Is it right that every solid matter in this universe including liquid and gas came from energy and in beginning there was no matter? ~2025-38835-24 (talk) 04:18, 6 December 2025 (UTC)[reply]

It is not correct. The origin of the universe, and of the Earth, is not known with any confidence. There are many theories as to the origin of matter and the origin of energy. Dolphin (t) 08:22, 6 December 2025 (UTC)[reply]

The definitions of the concept of matter vary considerably; see Matter § Definition. But all agree on the involvement of elementary particles. According to the prevailing physical theory of cosmogenesis, during the Planck epoch – the time within 10⁻⁴³ seconds of the Big Bang – energy was so concentrated that the concept of particle of quantum physics breaks down. So there was nothing yet that could be understood in current physics as being a particle. Compare this to what happens when you strike a bell with a hammer. If it is a good bell, you'll hear a clear musical tone, which is actually not a pure tone but more complex. But for the first millisecond after striking the bell, there is a loud (very energetic) sound, but nothing yet that could be called a tone. It takes some time for a tone to emerge from the initial chaos.  ​‑‑Lambiam 12:20, 6 December 2025 (UTC)[reply]

I drew this diagram based on http://nesdis.noaa.gov/news/hurricanes-cyclones-and-typhoons-whats-name but am unsure where to put the boundary between cyclones and typhoons in the South China Sea (so made it a gradient). Would anyone here know exactly? Thanks, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 01:51, 7 December 2025 (UTC)[reply]

Sources agree that typhoons are associated with the Northwest Pacific, and the South China Sea is part of the Pacific Ocean. Nothing in the NOAA diagram contradicts that as far as I can see, and the article that goes with the diagram consigns cyclones to "the western South Pacific and Indian oceans" (no mention of North Pacific).

On an unrelated note, there doesn't appear to be a definition for storms in the South Atlantic; the NOAA says they're hurricanes, whereas we have South Atlantic tropical cyclone. Clarityfiend (talk) 09:45, 7 December 2025 (UTC)[reply]

Thanks very much, @Clarityfiend: I would've thought so, but for Cyclone Senyar which has a track on the South China Sea. Is there any reason for this exception? cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 14:40, 7 December 2025 (UTC)[reply]

When the Joint Typhoon Warning Center issued a warning for Cyclone Senyar, at the time referred to as as Tropical Cyclone 04B but named "Senyar" a few hours later, it was centred over the Strait of Malacca, in the "Cyclone" area. When the system crossed into the "Typhoon" area, it was as a tropical depression. I don't know if it would have been renamed if it had gained storm strength again.  ​‑‑Lambiam 16:06, 7 December 2025 (UTC)[reply]

Thanks, I'll use the Malay Peninsula as the boundary then.  Done cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 20:39, 7 December 2025 (UTC)[reply]

@Cmglee: That leaves the South China Sea south of the equator in a sort of naming no-man's-land. Clarityfiend (talk) 09:53, 7 December 2025 (UTC)[reply]

It is very rare for a tropical storm to develop that close to the equator.^[1] List of tropical cyclones near the Equator does not list any from that area.  ​‑‑Lambiam 16:26, 7 December 2025 (UTC)[reply]

The first line of our article Typhoon states "A typhoon is a tropical cyclone that develops between 180° and 100°E in the Northern Hemisphere and . . . ."

As a former resident of Singapore (103°E) I can confirm that the name 'typhoon' is used there. [Edited to add} This would extend the 'pink zone' to halfway up the Strait of Malaysia. {The poster formerly known as as 87.81.30.195} ~2025-31359-08 (talk) 09:58, 7 December 2025 (UTC)[reply]

The designation Typhoon is also used in England,^[2] while Cyclone can be found used in Singapore, like for Cyclone Senyar mentioned above.^[3] The choice of designation does not depend on the location of the utterer but on that of the storm, in particular its location at the time it developed hurricane-level strength and was named by a warning centre.  ​‑‑Lambiam 14:52, 8 December 2025 (UTC)[reply]

In Kansas it is a cyclone - or so 'The Wizard of Oz' describes it! NadVolum (talk) 21:55, 7 December 2025 (UTC)[reply]

That's talking about a tornado, not a hurricane. --~2025-39159-56 (talk) 00:50, 8 December 2025 (UTC)[reply]

An old-fashioned term, in common American usage in the 19th century. It's how Cy Young got his nickname. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:58, 9 December 2025 (UTC)[reply]

Thanks for the insights, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 01:04, 19 December 2025 (UTC)[reply]

Do they occur naturally or are they only synthesized? --Preceding unsigned comment --Preceding unsigned comment 01:27, 9 December 2025 (UTC)[reply]

If you take an indole molecule, whose structure is shown here, and replace one of the carbon atoms in its benzene ring by a nitrogen atom, you get an azaindole. Or this can mean, more generally, a compound derived from a simple azaindole by replacing one or more of the hydrogen atoms by substituent groups.

