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February 15

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

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

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

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

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

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

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

Magnetic compass

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

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

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

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

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

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

That's

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

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

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

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

Increasing the needle's mass M by increasing its width or thickness increases its magnetic torque by the same factor, which does not help if the stickiness is due to pivot Stiction (static friction) whose resistance to motion is proprtional to the weight of the needle. Increasing needle mass instead by increasing its length moves the effective centers of application of torque further from the pivot, thereby adding a further increase in the moment of the torque to give faster settling. Another way to overcome stiction is to vibrate the compass. Philvoids (talk) 14:14, 17 February 2024 (UTC)

February 16

Tertiary color

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

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

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

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

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

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

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

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

If you go back to that table you'll see that black is an absence of color. Mixing pigments removes color components. Additive colors are probably more what you want if you want to talk about a 'tertiary' color and that would be white or a pale color. NadVolum (talk) 23:01, 16 February 2024 (UTC)

A tertiary color is an intermediate color resulting from an even mixture of a primary and a secondary color, i.e. a mixture of the primaries in 3:1:0 proportion resulting in a less saturated form of the dominant primary color of the mixture. Philvoids (talk) 01:41, 17 February 2024 (UTC)

Note that this is given as a more recent "alternative definition" conflicting with the more traditional (and quite different) one of a mixture in a 1:2:1 proportion.  --Lambiam 09:34, 17 February 2024 (UTC)

Using the RGB additive colour model, here are the three maximally light 1:2:1 mixtures:

     

and here are the six maximally light 3:1:0 mixtures:

           

Brown is more like a (not maximally light) 4:2:1 mixture:  .
 --Lambiam 09:56, 17 February 2024 (UTC)

What about yellow? Zarnivop (talk) 10:34, 17 February 2024 (UTC)

As an even mixture of R and G, it is a secondary colour in the RGB model.  --Lambiam 13:28, 17 February 2024 (UTC)

  Yes, or a primary color in subtractive color systems or 575–585 nm spectral wavelength. Philvoids (talk) 13:50, 17 February 2024 (UTC)

February 19

Why the physical and chemical properties of element 0 is very hard to find?

Why the physical and chemical properties of element 0 (neutronium) is very hard to find (e.g. it should be as noble as helium and neon, and should be gas in the room temperature, maybe neutronium is ideal gas)? The half-life of the free neutron is longer than the half-life of the longest- lived isotope of the elements with atomic number >= 107, and some physical and chemical properties of the elements with atomic number >= 107 are already known. 61.224.147.34 (talk) 07:50, 19 February 2024 (UTC)

If you look at our article your will read it is a "hypothetical substance". But see Neutron#Neutron compounds, the conditions to form these are not available in the Solar System, so anything known would be hypothetical. Perhaps some clues can be gleaned by neutron star collisions. Neutrons stored in a bottle would be a dilute gas. They do not interact with light, and so is transparent. Graeme Bartlett (talk) 08:42, 19 February 2024 (UTC)

I don't think neutrons can be stored in a bottle. Since neutrons are neutral they do not feel the electromagnetic Coulomb barrier that keeps atoms and molecules in the bottle. For the same reason it makes little sense to talk about "chemical properties". These are determined by the electrons in atoms; neutronium has no electrons, and neutrons cannot capture electrons to form negative ions. --Wrongfilter (talk) 09:03, 19 February 2024 (UTC)

See ultracold neutrons#Reflecting materials for what to make your neutron storage bottle from. It appears that beryllium is best. (But it's not transparent) Graeme Bartlett (talk) 09:37, 19 February 2024 (UTC)

That's very interesting, thanks! Pity the article doesn't provide enough information to understand how that works. --Wrongfilter (talk) 09:52, 19 February 2024 (UTC)

Apparently it works by the strong interaction, as expected. Which basically illustrates why element 0 doesn't really have chemical properties; it has no electrons, which would be needed for such things like forming bonds and a normal condensed phase. Double sharp (talk) 10:24, 20 February 2024 (UTC)

Also, neutrons are only stable when inside atomic nuclei: free neutrons (which 'neutronium gas' would consist of) decay with a mean lifetime of about 14¹/₂ minutes and a half-life of a little over 10 minutes. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 13:54, 25 February 2024 (UTC)

That's not the problem. As noted by the OP, all known bohrium and hassium isotopes have shorter half-lives than free neutrons, yet we know some chemistry of those elements. The problem is that a free neutron doesn't have electrons and therefore can't have chemistry. Double sharp (talk) 18:16, 25 February 2024 (UTC)

Has any gluon's velocity ever been measured?

I'm asking, because for the time being, only two particles, considered to have no mass, have been detected, one of which has empirically turned out to have a velocity, being (or sufficiently close to) the well known constant velocity C, if that particle (namely photon) is in a medium, being (or sufficiently close to) an absolute vacuum. So my question is: what about the other particle (namely gluon), as far as its measured velocity (especially in a vacuum) is concerned? HOTmag (talk) 13:04, 19 February 2024 (UTC)

The scale at which gluons operate (see Strong interaction § Behavior of the strong interaction} is less than 0.8 fm = 0.8×10⁻¹⁸ m. At the speed of light, the time scale is less than 3×10⁻²⁷ s or 3 rontoseconds. The inverse of the hyperfine transition frequency of ¹³³Cs used in atomic clocks is about 0.1 nanoseconds or 0.1×10⁻⁹ s, 16 orders of magnitude larger. I do not think technology is up to measuring such short time spans. Apart from technological limitations, I think there are also fundamental limitations baked into the laws of physics as we understand them, such as quantum uncertainty. I can't think of any kind of experimental setup that might clock a gluon as being at some definite position. Bounds on gluon mass will have to be deduced from energy budget calculations of high-energy collisions.  --Lambiam 17:59, 19 February 2024 (UTC)

Thank you for your (disappointing) reply. I hoped the gluon's velocity could be measured somehow, because I wanted to make sure that the well known hypothesis - stating that no massless particle can be a tachyon - could be emprically proved also for particles other than photons, but according to your reply I understand we will have to stay in the boundary of photons alone for emprically proving this hypothesis - which I suspect is not sufficiently reliable while it has been empirically proved for photons only. That no massive paricle can be a tachyon, is a well established fact, or rather a mathematicllay proved fact - deriving from the relativistic equation of momentum, but this fact cannot be mathematically proved for massless particles, for which we can only rely on experiments, which have actually been carried out for photons only, unfortunately, as I understand from your reply. What a pity... HOTmag (talk) 08:00, 20 February 2024 (UTC)

