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February 3

Contradictory views

Are there any studies on the effects on people when they hold inconsistent views?
I suspect it creates a certain amount of psychological distress. 2A02:8071:60A0:92E0:0:0:0:B32D (talk) 13:56, 3 February 2024 (UTC)

IP editor. The article on cognitive dissonance has many references. Mike Turnbull (talk) 14:25, 3 February 2024 (UTC)

See also wishy-washy. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:23, 3 February 2024 (UTC)

It only creates distress when people are aware of the inconsistency. I've heard speakers demand that critics be silenced, arguing that the criticism is an attack on their right to free speech.  --Lambiam 22:13, 3 February 2024 (UTC)

Ha ha, that's really quite a good one. :-} NadVolum (talk) 00:47, 4 February 2024 (UTC)

You might find [1] interesting. NadVolum (talk) 00:58, 4 February 2024 (UTC)

Tribal thinking AKA groupthink is polarizing. However, this study "No association between numerical ability and politically motivated reasoning in a large US probability sample" and most others (see its introductory section) failed to replicate the original study that appeared to "...show that deliberative analytical... reasoning is hijacked by identity...". Instead, it affirmed that political alignment affects everyone's accuracy, but those with better analytical reasoning had better scores regardless. Modocc (talk) 02:23, 4 February 2024 (UTC)

Well that's a bit reassuring - or perhaps the statisticians results were affected by their tribal identities ;-) NadVolum (talk) 12:34, 4 February 2024 (UTC)

February 4

Planck'law and stefan-Boltzmann law relation

In a forum I found that the relation between the Stefan-Boltzmann law and the Planck law in the black body cavity hole is:
$B=T^{4}\sigma =\int _{\lambda =0}^{\infty }\pi B_{\lambda (\lambda ,T)}d\lambda$
Is that true, or is there a different equation ?
Malypaet (talk) 13:42, 4 February 2024 (UTC)

See Stefan–Boltzmann_law#Derivation_from_Planck's_law, where the integral is written in terms of frequency rather than wavelength. --Wrongfilter (talk) 13:52, 4 February 2024 (UTC)

Thanks for the link, I understand better. Malypaet (talk) 09:21, 5 February 2024 (UTC)

Organic Nomenclature

I've been unable to find the name of the CH2= group, I suspect it might be methenyl. So what would be the preferred IUPAC name for 2-ethylbut-1-ene. Taabibtaza (talk) 14:34, 4 February 2024 (UTC)

Hey, I'm not a chemist, but AIUI IUPAC rule P2 gives methylidene for CH₂=, so I'd suggest 3-methylidene pentane for CH₃.CH₂.C(:CH₂).CH₂.CH₃ which is what I think you're after for CH₂:C(C₂H₅).CH₂.CH₃ HTH, Martin of Sheffield (talk) 15:12, 4 February 2024 (UTC)

@Taabibtaza IUPAC naming conventions are a minefield. For this sort of question, I'd use Chemspider and look at the various possibilities. See this entry for 2-ethyl-1-butene which they say is alternatively called 3-methylenepentane. IUPAC wouldn't use methylidene in the case of simple alkenes where the double bond is the only functional group. Mike Turnbull (talk) 17:13, 4 February 2024 (UTC)

February 5

I have requested change

Recently, I have requested change in Talk:Potassium but no one replied. Hope someone jump in and let me know! 2405:4802:64C7:BF70:B50C:773B:1A40:16BA (talk) 02:56, 5 February 2024 (UTC)

Regardless of what the time was in Vietnam, you made your request on a Sunday morning in Europe, and in the early hours of Sunday in the USA, where the bulk of English-language Wikipedia editors live. Please remember that Wikipedia does not have a 24-hour duty roster of paid editors – we are all unpaid volunteers, who do everything in the spare time available from our everyday jobs and lives.

On the Help desks, you might reasonably expect a response in a few hours (less if you're lucky); on the Reference desks perhaps longer; on an article Talk page, you should allow at least several days. I'm sure an editor interested (and competent) in this type of correction (so not me) will evaluate your request within a week, and fulfil it or not according to what they think appropriate. {The poster formerly known as 87.81.230.195} 90.199.208.215 (talk) 03:09, 5 February 2024 (UTC)

I'm sorry, as far as I'm concerned, the Help desk is suitable only for questions involving editing and using Wikipedia, not demanded for questions about general knowledge and questions that need broad agreement. I'm afraid this doesn't fall into those categories, thus not count as a proper enquiry and may need consensus before performing the edit by my request. As a result, I seek for help in this page, reference page. 2405:4802:64C7:BF70:20BF:895B:4915:B58A (talk) 04:39, 5 February 2024 (UTC)

I wasn't advising you to ask on another Desk. I was advising you to be more patient, and wait longer for a response to your Talk page request. {The poster formerly known as 87.81.230.195} 90.199.208.215 (talk) 10:08, 5 February 2024 (UTC)

Is 9.8 meters per second squared really accurate?

Is the figure 9.8 m/s² really accurate for the gravitational acceleration on Earth? It seems overly simplistic to me. I figured that the actual value would be more variable based on an object’s mass and distance from the center of the Earth (as both are factors in calculating the force of gravity objects exert on each other). Primal Groudon (talk) 04:49, 5 February 2024 (UTC)

The article Gravity of Earth discusses the variation of g depending on various parameters. --Wrongfilter (talk) 04:55, 5 February 2024 (UTC)

Mass doesn't matter, but distance from center of the earth does. EvergreenFir (talk) 05:05, 5 February 2024 (UTC)

Yeah, now that I see it, the object’s mass cancels itself out, leaving us with just the UGC times the mass of the Earth divided by the square of the distance from the center of the Earth. Primal Groudon (talk) 06:56, 5 February 2024 (UTC)

It feels like we should include a brief review of accuracy and precision; "9.8" is accurate (with respect to the standardized gravity of Earth) to a few parts per thousand, and it's precise to two significant figures. Beyond those special adjectives, we might also say that one is engaging in a category mistake when they conflate "inaccuracy" of the standard parameter with well-understood variation in magnitude of the measured acceleration due gravity caused by location and realities of the shape and mass-distribution of Earth. This is also - yet again - categorically distinct from any theoretical question about whether such measurements ought to be sensitive to the object's mass (...it ought not be, based on pretty well-established physical relationships, but the effort to re-establish this certainty would be an interesting exploration of experimental and theoretical physics).

All of these ideas raise real and valid concerns; but these concerns are best described using words other than "accurate" or "inaccurate." Nimur (talk) 13:50, 5 February 2024 (UTC)

The section Gravity of Earth § Comparative values worldwide gives the reader a good idea of the variation one can expect while remaining earthbound. It may be assumed that the figures in the table are accurate within the precision with which they are presented.  --Lambiam 14:37, 5 February 2024 (UTC)

Exactly, and therein comes both the issue of accuracy vs precision and significant figures for the "9.8" value. According to that article, surface varies from 9.7639 m/s² to 9.8337 m/s². Of course, one shouldn't just assume that the average surface gravity is halfway between those two numbers, but we can say with a high degree of accuracy that the surface gravity is 9.8 m/s², which for most applications is going to be more than good enough. With far more measurements taken around the globe (which might already have been done), we could give that average with greater precision, but I imagine any application that would benefit from that higher precision would need to know the local surface gravity to a high precision and not the global average surface gravity. I would imagine that once you are out at a distance where a global, rather than local average of high precision is of use, you are also at a distance where you shouldn't be calculating gravity based on surface values anyways.