There are both naturally occurring and synthetic azaindoles.^[4]  ​‑‑Lambiam 02:20, 9 December 2025 (UTC)[reply]

Also many are known as pyrrolopyridines, see Category:Pyrrolopyridines for some articles here. Graeme Bartlett (talk) 03:35, 9 December 2025 (UTC)[reply]

✔️ Thank you. --Preceding unsigned comment 23:38, 9 December 2025 (UTC)[reply]

I need to find a video for a fresh cadaver. Starting from the straight beginning of skin removal (including cut of the bone).

This would be for psychatric use. ~2025-39568-62 (talk) 07:28, 9 December 2025 (UTC)[reply]

https://www.youtube.com/results?search_query=skull+cadaver+dissection ~2025-39358-87 (talk) 00:58, 10 December 2025 (UTC)[reply]

Let's say you've got a relativistic starship, i.e. it can go very close to the speed of light but not faster than light. You point it "out there" and floor the accelerator, moving further and further from Earth, sending telemetry back by radio continuously. The telemetry will become more redshifted as the ship speeds up by its own engine power, and also because of the expansion of the universe.

Is it possible for the ship to travel far enough that its proper distance from Earth is increasing faster than light, which means that telemetry being sent from then on can never reach Earth? I haven't been able to find an astronomy term for this, but basically the ship in a sense becomes topologically disconnected from Earth's region of space. Of course(?) since the ship moves slower than light in its local frame, even if it can somehow reverse direction, it can't become visible again.

Is there a name for this phenomenon? Observable universe mentions "reachable universe" which is related, but surprisingly there's no article about that. Thanks. ~2025-39358-87 (talk) 01:09, 10 December 2025 (UTC)[reply]

Added: cosmic event horizon is pretty close to what I mean. I'm asking if there is some obstacle to crossing it with a slower-than-light spaceship. ~2025-39358-87 (talk) 01:37, 10 December 2025 (UTC)[reply]

I think not. According to the mathematics of Special relativity, velocities cease to be fully additive (if I can put it that way – see Special relativity#Lorentz transformation of velocities) to a greater and greater degree as the speed of light c is more closely approached, such that c itself can never be reached, much less exceeded. Thus the ship will never appear to reach c from the point of view of Earth, and vice versa (although extreme redshift may make it unfeasible for either to continue detecting the other). ~2025-31359-08 (talk) 04:34, 10 December 2025 (UTC)[reply]

If "reachable universe" means, "all parts of the universe that can be reached", then it is impossible, by definition, for a starship to reach a point in spacetime outside the reachable universe. Unless, of course, it uses a Heineken drive.  ​‑‑Lambiam 09:12, 10 December 2025 (UTC)[reply]

You can probably reach a point from where you can no longer reach Earth, but this depends on the future rate of expansion of the universe, which is unknown. For two extremes, in a Big Crunch you will return to Earth, no matter how hard you try no to, and in a Big Rip there will be a time where your left hand goes out of reach of your right hand. PiusImpavidus (talk) 09:39, 10 December 2025 (UTC)[reply]

Thereby fulfilling the requirement of Matthew 6:3.  ​‑‑Lambiam 00:52, 12 December 2025 (UTC)[reply]

Lambiam, stuff leaves the reachable universe all the time, due to cosmic expansion, from what I understand. It looks like a point around 4GPc (gigaparsec) from here will be receding at close to c, so anything past that is unreachable. So a star that's barely reachable today could become unreachable tomorrow. PiusImpavidus's point about a Big Rip answers my question, I think. If the expansion of the universe is fast enough, stuff becomes unreachable without having to become terribly distant or fast. But with today's expansion rate, I guess it will take the starship > 13 billion years to get there, a long enough trip that unknowns might predominate. Thanks everyone. — Preceding unsigned comment added by ~2025-39770-07 (talk) 12:24, 10 December 2025 (UTC)[reply]

Jens E. Matthiesen writes "An increasing number of galaxies become invisible to astronomers over time, because the universe is expanding. In the future, the light from distant stars will never reach us, so no matter how good our telescopes get, the universe will become ever darker." in "Why we will never see the end of the universe: space is expanding". I can imagine that some creatures on one side of that distant galaxy built a starship that took a few million years to slowly travel to the other side of that galaxy (perhaps at 0.01 c), at a time where if we're lucky (unlucky?) we on Earth may catch photons from the beginning of it's journey, but when the starship stops at it's destination, the photons signaling its arrival would never reach Earth. Since that starship can cross the cosmic event horizon, it seems plausible that a much faster starship should also be able to cross the cosmic event horizon. --DavidCary (talk) 19:49, 11 December 2025 (UTC)[reply]

When stuff "leaves the reachable universe", and "all the time" at that, it is not a starship that we have built. Like each observer has their own observable universe, of which each one occupies its centre, so does each observer have their own reachable universe, a subset of spacetime that only shrinks as time goes by. Stuff may leave our reachable universe, but in doing so it remains in its own reachable universe.  ​‑‑Lambiam 01:02, 12 December 2025 (UTC)[reply]

