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March 13

Lake Kakhovka

What's the situation at the former Lake Kakhovka from a "rewilding" point of view? All satellite images I've been able to find online are close to the moment of the destruction of the dam, but I was curious to know if trees and plants are starting to grow and in general how nature is recoverinf there. Thank :) --LametinoWiki (talk) 00:42, 13 March 2024 (UTC)

Given that the dam was destroyed less than a year ago, we may not have much significant regrowth on "tree" scale yet (especially given that a significant amount of that time period was winter). Additionally, this is a war zone or territory illegally occupied by Russia, so I doubt a ton of scientific teams are able to properly investigate the location. Both the war and occupation conditions could also be an impediment to significant regrowth. --OuroborosCobra (talk) 02:28, 13 March 2024 (UTC)

Two beams of light start moving perpendicular to the initial distance between them. Will it become shorter, because of any gravitational curvature caused by them in spacetime?

I'm asking from an empirical point of view, rather than from a theortical one (as predicted in General Relativity).

I was recently shocked to hear from a local astronomer, that beams of light frequently arriving from pulsars, are known to prove that the distance does not become shorter. As far as you know, is there any evidence that supports - or refutes - what I've heard from that astronomer, emprically speaking? HOTmag (talk) 11:04, 13 March 2024 (UTC)

That's not measurable. I don't know what your local astronomer was talking about but they may have thought of the pulsar timing array. --Wrongfilter (talk) 11:23, 13 March 2024 (UTC)

Thanks. Does the pulsar time array have anything to do with a (constant?) distance between two beams of light arriving from pulsars? HOTmag (talk) 11:43, 13 March 2024 (UTC)

Your astronomer friend is correct from the theoretical point of view though I don't know what they were referring to. Think of having an object going at near the speed of light alongside some light. The light would be deeply red shifted and have very little energy. NadVolum (talk) 12:52, 13 March 2024 (UTC)

As for your first point: Why was the astronomer correct from the theoretical point of view? Doesn't General Relativity predict both beams of light will curve spacetime because of their energy?

As for your second point: Please notice that according to General Relativity, the impact of a gravitational curvature on a given particle's momentum - does not depend on the particle's energy, whereas a gravitational curvature caused by a given particle - does depend on the particle's energy. That's why my current question is only being asked about the curvature (if any) caused by light, rather than about the impact of that curvature on light. HOTmag (talk) 15:29, 13 March 2024 (UTC)

I don't have the maths for that anywhere near my finger tips but it seems to me that such a prediction would lead to all sorts of problems. Intuitively I would guess that it's difficult to impossible to construct a non-zero curvature field that transforms covariantly. However, in another one of your threads somebody posted a link to a paper demonstrating some sort of attraction between light rays in the lab, I don't recall the details. Be that as it may, any effect is certainly much to small to be measurable in the presence of much larger perturbations that are ubiquitous and unavoidable on Galactic scales (this also precludes an empirical demonstration of a zero effect to any level that would satisfy you). --Wrongfilter (talk) 15:44, 13 March 2024 (UTC)

This is the link you've just mentioned. It mentions no "lab" nor any other empirical clues. All was "theoretically inverstigated" (as clearly stated in the last sentence of the abstarct). That's why my current question is being asked "from an empirical point of view".

As for the maths bothering you, please see our artice pp waves, indicating: The pp-waves solutions, model radiation moving at the speed of light. This radiation may consist of electromagnetic radiation, gravitational radiation, massless radiation...or any combination of these, so long as the radiation is all moving in the same direction....Penrose also pointed out that in a pp-wave spacetime, all the polynomial scalar invariants of the Riemann tensor vanish identically, yet the curvature is almost never zero. [They vanish]...because in four-dimension all pp-waves belong to the class of VSI spacetimes. However, this curvature "is almost never zero" because (as stated in our article about VSI) the solutions of the stress–energy tensor usually have also non-polynomial or non-scalar invariants, which actually do not vanish. HOTmag (talk) 15:59, 13 March 2024 (UTC)

"...light causes no acceleration to a co-propagating test ray." emphasis added. Thus co-propagating rays, like the light arriving from distant stars and pulsars, are unaffected. Modocc (talk) 16:17, 13 March 2024 (UTC)

Not in all cases. See again the last sentence of the abstract. Additionaly, please notice you are the person who presented this link for proving the photons "gravitate" (as you wrote). HOTmag (talk) 16:32, 13 March 2024 (UTC)

Yes not in all cases, nevertheless Wrongfilter and the astronomer are, in theory, both correct. Modocc (talk) 16:54, 13 March 2024 (UTC)

Please notice I asked user:NadVolum "why" the astronomer was correct. Doesn't General Relativity claim every object carrying energy curves spacetime? HOTmag (talk) 17:02, 13 March 2024 (UTC)

Yes and its gravity (GR's or any othr model's) depends on invariant mass. Co-moving photons appear to not have that with respect to each other. Modocc (talk) 17:34, 13 March 2024 (UTC)

Please notice, that General Relativity ascribes - the gravitational curvature - caused by a given particle, to the particle's energy, rather than to the particle's invariant mass. For more details about what properties of the particle curve spacetime, see Stress–energy tensor. HOTmag (talk) 17:44, 13 March 2024 (UTC)

Of course all particles gravitate (I've been consistent on this point) thus each photon contributes to the field (warpinging GR's spacetime). With a photon cloud their relative velocities also contribute to invariant masses which determine to what degree there are interactions between them if any. Modocc (talk) 17:59, 13 March 2024 (UTC)

Since "all particles gravitate" as you emphasize, was the astronomer correct?

I recall the astronomer has claimed - emprically [factually] speaking, that the initial distance between two parallel beams of light arriving from pulsars, does not become shorter, even after finishing the long journey to Earth. According to that apparently empirical [factual] claim, those beams of light don't curve spacetime, do they? HOTmag (talk) 18:13, 13 March 2024 (UTC)

Your astronomer cannot measure that. --Wrongfilter (talk) 18:30, 13 March 2024 (UTC)

I'm not arguing about whether the astronomer can measure it. Yet, the question still remaing is whether the astronomer could've been correct, had the astronomer been able to meausre it. Apparently, what the astronomer claims, from an apparently empirical [factual] point of view, contradicts General relativity, which claims that any given particle having non-zero energy (hence a non-zero stress–energy tensor) is supposed to curve spacetime. I thought, the astronomer's apparently empirical [factual] claim, really contradicted General Relativity, but I wanted to be sure, and that's why I posted this thread. HOTmag (talk) 18:58, 13 March 2024 (UTC)

Apparently you have no idea what "empirical" means. --Wrongfilter (talk) 19:14, 13 March 2024 (UTC)

Sorry for replacing "factual" by "empirical". Anyway, as I have already stated, "I'm not arguing about whether the astronomer can measure it". Whether because I "have no idea what 'empirical' means" (as you claim), or because - even if I do know what it means (me meaning "factual" rather than "empirical") - I don't want to ask about whether the astronomer can measure it, but rather about whether the astronomer could've been correct - had the astronomer been able to meausre it. I thought, the astronomer's claim contradicted General Relativity, but I wanted to be sure, and that's why I posted this thread. HOTmag (talk) 19:26, 13 March 2024 (UTC)

(ec Stop making all these microedits) Maybe the astronomer just wanted to end the conversation... You wanted empirical evidence one way or the other, now don't change that to "factual" (I don't even know what that's supposed to mean). I say the effect would be unobservable, even if there was an effect (I don't know - there are lots of subtleties in GR if one dives in deep), hence the question is empirically undecidable. But you will never be sure, no matter what anyone here says. --Wrongfilter (talk) 19:46, 13 March 2024 (UTC)

Just to be clear:

1. By mistake, I replaced "factual" by "empirical" when I referred to the astronomer's claim about the beams of light arriving from pulsars.

2. However, my original question was about any empirical evidence, any one of you knew of. Yes, I used the word "empirical" in my original question, on purpose, meaning empirical, because I do know what it means.

3. Before I read your response about the immeasurability, I posted this thread asking whether any one of you - knew of any empirical evidence - which could support or refute the astronomer's factual claim about the pulsars, because I wondered whether the astronomer could be correct - assuming the astronomer's factual claim about the pulsars could be resolved by measurement. I thought, this astronomer's factual claim contradicted General Relativity, but I wanted to be sure, and that's why I posted this thread asking about any empirical evidence you knew of - which could support or refute the astronomer's factual claim about the pulsars.

4. In your response to my original question, you've claimed all of that can't be measured. As I have already stated, I'm not going to argue about that. However, I still wonder whether the astronomer could have been correct - had the astronomer's factual claim about the pulsars been resolved by measurement.

