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July 8

Is that true that things which plugged in consume electricity while they don't work?

I've been told that things that are plugged in, consume electricity even while they don't work. For example kettle, oven, or even just a phone charger which plugged in, while it's without the phone. Is that true? If it is, then for me it doesn't make sense and I would like to know the explanation behind it. 5.102.253.81 (talk) 03:05, 8 July 2017 (UTC)[reply]

Many electronic devices such as televisions, do draw a trickle of current to maintain readiness when switched on. But 'dumb' devices such as toasters, electric ovens and so forth generally don't. You can buy a current tester at the hardware store to check how much electricity is flowing when a device is nominally 'off' if you worry about these things. Abductive (reasoning) 03:34, 8 July 2017 (UTC)[reply]

See our standby power article for more information. DMacks (talk) 03:35, 8 July 2017 (UTC)[reply]

And such "vampire" devices which use standby mode are an increasingly large proportion of all consumer electronics. Unless all current is cut to the device or a diagnostic forces a restart, most such items never de-power their built-in computers--which were traditionally very simple but which now have increasingly complicated tasks, not the least of which today is networking with other devices and (increasingly), the internet of things. Most consumer electronics are moving towards a design where they are almost never completely "off", nor even off in a significant sense. Many still give an impression that something drastic changes when you hit the power button, but the reality is that the difference between being "on" and being in "standby mode" for some devices is no more significant than the status of the indicator lights on the device's exterior: the device is still communicating with other devices and/or a network or even an internet connection, and is just as prepared to do what it is designed to do as it would be at any other time. Still, for larger appliances, these standby functions of the computer and antennae are likely to represent a small amount of power consumption compared to the energy utilized when the appliance is performing it's nominal function. Even so, there are concerns about the aggregate amount of extra power that is being utilized (and arguably wasted in many cases) by vampire electronics as they become more and more prevalent. Snow ^{let's rap} 18:23, 8 July 2017 (UTC)[reply]

'dumb' devices such as ... electric ovens .. generally don't. — Very old electric ovens don't, but I suspect that any electric oven built in the last few decades includes an electric clock, which always draws some current. Mitch Ames (talk) 12:42, 8 July 2017 (UTC)[reply]

And very soon, I doubt there will be an oven built for mass manufacture anywhere in the world that doesn't have a (relatively) sophisticated computer. I imagine an oven is one of those devices that manufacturers will begin to add network capabilities to last, because of liability concerns (the idea that an oven can be turned on remotely, and thus be susceptible to hacking or command errors must be a concern for even those companies which are moving aggressively on making their devices network enabled by default), but (at least when in the first world) I have rarely seen an oven in the last decade that does not have some degree of programmable function. Snow ^{let's rap} 18:36, 8 July 2017 (UTC)[reply]

(ec)To use electricity there has to be a completed circuit. Any appliance with something going on while powered off, such as a clock, will consume electricity. More details are in Standby power. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:36, 8 July 2017 (UTC)[reply]

Charging devices often draw some current even when nothing is connected to them. This happens because they contain transformers, which allow for a completed circuit on the input side even if there is not a completed circuit on the output side. However the null current is usually very small, in the milliwatt range. Looie496 (talk) 14:41, 8 July 2017 (UTC)[reply]

... and the same applies to modern switch-mode power supplies used for charging. Dbfirs 16:11, 8 July 2017 (UTC)[reply]

Plugging in even an appliance with its mains switch at OFF causes a small AC current to flow in the capacitance between its power conductors. This happens throughout the cables of a network and AC generators experience a significant capacitive load when driving long distribution lines. However this reactive current leads the voltage cycle by 90 degrees and therefore does not contribute to the measured "true" AC power consumption given by P=VI cos φ, see Wattmeter. An electricity supplier charges only for true power consuption but could in principle detect whenever you plug in an appliance by time-domain reflectometry. An advantage of a DC distribution system over AC is that there is no continuous current in cable capacitance, only a one-time inrush at start-up. Blooteuth (talk) 18:01, 8 July 2017 (UTC)[reply]

One rule-of-thumb is that if the plugged-in device gets hot, or even warm, when not in use, then it's using lots of electricity. The temperature is measured relative to the surroundings. StuRat (talk) 18:15, 8 July 2017 (UTC)[reply]

If humans never cut their hair, then will the hair insulate the body like fur?

Non-human animals don't wear clothes. Mammals may have fur to keep warm. If humans never cut their hair, then will the hair insulate body heat and cover the genitals? In other words, can long hair do away with the need to wear clothes? Also, I remember watching a King Kong movie and Jumanji, and I noticed the bearded men just shaved off their beard. What's the deal? Why do humans shave off their hair? 50.4.236.254 (talk) 04:04, 8 July 2017 (UTC)[reply]

1) No, human hair is just too thin to provide insulation anything near what clothes can provide. We don't have polar bear hair.

2) Shaving a beard may make hygiene easier (no food collecting in the beard) and make emotions more obvious if smiling and frowning weren't visible before. It also can make the man cooler in summer and avoid icicles forming in it in winter. Then there's just style and cultural or personal preference. StuRat (talk) 05:30, 8 July 2017 (UTC)[reply]

Did you actually bother to read Body hair or Beard before coming here, 50.4.236.254? {The poster formerly known as 87.81.230.195} 90.211.129.9 (talk) 05:40, 8 July 2017 (UTC)[reply]

It does not talk about fur. It does mention furry mammals, but still, humans today still have hair. It may not be as numerous as before, but it's not labeled as fur. 50.4.236.254 (talk) 11:48, 8 July 2017 (UTC)[reply]

Humans don't have fur. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:17, 8 July 2017 (UTC)[reply]

A thick head of hair does provide a bit of protection against cold or against sunburns of the scalp, although not substantially... As for the rest of body hair, it's less effective than the adipose/fat tissue under the skin. —Paleo Neonate - 12:50, 8 July 2017 (UTC)[reply]

Does this mean that humans with more adipose tissue have an easier time staying warm during the winter but harder time staying cool during the summer than humans with more muscle mass than fat mass? 50.4.236.254 (talk) 14:10, 8 July 2017 (UTC)[reply]

They said hair provided some protection from sunburn so there is no implication that way in what has been said. Please read adipose tissue which discusses adipose/fat tissue further. Dmcq (talk) 14:58, 8 July 2017 (UTC)[reply]

PaleoNeonate did also say "As for the rest of body hair, it's less effective than the adipose/fat tissue under the skin" so they did raise adipose/fat tissue insulation issues although didn't imply insulation works the way the IP seems to think it does. I agree the IP really needs to learn to read more rather than just asking whatever crap comes to their head, presuming they really aren't trolling which is getting harder and harder to believe. Nil Einne (talk) 15:36, 8 July 2017 (UTC)[reply]

What kind of questions do you expect or want to see? 50.4.236.254 (talk) 16:04, 8 July 2017 (UTC)[reply]