I feel a song coming on: "To reach the unreachable star. This is my quest." Clarityfiend (talk) 23:10, 12 December 2025 (UTC)[reply]

Clarityfiend, haha, that song needs a cosmology-based parody ;). Lambiam, that's what I'm getting at (or asking, or trying to confirm). Let's say point Y is within the reachable universe of point X at time T1 (I guess that means spacetime event (T1,X)). Starship leaves (T1,X) and eventually arrives at (T2,Y). On arrival, it sends a radio message M in the direction of point X. Question: are we sure that M can in principle reach X? Or can cosmic expansion mean that Y is now beyond X's cosmic event horizon and M could only reach X by travelling faster than light? I don't know what X's time coordinate would be in that picture. I don't understand relativity well enough.

I also don't see why the expanding universe is necessarily getting darker, if new stars can form in the newly expanded space. I've heard that mass and energy conservation happen because of Noether's theorem and symmetrics in the Lagrangian of stuff moving around in space. But if space is expanding, the symmetries don't hold in that context, so the conservation laws don't hold either, and energy (which eventually condenses into matter) can appear out of nowhere. Maybe the expanding universe really does get darker, but that seems like a nontrivial consequence of some messy calculation, rather than something obvious. ~2025-40343-80 (talk) 03:53, 13 December 2025 (UTC)[reply]

New stars can form, concentrating interstellar gas and dust into themselves, but although some of them eventually end in novae, returning some of that material to space, and a smaller number in supernovae returning a larger proportion of it, most do not and eventually cool into inert bodies, resulting in the long term in less material being available to form further new stars.

Any quantum fluctuations that may result in 'new energy' (see Vacuum energy) seem to do so in insignificant amounts compared to the existing matter and energy in the universe (as far as I know).

The net result is that as space expands and material crosses the boundary of the visible universe, that visible universe's matter-energy density falls, until eventually only the material in Galaxy clusters that are Gravitationally bound will remain visible to any observer within them.

This is assuming that the expansion always continues (though it may slow asymptotically) rather than stopping, or even reversing in a 'Big Crunch' scenario. Determining which of these possible outcomes is going to happen is a major goal of Cosmology. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 09:10, 13 December 2025 (UTC)[reply]

Assume a photon can travel from A to B, where B is at the far extreme of what is reachable from A. Can a mirror at B send it back to A, or has A now become unreachable? A problem in discussing is, is that A and B do not share an inertial frame of reference. In the narrative, "A" and "B" do not stand for points in spacetime but for world lines. Take a view on the universe in which A and B are each other's mirror image in a spatial mirror symmetry. Let A₀ be the point in spacetime where the photon commences its long journey to the (barely reachable) word line B, and let B₀ be its mirror image. In this view, by symmetry, world line A is at the far extreme of reachability from spacetime point B₀. By the time the photon reaches B, at say spacetime point B₁, it is later at B than it is at B₀. In the meantime, the ever-expanding universe has expanded further and the area of reachability has correspondingly shrunk, so, indeed, world line A is unreachable from B₁. So don't hold your breath for an "I safely arrived" message.  ​‑‑Lambiam 14:55, 14 December 2025 (UTC)[reply]

if new stars can form in the newly expanded space

They can't unless matter eventually travels into that new space from elsewhere. Space expands only where it's essentially empty. And expanding space doesn't create new matter in the new space, just more space between already-existing matter. -- Avocado (talk) 15:45, 14 December 2025 (UTC)[reply]

Hi all, I can do a number of things, but maths isn't one of my strong points. Could someone please check my answer?

If an engine produces 320 PS (metric horsepower), and is fitted with lower compression pistons to reduce its output to 265 PS, what is the power loss expressed as a percentage? I make it around 17.2%, thus:

265 / 320 = 0.828125
1 - 0.828125 = 0.171875
0.171875 x 100 = 17.1875 or about 17.2%.

Is this right? Cheers, MinorProphet (talk) 06:17, 10 December 2025 (UTC)[reply]

• Yes. --~2025-39752-49 (talk) 07:28, 10 December 2025 (UTC)[reply]

Brilliant, thanks. MinorProphet (talk) 09:35, 10 December 2025 (UTC)[reply]

Another way to think about it (and the way that I would have done it) is:

320-265 = 55 (power loss of 55 PS)

55/320 = 0.171875 (ratio of loss to original value)

And then, as you did above, multiply by 100 to get the percent. -- Avocado (talk) 12:59, 10 December 2025 (UTC)[reply]

Aha! Reduction of number of steps by 33.3%! Can I accuse you of sounding like a C [etc.] programmer? Or a bash aficionado/aficionada? Hmmm, Avocada? MinorProphet (talk)

Lol! I'm a programmer, but haven't touched C in... well, longer than I've had this account. Fwiw, I do think it's the same number of steps (I omitted the decimal to percent conversion step that you'd already done above). I just find it easier to reason about this way around. -- Avocado (talk) 20:08, 10 December 2025 (UTC)[reply]

Since the engine performance was deliberately changed the calculated quantity is more correctly a 17.2% reduction than a loss. ~2025-32374-12 (talk) 13:57, 10 December 2025 (UTC)[reply]

Actually, its wasn't the engine manufacturer's choice: it was foisted upon them. See my draft User:MinorProphet/Draft subpages/Engine power which attempts to make sense of this restriction.