Am I clearer now? HOTmag (talk) 20:43, 13 March 2024 (UTC)

The starlight is co-moving like racers in a dead heat so according to the citation the photons gravitate, but not each other. Modocc (talk) 18:27, 13 March 2024 (UTC)

Since "the photons gravitate" (me using your words), so they curve spacetime (don't they?), so how can they avoid gravitating "each other" (me still using your words), as long as the spacetime is being curved by them? HOTmag (talk) 18:58, 13 March 2024 (UTC)

Why they (the comoving photons) don't gravitate each other according to the research paper? Perhaps someone else better versed in GR can answer that question, but co-moving photons do not appear to have invariant mass with respect to each other (although their energies obviously contribute to the invariant masses of larger systems). Note that in reality no photon within our universe exists in isolation from it so the notion that they all contribute mass-energy to the field holds. Modocc (talk) 20:13, 13 March 2024 (UTC)

Plesse notice that also massless photons are influenced by the curvature of spacetime, that actually deflects their trajectory, e.g. when they approach the sun, so the question still remains: Why is neither of them influenced by this curvature when it's caused by the other photon, while you agree "they gravitate" - hence curve the spacetime. HOTmag (talk) 20:51, 13 March 2024 (UTC)

Their path's curvature (or spacetime's curvature) are dependent on the sun's mass. As for their affecting each other, that, will depend, according to the paper I cited, on their relative velocities which are vectors. As for measurement, I assume the strength of their combined field is proportional to their invariant mass and inversely proportional to the distance between them squared: in accordance with other masses, and subject to the same caveats (Newtonian approximation, quantum limits, MOND, etc). Comoving photons have zero invariant mass, but also add mass to larger bodies. Modocc (talk) 23:07, 13 March 2024 (UTC)

Two photons move in parallel paths. There is an electron between them. Do the photons curve spacetime (in addition to the curvature caused by the electron)? If the answer is positive, then do they gravitate each other (in addition to the gravitation caused by the electron)? HOTmag (talk) 01:35, 14 March 2024 (UTC)

At infinity the photons don't gravitate each other, according to the citation, but the electron is a gravitational lens causing their convergence: thus the photons' convergence should have an invariant mass that gravitates them, also in accordance with the citation. Modocc (talk) 02:01, 14 March 2024 (UTC)

What do you mean by "the photons' convergence should have an invariant mass that gravitates them"? I remind, we are talking about two photons co-moving in the same direction, so how can the photons' convergence have an invariant mass? HOTmag (talk) 08:23, 15 March 2024 (UTC)

Convergent rays carry energy and momenta towards a distant focal point (with a component velocity less than c). They are no longer propagating parallel to each other. Modocc (talk) 12:04, 15 March 2024 (UTC)

Can you think of any theoretical situation in which the invariant masses of two given photons will be zero, assuming that the whole world contains massive matter as well? HOTmag (talk) 13:50, 15 March 2024 (UTC)

As long as photons are moving along similar parallel geodesics they have energy, but, of course, it is not invariant. Invariant mass: "The invariant mass, rest mass, intrinsic mass, proper mass, or in the case of bound systems simply mass, is the portion of the total mass of an object or system of objects that is independent of the overall motion of the system." Modocc (talk) 17:37, 15 March 2024 (UTC)

Photons have energy and therefore, according to the Einstein field equations, curve spacetime just like other entities with energy. The predicted effect of two photons on each other is too small to be measured.  --Lambiam 21:06, 13 March 2024 (UTC)

You've responded to a response of mine that was, both addressed to user:Modocc only, and assuming user:Modocc's assumptions only. Do you think you can also respond to my original question under the title of this thread? HOTmag (talk) 22:02, 13 March 2024 (UTC)

You have restored a contribution of mine that I had removed. That is a big no-no. When I originally posted it, I had overlooked the fact that the question had been changed: in the OP they were moving in orthogonal directions, not parallel. The latter is much more tricky, one immediate issue already being that the notion of geodesics being parallel in a non-flat continuum is ill-defined.  --Lambiam 10:37, 14 March 2024 (UTC)

Wow, I'm quite surprised now. Of course I was not aware of the restoration made by me. I still wonder how it was actualy done, maybe because I had not received a warning about an ongoing edit conflict? Anyway it was not done on purpose. I guess you forgive me for this big mistake, for which I apologize from the bottom of my heart, don't you?

As for your second remark, I'm surprised again: What do you mean by "the question had been changed: in the OP they were moving in orthogonal directions, not parallel". As far as I remember: from the very beginning, the question indicated in the title (that has never changed) has always been formulated as follows: Two beams of light start moving perpendicular to the initial distance between them. Will it become shorter, because of any gravitational curvature caused by them in spacetime? In other words, from the very beginning the question has always been about two beams of light moving in parallel directions, hasn't it? HOTmag (talk) 08:09, 15 March 2024 (UTC)

Suppose one photon's trajectory is given by its position as a function of time as $(ct,0,0)$ while the other has $(0,ct,\delta ).$ Then, when $t=0,$ each moves perpendicular to the line segment from $(0,0,0)$ to $(0,0,\delta )$ connecting them, and in orthogonal directions with respect to each other. I now think that also in this case they would not gravitationally attract each other. It is difficult to find reliable sources handling the issue, as any effect would be too small to measure, but nevertheless I found this: "This conclusion explains ... why photons do not attract each other during the long trip from the distant galaxies and collapse into a single clump, despite of the fact that photons have gravitational mass." ^[1]  --Lambiam 19:00, 15 March 2024 (UTC)

Careful: that quote is from a paper published in Physics Essays, which looks like a rather dodgy journal. --Wrongfilter (talk) 19:19, 15 March 2024 (UTC)

I was trying to make it simple. There is no reason to invoke General Relativity, Special Relativity is quite enough to see why the rays don't attract each other. If two particles have the same direction and speed as each other then they are at rest relative to each other and only their rest mass counts for the gravitationl attraction between them - even if they are going at near the speed of light relative to us and have a huge mass relative to us. Now think of a photon as the limit of a particle getting a smaller and smaller rest mass but going faster and faster so the relativistic mass stays the same. in the limit we have a particle with zero rest mass but the same energy. Now two of those going along together won't attract each other because their rest mass is zero. NadVolum (talk) 20:57, 13 March 2024 (UTC)

Following your considerations, let's look at the whole picture - taking into account new considerations you didn't take into account, after making a distinction - as you've made - between two cases: A pair of massive particles, and a pair of photons.

1. Two massive particles start moving, in the same direction, perpendicular to the initial distance between them. In their reference frame, the initial distance between them will soon become shorter, because of the gravitational curvature caused by their rest mass (as you have indicated). However [and this is my new consideration you didn't take into account], the distance will become much shorter, in any other reference frame, because the curvature is bigger in that reference frame, because the curvature in that other reference frame is also influenced by their kinetic energy (being their "relativistic mass" in that other reference frame, provided that we allow to use this concept, and I allow, and I assume you allow as well, but even if we don't allow we can still adhere to their kinetic energy instead).

2. Two photons start moving, in the same direction, perpendicular to the initial distance between them. Had they had a reference frame in which they could have been at rest, then: In their reference frame the initial distance between them would have not changed, because the gravitational curvature caused by their zero rest mass - would have been zero as well - as you've indicated. However [and this is my new consideration you didn't take into account], the distance will become shorter, in any other reference frame, because the curvature is bigger in that other reference frame, because the curvature in that other reference frame is also influenced by their kinetic energy (being their "relativistic mass" in that other reference frame, provided that we allow to use this concept, and I allow, and I assume you allow as well, but even if we don't allow we can still adhere to their kinetic energy instead).

HOTmag (talk) 22:02, 13 March 2024 (UTC)

I'm having difficulty with what you said but apparent contraction of lengths only happens in the direction of motion, not at right angles to it. A super speed locomotive may seem shorter but the wheels will remain the same distance apart and stay on the track. Also while going any particular distance the particles will experience less time and a shorter distance than an observer and therefore will not approach each other even the amount one might assume from Newtonian gravity on their rest mass never mind with the extra energy. NadVolum (talk) 00:17, 14 March 2024 (UTC)

I was not referring to the Special relativistic effect of length contraction due to velocity, but rather to the General relativistic effect of difelcting a given photon's trajectory when the photon is in a gravitational field, e.g. when it moves near the sun, so the distance between the photon and the sun becomes shorter because of the effect of gravitational lensing, regardless of the Special relativistic effect of length contraction due to velocity.

That said, do you agree that when an electron and a planet start moving perpendicular to the initial distance between them, the direction will soon become shorter, according to the following three principles:

1. The faster the reference frame moves relative to the planet, the bigger kinetic energy the planet has (hence the bigger relativistic mass the planet has).

2. The bigger kinetic energy (hence relativistic mass) the planet has [in a given reference frame], the shorter the distance between the electron and the planet will become in that reference frame.

Logically, the following principle follows from the combination of the previous ones:

3. The faster the reference frame moves relative to the planet, the shorter the distance between the electron and the planet will become in that reference frame.

If you agree, then try to read again my previous response, but now you should interpret the effect as a General relativistic one, due to gravitational lensing effect that obeys the third principle mentioned above (rather than as a Special relativistic effect of length contraction). HOTmag (talk) 01:22, 14 March 2024 (UTC)

Changing reference frames, in general, does not, I repeat, does not change physical interactions. For example, did you know that the Earth has arbitrary KE! Pick any train, rocket, plane or the like, the Earth has different KE due to the velocity of your chosen reference frame. Instead, g depends simply on the reference frame independent invariant mass of the gravitating body. That said, accelerations of objects indeed changes interactions: such as how Doppler is measured. In other words, physically changing one's reference frame (as opposed to only applying a Galilean transformation) is not trivial. For instance, surfers that ride the crests of ocean waves and then glide off them physically change their reference frames and the frequency of wave interactions increases. Hence, both kinds of change happen at the same time. Modocc (talk) 02:30, 14 March 2024 (UTC)

AFAIK, in General Relativity, gravity is considered to be caused by curvature in spacetime, while this curvature is caused by the stress–energy tensor, while this tensor is defined by the relativistic energy E (and the momentum p along with some other components relating to these energy and momemtum) - of the entity (e.g. a planet) responsible for the gravity, while this relativistic energy E is equivalent to the relativistic mass E/C², rather than to the invariant mass E/C² devided by the gamma factor. HOTmag (talk) 07:52, 15 March 2024 (UTC)

Relativistic masses each contribute to the invariant masses of larger bodies, which in turn, gravitate in accordance with their invariant masses. So yes they gravitate. :-) Modocc (talk) 13:15, 15 March 2024 (UTC)