From you? Probably none would be the best solution. If you can't help it, at least put some basic thought and preferably reading/research before asking stuff like "Is there a governing body for astronauts? Are they excused from paying taxes? How are food resources used? Do they have families in space too?" Nil Einne (talk) 17:40, 8 July 2017 (UTC)[reply]

Long hair is nice but it doesn't extend indefinitely - there is a catagen cycle where hairs drop out, which usually doesn't let it get to storybook lengths. But we do have a picture in that article of a German model with some really long hair -- I don't know how she does it. Wnt (talk) 15:37, 8 July 2017 (UTC)[reply]

We should discuss some of the advantages of clothes over thick fur/hair. They allow us to select whatever clothes are appropriate for that day or even portion of the day, while fur/hair can't be changed nearly so quickly. So, while animals may have a summer coat and a winter coat, we can adjust our level of insulation based on the daily weather, can select waterproof or water-resistant clothing when rain is expected, and can clean or replace our clothing when it's dirty or damaged. Compare this to how a cat cleans it's fur with it's tongue, regularly resulting in hairballs, and you can see how much better clothing is. Also, fleas and ticks are more of a problem for animals with thick hair and fur. Sweating is more effective for cooling with bare skin, although some animals, such as horses, sweat with thick coats of hair. Military camouflage can also be changed quickly when it's in the form of clothes, versus hair or fur. Note that the most flexible camouflage in nature is from bare-skinned animals, such as cuttlefish and chameleons. There's also a range of special-purpose clothing, like a dry suit, chem suit, and astronaut's suit. StuRat (talk) 20:15, 8 July 2017 (UTC)[reply]

There is a myth among short-haired people that long hair is hot in the summer, but nothing could be further from the truth. The way hair lays allows an outer layer to take the brunt of the sunlight, while a lower layer wicks sweat from the body and stays cool. (In many other animals these layers are specialized, with guard hairs for the former function). Without hair, the sweat just drips off all over, a mostly useless waste of water. Wnt (talk) 10:05, 9 July 2017 (UTC)[reply]

Depends a bit on the type of hair. Afro-textured_hair#Evolution discusses how this is considered to be the type of hair all our ancestors had, and indeed provides a cooling effect. Various other types of hair developed in colder climates most likely have subtle differences in total insulation/air circulation effects. SemanticMantis (talk) 15:06, 9 July 2017 (UTC)[reply]

Digital audio: any advantage for whole number of cycles?

In digital audio, is there any advantage -- other than the computational optimization when synthesizing -- to using sound waves whose period is a whole number of samples (e.g. 441 Hz instead of A440 on a CD, since the sample rate is 44100 Hz)? I ask because 44100 is 2²3²5²7², which suggests a particular septimal tuning if the answer is yes. Neon Merlin 13:52, 8 July 2017 (UTC)[reply]

Theoretically there is no advantage —– the Nyquist–Shannon sampling theorem says that a sampling rate of 44100 Hz will perfectly capture any signal made up of frequency components below 22050 Hz. However if the program cuts corners in its signal processing algorithms, it is hard to rule out effects. Looie496 (talk) 14:32, 8 July 2017 (UTC)[reply]

A very simple monophonic tone Synthesizer for Electronic music is based on frequency dividers fed by a single oscillator. This allowed an early use of the speaker in the IBM Personal Computer to generate waveforms using the programmable interval timer. However division by whole numbers of clock cycles restricts the available frequencies so that Septimal tuning is possible but the conventional (in the West) Equal temperament tuning where successive note frequencies are in the ratio ¹²√2 ≈ 1.059463 can only be approximated. Modern digital music synthesizers use DSP techniques to calculate the amplitude of each sample in the time domain, and the stream of digital samples passes to a Digital-to-analog converter followed by a low-pass Reconstruction filter that removes the spurious effects of whatever finite sample frequency was used, see Whittaker–Shannon interpolation formula. Blooteuth (talk) 16:16, 8 July 2017 (UTC)[reply]

Lizard Identification

Looking to identify this lizard spotted on a tree in Home Depot parking lot in Stuart, FL. I suspect it is invasive / exotic pet. https://www.flickr.com/photos/somedumbaddress/35669273661/in/datetaken/ Thanks! — Preceding unsigned comment added by 2602:306:C4D5:DF60:FD57:922B:4DC7:DB82 (talk) 18:53, 8 July 2017 (UTC)[reply]

Looks like a Knight anole, which are native to Cuba but have been widely introduced to South Florida, which includes Stuart, Florida. StuRat (talk) 20:28, 8 July 2017 (UTC)[reply]

Yep, looks like that to me too. Certainly some type of anole, and the size and head shape do look like the Knight. SemanticMantis (talk) 15:02, 9 July 2017 (UTC)[reply]

Thanks, I think your ID is correct. I hadn't checked through the Anole family because I assumed they were all much smaller, like the more common Brown and Green Anoles. — Preceding unsigned comment added by 2602:306:C4D5:DF60:FD57:922B:4DC7:DB82 (talk) 17:13, 9 July 2017 (UTC)[reply]

You're quite welcome. StuRat (talk) 15:50, 11 July 2017 (UTC)[reply]

Resolved

StuRat (talk) 18:59, 10 July 2017 (UTC)[reply]

July 9

Non-human neurobiology, memory cues, and episodic memory

For some reason, certain objects around me trigger a specific episodic memory. The house always reminds me of the time when I literally drew a picture of it with colored pencils. When traveling to a destination by car, my spatial memory allows me to find a specific location. Then, I remember how elephants can migrate and remember exact locations, provided that humans allow them to migrate beyond the borders of nature parks and countries. Could it be that episodic memory evolved because it enabled creatures to remember specific things, like a watering hole or abundant food source during a specific time of year, and to recognize patterns? 50.4.236.254 (talk) 02:23, 9 July 2017 (UTC)[reply]

Absolutely. The memory of "what did I do the last time I was in this situation and was it the best choice" is vital to survival, and thus passing down genes. Of course, other animals won't think of it in those terms, and may only recall fear when they encounter an animal that wounded them in an earlier encounter, and then back off from the confrontation because of that fear.

As far as experiments to show this, there's the ever popular mouse running a maze experiment. The more they do a particular maze, the more they remember of which way to turn at each intersection, etc., and the quicker they get to the food at the end. Then there's Ivan Pavlov's dogs, which learned to associate a bell with food.

Incidentally, this type of memory may take up lots of space, so animals with larger brains may have substantially more of it, perhaps more than humans. StuRat (talk) 02:28, 9 July 2017 (UTC)[reply]

Also interesting: you may remember smelling when winter or spring comes and being overwhelmed by pleasant memories. There are strong links between olfaction and memory (also more at Olfaction). Other animals also obviously rely much more on scent than we do. —Paleo Neonate - 14:29, 9 July 2017 (UTC)[reply]

Can a person distinguish Greenwich mean time and coordinated universal time when setting a clock manually?