TLDR the needed precision depends upon the needs of the application, and the highly accurate 9.8 m/s² is precise enough for most everyday applications and, when it isn't precise enough, a highly precise global average would probably have a too low a local accuracy to be useful to said application. --OuroborosCobra (talk) 14:51, 5 February 2024 (UTC)

Naming of electromagnetism and optics standards

In Wikipedia and elsewhere we find different names like $I_{\nu (\nu ,T)},B_{\nu (\nu ,T)},B_{\nu T},M^{0}$,etc... for things that seem to be the same.
Is there a naming standard referenced somewhere?
Malypaet (talk) 09:30, 5 February 2024 (UTC)

I don't think there is, at least not among physicists (but there may be engineering regulations that I'm not aware of). Having said that, I believe that I've seen the convention that B refers specifically to a Planck spectrum, whereas I is the physical quantity "intensity" regardless of whether the spectrum is Planck or something else. --Wrongfilter (talk) 09:37, 5 February 2024 (UTC)

In the last link on wikipedia you gived me, one find:

$M^{\circ }=\sigma T^{4}$

I find also:

$P=\sigma T^{4}$

${\frac {P}{A}}=\sigma T^{4}$

$B_{\nu }(T)$ or $B_{\nu (\nu ,T)}$,etc,...

So, in fact there is no rule, it is like "Happy hour"...

I have just to give the naming convention at the beginning of my text!
Malypaet (talk) 13:39, 5 February 2024 (UTC)

Where do you see $B_{\nu (\nu ,T)}$? To me this notation looks confused. I think this ought to be $B_{\nu }(\nu ,T),$ in which $B_{\nu }$ is the name of a function and $(\nu ,T)$ is its argument. So the subscript $\nu$ is not a variable; the second $\nu$ is.  --Lambiam 14:25, 5 February 2024 (UTC)

What branch of physics discusses a given stationary system containing moving bodies?

Maybe Thermostatics? Maybe Hydrostatics? Maybe Aerostatics? I guess almost all branches of physics may discuss that, but I'm asking about the main branch. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 09:31, 5 February 2024 (UTC)

Easy - physicists study stationary processes! Nimur (talk) 13:34, 5 February 2024 (UTC)

The adjective stationary means that something remains invariant as time progresses. What kind of system do you have in mind and which are the unchanging aspects that lead to its being characterized as "stationary"? Ergodicity with respect to its phase space? This is something studied in dynamical systems theory – not by itself a branch of physics, but applied in many different branches.  --Lambiam 14:15, 5 February 2024 (UTC)

If anything is moving inside the system, then "-statics" is not quite appropriate. Galaxies are stationary systems (approximately) even though all the stars in a galaxy are swarming around like crazy. In a way even the Solar System is stationary. Stationary systems appear in all many branches of physics; which branch is appropriate for describing any particular system depends on the characteristics of the system, for instance how many particles make up the system, which forces act between the particles, etc. --Wrongfilter (talk) 14:23, 5 February 2024 (UTC)

By "stationary" I mean "at rest". Actually, my aim is to solve a contraction I'm coping with, regrading systems at rest, so I'd like to know where the whole issue may be dealt with. See below my new thread. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 20:35, 7 February 2024 (UTC)

February 6

Should backs of solar panels sometimes be black?

I noticed that the backs of solar panels tend to be silver or white but I was thinking, in the UK, where the climate is cold most of the time, it would be useful if solar panels were radiating as much heat as possible into the building they are attached to. Is a black coating better for this? Are passive daytime radiative cooling coatings relevant given that reflection isn't needed? It would be useful to passively cool the panels and heat the building. Doing that actively is not viable in the UK. 92.7.46.126 (talk) 00:32, 6 February 2024 (UTC)

Says here that the optimal temp for solar panels is 25°C, and hotter is bad. Abductive (reasoning) 05:20, 6 February 2024 (UTC)

Heat can only passively flow hot to cold, so you can only radiate net heat from the solar panels into the building if the solar panels are hotter than the interior of the building. That won't happen when you want to heat the building. Cooling the panels is useful, but radiative cooling isn't very important. Most cooling happens by conduction into the air, which then takes the heat away with convection. Liquid (evaporative or run-off) cooling can be effective too. Solar panels tend to perform best in unstable spring weather here in Europe: not too hot, bright sunshine, alternating with short showers to cool the panels. I don't think a black back coating will have any significant effect.

There are solar panels that are both PV and heat collectors. As most mixed devices, they're not the best at either of them: they have to be designed to run hotter than ordinary PV panels. PiusImpavidus (talk) 11:58, 6 February 2024 (UTC)

"idiopathic?" kidney damage and date rape drugs

What is known about the number of criminal convictions for using date rape drugs that have caused kidney damage to the victims?Rich (talk) 02:28, 6 February 2024 (UTC)

If kidney damage is known to have been caused by a drug, it is (by definition) not idiopathic. While the use of MDMA can lead to acute kidney failure (PMC 4220747), I suspect that the incidence of such convictions (where I assume the defendant stands accused of administering the drug surreptitiously to a victim) is too low to warrant keeping statistics.  --Lambiam 11:35, 6 February 2024 (UTC)

Nitrogen microwave heating

Nitrogen (N2) molecules are non-polar and have no molecular dipole moment. Microwave radiation induces rotational motion in molecules with a dipole moment, thereby increasing their kinetic energy and temperature. It seems that nitrogen would not be susceptible to microwave heating, yet nitrogen is heated with microwaves in many practical applications: [2], [3], [4]. How is this possible? 2600:1700:6D30:49A0:89F5:4D2B:8BCD:3F6C (talk) 11:20, 6 February 2024 (UTC)

All molecules (and atoms) have a non-zero "instantaneous" dipole moment. Although the electron cloud is symmetrically distributed, it undergoes spontaneous quantum fluctuations that break the perfect symmetry. See also London dispersion force.  --Lambiam 11:48, 6 February 2024 (UTC)

Thanks for your response. So are molecules with no permanent dipole moment heatable with microwaves in an efficient fashion in general? Or is nitrogen specifically susceptible? 2600:1700:6D30:49A0:89F5:4D2B:8BCD:3F6C (talk) 15:06, 6 February 2024 (UTC)

I don't think "efficient" (measured in power consumption) is the right term, but it is an effective method that is very easy to control.  --Lambiam 20:19, 6 February 2024 (UTC)

February 7

Planck's law 1901 article entropy and integration

In his 1901 article, Planck go to the following equation:
$\vartheta =\nu f({\frac {U}{\nu }})$
and
${\frac {1}{\vartheta }}={\frac {dS}{dU}}$
to
${\frac {1}{\vartheta }}={\frac {1}{\nu }}f({\frac {U}{\nu }})$
And integrated (that is what he writes):
$S=f({\frac {U}{\nu }})$

For me, it's too short, not clear.
How can it go from ${\frac {1}{\nu f({\frac {U}{\nu }})}}$ to ${\frac {1}{\nu }}f({\frac {U}{\nu }})$?
And what operation describes the integration that follows?

Malypaet (talk) 13:50, 7 February 2024 (UTC)

I think he uses f as a general symbol to denote unknown functions. The form of the function may differ for the various quantities, but the important thing is that the argument is $U/\nu$ or $\vartheta /\nu$ in this combination. Not particularly good style, but he's Planck and we're nobody. --Wrongfilter (talk) 18:16, 7 February 2024 (UTC)

Same as Wrongfilter, but I went to the effort of typing this out, so I'm darn well gonna post it.

I don't think you can make the assumption that the two appearances of ${\textstyle f({\frac {U}{\nu }})}$ are the same function.

It's clearer if you keep the derivative on the left:

${\frac {dS}{dU}}={\frac {1}{\vartheta }}$

with

$\vartheta =\nu f({\frac {U}{\nu }})$

clearly leads to

${\frac {dS}{dU}}={\frac {1}{\nu }}f({\frac {U}{\nu }})$

as long as you don't require the two "f"s to be the same.

The same holds for the integration - after you integrate dS with respect to dU, you have another ${\textstyle f({\frac {U}{\nu }})}$, with no reason to believe they are the SAME function. PianoDan (talk) 18:29, 7 February 2024 (UTC)

It's easy when you know the result in advance to use magic functions to build a demonstration. I did notice that he used $f$ for different functions (at least five), but until those there was a logic that I understood and that's what I'm looking for here.

