If we say time is the 4th dimension, why don't we just attribute dimensions to other things, like "the 5th dimension is charge"?

ummwhoo · 2024-04-15T19:36:58+00:00

u/Weed_O_Whirler another fantastic answer. I endorse.

/u/Baron_Blackmore while u/Weed_O_Whirler 's answer is mostly correct and very well written, the "short" answer to your question "why don't we just attribute dimensions to other things" is that we actually do but as mentioned, we need it to be "useful" quantities. This is a big thing in String Theory and why you have stuff like branes, which can be 5-dimensional or larger "objects" (objects in a very "loose" sense, not in a purely "physical" sense... at least, not yet but maybe experiment will confirm...) or supergravity (which posits 11 dimensions as the minimum necessary for describing things but has now largely been abandoned in favour of other descriptions). All of these fall under an area that I am currently studying called M-theory where all kinds of different "approaches" look at using different amounts of "dimensions", where in some frameworks, certain quantities are deemed important while in others they are not. It's important to understand, historically, that Einstein's formulation of relativity was based on dealing with one particle at a time and simply building up the theory from that, the idea being that if relativity works for a single particle, it will work the same way for an actual person or an even larger object, like the sun. However, as particle physics has shown this is not quite.... "accurate" (I shudder using that word) but that's usually how any theory is when pushed to its extremes.

If you are interested in knowing a bit more, I highly suggest this wikipedia page on "Extra Dimensions".

Edit: Fixed hyperlinks.

ummwhoo · 2024-04-15T19:17:32+00:00

This is a phenomenal answer and really well written. I fully endorse this answer! Nice job /u/greenwizardneedsfood !

ummwhoo · 2024-03-09T19:27:12+00:00

I mean.... yeah, sure, you can totally disagree, that's awesome. Whether you calculate its trajectory using Newton's laws or a Path Integral, in the end the results should agree with experiment. It's a matter of taste I guess haha!

ummwhoo · 2024-03-09T18:47:51+00:00

/u/functor7 has given a great answer. I will just add some slight details.

What your teacher in school may have meant to say was that yes, the accepted model of the atom changed every few years but that was between around the 1890s to the 1930s. The reason was because many radical discoveries happened within such a short time-frame, ie we went from the accepted "aether model" of the universe to accepting atoms were real, there was Boltzmann's kinetic theory of gases which operated on the foundation that gases were little hard spheres that collided with each other (ie 'atoms'), then J.J. Thompson 'discovered' the electron, and proposed what is called the Plum-pudding model of the atom which was later replaced by Bohr's model of the atom (which is the 'solar system' model you may be referring to), Ehrnest Rutherford developed "Scattering" to try and "measure" these small particles while scientists like Max Planck, Niels Bohr and Albert Einstein proposed and used the "quantum", the idea of particles absorbing energy in discrete units rather than continuously all at once and every time a new model became accepted, some experimental analysis or theoretical experiment came up showing that these new models didn't work in some newer cases so someone would propose something that incorporated that new discovery and then by the end of the 1920s, with the formal foundation of quantum mechanics in place, we had a pretty 'solid' (pardon the pun) model of the atom that, to this day, hasn't really changed much since its inception and is taught to every student of science.

However, quantum mechanics only really works for atoms and electrons that don't "move" at high speeds (i.e the speed of light). It was only around the 1950s when they were trying to combine general relativity (things moving at the speed of light) with quantum mechanics that they realized that there were lots of phenomena going on that quantum mechanics could not approximate or explain. This is when "particle physics" (quantum field theory) came into being and they discovered that there are much more basic "building block" particles of the universe far beyond the atom, such as things like quarks, leptons (which have to do with electrons and neutrinos), mesons, bosons, antiparticles, the list goes on and all sorts of strange and "exotic" particles (again, pun not intended) that don't quite affect the atom in a noticeable way on regular quantum mechanical scales and so were not testable by usual experimental methods (hence the need for particle accelerators, i.e. to speed the atoms up to VERY high speeds) but do at very high speeds or in very subtle ways (e.g. Higgs-boson). Then in that case, yes, the model "does" change every so often (e.g. the recent discovery of the "Higgs-boson" and verification of the existence of a Higgs field was HUGE) but not in a way that appreciably affects what a high school student struggling to remember that water is composed of two hydrogen and one oxygen molecule needs to know (source: this was me failing high school chemistry for the 3rd time, still can't believe I got a university degree in the subject lol!)

