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December 11

What's the product of a collision between a neutrino and an anti-neutrino, when their combined energy is not sufficient for creating the Z boson's rest mass?

Please notice the product of that collision can't be any massless particle, i.e. neither photons nor gluons, because neutrinos only react to the weak force and to the gravitational force, whereas gravitons has never been detected so their very existence is only theoretical for the time being.

Really, I'm not sure, but I do suspect, this kind of collision between matter and antimatter will have no effect: It will be like every "regular" elastic collision between two ping pong balls. Am I right? If I am, then this "regular" collision will be an extraordinary collision between matter and antimatter, right? HOTmag (talk) 01:19, 11 December 2023 (UTC)[reply]

Does it have to be a Z boson? Can't the energy be carried off by photons, possibly with an intermediary step of the creation of a virtual electron–positron pair?  --Lambiam 10:21, 11 December 2023 (UTC)[reply]
What about a collision between a neutrino and an anti-neutrino whose combined energy is not suffcient for creating the new pair you suggest? HOTmag (talk) 10:38, 11 December 2023 (UTC)[reply]
The annihilation cross section for the process suggested by Lambiam will be very small, but there is no lower energy threshold when virtual particles are involved. --Wrongfilter (talk) 10:56, 11 December 2023 (UTC)[reply]
Thank you for this clarification. By mistake, I overlooked the word "virtual" in Lambiam's response. Anyway, I still wonder if the first step after any collision between matter and antimatter may be particles that are not gauge bosons. HOTmag (talk) 11:13, 11 December 2023 (UTC)[reply]
The "first step" can instead be virtual Higgs bosons. There could also be neutrino-neutrino scattering by coupling to hypothetical particles outside the standard model, like in [1]. --Amble (talk) 18:29, 11 December 2023 (UTC)[reply]
But the Higgs particle is not a gauge boson. I still wonder if the first step after any collision between matter and antimatter may be particles that are not gauge bosons in the standard model. If the answer is negative, i.e. the first step of any collision between matter and antimatter must be gauge bosons in the standard model, then the question in the title still remains. HOTmag (talk) 08:02, 12 December 2023 (UTC)[reply]
"But the Higgs particle is not a gauge boson." Because you asked for an example that's not a gauge boson. --Amble (talk) 14:23, 13 December 2023 (UTC)[reply]
Oh, I didn't [ask for hypothetical collisions in which the first step is not a gauge boson]. I remind that you first responded to a response of mine in which I wondered whether "the first step after any collision between matter and antimatter may be particles that are not gauge bosons", i.e. whether there may be any collision between matter and antimatter [e.g. between a proton and an anti-proton or between an electron and a positron or between a neutrino and an anti-neutrino and so forth], for which [it has empirically been proved that] the first step of that collision was not gauge bosons. So you responded: "The 'first step' can instead be virtual Higgs bosons". That's why I answered you that [indeed] the Higgs particle was not a gauge boson [but it had never been empirically detected as the first step of any collision between matter and antimatter], i.e. it was not the particle I'd asked for in my response you had responded to (and I fully apologize from the bottom of my heart if it was not clear from the very beginning what I actually meant when I first wondered whether there might be a "first step" of this kind in any collision between matter and antimatter). Anyway, thank you for suggesting the Higgs boson, as a hypothetical answer [yet not an empirical one, though]. HOTmag (talk) 15:33, 13 December 2023 (UTC)[reply]
"Anyway, I still wonder if the first step after any collision between matter and antimatter may be virtual particles that are not gauge bosons." Yes, you did ask about an example that's not a gauge boson, and the answer was provided. Honestly, these threads of yours are incredibly tedious. You generally refuse to accept any answers you are given, refuse any corrections to incorrect (or at least partially incorrect) preconceptions or conditions in your stated questions, and also argue about whether answers given to you are even what you are asking about. Please, try to remember that we are all volunteers here, we aren't paid, everyone interacting with you is giving up their time to do so. With all of these various threads of yours, what is your overall intention? What are you trying to get at with these high level quantum mechanics and/or relativity questions? Maybe our knowing that would be a better place to start. --OuroborosCobra (talk) 16:01, 13 December 2023 (UTC)[reply]
"Yes, you did ask about an example that's not a gauge boson, and the answer was provided".
Only misinterpretation, probably because of the word "may" in my response to Amble (and to a previous user), which you interpreted as "may theoretically", but actually I've intended to ask Amble (and a previous user) if there may be any kind of collision between matter and antimatter (e.g. between a proton and an anti-proton or between an electron and a positron or between a neutrino and an anti-neutrino and so forth), for which it has empiraclly been proven that the first step of that collision is not a gauge boson. Actually I used the word "may", just to indicate I wondered whether any particle, which was not a gauge boson, was empirically detected as a first step in any collision between matter and antimatter: Since I referred to any collision, rather than to every collision, I found the word "may" appropriate for that purpose. Sorry for not making it clear when I used the word "may".
"Honestly, these threads of yours are incredibly tedious".
You are exempt from responding to threads you find tedious.
"You generally refuse to accept any answers you are given".
I did accept all answers that did answer my questions. This also Includes this thread.
"You...refuse any corrections to incorrect (or at least partially incorrect) preconceptions or conditions in your stated questions".
I'm pretty sure you can't quote any "incorrect (or at least partially incorrect) preconceptions or conditions" in my stated questions.
"You also argue about whether answers given to you are even what you are asking about".
I only do that when I'm sure answers given to me are not about what I intend to ask about, e.g. when those who answer me misinterpret my question (e.g when they misinterpret the word "may" I've used, and please notice I have already apologized for not clarifying - from the very beginning - what I'd really meant by that word).
"Please, try to remember that we are all volunteers here, we aren't paid, everyone interacting with you is giving up their time to do so".
I know that, and that's why I thank whoever gives up their time to interact me, including you.
"With all of these various threads of yours, what is your overall intention? What are you trying to get at with these high level quantum mechanics and/or relativity questions? Maybe our knowing that would be a better place to start".
Let's focus on the current thread (because it has nothing to do with other threads). So, I actually need the answer to the question I explictly presented in my first response to Amble (rather than my second response to them), i.e.: [Empirically speaking], must the first step of any collision between matter and anti-matter be gauge bosons? If it must, i.e. if the first step [in all experiments] of collisions between matter and antimatter was always gauge bosons, then (as indicated in the title of this thread) what's the first step [empirically received] in a collision between a neutrino and an anti-neutrino, when their combined energy is not sufficient for creating the Z boson's rest mass?
HOTmag (talk) 17:03, 13 December 2023 (UTC)[reply]

