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June 14

Long run equilibrium of infinite-dimensional open quantum systems

To start off, consider the simplest quantum system: a two-dimensional Hilbert space, or "qubit". The classical states are the 0 and the 1. If the system is isolated, then there exists a unitary time evolution operator which can be represented as a 2x2 matrix, e.g.

$U={\begin{bmatrix}a&b\\-b&a\\\end{bmatrix}}$

which acts on $|0\rangle ={\begin{bmatrix}1\\0\\\end{bmatrix}}$ and $|1\rangle ={\begin{bmatrix}0\\1\\\end{bmatrix}}$ and linear combinations thereof. The state of the system after n time steps is found by applying $U^{n}$ to the initial state.

For example, $U^{2}|0\rangle =(a^{2}-b^{2})|0\rangle -2ab|1\rangle$ , and subsequently the probability of observing the system in the 0 state is $|a^{2}-b^{2}|^{2}$ .

When the system is "open", the operator U "keeps track of the history". Denote the history of a state as a sequence of 0s and 1s, for example, $|*0\rangle$ is a sequence whose most recent state is the 0 state. Denote the new operator $U_{o}$ .

Now, for instance, $U_{o}|*0\rangle =a|*00\rangle -b|*01\rangle$ , and ${U_{o}}^{2}|*0\rangle =a^{2}|*000\rangle -ab|*001\rangle -b^{2}|*010\rangle -ab|*011\rangle$ .

The difference is that $|*000\rangle$ and $|*010\rangle$ , although both representing the 0 state, have different histories and therefore the probability of observing the system in the 0 state is now $|a^{2}|^{2}+|-b^{2}|^{2}$ .

It can be shown that as you take $\lim _{n\rightarrow \infty }{U_{o}}^{n}$ , the probability of observing the qubit in the 0 state (the sum of norm-squared probability amplitudes of all sequences whose most recent element is the 0 state) converges on ${\frac {1}{2}}$ .

In fact, for a quantum system described by an m-dimensional Hilbert space, it can be shown that the probability of observing the system in any given state approaches ${\frac {1}{m}}$ --all states are equiprobable in the long run.

What I want to know is what happens when the quantum system is described by an infinite-dimensional Hilbert space, such as a quantum field. Is the long run probability distribution of states still uniform in that case? My guess is that the answer is no, but that there is a long run equilibrium which is different from the uniform distribution over possible states.--49.183.144.209 (talk) 11:57, 14 June 2019 (UTC)[reply]

It depends on the system: you're asking about the long-range stability of a dynamical system, e.g. its Lyapunov stability; and this is a qualitative property that we can quantify only if we know the system equation - in other words, the time-evolution operator.

Whether the system is quantized, or not-quantized, you want to conduct stability analysis on the system equation; the quantization might introduce non-linearith into this analysis; but the methods are the same. It's a fascinating field, and there are a lot of open research questions. Wikipedia has articles on asymptotic analysis, a list of complex analysis topics; and so on that might point you in the right direction. I'm trying to think of a specific quantum mechanics book that has a section on stability-analysis - maybe I can recall a good one after checking the bookshelf. But I don't think you even need a quantum-physics book; even a book on linear systems - like Proakis & Manolakis - presents extensive mathematical analysis tools to help you prove whether an arbitrary system, representable as an operator with an infinite impulse response, converges or diverges asymptotically.

Unless the your operator implies that the individual elements interact with each other's values, it won't matter whether your state-space has one, two, or an infinite number of elements; their long-range stability depends on the operator only.

If the elements do interact, you've entered that branch of mathematics where the solutions depend strongly on tiny perturbations of the initial conditions.

Nimur (talk) 15:26, 14 June 2019 (UTC)[reply]

Thanks for your answer. I think the most important point is that the long range stability depends on the operator only, not the initial state. If I understand you correctly, you can form a unistochastic matrix from the 2x2 unitary matrix, and this unistochastic matrix can be interpreted as the transition matrix of a Markov chain, and it has an equilibrium distribution that doesn't depend on the initial state, and that remains true even if the state space is infinite.

What do you mean by "elements interacting with each other's values"?--49.183.144.209 (talk) 16:46, 14 June 2019 (UTC)[reply]

By "interactions" between elements, I mean to say that the system is coupled, e.g. a "coupled system of differential equations," or, in other words if the next transition of a state-variable depends on the value of a different state-variable. In mathematical jargon, this is true if and only if the matrix representation of the state-space transition equation contains non-diagonal elements. Here's Decoupling Systems, part of the Linear Systems class from MIT's math department; they describe a few methods of complex analysis that allow you to diagonalize any system of ODEs, or to determine that it is non-diagonalizable using only real (non-complex) solutions.

Then, to prove whether the system is stable or unstable, we can simply check if the magnitude of the real part of the diagonalized matrix is strictly less than unity.

This whole process has some kind of passing analogue in the mandelbrot fractal generator, which is itself an example of a test of convergence- or divergence- on one particular system's time-evolution operator. In that case, a system with two initial conditions (the real- and the imaginary- parts of c at time zero) can behave quite in a complicated manner, even though the time-evolution operator is almost trivial.

Nimur (talk) 22:30, 14 June 2019 (UTC)[reply]

I have another question:

A unitary matrix can be transformed into a unistochastic matrix by norm-squaring each element of the matrix. A unistochastic matrix is a stochastic matrix and can therefore be interpreted as the transition matrix of a Markov chain. This transition matrix has an equilibrium distribution which does not depend on the initial state (and is, in fact, always the uniform distribution over the states).

Can a similar analysis be performed beginning with a unitary operator on an infinite-dimensional Hilbert space? Is it the case that the equilibrium distribution is still uniform over the states (my guess is no)? Is it still the case that the equilibrium distribution does not depend on the initial state (my guess is yes)?--49.183.144.209 (talk) 04:19, 15 June 2019 (UTC)[reply]

This sounds like a hard question... I think the same analytical methods apply; I think the equilibrium distribution depends on the transition matrix, and I am not sure if the additional qualification that it is unistochastic puts constraints on the outcome (but I bet it can be proven one way or the other!); and I can't answer the last part of your question because I was not able to answer the earlier bit. Mostly, though, I'm looking for a good book to help you make progress on these topics...

After perusing MathWorld for a bit, I found Horn's Theorem, which provides a strict condition for the existence of an Hermitian matrix with specified eigenvalues (... and the physicist in me is leaping to the end-goal here, which is to say that these are the asymptotal or steady-state solutions - there's at least a passing relation to what you're seeking);... our encyclopedia also has an entry on the Schur–Horn theorem, but when the mathematicians start introducing symplectic geometries, that all becomes a bit unfamiliar to me...