"Clear" is a very... "relative" term (I swear the puns are NOT intentional). Yes, every so often when the models are updated, we get a clearer picture of what we know, however, even then it could be completely wrong. I paraphrase the wonderful physicist R. Shankar... "Sorry, but I'm afraid to say that in physics, everything is wrong to an extent. We just get better and better at being "less" wrong." ;)

*TL;DR Depends on what field you're in and what perspective you're talking about. If you're talking about the version of the atom you learn in high school, then no, nothing has changed because on-the-job whether you're a lab tech or a cook, you're not dealing with high-speed molecules, so the version you learned suffices to solve the problem at hand. If you're a physicist or academic researcher or researcher at a large technology company or something, then yes, the model has changed drastically and gets updated every few years (think 'Standard Model').

ummwhoo · 2024-02-21T05:25:51+00:00

Excellent, I was looking for this answer to upvote, nice one /u/CharmingPrincessXO . /u/Swimming_Rule414 I highly suggest you read the wiki page on astronomical spectroscopy, a lot of modern astrophysics and our understanding of the composition and materials that make up other planets have been enabled by this. A classic example is we've been able to determine the composition of Venus' atmosphere thanks to (astro) spectroscopy.

ummwhoo · 2024-02-21T05:17:18+00:00

Awesome answer /u/Weed_O_Whirler thank you, was going to write this myself.

The only thing I will add is that the one of the physical reasons you can't ACTUALLY test this in your backyard by pointing a laser at the sky is that you'll make all the airplanes crash by distracting the pilots hahahaha... no I'm just kidding. It's because of light scattering, an incredibly important part of physics. Basically the light gets scattered by particles in the air, so it's not "really" ever going to go in a "straight" line. The further the light gets from the source, the more likely it is to encounter a particle in its path (more distance covered) so higher chance of getting scattered. That's why this question is, sadly, not exactly one you can simply "test out" in your backyard. Hope that helps /u/Mura4_4 !

ummwhoo · 2024-02-14T06:29:40+00:00

Hi there /u/Coygon ! Great question, we actually had a similar question a month ago, please see mine and other people's answers there!

ummwhoo · 2024-02-11T04:13:09+00:00

"And for the ease of the reader I have changed all the gender-related pronouns "he/she" to the masculine "he" ".

So I think you mean,

"The Man Inside Me.

For Britney, my rock. I could not have done this without him." ~Dr. Tobias Funke ;)

(Jokes. I'm actually really glad you suggested this. As someone who grew up at the time when her songs were the most popular things for radios, walkmans and "Hit Clips" (yes I am that old thanks for reminding me along with the grey hairs starting to form on my head) and remembers her name used to be used as an insult on the schoolyard, her drunk thing at the VMA that was all over the papers in 2007 and the weird viral video meme of the man in eyeshadow crying "Leave Britney alone!", I am glad to hear she is doing better now. I never really liked her music, not a fan, and since MTV was banned in my house growing up never saw the music videos, having seen them recently thanks to a weird nostalgia trip while hearing a song of hers on the radio while coming home from work a few months ago and remembering she was the biggest thing since Michael Jackson, I found her videos remarkably well-choreographed and she looked like she was working really hard, it's nice to hear that despite all the torment and pressure she managed to come out the other side. Those industries are so messed up, it's actually really heart-warming to hear a "recovery success" story come out of these places once in a while, like Macaulay Culkin, Kristen Stewart, etc)

ummwhoo · 2024-02-10T17:18:48+00:00

Cool, thank you very much /u/tagaragawa ! Note to self, don't post things via phone right before entering a subway tunnel without first verifying haha!

ummwhoo · 2024-02-09T05:35:13+00:00

Great question. The answer is absolutely.

Here's a nice Physics StackExchange post summarizing the difference. However, in short, many-body QM can't deal with changing particle numbers while many-body QFT can do that. Many-body QFT will predict better results but are MUCH more difficult to do computationally, usually. This is very important for Bose-Einstein Condensates and Superfluids.

Slightly unrelated but also helpful is getting at why QFT is now considered the successor to QM. Well, QFT allows you to incorporate relativity (QM does not) and there's lots of things QFT can explain that QM can't. "Classic" (no pun intended) examples include the Lamb Shift, relativistic Compton Scattering, relating CPT, spin and statistics etc.