February 20

The increasing role of relativistic effects in heavy elements, since the speed of the 1s electron is not far less than c (speed of light)

The increasing role of relativistic effects in heavy elements, since the speed of the 1s electron is not far less than c (speed of light). Thus if the reciprocal of the fine-structure constant is 299792458 rather than 137, then the physical properties and the chemical properties of the elements in the period 8 will be completely different? 125.230.19.122 (talk) 07:01, 20 February 2024 (UTC)

I suppose by changing α you really mean changing c, implicitly using atomic units. (See this Physics Stack Exchange answer for the subtleties involved in such "alternate-physics" discussions.) So this really amounts to increasing the speed of light until relativistic effects are no longer important. In that case, not only is period 8 probably completely different, but also differences should be noticeable in periods 6 and 7. Without relativistic effects, mercury would be solid at room temperature, and lead-acid batteries would not work (because relativistic effects stabilise Pb^II relative to Pb^IV; without them, Pb would be much more like Sn than it really is). Double sharp (talk) 07:56, 20 February 2024 (UTC)

steel flexture lifetime

I was watching this video [1] when I noticed that at 0:30 they claim: "Infinite lifetime", "no friction, no wear".

I think this is a flexure mechanism, and I'm not too familiar with them. But I do know a little bit about springs, and springs have a finite design lifetime, because every time a spring is compressed fatigue is introduced. Springs "wear", in other words.

Since springs have a finite lifetime, and undergoes wear during operation, wouldn't the same principle apply to the device shown in the video? OptoFidelty (talk) 21:33, 20 February 2024 (UTC)

They also advertize their gear as "frictionless" on the very first frame, which I think is not physically possible.  --Lambiam 01:22, 21 February 2024 (UTC)

According to Plasticity (physics) however, load should not exceed the yield strength associated with a given material for it to retain its plasticity. It's a simple rule that may be perhaps not practical to observe in most real-life situations; and it always ended in failure at some point in all situations I studied, but that might be a matter of scale. Then the video is about a product intended to be used in a very sterile, quiet, and, finally, previsible environment. Thus, the notion of fatigue as we know it must also have a subjective dimension in it in some way. --Askedonty (talk) 01:27, 21 February 2024 (UTC)

Steel has a so-called endurance limit for stress, and according to this there is no limit to the fatigue life if you stay under it. According to an experienced test engineer, this is more a result of poor test technique than physics. Ultimately if you keep flexing steel, migration of discontinuities will continue, they will form cracks and then it breaks. Sorry, no refs. Greglocock (talk) 08:24, 21 February 2024 (UTC)

I would be very cautious about any claims within this advert since, as far as I can see, it indirectly claims to be breaking the laws of physics. Electric fan and direct radiant heaters already convert well over 90% of their drawn electrical energy to heat, so the inferences that these devices can heat a room much more quickly ("in seconds" – how many seconds?) much more cheaply ("for pennies" – how many pennies?) than existing devices seem dubious. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 15:21, 22 February 2024 (UTC)

What advert are you talking about? The OP posted a video about a steel flex mechanism with no electric heater mentioned. Philvoids (talk) 21:06, 22 February 2024 (UTC)

That's odd! When I first clicked on it, I saw a YouTube ad I'd seen before for what is essentially a small electric hot-air blower that mounts directly on to a wall socket. I guess it must have been prefixed by YouTube to the actual video the OP was referencing. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 21:57, 24 February 2024 (UTC)

Plus heat pumps are far better at heating than directly converting electric currents into heat. NadVolum (talk) 22:50, 22 February 2024 (UTC)

February 22

Aliens have killed themselves

There are reasons to think that extraterrestrial life should be common, but so far we have never find any. The Fermi paradox provides several possible explanations for it, such as that aliens may self-destruct after discovering atomic power or other advanced technologies. But today I was thinking, should that be an obstacle? Let's say that the Klingons started an atomic war and killed themselves before we could get anywhere near a Star Trek situation. Shouldn't we still be able to detect anyway an exoplanet with an atmosphere polluted with high levels of radiation, that could not be explained by natural processes? Wouldn't other ways of an alien civilization to self-destruct (chemical war, mere pollution, etc) leave detectable clues as well?

I don't think I need to explain that such a discovery (a world where life existed, and more, intelligent life!) would still be a revolutionary one, even if there was nobody left by now. Cambalachero (talk) 18:15, 22 February 2024 (UTC)

Well, for one, the Fermi paradox presumes that civilisations spread through the galaxy. If they self-destroy before reaching that phase (as we very well might still do, assuming we even are civilised to begin with), then they are much less frequent, and hence much harder to detect. Also, we can sometimes get some spectroscopic data for some exoplanet atmospheres, but I doubt we can distinguish between radioactive and other isotopes that way. --Stephan Schulz (talk) 18:54, 22 February 2024 (UTC)

That depends on the method of spectroscopy, element, and compound. Infrared techniques and microwave techniques, or redshifted forms of them, are molecular spectroscopy, see infrared spectroscopy and microwave spectroscopy. These are tools for measuring molecular vibrations and molecular rotations, respectively, and the masses of the atoms within the compounds play a major factor in the spectral signature. The frequencies of their absorptions are inversely proportional to the square root of the reduced mass of the system. In a diatomic molecule, for example, a significant change in the mass of of the atoms by isotopic substitution is easily detectable, and is often used as a tool in studying molecular systems by intentionally substituting for isotopes. Take the case of HCl; it's infrared spectrum is made up of doublets due to the presence of both chloride-35 and chloride-37 isotopes, which exist in known ratios (under natural conditions) reflected in the relative intensities of the doublet peaks. If that ratio were to be different, that would at least indicate something unusual had happened. The difference in frequency there is fairly small, but the difference in frequency between HCl and DCl is quite large and incredibly easy to detect. While it would be harder to detect than chlorine-37, HCl composed of chlorine-36 could, if detected, be an indicator of the use of nuclear weapons, since our underwater testing of nuclear weapons produced a great deal of chlorine-36. That's just one example, though. I'll bet a careful examination of the isotopic products of a nuclear exchange was done, others examples could be found, and a potential model for IR/MW examination for extra-terrestrial nuclear weapons use could be designed. --OuroborosCobra (talk) 19:29, 22 February 2024 (UTC)

