 2018 (UTC)

Your reasoning is incorrect, conservation of momentum needs to be taken into consideration when considering where the energy is directed. The amount of mass ejected into outer space, free from Earth's gravitational pull, is all but negligible, so in fact the vast majority (far greater than 90%) of the blast energy is directed at the air above, rather than the earth. Also, there is a substantial deceleration from the initial detonation velocity by the time any air gets ejected from the atmosphere so the impulse imparted to the earth is basically nothing. Radiation pressure also cannot be thought as acting only on the earth; all forces come in pairs per Newton's third law.--Jasper Deng (talk) 10:59, 22 April 2018 (UTC)

Imaging the all energy of E=2.1×10¹⁷ went into accelerating some mass $m=2E/v_{2}^{2}$ to the escape velocity of $v_{2}=11.2$ km/s. Then the kinetic energy of Earth would have changed by $\Delta E=(mv_{2})^{2}/(2M)=2(E/v_{2})^{2}/M$, where M is the mass of Earth. The result is $E\approx 27$ J. The change of the Earth's velocity would have been $E\approx 3\times 10^{-12}$ m/s – a negligible value. Ruslik_Zero 17:51, 22 April 2018 (UTC)

• It's better to treat the earth as a closed system. The explosion cannot affect the center of mass of the system unless it causes mass to leave the system. Blasting mass into orbit does not do this. The only mass that the explosion pushes out of the system is a mass of any photons that leave the system, and this is inconsequential: it's tiny by comparison to the reflected sunlight. There are several even smaller effects: mass thrown into orbit will increase the system's capture of solar photons and solar wind This is tiny compared to changes caused by the atmosphere expanding due to anthropogenic warming. Yet another tiny effect: if the explosion temporarily affects the earth's albedo, the sun's photon pressure changes as more (or less) photons are reflected back into space. This is is even smaller than the effects on albedo caused by weather, fires, volcanos, agriculture, contrails, etc. -Arch dude (talk) 04:02, 23 April 2018 (UTC)

Note that large releases of energy, like earthquakes, can alter the Earth's rate of rotation, by moving material around and thereby changing its moment of inertia, but yes, this isn't the same thing as its orbital position. Earth is a big old ball of rock and metal. It takes a lot to move it. --47.146.63.87 (talk) 09:18, 23 April 2018 (UTC)

Bass cabinet

How do gallien kreuger get such good bass response from rhier small cabinets?86.8.201.80 (talk) 13:31, 22 April 2018 (UTC)

See Gallien-Krueger for our article. I'm sure that the marketing literature available from their website will address the issue, if not in an entirely unbiased manner. Tevildo (talk) 15:52, 22 April 2018 (UTC)

I have looked at the website and they do not say how they achieve the low frequency response. That's why I'm asking here.86.8.201.80 (talk) 17:00, 22 April 2018 (UTC)

Well, for example, on the website we have "[The enclosures] utilize special bracing to eliminate standing waves while providing rock solid structural support to reduce cabinet resonance ... tuned ports, GK’s innovative Horn Bi-Amp System, passive crossover with attenuator for full-range operation ... [the] enclosures are equipped with proprietary neodymium drivers...". Any more detailed information will be proprietary to GK, and not available publicly. See trade secret. Tevildo (talk)

The Wikipedia articles Loudspeaker enclosure and Guitar speaker cover construction techniques that have not changed over many years. DroneB (talk) 10:34, 23 April 2018 (UTC)

What is the exact reason that woman can't get pregnant while breastfeeding?

What is the exact reason that woman can't get pregnant while breastfeeding? Maybe you can refer me to the relevant article here on Wikipedia and I'll read it.93.126.116.89 (talk) 23:10, 22 April 2018 (UTC)

Breastfeeding#Mother. Ian.thomson (talk) 23:13, 22 April 2018 (UTC)

Lactational amenorrhea. Carbon Caryatid (talk) 23:16, 22 April 2018 (UTC)

The second article (also linked in the first) is an important read. It's definitely not impossible for pregnancy to occur despite breastfeeding. Nil Einne (talk) 09:12, 23 April 2018 (UTC)

The baby wakes up, starts crying, bites her nipple and the woman is definitely no longer in the mood. Besides, the position is going to be pretty awkward, no matter how you do it, especially if there is no baby involved. (Unless of course it is one of the women who is doing the breastfeeding.) Real life is not a weird kinky porn movie. :-)John Z (talk) 00:12, 28 April 2018 (UTC)

Women definitely CAN get pregnant whilst breastfeeding! The article on lactational amenorrhoea referenced above will tell you more.RichYPE (talk) 14:12, 28 April 2018 (UTC)

April 23

How does solotone cause to strong smell through sweat glands?

I've read that "fenugreek contains an aromatic compound called solotone, which is responsible for the sweet-smelling "perfume" your sweat emits.". Now my question is how this compound arrive to the sweat glands in the armpit exactly? Is it by the blood or other way? 93.126.116.89 (talk) 04:14, 23 April 2018 (UTC)

That's sotolon, or sotolone, not solotone, which seems to be a woo-woo multi-vitamin supplement available in Nigeria. Sotolon is a lactone, so it would be absorbed in the small intestine and circulated in the blood. - Nunh-huh 04:49, 23 April 2018 (UTC)

Yes, the blood, the same way pretty much anything moves around your body. Some lipids do initially get dumped into lacteals instead of the blood, but the lymph returns to the blood eventually, so they still wind up there. Things you ingest only don't wind up in your general circulation if a) they aren't absorbed from your digestive tract at all, and therefore stay there and are excreted; b) if the liver (which is the first stop for blood coming from the digestive tract) stores, modifies, or excretes them back out via the enterohepatic circulation. This is a big deal for any medication taken orally, because the liver can modify it so it's no longer effective, in what's called first-pass metabolism. This is one reason why some drugs can't be or aren't ideally given orally—the other is because most large molecules, like peptides, get digested, which is why things like insulin have to be administered parenterally. --47.146.63.87 (talk) 09:12, 23 April 2018 (UTC)

It's an interesting question. Eccrine sweat glands work by a merocrine mechanism, which is to say, the vesicles that become sweat are filled inside cells (as opposed to transcytosis). That would appear to imply (assuming no specific transporters) that the lactone is able to cross phospholipid bilayers unaided, which given its structure seems pretty believable. In which case it might more or less suffuse the entire body, intracellular and extracellular, and all secretions from it. Wnt (talk) 12:27, 24 April 2018 (UTC)

For my grandfather.. planetary motion and ultimate conclusions

My great grandfather (of 91 years and still standing) would like to understand why with all the eccentric orbits of all the moons and planets around us, that we aren't going to be bereft of interplanetary cousins in due course. In other words, with the eccentric nature of everything stellar around us, why aren't we tending to see things move away (or closer) on a cosmic scale? The Rambling Man (talk) 22:10, 23 April 2018 (UTC)

Eccentricity simply means that the shape of the orbit is sort of like an "oval". Those "ovals" are still stable - they are not tending to eject, or decay, or collide.

