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This question is mostly theoretical, because there's no feasible way to create such heavy neutron-rich isotopes at present. But: what predictions are there on the N = 126 shell closure at low proton numbers? In particular, is 176Sn (Z = 50, N = 126) expected to be doubly magic, or will this shell closure disappear that far from the valley of stability (like N = 20 does)?
Yes of course, since 78Ni is also quite unstable to beta decay. What I'm curious about is (1) whether 176Sn should exist and (2) whether it does close the neutron shell, or if the energy gaps are expected to change in this extremely neutron-rich region. Double sharp (talk) 04:33, 25 August 2024 (UTC)[reply]
The heaviest isotope of tin known is 140 50Sn , which lives less 50 ms and already drips neutrons. The existence of an isotope as heavy as 176 50Sn seems unlikely. Magic number itself does not mean that the nucleus exists in any meaningful way. You can look at 10 2He . Ruslik_Zero20:48, 26 August 2024 (UTC)[reply]
That's beta-delayed neutron emission, so the drip line hasn't been reached yet, as expected.
It seems then that the best we could find at the moment are the papers Nucleus hydro elemon found at first, which suggest that 176Sn should be more or less on the borderbetween being bound and being unbound. Those two papers suggest N = 126 is still magic (because the two-neutron separation energy has a big jump going from 176Sn to 178Sn), but this one makes it less clear. Since this is so far from what's currently known, it's probably not possible to do better at the moment. I'd guess, therefore, that the best possible answer to my question at the moment would have to be "no one really knows; could be either way". Double sharp (talk) 06:27, 27 August 2024 (UTC)[reply]
I found a reference by Fang et al. about beta decay of 176Sn. Its β− decay energy is around 22 MeV (comparable with 29F 21.7 MeV) and has a half-life of <1 ms. 176Sn should undergo β−n instead of only β−.
The calculated mass excess of 176Sn is 217.59 MeV, as predicted by KTUY. Mass excess of 176Sb is predicted to be 195.49 MeV, thus the β− decay energy will be 22.10 MeV, not far away from Fang et al. Somehow the decay process β−,23n to 153Sb+23n is actually possible with decay energy 2.47 MeV.
KTUY predicts S2n of 174,176,178Sn are −0.42,−0.52,−2.69 MeV, implies all of them can possibly undergo 2n emission. The big jump from 176Sn to 178Sn suggest N = 126 is still magic.
I think 176Sn wouldn't get seriously affected by 2n emission, due to some trends related to atomic number. There is no heavy 2n emitters (the heaviest is 26,28O with Z = 8), so I just show it with 2p emitters. Despite 12O (S2p = −1.737 MeV, Z = 8) has a lower decay energy than 67Kr (S2p = −2.89 MeV, Z = 36), it decays much faster (8.9×10−21 s vs 7.4 ms). If the trend follows, then 2n emission of 176Sn is just not important compared to beta decay.
Not only would the signal-to-noise ratio be minuscule, but due to its orbit around the galactic center the signal would have to be aimed at where the Messier 13 cluster will be 25k years from now. 136.54.237.174 (talk) 18:05, 25 August 2024 (UTC)[reply]
From the data given in the article, I arrive at a minimum diameter of the receiving antenna of 2 kilometers. The diameter of the Arecibo dish is meters, the wavelength of the signal is cm. The beam divergence angle is then radians. Because is very small, the solid angle is to a good approximation , the exact formula is . The area of the beam after traveling a distance of r is then . Then with the power of the beam of 405 kW, at a distance of lightyears, the flux of the signal will be Watt/m^2 at M13. This signal can then be detected using one or multiple antennas. If the total area of the antennas is A, then the received power is F A. If we assume that the temperature of the antennas and receivers are T = 20 C = 293.15 K, then the noise power will be where is the bandwith, that in this case must be 10 Hz or larger, as this is the frequency shift used to modulate the signal. The signal power must be larger than the noise power. If we then equate F A to P and solve for A and then assume a single antenna is used, and put then the diameter of the receiving dish is 2 r and if I didn't make any mistakes, this yields a minimum diameter of approximately 2 kilometers. Count Iblis (talk) 19:25, 25 August 2024 (UTC)[reply]
There's also the matter of integration time. Noise adds incoherently, signal adds (hopefully) coherently, so with a longer integration time, the signal may rise above the noise. In this case, the integration time is limited to no more than 100 ms by the 10 Hz bitrate. The difference between the 0 bit and the 1 bit was only one wave, so a longer integration time doesn't help to decode the signal, but it may still help to detect the carrier wave.