Nimur (talk) 16:10, 15 June 2019 (UTC)[reply]

Weight lifting question.

Is it bad to weight lift 2 days in a row? A lot of professional weight lifters like Bruce Lee do a Mon/Wed/Fri schedule or Tu/Th/Sat schedule. The logic is that muscles need 48 hours to heal. But what is the argument against weight lifting 2 days in a row? Because I also hear, weight lifting just makes muscle cells bigger, bu if you want new muscle cells to be created, you need to lift when you're sore, so that could be M/T and Th/Fr schedule. Shrug. 67.175.224.138 (talk) 15:41, 14 June 2019 (UTC).[reply]

This depends on how close to the limit you're exercising. If you're using the shocking method to lift yourself out of a plateau as Schwarzenegger explains here, then you'll need a larger resting period because a single bout of such an exercise session is going to be extremely intense. The next day you should exercise other muscles to avoid injury as the muscles you used yesterday won't be able to handle anything close to what you are normally used to. Count Iblis (talk) 16:22, 14 June 2019 (UTC)[reply]

I don't think that this question qualifies as medical advice, but it's nevertheless the kind of thing that you should consult a qualified person on. Get ahold of a professional trainer (pretty much every gym has them) to help direct your progress. You can definitely hurt yourself lifting incorrectly, but even without that, it's very likely that you won't get the desired outcome without the proper advice. I know a few semi-pros and they all swear by the advice trainers gave them. Matt Deres (talk) 22:35, 14 June 2019 (UTC)[reply]

Incandescent bulbs shining brighter

I stayed in a dorm when I went to university and each room had two desk areas, each with a desk lamp attached to the wall by way of an articulated arm. The bulbs were incandescent (this was in the 90s). Someone had heard of a neat trick involving such set ups, which turned out to be true. If you smack the shade of the light hard enough so that the entire system is shocked, the bulb will burn brighter - quite a bit brighter actually. None of us had access to a light meter, but I'd bet they were easily twice as bright. Unsurprisingly, some of the bulbs died shortly afterwards (I'll say, over the course of the next week or two), while others kept working fine for the rest of the semester. For whatever reason, that "experiment" recently came to my mind. What the heck was actually happening? I described the set up as best I can recall, in case that had something to do with it, but I presume the effect was caused by shocking the bulb itself...? When it gets right down to it, none of it makes much sense, but we - ahem - conducted several iterations of the test, almost all of which worked. Matt Deres (talk) 22:31, 14 June 2019 (UTC)[reply]

An Incandescent light bulb often has a coiled coil tungsten filament. When hot a mechanical shock could cause turns to weld together resulting in lower resistance and higher power since supply voltage is constant. DroneB (talk) 23:05, 14 June 2019 (UTC)[reply]

This question reminded me that as a kid I would try to "repair" clear glass, incandescent bulbs with broken filaments by turning them around and around until the broken ends crossed over. Applying power at that moment would cause a weld to occur, and the globe would work again. Never noticed whether the repaired ones were brighter or not. These would normally be single, "straight" wire filaments, not coiled ones. HiLo48 (talk) 00:13, 15 June 2019 (UTC)[reply]

Neat - thanks! That actually makes sense. Matt Deres (talk) 15:15, 15 June 2019 (UTC)[reply]

June 15

Principle quantum number for lanthanum

What is its value? Sandbh (talk) 19:33, 15 June 2019 (UTC)[reply]

Principal quantum number is a property of an electron. From the electron configuration [Xe] 5d¹ 6s² you can see the maximum value is 6 for the ground state atom. For an excited atom it could be higher. Graeme Bartlett (talk) 22:47, 15 June 2019 (UTC)[reply]

Is it 5 for the 5d differentiating electron? And is it 4 according to the Madelung approximation, which erroneously predicts lanthanum should have a 4f differentiating electron? Sandbh (talk) 20:07, 16 June 2019 (UTC)[reply]

For the "5d differentiating electron" the principle quantum number of the electron would be 5 as you can see by the "5". But as for the principle quantum number of the element, it would still be counted as 6, (it is in Period 6 in the periodic table. Similarly the 4f electron is of principle quantum number 4, but the element is still principle quantum number 6 thanks to the 6s electrons. Rearranging the electrons between D and F subshells may excite the atom a bit but it does not change its principal quantum number until you can get enough energy to get a 7s electron. Note that an La³⁺ ion would go down to principle quantum number 5. Graeme Bartlett (talk) 00:31, 17 June 2019 (UTC)[reply]

Thank you. I understand what you mean by principle quantum number (PQN) = period number, in the case of the conventional periodic table. Does this “rule” fail in the case of palladium, with its 4d¹⁰ outer configuration? It is in period 5 but it’s PQN is 4. More broadly, there appears to be some terminological sloppiness in the literature. For example, the PQN is sometimes associated with the period number (per your answer) whereas in the case of the Madelung “rule” diagram the PQN is the idealised differentiating electron. Have I interpreted this correctly? In the case of La then, the PQN can be 6, 5, or 4 depending on the electron being referred to i.e. the 6s of the highest occupied orbital; the 5d as the actual differentiating electron; and the erroneous 4f as the differentiating electron according to the Madelung rule diagram. No wonder there is so much confusion around this topic. I intend to improve some of our own articles, once I have a clear understanding. Sandbh (talk) 12:22, 17 June 2019 (UTC)[reply]

Each electron has a PQN. An atom has (usually) many electrons, each with its own PQN. Atoms (elements) don't have PQN in the same way because electrons can come and go without changing the elemental identity. Elements have a fixed period (chemistry) based on the sequence of protons, but the jump back to a lower PQN, for example in lanthanum the "next electron added after 6s² is 5d", is still a linear increase in nuclear proton count, not a backward change. Lanthanum is still in Period 6. In keeping with your questions and preceding explanations, lanthanum does have its "last" electron added in n=5 but its highest populated electron shell is still n=6. DMacks (talk) 14:58, 17 June 2019 (UTC)[reply]

Thank you. Yes, the PQN applies to the electron in question, not the atom per se. I don’t understand what you mean by “elements having a fixed period based on the sequence of protons”. In a conventional table, Sc is in period 4; in a left-step table, Sc is in period 5. Even in a conventional table this would not appear to work given Pd in period 5 has a PQN, based on its outer configuration of 4d¹⁰, of 4. There seems to be much confusion surrounding this concept, including on my part. For example this source gives the quantum number of La as 4 (presumably based on the aufbau approximation or Madelung rule diagram). Sandbh (talk) 10:01, 18 June 2019 (UTC)[reply]

June 16

Does the universe have an electric charge ?