In physics, we often find stability in complex systems: this branch of mathematics is frequently considered one of the most elegant and complicated branch of mathematics.

Nimur (talk) 23:42, 23 April 2018 (UTC)

The long term stability of the solar system is an active area of research, in large part because it is a very difficult problem to solve. Exact solutions regarding the stability of planetary systems have only been achieved for relatively simple theoretical systems, such as one-star two-planet systems. Most everything else gets simulated due to complexity. These simulations have shown that solar systems of many planets could in principle be stable over billions of years[11], but it's not known for certain this is the case with our own solar system from this point forward. Someguy1221 (talk) 03:00, 24 April 2018 (UTC)

If the Solar System wasn't fairly stable, we probably wouldn't be here to observe it. Moreover, what we see now is the result of over 4 billion years of evolution. The early Solar System was quite chaotic. All the terrestrial planets are probably the results of dozens of planetesimals crashing into each other. It's increasingly thought that the outer planets migrated after their formation, in the process turning the whole Solar System into a shooting gallery. Anything that didn't collide with something else or get swept into a stable orbit was ejected. This may have included an entire extra ice giant (and we may have possibly found it, though this isn't confirmed). Even since then, comets periodically have been perturbed by galactic tides and passing stars, either yanking them away from the Solar System entirely or causing them to fall towards the Sun, which is why they show up here. Also, Triton and Mars's moons are in unstable orbits, and will eventually collide with their primaries, get ripped apart into ring systems, or fly off. It's just that these things take a long time relative to our puny human timescales. --47.146.63.87 (talk) 07:18, 24 April 2018 (UTC)

• The other thing to consider is that the systems are only metastable, meaning that on timescales open for active observation, they aren't varying much. When you expand your perspective from using days or years as your base unit to billions of years as your base unit, the system is highly unstable and stuff has been smashing and careening off in unpredictable directions quite a lot. It's all perspective. --Jayron32 12:24, 24 April 2018 (UTC)

The rule of eccentricity is that what goes up must come down ... caveat being, not all orbits intersect the ground. Throw a baseball in the air and its orbit will be highly eccentric; given the chance it would pass within perhaps a few kilometers of the Earth's core at great speed. For a classic Keplerian orbit to happen all the Earth's mass would have to be at its center, and we'd neglect any relativistic frame dragging etc., in which case the ball would return in due course to the pitcher's hand. But, the pesky ground is in the way. Throw the same baseball from a very high tower (that extends far above the atmosphere), with a very good pitcher (like kilometers per second good), and it might pass so far from the Earth's center that it never hits the ground at all. Then it simply goes down, and up again. At the highest point it is moving too slow to avoid falling, and at the low point it is moving too fast to stay in a stable loop around the planet. Wnt (talk) 12:36, 24 April 2018 (UTC)

Food poisoning

In the most recent news, romaine lettuce is tainted by Escherichia coli in one farm in which the romaine is grown. Somehow, the illnesses are tracked/logged throughout the entire country, and everyone is staying away from romaine lettuce. Some people seem to have a full recovery after a week. Others take several weeks of hospitalization because of kidney failure and diagnosis of HUS. No one has died, though. It just has made the infected people miserable. First of all, I'm curious how the plant is handled at home. I think romaine lettuce is usually eaten raw. Running tap water to wash the heads may not be enough, or maybe the tap water is contaminated. Second of all, what food-borne pathogens are resistant to heat, and how much heat is needed to theoretically kill the pathogen? Third, and this question may be a little off-topic, is eating unwashed fresh produce or raw meats a good way for a person who wants to commit suicide? Or will the pathogens just make the person miserable or make the person lose so much water that he/she dies of dehydration instead? SSS (talk) 23:30, 23 April 2018 (UTC)

Also interesting to know is whether the use of PPIs plays a role here. Count Iblis (talk) 01:59, 24 April 2018 (UTC)

The FDA and USDA make manufacturers and growers put lot numbers on their products so that they can be traced. The system is weak for produce, as evidenced by the fact that the source romaine lettuce for this latest outbreak has not been pinpointed. The typical raw produce outbreak starts with either improperly composted manure, or overworked farm workers forced to defecate in the fields by the landowners. No food-borne pathogen is resistant to heat, but some are better able to survive a given temperature. Bacillus cereus spores can survive boiling for as long as it takes to cook rice, if you are looking for a dangerous example. Abductive (reasoning) 05:12, 24 April 2018 (UTC)

I am reminded of this video. As a cultural note, this behavior of not drinking tap water is not only shared by Chinese people in China, but also overseas Chinese people who recently settled abroad. Foreigners from wealthy countries with safe water pipes and well-treated water come to China to find that they can't drink the tap water. Also, it is not really a cultural practice to eat salads; even the word salad in Chinese is a loanword. Cooking the water and the vegetables may kill the pathogens. SSS (talk) 10:54, 24 April 2018 (UTC)

Today, eating in China is a health hazard. Count Iblis (talk) 04:33, 25 April 2018 (UTC)

In any case, the lack of salad consumption doesn't mean raw vegetables aren't consumed. For example, while not something common in China, Popiah is ethnic Chinese cuisine and AFAIK raw lettuce (not normally romaine) can be a part. (Although the mung beans are I think normally blanched.) Likewise AFAIK raw lettuce is a not uncommon ingredient in fresh Vietnamese spring rolls (often called summer rolls.) And there are so many varieties of spring rolls in China including plenty of fresh ones I would be surprised if there aren't some where raw lettuce is common ingredient. Lettuce can also be used for presentation especially for professional prepared meals (e.g. our article includes one from HK) and these may not be cooked in any way. The lettuce itself may or may not be eaten but in any case, there's no guarantee the food itself will be hot enough to prevent the pickup of contaminants. To put it a different way China produces about half of the worlds lettuce per Lettuce production in China and lettuce, and most of this is for domestic consumption. Our article is unclear whether this figure is including stem lettuce, but it's still a lot. While a fair amount of this may be cooked, it's unlikely all of it is, well even if we put aside modern usages and only consider traditional ones (and a lot of traditions are more recent than people tend to appreciate). Or to put it a different way, China is such a big and diverse place it's a little silly to talk in absolutes. In other words, raw lettuce in China may be a lot less common than in some other places, but it doesn't mean it's unheard of. In any case, whatever the traditions, things are changing [12]. Nil Einne (talk) 10:45, 27 April 2018 (UTC)

• Regarding Romaine lettuce, its primary (probably only significant) use is in the U.S. in salads or as a garnish in sandwiches, it is (WP:OBPERSONAL warning) probably the second most common salad green after iceberg lettuce; it's most famous use is in Caesar salad, but it's a common enough green for use in many salads. --Jayron32 12:17, 24 April 2018 (UTC)