But if maximum performance is the aim, shurely some fiend of entropy must enter stage left, unbidden. MinorProphet (talk)

SFX: < Offstage trumpets, ff >

Nevertheless, if there were one and only one way to do anything and everything, there would be no need for the hallowed halls of the Ref Desks at all. :> MinorProphet (talk) 19:00, 10 December 2025 (UTC)[reply]

Taking the current US president as my shining example, when I come to power I will decree that henceforth there will be one and only one way to do anything and everything, viz. the way I decree. Anyone who complains, objects, or refuses to comply, will be shot as a traitor. What could be simpler, fairer, or more patriotic than that? -- Jack of Oz ^{[pleasantries]} 20:51, 10 December 2025 (UTC) [reply]

I understood the OP's phrasing "is fitted with lower compression pistons [in order] to reduce its output" to imply a conscious deliberate choice was made. An irreversible increase in entropy, meaning that energy in work-capable form is reduced, occurs in both versions of the combustion engine and more so in the less efficient version that must dissipate more of the fuel energy as heat. ~2025-32374-12 (talk) 20:14, 10 December 2025 (UTC)[reply]

I don't know what "lower compression pistons" are and how they work. Is more fuel consumed for the same amount of work, or is fuel consumption reduced, thereby reducing power?  ​‑‑Lambiam 00:22, 11 December 2025 (UTC)[reply]

Sorry, I'm going to be busy for several days. Check out lowering the compression ratio of an engine, e.g. if you're intending to fit a turbo. Dished pistons, etc. The piston doesn't even have to rise to the top of the bore at top dead center. In my case it's about coping with a reduced octane rating as a wartime measure. Engines with high compression ratios need a high octane rating to avoid 'knock', and there are various ways to reduce the compression ratio in a cylinder. For example milling out the 'bowl' in the piston crown, fitting shorter conrods etc., fitting thicker head gaskets, increasing the size of the combustion chamber in the cylinder head. These are relatively drastic measures. Sleep beckons... MinorProphet (talk) 04:18, 11 December 2025 (UTC)[reply]

The HZ (habitable zone) of the star TOI-700 seems to be occupied only near the inner edge by two known planets "d" and "e". This leaves space in the HZ for any additional planets, which might include Earth-like planets. Where is the region of stability (in terms of semi-major axis/ AU) for these planets, so that their mass does not perturb the orbits of four known planets around TOI-700? (Assuming these hypothetical planets are Earth-like in mass)SoojinHD219134star (talk) 15:59, 10 December 2025 (UTC)[reply]

I am not sure that anybody has done such calculations. Ruslik_Zero 20:04, 11 December 2025 (UTC)[reply]

The page Indigestion says "Indigestion is relatively common, affecting 20% of people at some point during their life". Is that really true? I thought everyone got it occasionally. ~2025-39925-85 (talk) 21:16, 11 December 2025 (UTC)[reply]

The key question would be whether the alleged sources of that statement elaborate as to "how they know that". ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:17, 12 December 2025 (UTC)[reply]

The article cites two sources, referenced with ^[1] and ^[5].

The first (^[1]) states,

At some point in their lives, one in five adults report epigastric pain, early satiety, postprandial distress, and other associated upper gastrointestinal symptoms, such as heartburn, regurgitation, or nausea. Although dyspepsia is not associated with higher mortality risk,^{2 3} ...

The cited sources ² and ³ are of longitudinal studies. It appears that the statements are supported by at least two solid studies.

The second (^[5]) states:

Dyspeptic symptoms are very common in the general population, affecting an estimated 20% of persons in the United States.¹

The cited source is:

¹ Heading RC. Prevalence of upper gastrointestinal symptoms in the general population: a systematic review. Scand J Gastroenterol Suppl. 1999;231:3–8.

I have not further examined the sources of this review, but it was published more than a decade before the above ² and ³, so it must be based on yet more studies.

One caveat is that the percentage may not reflect a global situation. Study ² concerns the UK, while ³ and the review article concern the US, both part of the affluent West with similar food habits. Also, the wording of the article citing the review study seems to suggest that the 20% estimate applies synchronically, rather than "at some point during their life", which would be the case for a higher fraction of people.  ​‑‑Lambiam 00:46, 12 December 2025 (UTC)[reply]

Isn't dyspepsia only a subset of types of indigestion? -- Avocado (talk) 01:10, 12 December 2025 (UTC)[reply]

It strikes me that the wording ". . . one in five adults report epigastric pain . . . [my emphasis]" is significant.

