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January 3

visibility of Venus by latitude and season

Venus is visible at most N minutes from sunset/sunrise. (For the purpose of this question, pretend that it is always at maximum elongation.) Does N vary with latitude and/or season? My geometric intuition fails me. —Tamfang (talk) 06:48, 3 January 2024 (UTC)[reply]

It does. Let's just pick an extreme case: At the North Pole the celestial equator coincides with the horizon, while the ecliptic (the apparent path of the sun and Venus [ignoring a small inclination] with respect to the stars) is inclined at 23.5 degrees with respect to equator/horizon. It can then happen that the sun is below the equator/horizon (that'd be northern winter) while Venus is above it. In that situation, the Sun never rises and Venus is always up. With the same constellation viewed from a lower latitude, there will be a finite time between the rising of Sun and Venus. Going through all possibilities is pretty tedious and exhausting, but in general, try to think of three circles: the horizon; the celestial equator, which at latitude φ is always inclined at 90°−φ with respect to the horizon, intersecting the latter due east and due west; and the ecliptic, which hosts both Sun and Venus, is inclined at 23.5° with respect to the equator and has variable inclination with respect to the horizon. That's not sufficient to deduce N, which should also depend on the declination of Sun and Venus, I believe (ideally we should find or produce a graph). I can think of another interesting situation that occurs at the polar circle (inclination of the equator 23.5°). There it can happen (once per day, actually) that the ecliptic coincides with the horizon so that all objects on the ecliptic rise or set simultaneously. --Wrongfilter (talk) 09:48, 3 January 2024 (UTC)[reply]

Apart from the question whether Venus is visible if just half a degree above the horizon (I've seen Venus very low in the sky, but that was from an aeroplane 30,000 feet over the Congo)...

Time of sunset depends on your latitude and on the declination of the sun, plus some small term to account for the non-uniform motion of the sun in right ascension, but this is the same for Venus. Time of Venus-set depends on your latitude, the declination of Venus, its elongation and the same small term. The declinations of Venus and the sun are correlated, but not the same. Venus, like all planets, stays close to the ecliptic. When Venus is visible at maximum elongation and the ecliptic makes a steep angle with the horizon, it must be at higher declination than the sun (assuming an observer in the northern hemisphere), making it visible for longer. Around the autumn equinox at sunset and the spring equinox at sunrise, the ecliptic makes a steep angle with the horizon, so that Venus is visible for longer; around the spring equinox at sunset or the autumn equinox at sunrise, it makes a shallow angle with the horizon, making Venus visible for shorter. At high latitude, this variation is stronger than near the equator. There's another correction for the fact that the orbital plane of Venus doesn't coincide with that of Earth; that's an almost seasonal effect – it changes with precession. Then we have some small corrections for the eccentricity of the orbits, another almost seasonal effect. PiusImpavidus (talk) 10:12, 3 January 2024 (UTC)[reply]

The radioactive of the natural occurring elements

e.g. natural potassium is ~10,000,000 times more radioactive than natural bismuth (32 decays/s per 1 g of K, ~100 decays/year per 1 g of Bi), and natural calcium and natural barium are weakly radioactive (from trace Ca41, and long-lived Ca48 and Ba130), thus for the elements 1 to 98 (they are the natural occurring elements, although some elements such as technetium and neptunium, are only trace), these natural occurring elements from more radioactive to less radioactive, is? 211.20.197.240 (talk) 09:24, 3 January 2024 (UTC)[reply]

The naturally occurring elements don't stretch to 98 these days, unless by "nature" you mean "anywhere in the Universe" rather than "on Earth", in which case they stretch beyond that every time an r-process event happens (relevant paper). All of Earth's elements beyond 94 have long since decayed since Oklo stopped functioning. Maybe traces of 96 (as ²⁴⁷Cm) could still be brought by supernovae, but no one found any (and good luck getting beyond that with the number of neutron captures required). See this paper.

