Wikipedia:Reference desk/Science

Welcome to the science section
of the Wikipedia reference desk.

skip to bottom

Select a section:

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

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

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

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

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

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

April 29

Planets

In the Planet article, it says that in order to be considered a planet, an astronomical body needs to be massive enough to be rounded by its own gravity. About how much mass is required for this? Is there some kind of "minimum mass" for a planet? 193.210.228.37 (talk) 11:43, 29 April 2020 (UTC)

This will depend on how strong the material is. If it is the very weak solid nitrogen, the smallest size would be much smaller than for an iron nickel asteroid. Graeme Bartlett (talk) 11:45, 29 April 2020 (UTC)

Here is an assessment in terms of radius. The radius is more useful for astronomers because it's easier to measure than mass (you need e.g. a satellite to measure mass). As expected the necessary radius depends a lot on the body's composition. 93.136.9.236 (talk) 15:30, 29 April 2020 (UTC)

Pluto is certainly round, but Tyson decided it wasn't a planet anymore, so it isn't. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:49, 29 April 2020 (UTC)

Because being round isn't the only qualifying factor to make something a planet. Tyson didn't decide anything, either. The International Astronomical Union decided on a definition. Tyson has been a proponent of said definition, but it wasn't his call. --OuroborosCobra (talk) 22:49, 29 April 2020 (UTC)

He championed it, so he gets the blame. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:57, 30 April 2020 (UTC)

That's not how this works. That's not how any this works. --OuroborosCobra (talk) 16:03, 30 April 2020 (UTC)

"Roundness" is not the only criterion. Pluto fails on the other tests for planetness. Or, to be more exact, if Pluto is a planet, then so are umpty-dozen other things in and around its orbit and in and around the asteroid belt. Which would make planet a less useful term. --Khajidha (talk) 17:36, 30 April 2020 (UTC)

How I Killed Pluto and Why It Had It Coming by Michael E. Brown is well worth reading. Mikenorton (talk) 17:53, 30 April 2020 (UTC)

Ceres with some more than 900 km and a mass of 9.5 × 1020 kg was the smallest celestial body thought to have been rounded by its own gravity. I think I read somewhere that another body with about 400 km could be also rather round and this would be probaly an example of minimum mass. But yes, mass alone is not all: a stony body covered with water ice will possibly get round faster than some irregular chunk of iron-nickel. 2003:F5:6F0A:C00:5CC0:1EE6:8417:F1F (talk) 20:17, 30 April 2020 (UTC) Marco PB

The self-styled experts changed the definitions. Pluto never changed. And Pluto will be around long after those who dissed it are gone. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:14, 30 April 2020 (UTC)

That's a very unscientific way of thinking. You do understand that the category "planet" is just a creation of the experts that is used to classify things so that we can understand them better? Also, experts had been debating Pluto's "planetness" basically since it was discovered.--Khajidha (talk) 22:31, 30 April 2020 (UTC)

And speaking of Ceres, when it was discovered in 1801, it was universally hailed as a newly-found planet. All decent science books from then on included Ceres along with Jupiter, Mars etc and all the other then-known planets. This situation obtained for almost 60 years (almost as long as Pluto was regarded as a planet), until it was reclassified as an asteroid after many similar-sized objects were found, and it dropped out of general consciousness so totally that most people these days have never even heard of it. That's unlikely to happen with Pluto, mainly because of all the hullabaloo after it was reclassified as a dwarf planet. I actually accuse most scientists of being closet astrologers whose equanimity was disturbed when they had to go back and recalculate all their horoscopes with Pluto left out. That's what this is really all about, mark my words. -- Jack of Oz ^{[pleasantries]} 23:47, 30 April 2020 (UTC)

I'm a supporter of Alan Stern's idea to call them "überplanets" ("main planets" sounds better to me). Pluto, Eris and Sedna are basically planet-sized and planet-shaped bodies that happen to be very far away from the Sun. Could've gone with that but that would mean avoiding a huge publicity grab... 89.172.8.118 (talk) 01:09, 1 May 2020 (UTC)

Baseball Bugs If you're interested, I strongly recommend this casual YouTube video from CGP Grey. Link here. --Puzzledvegetable^{Is it teatime already?} 00:00, 1 May 2020 (UTC)

What exactly does it mean for a photon to have an amplitude?

Given the extent to which photons cannot be described as classical waves, what exactly does it entail to measure the amplitude of one? As in, what are you physically measuring and called the 'amplitude', if the photon itself does not truly possess a 'height' the way a classical wave might. Apologies for any misconceptions inherent to the question itself. — Preceding unsigned comment added by Opossum421 (talk • contribs) 11:50, 29 April 2020 (UTC)

Amplitude in quantum mechanics refers to the level of probability of making that measurement at that value. See Probability amplitude (though that article like many at Wikipedia is probably too technical for someone unfamiliar with the field. Le sigh.) The basic way to think of the wave function is that it isn't a wave in space-space, it's a wave in probability space. Only the position wave function includes space-space in it, and even that isn't telling you where the particle is, it is telling you where the particle is likely to be if you were to try to interact with it. The particle isn't anywhere in particular until you interact with it, then it's wave function changes to a more narrow shape showing the increased probability that the particle is located where you interacted with it. There are lots of other wave functions, however, that do not have space or position in them, such as momentum or spin orientation or the like. If there was one thing about QM that most people get wrong (often because teachers either themselves don't get this, or don't make it explicit enough) is that quantum mechanics abandons the notion that a fundamental particle is an object in any sense we understand the term "object" to mean. Some properties of quantum particles are definite and well-defined, like electric charge and spin magnitude and things like that, but other properties (including position and momentum) are not well defined, meaning those values can only be described by giving the probability of finding the particle with that value for the measurement. If we graph all of those probabilities on a coordinate system, where the Y axis is defined as the probability of finding the particle with that value, and the amplitude is just the maximum probability of finding the particle with that value. The mathematics of the graph we draw obeys the mathematical rules of wave mechanics, which is why it is called a wave function. But again, it isn't a wave in the real world. It's a wave in probability space, which is a mathematical construct where the property being measured is plotted on some of the axes, and one of the axes is a probability. (This is a massive oversimplification, and ignores the difference between, for example Ψ and Ψ² and a WHOLE lot of messy math, but it captures the spirit, which is all we need here). I highly recommend a few channels on YouTube that explain this even better than I can through text are Science Asylum (Nick Lucid) and 3Blue1Brown (Grant Sanderson). Science Asylum does a fantastic job of explaining complex physics phenomena in a short, informative, and easy to digest way (pretty much the OPPOSITE of Wikipedia articles on the subject) while 3Blue1Brown covers complex mathematics in a slow, methodical, but still fantastically clear manner that also makes these things easy to understand. This video that Science Asylum did on the Wave Function is probably most germane to our discussion, but his entire series on QM would be useful. I don't have a specific 3Blue1Brown video to recommend on this direct topic, but he's done lot of videos the mathematics useful to understanding QM (such as this video on the uncertainty principle [1] which starts off his whole series on the Fourier transform, which is a mathematical tool with many wide-ranging uses, but which is SUPER important to understanding the mathematics of QM). I hope that helps some. --Jayron32 13:19, 29 April 2020 (UTC)

Like many simple questions in physics, there is a long and convoluted answer! To summarize shortly, whomever spoke of the photon's amplitude was being sloppy and imprecisely using terminology - and this sort of nonsense is very common amongst practicing physicists, because as a community, they've generally proven to themselves that they can be precise when they need to be, and they know that they don't usually need to apply that precision, because being excruciatingly correct all the time becomes very tiresome.