I suspect that many or most people who experience indigestion do not 'report' it to a physician or anyone else, but (like myself) either endure it silently, or purchase non-prescription remedies, or use home remedies. A well-conducted survey of the general population asking if people had ever suffered indigestion would likely return a much larger figure than 20%. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 14:51, 12 December 2025 (UTC)[reply]

I suppose that the cited studies used questionnaires with such questions like, How often do you experience a burning sensation behind your breastbone? □ Almost never | □ Sometimes | □ Regularly | □ Often.  ​‑‑Lambiam 22:03, 12 December 2025 (UTC)[reply]

I have supposed that they didn't, you have supposed that they did. Anyone else care to settle the matter? {The poster formerly known as 87.81.230.1095} ~2025-31359-08 (talk) 09:13, 13 December 2025 (UTC)[reply]

You questioned the significance of the quoted statement, since the "one in five" statistic becomes meaningless if based on contemporaneous reporting of the complaint to a physician, under the somewhat unlikely assumption that such a statistic can be compiled from the records of consulted physicians. If you desire to delve deeper into the matter, here is a link to an open-source version of one of the cited studies:

³: "Impact of Functional Gastrointestinal Disorders on Survival in the Community".

 ​‑‑Lambiam 11:45, 13 December 2025 (UTC)[reply]

I wish to illustrate how fluoride ions replace hydroxyl ones in water fluoridation. I've found two different structural formulae of hydroxyapatite, Ca₁₀(PO₄)₆(OH)₂ on https://www.researchgate.net/figure/Molecular-Structure-of-Hydroxyapatite-Ca10PO46OH2-Created-by-ChemSketch_fig3_380182120 and https://www.researchgate.net/figure/Chemical-structure-of-hydroxyapatite-Ca-10-PO-4-6-OH-2_fig8_47933950 – which, if either, is the one in tooth enamel?

Thanks, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 01:46, 13 December 2025 (UTC)[reply]

the diagrams in https://www.researchgate.net/publication/47933950 are far too simplified and does not even have the hydroxy included. The figure in https://www.researchgate.net/publication/380182120 shows how the phosphate arranges itself around one calcium ion, and does show hydroxy groups, but does not indicate the 3D structure. But you should be able to use the latter to see the substitution of OH with F. Both these anions are similar in size and charge. Graeme Bartlett (talk) 10:58, 13 December 2025 (UTC)[reply]

@Graeme Bartlett: Thanks for your reply. After drawing a diagram based on https://www.researchgate.net/figure/Molecular-Structure-of-Hydroxyapatite-Ca10PO46OH2-Created-by-ChemSketch_fig3_380182120 and tidying the angles as on the right, I found that two of the PO₄ are replaced by PO₄H₂ giving Ca₁₀(PO₄)₄(PO₄H₂)₂ instead. Where did I go wrong? cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 16:08, 13 December 2025 (UTC)[reply]

P.S. Why does the middle Ca²⁺ have 6 bonds but outer ones have two each? cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 16:10, 13 December 2025 (UTC)[reply]

The structure is infinite and three dimensional, so a diagram just can represent only part of the structure, perhaps just the elements in the formula. It also has to squash it to 2 dimensions. Your diagram is missing the charge on the phosphate, and the hydroxy ions. The diagram here: https://www.researchgate.net/publication/265731261_Synthesis_of_Hydroxyapatite_from_Oyster_Shell_via_Precipitation/figures?lo=1 attempts to show it in 3D. Graeme Bartlett (talk) 20:46, 13 December 2025 (UTC)[reply]

Thanks, @Graeme Bartlett. I'll trace the unit cell in the link.

Resolved

cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 12:34, 14 December 2025 (UTC)[reply]

If my calculation is right, from the constants given in Expansion of the universe, the distance from the earth to the moon increases by about 3cm per year from pure cosmic expansion, ignoring more local effects.

Can that be observed and measured, using those laser reflectors on the moon? Lunar distance doesn't mention this. It says the distance is increasing by about 1.5 inches per year, and Lunar Laser Ranging experiments mentions relativistic effects but says nothing about cosmic expansion. Any idea? Thanks. ~2025-40343-80 (talk) 05:43, 13 December 2025 (UTC)[reply]

Expansion of the universe only applies to gravitationally unbound features of the universe, and the moon is definitely gravitationally bound to the Earth. --~2025-40278-75 (talk) 06:20, 13 December 2025 (UTC)[reply]

Mrs. Singer is right, "Brooklyn is not expanding."^[5]  ​‑‑Lambiam 11:16, 13 December 2025 (UTC)[reply]

The Local Group is (mostly) gravitationally bound together. Does this imply that the group (and every galaxy/star/solar system/moon within it) is not expanding with the expansion of the universe, but that galaxies not gravitationally bound to the group will be receding from the group due to universal expansion? -- Verbarson  ^talk_edits 18:27, 13 December 2025 (UTC)[reply]