IQ Bell Curve

It has been argued that IQ follows a Gaussian bell curve, due to the central limit theorem.
However, this requires that the individual test batteries are independent of each other.
Isn't that a problem with the theory of g? 2A02:8071:60A0:92E0:FDE1:CCA5:77F9:C756 (talk) 17:19, 11 December 2023 (UTC)[reply]

I don't see the problem with g. You just need a reasonable amount of independence and a lot of them. Anyway what has happened is that the early tests poroduced a curve that matched a normal distribution pretty well - but modern tests are set up so they actually conform to a normal distribution closely and have a mean of 100 and a standard deviation of 15. They follow that curve by design. NadVolum (talk) 17:56, 11 December 2023 (UTC)[reply]
Normalized IQ distribution with mean 100 and standard deviation 15

IQ scoring is intended to yield a dimensionless positive number based exactly on the Gaussian normal distribution because:

  1. the mathematical model makes no discrmination about causes or correlations among groups of the population, avoiding controversial influences that are discussed here
  2. by theory, if in fact individual performances on the pre-chosen tests were truly random then average IQ would be perfectly predictable for large populations
  3. the degree to which accumulating IQ scores match the theoretical Gaussian distribution supplies a rating of the method's consistency. In practice, scores do fit the normal bell curve to a useful degree which encourages continued use of the model. However the model needs to be calibrated from field data with a standard deviation, and occasionally be completely recast to negate the observed Flynn effects of slow changes in average IQ recorded for some populations. Philvoids (talk) 02:57, 12 December 2023 (UTC)[reply]
    @1 doesn't the model is designt in a way that males and females have the same average?
    @2 Isn't it the case? 2A02:8071:60A0:92E0:4D2C:9077:8CBB:9452 (talk) 11:58, 12 December 2023 (UTC)[reply]
1. No because applying any gender weighting, besides being politically incorrect, would ruin IQ testing as a tool to investigate gender performance difference. Care is taken to avoid gender bias in the question design. See in the article cited for more about gender considerations.
2. No, the distribution shape of real scores cannot match exactly the model's smooth curve that has infinitely extending upper and lower tails. Philvoids (talk) 14:39, 12 December 2023 (UTC)[reply]
Modern tests are slightly tweaked in their question selection to give an equal average IQ of 100 for men and women. A different selection would favor one or the other. The standard deviation of the scores for men however is a little larger than that for women - they have more at the ends of the range. NadVolum (talk) 23:03, 12 December 2023 (UTC)[reply]

December 12

How old is a molecule of water

If you have some water, how old are the molecules, typically? Bubba73 You talkin' to me? 04:30, 12 December 2023 (UTC)[reply]

Same age as the earth, one would think? 67.169.17.27 (talk) 04:31, 12 December 2023 (UTC)[reply]
Not necessarily. According to "Origin of water on Earth" it may have been formed on Earth after the Earth's formation, or it may have come from somewhere other than Earth. Mitch Ames (talk) 04:53, 12 December 2023 (UTC)[reply]
Ceccarelli, Cecillia; Du, Fujun (June 2022). "We Drink Good 4.5-Billion-Year-Old Water" (PDF). Elements. 18 (3). doi:10.2138/gselements.18.3.155.
Given water is produced by many chemical reactions and destroyed by many chemical reactions, that IP's answer is obviously very wrong. And even if you have a bucket of water that is not being produced or destroyed by other reactions, water is a chemically dynamic environment, not static independent molecules that happen to be near each other. The specific H2O molecules themselves have a finite lifetime, due to the self-ionization of water and general proton-transfer in the hydrogen bonded network. The former article notes "Random fluctuations in molecular motions occasionally (about once every 10 hours per water molecule) produce an electric field strong enough to break an oxygen–hydrogen bond, resulting in a hydroxide (OH−) and hydronium ion (H3O+); the hydrogen nucleus of the hydronium ion travels along water molecules by the Grotthuss mechanism and a change in the hydrogen bond network in the solvent isolates the two ions, which are stabilized by solvation. Within 1 picosecond, however, a second reorganization of the hydrogen bond network allows rapid proton transfer down the electric potential difference and subsequent recombination of the ions..." (refs and links omitted). So dissociation/association is fast, and happens often. I don't have the actual reaction-rate data at hand, but if one mixes 2H2O with regular water (mostly 1H2O), one quickly gets 2H1HO, demonstrating that any given water molecule does not retain its specifc two hydrogen atoms and one oxygen atom composition for very long. DMacks (talk) 05:02, 12 December 2023 (UTC)[reply]