So - to the question: what is the amplitude of the photon?

To recite the rote-memorized answer, that would be a quite trivial question! Why, the amplitude of "the wave function", of course.

Now, the thing that's funny about quantum physics is that the books all speak in grandiose terms about the wave function; they have a special equation for the function; and they even have a special-purpose symbol, $\Psi$, which is exclusively reserved to represent the function. And if you read any book on the topic, the amplitude of $\Psi$ is a scalar; and it is an abstract quantity that can be used to predict the probability and the time-evolution of measuring some other physical observable. But what is it?

If you read a little deeper, and learn what the function is, you'll discover that it's actually unique to every single physical system. There are actually many different wave functions; the part that is common to all of them is that they must satisfy certain mathematical and physical requirements.

In the case of the quantum-mechanically-correct description of the photon, the actual equation that describes the wave function is not so surprising: it is derived from the very same relationships that describe the classical electromagnetic wave; and that wave is easily represented using the classical wave equation of conventional electrodynamics. That wave is a pair of coupled vector-quantities; the two components - a magnetic field and an electric field - have position and direction, but they have no extent.

By convention, we often simply work with the amplitude of the electric field and we ignore the amplitude of the magnetic field; but it doesn't really matter, because we can equally convert to calculate the amplitude of the magnetic field. Because the two fields are related by a very well-behaved governing-equation, we can easily switch between electric- and magnetic- field amplitude; this is simply a change of units, with a little extra-special mathematical care in the case the wave is propagating inside an imperfect material.

To make the equations for these amplitudes quantum-mechanically-correct, we just need to make sure that the time-evolution of the system abides by any quantized interactions; but in the case of the electromagnetic wave in a vacuum, there's no change from the classical case: the photon is the quantum particle, and its quantum mechanical wave function is the classical electromagnetic wave function.

The extra work to manipulate the equation so that it looks like the Schrödinger equation is really just a rearrangement of variables - and when you succeed in doing that, you've simply rearranged the same thing into a format that is particularly inconvenient for expressing macroscopic interactions between the photon and the outside world. But that form is useful for describing special cases in physics - things like photon-material interactions; or making statistical predictions about the probabilistic behaviors of individual photons; and so on. For those cases, the amplitude of $\Psi$ is used to predict the amplitude of E and B (the electric field and the magnetic field).

And of course, because the actual values of those field amplitudes are themselves a probabilistic attestation deduced from the amplitude of $\Psi$, we expect that $\Psi$ should evolve predictions that, over time, change in a manner that is statistically indistinguishable from the time-evolution of the amplitudes of E and B, in other words, the classical wave equation.

Nimur (talk) 13:27, 29 April 2020 (UTC)

Nimur's answer is also really good here, as it draws comparison to "classical" wave mechanics and quantum mechanics. I just want to point out that his analogy between the the classically defined EM wave equation and the quantum wave function is really useful here, but maybe not in the direction one thinks; for example, when one asks "what is waving" in the EM wave, we say "the electromagnetic field". This is similar to saying that the thing that is waving in an ocean wave is the ocean, except that the ocean is a real physical stuff we can pick up and touch and is an object. The electromagnetic field is not a stuff. Like all fields (classical or quantum) it is a mathematical construct. A field is just a set of numbers (which can be scalar, vector, or tensor) that we attach to space itself, describing a specific property that space has. The "manipulation of variables" and the other things Nimur talks about relating the quantum wave function to the EM wave itself are just a set of mathematical transformations. Just as you can do math to convert one function to another (for example, the way that the exponential function e^ix can be used to convert between circular motion and wave motion, there are also trivial mathematical tools we can use to convert between EM waves in the EM field to quantum wave functions in probability space. Such mathematical tools change our perspective on a phenomenon, but don't fundamentally change the phenomenon itself). So, when we say "what is waving" in an EM wave, it's the values we assign to the various points in space that we call the EM field, and when we say "what is waving" in the quantum wave function, it's the probability of the particle having the values in question. And because those functions are describing the same phenomenon from different perspectives (both describe light), there must be some mathematical tool to relate them to each other. --Jayron32 13:55, 29 April 2020 (UTC)

Thank you, Jayron. Your post has helped me to realize something: I am just assuming that everybody already knows what electric fields and magnetic fields are. Of course, if our readers are not intimately familiar, they can review those articles.

If I may coopt Jayron's terminology - and simplify by glossing some of the details - the "thing" that is "waving" is the Electric field (and the B-field, per our earlier discussion). The E-field is not "stuff." It is a set of numbers that tell us what force would be felt by a charged particle if it were at a specific place. Those numbers are what are waving up and down. If we had a way to measure those numbers, we could see them fluctuating in a more-or-less perfectly sinusoidal fashion. But the only way we can measure those numbers is to place a test-particle at some specific position, and watch how it moves.

The really tricky part is that if we actually put a charged particle there, the waving numbers that tell us what force the particle should feel would change because they would interact with the charged particle; it would have its own fields; and it would also be moving as the wave interacts with it. So these wave-equation representations are mathematical idealizations - they help us make useful predictions, and they only make helpfully-valid predictions when we consider statistically-large numbers of individual particles, so long as we are willing to ignore the effects of any individual particle.

It isn't necessarily obvious - great scientists became famous for finally realizing this very subtle detail! - but the E- and B- fields are deeply deeply related to the photon. Early physicists described electricity, and magnetism, and light; and for a lot of centuries, we thought of them as three separate things; but ... we now understand that all of them are actually the exact same thing under different conditions. Photons interact with E- and B- fields - but even more importantly, photons are actually made of E- and B- fields. There's nothing else inside of a photon except those fields - and those fields are not made of "stuff"! And if you're a real mathematical hot-shot with access to a good book, you can play some advanced vector-calculus-trickery to demonstrate that the E-field and the B-field are the exact same thing, too: they're not merely related or coupled fields - they're the same exact darned field. And this is the equally important bit, that's less-often shouted at students: if photons are made of E- and B- fields, then .... every E- or B- field you can conceive can be expressed as a photon. It might be a pathologically poorly-behaved photon - it might have a ludicrously useless wavelength; but ... there you have it. Photons-the-size-of-mountains, emitted by electric-wires-strung-between-peaks.

So if you're trying to figure out what "stuff" is "waving" when an electromagnetic wave ... "waves"... well, we can quickly find ourselves traveling down the rabbit-hole of physical definitions: what is "stuff"? Which physics-ese techno-jargon word corresponds to the plain English word "stuff"? (Matter? Mass? Momentum? Quantum state exclusivity?) And I am pretty sure our only conclusion will be that natural human languages are not really expressive enough to describe some of the physical realities unless we throw the weight of a lot of words at it.

