List of common misconceptions about science, technology, and mathematics#Human body and health states:

• Half of body heat is not lost through the head, and covering the head is no more effective at preventing heat loss than covering any other portion of the body. Heat is lost from the body in proportion to the amount of exposed skin.^{[1] [2]} The head accounts for around 7–9% of the body's surface, and studies have shown that having one's head submerged in cold water only causes a person to lose 10% more heat overall.^{[3] [medical citation needed ]}

1. If "the amount of exposed skin" is where heat is lost, then in a situation where being cold is a problem (so, clothed), would a person not lose about half their body heat from their head (face and neck [or more if bald]), with the other half from their hands?

2. Would a mask like a balaclava (clothing) not reduce loss by about half?

3. If the hair on the head doesn't expose skin on the top of the head, does that mean wearing a normal hat doesn't help at all in preventing loss of body heat? -- Jeandré (talk) 2025年10月31日 06:38z

1. If you read the sources cited, yes, that's what they claim. When warmly clothed for winter, a bit under half the heat lost is from the head -- and the research demonstrating that appears to also be how the "half" claim originated. The other half one could guess would be from hands and due to the fact that winter clothing doesn't totally prevent heat loss, only attenuates it. That information could conceivably be added to our article.

2. Probably somewhat less than half (since it will slow, not prevent heat loss from the head, and less than half the heat loss is from the head). But a significant amount, yes.

3. If you're already wearing a long sleeve shirt, does adding a winter coat not reduce your heat loss? A hat covers the ears (high surface area to volume ratio) and adds another layer of insulation to keep heat in. -- Avocado (talk) 12:58, 31 October 2025 (UTC) [reply ]

Thermal insulation is never 100%, so any covered skin areas will nevertheless lose heat to a colder environment. If the surrounding air is completely stagnant, the air itself will form an effectively insulating layer. Assuming a light air flow that swiftly carries away the heated air, given a thermal conductivity reducing factor ${\displaystyle \lambda }${\displaystyle \lambda } of the skin covering (where ${\displaystyle \lambda =0}${\displaystyle \lambda =0} would mean 100% effectiveness), the heat loss as a function of the fraction ${\displaystyle c}${\displaystyle c} of covered skin area is proportional to

${\displaystyle c\lambda +(1-c),}${\displaystyle c\lambda +(1-c),}

in which the first term covers the contribution to the heat loss by the covered skin, while the second term corresponds to the contribution by the exposed skin. If ${\displaystyle c=0.9,}${\displaystyle c=0.9,} meaning that 10% (say head and both hands) is exposed, for the contribution of the exposed skin to be at least 50% we get

${\displaystyle \lambda \leq 0.111...,}${\displaystyle \lambda \leq 0.111...,}

which means quite effective insulation, as for an Arctic expedition.

Thick hair reduces air flow and thus helps to keep one's head warm in cold weather. ​‐‐Lambiam 15:53, 31 October 2025 (UTC) [reply ]

This would need verification for accuracy but since it's somewhat related I thought it would be appropriate here. Apparently, because the brain and heart are vital organs, in case of hypothermia, other than the fact that limbs dissipate more, the body might limit energy and blood flow to them, in extreme cases potentially also resulting in cramps. I often heard the assertion that because of this, covering the head also helps to increase general comfort (including of the hands and feet)... 206.248.143.75 (talk) 20:03, 1 November 2025 (UTC) [reply ]

The publication of Madhusudhan, Nikku; Constantinou, Savvas; Holmberg, Måns; Sarkar, Subhajit; Piette, Anjali A. A.; Moses, Julianne I. (20 April 2025). "New Constraints on DMS and DMDS in the Atmosphere of K2-18 b from JWST MIRI". The Astrophysical Journal Letters. 983 (2): L40. arXiv:2504.12267v1 . Bibcode:2025ApJ...983L..40M. doi:10.3847/2041-8213/adc1c8 . has raised a storm because some principles for scientific publications were violated, such as not considering existing theories, using statistically irrrelevant data or drastically limiting the number of molecules considered.[1] (See our articles K2-18b and Nikku Madhusudhan.)