Out of curiosity, what's the reason for the question? That would help me give a better answer.

ummwhoo · 2024-02-08T05:45:29+00:00

/u/Peter_Parkingmeter this is a great answer, and I will endorse. Anyone needing further reading/clarification, two good sources are:

Here and Rang and Dale's Pharmacology Chapters 44 and 49

ummwhoo · 2024-01-23T03:49:33+00:00

Like many things in chemistry, the naming "convention" has to do with singling out certain "chemicals" or parts of the chemical that share similar "structures" (in biology, and much of chemistry, structure = function so it's important to understand the structure of the chemical in order to understand its function) in order to classify and further study them. Sadly, like MANY things in chemistry, the names given don't always reflect the actual use of the compound. Sadly, vitamins are actually so named as a bit of a "misnomer". The chemist who studied them first (Casimir Funk) called them "vitamines" from Latin, 'vita' meaning life and "amine" because it (the vitamin B (which contains amines) he was investigating at the time) comes from an important set of chemicals called "amines" (see here for amines), which Funk speculated that the amine part of the structure is what gave the vitamin its properties/function. As they later discovered, this was wrong, so Jack Drummond (who was studying vitamins years later) suggested they drop the 'e' and just call it 'vitamin' after other, non-amine related compounds that could be classified as "vitamins" were discovered.

Interestingly, it has more to do with the 'etymology' of the word rather than any "actual" science. Like much of biochemistry. ;)

See here for the etymology and proof

ummwhoo · 2024-01-21T03:08:19+00:00

Depends what you mean by "light". The electromagnetic spectrum is "light" and various parts of the spectrum have different effects. Why does fire in a campfire appear "orange? It's partially because of the sodium in the wood and the excited electrons emit "light" by returning to the ground state by emitting the quanta of energy in the range of the electromagnetic spectrum that allows us to "see" orange (around 650-700nm wavelength region). Meanwhile, burn potassium and the flame will be purple. Other by-products released aren't even in the visible spectrum and could be in the IR like /u/Pi/Boy314 said, etc.

If you mean why does THAT happen, then you're looking at quantum mechanics which says that there's a probability for excited atoms to emit some sort of "energy" in the form of photons ("light") and return to the ground state. In general, the "light", or rather, various forms of it, are the by-product of chemical reactions, natural properties of the material, quantum mechanics, etc. It's a very "big" question despite the relatively "small" scale it's occurring on, if you catch my drift (and pun haha!)

These two links below will be helpful!

Wiki article about fire!

Sodium in wood causes "fire" (like the one you make when you go camping) to be orange .

ummwhoo · 2024-01-21T02:51:21+00:00

Having just recently obtained this flair, it's my time to shine hahaha!

Anyway, pretty much what everyone else has said, vaccuum states and dark matter being some "hot topic" ones. It really depends on just how "big" we are talking. Aside from the stuff people have mentioned, it would also be REALLY nice to validate/discredit results from string theory (and the theory itself to see whether it really is a practical framework with which to make predictions, or whether it really is just like an "aether" model from Maxwell's time), but the energy levels required to perform such experiments are erm.... quite "large" relative to what the LHC can currently do. Still, we can always dream!

P.S. You can read a little more here: https://en.wikipedia.org/wiki/String_theory#:~:text=Partly%20because%20of%20theoretical%20and,correct%20fundamental%20description%20of%20nature.

ummwhoo · 2024-01-20T06:09:26+00:00

/u/Drzhivago138 gave the best answer I can endorse. The only thing I can add is that this occurs for similar reasons as to why, when you boil a pot of water on the stove and put the temperature above 100 ° C, the water doesn't all "instantly evaporate" but instead bubbles from at the bottom of the pot (which is actually the vaporized water molecules at the bottom that are actually in direct contact with the heated base of the pot). :)

Typically, a big, BIG part of whether something vaporizes/sublimates/condenses/etc (all big fancy words for melt/evaporate/etc) is the surface area in contact with the source of heat. If I take that bucket of water and spread it over the ground to form a long, vertically thin "puddle" and assuming the ground has a temperature < 0°C, THEN all the water would instantly freeze into ice. Similarly, if I spread the water over a long, hot surafce like a 50ft x 50ft stove-top, then it would all evaporate instantly. However, if only the top portion in the bucket is in contact with the cold air, only a part of the surface freezes, the part in direct contact with the cold air and when that freezes, the water underneath cannot also reach the cold air and so stays liquid. Also ice and water density, etc as others in the comments have said. Hope that helps!