Good point - I had only thought about atomic spectra. --Stephan Schulz (talk) 19:39, 22 February 2024 (UTC)

Also, going by our own example, it wouldn't take much to cause the collapse of a single-system technological civilization at our own level or beyond (which needn't cause total extinction), and any effects detectable at long range would be short lived in astronomical terms. Such civilizations might arise not uncommonly Galaxy-wide, but too far separated in time for their technological eras to overlap. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 19:27, 22 February 2024 (UTC)

There is no way of establishing ferm lower bounds on most of the variables featuring in the Drake equation. If a technologically advanced civilization arises about once every million years in our galaxy and then inevitably blows itself up, the signature of the calamity will have disappeared by the time we examine the planet.  --Lambiam 22:49, 22 February 2024 (UTC)

The Fermi Paradox has always struck me as being misnamed. The idea that the galaxy should be full of aliens rests on so many assumptions that it us not having found evidence of any (let alone evidence that they have visited us) hardly seems to constitute a "paradox". Iapetus (talk) 10:55, 23 February 2024 (UTC)

It has that in common with many puzzling facts named "paradoxes": Antarctic paradox, Denny's paradox, Elevator paradox, Potato paradox, Willpower paradox. Informally, the term paradox is often used to describe a counterintuitive result. Case in point: our page Counterintuitive redirects to the article Paradox.  --Lambiam 13:52, 23 February 2024 (UTC)

Our article Date of Easter talks about the "Easter paradox", meaning that it is not observed on the Sunday after the full moon, or it is observed on the Sunday after the "wrong" full moon. Kepler gave the answer to that: "Easter is a feast, not a planet." Incidentally, following the introduction of a new table acceptable to everyone Special:Permalink/1188536894#The Reichenau Primer (opposite Pangur Bán) it is anticipated that this year's divided celebration (split between 31 March and 5 May) will be the last. It was the 31 March celebration ordered by the Romanian Orthodox Church which led to the peasants' revolt and the disappearance of the Gregorian Easter from Orthodox eastern Europe (barring Finland, where the government offers the Orthodox church there money to keep the Gregorian Easter, to which their response is "Why not?") 81.154.229.214 (talk) 17:19, 23 February 2024 (UTC)

It may be that the Earth was one of the first planets where life could develop well. We'd really need to know the distibution of kilonovas through time to estimate when that happpened first, without enough of them a rocky planet would have very little of the heavy elements and having life evolve and a technological civilization develop could be quite difficult. Not quite so heavy elements come from the death of stars like our sun but that would also be billions of years after the big bang. NadVolum (talk) 22:24, 23 February 2024 (UTC)

Alpha Centauri A and B are kind of like the Sun a bit bigger and smaller but hundreds of millions of years older and about 2 decibels higher iron:hydrogen ratio which is a proxy for non-first 2 elements content. Sagittarian Milky Way (talk) 00:13, 24 February 2024 (UTC)

Yes: the number of a specific star's preceding 'stellar generations' is not well correlated with its absolute age (or rather youth), because stars' 'life cycles' are so heavily dependent on their masses, and the interstellar medium is variable in its composition. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 13:31, 24 February 2024 (UTC)

February 23

Does a photon measure set its velocity as plus C, i.e. from the left to the right, or as minus C, i.e. from the right to the left?

Two observers looking at each other, see a photon move between them on their axis of right/left. Observer A sees the photon move at plus C, i.e. from the left to the right. So, observer B, who looks at observer A, must see the photon move at minus C, i.e. from the right to the left. But what about what the photon sees when measuring setting its velocity? Will the photon set its velocity as plus C, i.e. from the left to the right, or it will set its velocity as minus C, i.e. from the right to the left? HOTmag (talk) 01:18, 23 February 2024 (UTC)

A photon cannot measure its own velocity, not just because it is not a physicist, but mainly because it has no proper time. Any two events along a lightlike curve have $\Delta \tau =0$ and the photon cannot derive a speed from that. --Wrongfilter (talk) 07:05, 23 February 2024 (UTC)

Thx.

I apologize, because I was probably wrong with choosing my words. Actually, instead of "measuring" I meant "setting". Anyway, please notice, that according to the principle of the constancy of the speed of light, the photon doesn't have to measure anything - e.g. its speed, because (even without being a physicist or a human being) the photon has already set its speed as carrying the absolute value of C, hasn't it? I assume you agree it has (without measuring anything), so the only question remaining is about whether the direction set by the photon is plus C - i.e. from the left to the right, or the other way around - i.e. minus C. I hope my question is clearer now. HOTmag (talk) 07:47, 23 February 2024 (UTC)

(ec) $\Delta \tau =0$ implies that there is no direction. The photon cannot say that it was first here, then there, or the other way round because it is here and there at the same time. --Wrongfilter (talk) 07:56, 23 February 2024 (UTC)

There is no physical concept of a particle "setting" its velocity. Velocity is a vector quantity; a description of the measure of a velocity requires a frame of reference. The velocity of a particle does not depend on the selected reference frame; it is only the description of its measure that does depend on it. Photons can do many things; selecting a reference frame for describing the measure of its velocity is not one of them.  --Lambiam 13:12, 23 February 2024 (UTC)

The entire Universe is split into the two half-universes occupied repectively by observers A and B. Any observer distinguishes "right" or "left" relative to a personal baseline such as the separation of their eyes. An observer can quantify the travel of a photon only after its arrival at a destination target is timed. Then if the source of the photon is known, occupants of A and B can agree about the photon's velocity C in vacuum but disagree about its left/right direction, however all are equally correct. On a later occasion an investigator may choose to walk the path of the long-gone photon; we do not need his opinion about right or left because the only path he follows is "straight ahead". Let's not complicate this with gravitational lensing or "conscious" photons. Philvoids (talk) 14:16, 23 February 2024 (UTC)

February 24

Special Relativity

I'm looking for a totally mathematical description of Minkowski space, and I'm unable to find one. The wikipedia page says that its a 4-dimensional vector space equipped with a bilinear norm. However, I want a mathematical definition of what that vector space is. Is it R^4? Taabibtaza (talk) 08:11, 24 February 2024 (UTC)