Beam size matters too. The wider the beam, the more noise from other sources like stars; the narrower the beam, the less likely those aliens pointed it well enough at Earth. PiusImpavidus (talk) 10:31, 26 August 2024 (UTC)[reply]
[1] says that the gain of Arecibo antenna at 2.38 GHz was 77 dBi, only 600 mdB short of Count Iblis's estimate from the physical diameter (an aperture efficiency × antenna efficiency of 87% if true). A receiver temperature of 20°C is a little pessimistic; usually the receiver would be cooled (it is not necessary to cool the antenna, assuming that it is low-loss). catslash (talk) 00:09, 28 August 2024 (UTC)[reply]
Regarding the integration time: when there are two possible symbols (0 and 1) represented by two orthogonal signals of equal energy ( = received power × time) then the bit error rate is something like
where is the complementary error function. This assumes (1) that it is known exactly what the two signals are - there is no random change in the phase between symbols, and (2) that the prior probability of each symbol is equal.
Without error-correcting codes it is impossible to reduce the error rate to zero, so it is necessary to decide what rate is acceptable before building the receiving antenna. catslash (talk)
[2] (Table 1: Legacy Arecibo Observatory planetary radar system.) says that the gain of Arecibo antenna at 2.38 GHz was 72.9 dBi, which seems more plausible. catslash (talk) 15:57, 1 September 2024 (UTC)[reply]
At the last point of the tabel or matrix about the far future, there is something like: "Because the total number of ways in which all the subatomic particles in the observable universe can be combined is
10
10
115
{\displaystyle 10^{10^{115}}},[152][153] a number which, when multiplied by
10
10
10
56
{\displaystyle 10^{10^{10^{56}}}}, disappears into the rounding error,"
My question is: How is the point about the rounding error valid? 2A02:8071:60A0:92E0:30AB:357:41DB:5492 (talk) 16:54, 29 August 2024 (UTC)[reply]
Or does the text try to say that is negligible compared to the rounding error in ? The intention is not clear to me. Why should these numbers be multiplied and what do rounding errors have to do with it? The argument should be, I think, that since there are "only" possible combinations, the specific combination that results in a repeat of the Big Bang is bound to occur sometime in the next years. However, this seems to assume that all combinations are about equally likely and ignores the effect of the expansion of the universe. I suspect SYNTH. --Lambiam19:57, 29 August 2024 (UTC)[reply]
When an electron collides with positron, and when a proton collides with the anti-proton, both situations they are transformed into 2 photons. Can those 2 photons be distinguished from the 2 situations? That is, can the 2 photons be traced to being formerly an electron or proton? They have different energy of initial states, different total spin? Thanks. 66.99.15.162 (talk) 17:46, 29 August 2024 (UTC).[reply]
See electron–positron annihilation and annihilation. At low energy, electron-positron annihilation will produce two photons, and you can be sure they didn't come from proton-antiproton annihilation because the total energy is less than the rest mass of two protons. The annihilation of a proton and antiproton, or an electron and positron with higher energy, can produce a variety of end states including baryons and weak bosons. A proton-antiproton annihilation to two photons would be a fairly rare process, see [3]. So we're talking about fairly uncommon end states. The high energy electron-positron annihilation and the proton-antiproton annihilation could produce the same types of final states, but with different probabilities, so you can make some statistical inferences, especially if you can observe multiple events. --Amble (talk) 19:15, 29 August 2024 (UTC)[reply]
And then, photons that were created from an electron moving up/down an orbital or shell, are obviously different than the above mentioned photons? Have different measurable properties? These properties (or just energy) are measured when a photon hits a solid, the energy measured by a photo-multiplier tube? 66.99.15.162 (talk) 20:22, 29 August 2024 (UTC).[reply]
Well these photons have lower energy than those created in annihilation. Also they are usually produced one at a time, rather than a pair or triple. Graeme Bartlett (talk) 00:28, 30 August 2024 (UTC)[reply]
But the concise answer to your question is: no. Once a photon is created, its only unique property is its energy. There is no difference between a 511 keV photon created from positron-electron annihilation or one created from any other source of 511 keV photons. See: indistinguishable particles. PianoDan (talk) 16:04, 30 August 2024 (UTC)[reply]
A friend of mine has a battery-powered clock. It stopped running. It was over a week before he got replacement batteries. When he went back to the clock, it was running again. Is there an explanation for this? Bubba73You talkin' to me?16:04, 30 August 2024 (UTC)[reply]
This effect is readily seen with a flashlight when the cells (battery) are approaching the end of their useful lives. After a short period of operation the light from the flashlight becomes dim. If the operator turns the light off and waits a few seconds, then turns it back on, the light is usefully bright again but only for a few seconds before it becomes dim again.