A request for an argument by a known sockpuppet, rather a genuine question.— Preceding unsigned comment added by Dbfirs (talk • contribs) 16:42, 10 July 2017 (UTC)[reply]

This article Reaction time suggests that they can't. 81.131.180.213 (talk) 07:29, 9 July 2017 (UTC)[reply]

Yes, because the difference is sometimes greater than the 200ms reaction time, and, with a bit of practice, I used to reduce the reaction time when setting my watch from the time signal to less than 100ms. I had to reset my accurate watch every time there was a leap second. I agree that most people are not that fussy about a fraction of a second, and most clockwork and quartz clocks cannot maintain the accuracy necessary. Devices that self-adjust from a GPS or long-wave time signal will certainly show a difference between mean solar time and Coordinated Universal Time. "GMT" can be used for either of these times, depending on context. Dbfirs 07:43, 9 July 2017 (UTC)[reply]

This seems to be something of an oxymoron. Greenwich mean time has been around since 1840. Coordinated universal time has been around since 1960. The various scientific terms have precise meanings - if your claim that GMT can be used for UTC is correct one would expect that UTC can be used for GMT. Do you have a reliable source for your assertion? 81.131.180.213 (talk) 08:02, 9 July 2017 (UTC)[reply]

UTC has a precise definition, unlike "GMT" which is used in two differing ways, though the difference is always less than a second and most people don't care about the difference. As timekeeping becomes more accurate, terms adjust their meaning, as stated in our article Greenwich mean time. I agree that for navigational purposes, GMT might retain its mean status. Dbfirs 12:26, 9 July 2017 (UTC)[reply]

The claim that GMT "is used in two differing ways" did not appear in your answer as originally posted. Neither of your responses provides references, although all refdesk answers are supposed to include them. The only reference you've provided is a Wikipedia article, which is not a reliable source. Are your answers simply your own personal opinion (something which is not allowed on the refdesk)? "Terms adjust their meanings" is another oxymoron - the adjustment is done by people. Science being a precise discipline I find it rather far - fetched that scientists, having gone to the trouble of providing precise definitions for different timescales would then use the names indiscriminately. Back in the 1920s "Greenwich Mean Time" did have two meanings - the civil meaning of "mean solar time at the Greenwich meridian counted from midnight" and the scientific meaning of "mean solar time at the Greenwich meridian counted from noon". In 1925 the civil timescale and the scientific timescale became identical (measured from midnight) and the astronomers started using the snappier title "Universal time", something which did not catch on with the general population, who continued to call the timescale "Greenwich mean time". In 1960 the civil atomic timescale was given the name "coordinated" to distinguish it from the rotation - based "mean". Are you saying the convention has changed? 86.159.235.7 (talk) 16:22, 9 July 2017 (UTC)[reply]

On the contrary, the reference desk is for directing enquirers to appropriate Wikipedia articles. I suspect that you are continuing an old argument, so I will not respond further. Dbfirs 20:24, 9 July 2017 (UTC)[reply]

This is becoming rather murky. On 24 April 2015 the lead included the statement

Coordinated Universal Time is kept within 0.9 seconds of GMT.

The statement was referenced, and having read the book I can confirm that it is properly sourced. On that day Dbfirs added the unreferenced claim

GMT is very close to Western European Time, and is often considered to be identical to Coordinated Universal Time in the UK.

The same evening he added this referenced claim:

GMT is now UT1 or UTC.

The reference did not support the claim (it didn't mention coordinated universal time, as you can see)[1] so the "BRD" cycle continued with the changes being reverted and an explanation provided in the edit summary. The next step should have securing agreement on the talk page. Instead Jc3s5h decided to tag - team by starting an edit - war backed by another reference which was not easy to verify as no URL was provided. When it was checked it nowhere stated that Greenwich mean time is now identical to coordinated universal time, as claimed. To add insult to injury, he removed a reference dating from 2014 with the spurious edit summary claim that it was 39 year[s old].

On the talk page it was pointed out that Seago, Seidelmann and Allen [a source cited in the article] confirm that

UT1 ... is a precise astronomical measure of the rotation of the Earth on its axis, synonymous with mean solar time at the meridian of Greenwich, sometimes known simply as Greenwich mean time (GMT).

Dbfirs didn't respond but instead made the counter - claim that the British government "seems to think that GMT has been the same at [sic] UTC since 1972". Examination of the source [2] shows him using synthesis to push his POV.

On 7 June 2015 the claim

Today GMT is considered equivalent to UTC for UK civil purposes (but this is not formalized)

was added, and the reference provided, when examined, was found not to mention Greenwich mean time at all. It appears that Dbfirs has some explaining to do - perhaps this is why he has disengaged from the discussion.

The great thing about Refdesk participation is that you don't have to give a damn about all these Wiki behavior issues and can just focus on the facts. The most basic one is that "GMT is a time zone and UTC is a time standard". [3][4] According to that site, GMT does not change with daylight savings silliness; instead, Britain (and presumably Greenwich ... what an indignity!) change to British Summer Time. Someone on stack overflow claims that computer systems may switch on GMT, but he got a downvote ... I'll take it as a faint warning. NOAA says [5] that UTC is simply the new name for GMT, and the time servers, by saying that GMT is a time zone and giving UTC+0 times for it, seem to be implying the same thing. This site [6] says that some astronomers still use a pre-1925 definition of GMT that is ... somehow... different than the current. So far, I'm not seeing the 0.9-second thing coming out in the top search hits, apart from Wikipedia and those derived from it, but I can't say it's not true. Wnt (talk) 20:21, 11 July 2017 (UTC)[reply]

According to StackExchange, one (Greenwich Mean Time) is derived from the sun - it's the one sundials show (after adjusting for the equation of time). It's the one marine chronometers (and hence ordinary clocks and watches) aspire to. The other (coordinated universal time) is measured from an atomic clock. The sun is not a perfect timekeeper (it runs slow - see Delta T) so to keep coordinated universal time from wandering too far from sun time the occasional leap second is added to rein it in. 81.151.129.198 (talk) 20:43, 11 July 2017 (UTC)[reply]

A good reference for discussions of GMT vs. Universal Time is an upcoming paper by Seago and Seidelmann with abstract at http://sot2016.cfa.harvard.edu/SoT2016/cgi-bin/TXT/Invited/Invited_SeagoJohnH.txt_N.html where they analyze the difference between mean solar time at Greenwich and Universal Time. With many more computations and diagrams they come up with the same answer that Sadler got in 1954 http://adsabs.harvard.edu/abs/1954ONRAS...3..103S Steven L Allen (talk) 22:37, 11 July 2017 (UTC)[reply]

Rare events predicted by a "theory of everything"

For example, most grand unified theories predict proton decay with a very long half-life, something not predicted by the standard model. Are there any extremely rare events predicted by a theory of everything that are not predicted even by a GUT, let alone the standard model? Surprisingly, I can't seem to find an answer in our articles on theory of everything, string theory, or M-theory. PeterPresent (talk) 14:36, 9 July 2017 (UTC)[reply]