Yes. See the first paragraph of the lede of Expansion of the universe. Whether or not any or all galaxies in the larger Local Sheet, the still larger Local Volume, the even larger Virgo Supercluster, the yet larger Laniakea Supercluster, etc., are also ultimately gravitationally bound is, I think, still undetermined, though recent analysis suggest the Laniakea Supercluster isn't. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 07:16, 14 December 2025 (UTC)[reply]

"Gravitationally bound" seems a fuzzy concept. Won't a bound cluster eject some massive object every few quintillion years or so, thereby at the same time reducing its gravitational binding force (see Future of an expanding universe § Stellar remnants escape galaxies or fall into black holes)? If the fate of the universe is a big RIP, Alvy was right all along.  ​‑‑Lambiam 12:16, 14 December 2025 (UTC)[reply]

Zero-velocity surface § Galaxy clusters has an explanation. If only I understood it... -- Verbarson  ^talk_edits 13:12, 14 December 2025 (UTC)[reply]

If we set the cosmological constant to 0, then the cosmic expansion only means two objects at a distance are moving apart with a speed depending on the distance (proportional, Hubble's constant ${\displaystyle H_{0}}$ is the proportionality constant). So if this was the case at some point between Earth and Moon, then this just makes the orbit of Moon around the Earth slightly different from what it had been. It doesn't add orbital energy. So I guess that the calculation yields the same order of magnitude as the measured value (explainable by tides) is coincidence.

For the general problem, consider the escape velocity at some distance ${\displaystyle a}$ from a mass ${\displaystyle M}$:

${\displaystyle v_{e}={\sqrt {\frac {2GM}{a}}}}$.

Compare this with the Hubble velocity between the mass ${\displaystyle M}$ and a test particle at distance ${\displaystyle a}$:

${\displaystyle v_{H}=a\cdot H_{0}}$.

Solving for the mass/distance relation where these are equal we get

${\displaystyle M={\frac {a^{3}H_{0}^{2}}{2G}}}$,

or

${\displaystyle a=({\frac {2GM}{H_{0}^{2}}})^{1/3}}$.

We have ${\displaystyle H_{0}\approx 70{\text{km}}/{\text{s}}/{\text{MParsec}}\approx 2.27\cdot 10^{-18}{\frac {1}{\text{s}}}}$. For 1 million light years, I get

${\displaystyle M_{10^{6}{\text{ ly}}}\approx 3.86\cdot 10^{40}{\text{ kg}}\approx }$ 19 billion solar masses (smaller than our galaxy by at least a factor of 10).

If the mass is above that, to a first approximation, no stars at this distance will escape. Though there is gravitational slingshot etc as mentioned by others.

Now, if we consider instead a cosmological constant, we can approximately say the universe expands exponentially, with a scale factor ${\displaystyle a(t)=a(0)e^{kt}}$, where ${\displaystyle k}$ is some constant, about ${\displaystyle {\sqrt {10^{-35}}}}$ seconds per second or so (see cosmological constant, ΛCDM model).

In this scenario we can have two equal masses staying at constant distance as they balance the accelerated cosmic expansion with the gravitational attraction.

We have as acceleration the second derivative of the scale factor at 0, with the scale factor set to the distance, and on the other hand Newtonian gravity (good enough here):

${\displaystyle ak^{2}={\frac {GM}{a^{2}}}}$.

Then, we get

${\displaystyle M={\frac {a^{3}k^{2}}{G}}}$.

For the 1 million light year example, we need two masses of about ${\displaystyle 1.6\cdot 10^{41}}$ kg each, still smaller than our galaxy.

For the Earth-Moon example, we'd need two masses at say 400000 km distance, each about 10 kg.

Icek~enwiki (talk) 18:11, 14 December 2025 (UTC)[reply]

I know that last example didn't answer the question. But the cosmological constant should add an additional centripetal force, so for circular orbits it only should change the relation between distance and the time of an orbit (slightly). It would be more interesting what it does to elliptical orbits. Icek~enwiki (talk) 18:19, 14 December 2025 (UTC)[reply]

Oh, that explanation is very interesting, thanks. (This is OP). I didn't realize the part about gravitational binding meaning that the gravitational attraction counterbalanced cosmic expansion. It sounds like it means that even in bound systems, gravity would seem a bit less strong than it would without the expansion. ~2025-40867-45 (talk) 08:10, 15 December 2025 (UTC)[reply]

That gravity would seem a bit less strong is the case only for nonzero cosmological constant. As observable effect, there should be a resulting perihelion precession, as the effective potential has approximately an extra term quadratic in the distance. Icek~enwiki (talk) 21:44, 17 December 2025 (UTC)[reply]

As it says at Einstein field equations#Cosmological constant, "The effect of the cosmological constant is negligible at the scale of a galaxy or smaller.". JRSpriggs (talk) 02:00, 18 December 2025 (UTC)[reply]