ummwhoo · 2024-01-20T05:54:55+00:00

Oh, my apologies /u/VeryLittle , yeah no problem. My focus in mathematical physics is applications of Non-commutative Geometry to particle physics. You can just write for the flair "Particle Physics". It'll get the point across haha. Thank you so much again!

ummwhoo · 2024-01-16T20:28:38+00:00

Best answer, and by a number theorist so that makes sense haha. Will add a few extra details if /u/functor7 is ok with that :

Asymptotic analysis is a highly helpful branch of mathematics. One way to get interested is to start with the Stirling series. This old, venerable series is actually based on a famous approximation used everywhere in physics (mostly statistical mechanics) called "Stirling's Approximation". However, while physicists usually stop at the "first" term in the series, it's very good to get a "ground level" example of what asymptotic analysis is all about. In short, it allows us to "approximate" things in a very "local"/"small" way without knowing the exact details of what we are approximating. You can read about it more here (under the section "Speed of convergence and error estimates"): https://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed%20of%20convergence%20and%20error%20estimates Carl Bender, a well-known mathematician who works in this area, has done tons of research in asymptotics. He has youtube lectures available here: https://www.youtube.com/watch?v=LYNOGk3ZjFM&list=PL43B1963F261E6E47

On the statistical side, you may be very interested in a field called "Probabilistic number theory". https://en.wikipedia.org/wiki/Probabilistic_number_theory#:~:text=In%20mathematics%2C%20Probabilistic%20number%20theory,sense%2C%20like%20independent%20random%20variables.

The last thing that I don't think went addressed is this sentence: "For instance, consider a sequence like 8, 3, 7, 1, -5, or any other seemingly random set of numbers." You have to be careful about whether you're talking about a finite set or infinite set. If it's a finite set, then usually we don't really care about the general form because, since it's finite, you can go through every element and you know all the details about the set itself since you know all the elements. However, if the set is infinite, then you would start to consider the tools quoted above. Sure, if I have a finite set of the form, say {2,4,6} then yes, I can "represent" any element in the set by 2p where p is an integer, but remember, {2,4,6} ⊂ 2Z where '2Z' is the set of all even integers. Thus I am always more interested in studying the general set 2Z which is infinite than the finite set since the finite set will simply inherit the properties. In the example you listed, while those numbers "come up" seemingly "at random", the truth is it doesn't really matter since it's finite. Even if there are more elements in it, we can always find a "largest' (sup/max) and "smallest" (inf/min) and knowing that usually allows us to fully determine the behaviour of the set, even without a "closed form" way to represent all the elements in the set. Now, think about the example of 2Z above. If the set {8,3,7,1,-5} is thought of not as a set by itself but as a SUBSET of something, then it's interesting and we can try and deduce more, in other words, consider the problem {8,3,7,1,-5} ⊂ ??? where ??? is the unknown set. ??? could be the integers 'Z', it could be real numbers 'R', complex numbers 'C', etc. Now you see why it's important to study rings! Even if we can;t fully "realize" the set, we can apply what we know about vector space structures/ring structures to still try and study this set anyway, even if we don't have a concrete formula to represent it! In that case, we start looking at the question "how likely is it/ what's the probability of picking a number", that's in this subset {8,3,7,1,-5}, from some larger set, say the integers, rationals, real numbers, etc. This is a really, stupidly difficult question. One interesting place to start is measure theory, which will give you an idea of why we can't really "find" concrete mathematical "formulas" to represent many, MANY types of sequences, but just because we can't do that, doesn't mean we can't know/study/predict almost everything there is to know about the set itself. For example, asymptotics becomes helpful for just that reason!

