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Polydactyly happens. But do people ever grow a finger with an extra joint? —Tamfang (talk) 05:44, 15 September 2024 (UTC)[reply]

Googling finger with an extra joint the first hit is our very own article Triphalangeal thumb. The first page of search results only mentions an extra crease on the little finger, with no underlying extra bones. But scroll down on the results, maybe you'll get lucky! 85.76.83.87 (talk) 14:54, 15 September 2024 (UTC)[reply]

Take a look at this video from a YouTube channel I follow. Steven the seagull injured her (yeah, Steven turned out to be female, but the name stuck anyway) foot somehow. Accident, fight, attack by a predator - whatever happened, the webbing between her toes got split. Does anyone know if that will grow back eventually? Iloveparrots (talk) 21:44, 15 September 2024 (UTC)[reply]

From Regeneration (biology) § Aves (birds): Owing to a limited literature on the subject, birds are believed to have very limited regenerative abilities as adults. I was able to find an issue of a waterfowl newsletter with a picture of a duck's foot webbing, apparently torn and partially regrown. Based on this, I think it's very likely to grow back somewhat, but perhaps less likely to grow back completely. I'm rooting for Steven though! jlwoodwa (talk) 01:52, 16 September 2024 (UTC)[reply]

She's been getting her foot doused with antiseptic and eats a REALLY good diet for an urban gull (fresh fish, fresh meat, mealworms daily), so she's in a better position than most. Iloveparrots (talk) 13:37, 16 September 2024 (UTC)[reply]

Presumably it could only heal if the split was mechanically sealed first. If the surfaces keep moving relative to each other or they get dirty I don't see how the split can heal. Shantavira|feed me 08:23, 16 September 2024 (UTC)[reply]

Reminds me of Paul Temple's wife! 2A00:23C5:E161:9200:D500:7967:3BC2:6E0B (talk) 10:04, 16 September 2024 (UTC)[reply]

I was thinking that it might grow back from the bottom outwards. Not really sure how it works in gulls, compared to us, if we split that thin bit of flesh between thumb and forefinger. Iloveparrots (talk) 13:34, 16 September 2024 (UTC)[reply]

My experience with wildlife rehabilitation, including occasional avians and waterfowl, but not gulls, suggests to me that it will not substantially regrow. In most animals with thin membranes of this sort, post-developmental regeneration is extremely limited once the semi-vascularized tissues are torn; the cells are of differentiated types which simply are not capable of interacting in the manner necessary to reconstruct the overall structure, instead prioritizing more localized closure of the wound site. In short, Steven will probably see an uneven healing pattern around the periphery of the torn segment with some thick scarification. I wouldn't expect more than 5-10% of the gap to fill. That said, while webbing is obviously a helpful adaptation for gulls, tears of this sort are not uncommon and typically not life threatening (either through initial infection or dehabilitation). Considering that Steven seems to have a much more robust support network than your average gull and lives in sheltered semi-urban, semi-natural conditions, she can probably be expected to have a fair long life by gull standards. Please feel free to update me about the foot if my conjecture proves wrong: I'd be interested in the outcome. S n o w Rise let's rap 00:22, 22 September 2024 (UTC)[reply]

HOTmag (talk) 13:05, 18 September 2024 (UTC)[reply]

Quarks are the only known particles whose electric charges are not integer multiples of the elementary charge. Therefore, in physical theories that accept both the Standard Model and the law of charge conservation, a quark cannot turn into another particle but a quark. But the types of quarks all have different masses, so all such quark–quark changes violate the law of conservation of mass. --Lambiam 17:57, 18 September 2024 (UTC)[reply]

If you are referring to a single elementary particle, so why didn't you mention the electron, besides the quark?

If that's because an electron colliding with a positron turns (together with the positron) into a pair of photons, then also a quark colliding with an anti-quark turns (together with the anti-qurk) into a pair of gluons.

Anyway, in my question I allow a given elementary particle to collide with its anti-matter for becoming another elementary particle.

More important: My question is theoretical, so it's not only about known particles, but rather about all possible particles, including those which haven't been discovered yet. HOTmag (talk) 18:21, 18 September 2024 (UTC)[reply]

If someone claims all swans are white, it suffices to debunk the claim by finding one purple swan. Maybe there also blue, brown or black swans, but it is not necessary to search for further counterexamples. Likewise, if some physical theory claims every elementary particle can turn into some other elementary particle, it suffices to debunk the claim by finding just one elementary particle that cannot turn into some other elementary particle. I just started with the top line of File:Standard Model of Elementary Particles.svg. There may be many other counterexamples (like the Higgs boson), but why bother to keep searching?

The as of yet undiscovered bunkon and trashon, whose properties are still unknown except that they are postulated to be different elementary particles, can turn into each other. A difficulty in finding them is that their properties are unknown, so experimental physicists don't know where to look. There may be many more such pairs, which may never be discovered. --Lambiam 10:17, 19 September 2024 (UTC)[reply]

By mistake, I thought you meant the quark was the only particle that couldn't turn into another particle but a quark, but now I see this was not what you meant, so I take my first sentence back.

However, I still emphasize that my question allows a given elementary particle to collide with its anti-matter for becoming another elementary particle.

Re. your senetnce: "The as of yet undiscovered...different elementary particles, can turn into each other": Is there any physical theory claiming what you've claimed in that sentence? Actually, this was my original question... HOTmag (talk) 10:44, 19 September 2024 (UTC)[reply]

I'm pretty sure that that sentence was a joke. Look at the names: BUNKon and TRASHon. --User:Khajidha (talk) (contributions) 11:25, 19 September 2024 (UTC)[reply]

No "maybe" about black swans. They were recorded by Europeans in 1697, possibly earlier. 2A00:23D0:F6F:1001:2D07:A712:8909:7D91 (talk) 11:56, 19 September 2024 (UTC)[reply]

All right. HOTmag (talk) 12:35, 19 September 2024 (UTC)[reply]

See also black swan and black swan theory. -- Jack of Oz [pleasantries] 18:37, 19 September 2024 (UTC)[reply]

OP may have in mind something more like the particles created from particle-antiparticle annihilation, such as those in the chart at Annihilation § Electron–positron annihilation, as opposed to something like the weak decay of quarks.