Please notice the current discussion is not about whether they gravitate but rather about your previous response in which you've emphasized: "Changing reference frames, in general, does not, I repeat, does not change physical interactions... g depends simply on the reference frame independent invariant mass of the gravitating body". In this previous response, you responded to a response of mine to user:NadVolum, in which I claimed "The bigger kinetic energy (hence relativistic mass) the planet has [in a given reference frame], the shorter the distance between the electron and the planet will become in that reference frame", so it seems your previous response meant to claim that gravity did not depend on relativistic masses. However, if you're claiming now that it does depend on them, then we agree, as far as our current discussion (about my response to user:NadVolum and about your previous response to my response) is concerned. HOTmag (talk) 13:50, 15 March 2024 (UTC)

I was careful to provide enough context with each of my points to avoid contradiction. Note that I said relativistic masses contribute to proper masses which constrains the problem of determining what curvature(s) are present within and outside the bodies under consideration. My first point was that unlike picking reference frames, any curvature present due to g is not arbitrary, and my second point was that for any fixed reference frame differences in invariant masses correspond to differences in curvature so yes we have to apply relativistic mass when calculating them. Modocc (talk) 21:59, 15 March 2024 (UTC)

There will be a gravitational wave associated with the two photons, it is generated when they are emitted. Like a single wave on water it has no effect if you're going along with it like a surfer but if you go across it has a larger effect the faster you cross it. NadVolum (talk) 13:23, 15 March 2024 (UTC)

Considering the indent of your current response, are you reponding now to my respose to user:Modocc? HOTmag (talk) 13:50, 15 March 2024 (UTC)

Irrelevant to what I said. You were not satisfied with a simple explanation using Special Relativity that fully explains what happens. As far as I can make out you want a General Relativity solution about why others would feel a gravitational attraction to the photons but the photons wouldn't feel any to each other. My description using a surfer is just descriptive, a fuller explanation would need you to read the papers and delve through the maths of General Relativity I believe. NadVolum (talk) 15:50, 15 March 2024 (UTC)

Hump

In a freight yard, is there any advantage to putting three lead tracks on the hump, as opposed to the more usual multiples of two? 2601:646:8082:BA0:8014:354A:6A03:A76 (talk) 13:34, 13 March 2024 (UTC)

I don't see anything about what you are saying at that article. Have you something describing what you are talking about? NadVolum (talk) 21:12, 13 March 2024 (UTC)

If you look at the photos in the article Classification yard, the hump in these photos always has two leads, never three -- my question is, is there any particular reason why this is the case? 2601:646:8082:BA0:8014:354A:6A03:A76 (talk) 02:13, 14 March 2024 (UTC)

I don't know how to see which tracks are leads, but in any case, a larger number requires a more extensive hump. If the hump is a constructed hill, as suggested in Rail yard § Freight yards, the extra cost and complexity may not be outweighed by an operational advantage.  --Lambiam 10:19, 14 March 2024 (UTC)

I looked at several examples of hump yards and found three layouts for the leads:

• A single lead going up the hump, after the downhill splitting multiple times into the bowl. For example in Cincinnati, Kansas City, Houston (all US), Odesa (Ukraine). This appears to be the dominant layout in the US.
• Two leads, merging into one on the downhill before splitting multiple times into the bowl. For example in Vienna (Austria), Hamburg, Mannheim (south side) (both Germany), Riga (Latvia), Antwerp (Belgium). This appears to be the dominant layout in Western Europe.
• Two leads with a scissors crossover on the downhill before splitting multiple times into the bowl. For example in Metz (France), Rotterdam (Netherlands), Mannheim (north side) (Germany), Shanghai (China).

All yards named after the nearby large city, not the name of the yard itself.

Now keep in mind the difference between Western European couplings and those used in other areas. Western Europe still uses screw couplings. Before humping a series of wagons, the train is first pushed slightly onto the hump, to compress the couplings. Then the couplings are loosened a bit, so that they can be unhooked. Then the wagons are pushed over the hump, whilst a worker uses a club to knock the link of the coupling off the hook, without stopping the train or getting between the wagons. This isn't needed elsewhere, as trains there use claw couplings. So having two leads is useful in Western Europe, as one train is pushed over the hump whilst the next is prepared.

Having a scissors crossover instead of a merge followed by a split allows sorting two trains at the same time, each into half the bowl tracks. In Metz, one can sort one train onto all 48 tracks or two trains onto 24 tracks each. You gain some flexibility for small jobs, at the cost of two sets of points and a diamond. I suppose that small jobs are more common in Europe, with shorter and more frequent trains, than in the US.

Three or more lead tracks gives diminishing returns, whilst rapidly making the required pointwork more complex. PiusImpavidus (talk) 11:02, 14 March 2024 (UTC)

Thanks! So, just as I thought -- adding a third lead gives you more than 50% higher construction costs (as well as operational costs, and per-capita aspirin consumption by the switchmen) in exchange for less than 50% higher throughput (and it gets worse from that point on)! (Incidentally, the only hump I've ever seen with three leads was a fictional one at Knapford Yards -- or was it Tidmouth Yards? -- in Thomas & Friends (one of the old classic episodes, "Old Iron" "Pop Goes the Diesel" I think) -- and I have to remark, although these yards as a whole look impressive on the screen, they are actually laid out pretty poorly, and the hump seems to have been added pretty much as an afterthought!) 2601:646:8082:BA0:0:0:0:2EB (talk) 13:55, 14 March 2024 (UTC)

And two lead tracks with a scissors crossover gives redundancy. Any of the sets of points can be taken out of service for maintenance, still allowing the yard to operate at half capacity. PiusImpavidus (talk) 10:48, 16 March 2024 (UTC)

Would some like to point out where the leads are in this diagram just so I know. It never occurred to me before that you need a sorting algorithm to build trains despite it being obvious and despite having probably gone past these kind of places numerous times. Sean.hoyland (talk) 11:19, 14 March 2024 (UTC)

It's the track from the approach yard to the classification yard, leading up to the hump. The hump is located just before the first retarder. PiusImpavidus (talk) 11:48, 14 March 2024 (UTC)

Thanks. So the red rectangles are retarders I guess. Sean.hoyland (talk) 11:54, 14 March 2024 (UTC)

Just saw the legend which says in clear large font Retarders...Sean.hoyland (talk) 11:55, 14 March 2024 (UTC)

arrest, dearrest / de-arrest, arrested

What's the meaning of "arrest and dearrest" on page 684 of https://archive.org/details/proteintransfero0000unse/page/684/mode/2up?q=dearrest and "de-arrest of arrested" on page 3 of https://archive.org/details/arxiv-1206.2024/page/n1/mode/2up?q=%22de-arrest%22? Mcljlm (talk) 21:42, 13 March 2024 (UTC)

The verb to arrest means, "to keep (something) from moving". What is kept in arrest in these cases are molecules, in the first case specifically proteins; in the second case, those of which a glass is comprised. The verb to dearrest means, "to release (something) from an existing arrest". The form de-arrest is a spelling variant.  --Lambiam 10:06, 14 March 2024 (UTC)

Thanks Lambiam. Until a few days ago I'd not come across dearrest/de-arrest at all {originally I was only intending to ask about that word}, and then only relating to people. Later I found it used relating to ships and goods in historical documents (the OED doesn't mention any use later than 1791). Is its use recent and common in a scientific context? Is the hyphen's use more or less frequent in scientific publications? Mcljlm (talk) 16:49, 14 March 2024 (UTC)

Regarding ships and boats: arresting a boat (in a legal sense) was certainly a thing up to the 1990s, and I would guess still is. In the early '90's my father, who worked at a legal firm, passed on to me an obsolete word processor (an IBM Displaywriter System) that had just been replaced at his firm; on the data disk were several form documents, including one for arresting a boat. 51.198.186.221 (talk) 00:43, 15 March 2024 (UTC)

It's dearrest/de-arrest which the OED doesn't have any mention for later than 1791.. Could it be that another word was used? Perhaps I should ask at the Humanities and/or Language reference desks. Mcljlm (talk) 04:52, 15 March 2024 (UTC)

The 1791 use actually uses the unhyphenated spelling dearrest. Here is a use of de-arrest, with a hyphen, from 1898.  --Lambiam 17:46, 15 March 2024 (UTC)

Since it's an edition of a 1334 document it would be interesting to know if the hyphen is in the original and if not why it is in this edition. Mcljlm (talk) 10:05, 16 March 2024 (UTC)

Until 1905, the so-called atomic theory – the notion that matter was composed of molecules which in turn were composed of elemental atoms – was a hypothetical concept. There was in fact considerable opposition to atomic theory. This only changed with the publication of Einstein's article on Brownian motion. So most of the use of the term in this scientific sense of a molecule being arrested is indeed relatively recent. Proponents of the atomic theory did use the term arrest, though, already in the 19th century (e.g, in 1874, "the heat of arrested motion"^[2]). Nevertheless, this is not a term of art but the use of a word in an accepted common sense applied to a specific context. The term dearrest is rather rare, and it is hard to tell which spelling variant is more common.  --Lambiam 20:58, 14 March 2024 (UTC)

Explained by the BBC (which appears to uses 'de-arrest' consistently): Who, what, why: What does it mean to be de-arrested? . -- Verbarson  ^talk_edits 13:49, 15 March 2024 (UTC)

I became aware of the use of de-arrest/dearrest relating to contemporary police action about a week ago Verbarson when someone quoted from a Metropolitan Police statement. I then noticed its use in the other contexts I mentioned. Mcljlm (talk) 15:38, 15 March 2024 (UTC)

Having been taken into custody many times, I can confirm that "I'm de-arresting you" is one of the sweetest things a police officer can say, although not having ACAB tattooed on my knuckles probably helps. MinorProphet (talk) 00:39, 26 March 2024 (UTC)