No. Minkowksi space is a vector space, period. It is defined by the vectors that live in it; to start with, these are abstract objects. They can be related to $\mathbb {R} ^{4}$ through an isomorphism, i.e. a mapping from those abstract vectors and operations to, I'll say, matrices ($4\times 1$ or $1\times 4$) in $\mathbb {R} ^{4}$. In fact there are (infinitely) many such isomorphisms distinguished by the choice of a vector basis; physically that corresponds to the choice of reference system. This distinction is in my view essential for a proper understanding of relativity: there is an absolute space-time, mathematically represented by Minkowski space, in which physical quanities, e.g. four-momentum, are given as vectors, geometrical objects, that are what they are independent of any observer. When we measure these things we measure vector components in a specific basis, our reference system, given by the experimental setup. These components (e.g. energy or one of the three spatial momentum components) are scalar products of the vector with one the basis vectors and they are relative in the sense that they depend on the choice of the basis vector, i.e. the reference system. --Wrongfilter (talk) 08:24, 24 February 2024 (UTC)

@Wrongfilter

I'm just starting to learn Minkowski space, so I'm assuming we have already chosen a coordinate system for Minkowski space such that it's origin corresponds to an inertial observer who has three axes attached to him such that the 3 spatial coordinates of any event correspond to the coordinates he will measure for that event using the axes that are attched to him, and such that the time at any event in spacetime corresponds to the time on our inertial observer's clock (whose clock was set to 0 at an arbitrary time) when that event happens. This is how the wikipedia introduces minkowski space (I'll learn how to shift to a different perspective later). Also, changing the basis while not changing the vector space won't won't allow us to change to an arbitrary frame of reference since our origin is still the same. Each vector space has a UNIQUE origin by the very definition of a vector space. The origin is the vector you get by multiplying any vector by the scalar "0". Anyway, now that we are done with the physical interpretation, think of Minkowski space mathematically.

Is that vector space R^4? If not, you have to give a definition of that vector space and its elements, you can't just say that vectors in that vector space are abstract objects. That's not how math works, even abstract objects have definitions. For example the elements of R^4 are defined as ordered quadruples of real numbers. Taabibtaza (talk) 11:01, 24 February 2024 (UTC)

First up, I'm a physicist with an incomplete and probably not too solid a background in the relevant mathematics, so I may be shaky in my use of terminology and incomplete in my perspective on these things. With that out of the way, back to pretending that I understand something. A vector is simply defined as a member of a vector space. It is defined by how it behaves, not by what it is. For instance you can show that the ordered quadruples (I called them matrices above, but ordered quadruples is better) with the appropriate definition of addition and multiplication by real numbers form a vector space. Therefore $\mathbb {R} ^{4}$ is a vector space and the ordered quadruples are vectors. Add the appropriate definition for the metric and you can show that $\mathbb {R} ^{4}$ with that metric $\eta (\cdot ,\cdot )$ is a Minkowski space. Next comes the physical assumption that space-time with events and separations between events can be described as a Minkowski space (if you insist in thinking about the origin, maybe as a class of Minkowski spaces because the origin is physically irrelevant by virtue of the relativity principle). Thus we have two spaces that are both Minkowski, therefore these two spaces are isomorphic, therefore we can map space-time to $\left(\mathbb {R} ^{4},\eta (\cdot ,\cdot )\right)$, and it will turn out that we can do that in infinitely many ways (through Lorentz transformations). Maybe more directly to your question: (1) Mathematically, a Minkowski space is an abstract space with certain properties. (2) Physically, in special relativity, Minkowski space is space-time, the vectors are separations between events. I guess that on this abstract level, four-momenta form yet another Minkowski space. (3) $\left(\mathbb {R} ^{4},\eta (\cdot ,\cdot )\right)$ is another, extremely useful, Minkowski space. This is of course the main mathematical structure to apply the theory.

I don't think, actually, that you need to worry too much about these things (subtleties?) while learning, but it's good be aware that they exist. Note incidentally that in quantum mechanics it is more common to say things like "Spin states form a Hilbert space", implying that Hilbert space is an abstract mathematical structure that can be filled with physical content in various ways. In relativity "Minkowski space" more uniquely associated with a particular physical context, viz. space-time. --Wrongfilter (talk) 12:05, 24 February 2024 (UTC)

@Wrongfilter

Can you please tell what η(.,.) means? Taabibtaza (talk) 14:07, 24 February 2024 (UTC)

See Minkowski space#Scalar product. Double sharp (talk) 14:58, 24 February 2024 (UTC)

How different would a 1st generation star be from a pure H1 star of same age+mass@ignition?

Obviously it'd only be pure hydrogen-1 before ignition. Assuming no mass is added by stellar companion or any other means.Sagittarian Milky Way (talk) 16:59, 24 February 2024 (UTC)

Helium would make it a bit more opaque by adding absorption lines. Also HeH⁺ would add opacity. More energy could be derived from pure hydrogen. A pure hydrogen star could form bigger than when mixed with helium. Graeme Bartlett (talk) 21:21, 24 February 2024 (UTC)

And note that some Helium (and possibly a trace of Lithium) was formed by the 'Big Bang' (the last I heard), so a completely pure Hydrogen star is never going to have happened. {The poster formerly known as 87.81.230.195} 176.24.45.226 (talk) 22:03, 24 February 2024 (UTC)

Yes I was aware. This does bring up the question of 1st generation stars vs. non-existent ones that are exactly the same except they ignited with exactly 100% isotopes of H and He in 1st gen star ratios. Could current tech tell them apart and from what distance? Sagittarian Milky Way (talk) 22:26, 24 February 2024 (UTC)

Can current technology tell existent and non-existent starts apart? That is an interesting issue.  --Lambiam 04:56, 25 February 2024 (UTC)

All stars have spectral lines and specs: Mass, luminosity and magnitude, radius, variability, isotope and element ratios, color indexes like blue minus V and U minus blue... Different starting compositions would have different specs or spectra at the same age and starting masses, the question is how easy could they be told apart with 6 orders of magnitude less non-1st row element content than the Sun vs infinite orders of magnitude less. Sagittarian Milky Way (talk) 06:46, 25 February 2024 (UTC)

Flying non-insect arthropods

Are there any flying arthropods that are not insects? —Mahāgaja · talk 22:13, 24 February 2024 (UTC)