It seems to me that OP could mean either: 1) a single elementary particle can spontaneously become another single elementary particle (with the help of another particle that remains unchanged), in which case I think the answer may be no for any particle; or 2) for two given particles, there's an interaction in some condition where it's meaningful to say that one specified particle is in the input, and it becomes in the output the other specified particle [Edit: which may include any number of other particles in the reaction doing anything else]. Not sure (I didn't do particle), but I think (2) might be considered more or less accurate (to the extent the fuzziness of the wording necessarily allows). SamuelRiv (talk) 16:10, 19 September 2024 (UTC)[reply]

I adopt your option 1# if it means Particle decay, and I adopt your option 2# if it means annihilation (i.e. by colliding with the anti-particle).

But contrary to the way you divided you answer: option 1# for all particles, or option 2# for all particles, I didn't exclude a third option which is: option 1# for some particles, and option 2# for the rest of the particles (without excluding particles that satisfy both 1# and 2#).

All agree, that some particles satisfy option 1#, and that some particles satisfy option 2#.

My question is about whether there is any physical theory claiming, that every particle (including any particle that hasn't been discovered yet), satisfies either 1# or 2# (or both). HOTmag (talk) 10:25, 20 September 2024 (UTC)[reply]

I was trying to interpret your question literally. #1 is one single particle becoming one different single particle with everything else unchanged -- this does not happen at all afaik, nor in general in theory (User:Lambian gives a simple example for quarks in their answer above). #2, the way I worded it, afaik can (and does) happen for all particles in the Standard Model.

What I'm trying to convey with this #1/#2 description is that it's not a particularly meaningful one, if you can claim "every elementary can turn into some other elementary particle" just by comparing one reactant to one product in a complex interaction.

(As maybe a sorta-ok example, consider the chemical reaction of a strong acid + base into salt + water: HCl + NaOH -> NaCl + H2O. Would you say the HCl (reactant) turns into salt? turns into water? Or does it change nothing at all because both the reactants and products are largely remain just free ions in aqueous solution? This is why I'm not sure what you're trying to ask is very meaningful.) SamuelRiv (talk) 17:24, 20 September 2024 (UTC)[reply]

I'm not asking about a single particle becoming one different single particle with everything else unchanged. Let's put it this way: Is there any physical theory claiming that all particles in the dark matter can turn into other particles? (whether by a decay or by annihilation or by any other way). HOTmag (talk) 01:30, 22 September 2024 (UTC)[reply]

There can never be such a theory. Why? Dark matter is named so because it never interacts with ordinary matter, except by gravity. As soon as it interacts in any other way it's not dark matter anymore but a new kind of ordinary matter. By interaction by gravity no distinction of particles is possible, only sum and distribution of mass is measurable. So dark matter can decay all it want, no human physicists are able to prove it or disprove it. By definition of the word. It may be that some particles, that we now subsume in dark matter, are later discovered to be not dark. And then we would know the conditions where they participate in the normal decay of ordinary matter. But that can never tell us about the real dark matter, as long as there's real dark matter. 176.0.153.105 (talk) 18:20, 23 September 2024 (UTC)[reply]

Most particles decay into other particles. The only "stable" ones are Electrons, Protons, Photons and (to a degree (see here)) Neutrinos. But even these can (under the right conditions) either combine or "destroy" and "create" with each other. 176.0.165.39 (talk) 12:20, 20 September 2024 (UTC)[reply]

These interactions can all be depicted in a Feynman diagram as lines meeting in a vertex. The lines correspond to particles. In a Feynman diagram, a vertex is always the meeting point of three lines. A particle → particle change would correspond to a Feynman diagram in which just two lines meet in a vertex. --Lambiam 22:47, 20 September 2024 (UTC)[reply]

Or two lines merging in two vertices that are connected by two lines in the form looking like an eye. 176.0.153.105 (talk) 14:24, 23 September 2024 (UTC)[reply]

Yes, I know that. but you're talking about partciles of the Standard Model, while I'm asking about a theory that claims that all particles, including those which haven't been discovered yet, can decay turn into other particles. HOTmag (talk) 01:30, 22 September 2024 (UTC)[reply]

In your question and first response you talked about a particle that "can turn into some other elementary particle", and people are trying to clarify what that can mean. But now you're asking about decay: "Particle decay" is where only one particle goes in, and some other number of different particles come out, but as others have said there are stable particles that do not decay. (There are of course experimental bounds to what we currently know of this, and there are interesting subtleties in the theory for example why a photon does not decay.)

There's no theory which can claim anything meaningful about all particles that have not been discovered yet. A theory predicts new particles, and the theory becoming successful may lead to building experiments to verify empirically the particles ("discover" them, although a "discovery" was equally done when the theory was written).

You could imagine another type of theory that might say all particle physics theories are really at a fundamental level part of X-theory, and in X-theory everything decays into X-dust in 20 billion years, but that's not a particle theory. SamuelRiv (talk) 05:48, 22 September 2024 (UTC)[reply]

Re. your first paragraph: I'm sorry for not being clear in my recent response. Thanks to your comment, I will make it clearer, as I've alraedy made it in my first post.

Re. your second paragraph: I mean, something like the supersymmetric theory, claiming something about all particles, including those which haven't been discovered yet. So, again, I'm asking whether there's a theory that claims that all particles, including those which haven't been discovered yet, can decay turn into other particles, whether by a decay or by annihilation or by any other way. HOTmag (talk) 12:45, 22 September 2024 (UTC)[reply]

Although the above discussion may imply that the words "hypothesis" and "theory" can be used interchangeably, a scientific hypothesis is not the same as a scientific theory.