That is, does the number of protons and electrons equal one another, or are there more protons, giving the universe a positive charge, or more electrons, leading to a negative charge ? And if the number is roughly equal, is there some regulating force that causes this, or is the reason unknown ? SinisterLefty (talk) 13:04, 16 June 2019 (UTC)[reply]

This question has been asked at Quora where various replies depend on assumptions about the size of the Universe. If the universe is finite, then it seems impossible to have a never-ending electric field line, so it seems that a finite universe must be charge neutral. If the universe is infinite, and has a uniform nonzero charge density through its infinite expanse, then it has an infinite net charge. Then every part of the universe is a source of some never-ending electric field lines. DroneB (talk) 13:26, 16 June 2019 (UTC)[reply]

standard cosmological model works like the universe has no net charge. If it had, the repulsion between same charge would need to be taken into account to explain expansion of the universe, which is not done AFAIK.

Also, the universe seems conductive enough to act like this (from Electrostatics#Electric_field):

Gem fr (talk) 16:19, 16 June 2019 (UTC)[reply]

So, are any theories proposed for what would explain this exact match in the number of protons and electrons ? SinisterLefty (talk) 19:22, 17 June 2019 (UTC)[reply]

Not sure you need to explain that, because this is the simplest solution you can imagine, and you'll some symmetry breaking for the number to be different. But don't trust my word on this, this math is far above my grade. Gem fr (talk) 19:54, 17 June 2019 (UTC)[reply]

There are other charged particles than protons and electrons so the net charge can be zero without a perfect match of those particular particles. And a proton is a composite particle composed of three valence quarks: two up quarks of charge +⁠23⁠e and one down quark of charge –⁠13⁠e. PrimeHunter (talk) 20:30, 17 June 2019 (UTC)[reply]

June 17

Using modern medical knowledge, what is Sarah chances of getting pregnant at the age of 90 if she had sex at her most fertile day?

“Shall a child be born to a man who is a hundred years old? Shall Sarah, who is ninety years old, bear a child?” (Genesis 17:17) Using modern medical knowledge, what is Sarah chances of getting pregnant at the age of 90 if she had sex at her most fertile day? Let's assume she never reached menopause. 49.177.234.140 (talk) 09:35, 17 June 2019 (UTC)[reply]

You're asking about a Bible myth. ←Baseball Bugs ^{What's up, Doc?} carrots→ 11:11, 17 June 2019 (UTC)[reply]

At those ages, the only real chance would be in vitro fertilization using frozen sperm and eggs from their younger days, implanted in a surrogate mother, and raised by an adoptive family, since they are unlikely to be able to bring the child to adulthood. (The surrogate mother and adoptive mother might be the same person.) SinisterLefty (talk) 11:14, 17 June 2019 (UTC)[reply]

Pretty good chances. Your "never reached menopause" assumption is the tricky one. Andy Dingley (talk) 11:44, 17 June 2019 (UTC)[reply]

Did you check menopause and Sarah#Historicity? Gem fr (talk) 15:12, 17 June 2019 (UTC)[reply]

You are using the modern definition of age as being 365.25 days long. That definition of a "year" didn't exist at the time of the Biblical stories. 68.115.219.130 (talk) 18:26, 17 June 2019 (UTC)[reply]

There was some speculation (see Methuselah#Mistranslation) that some Biblical ages were actually their ages in months, but I don't see how that could apply here, as that would mean Sarah was like 7-8 years old. SinisterLefty (talk) 19:20, 17 June 2019 (UTC)[reply]

Besides, the surprise of the pregnancy comes from her being old already. And she died at 127 ″years″. Now, for 2 cents I can add that, if those years were, say, ~6 month long, her pregnancy would had occurred at ~45y (which could be considered old), and she would have died at ~63 (a respectable age). Gem fr (talk) 19:40, 17 June 2019 (UTC)[reply]

But why would anyone measure ages in 6 month increments ? The use of months is common in hunter/gatherer societies, where the phases of the Moon are important, allowing for night hunting when the Moon is up and nearly full, and of course relating to women's periods. Annual cycles are more important to farmers, as they control the time to plant and harvest. SinisterLefty (talk) 19:51, 17 June 2019 (UTC)[reply]

well, this can happen if you mistranslate "season" into "year" in a context where a year is made of 2 seasons, which happens quite a lot, including for herders (of which Abraham was). (caveat: this is just, then again, my 2 cents). Gem fr (talk) 20:04, 17 June 2019 (UTC)[reply]

When I took ancient history (focusing on Egypt, Greece, and Rome), it was important to note that a "year" was not 365 days long. There were generally 10 months in a year. Sometimes a year had more. Sometimes it had less. A month was not a "moonth" as expected. It had been transformed into a tax cycle. Every month, people had to pay taxes. Every year, you had yearly taxes. If the government wanted more money, it was a new month. If they needed more money, it is a new month again. During war, years wizzed by faster and faster. During peace, years slowed down and took much longer to pass by. Then, with all this nonsense, the Romans wanted to normalize the calendar. That wasn't until after 400AD. So, it was hundreds of years after the stories from Genesis. If you take the variable concept of a year and add exaggeration of oral tradition, it is very easy to get people who are hundreds of years old. 68.115.219.130 (talk) 13:03, 18 June 2019 (UTC)[reply]

When you look better, it appears that romans were, at that time, just barbarians. Abraham came (so they say) from Ur, which had very solid foundations in mathematics, astronomy, etc. and used an elaborate calendar, just as good as the Julian. There is no way your explanation can hold Gem fr (talk) 19:14, 18 June 2019 (UTC)[reply]

My take is: not zero chance so it can happen, but so low chance you would make a story out of it, still remembered after millennia. Gem fr (talk) 20:13, 17 June 2019 (UTC)[reply]

Are glucocorticoids and catecholamines regulated by different systems of the body?