Assuming you mean powered flight, then no according to Flying and gliding animals. If you accept gliders, then see the bridge-spider as an example of a spider which uses threads in the wind to travel.-gadfium 23:03, 24 February 2024 (UTC)

Thanks! I do indeed mean true flight (like bats) not gliding (like flying squirrels). —Mahāgaja · talk 08:24, 25 February 2024 (UTC)

The human pelvis that has an anteroposterior diameter exceeding the transverse diameter qualifies a human as an anthropod. Human-powered flight has been possible since 1961 or possibly much earlier. Philvoids (talk) 05:06, 26 February 2024 (UTC)

An anthropod is not an arthropod. ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:00, 26 February 2024 (UTC)

But there is a good chance that spiders are hitching rides on helicopters and planes. Graeme Bartlett (talk) 09:20, 26 February 2024 (UTC)

BB is correct. An arthropod is invertebrate animal of the phylum Arthropoda, characterized by a chitinous exoskeleton and multiple jointed appendages. Humans do not qualify. Philvoids (talk) 12:37, 26 February 2024 (UTC)

The first words of our Insect flight article: "Insects are the only group of invertebrates that have evolved wings and flight". Alansplodge (talk) 21:29, 26 February 2024 (UTC)

ah but it doesn't say no other has evolved wings or flight. Loophole! —Tamfang (talk) 20:45, 27 February 2024 (UTC)

no, unless you count spiders (theyre arachnids, not insects) 'ballooning' themselves in the air. :P mushi^{( ? )} 16:40, 29 February 2024 (UTC)

February 26

Relativistic doppler effect

Is the doppler effect effected by any acceleration of the reciever and/or the source, according to special relativity. In other words, let's say you have two point objects S and R (for source and reciever) which will only move along a straight line. Let's say that at t=0 (from the perspective of R) S emits a photon of wavelength l. So, when the photon will reach R, will the photon be of wavelength l(1+ v/c)γ where γ is the lorentz factor, and v is the velocity between the inertial frame of refrence in which R is stationary when it recieves the photon and the inertial frame of reference in which S was stationary when it released the photon. Taabibtaza (talk) 19:43, 26 February 2024 (UTC)

When it receives the photon. So, when there's a stream of photons being emitted from S, their wavelengths as observed by R will change as the velocity of R changes. Also: "Is the Doppler effect affected...". --Wrongfilter (talk) 20:27, 26 February 2024 (UTC)

@Wrongfilter

I'm only talking about one photon, can we use the same formula for it, as what we use for the relativistic doppler effect if there is 0 acceleration. Taabibtaza (talk) 02:13, 27 February 2024 (UTC)

Our article on the relativistic Doppler effect has a section on the relativistic longitudinal Doppler effect, which corresponds to your scenario.  --Lambiam 21:54, 26 February 2024 (UTC)

I literally learned the effect and the formula I presented from that article, I'm just wanting to know if we can use the same formula if there is acceleration (and what would v mean in the formula in that case). Taabibtaza (talk) 02:17, 27 February 2024 (UTC)

Same formula, with v being the instantaneous velocity of R with respect to S at the moment that the photon is being detected by R. --Wrongfilter (talk) 12:09, 27 February 2024 (UTC)

February 27

Bulletin of the Seismological Society of America in online libraries?

I'm trying to make the cited sources in White Wolf Fault as accessible to readers as possible but I haven't been able to find this journal in the Internet Archive, JSTOR or Wiley. HathiTrust only has the issues from 1911 to 1926. The Seismological Society of America's website has has the issues cited, but it requires library or institutional access. If there are other journal repositories in our Library that have it, I've missed them. I'm stumped. If someone can help out, future readers will be better off...

I do not at all mind having to hunt down the article in the proper repository (keeps me from getting lazy, lol). Just point me in the right direction and off I'll go! Oona Wikiwalker (talk) 07:19, 27 February 2024 (UTC)

The cited articles are also available online for a (hefty) fee at GeoScienceWorld; see here and here. The fee imposed suggests that the publisher enforces its copyright.  --Lambiam 11:38, 27 February 2024 (UTC)

But Library access is within copyright restrictions because they pay for access for their patrons. That's what I'm inquiring after; did I miss a collection within our library that contains this journal?

Oona Wikiwalker (talk) 23:01, 27 February 2024 (UTC)

I've checked the Wikipedia Library and it doesn't provide access. I have added the doi and bibcode to the cite to make it a bit easier for readers to click to the journal. Mike Turnbull (talk) 16:48, 28 February 2024 (UTC)

It's long been an issue for those of us in Wikipedia:WikiProject Earthquakes, particularly for US earthquakes. You either pay the money or take a trip to a library that stocks that journal, which is generally easier said than done. Mikenorton (talk) 17:28, 28 February 2024 (UTC)

A mistake in our article Four momentum

The third paragraph states: p⁰ = E/c², with p⁰ being the momentum in the time dimension.

I'd fixed it: p⁰ = E/c, but somebody reverted. Would you like to give your opinion? HOTmag (talk) 18:06, 27 February 2024 (UTC)

The version with c² is formally correct. Another matter is whether that should be in the lede of the article (and who is "some authors"?). I don't think I've ever seen that convention — whenever I've seen $x^{0}=t$ it was due to setting $c=1$, in which case $p_{0}=E$. --Wrongfilter (talk) 18:13, 27 February 2024 (UTC)

I admit I'm quite surprised. The four momentum is (p⁰, p¹, p², p³), traditionally written also as (E/c, p_x, p_y, p_z), hence: p⁰ = E/c. On the other hand, both sides of the current expression (in the lede): p⁰ = E/c² don't have the same units, so how can it be correct? HOTmag (talk) 18:29, 27 February 2024 (UTC)

I wrote "formally correct", in some weird unit system. I don't think it makes much sense, which is why I removed it from the article. --Wrongfilter (talk) 18:38, 27 February 2024 (UTC)

Which one of the following three statements do you find "weird"?