The "theory of everything" does not exist yet. --AboutFace 22 (talk) 17:55, 9 July 2017 (UTC)[reply]

https://arxiv.org/abs/hep-ph/0703221 Count Iblis (talk) 20:09, 9 July 2017 (UTC)[reply]

Thanks for that paper! So, according to that paper, a prediction of string theory is that the universe will transition into an "exactly supersymmetric" universe with different properties. Is this actually a novel prediction of string theory though? Aren't there GUTs or other theories (e.g. Minimal Supersymmetric Standard Model) that also predict this phenomenon? PeterPresent (talk) 01:50, 10 July 2017 (UTC)[reply]

I think that within string theory it's harder to get around the exact supersymmetric ground state than within effective field theories such as the Minimal Supersymmetric Standard Model where you can introduce all sorts of terms in the Lagrangian to change the predictions. Count Iblis (talk) 21:27, 10 July 2017 (UTC)[reply]

The biggest criticism of string theory / M-theory is that it hasn't offered any testable predictions that distinguish it from other theories. There are a few possibilities that aren't very testable. Some versions of M-theory allow for other "universes" parallel to our own but whose gravity can still affect our universe (and sometimes identified with dark matter). Finding evidence of that would be exciting, but you can also have M-theory without that element (or with other universes so distant as to be completely undetectable). Similarly, string theory allows for compact extra dimensions. Evidence of such extra dimensions could show up in particle accelerators, or the extra dimensions could be a billion billion times too small to be seen with current accelerators. So, if you find them, it would be good evidence, but not being able to find them proves nothing. String theories can also give rise to a wavelength-dependent variable speed of light and time-variation of fundamental constants, though neither concept is exclusive to these theories, nor are the effects required to be remotely large enough that humans are likely to be able to observe them any time soon. These theories sometimes also predict non-trivial internal structures for black holes (e.g. fuzzball (string theory) and might have implications for Hawking radiation, but it is unclear if those ideas are testable even in principle. Dragons flight (talk) 08:08, 10 July 2017 (UTC)[reply]

Sometimes, a theory can make a prediction that some extremely rare event may occur. For example, although most predictions of GUTs would in practice require experiments in the GUT scale, most GUTs predict that protons decay - something that could occur not at all in the GUT scale - but only very rarely. I'm interested in predictions like that. Does string theory / M-theory / TOE predict some extremely rare phenomenon - even on normal energy levels - that is not predicted by our current theories of even GUTs. PeterPresent (talk) 08:24, 10 July 2017 (UTC)[reply]

A theory of everything is needed to unite gravity with quantum mechanics, which is believed to be the last stage in connecting the various known forces in the universe. Pretty much any particle physics process imaginable, such as proton decay, can already be predicted in some GUT theory without the need for gravity. Where a TOE makes new predictions, to the extent that any of them do, is in areas like gravitation, dark matter, black holes, etc. I already gave you examples of those. Dragons flight (talk) 08:53, 10 July 2017 (UTC)[reply]

Our article on gravitons (which emerge in many quantum gravity and string theories) has a quite entertaining discussion of how rare the interaction between gravitons and matter would be, and how large a detector you'd need to see one (spoilers: the required neutrino shield would be so large it would immediately collapse into a black hole). Smurrayinchester 08:59, 10 July 2017 (UTC)[reply]

Learning/teaching systems thinking

How can we learn or teach systems thinking? --Hofhof (talk) 19:30, 9 July 2017 (UTC)[reply]

First it needs a definition. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:52, 9 July 2017 (UTC)[reply]

It seems to be mainly founded on logic. Learning logic is a mix of understanding the principle and then practice repeated application to achieve a lasting imprint in out memory. --Kharon (talk) 09:07, 10 July 2017 (UTC)[reply]

That makes no sense Kharon. According to the corresponding articles - systems thinking and systems science - and to a trivial logical analysis systems include any kind of field that has an effect on an issue: cognitive, society, science, math. I'd bet a good 'systems thinker' has broad general knowledge about the involved issues. --Clipname (talk) 16:22, 10 July 2017 (UTC)[reply]

irrelevant comment
Looks to me like you dont know how much of our world is based on logic. Maybe i can convince you with irony: Ask an economist! They are said to teach the worst in both, bending all reality to their "logic" and "making sense". --Kharon (talk) 18:00, 10 July 2017 (UTC)[reply]

This is WP:NOTAFORUM for abstract debate, my friends. The OP's question is arguably too broadly phrased for us to answer without engaging in extensive speculation (which is not the point of this space); the thread can stand for the purpose of providing references that might be useful, but not our personal opinions. On that subject, the OP may be interested in the works of Fritjof Capra, who has made a decades-long endeavour out of trying to find ways to encourage systems thinking in the mainstream. In particular, The Turning Point may be off interest, as might Mindwalk, a quasi-narrative film that is based on his works. Snow ^{let's rap} 23:42, 10 July 2017 (UTC)[reply]

By not shying away from hard core reductionism. E.g. Biology = systems chemistry, see also here and here. Count Iblis (talk) 02:07, 11 July 2017 (UTC)[reply]

What temperature range can feral pigeons survive in?

At what temperature do they start feeling uncomfortable? Do these numbers change with the seasons? If so, what are the seasonal adaptations and what triggers them? Does the climate of an area feral pigeons are introduced to selectively concentrate the subspecies and sub-subspecies most fit for that climate? (similar to how the European climate selectively concentrated white people) Sagittarian Milky Way (talk) 20:11, 9 July 2017 (UTC)[reply]

I can attest that below −20°C they start feeling uncomfortable (seeking warm place). Ruslik_Zero 20:36, 9 July 2017 (UTC)[reply]

For the last bit, sure. We have shitty sub at species sorting but if you search that and "environmental filtering", you'll find lots of research and information about how species' abundances are tied to adaptation to local climate, resources, etc., and how that ties in to natural selection Careful about using the term "most fit" though, it's not a very good term, and tends to be used by people spouting discredited social darwinism. It may be that the feral pigeons of Minneapolis have slightly different traits than those of Miami, due to natural selection, but by virtue of stably persisting, they are all equally fit as populations. I'd WP:OR not be surprised if a box of pigeons moved from WI to FL had slightly less reproductive success than their native-born counterparts, but I'm not aware of any studies on the topic, nor could I find any. This [7] recent paper has some nice discussion of the genetic variation, convergence, divergence, etc. of feral pigeons. There are also some comments on how/why local populations may differ. I only skimmed it, but if you read it carefully and look at some of the refs (and look at papers that cite this one [8]), you'll know more than you even wanted to about pigeon genetics and spatial distribution! SemanticMantis (talk) 19:10, 10 July 2017 (UTC)[reply]

Oh, and this paper [9] is all about geographic variation in pigeon size, and explicitly invokes natural selection by environmental variable acting on heritable size traits. It's surprisingly right on target for explicitly answering your question, that's rare, so enjoy :) SemanticMantis (talk) 19:13, 10 July 2017 (UTC)[reply]