Looking at the article it seems to me that the mistakes by the authors were honest mistakes, understandable if one considers that it's only human to be biased towards one's own ideas. They honestly wrote down plenty of their assumptions, beginning (in the abstract section) with the dismissal of 3.4-σ significance in favor or a 3-σ significance. But why did The Astrophysical Journal Letters publish this despite such obvious blunders? The only possible explanation I can think of is that they were blinded by sensationalism; so much so that they cast aside any professionalism. Does anyone have a more satisfactory explanation? ◅ Sebastian Helm 🗨 19:28, 6 November 2025 (UTC) [reply ]

Is anybody accusing them of wrong-doing? Without having looked at this in very much detail, my impression is that this is a vibrant but otherwise normal scientific debate about data reduction and analyis methods and the statistical significance of potential results drawn from the data. The referee and the editor of ApJL did not see any obvious blunders − it is not their job to redo the analysis with the same or different methods. The results from the paper may not survive but there was presumably no reason not to publish it. The press release may be more problematic in that it overstated the significance of the results. Unfortunately that is not uncommon. — Why do you say "dismissal of 3.4-σ significance"? As I understand it they accept that the spectrum is not featureless and confirm this at 3-σ in independent data. --Wrongfilter (talk) 20:46, 6 November 2025 (UTC) [reply ]

Thanks for your answer. You're right about the "dismissal of 3.4-σ significance"; that was my mistake. It's not that they're refuting any article that claims a featureless spectrum. Also, good point about the press release; I haven't looked at that yet.

As for "wrong-doing": It depends on what you mean by "wrong-doing". Of course nobody is accusing them of any illegal act. But it's being criticized in public broadcasting such as here to great effect. That can't be good for the journal's reputation. ◅ Sebastian Helm 🗨 08:17, 7 November 2025 (UTC) [reply ]

I don't have time to watch all 20 minutes of that but I hope I've seen the important parts. Lesch is clearly critical and says he would have rejected the paper. The referee obviously came to a different conclusion, and that is — assuming good faith — perfectly fine. It might be interesting to see the referee report but we will not get it. As it stands the paper should be seen as a contribution to the scientific process and to the scientific debate and as such it is of course not above criticism — no scientific paper is. What is, however, definitely harmful (although not for ApJL) is the public hype triggered by the press release. --Wrongfilter (talk) 08:53, 7 November 2025 (UTC) [reply ]

Sorry for making you spend so much time on it. Yes, you have seen the important parts for this question; and the statement you refer to, at 11:35, is (along with his accompanying facial expression) the clearest sign of his disapproval. I see why you consider it less harmful for ApJ than for the public: Scientists will rather see it as just another contribution to the scientific process, while the general public may (further) lower their respect for science. So, since you already watched so much of the video, let me ask you if you think it rather helps or harms science's standing? ◅ Sebastian Helm 🗨 13:33, 7 November 2025 (UTC) [reply ]

Actually, I think there's an easy answer to my question: The replies in the discussion section express a strong sense of feeling encouraged by the video, which for me means that public reception of science is helped. ◅ Sebastian Helm 🗨 14:07, 7 November 2025 (UTC) [reply ]

Usually there are several referees; a paper like this would not have been an exception to the rule. The paper was accepted one day after it was returned with a revision, its first and only revision. This suggests that any issues the referees raised were relatively minor, so that the acting editor could readily see that they were satisfactorily addressed. ​‐‐Lambiam 10:23, 7 November 2025 (UTC) [reply ]

The astronomy journals that I've refereed for had only one. I'm not sure about ApJ, but I'd be surprised if the letters section (which goes for relatively fast publication) had two. --Wrongfilter (talk) 10:28, 7 November 2025 (UTC) [reply ]