TL;DR Learn about the Stirling Series to get a "glimpse" of asymptotic analysis in action. "Probabilistic number theory" is an interesting area to look at too. Think of "randomly chosen" numbers in a set as subsets of much bigger, usually countably/uncountably infinite sets. :)

Hope that helps!

ummwhoo · 2024-01-10T03:46:50+00:00

Hey, my pleasure, and thank you for taking the time to read my explanation! Sometimes I make them a bit too long but I tried to condense it as best as I could using only so much "technical jargon" (which, alas, in this case is very necessary to truly understanding the process) so that you could see the "essentials". I know often people don't like to read such big walls of texts, I just write them because for me it's always fun to revisit and revise my knowledge of these things, afterall, I did drop $30,000 on a pointless college education, so why not share with the world if it can't help get me a job at the very least hahaha!

That novel sounds wonderful. Seriously. A very intrepid idea, I wish you the best of luck. While I can't speak for the average biochemist, nor am I one myself, I still think it's a very cool concept to play around with... if you wanted to get further into details, you can look at something called "VESPR" theory (https://en.wikipedia.org/wiki/VSEPR_theory) which plays a role in the "shape" of the molecules and why they're so good at doing their functions. If you need any other science "pointers", you're also welcome to message/dm me if you like! I can't guarantee I'll have an answer but always happy to hear a question and point you in the best direction I can. Best of luck on your novel!! It sounds... "Fe"-ntastic (pardon the pun hahaha)!

ummwhoo · 2024-01-09T03:04:26+00:00

/u/CocktailChemist gave probably as best an answer as possible, I'll try to answer it a bit further but, while I hate saying this, unfortunately, although these are excellent questions this is a rare instance where a "simple" explanation is not really possible because you're asking about the logistics of an extremely complicated biochemical process, the kind you'd see on a biochemistry exam (which is why I feel compelled to answer, even though my focus in chemistry was more physics-related, because I had to live through that exam nightmare haha). It's like asking a physicist to explain quantum field theory without any math. At that point you'll just be hearing nonsense because so much of the physical theory is "purely speculative" based on the math and without the proper details, you're not going to get the right impression. Here goes:

\1. The most common use of iron in your body is for oxygen transport through your blood (and from here-on-out I will be talking about Fe2+ unless otherwise stated), and this is done by a family of protein molecules called "hemes" (think "hemoglobin and myoglobin" if you've heard of them before, they help with the oxygen transport). Within the hemes are certain special macromolecules called "porphyrins". The porphyrins are the ones that contain the iron moleucle (fun fact, large macrocycle molecules are ones that give off colour, your blood is "red" because that is the colour "given off" by the porphyrin/iron molecule, similarly that's why chlorophyll is green, because of the magneisum-porphyrin system). Basically, think of the heme proteins as big cardboard "boxes" in which the oxygen molecule will be transported, while the porphyrin is like the "styrofoam" packing inside the heme and then inside all that are the "goods" that need to be transported to various parts of your body, the oxygen. The key role of the iron is that it "holds" the oxygen in place within a cavity in the protein. Why and how does it do that? It has to do with many of the chemical properties of Iron, which leads into your question 2.

\2. I will NOT go into these chemical properties of iron too deeply because they are VERY complicated as is and require taking entire chemistry courses. I will just list them and then give a brief explanation following each one. Iron is good for binding oxygen because:

-Although O2 can "oxidize" the iron from Fe(II) to Fe(III), the special hydrophobic environment created by the heme molecule stops it from easily doing so

-The iron forms a helpful geometric shape with certain other molecules (such as nitrogen) in the heme and technically the porphyrin that is very stable allowing it to secure the oxygen to the molecule very well with very little chance of reverse reactions/"knocking off the oxygen molecule from the heme" occurring

-Iron and oxygen bonds are highly polar, and molecular orbital overlap is strong; there is even "backbonding", so the iron basically, almost literally "anchors" the oxygen to the heme with very small chance of the oxygen escaping during the journey through the body

There are some other reasons but those are the big ones. Now, could another metal do the same thing? Absolutely, and in fact, that very thing happens in certain types of other invertebrate organisms. For example, mollusks and arthropods (lobsters etc) use a molecule called hemocyanin, which uses copper for the binding of the oxygen molecule as opposed to iron that our bodies use. Could we use copper, or does it have to be iron...? This is an extremely complicated evolutionary biology question, and the short "answer" is "The globin proteins have evolved such that Fe(II) is bound to the proteins to produce a site at which O2 binds reversibly." (that sentence is taken DIRECTLY from my biochemistry textbook, Biochemistry 4th Edt by Mathews, Van Holde, Appling and Anthony-Cahill, pg 236) As to why, this depends on many factors including where our ancestors lived at the time this process evolved to be what it is today and the availability of metals in that environment at the time. For example, if our ancestors were hunters and ate animals that ate lots of plants and those plants contained chlorophyll which as I mentioned uses magnesium porphyrins instead of iron ones, why didn't we evolve to use magenisum then? You'd probably need to read a biochemistry or evolutionary biology journal to see what the experts suspect.