Thanks. Sagittarian Milky Way (talk) 19:28, 10 July 2017 (UTC)[reply]

Also of interest may be adaptive radiation (which is very general, but also affects birds in large continuous ranges). —Paleo Neonate - 19:38, 10 July 2017 (UTC)[reply]

July 10

Ground locomotion of birds

Today I saw a sparrow hopping and a pigeon walking. That got me to wondering—is there a pattern in bird taxonomy that predicts which ones hop and which ones walk? Or is there some other determining factor such as weight or the general environment they're adapted to? I couldn't find this out from our articles on locomotion or birds. Loraof (talk) 01:38, 10 July 2017 (UTC)[reply]

Then there's the odd-looking method of walking some birds use, where they hold their head stationary while walking, then jerk it forward, and repeat the cycle. This seems to be necessary because their vision system can't cope with movement of the eyes particularly well. In particular, their ability to detect objects in motion is compromised when everything is in motion, relative to them. As for hopping versus walking, is it just a matter of size ? It seems to me that small birds tend to hop while larger birds walk or run. (Of course, there are large mammals which hop, like kangaroos, but they have massive leg muscles, which would interfere with flying in a bird.) StuRat (talk) 05:49, 10 July 2017 (UTC)[reply]

What birds are you referring to that keep their heads still and move their bodies forward. Richard Avery (talk) 07:37, 10 July 2017 (UTC)[reply]

Pigeons, for one. Matt Deres (talk) 15:48, 10 July 2017 (UTC)[reply]

And chickens. (I got to watch them a lot this weekend.) Dad remarked that all their head motions are jerky, even when their feet are not moving; I said well, I guess, when your head muscles are adapted for staccato … —Tamfang (talk) 16:45, 10 July 2017 (UTC)[reply]

Yes, the same problem applies when they turn their head. If they did it slowly, they wouldn't be able to detect motion during the turn, so they do it quickly, so as to minimize the time when their vision is impaired. Sort of like how we blink, for the same reason. But apparently one thing a bigger brain does get us is the ability to detect relative motion even when our heads are moving. And not just us, but birds with larger brains seem able to do this, too. StuRat (talk) 17:25, 10 July 2017 (UTC)[reply]

Stu. Don't post your opinion and intuition as fact, please. This is all quite wonderful and fascinating speculation to you, I'm sure. But it's also all amateur guesswork, and completely inappropriate for a reference desk. For references that actually describe our knowledge on the topic, look at chicken cam image stabilization, e.g. here [10] or here [11] or even look at the scholarly literature on head-bobbing in chickens [12], [13] . Why make thing up? Especially when even a casual search on google scholar gives plenty of good references on the topic. Please research and include references when posting. Your guesses are well-intentioned I'm sure, but not helpful, and in this case misleading. You know better by now. You can do better. I know you can ;) SemanticMantis (talk) 01:23, 11 July 2017 (UTC)[reply]

Almost all birds can and will do both, walking and hopping, depending on their mood and situation. Some birds even dance sometimes! Some dance formal and slow paced, some like going wild. I think the determining factor is their default stance/degree of aggression and that shure fits our view of the peacefull pigeons. --Kharon (talk) 08:36, 10 July 2017 (UTC)[reply]

It has to do with energy conservation. Birds do whatever gets them what they want with the least expended energy. Smaller birds tend to hop because it uses very little energy and gets them where they want to go quickly. Larger birds tend to walk because it requires less energy to do so. There are exceptions. Further, birds who spend most of their time foraging in trees tend to hop because walking from branch to branch is hard. Birds who spend most of their time foraging on the ground tend to walk. [14] [15]. I personally believe it has to do with society as well. I live in South Carolina. We have sandpipers. They run quickly from place to place. When I was in Bermuda, I noticed that their sandpipers hop instead of run. I'm certain it is a different type of sandpiper, but my gut tells me that the Bermuda ones hop because they see all the others hop. The Carolina ones run because they see all the other ones run.209.149.113.5 (talk) 12:30, 10 July 2017 (UTC)[reply]

Great answer and references, thanks. But I have to weigh in on your last comment "...because they see all the others hop": I can find no evidence of any bird where a hopping species can be trained to walk, or a walking species can be trained to hop. Hopping vs. walking seems to be purely set by species, not culture/learning, as far as I can tell, and that is the perspective of the research I link below. SemanticMantis (talk) 14:11, 10 July 2017 (UTC)[reply]

I intended to make it clear it was my opinion and my opinion was completely unfounded. After searching, I found numerous resources claiming that sandpipers hop on one leg in colder climates. I was Bermuda in February, so it was colder than the South Carolina beaches. Perhaps if I go to North Carolina, I will find hopping sandpipers. 209.149.113.5 (talk) 14:51, 10 July 2017 (UTC)[reply]

All good, I was just offering a contradictory viewpoint. No need to strike out your text, you did make it clear it was speculation, and it might even be true! I'm a bit chagrined that I cannot (yet?) find a good reference that explicitly addresses my claim that that walk vs. hop is almost totally genetic, but I'm getting close. Here [16] is the best work I can find so far on multiple gaits within a single species, discussing walking, running, and (out of phase) hopping in the magpie. It does casually mention that "most species use only one or two gait types," viewing these magpies as a bit of an outlier. This paper [17] has a very good reference list in the introduction, and says that "patterns of walking, running, and even hopping are conserved among birds". This would indicated that gait is a conserved trait taxonomically speaking, i.e. that we'd generally expect many members of a given bird clade to share the same gait, though this does not rule out some members using derived gaits. SemanticMantis (talk) 15:28, 10 July 2017 (UTC)[reply]

Here [18] is a nice analysis of how bird morphology influences striding gaits. Hopping is not considered, but it does give good background on energetics, morphology and gait. As for hopping vs. walking, this recent research (2013) [19] finds that "Cervical movement capability is perhaps the single most deterministic factor in the bird's choice of terrestrial gait. [i.e. hopping vs. walking]" SemanticMantis (talk) 14:11, 10 July 2017 (UTC)[reply]

It would be interesting to see a proper study of the difference. Astronauts on the moon found it easier to move around by hopping. Dmcq (talk) 15:01, 10 July 2017 (UTC)[reply]

See the article Bird feet and legs. After they evolved the ability to fly, other kinds of locomotion such as walking, running, hopping, climbing, swimming became of secondary importance and these abilities evolved differently in various bird species. An essential hindlimb function is to accelerate when taking-off and to absorb the shock of landing. Further adaptations in specific bird species are to use the feet as "hands" (the forelimbs being reserved as wings) for manipulations such as grasping a perch or prey, pulling food apart, scratching the ground, building a nest, turning eggs, self preening, etc. Efficient ground locomotion became critical for survival only in the cases of birds losing the ability to fly which has occurred in many different birds independently, generally in situations of no ground predators or deliberate selective breeding by man. They include flightless domesticated Fowl (chicken, turkey, duck), penguins, and Ratites (ostrich, moa, emus,...). The pectoral muscles for flight decrease and the pelvic girdle for running enlarges. Blooteuth (talk) 15:05, 10 July 2017 (UTC)[reply]

may be we need some article, ground locomotion of birds, or at least some material in bipedalism specific to birds?