How the plant is handedld at home: First the housewife normaly removes the tougher outer leaves to expose the more tender, sweeter head. This faily good at removing most of the Escherichia coli that is ubiquduse and existes everywere in dilute quantise too small to cause infection. See: Infectious dose. Problem arises usually with prepared salads. The whole head, Escherichia coli and all go into the washing tanks and the lot gets contaminated, then copped up and put into little convent-to -use-bags of salad – were it continues to multiply. Note: that the authorities are tying to track down the source but not the cause. The cause has been understood for many decades. It is from the use of sewage sludge as a fertilizer which by its nature has very high concentrations of Escherichia coli. When it rains heavily, the drops cause heavily contaminated soil particles to splash up onto the outer leaves. So it is important to wash them... in clean fresh water and not recycled water as used in food processing factories. One may be able to get away with rewashing prepared salad in vinegar/table salt/sugar solution [13] but the extra time and effort involved make traditional home preparation easier, quicker and cheaper than prepared salad. Also, prepared salad bags are fill with a gas so the the cut (wounded) vegetables don't visibly degrade whilst on the supermarket shelf. Once opened the gas leave and by the next day and the rest of the unused bag has to be thrown away as it is going rotten. Were as fresh heads last longer. So, prepared is false economy as well as a potential danger to health. Think about the statics, over ones life time most people get food poisoning at least once so why invite it in? Still, to end on a positive note. One can only die from food poisoning once. Aspro (talk) 15:14, 24 April 2018 (UTC)

Food poisoning in produce is complicated because of how quickly things like lettuce take up water. Let's say you have a head of lettuce that's perfectly fine. It's picked and dirty water is splashed on it inadvertently. Washing would likely remove enough of that bacterial load to make it safe to eat raw afterwards, though there would always be the risk that enough germs "held on" to make the eater sick. However, if the lettuce is still in the ground and dirty water is splashed on it - say, from the pickers relieving themselves in the fields - the bacteria would get into the soil and thence into the very fibre of the plant. No amount of washing will have an impact there because the germs are inside the cells of the lettuce. In that sense, it's similar to infected chicken - rinsing the chicken would not make it safe to eat raw. Lettuce is particularly susceptible to these kinds of contamination because it's essentially only ever eaten raw; if it was contaminated when you brought it home, there's nothing you can do to render it safer to eat. Regarding safe temperatures to kill germs, see Danger Zone. Matt Deres (talk) 17:03, 24 April 2018 (UTC)

Matt Deres, if bacteria can get inside the cells of lettuce and other food eaten raw, then shouldn't we boil our lettuce and fresh fruit before eating it? Dbfirs 17:12, 24 April 2018 (UTC)

Ditto. Don't think your right about Escherichia coli existing inside a plant cell. Wrong environment.Aspro (talk) 17:30, 24 April 2018 (UTC)

According to this, they cannot. Per my source below, and per that source, they can be taken up into the roots of plants, though this seems to be as part of the vessels carrying the water, and I see no evidence that the bacteria can cross the cell membrane and be taken up inside of cells. It'd be like fitting a volkswagen through a peephole. It may be smaller than the inside of the cell, but the I don't know how such a thing would cross a membrane. What I think is unimportant, however, since we have two sources (that one and below) which clearly state "However, after looking at whether any of the E. coli strains traveled past the roots and up into the plant’s interior structures, the researchers concluded that that sort of internalization appeared to be unlikely. “We wanted to investigate this, because it was one of the questions out there,” USDA microbiologist Manan Sharma told Food Safety News in an earlier interview. “We’ve taken something that has been of concern for eight or nine years and put it to rest.” (quote from source at start of this comment). --Jayron32 17:48, 24 April 2018 (UTC)

And just to quantify my flippant "fitting a volkswagen through a peephole" comment, e. coli are about 1000 times the diameter of the largest pores in a cell membrane. See Escherichia coli which notes that the size is on the order of 1-2 micrometers (10^-6 meters) while pores in cell membranes seem to have a maximum diameter of about 1-2 nanometers (10^-9 meters). See here. A volkswagen is 2-3 meters wide, and a peephole is 2-3 millimeters wide. --Jayron32 17:56, 24 April 2018 (UTC)

I stand corrected. Thank you! I picked that up somewhere that seemed reliable and have been repeating it for years. Matt Deres (talk) 01:50, 25 April 2018 (UTC)

As an interesting aside, I came across [14] which is an experiment using plant xylem from coniferous trees for water filtration for human consumption. However the source mentions xylem from gymnosperms are much more suited from this than angiosperms although the only real reason relevant to this question is the longer vessel length and my vague memory suggests to me a single vessel from the roots to the leaf stalks is unlikely. Nil Einne (talk) 07:50, 25 April 2018 (UTC)

• This article is an interesting read, and directly relevent to the subject. --Jayron32 17:17, 24 April 2018 (UTC)

Thank you for that link which reassures me that I can continue to eat lettuce and fruit uncooked if it is well washed. The article states that the bacteria exist in the rhizosphere, that is, in the "region of the soil in the vicinity of plant roots, considered as a microenvironment in which the chemistry and microbiology is influenced by root growth, respiration, and nutrient exchange" (OED). Dbfirs 17:49, 24 April 2018 (UTC)

If human or animal poop containing a particularly bad strain of E coli was on the hand of one worker who picked one day a limited number of heads of Romaine would be contaminated and a few families would be affected. It seems more likely that the processing involves washing the product with water which becomes contaminated or that a farm has an irrigation process which contaminates their lettuce. Is a sensor commercially available, such as an analysis chip which could monitor in real time the water used to wash a farm product such as the Romaine, sounding a warning to stop the production line when significant contamination is detected? Then perhaps the lots of produce from various suppliers could be tested to find the bad guy. As is, the detection method is waiting for consumers to be very ill, then trying to analyse what they ate, just like they did about the year 1900. Is real time detection of deadly pathogens in the foodstream possible today? Edison (talk) 04:53, 25 April 2018 (UTC)

April 24

Galaxy Rotation Curves

Rotation curve of a typical spiral galaxy: predicted (A) and observed (B). Dark matter can explain the 'flat' appearance of the velocity curve out to a large radius.

Models of rotating disc galaxies in the present day (left) and ten billion years ago (right). In the present-day galaxy, dark matter - shown in red - is more concentrated near the center and it rotates more rapidly (effect exaggerated).

There was a thread here in the Science Desk recently about the dark matter. It spurred my interest and I've come upon this video. It depicts two galaxies: the one on the left is a modern galaxy and the one on the right is from a distant universe, perhaps 10 billion years ago. I really have a bunch of questions about the phenomenon. Click on the triangle and the galaxies start rotating. One can see that the contemporary galaxy rotates much faster than the distant one. Where did it get this angular momentum? I understand it is not necessarily the same galaxy, but hopefully a typical galaxy as it is.

Look at the left galaxy carefully. There is something strange about it. The spiral arms start at the same angular position, in case of the "right" arm at 12 o'clock position as we see it. It should rotate, and it does but there is a complicated movements of stars inside the galaxy ecliptic. When this "right" arm moves slightly, new stars form the missing portion and after a short time the external arm still starts at 12 o'clock. The same is true for the other arm.

Where do those stars come from? It seems they are thrown by a centrifugal force from the outskirts of the galactic bulge, then they drift to the periphery to be used for reconstitution of both arms! How come other stars in the bulge do not follow them?