The active materials in the cell migrate to get to their electrode and that takes time. Similarly the exhausted product migrates away from its electrode. Allowing the cells to rest for a short while allows these materials to migrate to/from the electrodes and improve the density of the active materials surrounding each electrode. Dolphin(t)22:13, 30 August 2024 (UTC)[reply]
Temperature also plays a role. Batteries near their end of life are more likely to fail if the room turns cold. If it warms up again the clock may restart. Shantavira|feed me09:04, 31 August 2024 (UTC)[reply]
An additional factor may be a slight irregularity in a clock's mechanism, such as a slightly mis-shaped or corroded cog tooth, that is overriden by the power of a good battery, but provides enough resistance to the weaker power of a near-dead one to stop the clock. If the clock is then moved or disturbed by vibrations, this may jar the tensioned mechanism past the obtruction, restarting the clock for a time. {The poster formerly known as 87.81.230.195} 94.1.209.45 (talk) 09:18, 31 August 2024 (UTC)[reply]
it might be that the relationship voltage-current is non-linear and/or hysteretic. when the clock doesn't get enough juice, it stops drawing any current, the battery "recovers" via the chemical mechanism above, the clock starts drawing current again. the system "clock -- battery" thus oscillates which manifests itself in just such intermittent operation Aecho6Ee (talk) 18:07, 3 September 2024 (UTC)[reply]
some devices do this (poweroff when voltage drops below a threshold and back on when it rises) intentionally as part of "brownout detection" or "undervoltage lockout" Aecho6Ee (talk) 18:14, 3 September 2024 (UTC)[reply]
In the United Kingdom, is room space and building height measured in metric units? And has rail transport seen any metrication in its history? --40bus (talk) 20:38, 30 August 2024 (UTC)[reply]
Rightmove are the UK's biggest online property agents. As you'll see from their website, they quote room space in sq ft but with conversions to sq m. I don't know about rail transport. Mike Turnbull (talk) 20:49, 30 August 2024 (UTC)[reply]
40bus: "On 5 May 1975, rail traffic switched to metric measurements for loads, capacities, tare weights and brake force. On the traffic side of the railway we are chiefly concerned with distance and weight. As to distance, no metrication is planned for the time being... (i.e. distance is still measured in miles, as are road distances). [5]Alansplodge (talk) 16:22, 31 August 2024 (UTC)[reply]
And are there any everyday things that are measured in metric units in the United States, as oppsoed to scientific things? --40bus (talk) 21:11, 30 August 2024 (UTC)[reply]
Ammunition is measured in mm. So are camera lenses. And mechanical pencil lead widths. And a lot of other things that are too small to be comfortably measured in inches -- such as jewelry components. Wrench sets are made for both metric and US customary units. Backpacks and other luggage are sized in liters. We run 5K and 10K races.