Nimur (talk) 17:22, 29 April 2020 (UTC)

All of that being said, I really do recommend the OP watches the videos I recommended. Seeing someone explain something in a video format, with pictures and animations, can be a lot more enlightening than reading the same information. --Jayron32 18:15, 29 April 2020 (UTC)

The energy and momentum of a Photon which is the elementary particle of the electromagnetic field depend only on its frequency or inversely, its wavelength. Increasing the amplitude of a beam of light or radio wave doesn't create "stronger" or higher-amplitude photons, it just creates more of them. DroneB (talk) 21:39, 29 April 2020 (UTC)

Thank you tremendously folks, I hadn't anticipated such a thorough set of responses. To Jayron32's suggestion - I'm quite familiar with 3Blue1Brown and will surely venture down that particular rabbit hole along these lines, and have not heard of Science Asylum but will give it a proper go as well. I appreciate everyone's helpful explanations here. Opossum421 (talk) 11:12, 30 April 2020 (UTC)

Rooibus

Where can I buy a rooibus plant to grow in my garden?99.145.194.98 (talk) 19:42, 29 April 2020 (UTC)

Possibly you mean the Rooibus tea plant cultivated in South Africa. The American Herbal Products Association has a website that may help you locate sources and there is an Australian supplier. DroneB (talk) 21:22, 29 April 2020 (UTC)

Postage to and from Australia is very delayed at the moment due to limited international flights, so if you are in the USA, buy from USA. Graeme Bartlett (talk) 23:56, 29 April 2020 (UTC)

To import plants or seeds from abroad you need a phytosanitary certificate or a small lots permit, to make sure you are not bringing non-native pests or pathogens into the US. Please check USDA APHIS website for what documentation is necessary in your case, and how to apply for it properly. Best regards, Dr Dima (talk) 06:19, 30 April 2020 (UTC)

Googling [buy “Aspalathus linearis”|"Rooibos" seeds] produces several outlets.  --Lambiam 14:04, 30 April 2020 (UTC)

Try at your local plant nursery. 89.172.8.118 (talk) 19:00, 30 April 2020 (UTC)

April 30

Weider weight bench

Are there Weider olympic weight benches available anywhere?99.145.194.98 (talk) 05:28, 30 April 2020 (UTC)

Googling ""Weider Pro 395" brings up several companies offering these for sale.  --Lambiam 13:57, 30 April 2020 (UTC)

This isn't really a science question. --OuroborosCobra (talk) 16:04, 30 April 2020 (UTC)

Thanks for the help and kinesiology is science!99.145.194.98 (talk) 22:09, 30 April 2020 (UTC)

Your question didn't ask anything about kinesiology. You asked where to buy an olympic weight bench. That'd be like saying asking about where to purchase a tickle-me-elmo was a developmental psychology question. --OuroborosCobra (talk) 15:25, 1 May 2020 (UTC)

Light sail

In episode 2 of Cosmos: Possible Worlds they talk about using a light sail to send interstellar probes. At first they are powered by lasers from Earth, then they carry lasers to fire photons into the sail to keep it accelerating. But wouldn't the backward force of the photons leaving the laser be at least as large as the force on the sail? Bubba73 ^{You talkin' to me?} 20:42, 30 April 2020 (UTC)

Yes, it's the equivalent of blowing in your own sails on a regular sailing vessel. However, efficient light sails must be made of a reflective material, so if the sails reflect the light from the probe's lasers it has the same effect as just aiming the lasers the other way, so it would still work (but the sails aren't really needed at that stage). - Lindert (talk) 21:21, 30 April 2020 (UTC)

By the way, Mythbusters tested "blowing in your own sails on a regular sailing vessel". It works. I was surprised when I saw the episode (they didn't explain it very well) until I realized that it was equivalent to the thrust reversers on a jet plane: directing the air backward is, again, similar to aiming the fan the other way. --76.71.6.31 (talk) 21:51, 30 April 2020 (UTC)

Thanks, that is what I thought about the lasers and light sail. A light sail is supposed to have a very low mass. Is it feasible to have lasers and (presumably) batteries to provide enough energy to accelerate to relativistic speeds? Bubba73 ^{You talkin' to me?} 22:25, 30 April 2020 (UTC)

To accelerate to relativistic speeds on battery power (taking the batteries with you), you need batteries with an energy content that's large compared to the rest mass of the batteries. That won't work with chemical or even nuclear batteries. PiusImpavidus (talk) 08:30, 1 May 2020 (UTC)

Ah, yes, that is right. Bubba73 ^{You talkin' to me?} 22:20, 2 May 2020 (UTC)

May 1

Human psychology during pandemics

From a human psychology point of view, why are so many aspects of the current situation divisive amongst people across the world regardless of their background? Have any studies been done in the past to look at this? Clover345 (talk) 14:46, 1 May 2020 (UTC)

Examples? I'm guessing you mean religious vs. atheistic viewpoints? Like, "punishment from God" to nothing peculiar at all? But what other answer could you be looking for besides genetics. 67.175.224.138 (talk) 14:55, 1 May 2020 (UTC).

Genetics is not at all likely to be the answer or even what they are asking for if they are asking a psychology question. --OuroborosCobra (talk) 15:23, 1 May 2020 (UTC)

I'm not sure our original question is specific enough to receive a proper response; but when I find myself wondering about the softer sciences, I start by browsing the collection at JSTOR, a free digital library collection that provides a lot of scholarly research.

Via JSTOR: The Crowd in History: Some Problems of Theory and Method (Social History, 1978). "What might be called crowd history as a field of study in its own right has, rather surprisingly, received very little critical comment." Perhaps the interested reader will be able to take a deeper look into further materials.

For regular readers of AAAS's journal Science - rather more likely to overlap with our regular contributors on the Science Reference Desk - you might find this article very exciting: The Lessons of the Pandemic (Science, 1919). Like many other Science articles in the older archives at JSTOR, it is available at zero cost.

If I may editorialize, a bit: well written scientific analysis, from any century, provides an astonishing clarity that - to me - proves beyond reasonable doubt that despite the progress of a few scientific minds in a very few select areas of specific understandings of our natural world; and in spite of amazing the proliferation of advanced technology in daily life - the overwhelming majority of the knowledge that we have about our natural world has truthfully not changed very much at all since the so-called Dark Ages - because the overwhelming majority of knowledge is, and always shall be, embodied in the minds of very scientifically-uneducated individuals who constitute the so-called crowd.

Nimur (talk) 16:22, 1 May 2020 (UTC)

One thing that can be observed is that some people who are materially better off do not appear to fully grasp the economic hardship on others resulting from the closures. Economic relief may be denied or only given grudgingly and in insufficient amounts. The question when to re-open in what phases depends on risk assessments; these will be based on the information one receives, which may be rather different in different bubbles. The issue also depends on the relative values assigned to human life and health versus material wealth, which also may be based on one's ideology.  --Lambiam 16:39, 1 May 2020 (UTC)

That's common to pretty much every economic downturn, including the Great Depression. 89.172.65.59 (talk) 04:25, 2 May 2020 (UTC)

Source of names for 4 mild human coronaviruses?

• Human coronavirus OC43 ?
• Human coronavirus HKU1 Hong Kong ?
• Human coronavirus 229E ?
• Human coronavirus NL63 Netherlands? -- Jeandré, 2020-05-01t20:17z

You should read discovery papers. Ruslik_Zero 21:00, 1 May 2020 (UTC)

In most cases these are just the labels used in the labs that isolated and described the viruses to label the samples from which the strains were isolated. Each research group had its own labelling method. The "OC" in "OC43" stands for "organ culture",^[2] while "43" is probably a meaningless sequence number. Described in: McIntosh K, Dees JH, Becker WB, Kapikian AZ, Chanock RM. "Recovery in tracheal organ cultures of novel viruses from patients with respiratory disease". Proc Natl Acad Sci USA 1967;57:933–40. "HKU" is the abbreviation for the University of Hong Kong where the strain was isolated.^[3] Strain 229E was isolated by Dorothy Hamre and John Procknow and described in: Hamre, D., and J. J. Procknow, Am. J. Epidemiol. 83:238 (1966). Perhaps their publication explains the label, but I suspect it was simply the label of one lab sample among many, labelled ..., 228A, 228B, ..., 229D, 229E, 229F, ... . And "NL" is obviously the Netherlands.^[4]

Does pure ammonia expand when it freezes like water does?