An answer is No, there is no scientific theory that just claims anything. A scientific theory offers a generalized explanation of how nature works described in such a way that scientific tests should be able to provide empirical support for it, or empirical contradiction ("falsify") of it.

The OP's question using the word "theory" to mean a claim that at best will remain unproven or speculative actually looks for a Hypothesis meaning an educated guess or thought about something that cannot satisfactorily be explained with the present scientific theories. The OP seems to be thinking aloud[1] a new hypothesis. Philvoids (talk) 10:37, 22 September 2024 (UTC)[reply]

See Supersymmetric theory. It claims something about all particles, including those which haven't been discovered yet, and it's still called a "theory". Anyway, I'm not focusing on terminology but rather on an idea: Is there any theory, or a well known hypothesis, or a well known conjecture, or whatever, claiming that all particles, including those which haven't been discovered yet, can turn into other particles, whether by a decay or by annihilation or by any other way. HOTmag (talk) 12:45, 22 September 2024 (UTC)[reply]

Wikipedia editors are careful about terminology. By "Supersymmetric theory" you link to the article titled Supersymmetry. Its first line clarifies that it refers not to a theory but a theoretical framework. Read further to see how it anticipates what might characterise a supersymmetry theory without specifying any one for attention. Merely calling a supposed bosonic superpartner to the electron a selectron hardly amounts to a falsifiable hypothesis and obviously says nothing about undiscovered particles. Beside our care with terminology, we are not in the business of prediction. Philvoids (talk) 09:17, 23 September 2024 (UTC)[reply]

I've linked to our article [page] Supersymmetric theory, which does exist in Wikipedia, so when I used the term "supersymmetic theory" I used a term used in Wikipedia as well. Indeed, it passes to the article Supersymmetry, but also this article does point out - in its second paragraph - that "Dozens of supersymmetric theories exist", whereas the first paragraph of this article - does point out that the suppersymmetry "proposes that for every known particle, there exists a partner particle with different spin properties. There have been multiple experiments on supersymmetry that have failed to provide evidence that it exists in nature."

Anyway, my original question was not about the terminology used in Wikipedeia, but rather about the very idea, and I allow you to call it: theory, theoretical framework, hypothesis, conjecture, proposal, suggestion, idea, or whatever, but the main idea still remains, as long as I understand you and you understand me (I guess this is the case). HOTmag (talk) 10:07, 23 September 2024 (UTC)[reply]

Anyway your claim "I've linked to our article Supersymmetric theory, which does exist in Wikipedia" is a false claim indeed. Philvoids (talk) 11:48, 26 September 2024 (UTC)[reply]

Try now to click again on Supersymmetric theory, and this time you will see that this article [page] does exist in Wikipedia, giving a link to the other article, but without passing to the other article. Actually the article [page] Supersymmetric theory exists in Wikipedia since 14 January 2006, as you can see in its history page. Generally, every article [page] that passes to another article [page] must exist in Wikipedia: If it hadn't existed, it couldn't have passed to the other article [page]. HOTmag (talk) 22:31, 26 September 2024 (UTC)[reply]

Please see WP:REDIRE. A Wikipedia redirect is a page that automatically sends visitors to another page, usually an article or section of an article. It aids navigation and searching but a redirect page is not itself a Wikipedia article nor is its mere existence a reliable source. The excuse of an honest mistake won't justify an attempt to argue black as white by misrepresenting links. Philvoids (talk) 11:47, 27 September 2024 (UTC)[reply]

My only innocent mistake was my replacing "page" by "article". Anyway, when I wrote "article" I really meant "page", and I was sure this would also be what you would interpret when you read the word "article" in my responses. But since I'm realizing now this was not what you interperted, I'm striking out every "article" and replacing it by "[page]".

As for what you call "a reliable source", I've already quoted the second paragraph of our article Supersymmetry: "Dozens of supersymmetric theories exist". This quote proves that there is no fundamental difference between linking "UK" to "United Kingdom" and linking "Supersymmetric theory" to "Supersymmetry": All of these four terms are legitimate. HOTmag (talk) 12:46, 27 September 2024 (UTC)[reply]

Now you only have to define what you mean by particle. For instance in Standard Model you have the everyday particles (Electron,Proton...). Nothing of that is stable in your definition. Then you have Quark. No quarks are ever single. So it is a question of the neighbouring quarks which reactions are possible. But even then there is no stability in your definition. But never is one particle turned into exactly one other particle. Even if that were possible you only have to look at a different level of abstraction and a group of particles would turn into a different group of particles. 176.0.158.114 (talk) 09:34, 23 September 2024 (UTC)[reply]

By elementary particle I mean what physicists mean by that term: quarks, leptons, gauge bosons, and also elementray particles that haven't been discovered yet, like axions. Anyway, I really meant what you suggested in yout last sentence: "a group of [elementray] particles would turn into a different group of [elementray] particles". HOTmag (talk) 10:07, 23 September 2024 (UTC)[reply]

Then I can give you a definite answer. Any group of ordinary matter particles (whether known or not) can turn into another group of particles if they encounter the right conditions (even if the conditions could not be achieved in a laboratory or somewhere near Earth). For dark matter particles that can not scientifically be said. And never will be possible to say with science. If someone says something about changing about dark matter it is and never will be science. That is part of the definition of the word "dark" in "dark matter". If that definition changes nothing about the future I have written will continue to be valid. 176.0.153.105 (talk) 19:28, 23 September 2024 (UTC)[reply]

Memorandum: My question was about all elementary particles, and by elementary particle I mean what physicists mean by that term: quarks, leptons, gauge bosons, and also elementray particles that haven't been discovered yet, like axions.