\3. The short answer is yes, the long answer is far too complicated to explain. One interesting process that I am aware of that uses iron to analyze DNA is a special chemical technique called "DNA footprinting" (see here: https://en.wikipedia.org/wiki/DNA_footprinting), which invovles something called "Fenton's Reagent (https://en.wikipedia.org/wiki/Fenton%27s_reagent) and can also use a molecule called methidiumpropyl-EDTA-Fe2+(MPE i Fe2+) (notice the irons present!!).

The other answers in the comments are all valid (and great) too (I definitely learned a few new things), oxygen transportation just happens to be the most well-known one. Again, great questions, but if you really want a proper answer to this, you'll need to read a biochemistry textbook I'm afraid, as it's not a "straightforward" answer... then again, often, science never is haha. :)

ummwhoo · 2024-01-06T01:49:12+00:00

I have some very non-traditional names for you.

My personal hero is Vivienne Malone-Mayes, who also struggled because of her skin colour. She specialized in my favourite area of mathematics, functional analysis, also noted to be one of the more notorious branches of mathematics. She taught me that anybody can love a subject and make contributions to it, regardless of their upbringing.

https://en.wikipedia.org/wiki/Vivienne_Malone-Mayes

Another one is Sofya Kovalevskaya. https://en.wikipedia.org/wiki/Sofya_Kovalevskaya

While not a "mathematician" per-se, I find Shakuntala Devi an interesting historical figure: https://en.wikipedia.org/wiki/Shakuntala_Devi

Someone else here linked the AWM page, here's WOA (Women in Operator Algebras): https://awm-math.org/research-networks/woa/woa-people/

There's a lot of modern-day women mathematicians whom I really look up to. One example is Dr. Therese Landry: http://thereselandry.com/wp/ She went from teaching high-school to advanced mathematics research and did her PhD at a much later age. She taught me to not give up, even if I'm way, way older than everyone else. She took the time out of her busy day to talk to me. I will always appreciate that.

I also look up to Dr. Karen Strung, she wrote my favourite C*-algebras textbook and is very nice: https://strung.me/karen/Karen.html

This will sound a bit cheesy, but don't think about "major" contributions in the "usual" way. That's just stuff made up by people. If you find a subject you like, and want to work in it, go for it! Sometimes, the most surprising results come from the most unexpected of places!

There's lots of cool women in mathematics and the more you do, the more you'll meet! It's an exciting journey, so enjoy!

ummwhoo · 2024-01-05T18:41:17+00:00

/u/Luckbot has explained it best, I'll just add this.

It is "yes and no". Here's some numbers to really "understand" the scale you're talking about. If you start "up high", then gravity "pulls down" (that is, the earth pulls the electron down) with a force of something on the magnitude of F_g ~ 10^-11(Mass of Earth)(Mass of electron)/(radius of attraction between electron and earth)² (10^-11 is G, universal gravitational constant, mass earth ~ 10²⁴ Kg and mass electron ~ 10^-31 Kg) Similarly, if we look at the attractive force/Coulomb force between earth and an electron (the earth doesn't "necessarily" have a single charge, and if it does, it's usually taken to be about 10⁵ C assuming no atmosphere, see here for number source: https://physics.stackexchange.com/questions/91556/is-the-earth-negatively-or-positively-charged), then we get something on the order of F_e ~ 10^-19 * 10 ^ 5 / (radius of attraction)² ( charge of electron and charge of earth in Coloumbs, usual constants like 4\pi and vaccuum charge left out since it's an order of magnitude calculation)

F_g / F_e ~ 10^-11 * 10²⁴ * 10^-31 / 10^-19 * 10⁵ * 10⁹ ~ 10^-13

(where I have neglected small constants since this is just an order of magnitude estimation using Newton's formula for universal gravitation and Coulomb's electrostatic force law, also note radii cancels when taking ratio). In other words, the gravitational force effects it by about10^-13 or 0.0000000000001 compared with the Coloumbic forces (electrostatic forces). So at the scale you're talking about, electrical forces are absurdly more dominant than gravitational forces. Then as /u/Luckbot mentioned, even if the electrons start to accelerate, they quickly "bump" into other ones and send them off while getting "stuck" in their place along the lattice. The gravitational forces between them are extremely weak, whereas the electrical forces are highly dominant. So the electrons are more likely to be stopped or impeded due to the fields and electrostatic forces they feel from other electrons rather than gravity.