The stars that travel, they are eventually brought back to the galactic bulge, so they end up again closer the the galaxy center. What is the force that moves them there? Can you imagine our Sun ending up near the Sagittarius A? According to the video, the galactic arms work as two brooms sweeping the stars from distant regions of the galaxy and bringing them to the bulge.

I wonder how realistic this video is? The curve on the top is the plot of rotation speed of the stars measured experimentally in a typical galaxy which is supposed to explain the videos, but does it?

Also if there is so much dark matter right beyond the circumference of the galaxy, the stars should gravitate to it. Instead they are repulsed, if you look carefully at the stars movement in the video.

Thanks AboutFace 22 (talk) 21:27, 24 April 2018 (UTC)

I can't specifically answer your questions on the simulations, but would point out that galactic arms aren't objects, they're density waves, and will not behave as a consistent body of the same stars. "centrifugal force" would not apply in the sense you're describing to a body of individual stars orbiting around a center of mass. Acroterion (talk) 00:20, 25 April 2018 (UTC)

Yes, see Crash Course Astronomy. --47.146.63.87 (talk) 07:47, 25 April 2018 (UTC)

The picture is misleading in that it implies the galaxy on the right would become like the one one the left when the dark matter clumped more round it. In fact what would happen is that it would become considerably smaller. There is no addition of angular momentum - the stars would circle closer to the center of the galaxy. Dmcq (talk) 10:31, 25 April 2018 (UTC)

@Acroterion, You are in error. Any object moving in a circle or over an arc of any curvature will experience a centrifugal force. AboutFace 22 (talk) 19:28, 25 April 2018 (UTC)

April 25

Matching a light source to a solar panel

I'm hoping someone can help me. I'm trying to make up a simple set-up to recharge a lot of solar powered decorative strings of lights. They each have a solar panel about 60 mm by 60 mm charging two AAA batteries. I'd thought that just putting them all under a light bulb would work, but now I realize that I might have to use a special kind of bulb that puts out the same light spectrum that the solar panels use. I can't find out much about what type of light these panels work best in. I'd appreciate any links or information anyone can give me. Thanks in advance 49.197.192.93 (talk) 03:26, 25 April 2018 (UTC)

From a little searching, it looks like anything that puts out visible light should work well enough. Sure, you might be able to increase efficiency a bit with very finely tuned light sources, but it sounds like that wouldn't be worth the time or money given what you're doing. Sure it's not better to just remove the batteries and charge them with a charger? You can find battery chargers that take lots of batteries at once. Or are the batteries not removable? Another suggestion is that, if this is something you do often, it might be worth it to wire charging plugs into the strings. Since they're battery-powered, this is low-voltage wiring: just add in DC plugs and get a suitable power supply to plug them into. --47.146.63.87 (talk) 04:24, 25 April 2018 (UTC)

(OP) Thanks for those thoughts. Taking the batteries out involves undoing 4 tiny screws, and we usually have 9 different strings on display in a market stall then need them recharged for next day. I hadn't thought of putting a charging plug into each one, that might well be the answer. I'll wire an ammeter into one of them and do some tests under an ordinary light bulb first.49.197.115.7 (talk) 06:44, 25 April 2018 (UTC)

An alternative would be to forget the batteries and solar chargers, and just wire the lights to a low voltage DC supply (probably 3v if the units use two 1.5v batteries). I've done this successfully with some of my solar lights. One good DC supply should run them all because the current will be small. It doesn't have to be smoothed, so a cheap mains transformer with a single diode on the output might suffice, though full-wave rectification might be better. Dbfirs 07:07, 25 April 2018 (UTC)

(OP) I've just hooked an ammeter into the battery circuit, and the lights use about 30mA when running. Late afternoon sun square on to the panel puts in 35mA, and putting the panel up very close to a selection of light bulbs gives 10mA at most. (So close that the solar panel gets quite hot very quickly). I need to display them pretty well exactly as a customer might buy them, so the charging socket idea seems the best one yet. 49.197.119.149 (talk) 06:45, 26 April 2018 (UTC)

What animals eat feces as a normal behavior?

I've read the following paragraph on educalingo site: "Coprophagia /kɒp.rə.ˈfeɪ.dʒi.ə/ or coprophagy is the consumption of feces. The word is derived from the Greek κόπρος copros, "feces" and φαγεῖν phagein, "to eat". Many animal species eat feces as a normal behavior; other species may not normally consume feces but do so under unusual conditions. Coprophagy refers to many kinds of feces eating including eating feces of other species, of other individuals, or its own, those once deposited or taken directly from the anus.". I'd like to know what animals most familiar with this phenomenon. I've heard about pigs something like that but I'm not sure. 93.126.116.89 (talk) 04:07, 25 April 2018 (UTC)

Have you read our Coprophagia article? Note that you're posting on Wikipedia, not Educalingo. That article gives many examples of animals that do so. What do you feel is not adequately discussed in that article? --47.146.63.87 (talk) 04:12, 25 April 2018 (UTC)

What is a "det. officer?"

(Moved to the Humanities Desk as per the original poster's intent. It's been answered, but it might as well be archived in the right place. --69.159.62.113 (talk) 00:31, 26 April 2018 (UTC))

Does Beano raise the effective glycemic index of beans?

I assume most people know that Beano is a product intended to allow you to eat beans (and some other foods) without experiencing certain inconveniences, or at least reducing them. According to our article, it works by breaking down oligosaccharides into simple sugars.

So does this raise the effective glycemic index of those foods, and if so how much? Is it a situation of "you thought you were being virtuous by eating beans, but to your pancreas, they look like ice cream"? --Trovatore (talk) 20:51, 25 April 2018 (UTC)

As you no doubt suspect, it will increase the digestibility of certain foods. This will possibly raise the glycemic index by losing fewer carbohydrate molecules to farts. Abductive (reasoning) 06:19, 26 April 2018 (UTC)

Thanks for responding, but I was hoping for something more substantive than "possibly", which I could figure out for myself. Any actual data or studies? --Trovatore (talk) 07:14, 26 April 2018 (UTC)

I once read a study of the calories lost to flatus (generally), but have been unable to find it again. Google is probably accidentally censoring it. Sorry. Abductive (reasoning) 23:19, 26 April 2018 (UTC)

I doubt "calories lost" is the key issue here. The point is that the oligosaccharides would be taken up more slowly than the simple sugars the enzyme breaks them down into. So the prompt response in blood sugar (and insulin) might be greater if you have the Beano, even if total calories eventually absorbed are about the same. That's the effect I was trying to find out about. --Trovatore (talk) 23:46, 26 April 2018 (UTC)

You are one of the top ten experts in this field, if one measures by characters typed. Abductive (reasoning) 03:44, 27 April 2018 (UTC)

????? I see plenty of discussions/speculation on this and related issue elsewhere [15] [16] [17] [18] [19]. Beano themselves while not really dealing with the glycemic index issue mention "As a result, it has been estimated that the use of beano® will produce an additional two to six grams of carbohydrates for every 100 grams of food treated by beano®." There is also the paper I highlighted below. I see zero reason to think the minor comments here somehow make them "one of the top ten experts in this field, if one measures by characters typed." Nil Einne (talk) 10:04, 27 April 2018 (UTC)