Water, H2O, expands when it freezes. It has a different structure than NH3, because it only has 2 Hs, but isn't it frequently like H3O+, which might resemble NH3 structure?144.35.20.92 (talk) 20:44, 1 May 2020 (UTC)

No as far as I know. Ruslik_Zero 20:55, 1 May 2020 (UTC)

Our article on Ammonia gives among its properties densities of 681.9 kg/m³ at −33.3 °C (liquid) and 817 kg/m3 at −80 °C (transparent solid). As the mass of a fixed number of ammonia molecules will not change as the stuff freezes, this implies that a liquid cubic meter of ammonia will take up a volume of 681.9/817 m³ = 0.834 m³ when frozen.  --Lambiam 22:41, 1 May 2020 (UTC)

I had asked this question before about hydrogen peroxide, which is probably more similar to water than ammonia. And our article doesn't have its density as a solid, or talk much about it. People say this is a result of hydrogen bonding, but hydrogen bonding alone doesn't answer the question. As H2O2 and NH3 are both examples of hydrogen bonding. On a separate topic, I did have a list of other compounds that are like water with its density, I'ma look for it. 67.175.224.138 (talk) 00:38, 2 May 2020 (UTC).

The phase diagrams in this article [5] (pdf) imply that solid ammonia is denser than liquid ammonia. Higher pressure favors the denser phase, and solid ammonia is above the adjacent liquid ammonia on the pressure-temperature phase diagrams.--Wikimedes (talk) 04:31, 2 May 2020 (UTC)

In addition to the above, it is worth noting that water is not "frequently like H₃O⁺. At room temperature, for example, the concentration of H₃O⁺ = 10^-7 M. For perspective, water itself has a "molarity" of about 55.5 M. That means that in a given volume of water, only about 1.8*10^-7% of the molecules will be H₃O⁺. -OuroborosCobra (talk) 19:34, 2 May 2020 (UTC)

There is a one-sentence section at Freezing#Expansion that lists just two substances that expand when they freeze. The other one is bismuth. But it doesn't say how rare or common the phenomenon is (I've formed the impression that it's pretty rare, but I don't remember where I picked that up from, so don't believe me), or what sort of chemical properties cause it, or anything. I've just flagged the section, rather appropriately, as needing expansion. --76.71.5.208 (talk) 23:18, 2 May 2020 (UTC)

Plutonium does as well. Plutonium#Physical_properties last paragraph. But it is a rare phenomenon.--Wikimedes (talk) 03:46, 4 May 2020 (UTC)

How are inorganic archaeological artifacts dated?

Of course, any object containing the appropriate elements can be dated via isotope dating, but how do archaeologists date inorganic objects with any specificity? If a stone tool or fragment of pottery is found, is it simply dated according to other, more easily determinable nearby objects? Is it as simple as assuming a constant rate of wear-and-tear? Are there other convenient isotopes which decay at a useful time scale? Surely there are sophisticated techniques I am unaware of, and would be interested in hearing about. Opossum421 (talk) 23:27, 1 May 2020 (UTC)

When objects are found in an archaeological excavation, their context is recorded, particularly where they lay in the site's stratigraphy. Other datable objects are used to establish the age of the different layers and thereby the objects found within them. Mikenorton (talk) 00:14, 2 May 2020 (UTC)

Radiometric dating discusses several isotopes that are used in dating.--Wikimedes (talk) 04:17, 2 May 2020 (UTC)

Chronological dating contains many dating methods. Luminescence dating is an interesting one.--Wikimedes (talk) 04:20, 2 May 2020 (UTC)

As others have basically said, there are a lot of dating methods, and (ideally) more than one is used so as to validate results. When no method for a specific artifact is available, they may have to make use of referencing something else. So, for example, you find an arrowhead on its own made of a particular material and particular design, but for some reason you can't specifically date that arrowhead. You then look to see if other arrowheads of the same material, same design, and in roughly the same region have been found. Perhaps one was found together with other items, including a material that could be dated (hair? wood? there are tons of materials that we can date). Those materials then form a reference to date the arrowhead they were found with, and the arrowhead you found off on its own can be assigned a similar date based on referencing. It's not perfect, but it is better than nothing. This type of reference dating is done with extreme care and, given other evidence, can be revised. --OuroborosCobra (talk) 17:09, 3 May 2020 (UTC)

May 2

Is there a minimum practical size (thermochemically) for a blast furnace?

I know there are minimum economic sizes, with considerations for fuel and maintenance costs. I'm not asking about that.

From a thermal/chemical standpoint, there can be other considerations. For instance, you can't scale a bloomery furnace much below 30cm in inside diameter for reasons of fuel element size, air injection, and thermal loss through the (~10cm) clay walls. That means that in practice, the minimal size creates maximum bloom sizes between 2 and 20kg

I'm wondering if there are similar low-end constraints on a traditional (tall, brick, cast-iron-producing) coal+iron ore(+ lime) blast furnace. Anyone know?

Riventree (talk) 01:18, 2 May 2020 (UTC)

Making a miniature version of a full production-capable Blast furnace would mean sacrificing all Economies of scale and require much effort to assemble special tiny fire bricks, flues, bellows, etc., the ingredients would need to be unusually finely powdered and metered, and it would not be easy to keep a very small quantity of coal burning. However there is no lower size limit on performing the essential Smelting operation on Iron ore in a laboratory Crucible. DroneB (talk) 18:57, 2 May 2020 (UTC)

Maths question on pandemic graphs

In the pandemic graph, with time (t) on the x axis and infected people (c) on the y axis, I understand that the area under the curve remains the same if you reduce R(t) on the exponential curve. So in other words, it’s modelling the spreading out of the same number of infections over a longer period of time, in order to lower and delay the peak which I’ll call C-max. So how do you model this for countries which have eradicated this virus? The countries which locked down early seem to have reduced cases to 0 after a short period of lockdown but unless I’ve misunderstood, number of infections would have continued to rise if lockdown had not been imposed and so I don’t see how infected number of people would be the same by the end. So when you model this, which variable do you change? Or is it that this model becomes less accurate at the lower numbers of c since it will just tend towards infinity? 90.196.238.188 (talk) 13:15, 2 May 2020 (UTC)

Is it simply that the model assumes that in both scenarios (restrictions vs no restrictions) the same number of infections will occur which in reality wouldn’t be true? Or that R(t) gets closer and closer to 0 as the virus finds less people to infect? If this was the case, would this be so for every country? That R(t) will start tending towards 0 if the same measures are kept in place? 90.196.238.188 (talk) 13:30, 2 May 2020 (UTC)