Are glucocorticoids and catecholamines regulated by different systems of the body? According to the book I read now "glucocorticoids secretion is regulated by hypothalamic - pituitary system" and some rows later "Catecholamines secretion is under the control of the central nervous system". Does it mean that the glucocorticoids are not regulated by the CNS? (I thought pituitary gland is a part of the CNS too. Isn't it? ) 93.126.116.89 (talk) 19:54, 17 June 2019 (UTC)[reply]

Yes and no. Glucocorticoids and catecholamines are both synthesized in and released into the bloodstream from the adrenal glands. However, glucocorticoid release is primarily regulated by the HPA system, as your book stated. Catecholamine release is controlled by the sympathetic nervous system. So, glucocorticoid release is controlled indirectly by the brain through endocrine signalling, while catecholamine release is controlled directly by action potentials transmitted through the sympathetic nerves. --47.146.63.87 (talk) 06:12, 18 June 2019 (UTC)[reply]

Our sun

What colour is it? 86.8.201.99 (talk) 22:22, 17 June 2019 (UTC)[reply]

The Sun is a yellow dwarf, but it is actually white. This is obvious when you see the Sun high overhead, through a layer of clouds (note: I make no representation as to how dangerous it might be to look at the Sun this way).

I think the reason most people think the Sun is yellow is that it looks yellow in the late afternoon (or early morning, I suppose, if you're that sort of person) when it is low enough in the sky to look at briefly without having a reflex to look away immediately. --Trovatore (talk) 22:33, 17 June 2019 (UTC)[reply]

It can also look red or orange when near the horizon, depending on atmospheric conditions. Looking at pictures of solar eclipses, the corona appears white. And using a primitive but effective cardboard box with a pinhole "projector", again the image looks white. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:09, 18 June 2019 (UTC)[reply]

It can even sometimes appear green. SinisterLefty (talk) 02:09, 18 June 2019 (UTC)[reply]

If it's millions of degrees why isn't it blue even in photographs? If a googolplex Celsius blackbody is blue why is lightning violet? Sagittarian Milky Way (talk) 00:20, 18 June 2019 (UTC)≈[reply]

Our article says the Sun's surface temperature is 5778 K, or 9941 °F, so not close to "millions of degrees". The core reaches millions of degrees, to be sure, but those photons don't make it to the surface. --Trovatore (talk) 00:38, 18 June 2019 (UTC)[reply]

I know that. The corona is extremely hot. If it wasn't orders of magnitudes less dense than the photosphere it'd be astoundingly bright. Is the visible light part of the emission lines distributed such that it isn't strongly colored? Sagittarian Milky Way (talk) 02:17, 18 June 2019 (UTC)[reply]

According to corona, "[t]he corona is 10^-12 times as dense as the photosphere, and so produces about one-millionth as much visible light." So for purposes of everyday viewing it doesn't matter what color it is, because you can't see it next to the Sun anyway. It can be viewed during a total eclipse, and if what you mean to ask is for an explanation of the perceived color of that, I guess I don't know, but it was hardly clear that that was what you were asking. --Trovatore (talk) 02:43, 18 June 2019 (UTC)[reply]

The article on Lightning says its color is kind of blue-white, not violet. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:01, 18 June 2019 (UTC)[reply]

The sky around the lightning might be violet, though, especially at dusk, so they might conflate the color of the two. SinisterLefty (talk) 02:04, 18 June 2019 (UTC)[reply]

Another factor to consider is our "mental white balance". That is, since we have a yellowish star, we tend to see that as white. If we had a bluish star, we likely would have evolved to see that as white. There's even a quick adjustment to the color. Try on different colored sunglasses, and the scene will seem to be different colors, but if you leave one pair on for a while, that will set the new white balance in your mind. SinisterLefty (talk) 02:22, 18 June 2019 (UTC)[reply]

We don't have a "yellowish star". It's white. As you note, that's essentially where our concept of "white" comes from, but nevertheless that's what it is. It just isn't yellow, period. --Trovatore (talk) 02:28, 18 June 2019 (UTC)[reply]

Our Sun article states that while it's a yellow dwarf, "its light is closer to white than yellow". So, in other words, it has a yellowish tint to it. SinisterLefty (talk) 03:06, 18 June 2019 (UTC)[reply]

Is there a source for that statement? ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:24, 18 June 2019 (UTC)[reply]

I don't see one, but I took it at face value. If that's inaccurate, it should be fixed. SinisterLefty (talk) 03:27, 18 June 2019 (UTC)[reply]

Finding a source, one way or another, would be the best option. Here's one from Stanford.[1] ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:00, 18 June 2019 (UTC)[reply]

I read that as "the Sun is often called a 'yellow dwarf', but it's really white." Out of the visible spectrum, the Sun actually gives off the most light in green. --47.146.63.87 (talk) 06:03, 18 June 2019 (UTC)[reply]

colour is subjective (and contextual), but one way to make it objective is to say that sun color is 5,777 Kelvin. Which translate as this (from black-body radiation), that is, in my eyes, slightly yellowish-white

Gem fr (talk) 07:07, 18 June 2019 (UTC)[reply]

I'm not altogether convinced by that graphic. You're right that 5800 K on that chart hits a yellowish patch (at least on my screen). But even a 5000 K lamp looks, if anything, slightly blue to most people, at least when used for indoor lighting. --Trovatore (talk) 17:54, 18 June 2019 (UTC)[reply]

The confusion may be due to that graphic not being the actual color of those objects, but rather their color "as seen in the night sky", so after passing through the atmosphere, which would diffract some of blues away, giving what remains more of a yellow hue. SinisterLefty (talk) 18:17, 18 June 2019 (UTC)[reply]

~~Rayleigh scattering is a small effect for objects high overhead, not enough to change their color significantly.~~ --Trovatore (talk) 19:26, 18 June 2019 (UTC)[reply]

They didn't specify the angle. SinisterLefty (talk) 19:44, 19 June 2019 (UTC)[reply]

Actually, after visiting that page, I'm no longer so sure it's insignificant. Still, I don't think the OP was asking what color the Sun is when you can comfortably look at it; that's not a question you really need to ask the refdesk. You just go outside and look. --Trovatore (talk) 20:44, 21 June 2019 (UTC)[reply]

Not if they're color-blind. Also, since there's a range of colors, it would take many observations to determine the full range. And it's rare that you can comfortably look at the Sun with the bare eye. There's sunrise and sunset, but it might still be too bright then. There's fog and overcast conditions. And there're certain total eclipses, but that's dangerous as you may suffer eye damage when the Sun peaks out from behind the Moon. Dark sunglasses help, but may also change the color. SinisterLefty (talk) 00:21, 22 June 2019 (UTC)[reply]

The color white is by definition that color that you see when you're outside looking at an object that reflects all wavelengths uniformly. Your brain will adjust the white balance to make it so. Count Iblis (talk) 10:39, 18 June 2019 (UTC)[reply]