References

March 15

Can't remember the name of this organic chemistry phenomenon where conjugation "extends" a functional group

I think I remember reading a Wikipedia article about this, but I can't find the name of this phenomenon at all. Pinging @Smokefoot: maybe you might know since you are an active editor of organic chemistry articles. Here I have drawn the concept:

This is the phenomenon in which a double bond "extends" a functional group by conjugation, so that it behaves similarly to the parent functional group. The bottom compound, an "extended" carboxylic acid, has similar chemistry as the top carboxylic acid, except the two parts of the functionality (C=O and -OH) are separated by an additional double bond. The reason for this similarity is because conjugation can transfer charges by resonance, e.g. the bottom compound is acidic like a carboxylic acid because the negative charge on the deprotonated -OH oxygen can move to the carbonyl oxygen by resonance. Michael7604 (talk) 05:50, 15 March 2024 (UTC)

See or search dicarbonyl or keto-enol equilibrium. @Michael D. Turnbull and DMacks:. --Smokefoot (talk) 14:37, 15 March 2024 (UTC)

@Michael7604 You already had the key concept in your title: see conjugated system and the articles linked from it. My favourite related topic is, of course, the Michael addition reaction. Mike Turnbull (talk) 16:28, 15 March 2024 (UTC)

A vinylogous group. DMacks (talk) 18:26, 15 March 2024 (UTC)

This is it, thank you Michael7604 (talk) 21:15, 15 March 2024 (UTC)

Fictitious force. Is the opposite phenomenon called a: "Real" force? "Physical" force? "External" force? "Natural" force?

I'm looking for the most useful term, intended to exclude fictitious forces. HOTmag (talk) 08:38, 15 March 2024 (UTC)

Each of the suggested terms is usable if you take away the ornamentation of Scare quotes. Philvoids (talk) 10:31, 15 March 2024 (UTC)

They should only be considered to be quotation marks. That's because I was looking for "the most useful term" (Btw now I'm quoting myself), i.e. a term that could be quoted from sources using useful terminology. HOTmag (talk) 10:45, 15 March 2024 (UTC)

A commonly used contrasting term is true force.^[3][4][5]  --Lambiam 17:25, 15 March 2024 (UTC)

Meh. Yes, true force has some popularity, sometimes even in quotes, but don't expect everybody to understand you when you mention true forces. And despite Newton's ideas on absolute space and preferred reference frames that only have constant velocity relative to it, in real life we deal with the equivalence principle, under which those fictitious forces are as real as gravity. Which is quite real in classical mechanics. In fact, they are gravity. In a free-falling lift, we just declare gravity zero. On a merry-go-round, we just declare the centrifugal force and Coriolis force part of gravity. Good luck describing orbital manoeuvring or accretion in compact binaries in an inertial frame. General Relativity (I think you'll love that) even makes it its basic principle. Turns out that the entire universe spinning around our stationary merry-go-round causes frame dragging, which exactly matches those fictitious forces. PiusImpavidus (talk) 19:57, 15 March 2024 (UTC)

That was Einstein's original view, but I think the matter has become somewhat controversial. We have an article on Mach's principle. --Trovatore (talk) 21:06, 15 March 2024 (UTC)

Just as a slight extension, in dynamics we invent a fictitious force m*a, called a D'Alembert force, and can then treat it as a statics problem. It isn't always useful. Greglocock (talk) 23:36, 15 March 2024 (UTC)

March 16

Weld; Syren

This 1878 report makes a passing reference to "Weld's Sound Experiment". What was that, and who was Weld?

It also mentiones "Syren and Galton's Whistle"; the latter is Francis Galton, but who was Syren? Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 21:28, 16 March 2024 (UTC)

My guess is a siren whistle. --Wrongfilter (talk) 21:49, 16 March 2024 (UTC)

Galton's whistle is similar to a boy scout's whistle . It differs in that it has a piston provided with a screw and a milled head... (it's an ultrasonic dog whistle).

From Oscillations and Waves p. 297. Alansplodge (talk) 21:56, 16 March 2024 (UTC)

The most promising Weld is Alfred Weld, who was a bit of a scientist, but I fail to find any indication that he may have dealt with sound. --Wrongfilter (talk) 22:02, 16 March 2024 (UTC)

And but nay. In duo together with Ernst Chladni, Alfred will make too much of a youngster, for those classicals. --Askedonty (talk) 01:33, 17 March 2024 (UTC)

What classicals? This is about the first annual meeting by the Midland Union of Natural History Societies. Also, they had no scruples looking at a toddler like Galton's whistle, which had been invented a mere two years before. But don't hesitate to suggest another Weld if you've got one. --Wrongfilter (talk) 07:29, 17 March 2024 (UTC)

Wiktionary:syren#Noun: "Obsolete form of siren". Alansplodge (talk) 22:03, 16 March 2024 (UTC)

For the meaning of "Galton's Whistle", see Dog whistle.  --Lambiam 00:53, 17 March 2024 (UTC)

Isaac Weld.^[6]  --Lambiam 10:24, 17 March 2024 (UTC)

Resolved

 – Thank you. all. Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 10:50, 17 March 2024 (UTC)

Length of a photon

A couple of questions about photons.

Firstly, am I right to feel slightly uncomfortable about the following passage? From the article photon:

However, experiments confirm that the photon is not a short pulse of electromagnetic radiation; a photon's Maxwell waves will diffract, but photon energy does not spread out as it propagates, nor does this energy divide when it encounters a beam splitter. Rather, the received photon acts like a point-like particle since it is absorbed or emitted as a whole by arbitrarily small systems, including systems much smaller than its wavelength, such as an atomic nucleus (≈10⁻¹⁵ m across) or even the point-like electron.

It's trying, I think, to make a point that a quantum-mechanical wavefunction is different from a classical wave, because it can suffer wavefunction collapse; and because it always represents a single whole particle (and also that a photon is a point particle -- which I understand to mean that its overall (extended) wavepacket can be evolved by integrating it forward via point to point propagators).

But am I right to feel uncomfortable with the argument about it not being distributed because it can interact with objects that have a very small scale? I seem to remember that in the classical treatment of radio waves and antennas, an antenna might be essentially just a single vertical pole; but may interact with a radio wave that may have a wavelength of hundreds of metres, if the antenna is part of a circuit that has the right resonant frequency. If I remember correctly, the radio wave induces a current in the circuit, which in turn can be thought of as exciting an induced radio wave of its own that has an opposite phase in the far field, thus effectively partially cancelling the original wave; and so, despite what may be its small physical size, the antenna can have an effective area ("aperture") that can be very considerably larger. (Did I remember that right? It's a long time ago since those lectures on EM ...)

Now I'm not saying that that is directly equivalent to quantum measurement, because measuring the location of a photon absorbs all of it, not just part of it (give or take whatever discussion we might have about decoherence...) But am I right that the antenna does maybe suggest that we should perhaps be just a little more careful before making a blanket statement that an object cannot be distributed if it can interact with a system much smaller than its own wavelength?

The remark that "the photon energy does not spread out as it propagates" also makes me feel a bit uncomfortable. If one thinks of gravitational lensing, surely it's reasonable to imagine a photon travelling both sides of the gravity well while still retaining self-coherence, with a corresponding energy flow associated with both sides?

Beyond that, the second thing I wanted to ask about (which was what led me to look at the photon article in the first place, to see if it anywhere touched on the subject) was this: I seem to remember, way back at school level, being told that a typical wavepacket of a photon of visible light (eg perhaps from a sodium vapour lamp) had a longitudinal length of the order of about a metre (maybe give or take an order of magnitude either way). Does that seem plausible / viable / about right ?

One constraint would seem to be the time <-> frequency uncertainty relation. The overall linewidth of the source would seem to set a limit to the maximum range of frequencies that could be associated with such a photon, and therefore a minimum to the length of time (and so therefore spatial length) that the wavepacket must persist. Presumably there is also a maximum length of time one pictures the emission event taking -- perhaps even measurable by trying to measure the coherence time of such a source using an interferometer at lower and lower light levels ? Is this sort of thinking at all on the right track?

Thanks in advance, Jheald (talk) 22:05, 16 March 2024 (UTC)

The following, unfortunately unsourced, passage is copied from Point particle § In quantum mechanics:

Nevertheless, there is good reason that an elementary particle is often called a point particle. Even if an elementary particle has a delocalized wavepacket, the wavepacket can be represented as a quantum superposition of quantum states wherein the particle is exactly localized. Moreover, the interactions of the particle can be represented as a superposition of interactions of individual states which are localized.

Does this help to ease your uncomfortability?  --Lambiam 00:48, 17 March 2024 (UTC)

Thanks @Lambiam:. No problem with what you've quoted, so far as it goes.

But I do think it can be useful (and meaningful) to try to give a sense of what the whole overall superposition - the whole wavefunction - may look like, when this is possible (for a typical environment or observational set-up).

So eg for an electron, it's a point particle, but it's also useful to be able to talk about the whole atomic or molecular orbital that may be a good representation of the whole wavefunction in particular circumstances.

And similarly, it seems to me, for a photon, it's useful to have an idea of what a whole photon wavefunction might be like in a particular circumstance; and in respect of an interaction, what shape the whole overall superposition of point interactions (as referred to in your quote) might take, to represent eg a whole molecular orbital interacting in typical way with the entirety of a photon wavepacket, and what basic picture that may give for a complete scattering (elastic or inelastic) or emission event. That's something I'd quite appreciate some input on, at least in respect of the second part of my question.