Many articles in ApJL (which used to be a section of The Astrophysical Journal but is now a separate journal) thank the "anonymous referees" (e.g. [2], [3], [4], [5], [6]). ​‐‐Lambiam 09:18, 8 November 2025 (UTC) [reply ]

Thanks, Lambiam, for pointing us to the article information. But if the issues were minor, why did it take 3 months for the authors to revise it? Couldn't that just as well corroborate Lesch's criticism in that the referee(s) abandoned scrutiny for the easier way out? ◅ Sebastian Helm 🗨 13:51, 7 November 2025 (UTC) [reply ]

Because they were busy with other work? Because the one tasked to do it was ill? I don't think the interval can be taken to corroborate anything at all.

The point of publishing any paper is not to provide a cut-and-dried, final, completely correct answer; it's to present (usually) new data and (sometimes re-)interpretations of them so that others can investigate further and confirm, correct, or refute the latter.

Press releases always overestimate the significance of papers: they are usually written by the publicity department(s) of the institution(s) that employ the researchers, who are _not_ experts in the disciplines in question, in order to drum up the institutions' reputations and hopefully attract more sponsorship money (from other, rich, non-experts). {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 07:49, 8 November 2025 (UTC) [reply ]

While the hype put in by the publicity departments is deplorable and IMO ultimately detrimental to the image of science in the court of public opinion, popular science media are also to blame, since they won't pay attention to research results if they are not presented ready-made hyped up. ​‐‐Lambiam 09:40, 8 November 2025 (UTC) [reply ]

Did it really take the authors three months to revise their paper? Perhaps it took the referee(s) two months to send in their reports – we don't know. ​‐‐Lambiam 09:18, 8 November 2025 (UTC) [reply ]

Thanks, everyone. With this, I feel we can consider this question closed. ◅ Sebastian Helm 🗨 09:38, 8 November 2025 (UTC) [reply ]

I understand that the metre was originally "one ten-millionth of the shortest distance from the North Pole to the equator passing through Paris", the gram "the absolute weight of a volume of pure water equal to 1 cm³ at the temperature of melting ice" and the second "⁠1/(×ばつ60)⁠ of a day".

Is it thus coincidence that Earth's gravitational acceleration is close to 10 ms⁻², or is there an underlying physical reason?

Thanks, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 00:41, 10 November 2025 (UTC) [reply ]

Yes, it's a coincidence. For one thing 9.7 or 9.8 is not 10. And in the old units 32 feet per second per second would suggest "close to " 10 yards per second per second. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:38, 10 November 2025 (UTC) [reply ]

Also, it varies by latitude and altitude. See Somigliana equation. Sean.hoyland (talk) 03:19, 10 November 2025 (UTC) [reply ]

Or let's say it varies a lot spatially if you are trying to use gravity gradiometer to do something, like model the subsurfacce to look for hydrocarbons or spot submarines etc. Sean.hoyland (talk) 03:35, 10 November 2025 (UTC) [reply ]

Even submarines? Amazing. According to Archimedes, the mass of a submarine is about equal to the mass of the displaced water, making it very hard to spot submarines by their gravity. PiusImpavidus (talk) 11:46, 10 November 2025 (UTC) [reply ]

Submarines generally have diving planes: hydrodynamic control surfaces which can exert force vertically if the submarine has sufficient horizontal motion. Ideally, a submarine will however adjust its ballasting when submerged to achieve neutral buoyancy once it meets its desired depth. As to whether it is possible to detect a diving/surfacing submarine (which may not be neutrally buoyant) with a gravity gradiometer, I rather doubt it, though I'm willing to be proved wrong. AndyTheGrump (talk) 23:25, 10 November 2025 (UTC) [reply ]

If we take the Earth to be a sphere of uniform mass, Newton's law of universal gravitation tells us that the force of gravity at its surface equals