Railroad question

How fast could an EMD E-unit go on one engine (assuming straight, level track, no speed restrictions and a train of typical size and weight for post-Great Depression, pre-World War 2 express trains)? 67.169.17.27 (talk) 04:37, 12 December 2023 (UTC)[reply]

It was normal to use multiple locomotives to haul a single express train, but let's assume this train is short enough that a single locomotive can bring it to its normal top speed. Each locomotive had two identical diesel engines. Running on one engine, power is halved. On level track, the engine works mostly against air drag, and power increases with the cube of speed. Half power means that speed is cut by 21%. With diesel-electric transmission, there shouldn't be a significant effect on efficiency.
Half the power also means half the climbing speed and half the acceleration, so the cut in average speed is worse than the cut in top speed. PiusImpavidus (talk) 14:07, 12 December 2023 (UTC)[reply]
In 1937 when EMD E-units were entering service they could be compared with the German DRG Class SVT 137 streamlined diesel that had a regular maximum speed of 160 km/h (99 mph) and had reached a world speed record of 205 km/h (127 mph). Philvoids (talk) 14:10, 12 December 2023 (UTC)[reply]
That must be the record for regular production trains. The absolute record was already faster and held by an experimental German railcar. PiusImpavidus (talk) 17:09, 13 December 2023 (UTC)[reply]
Yes, but there's a qualitative difference between a locomotive designed and used to haul trains, and a self-contained railcar that was not intended to haul anything, and because of its design could not effectively do so even in principle. {The poster formerly known as 87.81.23-0.195} 90.199.215.44 (talk) 01:36, 14 December 2023 (UTC)[reply]
You're quite right that the Schienenzeppelin wasn't scalable and therefore a technological dead end. At best, it could be used for research or VIP runs. But you don't need a locomotive to run useful fast trains. The DRG Class SVT 137 was no locomotive, but, like most fast trains, both now and back in the 1930s, a multiple-unit trainset. The experimental three-phase railcar of 1903 (210 km/h (130 mph), also German) was scalable. Its current collection method just wasn't very practical. PiusImpavidus (talk) 14:38, 14 December 2023 (UTC)[reply]
From the EA and EB models built 1937–8 to the E9s built 1954–64, the E-units had 5 successively more powerful models of Prime movers (engines), whose maximum power outputs (in pairs) rose from 1,800 to 2,400 hp. The answer will in part depend on which engines are in question, though there are likely other factors, such as the Traction motors used on given models (about which the article is largely silent, but which may not have been altered from model to model) and other equipment variations such as gear ratios, which this site discusses. {The poster formerly known as 87.81.230.195} 90.199.215.44 (talk) 14:21, 12 December 2023 (UTC)[reply]
Combining this all gives an estimate for the maximum speed of 312 × (2400/1800) × 160 km/h140 km/h, or, using the earlier speed record of 205 km/h, a top speed of about 180 km/h.  --Lambiam 19:08, 12 December 2023 (UTC)[reply]
A pre-WW2 express train (as specified in the question) would have had one of the earlier models of the E-unit, but yes, a top speed of about 140 km/h (87 mph) sounds about right. Considering that the Pioneer Zephyr (of the same period) on its famous run in 1934 reached a top speed of 112.5 mph (181.1 km/h), although it was mostly noted for its high average speed (1,015.4 miles (1,634.1 km) from Denver to Chicago in 13:04:58 hours, giving an average of 77.61 mph (124.90 km/h)). PiusImpavidus (talk) 17:09, 13 December 2023 (UTC)[reply]
Thanks! I was guessing about 80-90, so that's pretty much what I figured! 2601:646:8080:FC40:3503:6143:E40E:14BD (talk) 04:05, 16 December 2023 (UTC)[reply]

UK metric system

Why the metric system did not spread to the United Kingdom as quickly as the other France's neighboring countries? So, why UK does not have long history of metrication? --40bus (talk) 20:19, 12 December 2023 (UTC)[reply]