One assumes that people have already searched for sources before posting their question. And reading those sources only confirms my suspicion that this discussion is as advanced as any held anywhere in the world, ever. Abductive (reasoning) 21:01, 28 April 2018 (UTC)

Your reply makes absolutely zero sense. Of course a good OP would have searched before asking. They also failed to find something that's why they're asking. And your comment makes zero sense. I did not claim these are even close to the best discussion held. There is zero reason to think they are. For starters, they completely missed the first RS I found below. Second, discussions held in forums are rarely great sources. When searching for answers, I frequently comes across crappy discussion threads. Sometime they are similar to those here. Rarely they are better than those here. How I often also find decent RS which have been completely missed by many of the discussions. The presence of a bunch of not very good discussions is no reason to think that they are the best. (And IIRC some of those sources considered stuff not yet mentioned in this thread anyway.) I guess you're new to the internet, but that's simply not how the internet works. A lot of the internet is full of terrible quality discussions. Sorting the wheat from the chaff is one of the key reasons why the RD exists. Another reason is to provide quality sources, often reliable sources which internet discussions aren't. Reliable sources were specifically requested in this thread, so I did not spend a great time trying to find quality discussions other than to look for RS. I didn't spend a great time looking for RS either since I'm only mildly interested in this. This is clearly something that people have considered before, enough so that someone actually did a placebo controlled examination of the effect of beano on postprandial glucose concentration. There is no reason to think anyone in this thread, is anything close to an expert or has considered this at anything close to the level considered before. I mean I'm guessing no one in this thread knew of the source below until I found it either and while it doesn't directly deal with glycemic index, it's fairly related. Nil Einne (talk) 22:36, 28 April 2018 (UTC)

I remember now that the beano FAQ response is one thing that was mentioned in one of the discussions. While true it's only of limited relevance to the question, I wouldn't say it's completely irrelevant and I also wouldn't be surprised if no one here knew of if until I brought it up. Many of the discussions I linked to are both highly flawed (I think one of them possibly even conflated alpha-galactosidase that beano provides and alpha-glucosidase that acarbose inhibits), and yet contain limited new info, as is common with such things including in cases where we can be sure there have almost definitely been informed discussions somewhere. To be clear, I'm not trying to blow my trumpet here. My sole point is that with my tiny about of research, I've uncovered stuff that probably wasn't know to anyone in particular the OP. It also fairly likely that someone who is an actual expert in related areas is likely to be able to come up with a far more informed discussion. (With no disrespect intended to the OP, their background makes me thing they are quite far from an expert in areas for which experts obviously do exist and who would likely to be able to understand this issue far better without having specifically studied it.) Especially if multiple experts in different areas collaborate. Whether this has happened I have no idea, but I also know it can't be ruled it. (By collaborate here I'm mostly not thinking of some sort of scientific collaboration. It could be some people worked together to write answer a question in some newspaper column. It could be it came up over dinner for some reason and they talked about it.) Ultimate point being the idea the OP was an expert in this particular issue, or this discussion was somehow highly informed on the issue compared to discussions that may have taken place all over the world (some of which could be documented no where) is unsupported, and very unlikely Nil Einne (talk) 23:20, 28 April 2018 (UTC)

But wouldn't someone who has so much gas complaints that he/she wants to use this have slower absorption of these sugars to begin with compared to normal people? Count Iblis (talk) 02:14, 27 April 2018 (UTC)

I came across this small placebo controlled study [20]. It didn't deal with the glycemic index but did look at the effect on postprandial glucose concentration (as well as appetite). The interaction between beano and acarbose may also be of minor interest [21]. Nil Einne (talk) 10:04, 27 April 2018 (UTC)

My indenting was totally fucked up by another editor here [22]. I've fixed it. Hopefully no one was confused in the mean time. Nil Einne (talk) 22:36, 28 April 2018 (UTC)

BTW, although you can get this even from the abstract or heck even the title ('Not Their Glycemic Response'), for the benefit of the RD I might mention the study found beano (called α-Galactosidase but the details mention they used some form of beano) had no statistically significant effect on postprandial glucose concentration for any of the meals tested, suggesting it possibly does not effect glycemic index. Of course far from proving it, especially given the tiny sample size of 12 people. (Admittedly this isn't the kind of thing you're likely to get the kind of funding to do a decent sample size test.) And this also contradicts the findings of the acarbose study. However we don't know the reason for that result. E.g. was it some sort of interaction between the activities of the two? Could it be it's something that only occurs in diabetics? Nil Einne (talk) 23:20, 28 April 2018 (UTC)

Statics: Which zero-force member can be removed without reducing structural strength?

I find it difficult to figure out this kind of problems. Can you name some simple rules? 161.185.160.21 (talk) 21:20, 25 April 2018 (UTC)

Our article is at Zero force member. Perhaps all of them can be removed. However if the loading is changed they would not be a zero force member any more. Graeme Bartlett (talk) 21:58, 25 April 2018 (UTC)

April 26

Mass of entropy, and so on

Entropy is a genuine source of energy. For example, the proton gradient in the mitochondria really does power the cell and keep us alive. But at a simple level, it seems very mysterious...

• Suppose I have two tiny evacuated chambers, linked by a pore, that contain a total of three helium atoms. There is only one way for all three to be on the left, but there are three ways for one to be on the right and two on the left. So there is an entropy of k_B T ln 1 in the first case = 0 and an entropy of k_B T ln 3 in the second case = (0.0000862 eV/K)(300K)(1.10) = 0.0284 eV at 300 K. By analogy with a mitochondrion, this gradient could be tapped at a gate in the pore to transfer that energy to some chemical reaction or mechanical process.

• I think the entropy there relates to Landauer's principle in that one bit of information has a negative energy value of k_B T ln 2 = 0.0179 eV. Putting one of the three atoms on the far side can be done three ways, generating, it would seem, (k_B T ln 3)/(k_B T ln 2) bits of information. I'll admit, I would have thought the choice of 1-in-3 would be 1.5 bits (1 bit if you choose A, 2 bits if you choose B-or-C, then B or C) but this works out to 1.59 bits, not sure why. It would appear that generating information creates energy out of thin air, though I would assume the total mass of the entire set of evacuated chambers and contents never varies.

• A basic Carnot heat engine would appear to work by equilibrating the lone helium atom at T_H, which then crosses the pore to generate usable energy of k_B T_H ln 3; it then is cooled to T_L before it crosses back to consume usable energy of k_B T_L ln 3. Heating and cooling the actual atom will transfer 3/2 k_B (T_H - T_L) of heat energy. The efficiency, therefore, is ... erm, well, I'm getting (2 ln 3)/3, which surely isn't right. That calculation seems maddeningly close to making sense.