And please, don't say that we are nobody in front of someone who found an equation before its demonstration and who never wanted to tell how, by accident?

So far, I have managed to rewrite his article while remaining in the context of power and replacing the resonator with an element of surface $\lambda ^{2}$, I just stumble on these last two magic functions.

For example, if you rewrite this article from the beginning staying in power context, you get:

$\vartheta =\nu f_{7}({\frac {cU}{\nu }})$ where ${\frac {c}{\nu }}\equiv {\frac {1}{h\nu }}$ in time dimension. Just by logic.

Until 2019: $h={\frac {1}{c}}...$

I'll still try to find it myself.

I don't want to learn physics at Hogwarts School. Malypaet (talk) 23:19, 7 February 2024 (UTC)

It's not a question of "magic functions". Planck doesn't give a full expression for f here, because he is trying to establish a general property of ALL possible "f"s.

The point of that section of the paper was simply to demonstrate that the entropy is ONLY a function of ${\textstyle {\frac {U}{\nu }}}$, and NOT a function of the value of c. The derivation successfully demonstrates that fact, and that is its only purpose. PianoDan (talk) 00:17, 8 February 2024 (UTC)

@Wrongfilter My name is nobody, it's for the giant Cyclops Planck!

It's easier to understand when you know the result:

$U=f_{1}(\vartheta$)	$U=ln(\vartheta$)	$f1()=ln()$
$\vartheta =f_{2}(U)$	$\vartheta =e^{(U)}$	$f_{2}()=e^{()}$
${\frac {1}{\vartheta }}={\frac {1}{f_{2}(U)}}$	${\frac {1}{\vartheta }}={\frac {1}{e^{(U)}}}$
${\frac {1}{\vartheta }}=f_{3}{(U)}$	${\frac {1}{\vartheta }}={\frac {1}{e^{(U)}}}$	$f_{3}()={\frac {1}{e^{()}}}$

"c" is a constant and that's not the problem here. If we consider that the radiation circulates in the vacuum between the atoms of the cavity wall, at the microscopic level its value remains unchanged, while at the macroscopic level the speed of the radiation is given by $c_{m}={\frac {c}{n}}$ (n for refractive index), voilà.

Thank you all, our discussions allowed me to find the answer quickly.
Malypaet (talk) 13:59, 8 February 2024 (UTC)

Why isn't rain consistent around the world?

The laws of evaporation physics should be consistent around the world, right? So how can some place rain a lot more than other places. And let's make this about large cities near lakes/oceans. In the U.S., southern California had to open a saltwater treatment plant in 2015 as they are sucking the Colorado River try, while being net to an ocean, which is true of Washington state next to an ocean but rains heavily. What is the missing variable?

Also, what is the lowest altitude for rainfall? Thanks. 170.76.231.162 (talk) 17:36, 7 February 2024 (UTC).

The surface of the Earth is not consistent, so why would we assume that the application of the "laws of evaporation physics" will result in a single, uniform result globally? Different latitudes have different temperature profiles (e.g. it's a lot warmer near the equator and a lot colder near the north and south poles). Geography itself has effects, such as rain shadow, where warm, moist air follows the breeze up a mountain range and cools as it goes up, resulting it it raining on the mountains, and dry air coming down the other side. --OuroborosCobra (talk) 19:16, 7 February 2024 (UTC)

Then there's oceanic currents, and the Earth's tilted axis, and its non-circular orbit around its star, and... Bazza (talk) 19:58, 7 February 2024 (UTC)

You are too close to the horse latitudes and the west side of continents exaggeration of the horse latitudes. This is why Afrique Nord to west India is dry and the rest of the Eurafrasian horse is wet. With a transition zone in between. The horse latitudes exist cause Earth has three pairs of convection cells (the average wind circuits of the atmosphere). The air sinks on average in the horse latitudes reducing rain. The southbound California current also reduces rain on that side by causing subsidence. While the northbound West Atlantic express boosts East Coast rain if anything. All mountains to the west of you reduce rain. So Mojave is dry, Arizona is dry. Also fossil fuels is making the driest latitude which is in Baja (not the end or Tijuana, somewhere in between) move closer to you and many other effects. Sagittarian Milky Way (talk) 23:09, 7 February 2024 (UTC)

Death Valley: 282 ft (86m) below sea level, 2.2 in (56mm) annual rainfall. DOR (ex-HK) (talk) 16:27, 13 February 2024 (UTC)

Lowest natural surface on Earth, Dead Sea c. minus 1,400 feet and drying up, has had rain. It is true that rain dries continuously in a desert though, sometimes evaporating completely before reaching the ground. Sagittarian Milky Way (talk) 17:20, 13 February 2024 (UTC)

Also the East Coast would have less rain if the Gulf of Mexico and Caribbean were full-blown land. Sagittarian Milky Way (talk) 17:36, 13 February 2024 (UTC)

laser cutting glass

This video[5] shows glass being cut by a laser. There appears to be some sort of "white smoke" during the process. I'm trying to understand what's going.

The melting point of Silicon dioxide is approximately 1713 °C, the boiling point is around 2,950 °C.

I'm guessing that it's a combination of the following factors:

1. The glass in the cut area is melted, and remains attached to the the two pieces. After the cut, the melted parts solidifies as a part of the two resultant pieces. The total mass of the two resultant pieces is the same as the original piece.

2. The glass in the cut area is melted and then vaporized, the glass vapour is removed by the fume extraction system. The total mass of the two resultant pieces is less than the original piece.

3. The glass in the cut area is melted, and micro particles of it (in either liquid or solid state) are carried away as a "white smoke", and micro particles of it are also splattered around the machining area. The total mass of the two resultant pieces is less than the original piece.

Which one of these factors apply in this case?

If it's a combination of two or more factors, which one is the dominant one? OptoFidelty (talk) 20:04, 7 February 2024 (UTC)

It's likely closest to #2 in your list. See laser ablation. It's not necessarily melting and then evaporating, however, it could be sublimating (direct from solid to gas). Yes, some amount of mass is being lost (or at least no longer useful to these pieces, as I suppose you could recover and reuse the smoke material for a nice atom economy with the fume extractor), although whether you say that the larger or smaller resulting pieces of glass is losing that mass (or some ratio between them) is basically up to where you designate the division between the pieces. In this case, they seem to want the smaller piece of glass that they cut out, so they would program the machine to cut it out such that the tolerances of the dimensions of the final piece are within required specifications. Rather than cutting "directly on the line," if you were, it would intentionally cut slightly outside of that line knowing the diameter of the "drill." The same would be done with a mechanical process. --OuroborosCobra (talk) 20:21, 7 February 2024 (UTC)

Thank you!

Regarding, Sublimation (phase transition), what determines whether "solid->liquid->gas" happens or "solid->gas" happens? It it related to the speed of the heating?

For example:

1. Iodine is heated at room temperature. Only "solid->gas" occurs.

2. Silicon dioxide is heated slowly above its boiling point. Only "solid->liquid->gas" occurs.

But in the situation in the video, it's not clear whether it's "solid->liquid->gas" or "solid->gas", or a mix of both. Is there a way to calculate the behaviour if all the processing conditions are known? OptoFidelty (talk) 20:45, 7 February 2024 (UTC)

It depends upon both temperature and pressure. See phase diagram. Even for pure substances (and glass is usually intentionally not pure), calculating those values is non-trivial as you end up having to factor in a lot of things, such as calculating intermolecular forces. With a lot of effort, you might be able to do that with computational chemistry software packages, but one is better off figuring out the phase diagrams for given substances experimentally, such as through various forms of calorimetry. That said, phase diagrams for common substances have largely already been figured out experimentally, so you can usually look it up. I don't know how reliable it is, but here is one for silicon dioxide. However, that gets into another topic, which is "what is glass?" Look over at glass, and you will see that it isn't any single thing that can be just called "silicon dioxide;" that's just one type (and probably not the type in the video). What specifically that glass is made of, what crystal phase it is in, whether there are additives (such as in borosilicate glass, these are all going to have an impact on the conditions for melting, boiling, and sublimation. --OuroborosCobra (talk) 21:28, 7 February 2024 (UTC)

(edit conflict) For a one-component three-phase substance it depends on the pressure $P_{\text{tri}}$ of the triple point. A three-phase substance can only be in the liquid phase at pressures greater than $P_{\text{tri}}$. If it is heated at a lower pressure, it will go directly from solid to gas.  --Lambiam 21:35, 7 February 2024 (UTC)

True, but unless you already know the triple point pressure, you still can't just calculate it easily. --OuroborosCobra (talk) 23:12, 7 February 2024 (UTC)

Ordinary glass doesn't really have a crystalline phase, nor a melting point. At least, not one that can be practically demonstrated. It does have a glass transition. And most ordinary glasses are pretty liquid at 1000°C; no need to go all the way up to the melting point of silica. Which is good, otherwise glassmaking would've been impossible in antiquity.