Please, see Metrication in the United Kingdom. Ruslik_Zero 20:37, 12 December 2023 (UTC)[reply]
Very simply and briefly: Napoleon Bonaparte. A large part of western Europe came under the control (directly or indirectly) of France during Napoleon's reign and accepted what they were told. The British were at war with France during this period and were unlikely to adopt anything which had a whiff of Napoleon or France about it. Indeed I can recall the phrase "Napoleonic or Christian units" being used humorously in model engineering circles (c. 1990) to refer to metric or imperial. Martin of Sheffield (talk) 22:48, 12 December 2023 (UTC)[reply]
"[They] accepted what they were told" sounds as if they had a choice.  --Lambiam 07:58, 13 December 2023 (UTC)[reply]
Similar to comedian Larry Miller, talking about a pub crawl: "We decided to leave, right after we were thrown out." ←Baseball Bugs What's up, Doc? carrots01:29, 16 December 2023 (UTC)[reply]
There was a choice. When Napoleon had left, the Dutch chose to revert to their old units, but very soon thereafter returned to metric, because it was simply better.
Thing is, the UK had been one country for a while, under one king, and had been able to standardise its units all over the island already a long time before. The continent didn't have any such central rule, so there were borders everywhere where the units changed, but nobody had the power to select one standard over another. Until Napoleon. PiusImpavidus (talk) 17:17, 13 December 2023 (UTC)[reply]
So, if I understand this correctly, the Dutch explicitly rejected what they had been told as soon as they had a choice, and later accepted the new standard, not because they had been told to, but because they deemed it better, perhaps more for mercantile reasons than for any other reason.  --Lambiam 15:20, 14 December 2023 (UTC)[reply]
Yes, the metric system was introduced in the Netherlands in 1795 (not really imposed; the Dutch had their own revolutionary government, at that point still somewhat independent of the French), abolished in 1814 because it was French and everything French was bad, and reintroduced as the only legal system in 1820. The French names of the units were still bad, so the traditional Dutch names were used, but now defined as the metric units. PiusImpavidus (talk) 16:38, 14 December 2023 (UTC)[reply]
Yes, the Système International has its roots in the French Revolution, of which we Britons did not approve (to put it mildly). We also failed to adopt decimal time and the French Republican calendar, and it took us until 1971 to decimalise our currency of pounds, shillings and pence. Give us time, we'll get there eventually. Alansplodge (talk) 17:23, 13 December 2023 (UTC)[reply]
"Give us time" - yes. We started to decimalise the currency in 1849 with the introduction of the florin. 122 years seems about right. It always seemed a bit odd as a lad that we had the two bob bit and the half crown, so similar in value and size. Martin of Sheffield (talk) 17:47, 13 December 2023 (UTC)[reply]
"Better" is not necessarily a fact. ←Baseball Bugs What's up, Doc? carrots19:22, 13 December 2023 (UTC)[reply]
Not only the pounds, inches and feet changed every 30 km, even the number of inches in a foot was variable, and the variations were well over 10%. And 11 (as in, 11 inches per foot) isn't a very convenient conversion factor. So many people (including the king) considered the metric system better. PiusImpavidus (talk) 16:17, 14 December 2023 (UTC)[reply]
That must be why they still use miles and miles-per-hour. ←Baseball Bugs What's up, Doc? carrots21:47, 14 December 2023 (UTC)[reply]
I believe PiusImpavidus was talking about the Netherlands. In the UK, there were some variations in units used locally or in different trades, but the system was unified by the Weights and Measures Act 1825 and fine-tuned by several Weights and Measures Acts in the following decades. I think our lingering use of a few Imperial measures such as mph is due to a lack of courage by politicians, nobody wants to be the subject of the outcry which would invariably follow. Maybe George Orwell had a hand in this; in Nineteen Eighty-Four, enforced metrication (specifically of beer measures) is equated with totalitarian rule. Alansplodge (talk) 11:11, 15 December 2023 (UTC)[reply]
He had a point! ←Baseball Bugs What's up, Doc? carrots01:23, 16 December 2023 (UTC)[reply]
PiusImpavidus said: "Thing is, the UK had been one country for a while, under one king, and had been able to standardise its units all over the island already a long time before." Hmm. Read the Second Report of the Commissioners for Weights and Measures (1820), published five years after the battle of Waterloo, and reconsider. For example, there are 12 definitions of the hundredweight, around 20 English and 16 Scottish definitions of the pound (mass), and over two pages of bushels. Even the Scottish inch and foot are different to tthe English ones. So, no, not really. MinorProphet (talk) 00:41, 14 December 2023 (UTC)[reply]
A lot of local, special purpose units of volume and mass, but at least the avoirdupois (or was it troy?) pounds and ounces appear to have been an official standard all over the UK, along with inches and most derived units of length. The difference between English and Scottish inches was too small to matter for most practical purposes and presumably the result of a copying error. It looks like a better situation than on the continent before metrication. PiusImpavidus (talk) 16:07, 14 December 2023 (UTC)[reply]
How much of British trade was with Europe vs how much was with their own empire (which would have used British units)? If 80% of imports and exports was in the same units as domestic trade, why would they bother changing? --User:Khajidha (talk) (contributions) 14:03, 14 December 2023 (UTC)[reply]
The history of weights and measures is a subject which only the certifiably insane should tackle. If you really want to boggle your mind, investigate History of measurement systems in India and Indian units of measurement (the greater part of the Empire), and consider that the Troy pound was the standard legal weight in Britain and its colonies (along with the yard and gallon) from 1826 to 1856, although the prototype pound was destroyed in the disastrous fire in the Houses of Parliament in 1834. For British trade in the 19th century, see Pax Britannica for starters. MinorProphet (talk) 15:05, 14 December 2023 (UTC)[reply]
It's not just the UK in Europe who have been dragging their Metric feet (excuse the pun), see Metrication in Sweden:
Giving the measurements [for timber] in inches was still common in the late 20th century, about 100 years after the metric system was implemented. Alansplodge (talk) 17:32, 15 December 2023 (UTC)[reply]
It's still common - at least Down Here - for newborn babies' weights and heights (?lengths) to be given in pounds/ounces, and inches, even though none of the mothers giving birth nowadays have ever lived under the Imperial system. -- Jack of Oz [pleasantries] 20:23, 15 December 2023 (UTC)[reply]
I believe that Australian hospitals stopped giving babies' weights in pounds and ounces some years ago, but grandmothers have maintained the tradition. HiLo48 (talk) 04:48, 16 December 2023 (UTC)[reply]
Here in the UK I have in front of me a 454g jar of honey, and a 907g tin of golden syrup. I am shocked, shocked I say, at any suggestion that Britain is not fully metrificated! -- Verbarson  talkedits 21:07, 15 December 2023 (UTC)[reply]
Suspiciously specific, as in 1 and 2 pound jars, yes? :) ←Baseball Bugs What's up, Doc? carrots01:23, 16 December 2023 (UTC)[reply]
Couldn't say. There are no imperial measurements marked on either container, so I can maintain my plausible shockability. -- Verbarson  talkedits 10:17, 16 December 2023 (UTC)[reply]
Don't forget the estimated sign , meaning that the containers' contents will be 440–468 g (15+12 oz – 1 lb 12 oz) and 892–922 g (1 lb 15+716 oz – 2 lb 12 oz) respectively (or thereabouts). Bazza (talk) 10:38, 16 December 2023 (UTC)[reply]
Ah yes, European short measure pint beer bottles (500ml vs 1pt), European short measure loose foodstuffs (100g vs 4oz) and to cap it all we don't even get full measure with the -cheat. Bah-humbug! :-) Martin of Sheffield (talk) 10:56, 16 December 2023 (UTC)[reply]
Anyone who describes 500 ml as "a pint" on their container is obviously wrong, but I would imagine manufacturers don't do that. This is why Imperial measures have survived - they are convenient. The half crown was discontinued from 1851 but was brought back by popular demand in 1874. In my supermarket you can ask for "a quarter" of something and get it (a notice tells you its equivalent in grammes and the appropriate quantity is weighed out on the metric scale). 2A02:C7B:103:7100:D984:512D:2947:50AE (talk) 12:12, 16 December 2023 (UTC)[reply]
Just waiting for JackofOz to enlighten us about Adelaide pints. Bazza (talk) 13:24, 16 December 2023 (UTC)[reply]
I'm afraid the plethora of Aussie beer glass naming conventions is as much a mystery to me as anyone else. I couldn't even tell you how it's done in my own state of Victoria. My pub days are a distant memory (they date from my post-teenage years in Queensland, and then I spent 25+ years in Canberra before moving south). -- Jack of Oz [pleasantries] 20:43, 16 December 2023 (UTC) [reply]
I think most of Europe uses the inch for TV screens. In the 80s of last century you could still ask for butter by the demi-livre in France. You'd get 250 grammes, which isn't far out. DuncanHill (talk) 21:17, 16 December 2023 (UTC)[reply]
Since they introduced the horsepower, you know... and look at this one! Could be used without a cord (do not know about the network it doesn't apply much to TV screens) -- Grand-Pressigny, farmers collecting them used to call them "livre de beurre" at least the same kind of nearness is to be found in relationship with capacity, so: the Demiard --Askedonty (talk) 19:50, 18 December 2023 (UTC)[reply]