• But if we have no gate in the pore, where does the energy go? The atoms will be on the same side 25% of the time, and opposite sides 75% of the time. Being at 300 K, each atom on average has 1/2 kT = 0.0129 eV of energy per degree of freedom, so I take it they have 0.0388 eV of total kinetic energy on average. If the entropy of being on opposite sides adds to the kinetic energy of the atoms, that would imply that they are moving much faster whenever they are apart. Does that make sense? Is there some force that can be described as?

• But how small does the pore have to be? Could the three atoms be in a simple spherical chamber where you might simply measure which side each is closest to? Do they still have higher energy when they happen not to be on the same side? Can you do that dividing top-vs-bottom, left-vs-right, front-vs-rear??? Since the energy depends on information I suppose it depends on which way(s) you are actually measuring the atoms, and this necessarily perturbs them?

• Now the amount of entropy from a single bit of information at 300 K is a whopping 0.0179 eV, which may seem small, but is huge compared to the possible masses of neutrinos at 10-4 eV or less. Simply knowing a neutrino exists, therefore, should create a bit of information. Is that factored into the neutrino's mass already? Does learning of the existence of a neutrino add to its momentum like those helium atoms above, causing it to massively accelerate? Can you do this with, say, an entangled neutrino by learning about the other particle created by a reaction? Is the "temperature" for this entropy determination going to be determined solely by the neutrino's KE, or something else, i.e. is there a way to erase information about a stationary neutrino and leave it with a negative overall KE?

Any low-level example of this sort you might care to raise should be useful for understanding... Wnt (talk) 07:04, 26 April 2018 (UTC)

I think the important point to make is that information is not a substance, any more than caloric is. Reducing the disorder of a system can be described as "cooling" or "adding information", but in neither case is anything physically added or removed from the system - only the energy distribution of the system's components is changed. See Gibbs free energy as a starting point for the actual calculations required in these cases. Tevildo (talk) 07:50, 26 April 2018 (UTC)

That's what I was doing above - taking T delta S to figure the delta H. Caveat being that at the individual particle level "T" is the energy in each degree of freedom of one particle, and "H" is energy that could be extracted at a gate, if one existed. (True, I was simply dumping that work back into the particle as heat in the 4th point above) Wnt (talk) 11:44, 26 April 2018 (UTC)

• Sorry, but all that is not even wrong. A careful reading of our article entropy might or might not clear what confusions you may have (that is one of our best articles in the topic, but it is still quite a hard read).

Pay attention in particular to the fact that temperature and entropy are statistical concepts: they apply only to systems with a large number of atoms. You can compute an "entropy by atom" or similar by dividing the system entropy by the number of atoms but it does not mean that that entropy is held by a single atom in any meaningful sense. Tigraan^{Click here to contact me} 12:18, 26 April 2018 (UTC)

Per Tigraan, entropy is an emergent property of a system; the entropy of a system of atoms divided by the number of atoms does give you a number, but that value is different than the entropy of a single atom in isolation. The number of distinct states of a system depends on the relationships of particles in that system. Similarly, the temperature of 1 atom is a meaningless concept. Temperature is a mass property of a system of a large number of particles. A single particle has no meaningful temperature. Also, the OP is confusing thermodymic entropy with information theory entropy. Entropy in thermodynamics and information theory discusses the difference, at a very abstract level, the limit of informational entropy is thermodynamic entropy, but to say that the information about a particle adds mass to it is beyond silly. It is broadly true that information-as-negative-entropy has some thermodynamic meaning in that somewhere, there must be some system that you remove entropy from to store information, with the caveat that information is not stored in the same system as the system it is describing (that is, information about a neutrino is not stored in a neutrino!) Also, entropy mass, as such, is just the mass of heat energy; that is the energy a system has from its random thermal motions, if you cool a system down it loses mass; that lost mass is the "mass of entropy". --Jayron32 12:42, 26 April 2018 (UTC)

@Jayron32: I'm sorry, but what I take from Entropy in thermodynamics and information theory seems almost directly opposite to what you say. There is an ln 2 conversion factor between the Shannon entropy h and the physical entropy S, but that's in the formulae I use above. They give an example there of a single particle in a box with partition being used to provide work to two pistons, which therefore must acquire relativistic mass.

I will admit that there is something to this "emergent property" bit that I don't necessarily understand, which is the meaning of the temperature of a system. If I have three particles at a given temperature, does that mean that the average energy per degree of freedom works out to yield that temperature by the Boltzmann constant, or does it mean that they are in equilibrium with that distribution of energies so that their summed KE's are necessarily not conserved? I'm not sure I can have information about an isolated system, because if information is energy, my knowledge of it means the system is not isolated???? Wnt (talk) 14:24, 26 April 2018 (UTC)

I found a paper on a single-particle Carnot cycle at [23]. But it is ... formidable. I am not sure how much of the complexity is truly essential to understanding, as I certainly do not at this point. I note very superficially that it has a graph of temperature for anywhere from 0 to 1000 particles in a simulated system. Wnt (talk) 14:40, 26 April 2018 (UTC)

There are two very different types of questions that need to be answered here, and I feel we're at cross purposes. The two questions are 1) Can you calculate X and 2) Does X have any useful meaning? Clearly, you already know the answer to 1) You've made calculations and gotten numbers. What Tigraan and I are trying to tell you is that 2) is more important here. Any of these values can trivially be reduced to 1 particle systems. But what does a temperature of a 1 particle system tell you about that system? What real, physical, actual thing is such a system supposed to model, or what application do I have for it? That's what Tigraan meant above when he noted "You can compute an "entropy by atom" or similar by dividing the system entropy by the number of atoms but it does not mean that that entropy is held by a single atom in any meaningful sense." The entire system of thermodynamics breaks down when you try to consider 1-particle systems. A 1-particle system isn't a thermodynamics problem, it's a dynamics, problem. That is, we've reduced our system to where we can analyze it's behavior as we can with any other single object (or small number of objects) which are colliding and interacting. The whole point of thermodynamics is that large systems of interacting objects produce behaviors which are predictable and modelable without reference to the individual motions of the particles, and that one can understand how the system works without ever needing to understand how any one particle works. Once you've gotten down to looking at the one-particle situation, sure, you can trivially do thermodynamic calculations on it, and its a fun academic exercise, but so what. --Jayron32 14:58, 26 April 2018 (UTC)

This is a learning exercise. I don't understand entropy well enough, so I was thinking if I can work out a direct correspondence, I'll get a better knowledge of what it means. I still don't understand where the energy would come from if splitting up the three particles on one side of a chamber actually does release it due to entropy. I still don't understand whether creating a lot of new information can create mass-energy from nothing like some new variant of steady state theory, or whether it is conceivable to move information from some high-temperature source into a very light structure that would have negative mass and so on. So if this is a "trivial" exercise, please, by all means lead on! Wnt (talk) 23:50, 26 April 2018 (UTC)