1. The four momentum is (p⁰, p¹, p², p³).

2. The four momentum is (E/c, p_x, p_y, p_z).

3. Hence: p⁰ = E/c.

HOTmag (talk) 18:51, 27 February 2024 (UTC)

Oh please, give it a rest. The statement in the article had a context that you keep ignoring. --Wrongfilter (talk) 18:55, 27 February 2024 (UTC)

You are allowed to avoid answering my last question, which has nothing to do with the article. Actually, regardless of the article, I was quite curious to know whether you also considered the third statement p⁰ = E/c (which had been my correction before it was reverted): as "weird", while I assumed you accepted the two statements it followed, and that's why I asked you my last question, regardless of the article. But again, nobody forces you to answer my last question. HOTmag (talk) 19:44, 27 February 2024 (UTC)

$p^{0}=E/c$ is standard, nothing weird about that. --Wrongfilter (talk) 20:15, 27 February 2024 (UTC)

So (back to article), my correction p⁰ = E/c in the article was right, and it didn't have to be reverted to p⁰ = E/c². That's what I wanted to know from the very beginning: whether the user who reverted my correction was wrong (as I thought). HOTmag (talk) 22:26, 27 February 2024 (UTC)

No you were wrong in that context. That bit has been removed now from the article because what was there isn't common and as shown here will just cause confusion. It is best to have all the elements using the same units. NadVolum (talk) 23:25, 27 February 2024 (UTC)

I haven't asked about whether the context justified mentioning this formula there, nor about why the whole formula was eventually removed. I've only asked about whether, my replacing the original formulation p⁰ = E/c² by the correct formula p⁰ = E/c, had been a better version than the original version p⁰ = E/c². In other words, I've wanted to know if the user who reverted my correction p⁰ = E/c back to the original version p⁰ = E/c² made the article worse. I think they did, and I've wanted to get your opinion about what I think. All of that is quite clear in my first post, that had actually been posted before the whole formula was eventually removed. HOTmag (talk) 08:07, 28 February 2024 (UTC)

The original statement was a conditional, with an "if... then..." structure. The formula that you changed was in the "then..." part, and was correct under the condition stated in the "if..." part. By changing the "then..." part without taking into account the "if..." part, you introduced an error, and the other user was right in reverting your edit. --Wrongfilter (talk) 08:38, 28 February 2024 (UTC)

I got it now. Thank you. I understand now that the original "then" part (before I changed it) would be correct assuming that the "if" part were correct. However, the "if" part really introduces a definition I've never come across, and should be omitted. HOTmag (talk) 09:52, 28 February 2024 (UTC)

Parachutes and terminal velocity

Hi, I'm currently doing a lab using a small model parachute. As I currently understand it, the terminal velocity can be calculated by equating the weight of the object to the drag force, given by the drag equation. My question is, how can I use this terminal velocity to estimate the time it would take an object to fall a set distance? Do I assume the object starts out at terminal velocity? Do I use a SUVAT equation? Thanks! ARandomName123 (talk)^{Ping me!} 23:05, 27 February 2024 (UTC)

the time to accelerate to the terminal velocity needs to be accounted for. RudolfRed (talk) 00:22, 28 February 2024 (UTC)

So one needs to integrate the varying acceleration. Poor ARandomName123. Zarnivop (talk) 02:18, 28 February 2024 (UTC)

Integration is the bane of my existence. I guess I'll just approximate it then. Thanks for the help! ARandomName123 (talk)^{Ping me!} 03:09, 28 February 2024 (UTC)

Take the height of the fall, divide by the terminal velocity. That gives you a rough answer. So if the falling object gets close to terminal velocity after just a short fraction of the fall (for a human with a parachute, around 50 metres) and the fall is short compared to the scale height of the atmosphere (around 7 kilometres on Earth; else the changing density changes the terminal velocity) and you don't care too much about the second digit of your answer, this may be good enough. Otherwise, you have to integrate the equations of motion. PiusImpavidus (talk) 11:18, 28 February 2024 (UTC)

Yes, that's what I ended up doing. It wasn't actually too far off from the experimental result, only a few fractions of a second. Thanks! ARandomName123 (talk)^{Ping me!} 13:37, 28 February 2024 (UTC)

February 28

Why are airports never an effing big Greek cross?

Two runways in series instead of in parallel (i.e. 30 Back for landing 30 Front for takeoff 12 Back for landing 12F takeoff if wind azimuth near 120) and the paths never have to cross even in the air. And another two in series at right angles that they switch to if crosswind's too strong. In extreme both engines exploded on fire emergencies they clear all runways much easier landing.

Also why'd it take so long to invent faster runway exits with slight angles? That seems like common sense. Sagittarian Milky Way (talk) 02:28, 28 February 2024 (UTC)

Space at existing airports within metropolitan areas is obviously one limiting factor. With completely new builds in rural areas, it could work. Airports are also always looking at ways of minimising noise on the ground. Your proposal would stretch the area affected by aircraft noise to twice the distance. HiLo48 (talk) 02:48, 28 February 2024 (UTC)

I assume you mean noise due to the flight paths of the airplanes. But besides noise, different flight paths from different runway directions may also limit construction height near the airport or simply not be possible due to current constructions. See also [2] [3] and consider whether this works so well under the proposal. Nil Einne (talk) 09:43, 28 February 2024 (UTC)

Even away from metropolitan areas flat land suitable for an airport may be at a premium and the shape of the suitable area limits the possibilities. The angles of runway exits were not much of an issue until the introduction of heavy wide-body aircraft, which take longer to reduce speed and don't turn as readily as lighter aircraft. If the space is available, an L-shaped two-runway pattern offers the same crosswind-avoidance advantage as a cross while having a decreased collision risk and giving an easier task in designing the whole airport plan.  --Lambiam 10:03, 28 February 2024 (UTC)

Also consider what happens on two inline runways when there's a departure on one and a go-around on the other: the aircraft going around will end up in the wake vortex of the departing aircraft. So the capacity of two in-line runways isn't really much better than that of a single runway. It's better to put them in a staggered configuration.