Gem fr (talk) 13:17, 11 July 2017 (UTC)[reply]

July 11

How hot can CCFLs get before breaking?

How hot can a CCFL get before sustaining damage? (note that CCFL is different from a CFL). I'd like to use about 50 of them (from broken laptop monitors) in a project as close together as possible and expect ~200 W waste heat. --145.255.245.88 (talk) 07:48, 11 July 2017 (UTC)[reply]

Same as a CFL - are you talking about the tubes, or their drive electronics? The electronics generate most of the heat (especially for CFLs, less so for CCFLs) but the tubes are prety much insensitive to this. Although they're a "cold cathode", that only means that they're not a directly heated cathode, it doesn't mean that theyir cathodes have to be kept cold. Andy Dingley (talk) 10:25, 11 July 2017 (UTC)[reply]

If this project is intended to be viewed by the public, then simply use an electric fan or two so as to avoid roasting the public. If the public feel comfortable so will the monitors. Also, it would help if you explained the spacial arrangement of said monitors. 75 degrees gives max efficiency. Ie. the most light for the energy converted into light. Aspro (talk) 10:39, 11 July 2017 (UTC)[reply]

In general, the CCFLs I've seen have operating temperature ranges that go up to 50 or 60 C. If you know what parts you are using, you can often find a specification online. That will tell you want the manufacturer recommends. As to how hot your installation will get, that will depend on the airflow, the mounting materials (a metal backing would remove heat more effectively than a wood one, for example), and other aspects. If 200 W is the total dissipation for 50 monitors, that is actually not very much energy, and unless they are completely enclosed and lacking in airflow, you are probably fine. Dragons flight (talk)

Thanks for everyone's responses. I'm talking about taking the CCFLs out of the monitors and lining them all up next to each other for maximum density. I expect to be able to fit all 50 CCFLs in a space about 25 cm x 35 cm. I was hoping to passively cool them by attaching to a thin (~4 mm) aluminium plate but the way each tube is terminated precludes physical contact between the tube and the metal plate. I don't want to use a fan because I hope to use this light for (non-professional obviously!) video recording. 145.255.245.88 (talk) 17:03, 11 July 2017 (UTC)[reply]

Hotel body soap, hand soap, and shampoo

I looked through the archives and didn't find what I was looking for.

In hotels, they have all three. I'm guessing they're all pretty much the same, but difference smells and colours. Is this about right? Anna Frodesiak (talk) 07:48, 11 July 2017 (UTC)[reply]

Soap and shampoo have a couple of differences: manufacture and use.

Their use is that "soap" ought to feel soapy and should lather well. A "good" soap is perceived as being the one that feels "luxuriant" on this basis. A difference with body soap (as a rarely made distinction) is that hands are washed quickly, bodies more leisurely. So a hand soap isn't quite so richly lathering, as you have to get it on and off without waiting around for it. Shampoo is different - you want the minimum amount of lather, as that has to be rinsed clean in turn. It takes longer to rinse hair than to wash it clean. Also, if conditioning the hair afterwards, any remaining shampoo will make the conditioning ineffective, so there has to be a four stage wash-rinse-condition-rinse cycle.

Their manufacture is that they can be either soaps or detergents. Most are increasingly detergents rather than soaps. Local product labelling laws may require non-soap detergents to be labelled as "handwash" rather than soap. It's rare to see a bath "soap" that's a detergent, as they just don't feel like luxury. There are though plenty of in-shower bodywashes based on detergents. These are mostly pitched as "invigorating" rather than "luxurious" and they sell on the basis of fragrances and spicy tingles (don't miss the 'flaps on fire' post (non-FB link) - I have this stuff, they're not joking). Liquid hand "soaps" are almost all detergents, because they're quicker to rinse. Shampoo is all detergent based, not soap, because although soap is usable for hair washing (and used to be, as there was little else) it takes a huge amount of rinsing.

Incidentally, the "two in one" shampoo and conditioner formulas work by mixing an easily rinsed shampoo with a lot of conditioner, and making this conditioner persistent against washing and rinsing. By the time you've rinsed the shampoo out and the conditioner can get to work, there's still enough left to do it. Andy Dingley (talk) 10:36, 11 July 2017 (UTC)[reply]

Andy, I don't know how you know all this, but I am well impressed. I had no idea. I cannot view 'flaps on fire' because facebook is blocked here in China. Thank you so much for the very educational answer! I am grateful. Anna Frodesiak (talk) 11:40, 11 July 2017 (UTC)[reply]

Note that soap is made from fats, via saponification, whereas detergents are not. StuRat (talk)

Back in the old days, chemists and pharmacists weren't as protective of their trade secrets as one might think. Sure, they'd have their own tonic recipe that they kept secret, but for the hundred mundane things they had to make and sell anyway, there were standard industry formularies like The Chemical Formulary and Spons' Workshop Receipts. These listed a great many standard perfumes, soaps and shampoos. Often bulk soap would be made in large factories (a smelly process, not popular in towns) and large blocks at different grades (from faces to nits to floors) would be sold to a retail chemist who would blend them and package them.

These days I sell at craft markets, which often involves talking to high-end soap makers. To sell artisan soap today, you have to find a way to make it special.

Really I don't know much about this stuff at all, but I keep a house chemist who does. There's little other work for an industrial chemist. Andy Dingley (talk) 12:20, 11 July 2017 (UTC)[reply]

and "flaps on fire" is a cautionary tale, unaccountably omitted from Struwwelpeter, about bathing in Mint and Tea Tree Shower Gel. The warning is addressed to women but, from the replies, is also applicable to men. Hopefully it isn't on sale in China. Thincat (talk) 12:35, 11 July 2017 (UTC)[reply]

Ah, I found news items about the facebook page. I will avoid such products.

Thank you all again. And I guess all soaps in liquid form are detergent here, but am not sure. Anna Frodesiak (talk) 15:40, 11 July 2017 (UTC)[reply]

Almost all liquid "soaps" that are transparent will be detergents. It is possible to make liquid soaps (that are soaps), but they're rare these days, because detergents are cheaper and also liquid soaps store badly, having a tendency to set in the nozzle or pump. Detergents are more efficient (less mass needed for a wash) and so are much cheaper to manufacture in the same apparent quantity. Most people use much more detergent than they really need.