To sum up: The question is whether, for every group of elementary particles, including those which haven't been discovered yet, there exist (what you call) "right conditions", under which this group can turn into another group of elementary particles. HOTmag (talk) 13:10, 24 September 2024 (UTC)[reply]

Then as I said all the way before, with my "#2", then if it can be called a "particle", yes you can always make it turn into stuff.

But you should understand what people here are trying to say: this becomes rather meaningless as your understanding of what a theory means, in the sense of the "discovery" of particles, is not very accurate, so when you're trying to force these incompatible notions into your question it makes it difficult to give meaningful answers. (Mine, as I said before, is not particularly meaningful.) SamuelRiv (talk) 13:28, 24 September 2024 (UTC)[reply]

Theories don't necessarily discuss discoveries, and a theory about elementary paricles doesn't necessarily discuss what you call "discovery" of elementary paricles, it can also say something about elementary paricles that haven't been discovered yet, e.g. gravitons, axions, electrinos, gravitinos, axinos, and the like. Anyway, my question is about all elementary particles, including those which haven't been discovered yet. Are you referring to all of them in your first sentence? HOTmag (talk) 18:26, 24 September 2024 (UTC)[reply]

YES! Even for particles that are not discovered yet, the right conditions are possible to determine. For an example see the article about the Axion. Even although the particle is not discovered yet, the right conditions for some way to turn into another group of particles are already known (strong magnetic field). Of course there will probably be other ways but one is enough to answer your question. And there is even a general answer. Most particles will turn at the surface of a Neutron star. If you are in doubt, there will almost always be the right conditions. Is there some particle that does not turn at the surface of a neutron star? Of course the neutron. But that turns at enough distance from the surface. And it is possible that the Alpha particle is a special Quantum state of twelve Quarks. Should that be the case a neutron would even be turned on the surface of the neutron star into an alpha particle. 176.0.147.163 (talk) 13:03, 26 September 2024 (UTC)[reply]

Have you got any source for your claim, that all elementary particles (other than neutrons), including those that haven't been discovered yet, e.g gravitons (and gravitinos and electrinos and axinos and likewise), will turn at the surface of a nuetron star? HOTmag (talk) 22:13, 26 September 2024 (UTC)[reply]

First my claim was for almost all particles, what you acknowledge by excepting the neutron. Second, no I don't have a verifiable source, because it is based on a personal communication. But it is easy to verify. Elementary logic and quantum mechanics are sufficient. Every physicist knows that quantum mechanics and high gravity or strong electric or strong magnetic don't fit together. As soon as a quantum particle interacts with something that needs relativity to describe, the equations begin to change. Of course the particles will change with them. I can give you an example, that does not need an exotic environment. If you read any Wikipedia article about a heavy element you will read about relativistic effects, that affect the electrons. And at the same time the core will go more and more unstable. Sometimes the elements are only "observationally stable". And the more affected elements are really radioactive. Until it comes to U238. There's a rule. Every number of charge has an associated ideal atomic weight. If the weight is too large neutron are not stable anymore and decay. If the weight is too small, coulomb force wins and an alpha particle, the least energetic particle in the core, will be ejected. Now to U238. According to its pattern of decay it's both too heavy and too light. Of course that is relativity playing havoc with the quantum mechanics. And now imagine what will go on where relativity really comes into play. Such as at the surface of a neutron star. 176.0.144.177 (talk) 02:36, 27 September 2024 (UTC)[reply]

PS. The claim about the surface of a neutron star is meant here. For this claim exceptions are possible. The surface of a neutron star is an example for the right conditions and a good bet,but not guaranteed. Further above was another claim about the right conditions. That claim is universally true, because there is always an environment where the wave function can be distorted beyond recognition. Which is the whole point of the claim. 176.0.144.177 (talk) 02:49, 27 September 2024 (UTC)[reply]

I deliberately asked about what you call "exotic environment" (e.g. gravitons, electrinos) because, while your example regarding U238 is about matter quite known to us - experimentally speaking, you're using a kind of generalization, from what we have already encountered (e.g. U238), to what - we haven't indeed encountered yet - but we can assume something about in (your own?) theoretical framework stating that: since (as you begin) "as soon as a quantum particle interacts with something that needs relativity to describe, the equations begin to change", so (as you conclude) "of course the particles will change with them". Of course your conclusion is possible and legitimate, but you haven't proved that it's the only one possible - theoretically speaking - mainly as far as (what you call) "exotic environment" is concerned.

To sum up: It seems like you're using your own theory. This is legitimate of course, but when I asked for a "theory" (see the title), I mainly meant: a (well sourced) theory - although I didn't add this condition in the title (but only in my previous response). HOTmag (talk) 08:49, 27 September 2024 (UTC)[reply]

How I said, there's private communication involved, so it's possible that through misunderstanding a private theory emerged, but the goal was to talk about the normal quantum physics. I never claimed that quantum physics is the only theory where universal decay is possible/mandated. That quantum physics and general relativity are mutually incompatible is well sourced. The example of U238 was only to show that the equations begin to change in the region of both relativity and quantum physics. And that the trend in this regard goes to less stability instead of more. Of course there are exceptions, particles that are normally not stable but can be stabilised by the exotic environment, better the changed equations in the exotic environment. Of course every exotic environment changes the equations in its own way. So for every needed change a matching exotic environment can be (theoretically) constructed. And then there is the well sourced Hawking radiation. If you want a random change an appropriately sized black hole will provide an exotic environment that is guaranteed to destroy any quantum state (well sourced No-hair theorem) and emit another random quantum state. 176.0.167.84 (talk) 00:16, 28 September 2024 (UTC)[reply]

I let you use any (well known) theory, including Quantum physics, for claiming that as soon as a quantum particle interacts with something that needs relativity to describe, the equations begin to change. I only asked whether your conclusion, that if the equations begin to change then the particles will change with them, must be concluded from this (well known) theory.

