The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
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 of promethium 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]
It seems that your last sentence contains some logical mistake. That's why I couldn't figure out your answer.
Anyway, don't you agree, that whenever any component of this tensor refers to the energy of a given body - this component actually refers to the energy divided by the speed of light squared - i.e. actually refers to the (relativistic) mass? HOTmag (talk) 18:49, 7 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]
IIRC all the alkali metal dimers are known in the gas phase (not sure how much of the vapour is monatomic vs diatomic, though). 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]
Radioactive elements and their isotopes release, for example, alpha particles of various energies. These ionise surrounding material and the colours come from the ions relaxing back to their ground states. There is a chart which gives the energies (zoom in to the isotope of interest). Mike Turnbull (talk) 12:16, 7 September 2024 (UTC)[reply]
Sunshine duration says that locations on the Arctic Circle get 4,647 hours of sunshine (disregarding the effects of clouds), the most of any location worldwide due to the effects of atmospheric refraction. Imagine that Surtsey-type eruptions produce a new mountain in an ocean location on the Arctic Circle, and it's extremely high — say, 5000m. Disregarding the effects of clouds, will it get much more sunshine than locations at sea level? I assume it will get some more than those locations, since the sun's above the horizon longer for an elevated location, but I don't know how much longer. Nyttend (talk) 21:49, 7 September 2024 (UTC)[reply]
It doesn't have to be on the Arctic Circle, you could build a (thought experiment) 100,000 km tall tower on the Equator, and as the Earth rotated, the top of the tower would only briefly be in Earth's shadow, if at all. Abductive (reasoning)05:14, 8 September 2024 (UTC)[reply]
Let denote the radius of the Earth (taken to be a sphere) and let stand for the obliquity of the ecliptic. Then, for the top of a tower erected at latitude to avoid the shadow cast by the earth day and year around, its height should be at least
On the equator, with this comes out at about a mere 700 Burj Khalifas stacked on top of each other.
At the poles, with we'd need much higher than Low Earth orbit, so the tower may be hit by satellites in polar orbit.
On the arctic circles, with the tower needs to be infinitely high, which is impractical from an engineering point of view :).
I dunno if that makes sense. Consider Saturn's rings; they are often (always?) partly in shadow. But a tower on the axis will stick out into perpetual sunlight at a height much less than the rings. Abductive (reasoning)14:55, 8 September 2024 (UTC)[reply]
The rules for the systematic element names, approved by IUPAC in 1978, were designed solely for the purpose of assigning temporary names to unknown or not-yet-named chemical elements. Since all such elements have an atomic number greater than 100, the system only caters for such higher numbers, up to 999 (Ennennennium). The extension to other natural numbers for elements with IUPAC-approved names, while obvious, does not carry the IUPAC stamp of approval. Note that Nilium (a potential name for proton-free muonium or neutronium) and Quadium have already been given away. --Lambiam08:31, 8 September 2024 (UTC)[reply]