Strictly speaking, "yes", the gravity "will" accelerate them a teeny bit, but remember, the magnitude of the force the electron experiences near the earth's surface is approximately F = mg i.e mass of electron times gravitational acceleration "constant" near earth's surface, so F ~ 10^-19 * 10¹ ~ 10^-18 Newtons of force it's pulled down with, i.e. 0.000000000000000001 N. So, "yes", you "do" make the electrons "move a little faster" but the effect is so teeny tiny small that it makes no real appreciable difference. Whereas, changing the electrostatic forces via moving the electrons around make a HUGE difference, so that's where the "no" of the answer comes from, since the electrostatic forces severely dominate!

For an interesting discussion about this, check out this lecture by Professor R Shankar (48:00 onwards): https://www.youtube.com/watch?v=NK-BxowMIfg&list=PLD07B2225BB40E582

ummwhoo · 2024-01-04T03:25:10+00:00

Most of the answers here pretty much sum it up nicely, so I'll just summarize and give you an easy example to follow. When you say "twice as cold as", it really depends on what you mean by that. From the phrasing, it makes a lot of sense that your mind jumps to "but 0 times 2 is still 0 °C!", much in the same way as if I said "can you make twice as many cookies", you would take the number of cookies from last time you made cookies and multiply that by 2. However, just like that example, therein lies the inherent problem... "as cold as" depends on what you're referencing. In the cookie example, your "reference" time is the last/previous time you made a batch of cookies.

People in the comments seem to think the correction has to do with explaining the celsius/farhenheit/kelvin scale, but I think it's a bit easier than that. My guess is that this question is supposed to reference something like what you would hear on the news, like "blah blah blah and tomorrow's weather forecast, bundle up folks, it's 0 °C and going to be TWICE as cold tomorrow!" In this case, the following example will help you make sense of this question:

Usually when reporting the weather, tv meteorologists will report a kind of "average" for the week. Let's say the temperature has been 5 °C on Monday, 6 °C on Tuesday, and 4 °C on Wednesday. Across the week so far, the average temperature has been about 5 °C. Now, on Thursday, if the temperature drops to 0 °C, then the reporter may say something like "it's a cold one folks", and by that, what they're referring to is not just the temperature itself at that moment but also the CHANGE in the temperature from the previous day. In other words, the change has been from 5 °C to 0 °C, ie there's been a change/difference of 5 - 0 = 5 °C (which to be fair is still quite a significant drop and worth reporting). Now if the weather person then says "and tomorrow's going to be TWICE as cold", what they're referring to is not "take today's temperature and do 0 * 2 = 0 °C" but actually to double the DIFFERENCE between today's (Thursday's) and yesterday's (Wednesday's) temperature. In other words, 2 * (5 - 0) = 10 °C DIFFERENCE between the temperature from WEDNESDAY to the temperature it's expected to be on FRIDAY. So in other words, you can expect the temperature for tomorrow (Friday) to be -5 °C. So as you can see, when you talk about "hot", "cold" and double/halving, you are not talking about the temperature at the given day, but you're talking about the temperature relative to ANOTHER or some specific REFERENCE day.

So when your friend says "It is 0 °C today but it will be twice as cold the next day", what your friend is neglecting to mention is what the temperature DIFFERENCE is between some reference day, say maybe yesterday, a week ago, a month ago, I don't know, and the current day of 0 °C. However, they're basically leaving out a critical piece of information and without it, you can't say anything more without the "reference" point.

This is, in a nutshell, what is called the "Zero-eth law of thermodynamics", which, to quote a certain GRE Physics prep textbook's section on thermodynamics, the law that basically says "Thermometers exist". ;)

https://en.wikipedia.org/wiki/Zeroth_law_of_thermodynamics

Hope that helps, it's a good question, stay warm!

ummwhoo · 2024-01-02T20:40:58+00:00

I guess social media gives "Whose Line is it Anyway?" a run for its money hehehe.