High-speed runway exits appeared in the 1950s, right at the start of the jet age with its high landing speeds and long runways. And at that time, air traffic wasn't so dense that they needed to land 25 planes per hour per runway. PiusImpavidus (talk) 11:38, 28 February 2024 (UTC)

Wouldn't some who could afford to fly then like time-obsessed execs like fast trips? Sagittarian Milky Way (talk) 16:31, 28 February 2024 (UTC)

Corporate execs want bragging rights: "My bizjet can do Mach 0.92," not "I paid the airport $600k for a high-speed taxiway at exactly the right spot, so now I save 30 seconds each time I visit HQ." PiusImpavidus (talk) 23:02, 28 February 2024 (UTC)

Fair enough. Sagittarian Milky Way (talk) 00:14, 29 February 2024 (UTC)

And don't non-inline runways in the same direction cause problems unless very separated? Which would increase taxi distance between runway and terminal. Sagittarian Milky Way (talk) 16:35, 28 February 2024 (UTC)

Depending on the quality of your surveillance system, and ignoring wakes, you need a horizontal separation of 2.5 to 3NM, or a vertical separation of 300m/1000ft. A modern jet climbs about 1000m/30000ft per minute, i.e. it has the necessary vertical separation in 20 seconds. Takeoff speed is 150knots or higher, so at least 2.5NM per minute, and rapidly increasing. So you would have the required horizontal separation in less than a minute. IIRC, the very best airports manage about one takeoff per minute, so separation is not a problem. --Stephan Schulz (talk) 21:14, 28 February 2024 (UTC)

Parallel runways usually have about 500m separation (like 7C/25C and 7R/25L at EDDF). How much you need depends on the operational flexibility you want; parallel departures require less separation than parallel arrivals. 500m additional taxi distance is less than a minute travel time. Or you could put the runways on opposite sides of the terminal (like at EGLL), so there's no additional taxi distance at all. If you have them slightly staggered (like at EDDB) instead of exactly side by side (like at EDDL), you save some taxi distance and you gain some safety during parallel approaches. BTW, LEMD gets close to your solution. It has two parallel departure runways and two parallel arrival runways, pretty much end to end, but with an angle between them. The new terminal 4 is where the runways almost join. PiusImpavidus (talk) 23:40, 28 February 2024 (UTC)

So when I see 9,000 feet on Google that won't help too often vs 500 m? We have 300 yards apart in KEWR slightly staggered and 250 yards in KSFO but those are compromises with limited land. I think KSFO might be the record for big planes in my country. LEMD looks pretty cool though I wonder why they didn't join them, Space Shuttle accidents and Gimli Gliders would have an easier land and maybe they could avoid a tire explosion or gear breakdown from trying to stop so hard. Sagittarian Milky Way (talk) 00:55, 29 February 2024 (UTC)

I suppose KEWR normally uses one runway for arrivals, one for departures, with no parallel arrivals. KSFO does parallel arrivals, but has some restrictions. The aircraft must fly side by side, so that neither can get into the wake of the other (wakes are angled in a crosswind) and the pilots must see the other plane before aligning with the runway, so they can only do parallel arrivals of planes with the same landing speed in good weather. My local airport (EHAM) has 3 parallel runways with 2.1–2.8km separation and 1.5–3km stagger, allowing uncoordinated parallel arrivals on centre and right (left is departures only, right arrivals only) even in poor weather. They didn't need all that separation for that, but the terminal had to fit in the east gap and the west gap was needed for noise reasons.

LEMD was extensively rebuilt about 20 years ago. It used to have 4 crossing runways, 3 of which were closed. The original 18/36 was almost in-line with the new 18L/36R, but was closed. I suppose interference of go-arounds with departures and available space (both land and noise) was the reason. A runway in-line with 14R/32L (the old runway left) to the NW would have led up to La Moraleja, a hill with some expensive housing. PiusImpavidus (talk) 11:41, 29 February 2024 (UTC)

Does Chicago's Midway Airport fit your description? ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:00, 29 February 2024 (UTC)

Does a given body's total rest energy, have any other meaning, besides the body's ability to totally vanish - becoming energetic radiation?

HOTmag (talk) 10:01, 28 February 2024 (UTC)

A "kilobar" gold bar, with a rest energy of some 90 PJ, has a market value of about €60,000 or $65,000, twice that of a goldbar with half that rest energy. Transformed into "usable energy", the market value of this energy would be much higher, but we do not have the technology for the necessary transformation. The main reason for physicists to use the concept is to avoid conversions in calculating the energy balance of particle interactions, being able to use eV units throughout.  --Lambiam 10:31, 28 February 2024 (UTC)

By a "meaning", I've meant: a physical meaning (rather than an economic one). HOTmag (talk) 10:47, 28 February 2024 (UTC)

The main reason to introduce the concept of rest energy (corresponding to the mass of the body) is that it is necessary to include it in the energy balance to satisfy conservation of energy. This is what enables the conversion into radiation, for example (in practice, other conservation laws may stand in the way of that process). In addition, it determines the inertia and the gravitation of the body (as mass, rather than rest energy). In addition, the internal energy (which is part of the rest energy in the case of composite systems) plays a role in the thermodynamics of the thing (where this is relevant). --Wrongfilter (talk) 10:43, 28 February 2024 (UTC)

Note I asked about total rest energy. I can understand how it's helpful in determining the value of energy of the radiation - the body has turned into, whereas the value of the body's mass is already known. However, does the body's total rest energy - also have any meaning - in determining the inertia or the gravitation (or anything else) of the body whose mass is already known?

To sum up: besides a Gedanken experiment in which a given body having a given value of mass - totally vanishes - becoming energetic radiation, can you think about any other Gedanken experiment, which involves a body having a given value of mass, and which shows how the body's total rest energy can be helpful? HOTmag (talk) 11:28, 28 February 2024 (UTC)

What makes you think that there's a difference between a body's total rest energy and its mass (times c²)? --Wrongfilter (talk) 12:00, 28 February 2024 (UTC)

After the formula E=m^2 was discovered, all of us know there's no difference between a total rest energy and a total mass (times c^2). However, before discovering the formula E=m^2, we couldn't predict the value of energy - of the radiation a given body turned into, even when we did know the value of total mass - the body had had - before it became energetic radiation. This fact proves, that the meaning of a given body's total rest energy, lies (for example) in the body's ability to totally vanish - becoming energetic radiation - whereas the value of the body's total mass is already known. That said, I still wonder if this is the only meaning of the total rest energy, whereas the value of the body's total mass is already known. HOTmag (talk) 12:24, 28 February 2024 (UTC)

The terms "rest energy" and "rest mass" are practically interchangeable, except that some physicists feel compelled to express the latter in units of MeV/c². Others have no such compunction and are happy to write, for example, "a K⁰ particle of a mass of about 650 Mev".^[4]  --Lambiam 16:05, 28 February 2024 (UTC)

Yes, the terms "rest energy" and "rest mass" are practically interchangeable, due to the discovery of the formula E=m^2. However, they are not conceptually interchangeable, as I explained in my previous response you've just responded to. HOTmag (talk) 18:01, 28 February 2024 (UTC)

You keep writing ${\textstyle E=m^{2}}$. The formula is ${\textstyle E=mc^{2}}$. The units don't make any sense the other way.