${\displaystyle g={\frac {GM_{\oplus }}{r_{\oplus }^{2}}},}${\displaystyle g={\frac {GM_{\oplus }}{r_{\oplus }^{2}}},}

in which ${\displaystyle G\approx 6.67430\times 10^{{-}11},円{\text{m}}^{3}{\cdot }{\text{kg}}^{{-}1}{\cdot }{\text{s}}^{{-}2}}${\displaystyle G\approx 6.67430\times 10^{{-}11},円{\text{m}}^{3}{\cdot }{\text{kg}}^{{-}1}{\cdot }{\text{s}}^{{-}2}} is the gravitational constant, ${\displaystyle M_{\oplus }\approx 5.9722\times 10^{24},円{\text{kg}}}${\displaystyle M_{\oplus }\approx 5.9722\times 10^{24},円{\text{kg}}} is the Earth mass and ${\displaystyle r_{\oplus }\approx 6371\times 10^{3},円{\text{m}}}${\displaystyle r_{\oplus }\approx 6371\times 10^{3},円{\text{m}}} is the Earth radius. Plugging in these numerical values, we compute

${\displaystyle g\approx 9.82,円{\text{m}}{\cdot }{\text{s}}^{{-}2}.}${\displaystyle g\approx 9.82,円{\text{m}}{\cdot }{\text{s}}^{{-}2}.}

For its closeness to ${\displaystyle 10,円{\text{m}}{\cdot }{\text{s}}^{{-}2}}${\displaystyle 10,円{\text{m}}{\cdot }{\text{s}}^{{-}2}} not to be a coincidence, given the historical definitions of the physical units involved, requires a conspiracy between the gravitational constant, the density of water relative to the Earth's density, the mathematical constant ${\displaystyle \pi }${\displaystyle \pi } and finally the speed of Earth's rotation. ​‐‐Lambiam 10:00, 10 November 2025 (UTC) [reply ]

@Cmglee: An older proposal for the definition of the metre was the length of a pendulum with a period of exactly two seconds. By that definition, ${\displaystyle g=\pi ^{2},円\mathrm {m/s^{2}} \approx 9.87,円\mathrm {m/s^{2}} }${\displaystyle g=\pi ^{2},円\mathrm {m/s^{2}} \approx 9.87,円\mathrm {m/s^{2}} }. That proposal didn't make it as it turned out g isn't constant. Defining that ${\displaystyle g=\pi ^{2},円\mathrm {m/s^{2}} }${\displaystyle g=\pi ^{2},円\mathrm {m/s^{2}} } at ${\displaystyle 45^{\circ }}${\displaystyle 45^{\circ }} latitude wasn't acceptable either. 1/10,000,000 of the distance from the equator to the pole is conveniently close that that older proposal. If g had been closer to ${\displaystyle 9.15,円\mathrm {m/s^{2}} }${\displaystyle 9.15,円\mathrm {m/s^{2}} } (in our units), the Earth having the same circumference, it might have been plausible that the metre had been defined as 1/2000 of a nautical mile (1/10,800,000 of the distance from the equator to the pole or half a fathom), again making g close to ${\displaystyle \pi ^{2}}${\displaystyle \pi ^{2}}. So I think that g being close to ${\displaystyle 10,円\mathrm {m/s^{2}} }${\displaystyle 10,円\mathrm {m/s^{2}} } is coincidence, but g being close to ${\displaystyle \pi ^{2},円\mathrm {m/s^{2}} }${\displaystyle \pi ^{2},円\mathrm {m/s^{2}} } isn't. PiusImpavidus (talk) 11:36, 10 November 2025 (UTC) [reply ]

Also, there's nothing really special about the number 10 other than that it's incidentally the number of fingers most of us have. -- Avocado (talk) 12:18, 10 November 2025 (UTC) [reply ]

Thank you, everyone. Yes, I suppose "close to" is subjective. Nevertheless, 9.8 is just 2% off 10, and if I understand correctly, statisticians use 5% (1 – 95%) as a measure of significance.

Mathematical_coincidence#Gravitational_acceleration concurs that it's related to π2 which happens to be approximately 10.