The glass particles in the white smoke (which could also contain contaminants) must be fairly solid. If they had been fairly liquid, they would be glowing yellow.

Glass is also brittle. Quick localised heating can break off tiny particles without making them significantly liquid or heating them to incandescence. You should be able to tell when viewing the smoke particles under a microscope. PiusImpavidus (talk) 10:31, 8 February 2024 (UTC)

Couldn't the "smoke" also contain vapourized SiO₂ (boiling point 2,950 °C)?  --Lambiam 11:27, 9 February 2024 (UTC)

Classical Mechanics: I wonder where my mistake in the following calculation is:

0. (This setion was added later). Here, I'm only referring to theoretical Newtonian situations involving no potential energy (i.e. no chemical and nuclear particle bonds, nor physical force fields, nor deformation of solids).

1. Hence, a given system's kinetic energy is equal to the sum of the kinetic energies of the system's parts, regardless of whether the system consists of two bodies or two photons or whatever.

2. Let two bodies, having the same mass $m$ each, move by the same speed $v$ in opposite directions.

3. Hence, each body carries kinetic energy of ${\frac {1}{2}}mv^{2}.$

4. Hence, per #1, the whole system, consisting of both bodies, carries kinetic energy of ${\frac {1}{2}}mv^{2}+{\frac {1}{2}}mv^{2}=mv^{2}.$

5. On the other hand, per #2, the whole system is at rest, so it's kinetic energy must be zero.

I wonder how I can settle the contradiction between #4 and #5. I guess the mistake is in #1 or in #5, but I can't find it yet.

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 20:31, 7 February 2024 (UTC)

What is your definition of a system being "at rest"? The concept applied to a collection of objects is usually understood to mean that each object in the collection is at rest. For a fluid it means that the particles of which it exists are all at rest. (See Hydrostatics.) A spinning disc such as a gyroscope is not at rest, even though for each infinitesimal part having some velocity v there is a corresponding infinitesimal equi-massive part having velocity −v.  --Lambiam 21:01, 7 February 2024 (UTC)

Don't you agree, that a given collection of two photons, that move in opposite directions, is a collection at rest? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 21:42, 7 February 2024 (UTC)

You're confusing energy and momentum. Momentum is a vector quantity, so your two particles cancel each other out. The system has zero net momentum.

Kinetic energy is a scalar quantity, and so direction doesn't matter when you add it up. PianoDan (talk) 22:24, 7 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:44, 8 February 2024 (UTC)

1-4 only apply to the system's parts because the KE of any massive object is entirely dependent on its speed. To clarify, consider another example: A) A newspaper is at rest on someone's doorstep: v=0, m₀v=0, KE=0 and it has rest mass m₀. OK, and... B) Its molecules vibrate, thus: v_i>0, m_iv_i>0 and KE_i>0 for every ith molecule. Descriptions A and B differ in granularity, and because mass and energy are equivalent the sum of its internal energies E_i divided by the speed of light squared equals its rest mass m₀ even though it is still at rest on the doorstep (v=0, m₀v=0, KE=0), thus 5 is correct. Also the classical formula is an approximation and does not hold for photons. Clarifying 1 to say "...a given system's [internal] kinetic energy..." would distinguish it from the kinetic energy of 5. Modocc (talk) 00:54, 8 February 2024 (UTC)

you have discovered temperature. Greglocock (talk) 02:40, 8 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:57, 8 February 2024 (UTC)

More precisely, they've discovered internal energy (there's a connection to temperature, of course). Energy is a book-keeping device. Depending on the level of description of a system, energy can be booked to different accounts. The kinetic energy of the individual particles contributes to the internal energy (and total energy) of the two-particle system; in the rest frame of the two-particle system, it does not contribute to the kinetic energy. In a many-particle system one can distinguish between the mean motion of the particles (which make up a gas flow, for instance) and the random motion (which determines the temperature of the flowing gas). --Wrongfilter (talk) 07:01, 8 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:44, 8 February 2024 (UTC)

Number 4, which should be formulated as "the whole system ... has a total (or internal) energy of ..." (assuming there's no interaction between the particles). --Wrongfilter (talk) 07:53, 8 February 2024 (UTC)

Thank you. Are you sure, the section - which is the first one you don't agree with, is #4 - rather than #1? Note #4 directly derives from #1, so if you agree with #1 you apparently have to agree with #4. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:57, 8 February 2024 (UTC)

Yes, I missed that. So in #1 it should be "The systems total energy is the sum of the kinetic energies of its constituent particles". This is true if there are no interactions between the particles, and if there are no further internal degrees of freedom apart from the translational motion of the particles (translate as appropriate for photons, where the term "kinetic energy" is also not quite right). --Wrongfilter (talk) 08:14, 8 February 2024 (UTC)

Thank you. I've just added section #0 above, which excludes potential energies. Could you point now at the section which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 08:43, 8 February 2024 (UTC)

I've already said all I have to say. --Wrongfilter (talk) 08:46, 8 February 2024 (UTC)

New question: Do you agree that #1 directly derives from #0? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 08:50, 8 February 2024 (UTC)

I've told you what I think is wrong with #1, this is independent of your #0. Please don't make these threads endless. --Wrongfilter (talk) 08:55, 8 February 2024 (UTC)

I only wanted to be sure I understood well (which sometimes needs further clarifications). So, as I understand now (due to your last clarification in your last response), internal energy is still different from kinetic energy, despite #0. Thank you. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 09:02, 8 February 2024 (UTC)

February 8

How good could rayguns be at demolition if the possibly intractable problems were trivial?

Like if you pick the particle type, flux, beamwidth (at least .010 or .001 caliber) and energy per particle right then could a (probably impossibly low waste-heat) man-portable or even pistol-like photon beam or particle accelerator (that could convert much of its mass to beam) demolish as much stuff per second as a dude throwing grenades? Is there a beam that won't dump too much energy into the air, can cause tactical nuke-level damage a few kilometers away in 3 seconds and yet not kill the user before he can finish? Preferably you'd not want to harm the user more than current common infantry weapons (maybe trading less naked ear hearing loss for more radiation bouncing back?) or have too much recoil to aim well, what would be possible with that restriction? Would neutrons be better or worse than photons or charged particles? (I don't know if neutron accelerators are even possible) Sagittarian Milky Way (talk) 00:16, 8 February 2024 (UTC)

Particle accelerators as conventionally defined cannot accelerate neutrons, as they work by applying an electric field to a charged particle. You can't accelerate something with no charge using an electric field.

The way neutrons are generated for research is fascinating - check our article on Neutron sources.