As Lambiam has pointed out, in some contexts, physicists will adopt a system of units where c = 1 (and sometimes h = 1, too), but you still don't square the value of the mass in that case.

I would argue that the terms "rest energy" and "rest mass" are wholly synonymous. The former indicates nothing other than, "what is the rest mass of this thing, expressed in units of energy?"

PianoDan (talk) 19:31, 28 February 2024 (UTC)

Note I'd written to Wrongfilter: "After the formula E=m^2 was discovered, all of us know there's no difference between a total rest energy and a total mass (times c^2)". So, you can figure out that E=m^2 was of course a repeated typo (in which I only kept omitting the "c" before the "2") in my reciting the best known scientific formula, which I was reciting (correctly) by heart - before I knew to count (and before I knew what it meant)...

As for your second remark: Well, it seems you didn't read all of my response (mentioned above) to Wrongfilter. If I'm wrong and you did read all of it, so let me put it this way: If Einstein thought like you that those terms were synonymous, he wouldn't feel he had to prove the equivalence between those terms (up to c^2). HOTmag (talk) 23:21, 28 February 2024 (UTC)

But he DID prove it, and it's been experimentally verified continuously since then. We don't have to behave as though that hasn't already been thoroughly established. PianoDan (talk) 23:26, 28 February 2024 (UTC)

Yes, he proved it. But this fact doesn't mean those terms are synonynous. Just as the terms "101+10" and "110+1" are not synonymous, even though it's provable that 101+10 = 110+1. Unless we disagree about what "synonymous" means. It seems that you're coming from a practical standpoint, whereas I'm coming from a conceptual standpoint. HOTmag (talk) 23:34, 28 February 2024 (UTC)

February 29

Gravitational potential energy

Our article gravitational energy, asserts in the first chapter:

For two pairwise interacting point particles, the gravitational potential energy $U$ is given by $U=-{\frac {GMm}{R}},$

without making any distinction between Classical mechanics and Relativistic mechanics.

I wonder if this definition leads to a vicious circle in Relativistic mechanics, since: According to the formula $E=mc^{2},$ the whole mass $m$ of a given body already contains also the body's gravitational potential energy $U,$ defined above by the whole mass $m,$ whose content contains also $U...,$ so isn't it a vicious circle?

Apparently, it's like when someone mentioning A - who is asked "what do you mean by A" - defines A by B, but when asked "what do you mean by B" - they define B by A. In our case, the whole mass $m$ depends (according to the formula $E=mc^{2})$ also on the energy $U,$ depending (according to the formula $U=-{\frac {GMm}{R}}$) also on $m,$ right? Unless I miss something here.

However, if our article (in the first chapter) did have to make a distinction between Classical mechanics and Relativistic mechanics, then how does the latter avoid the vicious circle mentioned above? HOTmag (talk) 08:42, 29 February 2024 (UTC)

See Mass in general relativity. Graeme Bartlett (talk) 10:55, 29 February 2024 (UTC)

Thank you for the link.

Having read it, let me put my original question this way:

1. Can General Relativity give a clear cut answer to any/both of the following questions?

A. Does a given body's gravitational potential energy contribute to the body's total mass?

B. Does a given body's gravitational potential energy depend on the body's total mass?

HOTmag (talk) 12:13, 29 February 2024 (UTC)

I think the answers are yes and yes. This is what makes General Relativity nonlinear and very hard. This is related to the question whether gravitational waves could exist in GR — Einstein showed that linear waves could exist in the weak-gravity limit (that's easy, and in this regime your questions are quantitatively irrelevant), but had doubts himself whether waves could actually be generated in the strong-gravity regime. This question was only settled in the 1960s. --Wrongfilter (talk) 12:17, 29 February 2024 (UTC)

If (as you say) the answers are yes and yes, i.e. a given body's gravitational potential energy, both contributes to the body's total mass, and also depends on the body's total mass, so it follows logically that the gravitational potential energy, both contributes to itself (it being a part of the total mass), and also depends on itself (for the same reason). Isn't it like Baron Münchhausen who saved himself from drowning by pulling up on his own hair? In other words, isn't it a vicious circle? All of that reminds me of the algebraic equation: x=x+1...

Indeed, sometimes defining an object by itself does not lead to any vicious circle, e.g. in the algebraic equation: 2x+1=x+2, but that's only because we can prove that this equation has a solution (actually a unique one). However, how can we be sure that the interdependece between the total mass and the gravitational potential energy leads to no contradiction, as opposed to the case of the interdependece of the sides of the equation x=x+1? In both cases, an object depends on itself !

This is my wonder from a logical viewpoint. But, let's put logic aside, and get back to physics. My practical question is: Can General relativity describe the gravitational energy, in such a way, that this gravitational energy will only depend on the mass's components other than its gravitational potential energy component? Just as we can do the same in the algebraic equation: 2x+1=x+2, by replacing it by a direct equation x=1, i.e. so that x will only depend on a pure number. For simplicity, let's focus on the two object case, assuming that the whole universe only contains two electrons alone (or any two uncharged objects alone). HOTmag (talk) 13:15, 29 February 2024 (UTC)

I've removed the equation from the title of this question, this seems to cause some problems and weird behaviour of the system. --Wrongfilter (talk) 12:19, 29 February 2024 (UTC)

Energy of one period for alternating current

In alternating current, as at the output of an amplifier for a dipole antenna, I cannot find a direct formula giving the energy over a period. Is it simply accepted that this value is given by dividing the power by the frequency? Or is there a direct formula and if so which one? Malypaet (talk) 12:10, 29 February 2024 (UTC)

Energy is power per unit time. So "energy over a period" is just power.

The formula for that isn't so much "accepted" as it is simply the definition of the terms. Energy defined to be power per unit time. (Power / unit time) * (time period) = power. Just basic dimensional analysis. PianoDan (talk) 15:32, 29 February 2024 (UTC)

I assume they mean "energy emitted over a period", i.e. power * (time period). For a monochromatic sinusoidal oscillation this would indeed be power / frequency. --Wrongfilter (talk) 15:39, 29 February 2024 (UTC)