We do not know enough to create a mathematical model with any certainty that it reflects the situation of the current pandemic, but an extremely simplistic model with a stationary population (ignoring demographics – so no births and no deaths from other causes) may help to clarify some things. Assumption 1 is that everyone is susceptible and may get infected if in contact with an infected person. Assumption 2 is that even vigorous contact tracing and isolation of contacts cannot entirely eliminate spreading (since they may have been infectious before they were identified and isolated). Assumption 3: no one (think people at high risk) can be kept indefinitely in isolation with zero chance of being infected. Under these assumptions, it is possible that the disease may get completely eradicated, like smallpox, before everyone is immune or deceased. But if we add assumption 4 that the disease will remain endemic with a low but non-zero lower bound on its incidence until no one is left to be infected, then eventually everyone will get infected, if not sooner then later, possibly much later. The total area under the curve of new cases against time (with linear scales and 0 at the bottom of the y-axis) is proportional to the total fraction of the population that has been infected. If that is the whole population, then all you can do is alter its shape. To explain the apparent paradox: the curve may have a thin but very long tail; its thinness is canceled out by its length. But if the disease is totally eradicated, that area may be genuinely smaller.  --Lambiam 15:36, 2 May 2020 (UTC)

Thanks, appreciate this is a very simplified model. So what would be the best way to represent that 4th assumption in this very simplified mathematical model? In my mind, if R(t) is less than 1, the number of infections will always tend towards 0? Do you basically make the assumption that R(t) will be just below 1? I know in reality, there’s a lot more to take account of but keeping with the simplified model. 90.196.238.188 (talk) 16:05, 2 May 2020 (UTC)

I guess representing what you’ve described mathematically would mean that the curve would tend towards infinity? So how do you vary R(t) to model this? — Preceding unsigned comment added by 90.196.238.188 (talk) 16:08, 2 May 2020 (UTC)

If we are working with proportions of the total population, so that the total solution corresponds to 1 (100%), the area under the curve is bounded between 0 and 1, inclusive. The effective reproduction number $R_{\mathrm {e} }$ at time $t$ can be found by multiplying the basic reproduction number $R_{0}$ by the fraction $S_{t}$ of the population that is susceptible at time $t$. This is described at Herd immunity#Mechanics; it ought to be mentioned also in the article on Mathematical modelling of infectious disease.

We can partition the total population into three compartments: $S$ for susceptible, $I$ for infectious, and $H$ for the rest, hopefully healed, but in any case the cumulative fraction that has at some time been infectious. At all times they sum up to the whole population: $S_{t}+I_{t}+H_{t}=1$. Individuals can transition from $S$ to $I$, and from $I$ to $H$. The transition rate from from $S$ to $I$ is given by $\mu R_{0}S_{t}I_{t}.$ The factor $\mu$ is needed because an individual's infectiousness lasts several days, and $R_{0}$ is the cumulative effect over that whole period. If we take one day as the unit of time, and we assume that an infected individual remains on the average infectious for 10 days, a reasonable value is $\mu =0.1$, the reciprocal of the infectious period. The transition rate from from $I$ to $H$ should depend on how long the members of compartment $I$ have been there, but for simplicity let us assume the continuous equivalent of a Poisson process and set the rate at $\mu I_{t}.$ Together this gives the coupled differential equations

${\frac {d}{dt}}S_{t}=-\mu R_{0}S_{t}I_{t};$

${\frac {d}{dt}}I_{t}=\mu R_{0}S_{t}I_{t}-\mu I_{t};$

We can eliminate $I$ by solving the first equation for $I$ and substituting the solution into the second:

${\frac {d}{dt}}\left({\frac {1}{S_{t}}}{\frac {d}{dt}}S_{t}\right)=-\mu \left({\frac {1}{S_{t}}}{\frac {d}{dt}}S_{t}-R_{0}{\frac {d}{dt}}S_{t}\right).$

I don't know if this can be solved analytically. I have simulated the process by a simple difference method: repeatedly replacing $t$ by $t+\Delta t$, $S_{t}$ by $S_{t}+\Delta S_{t}$, and $I_{t}$ by $I_{t}+\Delta I_{t}$, where

$\Delta S_{t}=-\mu R_{0}S_{t}I_{t}\Delta t,$

$\Delta I_{t}=~~(\mu R_{0}S_{t}I_{t}-\mu I_{t})\Delta t.$

I set $\Delta t=0.01$; smaller values give negligeably different results. To start the epidemic, we need a value for $I_{0}$. I used $I_{0}=0.0002$; the final outcome is not particularly sensitive to this, except for very small values of $R_{0}$. $H_{0}=0$, so $S_{0}=1-I_{0}$. In the end, $I_{t}$ tends to 0. I stopped the process 1000 steps (10 virtual days) after its value dropped below $5{\cdot }10^{-9}$.

Other than I had expected, $H_{\infty }$ was substantially lower than $1$ for moderate values of $R_{0}$; for example, $R_{0}=1.2$ resulted in $H_{\infty }=0.31448$, reached in 1201 days. For large values of $R_{0}$, such as $R_{0}>4$, I saw that $H_{\infty }\approx 1-\exp(-R_{0})$. For very small $R_{0}$, on the other hand, $H_{\infty }\approx I_{0}(1+R_{0})$. I have not attempted to see if these observations have a theoretical explanation.  --Lambiam 18:06, 3 May 2020 (UTC)

Thanks for the detailed answer. I see for simplicity you used differential equations. I’m guessing in reality, such complex modelling would require the use of matrices. Do you know if any of the studies have released the source code? 90.196.238.188 (talk) 19:07, 3 May 2020 (UTC)

On YouTube there are several videos explaining mathematical models and simulation models for epidemics. Looking at the thumbnail search results I saw that the toy model I developed above is actually a well-known one (not surprising, in view of its simplicity), and is known as the SIR Model, S and I as above and R for Recovered (or Removed) instead of H as I used. (I rejected the obvious R because of potential confusion with the R of reproduction number.) What is particularly unrealistic about this model is the assumption of homogeneous mixing. The major axes of refinement, needed to make this type of model more realistic, are demographic segmenting (some groups of people are more at risk than others) and spatial segmenting (network models representing different areas and travel between them, as well as different behaviour patterns inn different areas). For the long run, you need to remove the assumption that there are no R → S transitions – immunity does not last forever. You then also need to throw in demographic dynamics, including births and deaths. One problem is that for many coefficients you don't have the data, so modellers need to use guesstimates, which they should vary to get a feeling for the model's sensitivity to variations in these guesses. Another one is that they depend on national policies or guidelines which may unpredictably change. If you have enough compute power, you may want to use stochastic models instead of the (deterministic) differential models, or even, with supercomputer power, run simulations of a particle model in which individuals are represented as moving particles that interact and change in time.  --Lambiam 05:58, 4 May 2020 (UTC)

Difference in assumptions in mathematical pandemic modelling?

Slightly related to my previous question, am I correct in understanding that Sweden’s model and the UK’s model simply makes different assumptions? So on the R(0) value, infection fatality ratio and whether there will be an intervention which can stop spread before it passes through the population (this last one seems to be the key one which is driving Sweden’s response)? 90.196.238.188 (talk) 21:54, 2 May 2020 (UTC)

The Swedish approach is based on a different strategy. Unlike what most other countries are doing, they are not aiming to stop the epidemic running its course. Rather, they are focusing on isolating the elderly from the rest of the population so that the epidemic can run its course in the non-elderly population and allowing the population to reach herd immunity. They have only taken rather mild social distancing guidelines to make sure the epidemic doesn't rip through the country too fast. They have argued that they don't need to take strong measures because their hospitals have more than enough capacity.