White balance has limits. Blood will never look white in full color vision (photopic). What blackbody is furthest from where your visual lobe can't white balance it to look white anymore? That is the whitest white blackbody possible. Now if you could do this in a holodeck and instantly raise the other wavelengths of that perfect blackbody to that of the brightest wavelength (one of the greens I think) wouldn't violet and red objects suddenly look brighter? If the green wavelengths are only 109% the brightness of the violet and red wavelengths in theory they'd get brighter but you'd barely notice if at all. Would objects with equal albedo between the world records of wavelength perception (at least 280-850nm) suddenly become so white they they suddenly look slightly green-tinged when you switch back to blackbody or is a blackbody of the right Kelvins already the whitest white the brain can perceive? Eclipse eyeglasses dim 100,000 times. Since the Sun is best photographed through a telescope at 1/125 to 1/1000 seconds or less to freeze the same effect that makes stars twinkle they make filters for cameras that dim only 10,000 times but those can damage your vision when looked through. If you were at the distance in space where the Sun was 100,000 times dimmer per cone (about 0.005 degrees) or whatever it is that doesn't saturate a cone but still bright enough for full color vision (so still bright by stars in Earth's night standards) then the Sun would look yellowish white (at least if you hold a rainbow of equally bright LEDs from 400 700 nm near it to tell your brain fuck you, stop white balancing) but from Earth it saturates all cones which would make it impossible to see any yellowness even if the brain didn't adjust white balance which it does. As the eye has microsaccades and averages brightnesses on the timescale of the persistence of vision the Sun probably can be more than 1/100,000th the square degrees of a cone cell before cones get saturated. Sagittarian Milky Way (talk) 13:57, 18 June 2019 (UTC)[reply]

Blood is fairly dark (nonreflective), so isn't going to look white, but does sometimes look black, as when viewed in dim light. SinisterLefty (talk) 14:17, 18 June 2019 (UTC)[reply]

Everything looks grayscale if it's dim enough (scotopic, too dark for cones to see anything). Then there's mesopic, rods not saturated yet but cones at the dim end of their scale. Thus blood doesn't look white or gray to the non-colorblind while the gibbous Moon is also fairly nonreflective (no glass bead retroreflection or complete lack of shadow hiding like full moon) and looks white at night when high up cause it's the brightest thing in view (sunlit ground) and close enough to white for the eye to white balance, unlike typical sunsets, blood, oranges, maple leaves, gas flames or monochromatic LEDs. Sagittarian Milky Way (talk) 15:49, 18 June 2019 (UTC)[reply]

FCS, guys, the OP just asked the color of the sun (which is, again, 5,777 Kelvin. Period.). look at your answers... Gem fr (talk) 09:53, 19 June 2019 (UTC)[reply]

The absolute color is certainly relevant, but since the OP is presumably here on Earth, it's reasonable to assume that they may also want to know what color the Sun appears to be on Earth (and they may not realize there's a difference). SinisterLefty (talk) 19:47, 19 June 2019 (UTC)[reply]

June 18

K9 breeding

Is it possible to interbreed different species of K9. I am aware that you can cross breed dogs but as Hyena's and foxes are K9s too,can you cross breed a large dog with a Hyena? I am not planning on doing this, I am merely curious if it is possible. has anyone tried this. The same goes for others in the K9 family, foxes, or wolves and hyena? Thanks Anton 81.131.40.58 (talk) 15:54, 18 June 2019 (UTC)[reply]

A quick Google search of the subject of what canines can cross-breed with, the answers are: wolf, yes; coyote, yes; fox, no; hyena, no. The reasons have to do with chromosome counts. And hyenas are not canines. They're more closely related to cats. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:23, 18 June 2019 (UTC)[reply]

I think it best to direct OP to crossbreed#Hybrid animals and canid hybrid. And Hyena. Gem fr (talk) 18:49, 18 June 2019 (UTC)[reply]

What about African Wild Dogs and jackals, can they be mixed with domestic dogs? — Preceding unsigned comment added by 81.131.40.58 (talk) 08:24, 19 June 2019 (UTC)[reply]

Jackals are several different species. There are hybrids between golden jackals and domestic dogs, see Jackal–dog hybrid, but the other species of jackal are far more distantly related, and I don't see any content about hybrids. African wild dogs are also much more distinct from domestic dogs than golden jackals, and I also see nothing about hybrids there. Someguy1221 (talk) 08:46, 19 June 2019 (UTC)[reply]

There are several dog wolf hybrids I believe dogs originated from the gray wolf and that is why it it easy to cross breed them especially with German Shepards and related breeds 64.222.180.90 (talk) 18:37, 20 June 2019 (UTC)[reply]

Electric engine speed or load monitor

I don't know what this is called, so I'm having trouble reading about it... Assume you have an electric motor. I doubt it matters if it is AC or DC. It rotates a spindle just like any other electric motor. Inside the motor there is a sensor. If the spindle is trying to spin faster than the motor is spinning, the sensor indicates that the spin is too fast or the load is too light. If the spindle is trying to spin slower than the motor wants it to spin, the sensor indicates that the spin is too slow or the load is too heavy. What would that sensor be called? I'm sure they exist, but I don't know the name of the sensor. 68.115.219.130 (talk) 18:43, 18 June 2019 (UTC)[reply]

Tachometer Andy Dingley (talk) 18:48, 18 June 2019 (UTC)[reply]

Types of Wheel speed sensor that may be used inside or outside electric motors include slotted-disk opto-isolators using Photodiodes and magnetic Hall effect sensors. DroneB (talk) 21:09, 18 June 2019 (UTC)[reply]

A bit off track from the question, but are there any motors that run on A/C? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:41, 18 June 2019 (UTC)[reply]

most did, actually. Induction motor Gem fr (talk) 23:48, 18 June 2019 (UTC)[reply]

Very few types of motor run on DC. Many run on AC. The two common types for DC are the permanent magnet brushed DC motor and the universal motor, which runs on AC or DC. Both of these use an internal commutator to convert the DC supply into an AC current within the motor, in phase with its rotation.