In respect of the first part of my question, while I feel comfortable with what you've just quoted from point particle, it doesn't seem to me to address the things I suggested gave me a bit of discomfort with the bit of text from photon. Do you think it should make me feel more comfortable with that text? Thanks, Jheald (talk) 14:00, 17 March 2024 (UTC)

I think you're quite right not to feel comfortable about its wavefunction. It is possible to get pictures that represent it. But thinking a particle was actually at any of the point when it has not been observed to be there - well ... have a look at Quantum Cheshire cat and marvel at its ... well whatever. NadVolum (talk) 15:51, 18 March 2024 (UTC)

• Just adding a holding message, to stop the thread being auto-archived while I continue to try to read into it. I do think it would be useful to try to establish where mainstream specialist thought is on this, as the variety of answers one can find on generalist sites like Quora, StackExchange, Reddit, PhysicsForums, etc and even preprints are all over the place. Even here on Wiki Reference Deska couple of years ago essentially the same question got shut down as not meaningful or not based on a proper understanding of the physics. (Pinging @Malypaet, PiusImpavidus, Jayron32, Nimur, and Rmhermen: from that discussion, if they'd like to come in).

Yet, despite the issues with setups like the Quantum Cheshire cat that User:NadVolum points to (and which I need to think some more about), or our article Wave packet bluntly making the statement that 'Physicists have concluded that "wave packets would not do as representations of subatomic particles"' without further explanation, it seems to me that a wave packet or evolving wavefunction is often a useful physical picture for a particle, at least until environmental interactions make considerations of decoherence or measurement unavoidable. They help us picture, for example, how single photons can exhibit interference effects from their own reflections, or between paths with different flight times in beam-splitter experiments. I need to read up more on Quantum Optical Coherence (eg [7]) as distinct from classical coherence, but it does seem to me that the coherence length of a source of incoherent photons probably can be reasonably pictured as the length of a photon, eg the 67mm for a low-pressure sodium vapour lamp given at Coherence_length#Other_light_sources, extending to six times this for such lamps cooled to liquid nitrogen temperatures.

Our article atomic electron transition seems once to focus on how short the time for atomic transitions can be (given very heavy environmental 'measurement' / decoherence) -- still a bit longer than zero, because (as I understand it) even then the electron wavefunction needs time to change without moving faster than the speed of light -- but little in that article that I could see as to how long in time the electron transition might be considered to take, in the absence of externally-induced decoherence, with measurement setups designed to maintain rather than destroy coherence. Perhaps there should be more of an investigation of this there ? Jheald (talk) 13:28, 24 March 2024 (UTC)

How long the transition can take instead of how short?, see Quantum Zeno effect ;-) NadVolum (talk) 16:45, 24 March 2024 (UTC)

@NadVolum: In fact pretty much the exact opposite of Quantum Zeno Effect. The Quantum Zeno Effect, as I understand it, arises if you are doing almost continuous measurement of the state of the test atom, destroying any superposition that may be beginning and forcing it into a state where either no transition has occurred or a very quick transition has occurred (and been completed) since the previous observation. The experiments to see how fast a transition could be observed were I believe developments of this set-up. The photons released from such a set-up I think have a very short coherence time (and coherence length), with a correspondingly very wide frequency line-width (like thermal broadening only very much more so), a much greater energy uncertainty made possible by the pronounced coupling of the system with the environment / measuring apparatus, that is used to achieve the decoherence.

In contrast I would like to see the articles make a bit more about how long the coherence times / coherence lengths of the emitted photons can be, corresponding to how long a time the transition superposition may have evolved over. Jheald (talk) 09:21, 25 March 2024 (UTC)

Ok @Jheald:, I'll give my opinion. Considering the orbital trajectory of an electron crossing a flow of electromagnetic radiation, how long will it receive enough radiation pressure energy added in this time to eject itself? If we consider the relation $E=h\nu$ as an energy over a second with $\nu$ not a frequency, but a number of cycles over this second, therefore with $h$ the energy of a cycle whatever the frequency, how many cycles of how many ray crossing this orbit will give their energy to eject the electron? I know, it's too simple to imagine with the analogy of an asteroid field crossing a satellite. It's not science fiction enough to enter the official domain of science. Maybe someday... Malypaet (talk) 22:40, 25 March 2024 (UTC)

March 18

Special Relativity. Is it possible to calculate the velocity of a system, composed of two bodies, we given their different inertial masses and velocities?

An equivalent question: What's the system's (relativistic) inertial mass, we bearing in mind that the conservation of (relativistic) mass, as a general rule, does not hold in Relativity Theory.

2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 16:58, 18 March 2024 (UTC)

There is an article on the general problem Two-body problem in general relativity. Good luck with it! NadVolum (talk) 19:08, 18 March 2024 (UTC)

Oh, I forgot to point out that my question was only about Special Relativity, the given masses being the inertial ones (rather than the gravitational ones). Due to your comment, I've just added above, that I'm only interested in finding the system's velocity (or inertial mass) in Special Relativity. 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 19:34, 18 March 2024 (UTC)

Given the four-momenta of the two bodies, $p_{i}=(E_{i},\mathbf {p} _{i})=(\gamma _{i}m_{i},m_{i}\mathbf {v} _{i})$ ($i=1,2$; the last expression only for bodies with finite mass), the combined four-momentum is $p_{\mathrm {tot} }=p_{1}+p_{2}$. The mass of the two-body system is given by $m_{\mathrm {tot} }^{2}=p_{\mathrm {tot} }\cdot p_{\mathrm {tot} }=(E_{1}+E_{2})^{2}-(\mathbf {p} _{1}+\mathbf {p} _{2})^{2}$. The centre-of-mass velocity is given by $(\mathbf {p} _{1}+\mathbf {p} _{2})/m_{\mathrm {tot} }$. (I've set $c=1$; spatial (three-)momentum is denoted by $\mathbf {p}$). For photons use $E=|\mathbf {p} |=h\nu$. --Wrongfilter (talk) 09:58, 19 March 2024 (UTC)

For being more explicit, let's assume that the bodies: have masses m,n, and move at velocities u,v (respectively) on the same line. What's the system's velocity, using m,n,u,v only? 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 11:13, 19 March 2024 (UTC)

I told you everything you need to know, you just have to restrict to one-dimensional motion and use your notation. Why the anonymity today? --Wrongfilter (talk) 11:29, 19 March 2024 (UTC)

if you suggest that I study maths, then I accept the idea. For the time being, could you calculate it for me (using m,n,u,v only) ? I'm bad at mathematics. (If you also suggest that I register, then you are not the first one to suggest that, but I'd rather avoid registering.) 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 11:54, 19 March 2024 (UTC)

SOS. Despite my bad maths, I think I've found a contradiction in your method. Please help.

I'm checking two cases, each of which involves two bodies, that have the same mass, and that move with the same speed in opposite directions, so according to your formula of the "combined four-momentum", we get:

First conclusion: $p_{\mathrm {tot} }=(\mathbf {p} _{1}+\mathbf {p} _{2})=0,$ right?

By combining, the first conclusion, along with your formula of "centre-of-mass velocity", we conclude:

Second conclusion: The velocity of the two-body system is $0$ as well, right?

By the second conclusion, which tells us that the two-body system is at rest, we conclude:

Third conclusion: The mass of the two-body system, $m_{\mathrm {tot} },$ is simply its rest mass, which is supposed to be invariant, right?

Fourth conclusion: In the first case checked out, the speed of both bodies is low, so $E_{i}$ is low for both bodies, so according to your formula of the "mass of the two-body system", along with the first conclusion, we conclude that $m_{\mathrm {tot} }$ (proportional now to the energy only) is small. While in the second case checked out, the speed of both bodies is high, so $E_{i}$ is high for both bodies, so according to your formula of the "mass of the two-body system", along with the first conclusion, we conclude that $m_{\mathrm {tot} }$ (again proportional now to the energy only) is big, right?

How can we avoid the contradiction between the third conclusion and the fourth one? Where is my mistake?

2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 14:27, 19 March 2024 (UTC)

The total mass of the 2-body system comprises the masses of the two components and their kinetic energies with respect to the common centre-of-mass rest frame. This is now to be considered as internal energy, i.e. energy in internal degrees of freedom. The total mass is invariant with respect to changes of reference frame. --Wrongfilter (talk) 14:45, 19 March 2024 (UTC)

Thank you. May I formulate it as follows: A rest mass is invariant in one-body systems only (e.g. an electron), rather than in multy-body systems, because of the internal energy that may change if several bodies are involved in the whole system? 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 14:52, 19 March 2024 (UTC)

In an isolated system the internal energy will not change. --Wrongfilter (talk) 14:54, 19 March 2024 (UTC)

Thank you. 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 13:28, 27 March 2024 (UTC)

Aliens: from cavemen to NASA

I have been reading many books about extraterrestrial life lately. Although there is no actual example to study, the usual approach is to study which things are required to happen or be there, which ones can be likely, which ones not so much, and the feasibility of achiving similar results in other ways (such as life being made of an element other than carbon, or with a solvent other than water). I have seen this approach used to see the chances of a given planet to turn into a Earth-like planet, and for life, once formed, to survive and evolve.