Transparent soap has an interesting history of its own, but I think it's solid, not liquid (I don't know of any transparent, liquid, true soaps). They're made from glycerine (although not all glycerines are transparent). Pears soap was the first such soap, the first branded soap and one of the world's first major retail brands. At a time when product adulteration was rife and some badly-made soap was still caustic, the transparency, brand identity and general high quality of Pears' were factors in making it an easily recognised premium brand. Andy Dingley (talk) 16:04, 11 July 2017 (UTC)[reply]

Yes, when you say "...much cheaper to manufacture...", it convinces me that hotel body soap, hand soap, and shampoo are all just detergent. Anna Frodesiak (talk) 18:14, 11 July 2017 (UTC)[reply]

Rotating an object with circularly polarized light

Suppose you have a beam of circularly polarized light, and it strikes an object. If absorbed, it should tend to rotate the object, though only to a very small degree as the Planck constant of momentum per photon is small: 6.6E-34 J*s (= N*m*s). Even a meter-long microwave photon carries relativistic mass-energy of 1.2E-6 eV = 2E-25 J, so I think to apply the equivalent of a middling 1 N*m torque screwdriver for one second (i.e. 1 N*m * s) you need 3E+9 J of energy, which our joule article describes as the energy of a one-ton vehicle moving at 100 mph. Times three, plus another factor of two because photons have *half* a Planck unit. Does that sound right to you?

However... photons don't have to be absorbed. They could reflect, and mirrors can be up to 99.999% reflective. That means that in concept you could recycle a photon 100,000 times, getting double the angular momentum transfer each time ... provided, that is, you can come up with a really good mirror, perhaps of some metamaterial, that works at the right (long) wavelength and which reverses the polarization of the photon striking it. Though I think this is actually the norm - not reversing it would actually be unusual! That would knock down the required photon energy to 1.5E+4 J, more like the solar radiation striking the Earth each minute.

I also get curious about the linear momentum (or impulse (physics) ... hmmm, is there a meaningful difference?) transferred in this system: photon says h/lambda (i.e. wavelength); for those one-meter microwaves that is 6.6E-34 N*s per photon. In the mirror system we needed 1.5E+4 J/2E-25 J = 0.8E+29 photons, so that would be still something like 5E-5 N*s of momentum, not much... wait, no... the mirror reduces the amount of energy needed, but not the force of each impact -- so actually it's the full 3E+9 J/2E-25 J = 1.5E+34 photons (that should be twice the inverse of the Planck number) that hit, carrying 2 N*s of kick, certainly not negligible. It could be made smaller with radio wave photons...

Anyway, the questions:

a) can you point to any good lab demonstrations of turning an object with circularly polarized light?
b) how good do low-frequency mirrors (microwaves, etc.) get?
c) with perfect lens technology or the like, could you confine the microwaves to a cavity without completely enclosing it? How far could you get? In other words, could you use them to transmit torque between two objects at a distance from each other?
d) the linear momentum issue is also interesting -- with that, the energy for "kick" is constant, so I think that any sufficiently "perfect" set of mirrors (any frequency, and not much better than those presently available!) should be able to physically levitate objects... at least, until a mote of dust gets in. Is anyone trying to do this?

Wnt (talk) 13:16, 11 July 2017 (UTC)[reply]

Couple should suggest why this is a tiny effect and hard to measure - but it is real. I think Poynting was the first to recognise this, around 1909, a couple of years after Einstein's annus mirabilis. It took until the 1930s before it was successfully measured though. Andy Dingley (talk) 13:31, 11 July 2017 (UTC)[reply]

Easier to measure the effects at microwave wavelengths, BTW. Andy Dingley (talk) 13:35, 11 July 2017 (UTC)[reply]

You are basically describing some sort of movement-out-of-nowhere mechanism: a single photon strike a mirror A, makes it rotates, and restart the other way with reverse polarization, strike another mirror B just in front of A, makes it rotates, too, than restart again toward A with just the same rotating power than the first time, so it increase the rotation, etc.

congratulation for your new perpetual motion device, first time i read about this one.

Gem fr (talk) 13:45, 11 July 2017 (UTC)[reply]

@Gem fr: The "perpetual motion" issue is something I thought about, but from the other end: it is clear that the energy for the rotation has to come out of the light (which as I explain above, contains considerable energy) but how? For one thing, bear in mind that the energy gained by the target object is not going to be simply proportional to the angular momentum - they're two different quantities. I was still thinking about that part, perhaps for a follow-up (I have to work out how energy and moment of inertia relate, etc.), but first I was hoping to hear more about the state of the art for photon energy transfer. Note that bouncing photons between mirrors is often addressed in the linear sense - there it is clear that bouncing off a receding mirror redshifts the photon, deducting the energy fee. Real ordinary mirrors do reverse circular polarization ... hmmm. I just ran into an issue thinking about that.

If I visualize the photon as a rotating disk striking a wall, the direction that the disk rotates relative to its forward motion does reverse when it hits the wall ... but the angular momentum doesn't change. So can a photon change angular momentum in order not to give up angular momentum? Would a mirror somehow rigged (metamaterials and handwaving...) to keep the same circular polarization actually be what gets angular momentum transferred to it? I realize I am quite confused about this. But I know that some angular momentum sure as hell has to go somewhere if you bounce the photon at a 90-degree angle, because it can't keep the same plane it came in with. So the question is valid, I just don't know what I'm asking. ;) Wnt (talk) 16:06, 11 July 2017 (UTC)[reply]

I think you need to check redshift. The energy of a photon is tied with the frequency of the photon (energy = h*frequency). So, since Planck's constant h cannot change, when energy is lost, frequency is reduced. Visible light becomes shifted towards the red end of the spectrum. The photon will keep shifting to a lower and lower frequency, losing energy. Since you can't have a mirror that reflects every possible frequency of electromagnetic radiation, the photon will eventually be absorbed (or pass right through the material). 209.149.113.5 (talk) 16:55, 11 July 2017 (UTC)[reply]

I didn't intend to disagree with any of that. There are, nonetheless, some things I'm presently confused about. Bear in mind that linearly, kinetic energy is obtained by integrating momentum, by which I suppose I mean impulse as I described before. It makes intuitive sense that the faster something is moving away, the more redshifted the photon is, because in the plane of the mirror the interaction needs to be symmetrical. A bouncing photon should be redshifted by the same amount - which depends only on that difference in velocity - so it will lose just as much energy to any mirror receding at a velocity v, regardless of its mass; it will also give just as much momentum to the receding object as any other mirror moving at v hit by a similar photon. And yet ... the object's energy is 1/2 mv^2, and it doesn't seem like that increases the same for large and small m given an equal change in momentum from the photon. Now rotational energy is obtained by integrating angular momentum in a very similar manner; the velocity along the radius of gyration changes in the same manner, and I have the same confusion. Wnt (talk) 20:02, 11 July 2017 (UTC)[reply]

It isn't the case that a photon will lose the same amount of energy to any mirror that is receding at velocity v - that is only an approximation for large mirror masses. In the general case, the velocity of the mirror changes as well, by a different amount depending on the mirror mass; and that change in velocity occurs at the same time that the frequency of the photon changes.