So, this is not really about a different judgement about parameters such as R0 and the infection fatality ratio, rather a judgment that a vaccine is going to come too late to stop the epidemic from running its course. This means that whatever the infection fatality risk is and whatever R0 is, all the deaths due to infecting the 1-1/R0 fraction of the population to reach herd immunity will happen anyway. There is then only the choice of limiting the infection fatality ratio by making sure the people who are at the greatest risk of dying don't get ill. And one has to make sure that hospitals should not face capacity problems. Count Iblis (talk) 00:45, 3 May 2020 (UTC)

While the Swedes are busy patting themselves on the back, their total death rate per million people peaked at more than 10 deaths/million/day averaged over 7 days, putting it in the bad boys club, Belgium(28) Spain(18) France (17) Britain(14) Ireland (14) Italy(14) Sweden (11) . In terms of total deaths per million they are on the same trajectory as the Netherlands and Ireland. Other countries have done far better than that, and some of course are worse. Greglocock (talk) 02:56, 3 May 2020 (UTC)

The Swedes have chosen to go through the epidemic faster than other countries, therefore their death rate per unit time should be higher. What matters is whether when it's all over, the number of deaths per head of the population will be lower. Count Iblis (talk) 03:07, 3 May 2020 (UTC)

and as I said, they are on the same trajectory as several other countries for overall deaths per million. At this stage we don't know if flattening the curve, or not, results in lower overall deaths, but Sweden doesn't stand out for either. I see no sign that their strategy is an outlier in death-related results. Greglocock (talk) 05:35, 3 May 2020 (UTC)

I see. What about Japan? I note they have a softer lockdown too although they seem to be somewhere in between Sweden and the rest of Europe? Are they making similar assumptions? 90.196.238.188 (talk) 07:30, 3 May 2020 (UTC)

Japan is the big outlier, way down in terms of confirmed cases (a problematical measure) and deaths (a slightly less problematical measure). I suggest you spend a bit of time exploring https://ourworldindata.org/coronavirus Greglocock (talk) 21:39, 3 May 2020 (UTC)

An added difficulty with making these comparisons is that, by looking just at policies, one may not be accounting for variation in compliance with policies in the different populations. So, population A with a light lockdown, but near 100% compliance, may have better numbers than a population with a harder lockdown, but only 25% compliance. --OuroborosCobra (talk) 22:07, 3 May 2020 (UTC)

so basically the outcome depends on the social characteristics of that country as well? 90.196.238.188 (talk) 08:49, 4 May 2020 (UTC)

May 3

How close can you be to a fighter jet passing you at mach 1 without sustaining injuries?

Some closes passes are shown here, but could the people have been injured if these planes had passed the public at, say, 10 meters distance? Count Iblis (talk) 01:22, 3 May 2020 (UTC)

Does getting your eardrums blown out count? ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:03, 3 May 2020 (UTC)

Wouldn't earplugs prevent that? Count Iblis (talk) 03:08, 3 May 2020 (UTC)

Maybe, if you knew it was coming and had them to put in. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:55, 3 May 2020 (UTC)

See Tourist killed by jet blast at notorious Caribbean airport.

If you get VERY close, This Sailor Got Sucked Inside a Fighter Jet Engine (and survived with minor injuries). Alansplodge (talk) 11:11, 3 May 2020 (UTC)

I mean, the scenario is slightly contrived. At some point you're very likely to be injured, though it's always possible to escape unscathed in freak circumstances. A jet traveling at Mach 1 is a large object moving very quickly and doing so using one or more turbofan engine(s). Air is a thing that has mass, and the jet compresses and heats the air in front of it as it travels. The engine also sucks in air and propels air+exhaust out the back. Close to the ground, this will kick up any debris lying around; foreign object damage is a major concern for jet aircraft. If the jet blast hits you directly enough, it will knock you over at least, per Newton's Third Law. The article linked by Alansplodge describes a worst-case scenario for this, and you can find plenty of videos of jet blasts knocking things over. In part this is probably standard human intuition misfiring. We kind of innately think of air as "insubstantial" since it doesn't seem to affect us much, but go stand in a hurricane and you'll see the power of lots of it moving at high speeds. --47.146.63.87 (talk) 19:37, 3 May 2020 (UTC)

Could the pressure of the sonic boom be harmful at close distances? Our article on sonic booms does not give data allowing an estimate of the pressure at Mach 1. Assuming that the pressure decreases with the square of the distance, the space shuttle at Mach 1.5 results in 19.44GNd⁻² at distance d. This formula cannot be used at extremely close distances, but should be reasonably valid (for the space shuttle at Mach 1.5) at distances of only a few times the length of the vessel.  --Lambiam 07:05, 4 May 2020 (UTC)

"The strongest sonic boom ever recorded was 144 pounds per square foot and it did not cause injury to the researchers who were exposed to it. The boom was produced by a F-4 flying just above the speed of sound at an altitude of 100 feet"; see Official United States Air Force Website - Factsheets - Sonic Boom. Alansplodge (talk) 11:19, 4 May 2020 (UTC)

About 6.4kPa at 30m. Then it is more likely that the effect is inversely proportional to the distance. Reportedly, the human body can withstand 50 psi (345000Pa) for a sudden impact. So a supersonic flyover at a height of 0.55m may be iffy :).  --Lambiam 20:30, 4 May 2020 (UTC)

DSLR

People say smartphone cameras are getting closer to DSLR quality but what are the main advantages of DSLR over smartphone? Is it night shots and more control over what’s in focus? 90.196.238.188 (talk) 14:35, 3 May 2020 (UTC)

DSLRs have much large aperture than smartphone cameras. Ruslik_Zero 17:21, 3 May 2020 (UTC)

Link: digital single-lens reflex camera. Also optical zoom and interchangeable lenses. There have been moves towards bringing these to phone cameras, and you can always jury-rig a lens attachment to a phone, but dedicated cameras have an advantage in being designed exclusively for photography and thus not needing to compromise for other things like form factor. Some cameras probably have an advantage in battery life, though this will vary widely, and some use swappable batteries while most smartphones these days have internal batteries. (You can still carry a power bank but this adds inconvenience.) --47.146.63.87 (talk) 19:45, 3 May 2020 (UTC)

Give it time. I recall 20 or 25 years ago when this same question was being asked about digital vs. film cameras. Digital has since become so good that film is kind of passé. And phone cameras are way much better than they were 10 or 15 years ago. So they're getting there. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:35, 3 May 2020 (UTC)

Control of focus, exposure characteristic (aperture and shutter speed), depth of field, ISO, processing style options, high-quality interchangeable lenses, use of filters, flash sync and tripods, multiple exposure capability, ergonomics, dedicated customized processors, fast control change options, raw file formatting, and vastly larger sensors. Full-frame SLR sensors are 24mm x 36mm - phone camera sensors are around 4.2mm x 5.6mm, so for the same resolution the phone camera pixel size is vastly smaller, leading to problems in low light with such high pixel densities. You can get really excellent pictures with a phone camera, but you need to have good timing and some luck, because you're only likely to get one chance. A dedicated camera lets you take multiple shots in quick succession, bracket exposures and fix problems in near real time. The shutter lag with phone cameras is a perennial problem. Acroterion (talk) 04:04, 4 May 2020 (UTC)

To go a little farther, it's analogous to film cameras - large-format view cameras beat medium-format cameras for image quality (at a sacrifice of time), and a medium-format Hasselblad will beat a 35mm SLR (again, as long as speed is not required). A 16mm Instamatic substantially underperformed a 35mm camera in all respects. Sensor output, whether in film or CCD, always improves with size, all other things being equal - spreading out photons on the film or the sensor produces a better image. And Instamatics were prone to motion blur because they sacrificed ergonomics to the form factor - just as phones aren't designed for steadiness or motion-free shutter release. Acroterion (talk) 19:44, 4 May 2020 (UTC)

Yes. I've been a SLR/DSLR user for 42 years. I have problems taking a photo with a phone without it moving and bluring the photo. And I think that instamatics didn't have a focus. It just set the focus to the hyperfocus distance to get things sort of in focus. I recently read that 4x5" film still beats the best digital. Bubba73 ^{You talkin' to me?} 19:59, 4 May 2020 (UTC)

My cheap 7 year-old crop sensor DSLR outperforms my brand new state of the art phone camera by a huge margin when taking night pictures. I can put my DSLR on a tripod, take a large number of pictures, and then do a large amount of processing such as averaging out the noise, make up for the crop sensor by stitching a panorama, forego doing demosaicing by interpolation and instead use sensor shifts to get to the actual gray values in the different color channels etc. etc. The resulting picture is of a quality that you would struggle to get using a single shot from a medium frame $100,000 Hasselblad camera.