But there is a missing step: once we get full ecosystems with diverse animals, how likely would it be for some animal to evolve into a human-like intelligence? Which would be the requirements and obstacles? And if one actually does, how likely would it be to go all the way from rocks and sticks to space programs? I'm not asking to discuss that ourselves, but to point me some author that did. The article Extraterrestrial intelligence is almost devoid of content and does not help with this search. Cambalachero (talk) 19:44, 18 March 2024 (UTC)

Are you familiar with the Fermi paradox? 41.23.55.195 (talk) 05:39, 19 March 2024 (UTC)

There is a strong bias in the question itself. Which is more intelligent: A species of humans who invented war and terrorism or a species of dolphins that chat and play most of the day? What if a planet of "intelligent" creatures stopped at having fun and didn't feel the need to progress into colonialism and repression? It leads to two common points on the topic. The first is that we will meet technology from other space-exploring societies before we meet the society just as others will see our satellites and transmissions long before they meet a human. The second is that any species that travels through space to meet us isn't doing so to be nice. They see our planet as a source of profit in some way and will want to exploit it as much as possible. 12.116.29.106 (talk) 18:46, 19 March 2024 (UTC)

“For instance, on the planet Earth, man had always assumed that he was more intelligent than dolphins because he had achieved so much—the wheel, New York, wars and so on—whilst all the dolphins had ever done was muck about in the water having a good time. But conversely, the dolphins had always believed that they were far more intelligent than man—for precisely the same reasons.” ― Douglas Adams, The Hitchhiker’s Guide to the Galaxy I'm betting you were making a reference, but figured this was a good chance to introduce the unenlightened to Adams. --User:Khajidha (talk) (contributions) 13:19, 20 March 2024 (UTC)

Dolphins probably never advanced towards a civilization because, as an underwater species, they have no access to fire, the one element that allowed all human technology in the first place. And the lack of versatile limbs, such as our hands with opposing thumbs, probably does not help either. Those two are the kind of points I'm sure some author must have thought and organized in a related book. Cambalachero (talk) 14:04, 20 March 2024 (UTC)

Why would dolphins need fire? ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:56, 20 March 2024 (UTC)

To cook all the fish we gave them. --User:Khajidha (talk) (contributions) 17:32, 20 March 2024 (UTC)

Maybe they prefer sushi. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:26, 21 March 2024 (UTC)

Good luck getting to the copper age or further without fire. Cambalachero (talk) 17:25, 21 March 2024 (UTC)

Why would dolphins need metals? ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:10, 21 March 2024 (UTC)

How else could they become our tyrannical overlords, forcing us to frolic and have fun every single day? Clarityfiend (talk) 00:51, 22 March 2024 (UTC)

Evolution of human intelligence discusses how some apes got human-like intelligence. It includes links to cephalopod intelligence and cetacean intelligence. Also, possible extraterrestrial life has no obligation to develop intelligence, space pollution or Rubik cubes. To believe otherwise is probably a misleading use of teleology in biology. --Error (talk) 23:34, 19 March 2024 (UTC)

March 20

Removing air from water to reduce your water bill

There are a number of companies out there that purport to be able to reduce your water bill by reducing the volume of air in the pipe ahead of the meter so that you "just pay for water". (Example site). Leaving aside any issues with the water company, does the science of these things make sense? Matt Deres (talk) 19:32, 20 March 2024 (UTC)

The example site you provided is based on little more than advertising hyperbole! I have read it in its entirety and I saw nothing that made technical sense. The great majority of water systems connected to a water distribution network don’t contain air (unless the advertisement is referring to air dissolved in the water.)

If this hardware actually achieves something worthwhile this advertisement fails to disclose what it is or how it works. Dolphin (t) 21:11, 20 March 2024 (UTC)

In my area of the UK, meters are installed under the tarmac of the public pavement (sidewalk) outside the residence, accessed by a very small hatch (they are read remotely), so to treat the water before it reached a meter would likely require excavating the pavement – not a practical or legal proposition – and installing the device (whatever it is) on the water company's pipework, which would likely also be illegal. Moreover, even if it worked, it would likely be a violation of the water company's contractual terms. {The poster formerly known as 87.81.230.195} 51.241.39.117 (talk) 06:39, 21 March 2024 (UTC)

What a wonderful advice: "trusting advice about Smart Valves from anyone other than experts like us is ill-advised." Experts "like us" can easily be recognized, because they wouldn't recommend any other valves. So where the ad admonishes, "Don't fall victim to outdated first-generation valves", I fully agree. I'd even go further and say, "Don't fall victim to any valves, whether first, second or third generation."  --Lambiam 10:00, 21 March 2024 (UTC)

Water metering#Problems addresses the issue of air in the water, giving a reference to regulations, but the cite is to an 'International Recommendation' rather than any national law. -- Verbarson  ^talk_edits 12:23, 21 March 2024 (UTC)

...and the recommendation specifically refers to 'liquids other than water'. -- Verbarson  ^talk_edits 12:27, 21 March 2024 (UTC)

Faucets are equipped with aerators specifically to introduce air into the water at the point of use, reducing water use through more effective wetting in the dispersed water, and reducing splashing. Maybe these marketers are trying to make people think that it happens in the supply lines and they're getting cheated? Acroterion (talk) 12:33, 21 March 2024 (UTC)

It all sounds like a scam. And doesn't dissolved air in the water improve the taste? Otherwise you'd be drinking distilled water, which is pretty bland. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:37, 21 March 2024 (UTC)

That would be minerals, at least up to a point. Air, unless it's carbonation, doesn't affect taste significantly. Acroterion (talk) 02:56, 22 March 2024 (UTC)

Our colleagues over at WikiHow (and many plumbing sites) have addressed this - "Water occasionally comes out of the tap with a cloudy or milky appearance. In most cases, cloudy water is caused by air bubbles in the water, and these will dissipate on their own if you let the water sit for a few minutes." HiLo48 (talk) 03:55, 22 March 2024 (UTC)

Here's another site offering the same kind of thing: [8], but searching for water valves that remove air to reduce water bills will give you no end of similar items. This one at least has a demonstration video of how it's supposed to work. Many, like the Watergater, tout that they are 'certified' and/or 'approved' which always strike me as weasel-words in this kind of context. Like, they were maybe approved in the sense that they wouldn't explode or poison the water or something, but meanwhile implying that an impartial researcher has confirmed their claims. But just because it sounds fishy doesn't mean that it is, so I guess my questions are: 1) does municipal water typically have significant air in it? 2) does that air actually impact the reading of a standard water meter and 3) is there any chance this doohickey could affect that in a meaningful way? Matt Deres (talk) 14:05, 22 March 2024 (UTC)

Regarding (1), the charts here indicate that at normal temperatures (40-80 F) and typical municipal water pressures (60 psi), the ratio of the volume of air to the volume of water is roughly 8% to 13%. This is more than I would have guessed. Whether a valve can change that in any significant way, without changing the pressure, seems doubtful to me but I don't know for sure. CodeTalker (talk) 02:58, 23 March 2024 (UTC)

There is a difference between an air and water mixture, and water containing dissolved air (properly, dissolved nitrogen and oxygen, being the significant components of air). To my understanding, dissolving gases in water does not greatly increase its volume (if at all), so it's important to be sure what is being measured: this reference discusses air dissolved in water. {The poster formerly known as 87.81.230.195} 51.241.39.117 (talk) 09:26, 23 March 2024 (UTC)

March 22

2 related questions about thought

I've read novels where a character says thinking too hard about relativity or some other difficult subject "made his head hurt." I've also had heard people say that, jokingly or not. Q1). Shoud I take it literally? Q2). Could there be microbes in the brain that cause inflamation and make hard thinking painful and exhausting? Rich (talk) 03:48, 22 March 2024 (UTC)

I think it is just idiom, like saying that something makes one's blood boil. Tension headache can result from being literally tense, for which stress can be a factor. So in a desperate last-minute study of a difficult subject for an impending exam, finding it inscrutable, a student may clench their teeth and at some time get a headache. But then it is not caused directly by the brain activity itself. The blood–brain barrier should defend the brain against invading microbes. If pathogenic microbes nevertheless manage to get in, as sometimes happens, a patient will likely develop encephalitis, which can be debilitating and life-threatening. Early symptoms include headache and confusion.  --Lambiam 10:38, 22 March 2024 (UTC)

I should add that the brain itself does not contain nociceptors. Brain surgeons poking around in the brain of a conscious and unanesthetized patient may cause all kinds of subjective sensations, but this is not by itself painful.  --Lambiam 14:44, 22 March 2024 (UTC)

Lapse rate between sub-adiabatic and inversion

Simplified graph of atmospheric lapse rate near sea level

I sketched this graph but am unsure what to label the teal vertical line. Does it have a special name? Thanks, cmɢʟee⎆τaʟκ 08:13, 22 March 2024 (UTC)

"Isothermal lapse rate" is used e.g. in the abstract of this article. --Wrongfilter (talk) 09:58, 22 March 2024 (UTC)

@Wrongfilter: Thank you very much. Updated... cmɢʟee⎆τaʟκ 13:13, 22 March 2024 (UTC)

The area between dry adiabatic lapse rate and wet adiabatic lapse rate is marked as neutral. It could be stable or unstable, depending on the humidity. Neutral is a single line, that must be in the area marked as neutral. PiusImpavidus (talk) 09:36, 23 March 2024 (UTC)

Fair point, thanks. It was too wordy to label ("region with neutral"?) so I just omitted the label. cmɢʟee⎆τaʟκ 12:29, 23 March 2024 (UTC)

Chemical X real or not?