Like the OP, I get (almost) 3 cm / year for the cosmic expansion of a stretch of space with a length equal to the radius of the Moon's orbit around Earth. At the scale of the diameter of the Andromeda galaxy, this is more like 10¹⁰ m / year. Are the effects of cosmic expansion on the Moon's radius and the size of the Andromeda galaxy negligible because they are actual but relatively small, or because these are gravitationally bound systems?  ​‑‑Lambiam 15:47, 19 December 2025 (UTC)[reply]

I do not think that being gravitationally bound has anything to do with it. The effect on the Moon is as if the mass of the Earth were reduced by a very small amount proportional to the volume of the sphere centered on the Earth and encircled by the Moon. If the volume of a sphere around the Milkyway Galaxy is not enough to be noticeable then the volume around the Earth is not going to make any measurable difference. JRSpriggs (talk) 18:12, 19 December 2025 (UTC)[reply]

Yes, the cosmological constant/dark energy certainly doesn't change the fact that Earth/Moon or the Local Group are gravitationally bound ("virialised" is the better term), but it should have a tiny effect on the equilibrium distance/size. I'm not positive about the sign of the effect, but the negative pressure might indicate slightly larger than without it. It is unmeasurably small: the effective mass of earth would be changed by something on the order of a kilogram or two if I did my sums correctly. --Wrongfilter (talk) 18:21, 19 December 2025 (UTC)[reply]

While the dark energy density has a small attractive effect, the associated negative pressure has a repulsive effect which is three times that (three spatial dimensions overwhelm one time dimension). So it nets out to a repulsion of twice the mass of the dark energy. JRSpriggs (talk) 18:48, 19 December 2025 (UTC)[reply]

Sure, that's how it works in the Friedmann equation. I guess it's plausible that it works like that in a Schwarzschild situation as well. --Wrongfilter (talk) 20:15, 19 December 2025 (UTC)[reply]

Margie Profet did some novel evolutionary biology work in the 1980s and 90s, won a MacArthur fellowship, and then disappeared while suffering mental health issues. She was reuinited with her family around 7 years later. Does anyone know if she has been involved in scientific or academic work since then? Not asking about her personal life or anything not public. Her story reminds me somewhat of John Forbes Nash's. She even changed subjects to mathematics after winning the MacArthur. Thanks. (Added: web searches that I tried didn't find much). ~2025-40867-45 (talk) 08:05, 15 December 2025 (UTC)[reply]

One thing I found is that she filed for a patent in 2016, which was granted in 2017.^[6]

I can't find the text of her 1988 publication in the defunct journal Evolutionary Theory, but several of her publications, from 1990 to 1993, give her affiliation as Division of Biochemistry and Molecular Biology, University of California, Berkeley. I do not find this affiliation reflected in the current version of the article, which instead suggests that Profet was outside of academia between her 1985 bachelor's degree and her return to school in 1994. The article in Psychology Today writes, "Amazed, the professor [Bruce Ames, up to 1989 Chairman of the Division of Biochemistry and Molecular Biology^[7]] hired Profet—who had degrees in physics and philosophy but no biology background—for a position more fitting her interests." It does not give a date, though, so this could have been any time from 1985 to 1989. The 1994 interview in Omni has, "Beginning in the mid Eighties", and "Bruce Ames ... read her allergy paper and offered her a part-time research job in his lab", not making clear which paper this is. It also states that Profet gave herself a sabbatical, a few months after getting the MacArthur. This is enough to source that she had a (part-time) job in the lab, but insufficient to give a clear time bracket.  ​‑‑Lambiam 16:41, 15 December 2025 (UTC)[reply]

Thanks, nice job finding the patent, so at least she returned to some kind of activity after re-appearing. That's interesting about the 1988 pregnancy sickness paper being hard to find online, since it's cited a fair amount. I looked for it too and found only [8] which says it's collected in a book about evolutionary psychology which might be easier to get hold of. There also are libraries with back issues of the printed journal. I don't feel likely to chase it down since I'm not particularly studying that topic. WP:RX might be able to furnish a copy though, if someone wants it. ~2025-40867-45 (talk) 23:28, 15 December 2025 (UTC)[reply]

The 1988 paper I referred to is:

Profet, Margie (1988). "The Evolution of Pregnancy Sickness as Protection to the Embryo Against Pleistocene Teratogens". Evolutionary Theory. 8: 177–190.

The 1992 book The Adapted Mind: Evolutionary Psychology and the Generation of Culture contains a much cited 40-page chapter by Profet , titled "Pregnancy Sickness as Adaptation: A Deterrent to Maternal Ingestion of Teratogens". This contribution may be called an extended version of her 14-page 1988 article, but it is not the same article. In the Acknowledgements section she writes: "This paper was completed in July 1989 and updated in March 1991." The Contributors pages of this book list her with the UCB affiliation. It is quite plausible that Profet had been hired before she made the extensive expansion completed in July 1989, but this is based on inference from indirect evidence, and even then we still don't know the year in which she was hired with any degree of certainty.