Answering this question would imply a whole lot of work finding the abundances of trace cosmogenics and fission products. :) Maybe helium is not a bad answer for the least radioactive natural element, as it only has four bound isotopes – two stable and two living under a second – and the two unstable ones are unlikely to be present significantly in nature. (Maybe lithium also comes close?) As for the most unstable natural element, surely it must be astatine. (The longest-lived At isotope is ²¹⁰At, and it is more stable than ²²³Fr, but it's also only synthetic. The longest-lived natural At isotope is ²¹⁹At, less stable than ²²³Fr.) Double sharp (talk) 11:25, 3 January 2024 (UTC)[reply]

Again - you appear to be asking others to do a great deal of work compiling information for you. "What is the naturally occurring element with the highest average activity?" is a reasonable question for this board. "Please rank ALL the naturally occurring elements in order by average activity." is not. PianoDan (talk) 17:22, 3 January 2024 (UTC)[reply]

Well, I just want the four tables like the four tables in List of elements by stability of isotopes (Elements with primordial isotopes, even/odd proton numbers, sorted by number of stable nuclides and primordial radioactive nuclides: Elements with no primordial isotopes, even/odd proton numbers, sorted by the half-life of the longest-lived nuclide), but for the neutron numbers rather than the proton numbers, also a table like this for the neutron numbers rather than the proton numbers. 61.224.143.214 (talk) 05:08, 4 January 2024 (UTC)[reply]

Sounds like you have the data (or leads to finding it). Time for you to make a table in your preferred format. Import copy/import into Excel and then you can process and organize it all sorts of ways. DMacks (talk) 18:22, 4 January 2024 (UTC)[reply]

Someone partially answered this question on ScienceMadness, FWIW (see the post from 17 February 2019 at 07:36). By activity the order goes U, Th, Re, Rb, Sm, Lu, K; everyone else is below 1 Bq/g. Though due to extremely low decay energies, practically speaking an amateur cannot detect Re or Sm.

Detection of ⁴⁰K, ⁸⁷Rb, ¹⁷⁶Lu, and ¹⁸⁷Re was done in this recent YouTube video. Double sharp (talk) 18:14, 21 January 2024 (UTC)[reply]

Collision between matter and antimatter

When an electron and a positron - or a quark and an antiquark - collide with each other, the whole massive matter involved can entirely vanish and become massless particles - e.g. photons or gluons respectively - as the final state.

Question: Empirically speaking: can this process - of entirely annihilating the whole massive matter - happen as a final state, also in any collision (at least one) between particles - some (or all) of which are stable massive particles other than an electron/quark (and than its respective antimatter)?

By stable massive particles, I also include - subatomic particles like protons/neutrons - as well as their respective antimatter. I don't exclude more composite stable matter (like stable atoms) either. Anyway, I'm only asking about facts, i.e. about what's backed - by empirical evidence - or at least by a well established conclusion derived from empirical evidence. HOTmag (talk) 17:26, 3 January 2024 (UTC)[reply]

The short answer is no - baryon number and lepton number are conserved quantities. So for example, if you fire two protons at one another (as in the LHC), you will need to end up with a final state that still has a total baryon number of +2. There are some exotic potential asterisks to both of those conservation laws, but if you want to completely annihilate your collision inputs, you need all of your conserved quantities other than energy and momentum to be net zero. PianoDan (talk) 19:45, 3 January 2024 (UTC)[reply]

I guess HOTmag wants to collide a proton with an antiproton, so the total baryon number is 0. See Annihilation#Proton–antiproton_annihilation, which confirms my suspicion that at best you get a quark-antiquark annihilation while the remaining constituents dance a complicated ballet. I guess it's not excluded that there is a channel that ultimately leads to just photons but the probability for that to happen seems very small. --Wrongfilter (talk) 20:09, 3 January 2024 (UTC)[reply]

You interpreted me correctly, and actually anybody else can very easily interpret me correctly as you did - once they read the title (or the first paragraph under the title).