Cheers, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 12:55, 10 November 2025 (UTC) [reply ]

Experimental scientists (who are usually at best amateur statisticians) may do this, but they shouldn't; see Misuse of p–values. ​‐‐Lambiam 23:12, 10 November 2025 (UTC) [reply ]

I am working an elevation diagram in vector editor InkScape for William Street tunnel and I am using a satellite image with overlays (diagrammatic plan) as a basis.

I need the basic length of the tunnel, and features along it.

I imported the image into InkScape but when I started measuring the distances seemed a little small. The plan says "Scale 1:2000 (@ A0)". In InkScape I can measure in mm, it has no idea what a mm is, but it a relative measurement. The 1:2000 didn't seem right, so I measured known lengths on Google Maps and OpenStreetMap. For the length of Murray St between Milligan and William Streets I got 1 mm : 22.7 m (InkScape : Reality). For the pool near the Mounts bay road label (via the adjacent tennis court) I got 1 mm : 22.64 m. Does this mean the plan should be "Scale 1:2270 (@ A0)" and they just ballparked it? I can see the sides of buildings, so the camera must not have been perfectly vertical; could that affect the scale this much? Also, what is the bar scale in the SCALE box about? It seems to go to 100 but equates to about 199 m on the map.

As an aside that I haven't websearched yet: how accurate are the Google Maps measurements? I used a tennis court near the pool and got length 23.3 m whereas a tennis court is 23.77 m in length. Commander Keane (talk) 11:30, 10 November 2025 (UTC) [reply ]

I just used Wikimapia's measuring tool and got 1523 m. Abductive (reasoning) 22:35, 10 November 2025 (UTC) [reply ]

This all looks very much like WP:OR to me, and accordingly doesn't belong in an article. AndyTheGrump (talk) 23:07, 10 November 2025 (UTC) [reply ]

Perhaps, but using an easier tool will at least get a better estimate, which may help a user choose which source(s) to believe. Oftentimes the sources play loose with the numbers. Abductive (reasoning) 23:58, 10 November 2025 (UTC) [reply ]

Fair enough, one can certainly discuss the validity of existing sources on talk pages (where WP:OR doesn't apply), if that is what is intended. As long as that is all that is intended, and measurements on Google maps made by a contributor aren't then used in the article itself. AndyTheGrump (talk) 00:14, 11 November 2025 (UTC) [reply ]

The image is 3425 ×ばつ 2422 pixels, while A0 is 1189 mm ×ばつ 841 mm. This means that the length of a pixel side, @ A0, is (rounded to 4 decimals) 0.3472 mm. If the scale is, as advertized, 1 : 2000, a pixel is supposed to represent a square of size 694.4 mm, or 0.6944 m, while the whole area of which this is an aerial photograph is then about 2378 m ×ばつ 1682 m, with a diagonal measuring 2913 m. If, however, the true scale is 1 : 2270, the area should be 2699 m ×ばつ 1909 m, with a diagonal of 3306 m. ​‐‐Lambiam 23:07, 10 November 2025 (UTC) [reply ]

The distance between two easily recognizable points on the satellite image comes out at 3332 pixel widths (the hypotenuse of a right-angled triangle with right sides of 1845 by 2775 pixels), which @ A0 should be 1157 mm. Google maps measure the distance as 2.02 km, which implies a scale in the range of 1 : 1742 to 1 : 1750. Or did I make a calculation error? ​‐‐Lambiam 14:21, 11 November 2025 (UTC) [reply ]

A tilt of 6 degrees would give a 10% error in scale in the worst direction. I vaguely remember that some google maps photos are from aerial surveys. Greglocock (talk) 23:33, 10 November 2025 (UTC) [reply ]

Rotation is an isometric transformation, so this cannot be, by itself, the immediate cause of a scaling error. ​‐‐Lambiam 13:07, 11 November 2025 (UTC) [reply ]

@Lambiam: I am drowning in pixels here, but at 1 : 1750, are you saying that the map (1 : 2000) is underestimating lengths; it is zoomed out, if you will? I am saying (1 : 2270) it makes everything too big, but I am generally confused. Commander Keane (talk) 16:37, 11 November 2025 (UTC) [reply ]

What I can say with some degree of certainty is that at least one of the following three statements is correct:

1. I made some error in the process of computation or measuring;
2. the "measure distance" of Google Maps can give results that are wildly off;
3. the statement "SCALE: 1 : 2000 (@ A0)" in the caption is wrong.