Blocked user. Matt Deres (talk) 16:53, 22 March 2024 (UTC)
The following discussion has been closed. Please do not modify it.
From cartoon references this chemical they can real or not? 2001:44C8:40E2:8DFE:54A:6AF:F2CA:27EA (talk) 08:18, 22 March 2024 (UTC) Chemical X is clearly fictional. Shantavira|feed me 09:44, 22 March 2024 (UTC) It is a plot device. Graeme Bartlett (talk) 11:09, 22 March 2024 (UTC) As with Roger Ramjet and his proton pills, and Mighty Mouse with his super cheese. ←Baseball Bugs What's up, Doc? carrots→ 11:20, 22 March 2024 (UTC)

Personality of science deniers

Have any studies been done whether, say, climate change deniers and flat earthers typically share specific personality traits?  --Lambiam 14:32, 22 March 2024 (UTC)

Somewhat related: The article Anti-vaccine activism may contain relevant links / references. --Cookatoo.ergo.ZooM (talk) 18:42, 22 March 2024 (UTC)

The article contains 180 wikilinks and 172 references with 183 external links, but I did not spot anything suggesting any studies of personality traits of anti-vaccine activists.  --Lambiam 20:19, 22 March 2024 (UTC)

Not sure that there's much on personality traits but this paper may be of interest. Mikenorton (talk) 20:31, 22 March 2024 (UTC)

Denialism is the Wikipedia article that should cover that, but it doesn't say anything much about it. Some of the external references though could be of interest to you. Quite a few of them are in it for the money from their audiences, but there's lots of perfectly reasonable people out there who seem to just get this urge to spend time learning more just so they can defend their denial and spread it to the world. NadVolum (talk) 20:57, 22 March 2024 (UTC)

The article Sectarianism may contain some interesting links, eg Collective Narcissim et al. It seems that S Freud (and, by implications, others) were speculating on these internecine dynamics. --Cookatoo.ergo.ZooM (talk) 07:17, 23 March 2024 (UTC)

PS: Using Google Scholar with suitable query terms seems to bring up useful references on psychological parameters. --Cookatoo.ergo.ZooM (talk) 14:05, 23 March 2024 (UTC)

March 23

Why do Blattella germanica cerci point up then out?

Not adult: they're in a vertical plane, adult: in horizontal plane. Sagittarian Milky Way (talk) 05:43, 23 March 2024 (UTC) Okay that's not very accurate but it sure does look that way from above. They have different orientations before and after sexual maturity. Sagittarian Milky Way (talk) 06:16, 23 March 2024 (UTC)

Courtesy links: Blattella germanica, cerci.-Gadfium (talk) 05:46, 23 March 2024 (UTC)

Hours of daylight vs latitude vs day of year

Hours of daylight vs latitude vs day of year

@SebastianHelm: modified a vectorisation I made of File:Hours_of_daylight_vs_latitude_vs_day_of_year.png. Yesterday, @Episcophagus: posted some questions which I couldn't answer:

1. Why are the borders to constant night and constant day labelled with ‘1 hour’ and ‘23 hours’, respectively?
2. Why isn't ‘equal day length’-latitude south of the equator, as the rate of change in the equation of time affects the daylength (the steepness of the ecliptic at the equinoxes make the sun to move slower in rectascension and thus the daylength is shorter than at the solstices)?

Can anyone help them, please? Thanks, cmɢʟee⎆τaʟκ 12:11, 23 March 2024 (UTC)

1. I don't know but it increases asymptotically "exponentially" cause geometry, the 1 and 0 hour lines would be very close at this scale. I don't know if it's an exponential function but the layman meaning of exponential. Half of 12 hours is most of the way to the polar circle 1 hour would be over 2 more halvings, at the polar circle refraction and Sun radius add at least 50 nautical miles of midnight Sun above simple geometric anyway. Very close it could rise after set while still going down or vice versa (at extremely glancing angle) just from refraction fluctuations. And with a Titanic-like extremely calm sea horizon you could float in the water and climb 5 feet when it set and it rises, float in the water again and it set repeat (eyeballs 5 feet above ocean lowers the horizon about 0.1X Sun radii with ever diminishing returns with height. 10 times more horizon dip than 5 feet takes hundreds of feet, 1,000 times more dip takes millions of feet and 2,000 times more dip takes over infinity light years cause the horizon can't dip more than 90 degrees). At medium latitude this period only lasts seconds.
2. At the equator the Sonnar goes up about 3 and one third minutes before the center would be risen without refraction. Daytime would be about 12 hours 6 minutes and 40 seconds at the equator if there was no equation of time.

Sagittarian Milky Way (talk) 15:57, 23 March 2024 (UTC)

Thanks, @Sagittarian Milky Way: cmɢʟee⎆τaʟκ 14:29, 24 March 2024 (UTC)

British physician who debunked Lourdes miracles

Hi, I've read in a book by Piero Angela (Italian science writer) an interview with William A. Nolen who said a British physician once wrote a book debunking alleged Lourdes miracles. Does anyone know who he was? Thanks.-- Carnby (talk) 12:39, 23 March 2024 (UTC)

Probably Donald J. West in Eleven Lourdes Miracles (1957). See this URL Mike Turnbull (talk) 12:47, 23 March 2024 (UTC)

Thank you!-- Carnby (talk) 12:57, 23 March 2024 (UTC)

The definition of relativistic momentum

As opposed to Newtonian Mechanics, which defines momentum as $P=mv,$ Relativity theory defines momentum as $P=\gamma mv,$ i.e. as $P=m(dx/d\tau ),$ but I wonder why $d\tau ,$ rather than $dt$ - thus sticking to the original Newtonian definition of momentum...

If the reason for preferring the "rest" time $\tau$ is because also the mass $m$ being referred to is the "rest" mass, then also the velocity referred to should be the "rest" velocity, i.e. $v=0,$ shouldn't it?

On the other hand, since the velocity being referred is not the rest velocity $v=0$ but rather the real veloctity, so isn't it usually defined elsewhere as $dx/dt,$ rather than as $dx/d\tau ?$

On the third hand, If the reason for defining the relativistic momentum as $P=m(dx/d\tau )$ is our desire to let Newton's second law $F=dp/dt$ remain invariant under the Lorentz transformations, then why do we replace the expected definition $P=m(dx/dt),$ by the less intuitive definition $P=m(dx/d\tau ),$ rather than by any other definition (e.g. $P=2m(dx/d\tau )$ or whatever), that lets Newton's second law remain invariant under the Lorentz transformations? 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 22:21, 23 March 2024 (UTC)

τ represents proper time, as distinct from coordinate time represented by t. See those two articles for more details, although to be honest, neither article does a great job of explaining the difference. CodeTalker (talk) 01:26, 24 March 2024 (UTC)

Yes, I know that, but how does that answer my question? isn't the proper velocity usually defined elsewhere as $dx/dt,$ rather than as $dx/d\tau ?$ 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 01:29, 24 March 2024 (UTC)

Four-momentum ${\vec {p}}=m{\vec {U}}$ and four-velocity ${\vec {U}}={\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }}$ have to be covariant, i.e. frame-independent, vectors; this is only possible by taking the derivative with respect to a frame-independent quantity, $\tau$, but not with respect to a frame-dependent coordinate time $t$. Proper time $\tau$ parameterises the space-time trajectory ${\vec {x}}(\tau )$ of the particle in question, and ${\vec {U}}(\tau )$ is the tangent vector to that curve. In the non-relativistic limit these definitions reduce to the standard Newtonian momentum and velocity, so there is no contradiction. --Wrongfilter (talk) 09:09, 24 March 2024 (UTC)

If the only reason for defining, the four-momentum as ${\vec {p}}=m{\vec {U}},$ and the four-velocity as ${\vec {U}}={\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }},$ is our desire to let the four-momentum and the four-velocity be covariant, then why do we replace the apparently-more-intuitive definition $P=m(dx/dt),$ by the less intuitive definition $P=m{\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }},$ rather than by any other less intuitive definition (e.g. $P=2m{\tfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} \tau }}$ or whatever), that lets the four-momentum and the four-velocity be covariant? I'm pretty sure there are infinitely many options for the four-momentum and four-velocity to be covariant, aren't there? 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 12:59, 24 March 2024 (UTC)

Most people would want the relativistic momentum to coincide with the Newtonian momentum in the non-relativistic limit. That puts a lot of restrictions on the relativistic form (not sure whether that makes it unique). Your factor 2, for instance, just wouldn't go away. --Wrongfilter (talk) 13:14, 24 March 2024 (UTC)

Well, your remark in the parentheses is the most important one so far, in my eyes. I'm really curious to know (even though you are not sure), if the common relativistic form of momentum is unique, for it to be frame independent and to coincide with the Newtonian momentum in the non-relativistic limit. If the relativistic form of momentum is really unique, then there must be a rigorous proof that derives this relativistic form somehow, so this proof is probably mentioned in textbooks, but it's not, AFAIK. On the other hand, if the relativistic form of momentum is not unique, then I wonder what other forms the relativistic momentum could have, while still being frame independent and coinciding with the Newtonian momentum in the non-relativistic limit. 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 13:39, 24 March 2024 (UTC)

This is the unique definition that (1) reduces to the Newtonian momentum for small speeds, and (2) makes conservation of momentum frame-independent. See [9]. --Amble (talk) 20:59, 26 March 2024 (UTC)

Thank you. However, I noticed that (3.6) and (3.7) in the chapter, are definitions rather than conclusions.