A lot of components of food and supplements (carbohydrates, cholesterol, sodium, caffeine, etc.) are measured in grams or mg, especially on standardized nutrition labels -- and they've crept into the vernacular from there. Nobody talks about how many ounces per day of protein or grains of caffeine one should consume. -- Avocado (talk) 02:03, 31 August 2024 (UTC)[reply]
Miles are still used in Norway, but those are metric miles of 10 kilometres. The traditional Norwegian mile was 11298 metres. English miles have obviously never been in common use in Norway, but the abovementioned race is an English mile long. PiusImpavidus (talk) 17:48, 31 August 2024 (UTC)[reply]
Oh, yeah, not saying we don't use customary units for those things at all. Just that we do commonly use metric ones for them, too (such as 9mm for ammo). -- Avocado (talk) 13:53, 31 August 2024 (UTC)[reply]
I hope that the US will metricate at least some everyday things in next 20 years so that US-related articles will use metric units first in 2044. --40bus (talk) 17:25, 31 August 2024 (UTC)[reply]
I don't know everything about British railways, but if I'm to believe openrailwaymap, it appears that some British lines have a speed limit in kilometres per hour and some have a limit in miles per hour that is an obvious conversion from a round number in kilometres per hour. Look at the high speed lines like High Speed 1 (not the older, improved lines with 125 mph limit), DLR and some of the more recent lines of the London Underground. So it looks like a conversion has started. PiusImpavidus (talk) 16:39, 31 August 2024 (UTC)[reply]
Tower Bridge is 130 years old. The sources (and the sources' sources) are therefore far more likely to use Imperial units. The response to the Grenfell fire is contemporary, so better reflects current practices. -- Verbarson talkedits05:06, 1 September 2024 (UTC)[reply]
Not mistaken at all, Verbarson! The operations were performed by hydraulic pressure, supplied by hydraulic accumulators, but "[w]ater at a pressure of 750 psi (5.2 MPa) was pumped into the accumulators by a pair of stationary steam engines" – see Tower Bridge#Hydraulic system. {The poster formerly known as 87.81.230.195} 94.1.209.45 (talk) 21:45, 1 September 2024 (UTC)[reply]
Ah, thank you. When I saw it was 'hydraulic', I somehow assumed it used the London high-pressure network (which was itself steam-powered). What makes it feel most like steampunk is the combination of ancient (Gothic turrets), more recent (suspension spans for the approaches) and up-to-date (steam-hydraulic powered bascules) styles and technologies. -- Verbarson talkedits22:17, 1 September 2024 (UTC)[reply]
Do bears have a cold nose like a dog? Wasn't able to find an answer with Google and finding out myself would be against the policies of my health insurance. --Question123Ka (talk) 04:46, 2 September 2024 (UTC)[reply]
Dog noses are cold because they are cooled by the evaporation of moisture on their noses. Dog noses are moist because dogs lick their noses when they get dry. So the question amounts to, do bears lick their noses, regularly? I suppose your health insurance covers the risks of an investigation if carried out in a zoo or using binoculars or a zoom lens. --Lambiam06:29, 3 September 2024 (UTC)[reply]
I don't go outside. That's where bears live.
But thanks, you helped me: If one specifically searches for bear and wet nose or rhinarium, one might find some results. At least one in Google Books: International Wildlife Encyclopedia, where they say brown bears have a wet nose.
Thanks, that's exactly what I found out too, that polar bears have a cold nose (means: colder than the rest of their body). I'm mostly interested how it would feel to touch a bear's nose. --Question123Ka (talk) 12:32, 3 September 2024 (UTC)[reply]
According to the book "Grizzly Bear Science and the Art of a Wilderness Life," all bears lose heat primarily through their paws, but also through area with limited hair such as their nose and ears. That makes the nose appear warmer than the surroounding face because it is radiating more heat. Checking "Ecology and Behaviour of North American Black Bears," it agrees that all bears lose heat through their paws, ears, and nose. It also states that the grizzy and black bears have a stronger sense of smell compared to other bears and they further accentuate it by using their tongue to force air into their nose in a motion that looks like licking. A side effect of this is that the water evaporates on their wet nose, cooling it down, and allowing more heat to escape. As a result, grizzly and black bears tend to have more and larger blood vessels just under the surface of their nose. Being so close to the Appalacians, we have plenty of books on bears, but most are aimed towards children. I didn't check those because I assume they won't have anything of interest and, usually, they don't have an index where I can quickly look for pages about the bear's nose. 75.136.148.8 (talk) 10:44, 5 September 2024 (UTC)[reply]
Quote from the article: "A potential efficiency loss of only 12.79% can be achieved, or 4.26 kW⋅h/kg out of 33.3 kW⋅h/kg." What is the reason for using 4.26 out of 33.3? Is there some significance to those numbers? I get 12.79 out of 100 but what is special about 4.26 out of 33.3? 196.50.199.218 (talk) 13:49, 2 September 2024 (UTC)[reply]