If I made no error and the distances given by Google Maps are fine, the last statement overestimates the lengths in reality; 1 cm on the plan is 17.5 m, not 20 m.

My plan is (if I find the time) to do all over from scratch tomorrow to see if I get the same result. ​‐‐Lambiam 00:27, 12 November 2025 (UTC) [reply ]

I suggest you draw it out, it takes 5 seconds. 2d projection onto an inclined plane is not isometric. Greglocock (talk) 05:11, 12 November 2025 (UTC) [reply ]

I now realise that my 1:2270 was ficticious (I stopped trusting the map a while ago due to the article previously using an incorrect source on length and confirmation bias did the rest). Using A0 dimensions and the outer black box (Note: not the jpeg dimensions 3425 ×ばつ 2422 pixels, I assume this is the correct way to do it) I get between 1:1900 to 1:1950. At this stage, the scale is sensitive to my measurements and the stated scale of 1:2000 may well be correct. However, as mentioned in my initial question, I still don't know what bar scale is trying to show. Commander Keane (talk) 08:23, 12 November 2025 (UTC) [reply ]

After repeating the exercise with two different recognizable locations, and taking Google's measurement to be gospel truth, I now find that the scale for the diagrammatic plan, printed at A0 size, lies in the range 1 : 2005 to 1 : 2020, which is close to 1 : 2000.

Printed at A0 size, 1 pixel in the jpeg image corresponds to 0.3472 mm; the bar scale in the caption, with tick marks marked 0, 10, 20, 30, 40, 50 and 100 mm should correspond to these lengths on paper when the image is printed at A0 size. ​‐‐Lambiam 19:25, 14 November 2025 (UTC) [reply ]

Thanks Lambiam. Commander Keane (talk) 07:58, 15 November 2025 (UTC) [reply ]

On a spectroradiometer measurement, the wavelength value is given as the middle of the bandwidth. But exept that we have a kind of gaussian shape in reality, as the curve in the bandwidth is not "horizontal" (=constant power) and not also a linear function, I am trying to find if the wavelength measured must be considered as the average wavelength ${\displaystyle {\bar {\lambda }}}${\displaystyle {\bar {\lambda }}} or the median wavelength ${\displaystyle {\tilde {\lambda }}}${\displaystyle {\tilde {\lambda }}}, any idea ? Malypaet (talk) 09:25, 12 November 2025 (UTC) [reply ]

The term "average" is an umbrella term for any of several ways to capture a dispersed value with a single number, which includes the arithmetic mean (or "arithmetic average") as well as the median. I expect the distribution of the measured wavelength of a monochromatic source to have a more or less log-normal distribution, in which case the expected median coincides with the expected geometric mean. Using the geometric mean (or the median) as the average has the advantage that the average of the frequency is proportional to the inverse of the average of the wavelength, which is not true for the arithmetic mean or the modus. Unless the spread is gigantic, however, the differences between these various averages should in practice be much smaller than the measurement error and may even vanish when rounding the values. ​‐‐Lambiam 12:32, 14 November 2025 (UTC) [reply ]

For a spectroradiometer the spectral bin (channel) width should be smaller than a spectral resolution element (the width of an unresolved line), so indeed it does not matter how the wavelength is defined exactly. In broad-band photometry one often uses the "effective wavelength" to characterise a filter; this is an average wavelength, weighted by the transmission ${\displaystyle T(\lambda )}${\displaystyle T(\lambda )} of the filter:

${\displaystyle \lambda _{\mathrm {eff} }={\frac {\int \lambda T(\lambda ),円\mathrm {d} \lambda }{\int T(\lambda ),円\mathrm {d} \lambda }}}${\displaystyle \lambda _{\mathrm {eff} }={\frac {\int \lambda T(\lambda ),円\mathrm {d} \lambda }{\int T(\lambda ),円\mathrm {d} \lambda }}}. --Wrongfilter (talk) 13:44, 14 November 2025 (UTC) [reply ]

I was trying to position the wavelength value on the curve given by Planck's law, taking into account the bandwidth and ensuring equal power in both portions of that bandwidth. But I realize that the measuring instrument, with its filter, complicates the problem and reformulates it, as you point out. Thank you very much for these explanations. Malypaet (talk) 21:40, 14 November 2025 (UTC) [reply ]

Not sure for the frequency, if you keep a constant wavelength bandwidth, the corresponding frequency bandwidth is a function of the wavelength:

${\displaystyle \Delta \nu ={\frac {c}{\lambda ^{2}}}\cdot \Delta \lambda }${\displaystyle \Delta \nu ={\frac {c}{\lambda ^{2}}}\cdot \Delta \lambda }

So I don't think it's a problem.

And indeed, compared to the accuracy of a measurement, it is a negligible problem.

Before the filter, it's theory; after, it's experimental physics.

Thanks
Malypaet (talk) 22:04, 14 November 2025 (UTC) [reply ]

The title says it all. Are there any general anaesthetics that don't cause dependence or addiction? ~2025-33937-00 (talk) 05:22, 16 November 2025 (UTC) [reply ]

Etomidate is not known to cause dependence or addiction, but there are definite indications that the potential for addiction exists.^[7] ​‐‐Lambiam 09:56, 16 November 2025 (UTC) [reply ]

Carbon dioxide is regularly used in entomology laboratories to anaesthetise insects: see doi:10.1016/j.jinsphys.200607001. There's no reason to think that it would be addictive. See also Diethyl ether#Anesthesia for some historical human anaesthetics. Mike Turnbull (talk) 14:53, 16 November 2025 (UTC) [reply ]

We have an article on ether addiction ! ~2025-34243-06 (talk) 11:28, 17 November 2025 (UTC) [reply ]

Wikipedia:WHAAOE indeed. Mike Turnbull (talk) 14:37, 17 November 2025 (UTC) [reply ]

Before Einstein, physicists were addicted to ether. ​‐‐Lambiam 23:25, 18 November 2025 (UTC) [reply ]

Unlikely to be a problem for normal indicated usage, then many are problematic for exended use. The reference desk cannot give medical advice, so consult your doctor about your concerns. ~2025-32692-02 (talk) 03:42, 19 November 2025 (UTC) [reply ]

This question does not express a concern of the questioner and can IMO not be classified as a question for advice, whether medical or otherwise. If you ask your doctor (who is most likely not specialized in anesthesiology) whether any general anesthetics don't cause dependence or addiction, they will probably need to do research to answer the question. In doing so, they may consult our reference desk. ​‐‐Lambiam 08:18, 19 November 2025 (UTC) [reply ]

The nitrous oxide article doesn't mention dependence or addiction, as far as I could tell from Ctrl+F; if you ignore navigational templates and references, "addic" finds nothing, and "depen" finds only amount released depends on which fuel (environmental impact of burning), This correlation is dose-dependent (damage from occupational exposure), and The ignition of nitrous oxide depends critically on pressure. If the concepts are mentioned, it's via other terms that I didn't think to check. Nyttend (talk) 19:52, 22 November 2025 (UTC) [reply ]

The dentist from Little Shop of Horrors would like a word. -- Avocado (talk) 20:03, 22 November 2025 (UTC) [reply ]