December 13

twins and paternity tests

Not seeking advice, this is about internet drama that I have nothing to do with. Basically someone has claimed that it was formerly impossible or extraordinarily difficult to do a paternity test on a pregnant woman carrying twins, as opposed to a with more typical one-fetus pregnancy of the type Jerry Springer used to test all the time. Also, supposedly, recent medical advances have made testing doable even with twins, which tripped up one of the participants of the drama. Our article DNA paternity testing doesn't mention anything about difficulty related to twins.

Is there something to it? What is the story? Should the article be expanded? Thanks. 2601:644:8501:AAF0:0:0:0:9A17 (talk) 02:15, 13 December 2023 (UTC)[reply]

What? You said in the content you removed that you "don't want to say who it is", so that's a sign enough that it has nothing to do with anything we should be editing a wiki article about. Remsense 03:30, 13 December 2023 (UTC)[reply]
Why would there be a problem doing a paternity test on twins? ←Baseball Bugs What's up, Doc? carrots03:50, 13 December 2023 (UTC)[reply]
Remsense, the idea is to put something in the DNA testing article about difficulty testing twins, assuming that difficulty is true and has good scientific sourcing. Nothing about the specific situation where I saw the topic come up. Bugs, your question is the same as mine. 2601:644:8501:AAF0:0:0:0:9A17 (talk) 04:37, 13 December 2023 (UTC)[reply]
In the case of heteropaternal superfecundation, seems like there would be a 50% chance of false-negative. DMacks (talk) 04:43, 13 December 2023 (UTC)[reply]
That would seem to apply to invasive prenatal paternity testing. With non-invasive testing, both fathers would likely test positive.  --Lambiam 07:53, 13 December 2023 (UTC)[reply]
According to the American Pregnancy Association [2], non-invasive paternity test cannot be peformed on twins. 97.82.165.112 (talk) 10:04, 13 December 2023 (UTC)[reply]
Good point. So either way, there's a problem. Albeit for a fairly rare type of pregnancy. DMacks (talk) 10:25, 13 December 2023 (UTC)[reply]
Reference Desk? fiveby(zero) 15:56, 13 December 2023 (UTC)[reply]