The section was introduced in this edit, you could try to contact the author (but they haven't edited since 2018). I cannot find those numbers in the reference given. Not sure what to make of that. --Wrongfilter (talk) 14:36, 2 September 2024 (UTC)[reply]
Can anyone explain how it is technically possible for a cat to be simultaneously alive and dead at the same time? I've seen this referenced in pop culture a few times, but I don't understand. 146.200.107.107 (talk) 00:59, 4 September 2024 (UTC)[reply]
If the cat in the box has a philosophical bend and knows Latin as well as the Schrödinger equation and wave function collapse, it might think, Miror si vivo vel non. Ergo vivo. But, actually, we as outside observers do not know whether the cat knows anything. If it does not know whether it is alive or dead, it is more likely dead. The cat may be in a superposition, though, of a state of knowing and a state of not knowing it is alive. --Lambiam08:43, 4 September 2024 (UTC)[reply]
The question is, does a cat count as an observer for quantum physical purposes? And can its quantum state be collapsed inside the box but not yet outside? {The poster formerly known as 87.81.3230.195} 94.6.83.137 (talk) 12:24, 4 September 2024 (UTC)[reply]
To make it clear to others, this was a thought experiment that Schrödinger described to illustrate his opinion that the Copenhagen interpretation of quantum mechanics led to (in his opinion) absurd conclusions: neither he nor any other physicist has proposed carrying it out with a real cat. For one thing, the cat would likely have to be listed as one of the authors of the resulting paper, which would be silly :-). {The poster formerly known as 87.81.230.195} 94.6.83.137 (talk) 04:30, 6 September 2024 (UTC)[reply]
Whenever quantum whatever is mentioned in pop culture it's probably better to just enjoy it as it is, because it's probably completely off the rails.
If you accept that reality is real, then a macroscopic event really happens in real life and the cat really lives or dies, one or the other. So the original trigger to the experiment (a beta decay, say) is probabilistic based on a superposition of states. Note we need to remain ignorant for as long as possible as to whether the particle has decayed if we keep the superposition across the system -- see Zeilinger's quote in the superposition article -- you can see how that would be a problem if you expect the cat itself to be entangled with the particle. The question is, when and by what nature does the system decohere? In some sense, we know how to describe the experiment from when it begins to some midpoint and how to describe it from some midpoint to when it ends, but somewhere in the middle the magician critically slips the other cards into his pocket -- so when did that happen (or maybe it wasn't a card trick to begin with)? SamuelRiv (talk) 18:21, 4 September 2024 (UTC)[reply]
I don't get the other people's responses, but I thought it had to do with multiple universes. That is, after an incident happens to a cat in a box, the scenario is split into 2 additional universes: 1 in which you open the box and the cat is dead, and the other where you open the box and the cat is alive. 66.99.15.162 (talk) 20:55, 4 September 2024 (UTC).[reply]
At the infobox for promethium(III) chloride we find two different sources for this compound's colour, one calling it purple and the other calling it yellow. Probably few have seen this compound (considering the short half-life of Pm), but what's the story here? Is it perhaps similar to how the colour of NdCl3 varies under lighting conditions (between purple and yellow as well)? Double sharp (talk) 04:40, 4 September 2024 (UTC)[reply]
I would expect most trivalent compounds have similar colour. But there may also be radiation damage to the material. Also perhaps the compound is not pure and contains neodymium. Graeme Bartlett (talk) 21:49, 6 September 2024 (UTC)[reply]
Why is it called "stress energy tensor", although its components intended to refer to the energy - actually refer to the energy divided by the speed of light squared - i.e. actually refer to the (relativistic) mass? HOTmag (talk) 13:14, 6 September 2024 (UTC)[reply]
The article about real-life dilithium has the potential to be quite interesting, but because of search result overlap with the Star Trek substance, it's difficult to nail down references on this topic. We could use some to establish - for example - if it's a gas at standard temperature and pressure, and if it's "stable", what its typical lifetime is. I would be interested if anyone could put their fingers on sources with this sort of info. -- Beland (talk) 08:22, 7 September 2024 (UTC)[reply]
Alkali metals tend to form dimers in the gas phase. But they will condense back into metallically bonded structures when cooled below the boiling point. Double sharp (talk) 08:25, 7 September 2024 (UTC)[reply]
@Beland The general way to find decent references is to use Google Scholar. This search is a start and removes some hits from articles about a piece of software called dilithium by searching for co-occurence of that word with "lithium". Mike Turnbull (talk) 11:05, 7 September 2024 (UTC)[reply]
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