There are good sources Nitrous oxide - Alcohol and Drug Foundation and Nitrous Oxide Addiction which suggest it has, at least, potental to cause psychological addiction or compulsive behaviour. These sources should probably be added to nitrous oxide (medication). Mike Turnbull (talk) 23:17, 22 November 2025 (UTC) [reply ]

The article says human physicists are making quark gluon plasma at CERN now. That's great but isn't the equivalent substance early in the Big Bang probably subjected to strong gravity, strong and complicated magnetic fields, and strong and complex angular momentum that would bring in more relativistic effects that the substance doesn't experience at CERN?Rich (talk) 01:40, 18 November 2025 (UTC) [reply ]

I gotta ask. What are the non-human physicists doing? Clarityfiend (talk) 04:15, 18 November 2025 (UTC) [reply ]

I was warned not to say, unless I want to breathe vacuum.Rich (talk) 06:58, 18 November 2025 (UTC) [reply ]

That is probably the case, but we have to start somewhere. Actually creating quark soup at all and observing its physical properties are steps forward, necessary before we can begin to figure out the significance of those other factors in the Big Bang (even assuming that's possible). {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 19:45, 18 November 2025 (UTC) [reply ]

Morning folks!! I was wondering if anybody knows what "healing by organisation" means in its context in 1960-1870. This is specifically for the Joseph Lister article which I'm writing. Its long I know. It will be split. I'm confused about this term. It seems to be healing by organisation in blood clots and relates to an address that Lister gave in "An Address on the Antiseptic System of Treatment in Surgery, delivered before the Medico-Chirurgical Society of Glasgow" on 17 April 1868. As a surgeon there is discussions about wound healing. I've discovered healing by "First Intention" means but not by healing by organisation. Would any medical historians have any idea?

• The collected papers of Joseph, Baron Lister Volume 2 pp=51-86. scope_creep ^Talk 08:33, 19 November 2025 (UTC) [reply ]

According to ChatGPT:

"Healing by organization" is an archaic medical term used in 19th–early-20th-century pathology and surgery.

It referred to:

Definition

Healing by organization meant that damaged tissue was repaired not by regeneration of the original tissue, but by the ingrowth of granulation tissue that later matured into fibrous (scar) tissue.

In other words, the body "organized" the injury by filling it with connective tissue.

Modern terminology

Today this process is simply called:

Fibrous (scar) formation

Healing by fibrosis

Granulation tissue formation and maturation

Secondary intention healing (when referring to open wounds that must fill in from the bottom)

Historical context

Before modern histology and wound-healing science, "organization" described how:

A blood clot or inflammatory exudate becomes infiltrated by fibroblasts and capillaries.

This tissue gradually becomes dense collagenous scar tissue. ~2025-34800-71 (talk) 09:04, 19 November 2025 (UTC) [reply ]

@~2025-34800-71: I know what is now, which I didn't before, but I need proper references for the definition, so I can check the book reference. scope_creep ^Talk 19:57, 19 November 2025 (UTC) [reply ]

Here you find a discussion of what Lister meant by the term. It appears that the term has not fallen into complete disuse: [8], [9], [10]. ​‐‐Lambiam 23:21, 19 November 2025 (UTC) [reply ]

Thanks @Lambiam: I don't know how I didn't see that last night. I must have been conked out when I was searching; on automatic. Thanks folks. scope_creep ^Talk 23:45, 19 November 2025 (UTC) [reply ]

Is this guy (found in New England) an Agelenidae coras? If so, what species? Jay Cubby 14:33, 19 November 2025 (UTC) [reply ]

Assuming the specimen is in the genus Coras , it may be hard to identify the species from these images, because the visual differences between different species may be subtle and not displayed in these images. Case in point: the caption for the first image in our article on the genus Coras has: "Coras species, probably C. medicinalis". Is the specimen even in this genus? If so, their anterior median eyes are larger than their anterior lateral eyes, but I doubt even this can be discerned from the available images. ​‐‐Lambiam 10:23, 20 November 2025 (UTC) [reply ]

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