December 14

Hippo in the sky

Emblem of the Task Force Hippo

How fast would the pictured hippo have to flap its wings to become airborne? (Disclaimer: Not seeking advice.) ◅ Sebastian Helm 🗨 19:59, 14 December 2023 (UTC)[reply]

With wings that small it would never become airborne. Lift depends on the size of the wings much more than their speed. Shantavira|feed me 20:01, 14 December 2023 (UTC)[reply]
Interesting opinion. I would expect the speed to be finite, since drag force (at least ideally) is proportional to
area · flow_speed²
. ◅ Sebastian Helm 🗨 20:25, 14 December 2023 (UTC)[reply]
Maybe it's part bumblebee. Especially if the illustration displayed here is life-size. ←Baseball Bugs What's up, Doc? carrots21:46, 14 December 2023 (UTC)[reply]
You have a point, Bugs. I just had assumed the “hippo” referred to hippopotamus, but that might be wrong. The wings rather resemble those of millimeter-sized flying animals. ◅ Sebastian Helm 🗨 22:20, 14 December 2023 (UTC)[reply]
In theory, if the wings were to be flapped fast enough, they could generate enough downward force to lift a full-sized hippo off the ground, but I'm thinking they would have to move at extremely high, possibly even relativistic speeds in order to do so. Gyatt W Rizz (talk) 22:28, 14 December 2023 (UTC)[reply]
I find it hard to obtain quantitative date, but in this video we see a bald eagle lift off for a brief moment with a full wing flap from 0:51 to 0:59. The initial upward speed was almost certainly achieved by a jump, but let's take this as a lower bound. The video is recorded at 1000 fps. Assuming it is played at 24 fps, this means one wing beat in (24/1000) × 8 s = 192 ms, which means a frequency of about 5.2 Hz. Now an eagle weighs about 5 kg while a hippopotamus is more like 1400 kg. The wings on the hippo seem to be about the same size as those of an eagle. Going with the theory that lift is proportional to the square of the flow speed, so flow speed is proportional to the square root of the required lift, we get a back-of-the-envelope estimate for a lower bound of
1400/5 × 5.2 Hz ≈ 87 Hz.
So we'd have a humminghippo. But intuitively this feels way to low to me, at least with this size of wing, so the theory may need some refinement. For comparison, a hippopotamus weighs more than the Airbus Helicopters H125.  --Lambiam 12:27, 15 December 2023 (UTC)[reply]
Excellent reply! And a beautiful video to boot. Yes, originally i would have agreed with your intuition, but i think your result is a good approximation and see no reason why it should be much higher, apart from some nonlinear effects that may not scale well.
That said, let's try another way: Since you mention a helicopter, and since this is the emblem of a helicopter task force, another way to compute this would be by comparison with a helicopter. For that, the H125 is a good choice, since, with a bit of cargo, it will weigh the same as a hippo. I don't know its rotation frequency, but if we assume (with [3])
,
and if we treat the three blades as one rotating at 3 times the speed, we get:
.
Now we know (from the specs in our article) that the rotor has a diameter of 10.7 m, with which we can estimate (from the picture) the area of one blade:
.
For the depicted hippo i'd estimate the area of both wings together:
.
With this (and with speed at the wing tip, length of wing resp. blade) inserting the drag equation gives:
Inserting
gives
This is an order of magnitude more than your estimate, but given how crude some of my assumptions are, i think the true value will lie closer to your result. Besides, i prefer the idea of a humminghippo over a soprano hippo. ◅ Sebastian Helm 🗨 19:38, 15 December 2023 (UTC)[reply]
Yes, at that frequency it would be more of a mosquippo. {The poster formerly known as 87.81.230.195} 90.199.215.44 (talk) 03:29, 16 December 2023 (UTC)[reply]
Maybe your hippo can run really, really fast to get airborne? There are quite a few papers and sites on aerodynamics of flapping wings for ornithopters such as "How Ornithopters Fly" with different models, but they all seem to be applicable for large heavy birds with high aspect ratios, large forward velocity, and low flapping frequency. Michael Dickinson says for insect flight wing sweeping is a bit like a partial spin of a "somewhat crappy" helicopter propeller which seems to validate your approach. Maybe something like "Wing Rotation and the Aerodynamic Basis of Insect Flight" could help? fiveby(zero) 16:18, 16 December 2023 (UTC)[reply]