The phone is far more useful to take quick pictures with. It's small, and handy, you can take it out of your pocket at a moment's notice and start shooting with it. The settings are easily adjusted in a few seconds. Also, even if you do have the time to get your DSLR out of the bag and make it ready for use, for certain tasks like shooting video from a plane it's not going to work out well, like these videos I made using my smartphone. Count Iblis (talk) 04:31, 4 May 2020 (UTC)

Here are some things I came up with:

• The more light you can get into a camera, the better the photo. DSLRs have much larger sensors and much larger lenses to gather much more light. The pixels on a phone are so small and the sensors are so small that they can't gather much light.

• DSLRs have much better lenses. Small imperfections are magnified on a phone.

• DSLR lenses are much sharper and have fewer distortions, like chromatic aberation, etc, and probably better contrast, flares, and ghosting.

• DSLR lenses can go quite a bit more wide angle and much more telephoto.

• With DSLRs, you can get lenses for macro photography or fisheye lenses. You can also get more specialized lenses like tilt-shift lenses that take out perspective distortion. (Phones have a huge amount of perspective distortion.)

• DSLRs have a much wider dynamic range - the difference between the brightest and darkest areas. Details in shadows are lost on phones.

• With DSLRs you can crop or blow it up and retain details.

• With DSLR viewfinders, you can easily compose the shot like you want it. You can follow a moving subject naturally.

• On DSLRs you can pick a point to expose properly and pick a point to be in sharp focus.

• You have control over the ISO, the aperture, and the shutter speed. Aperture control lets you decide how much in front of or behind the subject is in focus. Controlling shutter speed lets you freeze a moving subject or let it blur, as you decide. Also with aperture control, you can control sun stars.

• DSLRs usually have a built-in flash; you can add an external flash; you can control a remote flash.

• I don't know about ISO on phones, but I suspect that DSLRs can go to much higher ISOs and take photos in much darker conditions.

• DSLR lenses generally have hoods to keep sunlight from hitting the lens from the side and causing flares.

• DSLRs can use filters for special effects.

Bubba73 ^{You talkin' to me?} 08:16, 4 May 2020 (UTC)

I could also list the changeable batteries and the changeable cards for storage. But a person who would have used an Instamatic, a Brownie, or an instant camera in decades past would be happy with a phone camera. Bubba73 ^{You talkin' to me?} 15:23, 4 May 2020 (UTC)

Some phones with high end cameras still have add on storage e.g. Huawei phones (albeit using their Nano Memory) and Samsung Galaxies (with microSD) so it's not really a point of difference. Changeable batteries is something that has basically disappeared from high end phones. Most high end camera phones also let you pick a point for exposure and focus; and manually adjust ISO, aperture and shutter speed. Although the interaction of the latter set of features with the semi-AI features they rely on to improve photo quality is complicated and given the nature of camera phones I think even most expert or professional photographers don't bother with too much tweaking. Pretty much all phones except for some very basic ones have built in flashes although of course for size and maybe other reasons even the best phones flashes are not comparable to a DSLR. (I suspect you can control a remote flash but it's not generally a useful feature with a phone.) Edit: Sorry. I am wrong about aperture. You cannot adjust this as it's normally fixed. (You can do "stimulated" adjustment as I mentioned below although this is often a separate feature from the pro modes where you can adjust shutter speed and ISO.) Samsung did try a real variable aperture for 2 generations but abandoned it with their latest phone, which IMO demonstrates it's not very useful for smart phone cameras. [6] [7] Nil Einne (talk) 06:14, 5 May 2020 (UTC) 11:33, 5 May 2020 (UTC)

The most serious difference is the size of the sensor. That is not DSLR-specific since there are also compact and so-called "mirrorless" cameras with equally large sensors. If you think of the sensor as a pizza, then pixels are individual slices of pizza. DSLRs and phone cameras are now equivalent in the sense that in both cases, the pizza is cut into say 10 million slices (10 megapixels), but in the DSLR, the pizza itself is 10 times bigger, so the slices are 10 times bigger! They can hold a lot more electrons, giving more dynamic range, lower noise, fewer diffraction effects, etc.. Yet for a long time (and even now), megapixels was an important marketing metric, like selling pizza by how many slices it was cut into rather than its diameter. It's almost as dumb as it sounds.

The main thing that has made phone cameras so good over the past decade or so is computational photography: basically the phone sensor takes a relatively crappy image, but then the chips in the phone are able to enhance it til it looks really nice, equal in some regards to a relatively unprocessed DSLR image, with some artifacts here and there. In principle you could apply similar processing to DSLR images, but this is rarely done because of some combination of older technology and having good enough quality already to not be willing to tolerate the artifacts. DSLRs themselves are getting displaced from the high amateur / mid pro market by "mirrorless" which are interchangeable lens cameras with electronic viewfinders (the "reflex" in DSLR refers specifically to a certain type of optical viewfinder where you see through the lens). I haven't followed this stuff too carefully though. I have an old DSLR that's very obsolete by today's standards but is still fine for my purposes. I bought it as soon as I could afford one, and I'm sticking with it. 2602:24A:DE47:B270:DDD2:63E0:FE3B:596C (talk) 09:55, 4 May 2020 (UTC)

I don't disagree with anything you've said but I'd suggest "take a relatively crappy image" could cause confusion. Depending on the specific example, the computational photography may involve using multiple crappy images in some fashion. This is often the case for "night modes" on phone cameras when multiple short exposures are combined see e.g. this which mentions Google's Night Sight [8]. Of course long exposures have been a part of cameras since the very beginning of film, but the classical long exposure is just capturing light for all that time on whatever. Meaning you camera needs to be very still or the light falls on different parts of the frame and you get a blurry image hence tripods etc. Image stabilisation helps but has its limits especially on a phone. So you instead take multiple images and combine them. (As mentioned in The Verge, the same thing had been done for HDR for a while.)

Another related example is bokeh on smart phones. These may rely on using the information from a depth sensor (generally a Time-of-flight camera) and then selectively "blurring" the image of the main sensor. Alternatively for phones without such a sensor, they may use the multiple cameras and other information from multiple images to try and determine depth information. And the "blurring" itself could be a completely stimulated out of focus effect, or perhaps it involves adjusting the focus or using multiple cameras to help produce the effect. You can also just use one image, like Google does. (In somewhat the reverse, I thought that some phones can also use Focus stacking but can't seem to find any discussion of this so I'm probably wrong.) Other fancy exposure modes like light trails, silky water etc operate by the same principle. See also [9].