If you're trying to understand entropy in a general sense, take the numbers out for a minute. Just look at it qualitatively: consider that you have 2 particles and are trying to use the 2 particles to move a 3rd particle to the right. Moving an object in physics is what is called work (physics). Now, imagine the 2 particles are moving at the third from the same side, and both hit it to the right. Now, that third particle will start moving in that direction. Your two particles have done work on the third. HOWEVER, in order to do that, you need to assure that the striking particles are always hitting the target from the same side. If the two particles are moving about randomly, in all sorts of directions, with no overall motion in any one direction, the sum total of all of their hits will result in no net motion of the target particle. The total energy of the system (that is the energy of all of the particles) is the same in situation 1 (concerted motion produces work) as in situatiomn 2 (random motion does not produce work). The difference between 1 and 2 is called "entropy". All entropy is is the difference between the total energy of a system and the actual energy used to do work. Of course, being energy that energy has an associated mass, but entropy is inherent to the system and because of that, you can't take the entropy out to find its mass effect on the system; you can calculate that number, but it has no physical meaning. The mass of the total energy in situation 1 is still the same as in situation 2. Entropy is not contained in any of the particles, and cannot be parted out physically, it is just a way to represent the notion that there can be energy in a system which is unavailable to do work. --Jayron32 14:24, 27 April 2018 (UTC)

@Wnt: Be sure to read Jayron32's post above, as that is a concise yet almost accurate of what entropy represents. I have a small nitpick about it (hence the "almost"), but be sure to understand the post before reading the nitpick.

Ok, here we go. The problem is that the sum total of all of their hits will result in no net motion of the target particle is only true in an average sense. Statistical physics only apply to large number of particles; in a three-particle system, entropy (as defined by information theory formulas) can increase by random deviations and you can extract work from a single-temperature source cycle (the average over many cycles will have nonpositive work, but you will see single-cycles where work is positive); however, the probability of such random deviations decreases exponentially with the number of particles involved. The second principle is basically the assumption that we never see improbable deviations in many-particle systems (under certain tedious hypotheses that are not worth mentioning here). Tigraan^{Click here to contact me} 15:11, 27 April 2018 (UTC)

Yes, I knew that inaccuracy, but was trying to contruct a simple, visual model to display entropy. Thanks for expanding on it with the corrections! --Jayron32 16:36, 27 April 2018 (UTC)

You should take serious what Jayron and Tigraan are saying here about how thermodynamics arises. If you are very careful about deriving the thermodynamic formulas of entropy of simple systems, see e.g. the book by F. Reif, you'll see that it depends on the level of coarse graining that you need to introduce. For an isolated system we introduce a function $\Omega (E)$ which counts the number of energy eigenstates between energies $E$ and $E+\Delta E$. Then $\Omega (E)$ will be proportional to the parameter $\Delta E$, but entropy needs to be defined as being proportional to $\log \left(\Omega (E)\right)$. In the thermodynamic limit where we consider the limit of infinitely large systems and consider the specific entropy, the choice of $\Delta E$ doesn't affect the result.

But for finite systems, the entropy will get a contribution proportional to $\log \left(\Delta E\right)$, which is still utterly negligible in practice, of course. This term that arises as a result of having to choose a coarse graining scale to define the entropy will make increasing negative contributions the more fine grained view of the system you take. In the limit where you can resolve the exact physical state, this term will exactly cancel out the ordinary entropy of the system. When you know the exact physical state, $\log \left(\Omega (E)\right)=0$, and therefore the entropy is zero. If you work at this fine grained level where the entropy is zero, then the laws of physics will ensure that it stays zero as they forbid erasure of information. If you work at a coarse grained level, then systems will tend to lose information that's visible at the coarse grained level to a more fine grained level (due to statistics). This is how the second law that says that entropy can only increase, arises. Count Iblis (talk) 02:43, 27 April 2018 (UTC)

Silkie chickens

Why do Silkie chickens have 5 toes on each foot instead of the usual 4?? Georgia guy (talk) 15:18, 26 April 2018 (UTC)

As described in our article on polydactyly § other animals, this is a genetic mutation that is now standardized and encouraged by the humans who breed this type of chicken.

I searched the online database of the Poultry Science library at North Carolina State University, and found this paper: Genomic Regions Associated with Dermal Hyperpigmentation, Polydactyly and Other Morphological Traits in the Silkie Chicken (2010). If you're interested in more technical reasoning for the multiple toes: "A single SNP in a highly conserved cis-regulatory region of Sonic Hedgehog was significantly associated with polydactyly..." which is biology-ese that means the multiple toes are caused by a single mutation, with lots of technicalities.

Nimur (talk) 15:41, 26 April 2018 (UTC)

April 27

Why aren't 3/8 inch and 1/4 inch common calibers?

But .38, .357 and c. .22 are. Decimal inches are less common than fractions of inches in most non-gun fields so why not? Sagittarian Milky Way (talk) 01:46, 27 April 2018 (UTC)

Although the article Caliber and especially the article .357 Magnum don't quite come out and say it explicitly, it seems like specific calibers were derived experimentally, for specific usages. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:36, 27 April 2018 (UTC)

.25 ACP 251/1000" was introduced in 1906 (patented 1905). .25 NAA 251/1000" was introduced in 2002. .380 Revolver Short and .380 Revolver Long 3/8" were introduced in 1860s. .375 Winchester 3/8" was introduced in 1978.
Sleigh (talk) 12:26, 27 April 2018 (UTC)

It is instructive to think about how these items are manufactured - and how they were historically manufactured. Firearms, and their ammunition, are one of the oldest artifacts that were built using machine tools and their manufacture was intricately linked to other developments of the industrial revolution. Modern commercial ecosystems carry a lot of the historical "inertia" - so even though today we may use CNC mills and robotic metal machines, we have a lot of backward compatibility to consider. (Vis: 2017 Annual Statistical Update from the ATF).

If we're building barrels using a CNC router - or if we're using computer simulations to study ballistic physics - then typing in a simple fraction or a simple decimal makes that part of the work easier.

But if it's 1855 and we're using a metal gauge to control the mill, it makes no difference whatsoever if the gauge is a nice rational number, or some part-of-an-inch or part-of-a-furlong-whose-value-cannot-be-expressible-using-conventional-algebras. Your manufacturer would just slide the metalworking tool down to the gauge and mill that much. The result is a perfect, reproducible, accurate, standardized metal piece, whose precise dimensions expressed in inches are irrelevant for many purposes.

If you spend two hundred years building armament in that fashion, then by the time you invent the precision CNC mill, you have to type in whatever inconvenient number is compatible with the billions and billions of bullets that are already on the marketplace, and your product has to be compatible with the stuff built by hundreds of thousands of other people who are building ammunition. Millions of people aren't going to change their way of life just because it'd be great if everyone wised up about simple facts!

Many times before, I have cited James Burke's excellent and very old history documentary series, Connections (1978). In several episodes - particularly, Episode 5 (40 minutes in), there is a great analysis of the interplay between proto-industrial manufacturing and tooling, and the embodiment of those historical technological conventions into the shape of modern artifacts. Half of that episode is about the significant transition from musket to rifle, and why this minor detail of technology is so significant to the culture and history of the United States. (To wit: the Rifleman's Creed is not a general confession of adoration for firearms at large, but of rifles in specific; the first shots of the war that founded our nation are conventionally attributed to precision home-made rifles, rather than muskets - though historians naturally quibble about the important details; and of course, the Americans were the first to prove that rifles could kill other Americans really effectively). In fact, historians call this transition of technologies, which irreversibly guided the course of the Industrial Revolution, the "American system of manufacturing" - in true form, largely guided by American innovation and our desire to build lots and lots and lots of guns.