As for the option of black holes, yes, this is what I'd thought, and I'm asking now a further question in my following thread, regarding this option. HOTmag (talk) 18:14, 28 September 2024 (UTC)[reply]

See Richard Feynman as he talks about Path integrals. If the formula changes, the sum will change. That is clearly mathematics. I don't know if somebody has written about these mathematical relations, because it is only article worthy if there is a whole theory in the conclusion, but that is not how far we know the formulas yet. But even as you begin to try it, you will see immediately that there is change imminent. And the example of U238 shows that the change is not an artefact of the mathematical formulation but is reality. And the ultimate weapon (the black hole) drives the point home. Only the exact relations are unknown today. 176.0.164.155 (talk) 20:24, 28 September 2024 (UTC)[reply]

"If the formula changes, the sum will change". IMO, it's still too obscure, because:

As long as the particles haven't been discovered, we don't know what properties those particles carry, so we don't know how the equations involving those particles look like, so we don't know how those equations are going to change, so we don't know whether this change may influence the issue of whether the particles involved in those changing equations may turn into other particles.

As I've already said, IMO it's still too obscure, as long as we don't know what particles we are talking about. HOTmag (talk) 22:39, 28 September 2024 (UTC)[reply]

Every particle, discovered or not, must obey the Schrödinger equation. Part of the solution of the Schrödinger equation is an Integral. One way to do it is the Path integral formulation. That formulation has the advantage that the Spacetime is part of the explicit input. Relativity Theory modifies space. Modified input results in modified output, according to Mathematics.

That is essentially the whole point of the "theory". 176.0.164.155 (talk) 00:15, 29 September 2024 (UTC)[reply]

"Modified input results in modified output". For claiming that, you must make sure that this equation reflects a one-to-one correspondence between input and output. HOTmag (talk) 01:05, 29 September 2024 (UTC)[reply]

That's not true. In the pure mathematical sense there may be a function where that applies. But never in physics. Even for the Sine and cosine function it's not true, if you really think about it. In the Schrödinger equation there is a double Differential operator. So for the same output you need the same value, the same first differential and the same second differential. And now you can imagine a function that is at least two times differentiable, has at two inputs the same output, has at least a third input with a different output (constant function do not count) and is not (theoretically) reversible in the limits of our universe. 176.0.164.155 (talk) 12:05, 29 September 2024 (UTC)[reply]

You did not prove that the Schrödinger equation reflected a one-to-one correspondence. You have only presented some kind of intuition for claiming that. I still wonder if this intuition is rigorously provable. HOTmag (talk) 17:41, 29 September 2024 (UTC)[reply]

I've seen the symbols M and X in chemistry, what do they mean? HAt 04:07, 21 September 2024 (UTC)[reply]

Well M could represent a metal, and X could represent a halogen in a chemical formula, eg NaX could be a halide salt of sodium. X is probably F, Cl, Br or I. Theoretically it could be At, but not in any visible amount. If the symbols are in italic text they may represent a number. Graeme Bartlett (talk) 06:04, 21 September 2024 (UTC)[reply]

Besides a link to our article electroweak interaction that mentions the B-boson without telling anything about it, we have no article about this boson. Why?

I'm eager to know some basic data about it, e.g. mass, electric charge, spin, isospin, stability, experiments trying to detect it, and likewise. But... nothing? HOTmag (talk) 16:27, 22 September 2024 (UTC)[reply]

You'll find it at B meson. Sorry, although the B meson is a boson, it is not the beast that's mentioned in electroweak interaction. --Wrongfilter (talk) 17:23, 22 September 2024 (UTC)[reply]

What a pity. HOTmag (talk) 08:57, 23 September 2024 (UTC)[reply]

The B-boson isn’t a well-known particle in the standard model of particle physics. Are you thinking of the Higgs boson that was discovered in 2012? Philvoids (talk) 08:47, 23 September 2024 (UTC)[reply]

No, I'm looking for basic data (e.g. mass charge spin isospin parity etc.) of the B-boson mentioned in our article electroweak interaction. HOTmag (talk) 08:57, 23 September 2024 (UTC)[reply]

The thing is that the B boson doesn't exist as an entity in our Universe. It "exists" before symmetry breaking, but since our world has a broken symmetry the observable bosons are the Z0 and the photon instead of the W3 and B. As the B cannot be observed, many of your questions are moot. --Wrongfilter (talk) 09:07, 23 September 2024 (UTC)[reply]

Got it. HOTmag (talk) 09:31, 23 September 2024 (UTC)[reply]

In theory, would it be possible to clone the extinct Carolina parakeet by taking whatever existing DNA we have of the species and filling any gaps with DNA from the sun conure (closest living relative and very common pet) and then putting that into a sun conure egg? I have read that it's very difficult to clone birds though. 146.200.126.178 (talk) 23:12, 24 September 2024 (UTC)[reply]

Here's an Audubon Magazine write up on the topic. It foolishly says that the Carolina parakeet, ivory-billed woodpecker, and passenger pigeon could be invasive, which is nonsense; only the Carolina parakeet could conceivably be successful beyond its former range. See monk parakeet for why. Abductive (reasoning) 23:55, 24 September 2024 (UTC)[reply]

Yes, there's been a lot of talk about removing monk parakeets from areas where they are non-native. But the public tend to get extremely angry about the idea of culling parrots specifically, even to the point of taking direct action when nests are removed. I believe there was even a politician a few years ago who said something along the lines of "absolutely not" when it came to the idea of removing that species being raised by his advisors. 146.200.126.178 (talk) 00:19, 25 September 2024 (UTC)[reply]

Would you then get an actual Carolina parakeet, or merely a hybrid that kind of looks like it? ←Baseball Bugs What's up, Doc? carrots→ 01:30, 25 September 2024 (UTC)[reply]

Yes, I was thinking that myself. How much of the Carolina parakeet's DNA can be replaced by another (very closely related) species before it can no longer be considered a Carolina parakeet? 146.200.126.178 (talk) 01:42, 25 September 2024 (UTC)[reply]