As far as the rest of it goes, given the amount of heavy power infrastructure attached to even small medical accelerators, the idea of something man-portable seems unlikely. PianoDan (talk) 00:23, 8 February 2024 (UTC)

I know, I was just wondering how powerful a theoretical particle (including photons) black box could be before you'd be too close to target and/or beam and what kind of particle is that. Obviously there could be orders of magnitude destructiveness difference between "OSHA-compliant" and "not possible even as a suicide attack (unless you can make it hold down the trigger after you're dead or something like that)". Sagittarian Milky Way (talk) 00:36, 8 February 2024 (UTC)

If there was such a thing, then why is not in use by armies? Probably a small rocket propelled tiny tactical nuke could do what you say. But that is a big particle. Or fire antimatter bullets or pellets at the target. Otherwise generating beams from a portable device would have to be extremely efficient, not to overheat the user, and impart all energy to the target. So it is not feasible. Graeme Bartlett (talk) 12:36, 8 February 2024 (UTC)

Yes but many cool science-fiction things can already be done with a different current tech. Like the Independence Day city ray. I'm not sure if antimatter bullets would not cause user damage either as air is matter. Even if you had a force field to delay matter contact till leaving the barrel. I am not sure what kind of particle beam (including different kinds of photons) has the highest ratio of impart energy to target to impart energy to air between raygun and target and would like to know. Also some kinds of impart energy to air are worse then others, i.e. X amount of energy as ionizing radiation or boom sound is worse than the same amount as heat. Sagittarian Milky Way (talk) 18:00, 9 February 2024 (UTC)

You want something like this [6] but beefed up and stuck in a gun? There's far better weapons already for everything you say, a rocket is needed for the three seconds and nuke business but really you don't want that sort of energy being produced in something you hold! NadVolum (talk) 13:16, 8 February 2024 (UTC)

There's a reason that the atomic hand grenade has never caught on. {The poster formerly known as 87.81.230.195} 90.199.107.217 (talk) 06:08, 9 February 2024 (UTC)

Also it wouldn't fit in a hand grenade. Lots of suicide terrorists would be willing to use them though if they could get them. There is an atomic grenade launcher at the limit of feasible though. Would not meet OSHA regulations on blast nearness and recoil. Sagittarian Milky Way (talk) 18:05, 9 February 2024 (UTC)

Possibly you missed the fact that I was making a joke [clue: small type], the point of which was that, assuming one could make an atomic device that small, one would of course not be able to throw it far enough to avoid being killed by its detonation.

Grenades are battlefield weapons: a suicide mission would not require a device small enough to be throwable, one would merely carry a suitcase bomb to the target.

Fortunately, construction of such devices is probably beyond the capabilities of most terrorist organisations, but there remain the possibilities of a rogue state supplying them with one, one being stolen from a state's arsenal, or the alternative of a small dirty bomb which would be less immediately destructive but possibly even more effective in the longer term. {The poster formerly known as 87.81.230.195} 90.199.107.217 (talk) 03:37, 10 February 2024 (UTC)

There was a Davy Crockett (nuclear device) really inconvenient for one person but much less so than any chemical energy way of carrying 20 tons of TNT equivalent. 1.25 mile range. Sagittarian Milky Way (talk) 04:20, 11 February 2024 (UTC)

If the problems of making rayguns with excellent demolition capabilities were trivial, only trivial problems stand in the way to make rayguns with excellent demolition capabilities. Assuming we can overcome these trivial problems – which is a reasonable assumption, since otherwise the problems are nontrivial – the inevitable conclusion is that the rayguns could be very good, nay, indeed excellent, at demolition. I hope this answers the question conclusively.  --Lambiam 14:43, 12 February 2024 (UTC)

Yes but if the beam follows the laws of physics starting at the muzzle and so does any protection (goggles? earplugs and earmuffs? lead-backed volcanology suit?) then it becomes more and more annoying to use the higher the power (for any given beam specification, target distance etc), eventually it either kills you almost instantly or has so much recoil you can't aim or stand up. I'm wondering what's the most damaging kind of beam a human could hold (not necessarily most watts as neutrinos may or may not have a good enough damage to miles away to damage to user ratio, at the extreme fluxes that damagr even user regardless of target range (through Earth's atmosphere, line-of-sight surface-to-surface)). Maybe a good particle is a specific kind of photon with sufficiently long mean free path for interaction with dinitrogen, dioxygen, argon and as many other common air substances and pollutants as possible while having a good interaction mean free path in as many common target substances as air transparency allows? Also a low percent of energy reflecting or scattering from target or its vaporization cloud, and low vaporization cloud shielding. An issue with publicly known lasers is they're slower than if the cloud of vaporized surface was transparent to that wavelength of infrared. Sagittarian Milky Way (talk) 17:09, 12 February 2024 (UTC)

February 9

Neurotransmitters, part 3

Does prolactin counteract the effects of dopamine (either through suppressing its release, through blocking its signalling and/or through exerting an opposite effect on the brain)? Does it counteract the effects of oxytocin? 2601:646:8080:FC40:B8B0:B42B:BB0C:97BA (talk) 22:18, 9 February 2024 (UTC)

Oxytocin stimulates prolactin secretion, while prolactin, once a threshold is reached, stimulates dopamine release, which has an inhibitory effect on prolactin secretion (PMID 22129099, PMID 18477617).  --Lambiam 14:15, 10 February 2024 (UTC)

Thanks! And a related question: does prolactin then inhibit D4 receptors? (Yes, this has to do with my personal research about introversion -- I'm trying to see whether or not increased prolactin levels would be pleasurable for an introvert from the deep end of the scale!) Also, does prolactin either stimulate or inhibit oxytocin? 2601:646:8080:FC40:9D4E:C237:28BF:6A78 (talk) 03:59, 11 February 2024 (UTC)

Autism vs. introversion

Two questions in one: (1) how can they tell for sure between autism and just an extreme case of introversion (given that 3 of the 4 diagnostic symptoms, of which only 2 are needed to diagnose autism -- in fact, all of them except stimming -- can also be associated with introversion)? And (2) how does autism specifically affect oxytocin signalling in the brain? 2601:646:8080:FC40:B8B0:B42B:BB0C:97BA (talk) 23:14, 9 February 2024 (UTC)

As to (2), here is a review article, "Oxytocin and Autism Spectrum Disorders", PMID 28766270. Several studies have reported a tendency of lower oxytocin levels in children on the autism spectrum, but the causal connection is not clear. In summary, autism is not well understood, and more research is needed.  --Lambiam 14:25, 10 February 2024 (UTC)

Not sure why the two should be confused at all, they're very different. Being an introvert is something they would diagnose for? Do they diagnose being an extrovert as well? What is the opposite of autism if they think that way? NadVolum (talk) 20:41, 11 February 2024 (UTC)

Do you not think a psychiatrist would never mistake an extreme case of introversion (a normal condition) for autism and misdiagnose someone when in fact the person doesn't actually have anything to be diagnosed with??? After all, as I pointed out, 3 out of the 4 diagnostic symptoms for autism can also be observed in introverts -- and now that I think of it, even stimming might not always be a reliable symptom of autism, in some cases (although uncommonly) what is thought of as "stimming" might actually be just a reflexive pain response (because, in introverts, over-stimulation -- such as being at a big loud party -- can in some cases cause physical pain, such as a headache!) Not to mention that there's also the possibility of malicious diagnosis (although that's probably another topic for another time)! 2601:646:8080:FC40:8443:5BCE:DED0:5463 (talk) 03:27, 12 February 2024 (UTC)

You appear to be assuming that autism is an all-or-nothing condition; but Biology is complicated and messy.

It's often thought (rightly or wrongly) that everyone is on a "spectrum", with most (the "neurotypical") at various low values on it and some at higher values which correlate with greater interactional (with the material world and/or other people) difficulties.