Niagara Falls through a straw

This question may have been answered by someone on the internet at some point in time, but I'm still going to ask it. What would happen if you tried to funnel Niagara Falls through a straw? What would be the result of taking a straw, raising it up to the Falls, and somehow getting all the water coming down to enter the straw? Gyatt W Rizz (talk) 22:21, 14 December 2023 (UTC)[reply]

Water moving through a more constricted pipe or channel tends to speed up. So the physics question would seem to be, how fast could the water move through a straw (of what size?) such that it wouldn't back up and spill over. ←Baseball Bugs What's up, Doc? carrots23:09, 14 December 2023 (UTC)[reply]
I've thought about that before, about how it can back up. Wouldn't the water coming from upstream push the water that's closer to the edge of the Falls forwards, preventing it from backing up? Think of it as a sort of vertical compression-like thing, where there's a small opening at the end for water to go through. It'll only go through it at a certain speed, but if you compress the water behind it, it pushes that water forwards, resulting in an increase of speed. Using this information, what would the implications be of water going into that straw? Gyatt W Rizz (talk) 23:18, 14 December 2023 (UTC)[reply]
See the article Choked flow.  --Lambiam 11:09, 15 December 2023 (UTC)[reply]
When a river backs up, the waterflow can get wider and deeper. It isn't confined in a straw to prevent that. This happens often. For example, what would happen if you took the Colorado River and forced it to flow through a few tubes (very large metal straws) that happen to be connected to some hydroelectric generators? You end up forming a lake (Lake Mead). What happens when you take the Columbia river and force it through a large metal straw in a hydroelectric generator? You form the Franklin D. Roosevelt Lake. What happens when you force the Osage river to go through a metal straw? You form the Lake of the Ozarks. Do you see a trend here? 97.82.165.112 (talk) 11:26, 15 December 2023 (UTC)[reply]

December 15

Biology -- corrections

I am not a biologist but have noticed two cases of evident inconsistency in pages related to the renal system. There is a portal for such things but I do not see how to pass my queries to it. I would be grateful if you would pass these points to the people responsible.

1. The following page links "Ludwig theory" to "Ludwig von Bertalanffy". I think this is a mistake -- from what I can see von Bertalanffy had nothing to do with the study of renal systems. https://en.wikipedia.org/wiki/Bowman%E2%80%93Heidenhain_hypothesis

2. The following page states seems to include a discrepancy between two estimates of filtration rate: "approximately 180 litres of filtrate per day", and (in next sentence) a "normal range" of "800 to 2,000 milliliters per day". The first quantity seems very large and I think it should be corrected to "1,800 milliliters". https://en.wikipedia.org/wiki/Kidney M E T Horn (talk) 04:19, 15 December 2023 (UTC)[reply]

1. The "Ludwig" of the theory is German physician and physiologist Carl Ludwig (1816–1895). Quoting the abstract and the final text section of a paper discussing the issue:[1]
Marcello Malpighi discovered the glomerulus that bears his name in the 17th century, but it was not until the middle of the 19th century, in 1842, that William Bowman in London published his studies of the histological structure of the glomerulus and proposed that urine formation begins with glomerular secretion. At nearly the same time in Marburg, Carl Ludwig, unaware of Bowman’s findings, proposed that urine formation begins with glomerular filtration followed by tubule reabsorption. The controversy lasted 80 yr.
...
The widespread advances in understanding urine formation were based on the decisive results of the experiments by Joseph Wearn and Alfred N. Richards in 1924 that resolved the controversy in favor of the filtration reabsorption theory of Carl Ludwig and launched the modern era of renal physiology.
Based on this article, it seems to me that the wording "This theory was later merged with ..." is incorrect, and should be, "This theory was later replaced by ...".  --Lambiam 10:36, 15 December 2023 (UTC)[reply]
2. The cited source states: "Approximately 180 L/day of glomerular filtrate is generated in an average adult human, the majority of this is reabsorbed by the highly water-permeable proximal tubules and descending thin limbs of Henle's loop." [my emphasis by underlining; --L.] Other sources state the same estimate; for example, ""Thus, in a 24 h period, as much as 180 L/day of plasma is filtered at the glomerulus. In other words, in one day kidneys filter an amount of fluid equal to 4 times the total body water."[2] The other figure, 800 to 2,000 milliliters per day, is the urine production – the fraction that is not reabsorbed.
Feel free to edit the article so as to make clear to the unaware reader that most of this enormous amount is reabsorbed.  --Lambiam 11:05, 15 December 2023 (UTC)[reply]

References

  1. ^ Jamison, Rex L. (December 2014). "Resolving an 80-yr-old controversy: the beginning of the modern era of renal physiology". Advances in Physiology Education. 38 (3): 286–295. doi:10.1152/advan.00105.2014.
  2. ^ Indu Khurana; Arushi Khurana (2015). Textbook of Medical Physiology (2 ed.). Elsevier Health Sciences. p. 491. ISBN 978-81-312-4254-4.