Has the genome of any of these extinct species been mapped? I don't think the article says. But that would be a way to maybe get closer to a real clone. And I understand what they mean by invasive. If a species wasn't someplace, and then appears in that place, by definition it's invasive. That doesn't necessarily mean it will be harmful to other species in that place, but it could be. ←Baseball Bugs What's up, Doc? carrots→ 07:18, 25 September 2024 (UTC)[reply]

See Ship of Theseus. -- Jack of Oz [pleasantries] 17:41, 25 September 2024 (UTC)[reply]

"Reconstructed", bit by bit. In the case of an organism, some careful selective breeding could eventually render the species closer to the original. ←Baseball Bugs What's up, Doc? carrots→ 00:55, 26 September 2024 (UTC)[reply]

We have an article on the concept De-extinction. It mentions the passenger pigeon example discussed in Abductive's link. It also briefly mentions the Carolina parakeet and claims a "full genome of the carolina parakeet was sequenced" in 2019 although I'd suggest reading the sources as I suspect this isn't a complete genome. Revive & Restore seem to be the ones most involved in this for birds or at least the ones who seemed to have received the most publicity. (Colossal Biosciences is also popular but although they did make noises about the dodo they seem more concentrated on mammalian species. And while I personally think anyone involved in this is fairly PR oriented, Colossal Biosciences chosen projects seem to be more flash.) At the moment, in terms of actual "de-extincting" species they seem to be concentrating on the above mentioned passenger pigeon which is their flagship project with also some work on the heath hen. Interesting they've set a goal of the "first generation of new Passenger Pigeons before 2025" but considering their progress page [2] hasn't been updated since 2019, I have doubts we're going to see this first generation in the next 3 months. In fact, even their 2023 report [3] doesn't seem to mention the passenger pigeon. With the heath hen project they at least seem to have been less optimistic about promising stuff by certain dates [4]. However that too doesn't seem to have had an updated since 2020 [5]. Possibly COVID-19 related disruption to their work hasn't helped, still it's been a while since it should have been a barrier to any lab work etc. BTW, I'd say Revival of the woolly mammoth and Thylacine#Research are perhaps the projects which have generally received the most historic interest so might be a good place to look beyond general discussions about de-extinction about the feasibility of such proposals and whether the result if we did succeed should really be compared with the original species. Although the research related to the Maclear's rat mention in our article might be another recent place to look in to. Nil Einne (talk) 10:09, 27 September 2024 (UTC)[reply]

What is the etymology of wheldone? The references doi:10.1021/acs.orglett.0c00219 and doi:10.1021/acs.jnatprod.4c00649 only described the chemical as a fungal metabolite isolated from the coculture of Aspergillus fischeri and Xylaria flabelliformis. --Leiem (talk) 09:00, 25 September 2024 (UTC)[reply]

The clue is in the acknowledgments section of the first paper: "This Letter is dedicated to the late Daniel (Dan) Clive Wheldon". And since Wikipedia has everything: Dan Wheldon. --Wrongfilter (talk) 09:57, 25 September 2024 (UTC)[reply]

I understand. Thank you! --Leiem (talk) 15:38, 25 September 2024 (UTC)[reply]

One might even say...well done. :P S n o w Rise let's rap 22:39, 25 September 2024 (UTC) [reply]

I'm trying to split Nut (fruit) back into separate articles for food and fruit (because currently a lot of the article content is about things that aren't botanical nuts), but there's not much left after you strip out the food content. Are there any botanists in the house who can recommend sources on botanical nuts and/or weigh in on what should be covered and isn't? AlmostReadytoFly (talk) 10:10, 25 September 2024 (UTC)[reply]

Not a botanist, but I think there's a bit of a problem for having a botanical nut. It is an even fuzzier idea than that of tree. But there's an article on trees so I suppose the could be one like that on nuts. You'd have to decide whether it is a very resticted topic on the botanical idea of a nut which doesn't contain most of what people mean by a nut or a wider one like the article on trees. Would you include pine nuts, coconuts peanuts, and the stones of drupes like almonds or pecans or plums? NadVolum (talk) 14:33, 25 September 2024 (UTC)[reply]

The plan is for them to go back to Nut_(food) as part of the unmerge. The unmerge discussion is here. AlmostReadytoFly (talk) 08:51, 26 September 2024 (UTC)[reply]

For SODIS water disinfection I'm interested in testing how much UVB and UVC passes through various types of plastic and other materials. I don't care so much about absolute levels as relative intensity compared to unfiltered sunlight. I'd also like to check out whether my cheap sunglasses are letting any UV through. I see this device says peak spectral response 300-350mm which is describes as UVA (I thought that was the A-B range). Is that sufficient, i.e. is something that transits UVA unlikely to block UVB/UVC? Thanks. 2601:644:8581:75B0:0:0:0:C813 (talk) 22:21, 25 September 2024 (UTC)[reply]

I doubt that it's any indication one way or the other, different materials can have widely different transmission spectra both within and outside the visible band; one would have to test each one specifically (if references aren't available). {The poster formerly known as 87.81.230.195} 94.1.171.3 (talk) 02:20, 26 September 2024 (UTC)[reply]

Body_roundness_index#Calculation gives the formula

364.2 − 365.5  × √(1 − [waist circumference in cm / 2π]2  /  [0.5 × height in cm]2)

which seems unnecessarily complicated. Why isn't it simply simplified to

364.2 − 365.5 1 − ( waist circumference in cm π × height in cm ) 2 {\displaystyle 364.2-365.5{\sqrt {1-{\Big (}{\frac {\text{waist circumference in cm}}{\;\;\;\;\pi \;\;\;\;\;\times \;\;\;\;\;\,{\text{height in cm}}}}{\Big )}^{2}}}} {\displaystyle 364.2-365.5{\sqrt {1-{\Big (}{\frac {\text{waist circumference in cm}}{\;\;\;\;\pi \;\;\;\;\;\times \;\;\;\;\;\,{\text{height in cm}}}}{\Big )}^{2}}}} ?