If this is so, inevitably the threshold value for "having autism" (or whatever) will in part be a matter of subjective judgement that may vary between medical practioners, and will evolve as medical science advances. Moreover, some people (including myself) may consistently exhibit some cognition and behaviour which, if more pronounced, would qualify as autism. Lastly, I doubt that anyone scores exactly the same "spectral value" all the time. 90.199.107.217 (talk) 07:29, 12 February 2024 (UTC)

Right, and that brings us back to the same question: where exactly is the line between just strongly pronounced introversion on one hand and high-functioning autism on the other? 2601:646:8080:FC40:99FE:A33D:618C:8AB5 (talk) 11:51, 12 February 2024 (UTC)

As indicated, there isn't a line - see idiot savant. Reticence is not an indicator of anything - for example, an individual may not speak for weeks on end (apart from things like asking for a newspaper in a shop) but may be active on social media. 2A00:23D0:EF3:2001:F9CC:1EC8:88FD:4354 (talk) 14:24, 12 February 2024 (UTC)

The big difference is that with autism people tend to lack understanding of other people's feelings and have to follow rules to get on with them, whereas introverts have no special problem that way but feel general socializing is a drag. If 'strongly pronounced introversion' is leading to real problems for a person it would practically always be because of some other problem that has developed from it like being very anxious in social circumstances. NadVolum (talk) 18:15, 12 February 2024 (UTC)

OK, I see your point, but there's still room for ambiguity: suppose a hypothetical scenario with a person who's not only on the deep end of the scale in terms of introversion, but also has a high score for assertiveness (unlikely because introversion tends to correlate with low assertiveness, but still possible) and low scores for cooperation and sympathy -- might a person with that combination of personality traits not be misdiagnosed with autism without actually having it? Or is this combination of personality traits in itself diagnostic of autism? 2601:646:8080:FC40:A0EA:E55B:2645:FEAF (talk) 03:14, 14 February 2024 (UTC)

Lots of company heads are like that. They can cope with meetings fine because they're achieving something - and they'll push to get things done instead of just waffling. Getting people to work with you when you don't understand feelings is a lot more difficult though. And understanding peoples feelings doesn't necessarily mean having problems firing them if needed and that can be helped by not getting too friendly. NadVolum (talk) 13:51, 14 February 2024 (UTC)

Socialising is an integral part of our lives. While some of us have no issues around it, for many it can be daunting. Social anxiety can cause symptoms such as feeling sick, trembling, or even dizziness. It is estimated that 12 per cent of the general population is diagnosed with the disorder at some point in their lives. If you are struggling and not sure how to get help, my support pack Social Anxiety has lots of advice. - Deirdre, Sun on Sunday, 28 January 2024. [7]. 2A00:23D0:5D3:5601:71F9:9EF3:5186:F60D (talk) 16:05, 15 February 2024 (UTC)

Planck's law 1901 article conundrum

In my personal popularization of Planck's 1901 paper on Planc's law, I have a sticking point. I am ok up to the equation (9):
(9)     ${\frac {1}{\vartheta }}={\frac {DS}{DU}}$
But between:
$S=k${$(1+{\frac {U}{h\nu }})log(1+{\frac {U}{h\nu }})-{\frac {U}{h\nu }}log{\frac {U}{h\nu }}$}
And
Substitution in (9) gives:
${\frac {1}{\vartheta }}={\frac {k}{h\nu }}log(1+{\frac {h\nu }{U}})$
Between these 2 equation I don't find?
(After to the end it is ok)
I arrive to this point:
$S=klog(1+{\frac {U}{h\nu }})+{\frac {kU}{h\nu }}log(1+{\frac {h\nu }{U}})$
Apparently, that suppose:
$S-klog(1+{\frac {U}{h\nu }})={\frac {U}{\vartheta }}$
But I'm stuck there and it is not clearer in Planck's following manuscripts.
Any idea ?
Malypaet (talk) 23:18, 9 February 2024 (UTC)

I think you've made a math error in there somewhere. Taking the derivative of S with respect to U gives:

${\frac {1}{\vartheta }}=k[{\frac {1}{h\nu }}ln(1+{\frac {U}{h\nu }})-{\frac {1}{h\nu }}ln({\frac {U}{h\nu }})]$

Which only takes a little poking with log identities to yield the desired result. PianoDan (talk) 02:27, 10 February 2024 (UTC)

An error, sure.

But how do you go from Planck:

$S=k${$(1+{\frac {U}{h\nu }})$$log(1+{\frac {U}{h\nu }})-{\frac {U}{h\nu }}log{\frac {U}{h\nu }}$}

to your formula "taking the derivative of S with respect to U"?

${\frac {1}{\vartheta }}=k[$${\frac {1}{h\nu }}$$ln(1+{\frac {U}{h\nu }})-{\frac {1}{h\nu }}ln({\frac {U}{h\nu }})]$

You cannot simply replace $S$ with ${\frac {U}{\vartheta }}$ to get your result, isn't it?
Malypaet (talk) 09:29, 10 February 2024 (UTC)

I think I've asked before: Do you know what the derivative of a function is and how to compute it? --Wrongfilter (talk) 09:36, 10 February 2024 (UTC)

I only ask to relearn, Planck integrates where pianodan takes the derivative? Malypaet (talk) 11:17, 10 February 2024 (UTC)

Planck isn't integrating here either. PianoDan (talk) 17:32, 10 February 2024 (UTC)

${\frac {1}{\vartheta }}={\frac {{\text{d}}S}{{\text{d}}U}}$

$~~~~={\frac {\text{d}}{{\text{d}}U}}~k\!\left((1+{\frac {U}{h\nu }})\log(1+{\frac {U}{h\nu }})-{\frac {U}{h\nu }}\log {\frac {U}{h\nu }}\right)$

$~~~~=~...$

 --Lambiam 14:39, 10 February 2024 (UTC)

$=k[(0+{\frac {1}{h\nu }})log(1+{\frac {U}{h\nu }})-{\frac {1}{h\nu }}log({\frac {U}{h\nu }})]$

$=\cdots$

Many thanks Lambiam, that’s all I was missing.
Now with Planck's 1901 paper, for me everything is clear from start to finish and in detail.

I will be able to simplify it and make it more realistic, without resonators.

(°—′)
Malypaet (talk) 18:51, 10 February 2024 (UTC)

February 10

Rotating a magnetic field

I've been thinking about a superconducting electric coil with some current, creating a stable magnetic field. Now assume an axis through the middle of that coil. My intuition tells me that rotating the coil around this axis should be inconsequential. But assume an axis perpendicular to the first one. If I rotate the coil around this axis (and, I assume, the magnetic field with it), a distant observer would see a regularly fluctuating magnetic field - and, I again assume, because auf the unity of electric and magnetic fields, this would appear as an electromagnetic wave. Is this correct? If so, where is the energy carried by the wave coming from? Is the magnetic field dissipating? Or do I need to put energy into the system to keep rotating the coil? Thanks! --Stephan Schulz (talk) 13:53, 10 February 2024 (UTC)

Assuming that the setup indeed creates a changing magnetic field, by Lenz's law this induces an opposing current which (as in an induction brake) should counteract the rotation.  --Lambiam 15:00, 10 February 2024 (UTC)

An ordinary permanent magnet is a cheaper and more convenient alternative to a super-conducting coil. Spinning the magnet about its transverse axis will induce eddy currents in any nearby metal objects, which will oppose the rotation of the magnet and remove energy by resistive heating of the metal - this is induction braking. In the absence of any nearby metal objects, then there will indeed be braking associated with the energy and angular momentum carried away by the radiated field. However it is necessary to spin the magnet very fast or be very far from metal for the radiation braking to be significant compared with the induction braking, requiring something of the order of ω = c / d where ω is the angular frequency of the rotation and d is the distance from metal. catslash (talk) 18:09, 10 February 2024 (UTC)

Thank you - and of course. I should have thought of that! --Stephan Schulz (talk) 05:07, 11 February 2024 (UTC)

An example of such a strong, rapidly spinning permanent magnet with no metal anywhere near would be a pulsar. There is, however, some interesting plasma physics happening in the pretty good vacuum of the near field, so although there are no metals, there are electric currents. And pulsars do spin down. PiusImpavidus (talk) 11:57, 11 February 2024 (UTC)

February 11

Collision analysis

A bus of mass 25 tons starts to travel along a straight road from A to B at 50 mph. At the same time a bee of mass 2.5 grams starts to fly from B to A along the same road at 5 mph. Eventually the bee and the bus collide.