December 16

Tree canopy formations

Are these witch's brooms on trees or something else? Particularly wonder why they are spherical rather than in irregular pattern (taken recently when most of the foliage fell due to winter). Brandmeistertalk 12:21, 16 December 2023 (UTC)[reply]

@Brandmeister: Mistletoe. Bazza (talk) 13:20, 16 December 2023 (UTC)[reply]
Take care with whom you stand underneath it. Alansplodge (talk) 13:49, 16 December 2023 (UTC)[reply]
What an unromantic way of being parasitic... Brandmeistertalk 17:10, 16 December 2023 (UTC)[reply]
Thanks, my second thought today became that, but wasn't sure. Brandmeistertalk 17:10, 16 December 2023 (UTC)[reply]


December 18

Organ pipe

If the air temperature rises with height due to poor church HVAC then would there be only one main pitch equal to putting the average speed of sound of the standing wave in the pitch formula or will there be two main pitches one for the speed of sound at each end or will the fundamental be smeared into an infinite number of pitches more or less equally loud all lying between a pitch based on the speed of sound of the most heated part and the speed of sound of the least? 166.199.7.54 (talk) 00:51, 18 December 2023 (UTC)[reply]

Are you asking whether a temperature gradient across a column of air causes its resonant standing wave to vary in frequency accordingly? Remsense 02:11, 18 December 2023 (UTC)[reply]
Correct, if a pipe is x Hertz when everything's 20C what sound will come out when it's 20 on average but the air has a temperature gradient? 166.199.7.54 (talk) 02:40, 18 December 2023 (UTC)[reply]
I've thought about this for about 15 minutes, and I think your "two frequencies at each end" guess is going to be my guess also. Ideally, there's only meaningfully a "pitch" in the direction that changes in air pressure are propagating (i.e. down the length of the tube) Remsense 04:38, 18 December 2023 (UTC)[reply]
As well as the standing wave producing the sound, there is also movement of air due to the action of the bellows. I suspect that quite quickly the air in the pipe will have a uniform temperature closely related to the intake temperature of the blower. Martin of Sheffield (talk) 09:20, 18 December 2023 (UTC)[reply]
Even if the temperature varies considerably in the pipe, there is only one pitch. The pressure waves will travel through the pipe with a varying velocity, the speed of sound, which increases with the temperature. But the pitch is determined by the standing waves, whose frequency only depends on the total travel time of the wave through the pipe. Compare this to a pendulum clock. The speed of the pendulum varies over its trajectory, but the clock ticks with only one rhythm.  --Lambiam 10:15, 18 December 2023 (UTC)[reply]
Right! I think my mistake was thinking about the standing wave as if it were a traveling wave, not the sum of traveling waves. Remsense 15:29, 18 December 2023 (UTC)[reply]
When the air temperature rises, the speed of sound in air increases proportionally to the square root of the absolute temperature and the resonant length of the metal Organ pipe expands. The sound pitch does not change as much as either effect alone would cause because they partially compensate one another. Philvoids (talk) 18:45, 18 December 2023 (UTC)[reply]
This is not the case: the expansion/contraction of the aerophone body caused by change in temperature is negligible proportional to the corresponding expansion/contraction of the column of air inside. Remsense 18:51, 18 December 2023 (UTC)[reply]
The length of the column of air inside the pipe is essentially the length of the pipe containing the air.  --Lambiam 19:47, 18 December 2023 (UTC)[reply]
Right, but the point I was making is that change in the length of the pipe due to the expansion or contraction of the aerophone itself is negligible. I suppose I misspoke when I said the air was expanding/contracting, I should've said it was becoming more or less dense. Remsense 19:49, 18 December 2023 (UTC)[reply]
That is correct. A pipe made of an alloy of tin and lead, as many church organ pipes are, will expand by about 0.05% when the temperature rises by 20 degrees. The speed of sound will increase by about 3.4%.  --Lambiam 20:29, 18 December 2023 (UTC)[reply]
I assume that the Fourier analysis in terms of wavelength of the waveform inside the pipe depends only on the shape of the pipe but not on the conditions of the air inside it. The lowest eigenmode (I guess that's what's meant by "standing wave" above) has wavelength twice the length of the pipe (if the later is closed at both ends and essentially one-dimensional). However, the conversion to frequency (the dispersion relation) depends on temperature, as demonstrated in the video linked above. With a temperature gradient, the air molecules should oscillate with varying frequency along the length of the pipe. The next question is how the oscillation inside the pipe is coupled to the air outside. If the coupling is homogeneous along the pipe then we should here a continuous superposition of frequencies over a range determined by the temperature range. If the coupling occurs at a single point (the labium or an open end (are organ pipes open?) come to mind) then the pitch will be determined by the temperature at that point. The actual situation will be somewhere inbetween these extremes, and we won't discuss the effect on timbre. --Wrongfilter (talk) 20:31, 18 December 2023 (UTC)[reply]
It feels like we should link to Q factor, a parameter that describes (among other things) the "perfectness" of a resonator - in other words, how ideally the resonating energy is exactly concentrated at one single harmonic frequency. Real physical resonators have a bit of a spread - a "bandwidth" - over which the energy is spread. Things like imperfections or nonuniform conditions along the extent of the resonant cavity (including temperature) will cause the resonator to have a lower "Q" factor - and consequently a wider bandwidth. Audibly, this means a tone that is less "clear" and it will sound different - perhaps even sounding "out of tune" in severe situations. Nimur (talk) 20:43, 18 December 2023 (UTC)[reply]