Rotary electric motors rely on some rotating magnetic field internally, generated by an AC current. The motor rotates at, or close to, some simple multiple of this frequency. The difficulty with AC motors in the first half of the 20th century wasn't in making them, but in controlling their speed. From the very earliest days of AC it was recognised that motors had great advantages in running from an AC supply, but they (mostly) would only run at one speed (Often a half or one-third speed option too, by some simple switchgear, but nothing to make a continuous change). DC motors, in contrast, could be speed-controlled almost from the outset. When these AC motors were used for the first electric railways, such as the Italian three-phase system, the motors would typically run at the synchronous speed, accelerating or decelerating to this speed quite effectively on their own, without any control gear being needed. This gave simple operation for freight services, but a rather rough (and speed-limited) ride for passengers, as the train drivers couldn't adjust their speed to the track conditions. Andy Dingley (talk) 01:56, 19 June 2019 (UTC)[reply]

I'm not sure exactly what you mean by the spindle is trying to spin faster than the motor is spinning, but governor (device) may be relevant. CodeTalker (talk) 03:49, 19 June 2019 (UTC)[reply]

Maybe a torquemeter? DMacks (talk) 12:31, 19 June 2019 (UTC)[reply]

It sounds like the motor and spindle are actually going the same speed, but you want to know whether the motor is driving the spindle or is a drag on it. The drag part could either happen if something else is driving the spindle, or if the inertia of the spindle keeps it spinning after the motor power is reduced. A strain gauge would tell you which is the case based on if it is in tension or compression. It measures microscopic stretching of the material (strain), which tells you how much of a torque is applied (stress). Or, alternatively, you could just measure the speed of the motor/spindle assembly, and compare that with the power being used by the motor. This info could be combined to figure out how much of a drag there must be on the motor (assuming it's working properly), or even if it was getting "help" from the spindle. SinisterLefty (talk) 14:27, 19 June 2019 (UTC)[reply]

That sounds like a shaft torquemeter (which isn't anything like a dynamometer). They're not usually very successful trying to use a strain gauge for this, as strain gauges are good at measuring 1-dimensional strain and this is more like a 2-dimensional shear. The usual measuring technique is more like a phase meter, measuring the torsion between two ends of a narrowed quill shaft. Put a pair of discs on the shaft, one with a radial slit, one with a spiral slit. Shine a collimated light beam parallel to the shaft and you'll see a narrow spot, once per revolution, where the slots cross. As the shaft twists, their relative position changes and the spot moves in and out radially. There are many variations on this idea. Andy Dingley (talk) 16:31, 19 June 2019 (UTC)[reply]

If a torquemeter and a dynamometer are nothing alike, the redirect from the first to the latter should be replaced by an actual article on torquemeters (or we could just redirect them to Torquemada and call it good). :-) SinisterLefty (talk) 01:20, 20 June 2019 (UTC)[reply]

The fundamental task of a dynamometer is to apply a braking load, secondly to record that force (or torque). The torquemeter only measures, it doesn't apply a brake. Andy Dingley (talk) 01:37, 20 June 2019 (UTC)[reply]

Solubility of salts in presence of other salts

Is the solubility of a salt like sodium chloride modified when potassium chloride is also present in various mole fractions up to saturation? A similar question for potassium chloride in presence of sodium chloride! A similar question for sodium chloride in presence of potassium bromide! How does the non-ideal nature of the mixture influence solubility of each salt in presence of the other? Thanks!--37.251.220.173 (talk) 21:25, 18 June 2019 (UTC)[reply]

If you mean "can I dissolve less NaCl in a KCl solution" then short answer: yes. Solubility#Solubility_of_ionic_compounds_in_water indicates why. You'll find more info in Category:solutions Gem fr (talk) 22:33, 18 June 2019 (UTC)[reply]

How do the ionic strength and activity coefficients influence the mixed salts solubility? Are there any predictive models? --37.251.220.173 (talk) 15:46, 19 June 2019 (UTC)[reply]

The most basic model is simply to solve for conditions that satisfy all relevant equilibrium constants simultaneously. Someguy1221 (talk) 16:23, 19 June 2019 (UTC)[reply]

June 19

Black holes

What is the internal temperature of a typical black hole? And why? 86.8.201.134 (talk) 01:00, 19 June 2019 (UTC)[reply]

If I'm reading Black hole correctly, it's very close to absolute zero, and it has to do with a lack of internal energy. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:06, 19 June 2019 (UTC)[reply]

You are not reading correctly. The low temperature mentioned (a few nanokelvin) is the temperature of its horizon, not its surface, and not the internal. The typical black hole sucks energy (=mass) out of the 2.7K background radiation more than it loses, and has tons (so to speak) of internal energy. Gem fr (talk) 09:47, 19 June 2019 (UTC)[reply]

My understanding is that black hole thermodynamics struggles to answer this very question, still open. Said otherwise: we cannot answer Gem fr (talk) 09:47, 19 June 2019 (UTC)[reply]

The issue is "where" in the black hole you look. As a corollary, consider the bottom of the ocean. Due to pressure, the water at the bottom of the ocean is forced to be as condensed as possible. When water is as dense as possible, it will be about 3C or 35F. The pressure is the primary factor in the resulting temperature. But, if you go up, pressure decreases and the temperature can be influenced by other factors. In stellar pressure systems like neutron stars and black holes, it gets very weird. Models of neutron stars are attempting to work out what happens when the pressure on matter is so high that electrons and protons merge to form neutrons. Then, everything is a neutron, commonly called neutron matter. What is the temperature of neutron matter? It is very close to absolute zero because any energy that exists is barely enough to keep the neutron from collapsing. In other words, the density of the neutrons inhibits energy which inhibits temperature. Inside a black hole, neutrons should collapse into quark matter. The energy keeping the neutrons stable is lost. With less energy, there is lower temperature. An expected model is that inside the event horizon, there is inward rushing matter that collapses around the singularity first as neutron matter. Inside that it collapses into quark matter. Inside that, it collapses further into a singularity, but bosons create a huge problem. They don't collapse (theoretically). So, is the singularity a mess of boson matter? If so, can it retain energy? What if bosons have nothing to do with it? What if it all collapses into string energy? Without mass, in the relative sense, what is the temperature of a singularity of pure energy? That makes no sense. Hopefully, it does make sense that there is a big hunch that once inside the event horizon temperature will drop quickly towards absolute zero. But, when you hit the singularity, temperature no longer has meaning. 12.207.168.3 (talk) 17:49, 19 June 2019 (UTC)[reply]

June 20

Likelihood of death from influenza

Hello. As a general question, if a person contracts influenza, how likely is it that he or she will die as a result? An approximate answer is OK. I gather that people are more likely to die of this condition if they are very young or very old. Is that correct? FreeKnowledgeCreator (talk) 00:50, 20 June 2019 (UTC)[reply]