As for your answer: Note that not only "photons" (as you say), but also gluons - as happens in a quark-antiquark collision. Additionally, not only "proton" (as you say), but also any stable massive particle/body other than an electron/quark (and than its respective antimatter), including neutrons, atoms, and the like. Anyway, as I've alraedy indicated in my question, I'm still looking for a well based "emprical evidence" for your hypothesis about this "channel" (as you call it). HOTmag (talk) 21:05, 3 January 2024 (UTC)[reply]

Please stop lecturing us on what you think we do and do not understand. I really don't know why I bother. --Wrongfilter (talk) 21:23, 3 January 2024 (UTC)[reply]

Please notice I've never said anything about what anybody "understands". I've only said that you had "interpreted" me correctly, and I said that because I realized that there was a (seemingly) disagreement about what I (being the OP) had meant, so I wanted future users to interpret me according to your interpretation rather than according to other interpretations. This was my only motivation for indictating that you had interpreted me correctly. Not any other motivation.

Anyway, in my view, there is no difference between all of us, as far as our "understanding" is concerned. In other words, all of us are equal, as far as our "understanding" is concerned. HOTmag (talk) 21:37, 3 January 2024 (UTC)[reply]

Per the Standard Model, ALL stable massive bodies are either made of quarks or leptons (or their anti-particles) - there aren't any other possibilities. "Neutrons, atoms, and the like" are still just collections of quarks and leptons.

So there's no such thing as a collision of stable massive particles that isn't either a collision of leptons, quarks, or their anti-particles, because there's literally nothing else stable that has mass.

As far as evidence goes - you could look at the reference list on the Standard Model page. The empirical support for the model is basically best described as "all of twentieth century experimental physics," so it's hard to point you to A single source. Any reputable textbook on Modern Physics would be a good starting point. PianoDan (talk) 22:16, 3 January 2024 (UTC)[reply]

To me, the Standard Model has always been a reliable source. But unfortunately it says nothing about whether, only collisions between an electron/quark and its antimatter, annihilate the whole massive matter involved. Check a neutrino: It falls under my original definition, since it's a stable massive particle other than an electron/quark, so what about it? I guess, this lepton is never annihilated when it collides with its anti-particle, but this is only my guess - which I could never prove from our well known Standard Model. Additionally, what about a proton/neutron colliding with its anti-particle? Indeed, as you say, a proton/neutron is made of quarks; However, subatomic particles like protons/neutrons are not elementary particles as quarks are: That's why, while a quark and an antiqurak - colliding with each other - may become gluons (alone), this is not the case when a proton and an antiproton collide with each other - as far as the empirical evidence tells us. This is why I can still ask, if also collisions between a proton/neutron and its anti-particle annihilate the whole massive matter involved (even though it doesn't become gluons alone). HOTmag (talk) 22:52, 3 January 2024 (UTC)[reply]

Whether neutrinos are their own antiparticles—antineutrinos—is currently an open question in physics. Nobel waiting for you there if you can answer that. If true this would mean neutrinos are Majorana particles, and could therefore be involved in processes that fail to conserve lepton number.

Below an extremely high temperature, free (anti)quarks are not possible, because of color confinement. Quarks exist only as bound states within hadrons, or as virtual quarks. Slowking Man (talk) 05:10, 4 January 2024 (UTC)[reply]

There have been and are a lot of experiments aimed at answering whether neutrinos are Majorana particles: see Double beta decay#Neutrinoless double beta decay. So far no evidence of that has been found. Double sharp (talk) 05:15, 4 January 2024 (UTC)[reply]

Regarding neutrinos: Yes, I'm aware of the dispute about whether the neutrino has an anti-particle. That's why I didn't mention the neutrino in my question. I've only mentioned it in my clarification to another user, as an aside remark, intended to clarify why the whole issue is not that simple as they might have thought it was.

Regarding quarks: Yes, I'm aware of that. Anyway, see our article annihilation:

"when a proton encounters an antiproton, one of its quarks, usually a constituent valence quark, may annihilate with an antiquark (which more rarely could be a sea quark) to produce a gluon..."

"fusion of two gluons (via annihilation of a heavy quark pair)".

HOTmag (talk) 09:25, 4 January 2024 (UTC)[reply]

There was actually proton-antiproton collider called Tevatron, which produced a lot of interesting staff including t quark. So no, hardron-anti-hardron collisions are highly unlikely to result in only light being emitted. Ruslik_Zero 14:40, 5 January 2024 (UTC)[reply]