It remains unclear what the "allergy paper" is that impressed Ames so much that he hired her, despite a lack of academic background in biology (and, one may assume, biochemistry and toxicology). In the introduction of the Omni interview, it is described thus: "An early article explores how allergies shield us from toxins in plants and venoms." Was this the 1988 paper?  ​‑‑Lambiam 09:24, 16 December 2025 (UTC)[reply]

Pretty sure the allergy paper must be the 1991 one cited in her biography. It argued that the function of allergies was to expel toxins from the body. ~2025-41519-59 (talk) 10:48, 18 December 2025 (UTC)[reply]

One of the references in the 1991 article is itself a 1991 paper, and Ames's help is acknowledged, so this cannot be the paper that Ames read in the mid eighties. And nothing in this paper identifies it as a revised version of an earlier article.  ​‑‑Lambiam 21:46, 18 December 2025 (UTC)[reply]

Hello all, does anyone know where I could find a website, directory, publication, book listing Alpine/high-elevation snail species? When searching for the terms I find scattered records, but I'd like a more comprehensive source... Barbalalaika 🐌 22:19, 15 December 2025 (UTC)[reply]

Are these species likely to be part of an identified clade - or is it merely descriptive? If it's a specific group of snails then Wikispecies may be helpful to you. If alpine just refers to any old snail above a certain altitude that link is less helpful; I didn't immediately see anything on their bibliography list that seemed germane. It certainly seems like marine species are more comprehensively covered when it comes to snails. Matt Deres (talk) 20:34, 16 December 2025 (UTC)[reply]

I've got this mental picture of the world's slowest slalom. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:23, 16 December 2025 (UTC)[reply]

It's merely descriptive, or rather related to the species' elevation of occurrence. Not a group, still grateful for your help! I got great guidance in the comment below yours too :) Barbalalaika 🐌 18:40, 17 December 2025 (UTC)[reply]

A good approach would be to look at works on the mollusc species of Switzerland:

Turner et al. (1998) Atlas Der Mollusken Der Schweiz Und Liechtensteins. Centre suisse de cartographie de la faune

Boschi, C. (2011) Die Schneckenfauna der Schweiz. Bern: Haupt Verlag.

Not all the species listed are alpine, but there is the information on altitudinal range included for you to filter them out.

For a species list you can also use the following, but there is no information on altitudes:

Rüetschi J. et al. 2012: Rote Liste Weichtiere (Schnecken und Muscheln). Gefährdete Arten der Schweiz, Stand 2010. Bundesamt für Umwelt, Bern, und Schweizer Zentrum für die Kartografie der Fauna, Neuenbu. Umwelt-Vollzug Nr. 1216. Available as free download.

There are some species that occur in the Alps but not in Switzerland. For those you would need to look at a book covering the whole of Europe, perhaps most conveniently:

Kerney, M.P, Cameron, R.A.D. & Jungbluth, J.H. (1983) Die Landschnecken Nord- und Mitteleuropas. Hamburg: Parey.

More up-to-date, but covering a wider area, so harder work to filter, is:

Welter-Schultes F. W. (2012) European non-marine molluscs, a guide for species identification. Göttingen: Planet Poster Editions.

Both the latter two works have crude maps, which would help you select the Alpine species. JMCHutchinson (talk) 08:19, 17 December 2025 (UTC)[reply]

That's great and exactly what I needed! Thank you so much. I was considering contacting an expert outside of Wikipedia, but it seems the expert found me :) Barbalalaika 🐌 21:39, 18 December 2025 (UTC)[reply]

An IP/temporary account added an unsourced value in this edit, I think we can keep it but I need a source for such a claim. Jo-Jo Eumerus (talk) 12:53, 17 December 2025 (UTC)[reply]

Would we need a specific piece of text stating this, or would an editor's measuring the distance on a good quality map (presumably itself a citable Reliable source) fall short of Original research? I feel sure this question must have been addressed before, somewhere. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 14:29, 18 December 2025 (UTC)[reply]

In Blum et al. (1996) it states "Tropic Seamount (23”53’ N, 20”42’ W at summit), also known as Carmenchu Peak atop the Tropical Bank (Emery and Uchupi, 1984), is an isolated, prominent submarine volcano, situated halfway between the Canary Islands and Cape Verde Islands, 260 nautical miles west of the coast of Mauritania/Westsahara". In Hassan et al. (2024) it says "It is located 400 km from West Africa's passive continental margin, halfway between the Canary Islands and the Cape Verde Islands". Using Google Earth, measurements to the nearest coast range from 440–480 km (240–260 nautical miles). There's no need to state which country, using Blum et al. (1996), you can say that it lies 260 nautical miles west of the African coast. Mikenorton (talk) 17:06, 18 December 2025 (UTC)[reply]

Which, I note, is 482 km, considerably different from the rather oddly worded ". . . and distance to Africa (Morocco/Western Sahara): approx. 250–300 km" that the temp account added. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 17:48, 18 December 2025 (UTC)[reply]

OR alert: The "measure distance" function of Google maps gave me 443 km.  ​‑‑Lambiam 21:29, 18 December 2025 (UTC)[reply]