Anyway, also the continuity of the function should be assumed, unless it's considered to be included in your first assumption (1). 2A06:C701:7478:1B00:1F2:1345:432B:F844 (talk) 13:30, 27 March 2024 (UTC)

March 24

Smallest electric current intensity

Can we consider that the smallest electrical intensity corresponds to one electron per second? Or is there another value? Malypaet (talk) 23:55, 24 March 2024 (UTC)

Surely one electron per hour is a smaller intensity than one electron per second. And one electron per day is smaller than that. I don't see that there's any meaningful definition of the "smallest electrical intensity". CodeTalker (talk) 01:53, 25 March 2024 (UTC)

Thanks, I hadn't seen things from that angle. The electron is a discrete quantity, but indeed the intensity also depends on time which is continuous. However, if we know the start date of this current, we still know the minimum intensity at a given date: an electron between these two dates. Malypaet (talk) 22:06, 25 March 2024 (UTC)

Imagine an infinitesimally thin plane the electron is passing through. Since it is a wave in the electromagnetic field, it is not possible to define when it passes through. But using the probability associated with the wave, we can consider the probability $p(t)$ that, at time $t$, it has already passed through. The intensity of the current would then be $e{\tfrac {{\text{d}}p(t)}{{\text{d}}t}},$ in which $e$ denotes the elementary charge. By slowing down the electron, this can be made (mathematically) arbitrarily small. Perhaps the wave packet started to move through when the first electrons came into being and has not made substantial progress since, the prognosis being that it will not even be halfway through at the end of time, $10^{10^{10^{.^{.^{.}}}}}$ eons into the future.  --Lambiam 12:00, 26 March 2024 (UTC)

My question was about physics, not math. An electron being a discrete quantity and matter also, in physics, an infinitely thin plane does not exist. Unless they are electrons in a vacuum. But since the measurement of intensity is done by material instruments, we return to discrete quantities. Physics is an experiment. Malypaet (talk) 21:55, 26 March 2024 (UTC)

Can you define the notion of the magnitude of the electric current represented by a single moving electron without appeal to mathematical concepts? If a regular stream of electrons moving at a certain velocity represents a certain current, an equally spaced stream at half that velocity will represent half that current. However you choose to define the magnitude of the current of a single electron, the same should hold for this single electron: halve its speed, and you halve its current. So now we should wonder, what is the smallest speed?  --Lambiam 09:22, 27 March 2024 (UTC)

March 26

If Neptune was named Janus, what will be the name of element 93?

The first suggestion of Neptune is Janus (by Galle), if Neptune was named Janus instead of Neptune, what will be the name (and the symbol) of the element neptunium (element 93), which was named after Neptune? 125.230.237.23 (talk) 04:05, 26 March 2024 (UTC)

We don't answer requests for opinions, predictions or debate. 41.23.55.195 (talk) 05:07, 26 March 2024 (UTC)

It's safe to say the element would have a different name. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:24, 26 March 2024 (UTC)

The elements named after astronomical objects named after mythological figures whose name in Latin ends in the second-declension suffix -us (Neptunus, Uranus) replace the suffix by -ium. This is also the most common suffix for other element names. An author writing a novel set in an alternate universe in which a planet came to be named Janus could believably use the name janium for a later discovered or synthesized element. In the novel's universe so many things may diverge from our universe (in fact, should, to make it a compelling read) that there is little reason for this thus-named element to be specifically that with the atomic number 93.  --Lambiam 11:27, 26 March 2024 (UTC)

Copper IUD end of life

The Copper IUD article mentions that they have a limited life time, but does not mention why the life time is limited.

I am imagining two possible ways in which this happens:

1. the copper slowly erodes. eventually the amount of copper left is too small to be effective

2. the amount of copper left is always sufficient, but a coating slowly develops over time. Eventually the amount of coating is so great that there is insufficient copper released.

Which is more correct in this case? Or is it a combination of both? Or is it some other mechanism? OptoFidelty (talk) 05:25, 26 March 2024 (UTC)

"Unlike an ordinary IUD which is left in the uterus indefinitely, the copper IUD may need to be replaced when the copper is exhausted, usually after about two years, in order to retain contraceptive efficacy."^[10]
    However, "The higher pregnancy rate among the women who continued to use their copper 7 device for a third year suggests that although the copper is still there, it is not available for contraceptive action. ... Examination under the microscope showed hard deposits of material on the surface of these copper 7 IUDs, which Gosden et a1 have shown is a layer of calcium, which would seriously interfere with the contraceptive action of the available copper. This may explain the highly significant difference in the pregnancy rates seen in our two groups-12 pregnancies in the continuation group during the third year of use and no pregnancies in the replacement group. Even allowing for possible differences in the fertility rates of these two groups, which were similar in mean age and parity, this striking difference is remarkable and seems to confirm that although the copper may not be exhausted by three years of use, it may not be available for contraceptive action."^[11]  --Lambiam 10:44, 26 March 2024 (UTC)

March 27

No-signaling and the infinite hat game

(posting in RDS instead of RDMath because the math is pretty clear; it's the alleged "physics" that is not making sense to me)

Recently someone added a note to our article on the axiom of choice a reference to an interesting paper that describes an allegedly counterintuitive consequence of AC. The first bit is fun but not super-surprising to a set theorist (which I am by academic training though not by current employment). The game involves a countably infinite collection of men (just to make the pronouns easy) lined up in the order of the natural numbers. They are given hats, either red or white. No one can see his own hat, but each can see the hats of every man in front of him. Each is supposed to guess the color of his own hat. They are allowed to agree on a strategy beforehand.

It turns out that if they choose and agree on a transversal of the equivalence relation of finite differences on sequences of red and white (that is, two sequences are equivalent if they differ in only finitely many places), then no matter how the hats are placed, all but finitely many of them can guess correctly. The proof of this is essentially trivial once you've understood the definitions. AC guarantees that such a transversal exists.

Now, the counterintuitive thing here is that no man can get any information whatsoever about the color of his own hat just from the colors of the hats in front of him, but nevertheless the strategy works perfectly after some point. I guess if you want you can take this as a point against AC, though from my perspective it doesn't really compete with the clear and simple intuition in favor of AC.

But where I think the paper might go off into the weeds a bit, or maybe I just haven't been clever enough to get what it's talking about, is when it tries to connect this with forms of the no-signaling principle. They identify a "probabilistic" and "functional" version of the NSP, and claim to show that, allegedly counterintuitively, the "functional" version is a stronger "resource".

This I cannot make heads or tails of. First, I don't see how either version of the NSP is a "resource", unless the word is being used in some way I'm not familiar with. As far as I can tell they are constraints, not resources. So my first clear question is, is there some physics-specific sense of the word "resource" that would make this make sense?

Then, in the "functional" version, they allow an oracle for a transversal of the equivalence relation. I don't think they allow that in the "probabilistic" version, though I haven't combed over every line of it, because if they did, you could just use it and ignore the probabilistic part. So isn't it the oracle that's the extra "resource" here? And then, second clear question, why should it be counterintuitive that you can do more with the oracle than without it? --Trovatore (talk) 05:19, 27 March 2024 (UTC) Pinging GiordanoB, who added the note. --Trovatore (talk) 06:35, 27 March 2024 (UTC)

Not relevant to your question, but I find this statement in the paper questionable: "After all, the whole of the theoretical apparatus of physics relies heavily on the use of set theory, thus on the AC." I doubt the theoretical apparatus of physics needs more than naive set theory; if formalized, I expect it can be accommodated equally comfortably in ZF+AD.  --Lambiam 07:46, 27 March 2024 (UTC)

The idea that the infinite hat game can tell us something about physics is IMO absurd. It should be completely obvious that the lined-up players cannot do better than one would expect from guessing in a purely Newtonian universe, so this has nothing to do with space-like separation. At any time, a player can only have information about a finite number of other hats, so good luck in forming the equivalence class. The equivalence class exists in a Platonic sense, but the player has no access to it. One might as well let player $k$ apply AC to the set $\{\{c_{k}\}\},$ in which $c_{i}$ is the hat colour of the $i$-th player. While the player cannot know this set of sets, it "exists", so in the logic of the puzzle one may apply AC. So if the player now does apply AC, we also apply some smoke, and boom!, paradox.  --Lambiam 08:05, 27 March 2024 (UTC)

Here, in a book titled Bell Nonlocality, the author refers to "a hypothetical no-signaling resource called a PR-box (Popescu and Rohrlich, 1994)". The reference is to: Popescu, Sandu and Rohrlich, Daniel (1994). Quantum nonlocality as an axiom. Foundations of Physics, 24, 379–385.  --Lambiam 08:45, 27 March 2024 (UTC)

Gah. And if you replace the Axiom of Choice with Axiom of determinacy which is supposed to make things simpler I think you'd probably still end up with this problem. NadVolum (talk) 10:07, 27 March 2024 (UTC)

A red photon and a blue photon are approaching each other in opposite directions. Is the whole system at rest?

On the one hand, it seems the whole system is apparently at rest, because they are approaching each other in opposite directions, by the same speed, while carrying identical masses. On the other hand, both photons have different momenta, so the whole system's momentum is not zero, so apparently the whole system can't be at rest. That's why I wonder, what's the correct answer to the question: Is the whole system at rest?

I guess it's not at rest, yet I'm not sure about what's the mistake in my first consideration, by which I've concluded the whole system is apparently at rest. 147.235.215.72 (talk) 14:03, 27 March 2024 (UTC)

At rest relative to what? There's no priviliged reference frame to automatically define "at rest" against. If the two photons are approaching each other, you can CHOSE a reference frame where the net momentum of the system is zero. PianoDan (talk) 14:23, 27 March 2024 (UTC)

Relative to whoever sees the first photon as read and the second photon as blue. 147.235.215.72 (talk) 15:29, 27 March 2024 (UTC)

The mass $m_{\lambda }$ of a photon in empty space is given by $m_{\lambda }=E_{\lambda }/c^{2}=(hc/\lambda )/c^{2}=(h/\lambda )/c.$ The magnitude of its momentum, $p_{\lambda },$ is given by $p_{\lambda }=h/\lambda .$ We see that the two are related by $m_{\lambda }=cp_{\lambda }.$ So if their momenta are not equal in magnitude, neither are their masses.  --Lambiam 14:31, 27 March 2024 (UTC)

Actually, there is a dispute over whether light has mass. My question assumes light has no mass, yet I know my question could have a clear answer if we assumed light had mass. 147.235.215.72 (talk) 15:28, 27 March 2024 (UTC)