That said, we shouldn't ignore the advances in sensor and other aspects of the camera. See e.g. this which compares Google's long/multiple exposure Night Sight with Huawei's short exposure night photography in the P30 Pro [10] [11]. I think it's clear this isn't coming just from computational photography advances. Note though even with such short exposures, I'm not sure if we know that they aren't also using multiple images as part of the process. (The P30 still has a long exposure night mode option which is sometimes also used automatically if you turn on the AI feature. I'm guessing the P30 Pro has the same based on [12].)

Of course with computational photography you also get interesting questions. For example, the P30 moon mode controversy [13] seems to me to be nonsense and the coverage in the English media was often terrible. But one thing which didn't seem to be well discussed, is that if you are using deep learning techniques to enhance your image, at least from my understanding of he these tend to work, you probably don't know that well what it's actually doing without a lot of analysis. Clearly it's easy to tell you're not just substituting a stock photograph, but at deeper level, the difference between "enhancing captured details" and "adding details you think should be there based on other captured details even though they were no where in your capture" may not be so clear cut.

Nil Einne (talk) 12:00, 5 May 2020 (UTC)

May 4

Palladium pentafluoride

I'm don't sure that PdF₅ will have dark red in color. So what is the color of this compound? Thanks for much (Sorry if you don't understand, because my English is not good). --Ccv2020 (talk) 10:23, 4 May 2020 (UTC)

If you have a reference for this compound even existing, please post here so we can track down other details. DMacks (talk) 10:30, 4 May 2020 (UTC)

It is expected to disproportionate in to PdF4 and PdF6 and there is no good evidence that PdF6 actually exists. IR spectra have been calculated but not UV-vis (TD-DFT isn't very trustworthy anyway).Pelirojopajaro (talk) 10:49, 4 May 2020 (UTC)

You've been asking a lot of questions about transition metal complex colors. Is there some specific goal we can help you with? It might help to know what you need this information for... --OuroborosCobra (talk) 15:07, 4 May 2020 (UTC)

These questions make a lot of sense if someone is developing a model for predicting the colours of chemical compounds in their various states of matter.  --Lambiam 08:35, 5 May 2020 (UTC)

The problem is, that isn't actually possible. It's the quantum equivalent of solving the many body problem for a quantum system. It basically involves solving the wave function quantitatively for a system consisting of several nuclei and dozens of electrons. We don't even have a good quantitative solution for the Helium atom, and that's only a 3-particle system. There are things like the Hartree–Fock method, which provides a decent enough approximation of the wave function for complex systems, but not at the fine detail necessary to pick out exact colors. --Jayron32 13:15, 5 May 2020 (UTC)

I think that may be understating modern capabilities a bit. I've done TD-DFT modeling of heme systems, and gotten pretty good estimations of at least the Soret band that were within about 10 nm of my experimental measurements. Granted, that was pi-pi* transition within the conjugated pi system of the ligand, but it was a far larger system than a helium atom. This was not a trivial calculation, mind you, and I needed access to a supercomputer cluster to do it, but it was possible. This is, of course, not the same as different colors arising from things like d-orbital splitting due to ligand interaction with transition metals... but sometimes, you don't even need TD-DFT for that. In a high spin ferric compound, for example, you can calculate the energies of the d-orbitals without examining excited states, as all of the iron 3d orbitals will be occupied by at least one electron in the ground state. Additionally, with enough similar experimental data of a particular transition metal and other ligands, you can make some decent qualitative guesses at color using spectrochemical series and applying ligand field theory. Is it perfect? No, but we are only talking qualitative. All of that said, I'm still unclear as to the goal of Ccv2020, as they seem to be randomly throwing out transition metal complexes (many of which haven't even existed at all) and asking us to guess the colors. You don't build a model by randomly throwing darts. --OuroborosCobra (talk) 14:57, 5 May 2020 (UTC)

I'll have to defer to you on this then. It is clear you've done more detailed work in the field, mine is a more general exposure to it. It's plain you have access to better sources of information than I do, and I will defer to your knowledge on these matters. --Jayron32 17:56, 5 May 2020 (UTC)

This is the existence of this compound (also PdF₆): [14] — Preceding unsigned comment added by Ccv2020 (talk • contribs) 07:42, 5 May 2020 (UTC)

That's not the compound PdF₅, that's the complex ion PdF₅^1-. Ions are not compounds. --Jayron32 13:08, 5 May 2020 (UTC)

May 5

Making use of earth gravitational model

There are many gravitational models for earth that extend to high accuracy, but they all apparently describe a surface of equal gravitational potential energy relative to a reference ellipsoid, which seems only useful on the surface and in a reference frame that rotates with the earth. How would one use such a model to calculate the direction and magnitude (3D vector) of Newtonian gravity force for an arbitrary 3D point in space relative to the center of earth, assuming that the instantaneous orientation of the earth is known. I'm ultimately only looking for a fairly low order approximate (4 - 10 constants) and don't care about longitudinal variations, but it must be good enough to capture the effect of the equatorial bulge on the moon and artificial satellites. 102.65.153.81 (talk) 12:18, 5 May 2020 (UTC) Eon

The model specifies the spherical harmonics. Each harmonics (l,m) depends on radius as $~r^{-l-1}$. By summing the harmonics your can calculate the gravitational potential at any point outside the solid Earth. Ruslik_Zero 21:03, 5 May 2020 (UTC)

Spirals and Einsteins equations

Are the equations for spirals (like under the logarithmic spirals) post in any way related to the Einsteins Field Equations under that post? Since the spirals are very evident in nature and even in galaxies, I wonder if there is some relation. The math is beyond me. Tofflet (talk) 17:18, 5 May 2020 (UTC)

...There exists at least one way to relate the equations for spirals to the equations that describe the geometry of space-time.

...If we showed you such a way to relate them, but the math were beyond you, would it enlighten you in any way? Or are you simply seeking a confirmation from somebody else who you think might understand it better than you, in the hopes that their confirmation could bring you some kind of solace? Because that is a logical fallacy called the argument from authority, and if you wish to use mathematical ideas, one of the first skills you should formalize is your ability to work within a logical framework.

Nimur (talk) 17:31, 5 May 2020 (UTC)

May 6

Naegleriasis

I am looking for information on the infection rates of naegleriasis in different countries, which occurs when naegleria fowleri consumes a host's brain. It looks as if most cases are found in the United States, but I question the accuracy of these findings. Medical evidence implies that Americans should have a biological immunity to this parasite, but the infection data does not reflect this. Has the data ever shown the infection rates for Americans with pre-existing conditions, such as heart disease or morbid obesity? Thank you CombustibleTaco (talk) 02:36, 6 May 2020 (UTC)

What are "ferrallitic" and "fersiallitic" soils?

I'm looking for info on mineral composition and pH of Bushveld soils, to understand what are the macro- and micronutrient requirements of the plants native to the Bushveld region. I'm looking at this map: [15] but I cannot find any good explanation as to what "ferrallitic" and "fersiallitic" soils are. What are they? And what is a good source of info on the macro- and micronutrient requirements of the plants native to the Bushveld region? Thanks in advance! Dr Dima (talk) 03:33, 6 May 2020 (UTC)

An explanation of "ferrallitic" can be found at Ferrallitisation. Fersiallitic is proving harder to find. A Bing search for "siallitic" turns up this paper which you might find interesting: [16] My McGraw-Hill Dictionary of Earth Science does not contain either term or likely derivatives. A lot of help that turned out to be.--Wikimedes (talk) 04:30, 6 May 2020 (UTC)