Quoting from our article Influenza, "Influenza spreads around the world in yearly outbreaks, resulting in about three to five million cases of severe illness and about 250,000 to 500,000 deaths." This implies a death rate of about 10% for those with "severe illness". The Spanish flu pandemic of 1918-1919 killed approximately 50 million people. It began in the final months of World War I and ravaged the planet. The virulence of the dominant strain of flu varies from year to year. The elderly are always very vulnerable especially if in poor health, but some strains seem to affect younger people more than others. On a personal note, my grandfather's first wife died in 1919 in that pandemic, and she was in her 30s. He remarried my grandmother a couple of years later, and I am the son of my grandfather's youngest child, my mother who was born in 1930. If it had not been for that pandemic, I would not be alive today. Every cloud has a silver lining. Cullen³²⁸ Let's discuss it 01:08, 20 June 2019 (UTC)[reply]

The different strains can have very different mortality rates, and the availability of hospitalization also has a big effect. Then there's difficulty in calculating just how many people actually have the flu each year, as mild cases and the many other conditions that produce flu-like symptoms are never distinguished for those who don't seek medical attention. And yes, young and old, and people who are otherwise immune compromised or just in poor health are at much greater risk. Our article puts the average number of deaths each year in the US at 36,000-41,400, which would put death in the range of 1 out of every 1000 cases (the death rate is much higher for those hospitalized with the flu). Also see [2]. However, some strains can kill millions, such as the 1918 flu pandemic (Spanish flu) mentioned above. SinisterLefty (talk) 01:11, 20 June 2019 (UTC)[reply]

Just to build on the above, it seems that some strains of flu trigger what was known as a cytokine storm, but which now is known by the much less cool name of "cytokine release syndrome". In instances such as that, the healthy young adults experience much higher mortality as their own immune systems spin out of control - the 1918 pandemic in particular has been identified with this. Matt Deres (talk) 14:38, 20 June 2019 (UTC)[reply]

Anti-venom

How is anti-venom made, in layman's terms? I know they need to extract or milk snakes to get venom to be used in the creation of anti-venom, but how? Thanks Anton 81.131.40.58 (talk) 10:47, 20 June 2019 (UTC)[reply]

Inject a survivable amount of venom into an animal, perhaps a rabbit, and then collect and purify antibodies against that venom out from the rabbit's blood. Alternatively, collect white blood cells from the rabbit and screen for ones that secrete antibodies against the venom, then grow those cells in a laboratory to make as much as you want without wasting more rabbits. Read more at Antibody#Research_applications. Someguy1221 (talk) 11:07, 20 June 2019 (UTC)[reply]

Snake antivenom. DroneB (talk) 12:21, 20 June 2019 (UTC)[reply]

Why do news stories about someone who was snakebit commonly refer to it as the synonym "antivenin?" Edison (talk) 14:34, 20 June 2019 (UTC)[reply]

It's French. It's also called antivenene. Matt Deres (talk) 14:54, 20 June 2019 (UTC)[reply]

In some English-speaking countries "antivenin" is the word. It's not a kind of venom, so "anti-venom" looks like an error. --76.69.116.93 (talk) 22:34, 20 June 2019 (UTC)[reply]

EO says "antivenin" dates to the 1890s and that the trailing "in" may be a chemical suffix.[3] ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:29, 21 June 2019 (UTC)[reply]

In Australia most of the anti-venenes are made using horses rather than rabbits. Treatment using this serum on a person for more than one episode can trigger an allergic reaction to horse serum on the second or third use. Hence most snake-handlers who get bitten try to not use the anti-venene unless/until it is absolutely neccesary. 49.197.107.84 (talk) 06:56, 21 June 2019 (UTC)[reply]

Are toothpastes exercises in branding?

I saw a new brand of toothpaste at the drugstore, parondontax, whose sole active ingredient is stannous flouride, or tin(II) fluoride. That's no different from, say, Crest Pro-Health, except maybe in price and manufacturer (gsk). What's the diff? 104.162.197.70 (talk) 11:56, 20 June 2019 (UTC)[reply]

Everything like that is an exercise in branding, more than anything. But stannous fluoride is a good thing, and not universal, even in fluoride toothpastes, as most of those still use sodium monofluorophosphate (MFP). Andy Dingley (talk) 12:13, 20 June 2019 (UTC)[reply]

I believe stannous fluoride is the least effective toothpaste and sodium fluoride in in the middle and sodium monoflurophosphate is the most effective of the 3 fluoride toothpaste ingredients available in the US not sure about non fluoride toothpaste but I do know in the US at least Crest mostly uses sodium fluoride and Colgate mostly uses sodium monoflurophosphate and stannous fluoride is in the sensitive products as it irritates teeth the least of the fluoride toothpastes 64.222.180.90 (talk) 18:33, 20 June 2019 (UTC)[reply]

Our articles on the relevant chemicals disagree with you about the relative effectiveness (for various meanings of that term) and have some citations to support. DMacks (talk) 19:16, 20 June 2019 (UTC)[reply]

Products are about more than just the active ingredients. Take alcoholic drinks. Alcohol is the only "active ingredient", but the other ingredients make all the difference in the experience. In the case of toothpaste, there's whiteners, breath fresheners, tartar reducing agents, flavors, etc. SinisterLefty (talk) 03:50, 21 June 2019 (UTC)[reply]

It's hard to recommend specific toothpastes without seeing the patient (so talk to your hygienist, this can actually matter). For most people, most good toothpastes are equally good. But there are some patients where there will be variations. If you suffer from a dry mouth (for whatever reason), then stannous fluoride is certainly preferable to the sodium fluorides, for its antibacterial effect compensating for that normally provided by saliva. If you suffer from staining after stannous fluorides (rare, but it can happen), then the better toothpaste formulations have ingredients to deal with that, which cheap ones might not. But it's a relatively rare problem, specific to some patients, and if you haven't got the problem, you don't need the solution. Andy Dingley (talk) 11:16, 21 June 2019 (UTC)[reply]

June 21

People with influenza drinking coffee

Hello. If a person contracts influenza, is it considered safe or medically desirable for him or her to drink coffee? FreeKnowledgeCreator (talk) 03:05, 21 June 2019 (UTC)[reply]

We don't do medical advice.

As to coffee and influenza, then advice is split anyway. Hydration is important, thus drinking something; but coffee can have a diuretic effect, thus the opposite. Paracetamol has been standard advice for years, although recent studies (Respirology, 2015) found no effect – but then placebos are still one of the most effective treatments for flu generally, even better than homeopathy. Many proprietary formulations of paracetamol or ibuprofen will include caffeine anyway, which has a useful effect for the rapidity of their effects on acute headaches, but no relevance to flu.