The point - if there is any - is that precision in manufacturing does not specifically require a convenient numerical representation; and, if history gives us any wisdom, we can predict that people are going to keep doing the same old thing, for many more years.

Nimur (talk) 17:12, 27 April 2018 (UTC)

Agree with User:Nimur, but the British Brown Bess musket had a internal barrel diameter of 0.75 (¾) inch, which required a ball diameter of 0.69 inch to stop it getting stuck inside. From there, we went to the 0.45 inch Martini–Henry rifle. So it seems that there has been a move away from nice round figures, perhaps because of the increasing complexity of the variables. Alansplodge (talk) 11:27, 28 April 2018 (UTC)

Has the 75% loss of flying insect mass been observed outside of Germany?

Ref.: [24] 174.16.98.178 (talk) 20:00, 27 April 2018 (UTC)

It was observed in England. Ruslik_Zero 08:02, 28 April 2018 (UTC)

Citation needed User:Ruslik0? The 75% figure was from a study of nature reserves in Germany. A study by Rothamsted Research comparing insect traps in England between 1970 and 2000 found no decrease in weight of insects in some areas, but a 60% reduction in Herefordshire. [25] Alansplodge (talk) 16:23, 28 April 2018 (UTC)

Insects are gone, the birds that eat them are gone also, fruit plants are not pollinated. A wonderful consequence of progress and development. AboutFace 22 (talk) 21:07, 28 April 2018 (UTC)

April 28

Deaths from opioid withdrawal

How often do people die due to complications of opioid WITHDRAWAL?

I'm hoping for a good reference that says it's rare, as a coworker has claimed that it is a common occurrence in the United States today. Thank you any help you can provide, and have a nice day. :) — Preceding unsigned comment added by 2600:100E:B010:D644:B13E:8816:6408:E067 (talk) 14:50, 28 April 2018 (UTC)

Yes, people can die from opiate withdrawal: "Death is an uncommon, but catastrophic, outcome of opioid withdrawal. The complications of the clinical management of withdrawal are often underestimated..."

Mortality risk during and after opioid substitution treatment: systematic review and meta-analysis of cohort studies.

Alansplodge (talk) 16:33, 28 April 2018 (UTC)

Thank you!

From the first link: "How could someone die during opiate withdrawal? The answer lies in the final two clinical signs presented above, vomiting and diarrhoea. Persistent vomiting and diarrhoea may result, if untreated, in dehydration, hypernatraemia (elevated blood sodium level) and resultant heart failure."

That strongly supports my belief that it is almost impossible to be picked up by an ambulance and die on the way to the hospital from opioid WITHDRAWAL. Which is what my coworker claims happens frequently in the US.

The latter link is a study of the mortality of patients on opioid replacement therapy, so there does not seem to be much mention of death from withdrawal, since anyone in the study experiencing withdrawal would presumably be having those symptoms under medical supervision where, again, they are extremely unlikely to die suddenly of dehydration.

2600:100E:B010:D644:1415:E13:2FA1:9E59 (talk) 19:07, 28 April 2018 (UTC)

That scenario is certainly not impossible. Rehydrating someone who is nearly dead of dehydration is not as simple as giving him a glass of water, and people attending might be afraid to call the ambulance until it's too late to save the person. Death by dehydration is also not a very painful way to go nor does it generally have quick onset and sudden symptoms like heroin overdose, so it might appear a less acute problem to untrained people. 78.0.233.56 (talk) 22:31, 28 April 2018 (UTC)

In adults, opiate withdrawal is uncomfortable but not inherently dangerous, which is why there are several trials that have used naltrexone to induce sudden long-lasting withdrawal in people addicted to heroin or methadone. On the other hand, neonatal withdrawal has a high mortality rate. Klbrain (talk) 23:39, 28 April 2018 (UTC)

References

April 29

Feynman Lectures. Exercises. Exercise 21-8 PNG

...

21-8. The gravitational force felt by a particle embedded in a solid uniform sphere, due to the mass of the sphere only, is directly proportional to the distance of the particle from the center of the sphere. If the earth were such a sphere, with a narrow hole drilled through it along a polar diameter, how long would it take a body dropped in the hole to reach the surface at the opposite side of the earth?

—  R. B. Leighton , Feynman Lectures on Physics. Exercises

I want to prove the statement that the force is proportional to radius inside the uniform sphere.
Consider a spherical shell 1-2 PNG.
The potential in point 1 is:
$\Psi _{1}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}+\Psi _{\text{1-2,at 1}}$
$\textstyle \Psi _{1}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}+\int _{R_{1}}^{R_{2}}-{\tfrac {Gdm}{x}}$
$\textstyle \Psi _{1}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}+\int _{R_{1}}^{R_{2}}-{\tfrac {G4\pi x^{2}dx\rho }{x}}$
$\Psi _{1}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}-G2\pi \rho (R_{2}^{2}-R_{1}^{2})$
The potential in point 2 is:
$\Psi _{2}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}+\Psi _{\text{1-2,at 2}}$
$\Psi _{2}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}-{\tfrac {G(M_{2}-M_{1})}{R_{2}}}$
$\Psi _{2}=\Psi _{\text{0-1}}+\Psi _{\text{2-3}}-G{\tfrac {4}{3}}\pi \rho (R_{2}^{2}-{\tfrac {R_{1}^{3}}{R_{2}}})$
Difference in potential:
$\Delta \Psi =\Psi _{2}-\Psi _{1}=-G{\tfrac {4}{3}}\pi \rho (R_{2}^{2}-{\tfrac {R_{1}^{3}}{R_{2}}})+G2\pi \rho (R_{2}^{2}-R_{1}^{2})=G\pi \rho [{\tfrac {2}{3}}{R_{2}}^{2}-{R_{1}}^{2}(2-{\tfrac {4R_{1}}{3R_{2}}})]$
If $R_{1}=R$,
$R_{2}=R+dR$
then:
$d\Psi =G\pi \rho [{\tfrac {2}{3}}{R_{2}}^{2}-{R_{1}}^{2}(2-{\tfrac {4}{3}})]=G\pi \rho {\tfrac {2}{3}}[{R_{2}}^{2}-{R_{1}}^{2}]=G\pi \rho {\tfrac {2}{3}}(R_{2}-R_{1})(R_{2}+R_{1})=G\pi \rho {\tfrac {2}{3}}dR(2R)=G\pi \rho {\tfrac {4}{3}}RdR$

$\textstyle ({\textbf {field}})=-{\tfrac {d\Psi }{dR}}=-G\pi \rho {\tfrac {4}{3}}R$

Is it correct? Username160611000000 (talk) 06:36, 29 April 2018 (UTC)