Moreover, as both numerator and denominator have length units, the unit doesn't matter as long as they are the same e.g. cm, mm or inches.

Thanks, **cmɢʟee**⎆τaʟκ 04:31, 26 September 2024 (UTC)[reply]

Probaly because the editor who added that didn't know how to render mathematical formulae. You're welcome to change it. Shantavira|feed me 09:16, 26 September 2024 (UTC)[reply]

Thanks. I don't mean the formatting but the division by 2π on top with multiplication by 0.5 below.

Presumably, they meant waist / (2π) instead of (waist / 2)π – BEMDAS is confusing! **cmɢʟee**⎆τaʟκ 10:17, 26 September 2024 (UTC)[reply]

Because it's left over from its derivation from the eccentricity formula for an ellipse, 1 − b 2 a 2 {\displaystyle {\sqrt {1-{\frac {b^{2}}{a^{2}}}}}} {\displaystyle {\sqrt {1-{\frac {b^{2}}{a^{2}}}}}}, where the semi-minor axis, b, is the "radius" of the waist and the semi-major axis, a, is half the body height. AlmostReadytoFly (talk) 11:05, 26 September 2024 (UTC)[reply]

Makes sense, thanks! **cmɢʟee**⎆τaʟκ 11:34, 26 September 2024 (UTC)[reply]

To demonstrate his law, Max Planck converted radiation power measurements per unit area into energy density ( × 4 π / c {\displaystyle \times 4\pi /c} {\displaystyle \times 4\pi /c}).
Is there a demonstration directly with the radiation power from the experiment without conversion into energy density?
I can't find any, but maybe I'm not seeing the correct wording for search engines.
Malypaet (talk) 21:28, 27 September 2024 (UTC)[reply]

Here are three accepted assumptions:

1. There are black holes.

2. A given black hole can-theoretically absorb any given material.

3. A given black hole can-theoretically evaporate, by becoming Hawking radiation.

Hence, logically, any given material can-theoretically become energy: Just let this material be absorbed by a black hole, and then let the black hole evaporate and become Hawking radiation.

Apparently, all of this is done without using Einstein's formula [ E = m c 2 ] {\displaystyle [E=mc^{2}]} {\displaystyle [E=mc^{2}]}. So, it seems that Einstein's formula is not needed for proving that any given material can-theoretically become energy, right? HOTmag (talk) 18:00, 28 September 2024 (UTC)[reply]

Hawking radiation is not all energy. It contains particles (and anti) too. To theorize about Hawking radiation you (or Hawking) need(ed) Einstein's equation. So you need it, but you need not write about it. 176.0.164.155 (talk) 19:55, 28 September 2024 (UTC)[reply]

1. Re. the particles contained in Hawking radiation: So why does the lead of our article Hawking radiation only describe it as "black body radiation", i.e. "electromagnetic radiation", without mentioning any "particles" contained in Hawking radiation?

2. Are you sure the formula E = m c 2 {\displaystyle E=mc^{2}} {\displaystyle E=mc^{2}} is needed for concluding that black holes emit Hawking radiation? HOTmag (talk) 22:09, 28 September 2024 (UTC)[reply]

1 See the first paragraph in Emission

2 See the first paragraph in black hole evaporation

In 1 you need to pay special attention to the word "particle". 176.0.164.155 (talk) 23:35, 28 September 2024 (UTC)[reply]

Re. 1: Yes I'd seen this paragraph, but it doesn't answer my previous question, so let me repeat it: Why does the lead of the article only describe Hawking radiation as "black body radiation", i.e. "electromagnetic radiation", without mentioning any "particles" contained in Hawking radiation? Are you claiming that black body radiation can contain particles (besides energy)?

Re. 2: Yes this paragraph really shows how Hawking uses Einstein's formula for concluding that the black hole, not only creates energy, but also becomes energy. However, this article indicates also that "some [authors] find Hawking's original calculation unconvincing" - because it uses an "infinite frequency" as well as "a wavelength much shorter than the Planck length", while these authors use techniques other than Hawking's one, so I still wonder whether Hawking's technique using Einstein's formula is necessary for concluding that the material in the black hole, not only creates energy, but also becomes energy. HOTmag (talk) 01:06, 29 September 2024 (UTC)[reply]

Particles are energy and electromagnetic waves are particles; they are two aspects of the same. It's just that at typical temperatures used for blackbody radiation, the only particles you can make are photons. (There's enough energy too to make neutrinos, but that requires some weak interactions, so it's unlikely to happen.) Once energies go to the MeV scale (temperatures of gigakelvins), your blackbody radiation will contain other particles.

Not sure what you mean by "creates energy" or "becomes energy". Energy cannot be created or destroyed; it's always there. It just changes shape. Mass is equivalent to energy, that's an intergal part of relativity. And "equivalent to" doesn't mean "can be turned into", it means "is an alternative view of". PiusImpavidus (talk) 09:34, 29 September 2024 (UTC)[reply]

The lede indeed says "black-body radiation" and I think that's misleading. It was introduced here. I've changed it to prevent misunderstandings. --Wrongfilter (talk) 11:44, 29 September 2024 (UTC)[reply]

Thank you for this important correction. HOTmag (talk) 17:29, 29 September 2024 (UTC)[reply]

the only particles you can make are photons. (There's enough energy too to make neutrinos, but that requires some weak interactions, so it's unlikely to happen.) Are you claiming, that the "particles" mentioned in the first paragraph of the chapter Emission only mean "photons" (or neutrinos but it's unlikely), for "regular" tempratures?

Not sure what you mean by "creates energy" or "becomes energy". When I wrote "this paragraph really shows how Hawking uses Einstein's formula for concluding that the black hole, not only creates energy, but also becomes energy", I meant that the first paragraph in black hole evaporation really showed how Hawking used Einstein's formula for concluding that the black hole, not only emitted energy, but also lost mass equivalent to the emitted energy.