So no-one knows. And if they did, we wouldn't tell you here anyway. Andy Dingley (talk) 11:11, 21 June 2019 (UTC)[reply]

The dose makes the poison.41.165.67.114 (talk) 11:47, 21 June 2019 (UTC)[reply]

Long-term complications from microfracture surgery?

I heard somewhere that for professional athletes, complications may develop approximately eight years after undergoing microfracture surgery. Is there any literature out there which supports this? 67.83.118.216 (talk) 05:38, 21 June 2019 (UTC)[reply]

This paper apparently on 1200 patients claimed worsening results in 5% of patients, but no control group and honestly this is a very weird paper, I wouldn't trust it. I get the sense this paper was written to promote the practice. [This much smaller paper https://journals.sagepub.com/doi/full/10.1177/0363546508316773] reported 6 out of 17 patients seen over 5 years saw their symptoms get worse after surgery. [This seems to be a quite comprehensive article https://journals.lww.com/jbjsjournal/Fulltext/2007/10000/A_Randomized_Trial_Comparing_Autologous.2.aspx] and compares to a group treated with autologous chondrocyte implantation. They found that the "failure" rate increases from 2 years to 5 years, going from about 5% to about 23% in both groups. Yet another paper also found a proportion of patients who did worse over up to 10 years of followup, though they get extremely technical and don't have just one value to report. They also mention in their discussion another randomized trial claiming that the chance of backsliding seems to be higher for microfacture treatment than for other treatments over a ten year followup. Concluding, though, I think in all of these cases it may be hard to tell unless you really dig into case reports whether the complication is caused by something actually going wrong as a result of the surgery, and something just deteriorating as it was going to regardless. Seems to be a forgone conclusion that as a population, people getting either treatment are better off on average than people who got nothing, but I didn't look back to find the articles that support that claim. The last paper I linked does describe how many patients required additional surgery, so maybe that helps give an idea of what you're looking for. Someguy1221 (talk) 06:01, 21 June 2019 (UTC)[reply]

Immune system and infectious diseases

Are the chances of a person contracting a serious infectious disease influenced by the strength of their immune system? So for example, if 2 people were exposed to the same virus or bacteria which causes a serious illness, would the chances of one of them developing symptoms and being affected by it be influenced by the strength of their immune system? I guess my question is why is it that not everyone is affected when exposed to the same bacteria, virus or other disease causing agent? 90.210.250.247 (talk) 21:51, 21 June 2019 (UTC)[reply]

Yes, the general strength of the immune system is important, but there's also specific immunity to specific organisms. For example, if you've previously been exposed to a disease, or had a vaccine, then your body may still have antibodies designed to target that organism, which gives your immune system a significant jump-start on fighting off a new infection. This immunity tends to fade with time, and some infectious diseases, like the flu, mutate every year, so immunity doesn't last. Also, there's genetic immunity. For example, those with one copy of the sickle-cell gene are immune to malaria. There are many other examples. SinisterLefty (talk) 23:32, 21 June 2019 (UTC)[reply]

Why do people have fingerprints?

One reason I can think of is that fingers are too slippery otherwise so grabbing objects would be more difficult, but that doesn't sound convincing. In fact, those kind of wrinkles are visible on the rest of the inside of hands as well. Maybe it's simply "cheaper" (evolutionary wise) to do have wrinkled fingertips where it would cost more to have a flat skin, but then I'd like to know what's so different about my nose, or the upper side of my hands that don't have this feature. Joepnl (talk) 23:18, 21 June 2019 (UTC)[reply]

Your first instinct was right, it's for improving grip. Our palms are also used to grip things, so they have such wrinkles, too. SinisterLefty (talk) 23:27, 21 June 2019 (UTC)[reply]

Also for (or as outcomes of) improving touch sensitivity through friction, maintaining the health of the Epidermis, and helping the Epidermis and Dermis to cohere. See Fingerprint#Biology and Dermis#Dermal papillae.

(This being a reference desk, let's all remember to give actual references, people :-). Sometimes we might learn/remember additional facts ourselves as well as better answering the querants.) {The poster formerly known as 87.81.230.195} 2.122.177.55 (talk) 00:07, 22 June 2019 (UTC)[reply]

At least Fingerprint#Biology is completely wrong saying "These ridges may also assist in gripping rough surfaces and may improve surface contact in wet conditions" linking to an article that says the very opposite. "Scientists say they have disproved the theory that fingerprints improve grip by increasing friction between people's hands and the surface they are holding." ... "This confirmed that fingerprints do not improve our grip, because they actually reduce our skin's contact with the objects that we hold." Which is what I was thinking. Not convinced of health reasons either. Even without clothes, my upper legs happen to experience friction when walking, and they can do without wrinkles. Joepnl (talk) 00:41, 22 June 2019 (UTC)[reply]

That source sounds wrong, then. Car tires have treads to increase traction. This prevents hydroplaning, and something similar can also happen between smooth skin and a smooth surface, with a layer of liquid between them, like oil and/or water. And the area of contact does not have a direct effect on friction. The formula for frictional force specifies that it is only a function of the coefficient of friction and the normal force. And that coefficient of friction is reduced if a layer of liquid is between the two objects, which is why it's important to provide ridges that extend above the troughs containing the liquid, whether on a tire or a finger.

As for your legs, I'm guessing they aren't very hairy, as hair seems to serve the purpose (among others) of reducing friction. This is why many land animals have hairless toe pads, on an otherwise hairy paw, to increase traction. SinisterLefty (talk) 00:46, 22 June 2019 (UTC)[reply]

Even if that's true, what would my great (^50) parent need to pick up that's slippery where it would help to develop special wrinkles to hold it? Joepnl (talk) 01:07, 22 June 2019 (UTC)[reply]

Re: hair: where the friction takes place it's hairy like a baby. Joepnl (talk) 01:09, 22 June 2019 (UTC)[reply]

Not surprisingly, other primates - gorillas, chimps, etc. - also have fingerprints, but the surprising one to me is that koalas have them too - https://www.abc.net.au/news/2018-07-01/biometrics-koalas-and-wood-glue-fascinating-fingerprint-facts/9920802?pfmredir=sm It's believed to serve the same purpose as in humans. Remember that when a koala grabs your smart phone. HiLo48 (talk) 00:20, 22 June 2019 (UTC)[reply]

A case of "sticky fingers" ? SinisterLefty (talk) 01:04, 22 June 2019 (UTC)[reply]