Assume the bee impacts a surface of the bus which is flat and at right angles to its direction of travel, such as the windscreen, and the bee sticks to this surface.

Then the bus will be slowed by a tiny amount reflecting the relative masses / momenta of the bee and bus. In addition, relative to A, the velocity of the bee changes from -5 mph to +50 (approximately) mph during the collision. So at some stage during the collision the bee must be stationary. This can only be when the bee is in contact with the bus.

The question is : Why is the bus not stationary at the same time?

Does it make any difference to the situation if the bee is replaced by either a perfectly elastic object, or a perfectly rigid object, of the same mass?

And does it make any difference if the collision is considered at the level of the individual atoms involved? Ionlywanttoknow (talk) 18:40, 11 February 2024 (UTC)

If you define the direction from A to B as positive, the bee is changing from a speed of -5 mph to +50 mph. Regardless of the exact nature of that change, in order to get from -5 to +50, at some point the speed must pass 0.

The bus's speed is changing from +50 mph to +49.999999... mph. That change does not pass through zero, so the bus is never stationary.

You could get into more details of the interaction, but that's the basic point. PianoDan (talk) 20:11, 11 February 2024 (UTC)

There's two explanations. If you're thinking of absolutely rigid bodies then the bee is never stationary, it immediately switches between going in one direction and the other. In the physical world there is always an bit of elasticity so the bus atoms can keep going forwards whilst the bees ones slow down and reverse as the bee is squashed against the windscreen. NadVolum (talk) 20:30, 11 February 2024 (UTC)

This follows from the Intermediate value theorem. Ruslik_Zero 20:35, 11 February 2024 (UTC)

There is no intermediate value of speed in the case of rigid bodies. NadVolum (talk) 21:06, 11 February 2024 (UTC)

Here is the argument: The intermediate value theorem tells you that the speed of the bee is zero at some points in time. By assumption, the bee and the bus move at the same speed when they are in contact. So the bus must move at speed zero at that point in time. So far, so good. But we are talking about points in time. Speed, being distance per time, is not well-defined for an individual point. You need an "expanse of time" to have speed. --Stephan Schulz (talk) 10:20, 12 February 2024 (UTC)

The intermediate value theorem is valid for continuous functions. For rigid bodies, the speed is discontinuous (acceleration infinite), so the IVT does not apply. As was said above, real bodies are always elastic to some point, which causes gradual changes of the speed and keeps the accelerations finite. --Wrongfilter (talk) 12:06, 12 February 2024 (UTC)

Why is the bus not stationary? Because it is moving at 50 mph! For the bus to decelerate from 50 mph to stationary in the brief duration of this collision would take a very, very large force. So large it could not possibly occur in a collision with a small object. Dolphin (t) 06:19, 12 February 2024 (UTC)

The meaning of "why" in the question is unclear. However, one of many ways to see that the bee+bus system cannot be stationary is that its kinetic energy would be zero, violating the law of conservation of energy. In the idealized analysis, the bee is treated as if it is a point particle. In a more refined analysis, the collision takes some time; the drama unfolds in about half a millisecond. Halfway through the process, the front part of the insect has already been squashed and splotched across the bus, co-moving with it at almost 50 mph. Shockwaves through its body may have caused parts of its now liquefied internals to erupt through its rear, moving even faster than the bus. Other parts of the bee are still moving in the original direction, towards the bus. If we zoom in to the elementary particle level, we run into the fundamental limit of quantum uncertainty: we cannot assert meaningfully that any part of the bee is stationary at any time.  --Lambiam 13:30, 12 February 2024 (UTC)

By the way you might like to read Windshield phenomenon about why this hasn't been happening much recently. NadVolum (talk) 17:01, 12 February 2024 (UTC)

According to the reincarnated bee (murmuring something about local inertial frames not accelerating) it was the bus that stopped and the road reversed and took the bus+bee to the land of honey. Modocc (talk) 14:18, 13 February 2024 (UTC)

February 12

Hospital Gangrene

Is there is Wikipedia on Gangraena nosocomialis or hospital gangrene, predominant up until about 1870-1880. I can't seem to locate anything although it might on under some different name. scope_creep^Talk 09:29, 12 February 2024 (UTC)

Here ? Heihaheihaha (talk) 10:29, 12 February 2024 (UTC)

Hospital gangrene redirects to Necrotizing fasciitis. The historical use of the term can perhaps not be precisely related to current terminology, including many cases that could have been treated non-surgically by suitable antiseptic wound dressing and administering antibiotics if they had been available at the time.  --Lambiam 13:08, 12 February 2024 (UTC)

Strange "mitotic phase" of oocytes in rabbit's primary follicle

Two months ago, I posted a question on StackExchange but no one answered me there. It's about a primary follicle (Inferred from the number of layers of granular layer cells), whose oocyte seems to be undergoing division, which is supposed not to happen at this stage by my textbook. I've asked my histology teacher and got no answer. My initial idea was whether it is possible to have multiple oocytes simultaneously in the follicles of rabbits. I've tried to google and searche on PubMed, I didn't find relevant evidence there. The full size picture is provided here, which may provide more information.

The question is: how to explain this phenomenon? I did not observe this phenomenon in other many slices. Heihaheihaha (talk) 10:26, 12 February 2024 (UTC)

Interesting. Following. Zarnivop (talk) 10:21, 13 February 2024 (UTC)

Rabbits have a couple of wrinkles not found in all mammals, induced ovulation and embryonic diapause, could these be related? Abductive (reasoning) 12:20, 13 February 2024 (UTC)

February 13

UV index

This thing interests me, and I don't know it (it's not homework): in which time of the year is the UV index highest in any given place? Is it arund the summer solstice, which is in June in northern hemisphere and in December in southern hemisphere, or is it in July in northern hemisphere and in January in southern hemisphere, when temperatures are highest? --40bus (talk) 15:58, 13 February 2024 (UTC)

See Ultraviolet index. I expect the UV Index at any geographical position to be at its annual highest value when the sun is at the greatest elevation - the summer solstice. Dolphin (t) 16:27, 13 February 2024 (UTC)

But the high-latitude ozone hole peaks in early spring, in low latitude the Sun height changes of Earth's axis cause smaller incoming solar radiation changes than elsewhere cause trigonometry so the strong 1-year wet and dry season cycle and milder seasonal ozone strength cycle might overpower the small axis tilt effect and make it not the time of highest noon Sun (which is not the solstice less than 23.44 degrees from the Equator. Sagittarian Milky Way (talk) 18:00, 13 February 2024 (UTC)

Sun Safety Monthly Average UV Index 2006-2023 from the United States Environmental Protection Agency shows similar results for June and July. Alansplodge (talk) 12:41, 15 February 2024 (UTC)

Some relevant considerations also are to be found in that article Record solar UV irradiance in the tropical Andes. As can be expected, peak values were measured between the end of December, and January (2004). --Askedonty (talk) 13:38, 15 February 2024 (UTC)

Peak rain time is afternoon, Andes might often pierce the clouds, the ITCZ usually lags the Sun zenith latitude and it only has to be sunny in late December once to be a record. Sagittarian Milky Way (talk) 14:12, 15 February 2024 (UTC)

Yes, that's why the team hit bonanza. But no residual glacier, which is a course other latitudes tend to be following lately [8]. --Askedonty (talk) 14:52, 15 February 2024 (UTC)

Also the Sun is closest to Earth in early January. Sagittarian Milky Way (talk) 14:57, 15 February 2024 (UTC)

However the zenith noon latitude is only December solstice at the Tropic of Capricorn, slightly norther it'll be zenith twice a year and the early January one will be at the nearest part of Earth's orbit but not even get peak doldrums. Also it's dry as fuck there. Atacama Desert, nothing to block UV not even tropospheric ozone pollution. Sagittarian Milky Way (talk) 15:05, 15 February 2024 (UTC)