Mass is equivalent to energy, that's an intergal part of relativity. Who said that that was not? I only said, that without Einstein's formula E = m c 2 , {\displaystyle E=mc^{2},} {\displaystyle E=mc^{2},} [you'd have had no special relativity, so] you couldn't have concluded: "Mass is equivalent to energy".

Energy cannot be created or destroyed; it's always there. Correct, but without Einstein's formula E = m c 2 , {\displaystyle E=mc^{2},} {\displaystyle E=mc^{2},} that paragraph couldn't have concluded that "When particles escape, the black hole loses a small amount of its energy and therefore [loses] some of its mass", because without Einstein's formula - one could imagine a body emitting energy - while the body's mass remains the same as before the emission - while the emitted energy does not disappear but is only released ousdise. HOTmag (talk) 17:29, 29 September 2024 (UTC)[reply]

You also need Einstein's equation to prove that black holes can exist (assumption 1). PiusImpavidus (talk) 20:28, 28 September 2024 (UTC)[reply]

I was referring to Einstein's formula, i.e. E = m c 2 . {\displaystyle E=mc^{2}.} {\displaystyle E=mc^{2}.} Are you sure this equation (=formula) is needed for concluding that black holes exist? HOTmag (talk) 22:09, 28 September 2024 (UTC)[reply]

You want to remove special relativity, but maintain General relativity? You can't. The latter relies upon the former. 2A0D:6FC0:767:D900:3439:2201:29C1:1A87 (talk) 08:31, 29 September 2024 (UTC)[reply]

Theoretically, one could consider General relativity without considering Special relativity: Combining both theories, gives us a Pseudo Riemannian manifold - and as a special case - a Lorentz 4D space. But Special relativity alone - would only give us a Pseudo Euclidean 4D space - and as a special case - Minkowsky space, while General relativity alone - would only give us a Riemannian 4D space. To sum up: Theoretically, one could imagine a Generally relativistic 4D space, that ignores Special relativity. The same is true for the issue of mass-energy equivalence you're talking about: Also without Special relativity, one could still consider the Einstein field equations of General relativity, so that the geometry of spacetime would be shaped by the density and flux of momentum and of energy according to these field equations, but without assuming anything about any relation between mass and energy.

But this is a side point. My main question to user:PiusImpavidus was about whether Einstein's formula E = m c 2 {\displaystyle E=mc^{2}} {\displaystyle E=mc^{2}} is really needed for concluding that black holes exist. So I'm still asking: Is it needed? HOTmag (talk) 17:29, 29 September 2024 (UTC)[reply]

Since Hawking radiation includes particles, no your asumptions don't logically lead to that any material can become energy. NadVolum (talk) 19:06, 29 September 2024 (UTC)[reply]

Yes.

Due to the current thread, the article Hawking radiation has just been corrected by user:Wrongfilter, so now it explicitly states (in the lede) that Hawking radiation includes also particles. But when I posted my original post, the lede of the article had only mentioned electromagnetic radiation. That's why I posted my original post. HOTmag (talk) 19:33, 29 September 2024 (UTC)[reply]

  1. 1 gallon gasoline = 127 megajoule (per the gasoline article) = 35KWH thermal energy
  2. If you can convert that to electricity at 28% efficiency (portable generator), that's 10KWH electric
  3. Ebikes can go around 50 miles on a 500 WH battery charge, so 100 miles per KWH
  4. So that's 1000 miles per gallon if you power the bike from a generator.

Questions: 1) Amirite? I.e. does the math above look ok? 2) Why are motorized bikes/mopeds so much less efficient? They typically get 100 mpg or so.

Thanks. 2601:644:8581:75B0:0:0:0:C813 (talk) 22:20, 28 September 2024 (UTC)[reply]

28% is way high for a generator. Tires, mass, are significantly different. Does your genny meet the emissions regs for a moped? Moped's aren't optimised for economy, bicycles are. Greglocock (talk) 08:03, 29 September 2024 (UTC)[reply]

E-bikes (which are basically light electric mopeds; I don't see why the law makes a distinction between those) typically cruise at about 6 m/s (22 km/h). A regular moped cruises at 12.5 m/s (45 km/h), twice as fast. That quadruples drag and energy use. Combine that with the low efficiency of small (but still oversized), two-stroke petrol engines and the much lower rolling resistance of bicycle tyres, in particular when compared to the tyres of motorscooters. PiusImpavidus (talk) 09:08, 29 September 2024 (UTC)[reply]

In other words, what rules out the following scenario?

1. A body, being right now a Schwarzschild black hole, starts emitting Hawking radiation.

2. However, the body's radius remains constant during the emission.

3. When the body has lost too much energy - along with its equivalent mass, the body's mass inside the constant body's radius becomes less dense, untill the body's current radius becomes bigger than the body's Schwarzschild radius - because of the stability (constancy) of the body's radius, so the body - which has just been a black hole - stops being a black hole and becomes a regular body.

What's wrong with this scenario? Is this really assumption #2 ? HOTmag (talk) 18:54, 29 September 2024 (UTC)[reply]

THe theory says #2 is wrong - black holes become smaller as their mass goes down. NadVolum (talk) 19:10, 29 September 2024 (UTC)[reply]

Yes, both our article black hole and our article Hawking radiation state that when the black holes emit radiation they "shrink", but how do you know that their shrinkage refers, not only to the body's mass, but also to the body's radius? This is the main question of this thread.

Is this because of the internal gravitation, which is the only "force" active inside the black hole? HOTmag (talk) 19:47, 29 September 2024 (UTC)[reply]

I think you're giving the word 'body' too much meaning. The radius is a gravitational result which depends on the mass.You wouldn't notice the surface as you fell through. NadVolum (talk) 20:33, 29 September 2024 (UTC)[reply]