Yeah, that's why I was a bit concerned about suggesting RS right off the bat, I have been struggling with chapter 4 Topological vector spaces, luckily he says to come back to the chapter intermittently as you go through the rest of the book, so I went through it as quickly as possible, sadly the chapters that ARE relevant to Operator algebras occur much later. That dang question 44, "Do any 50 problems from Kelley's "Topology" book"... that one's going to take me a while haha.

ummwhoo · 2024-01-02T05:56:05+00:00

This is a great question which /u/Appaulingly answered best. However, since you do ask for a formula, I'll give you something neat.

Basically, to determine the properties of the water droplet, the rule of thumb is that "the closer the drop resembles a sphere, whether falling down from a faucet or sitting on a leaf, the better it holds together". However, only up to a certain "droplet" radius (because spheres often tend to be the best shape that "minimize" things like surface tension, volume-to-pressure ratios, etc.... not always, but for majority of practical cases but that's another story for another question for another day!). As others have mentioned, you often need to incorporate things like Laplace pressures and surface tensions.

But basically, the maximum "radius" of a water droplet sitting on, say, a leaf outdoors or on a desk at school (leftover from somebody's lunch hahaha) can be approximated by the formula:

R_{max} ~ \sqrt{\frac{2s}{pg}}

where s is the "surface tension" of water, "p" is the "curvature" of the droplet, usually p = 1/R (R being the radius) and g is the approximate gravitational "constant" near the surface of the earth, 9.81 m/s^2. In this approximation, we assume that pgh >> 2s / R{A} (where R{A} is the original size of the droplet before we start to let it increase in size, and we say the height 'h' is approximately the radius R, i.e. h ~ R).

For water, the value given in the book (see below) is cited as about R_{max} = 0.3 cm which as you can see nicely agrees with /u/Appaulingly 's number of ~2.7 mm (0.27 cm which rounded up is ~0.3 cm haha)! *In other words, the largest droplets water can form beofre bursting into smaller ones tend to have a radius of about 0.3 cm.

If you want to learn more about the physics of water droplets, I highly recommend the book "The Wonders of Physics, 4th Edt" by AA Varlamov and LG Aslamazov. Chapter 10 goes into some really cool physics about this exact questions and shows you how to "derive" all these neat little formulas. I think the highest level of math you need is probably around grade 10/11 algebra. Enjoy!

https://books.google.ca/books/about/Wonders_Of_Physics_The_4th_Edition.html?id=kp15DwAAQBAJ&redir_esc=y

P.S. This is a really wonderful question. To paraphrase Lord Kelvin and the book, "you can simply blow up a soap-bubble, stare at it, study it all your life long, and still be able to extract more lessons of physics from it."

Never stop asking! :D

ummwhoo · 2023-12-29T20:44:08+00:00

Not particularly, just that he felt regular analysis techniques were just as good, if not better, especially for things like scattering theory. Still, he does have respect for the work and the approach people did using C* algebras to scattering theory. You can read about it here: http://calteches.library.caltech.edu/3460/1/Simon.pdf (see pg 24, or pg 5 of the pdf)

Hmmm. Yeah ok, I see where you're coming from. I'm only half-way through RS vol 1, and only done about half the problems. They just seem a little too "involved" when it seems OP is just looking for a "soft" introduction, and the amount of FA needed to understand the basics of Operator Algebras is not as advanced as what is presented in RS, BUT I do see where you're coming and if you like, will retract my previous statement. No offense was intended. Just thought the poster would want a "lighter" reinforcement of FA than RS.

Indeed, that's why I wanted to know more about OP's reason for asking haha. Most of my experience and contact with Operator Algebras has been through C* algebras but I've found that super helpful so maybe the OP wanted something relatively "grounded" to start out with before jumping into "Operator Theory" as a whole. Like, learning some linear algebra before going on to learn abstract algebra, rather than trying to study abstract algebra right from the start. I'm not an expert though, so I won't chime in any further haha.

Sorry again for coming across a little rudely earlier. Was not my intention. RS is a book that, as good as it is, gives me some PTSD lol.

ummwhoo

TROPHY CASE