In NYC the summer month percent of possible sunshine is 57 59 63 (frontal thunderstorms decline after June, frontless land heating thunderstorms likely peak July). Avg UV index 7 8 8 8 6 MJJAS according to Wikipedia, maybe enough of the country is similar to make June and July the national average peak. Sagittarian Milky Way (talk) 14:19, 15 February 2024 (UTC)

Wilson-Effect

Does the Wilson-Effect apply to other properties or just hight (how large somebody is)? 2A02:8071:60A0:92E0:C1E2:268B:CB67:A6BA (talk) 22:07, 13 February 2024 (UTC)

What Wilson effect? There's one concerning sun spots: Wilson effect. There's one concerning inheritability of IQ: [9]. You seem to be talking about yet another one. Please try to ask a proper question. --Wrongfilter (talk) 22:22, 13 February 2024 (UTC)

This refers to an increase in heritability of IQ with age, not of height:

Thomas J. Bouchard (October 2013). "The Wilson effect: The increase in heritability of IQ with age". Twin Research and Human Genetics 16(5):923–30. PMID 23919982

Reportedly, looking at a proxy for general cognitive ability instead of IQ, some researchers found the opposite effect for the age bracket of 50–69 years:

Matthew A. Sarraf, Michael Anthony Woodley of Menie, Mateo Peñaherrera-Aguirre (February 2023). "The anti-Wilson effect: The decrease in heritability of general cognitive ability, as proxied by polygenic expressivity, with advanced age". Personality and Individual Differences 202:111969. doi:10.1016/j.paid.2022.111969

 --Lambiam 09:22, 14 February 2024 (UTC)

I've known about that for a long time and am surprised it still is just a citation in Heritability of IQ and not even named there. Yes I would expect height to follow a similar pattern but haven't seen anything about it. NadVolum (talk) 14:12, 14 February 2024 (UTC)

Children develop at different rates for all sorts of reasons, including non-genetic reasons like nutrition. So children of a particular age are more variable in IQ and height than older age groups that have reached full adulthood. This is what causes the heritability to be higher in older age groups; there is less non-genetic variation, so the proportion of total variation explained by genetic variation is higher. Hence I would expect the Wilson effect to exist in most traits that show developmental changes with age during childhood. A character like baldness might show the opposite pattern. JMCHutchinson (talk) 22:18, 14 February 2024 (UTC)

February 15

Is salted whipped butter more friable and/or sticks less to bread?

If so then why? Solid pieces of unsalted whipped butter seem to fall off less often.Sagittarian Milky Way (talk) 01:06, 15 February 2024 (UTC)

Than what? Salted whipped butter? Unsalted unwhipped butter? Salted unwhipped butter? {The poster formerly knwn as 87.81.230.195} 176.24.45.226 (talk) 12:06, 15 February 2024 (UTC)

Salted whipped butter vs unsalted whipped butter Sagittarian Milky Way (talk) 14:22, 15 February 2024 (UTC)

Take it with a grain of salt: my short answer without references. When pressed and/or placed onto bread, some of the butter liquefies and is wicked into the bread giving it additional contact and surface tension (see Capillary action) causing the butter to stick. At the point the bread becomes saturated, the butter tends to slide off because a fluid film forms between the saturated bread and the intact butter solids. Added salt suppresses the melting points of solids such as ice, so it likely significantly speeds up the end result.

Modocc (talk) 14:50, 15 February 2024 (UTC)

Taken unsalted, your answer may stick better.  --Lambiam 17:17, 15 February 2024 (UTC)

Magnetic compass

Not homework but I'd like to know how to answer this at the level of an introductory E&M physics class or that sort of thing. Basically a magnetic compass is a magnetized needle with a pivot in the middle, sitting in the Earth's magnetic field. The needle has mass M and length L and I guess we can ignore most subtleties.

My question is, how do you calculate the torque around the pivot, at least dimensionally? My first thought was that it would be quadratic in L (by integrating along the needle) but maybe that's wrong, and I just don't understand magnets well enough.

Oh yes, I guess the needle material itself needs to have some physical magnetization parameters specified. How would I find those, for whatever permanent magnet material is generally used in not-fancy compasses? Do fancy ones use fancier materials like rare earth magnets?

Motivation for asking: if I get a small cheap compass, say 1 inch in diameter, it will tend to get stuck easily, because the torque from the magnet isn't enough to overcome the friction in the pivot. Compasses with better (lower friction) pivots cost more. If I get one of similar quality that's 2 inches diameter, it will have 2x the friction in the pivot (because the needle is twice as heavy) but I wondered if it would have 4x the torque, similar to a moment of intertia calculation. That is, I'm wondering whether big cheap compasses work better than small cheap compasses.

Someday I'll try to work through a textbook on this magnetism stuff. Thanks. 2601:644:8501:AAF0:0:0:0:2F14 (talk) 03:55, 15 February 2024 (UTC)

The torque depends on the magnetic moment, which for a permanent magnet is the remanence times the volume (I think; you should read the articles to check). catslash (talk) 11:13, 15 February 2024 (UTC)

That's

${\boldsymbol {\tau }}={\frac {\text{vol}}{\mu _{0}}}\mathbf {B} _{\text{rem}}\times \mathbf {B} _{\text{earth}}$

catslash (talk) 12:08, 15 February 2024 (UTC)

I asked about this in 2019 and didn't get a good answer. Sagittarian Milky Way (talk) 14:29, 15 February 2024 (UTC)

Since the earth's magnetic field and the permeability of free space are outside our control at present, it seems we just need to maximize the volume of the compass magnet and the remanence of it's material. catslash (talk) 14:43, 15 February 2024 (UTC)

February 16

Tertiary color

There are 3 primary colors and (3*2)/2 = 6/2 = 3 secondary colors. Then the number of tertiary colors should be (3*2*1)/6 = 6/6 = 1. Wikipedia's article List of colors by shade says that in theory this color should be black, but that in practice it is brown because blue pigments are so weak. Why is blue so weak?? Georgia guy (talk) 17:09, 16 February 2024 (UTC)

Color is such an interesting topic, because there are so many different ways to approach this complex subject!

If we take a very bland and scientific approach - like the one you might find in the main article about color mixing - you'll see a different (and more precise) way of describing the phenomenon. It also includes a link to a few useful citations.

We have to be careful to distill your question to its core, practical application, and to make sure we lay out some reservations: the observation is not universally true; but it manifests in some examples like mixing paints (especially the kind of paint you might buy in an art store). I'm trying to avoid weasel-words, and also trying to avoid being unnecessarily technical here, but it's important that we're scientifically accurate!

If somebody - like an artist who mixes paint - describes "blue" as "weaker" (... in the context of how it qualitativelt affects the outcome when mixing paints) - then they are probably dancing around the topic of opacity as it affects mixing pigments.

If this is what they mean, we can study scientific explanations for why the paint has this particular opacity. I'm reluctant to use a word like "weak" or "strong" here - we might disagree on which word is more apt - but it's helpful to know that most paints (like the ones you find in an art store) contain a chemical mixture of pigment in a binder (or sometimes a solvent). The material is not "pure" pigment - it's a mixture. The color of the mixture can be made more- or less- opaque by adjusting the concentration of the pigment. Bafflingly, we might say that blue pigments are stronger, so paint companies can use less pigment in their paint to obtain a qualitatively equivalent amount of color , ... which makes for a bizarre reversal of qualitative language use..., which is confusing, and it's why we probably shouldn't use words like "weaker" or "stronger" when we're being scientific in our discussions about color.

All this aside, it would be misplaced if I attacked the fundamental premise here (as somebody who spends a lot of time scientifically studying color, and human perception of color, it's very tempting to shout ..."there's no such thing as a primary or secondary color...!" But ... human perception is involved here, so ... it's more complicated than that, too!) Rather, I think what I would say is - "this simple model of color-primaries is useful only in very simplified cases, like kindergarten-level paint-mixing examples." We do not have to work very hard to find examples where the simple model is insufficiently detailed to describe behaviors that are plainly visible to (most) everyone! In fact, you found such an example!

Nimur (talk) 17:32, 16 February 2024 (UTC)