Basically it's just a totally elastic collision of a photon and a mirror (what is conserved are the relativistic versions of energy and momentum). Icek~enwiki (talk) 20:24, 11 July 2017 (UTC)[reply]

@Icek~enwiki: That's true - the math is gone into here. But what troubles me is that if you bounce a photon off a 1-ton mirror, the delta v will be twice what it is if you bounce it off a two-ton mirror, which means that the "delta 1/2 mv^2" will be twice what it is for the two ton mirror. But the term in the paper I link for the change in photon frequency, as you'd expect, is a small inverse term for mass that is negligible compared to the velocity term. So I'm not convinced this actually is relevant to the issue. Wnt (talk) 21:31, 11 July 2017 (UTC)[reply]

For the change in kinetic energy of the mirror, we have

\Delta {\frac {1}{2}}Mv^{2}={\frac {1}{2}}M(v+\Delta v)^{2}-{\frac {1}{2}}Mv^{2}=Mv\Delta v+{\frac {1}{2}}M(\Delta v)^{2}\approx Mv\Delta v

neglecting the small quadratic term.

This gain in energy, linear in

\Delta v

, is balanced by the loss in energy described by the middle term of the right hand side of equation (6) in the article that you linked to.

But your question pertains to this neglected quadratic term, I guess. It is cancelled by the rightmost term of the same equation, just that in the next step the author neglects that term as well.

Icek~enwiki (talk) 22:17, 11 July 2017 (UTC)[reply]

@Icek~enwiki: I shudder to get into math markup, but that equation 6 is:

Change in photon energy =

\hbar \omega -\hbar \omega '={\frac {\hbar v}{c}}(\omega cos\alpha +\omega 'cos\beta )+{\frac {\hbar ^{2}}{2Mc^{2}}}(\omega cos\alpha +\omega 'cos\beta )^{2}

... for our purposes,

\alpha

and

\beta

can be 90 degrees; they are the angles light moves relative to the mirror taking into account that the incidence = reflection only in the mirror's rest frame. Now, there's a problem with this equation that it's obviously not solved for

\omega -\omega '

. Still, we know that its limit as M tends to infinity is

\omega '=\omega {\frac {1-\nu cos\alpha /c}{1+\nu cos\beta /c}}

. Now your expression

Mv\Delta v+{\frac {1}{2}}M(\Delta v)^{2}

can, with a slow-moving heavy mirror, be written under the assumption that the momentum is conserved - without angles this is

{\frac {\hbar (\omega +\omega ')}{c}}=M\Delta v

... Well, I can't say phooey, because now it seems like it works out, and I'm not sure why it didn't before. I am not getting top marks in math today. Wnt (talk) 23:45, 11 July 2017 (UTC)[reply]

I scanned to see if anyone had made this point; apologies if I missed it.

User:Wnt, assuming you're right that photons flip their circular polarization when they reflect, that actually means they don't transfer angular momentum.

That's because the circular polarization of light is defined by whether the angular momentum is in the same direction as the velocity (applying the right-hand rule), or the opposite direction.

So if you shoot, let's call it a "right-handed photon" headed at your mirror, that means that it's "spinning clockwise" (of course there's nothing that really spins AFAIK) as you look at it in its direction of motion.

Now it reflects, and its direction of motion is reversed, but it's still spinning in the same direction as you look at it. But now it's coming back at you rather than going away from you, so now that same spin is counter-clockwise for someone looking towards it in its direction of motion, and hit has become a left-handed photon.

And yet no angular momentum has been transferred. --Trovatore (talk) 20:19, 11 July 2017 (UTC)[reply]

@Trovatore: Well, as I said above, you can surely... maybe... make them transfer some angular momentum by bouncing them at a 90-degree angle. Because then the angular momentum out isn't what came in, but in a whole different plane. True, that seems to be transferring partial Planck units, which is a no-no, but what I guess it to mean is that you must end up racemizing the photons to a random mix, with the momentum necessarily going into the mirror. Wnt (talk) 20:25, 11 July 2017 (UTC)[reply]

@Wnt: Angular momentum can be a little painful to deal with at the quantum level. The problem is that the operators corresponding to the components of angular momentum in different directions do not commute with one another. So in general, a particle won't have well-defined x-, y-, and z-components all at the same time. I think maybe for a photon they can, if z happens to be the direction of motion, because the other two are zero so it doesn't matter that they don't commute. (For particles with half-integral spin, like the electron, they can basically never have well-defined angular momentum components in all directions simultaneously.)

See Clebsch–Gordan coefficients for more information. Not exactly a light read, and I don't have time right now, though I would like to get this straight in my head. --Trovatore (talk) 23:39, 11 July 2017 (UTC)[reply]

@Trovatore: As I recall, photons actually have angular momentum of sqrt(2) Planck constants. Like in a lot of systems, a smaller amount (1 Planck unit) can be measured in a given direction. Intuitively I assume that if you know a photon has a spin along its axis, it must have a "topspin" also in some unknown direction, sort of like an electron in a p orbital that really has an angular momentum of sqrt(l(l+1)) = sqrt(2) in that case. You know it is going round and round the nucleus in the complex solution, but it's also doing something else you don't know (there's a picture in magnetic quantum number). Wnt (talk) 23:53, 11 July 2017 (UTC)[reply]

I think Wnt has made this point himself already. As both of you say, an ordinary mirror does not change the direction of rotation of the circularly polarized wave, i.e. it does change the polarization (the relation between the rotation and the direction of propagation).

But if you have an interface between two materials of which at least one is birefringent, you can have a reflection without a change in polarization: The circularly polarized photon can be represented as superposition of 2 linearly polarized states, say a vertically polarized state and a horizontally polarized state, with a phase shift of π/2 between them.

When light would pass from a material A to a material B and is instead reflected, the phase of the light changes by π if the refractive index of B is larger than that of A.

What we need is that for the vertically polarized light, the refractive index of B is larger than that of A, and the other way around for the horizontally polarized light. Then the direction of rotation of the reflected light is inverted, and thus the polarization stays the same.

Because we want to make it easy to measure the angular momentum transfer, material A should be a vacuum; so we need a material that has a refractive index larger than 1 for one linear polarization and smaller than 1 for the other.

Icek~enwiki (talk) 20:24, 11 July 2017 (UTC)[reply]

bottom line: as stated above by @Trovatore:, NO momentum is transfered in the reflection process. The photon just reverse course, but keeps the very same angular momentum. Same if reflected with an angle.

there is no real energy / momentum issue, anyway: a mirror is a macroscopic object with close to infinite energy(thermal, etc.) relative to a photon. The reflected photon is just a minute contribution to the regular emission of this macroscopic object, that is, it makes no process difference in energy /momentum for the macroscopic object, whether it reflects incoming light (white body), or absorbs it then emits it as per Planck's law ([[black body}}), or anything in between (grey body): in any case, Radiation pressure applies, but the special photon considered is just one in the ocean of photons that the mirror floats in/contributes to.

Basically, you cannot cope with your question with the particle avatar of the Wave–particle duality of incoming light.

Furthermore, entropy prevents the small photon to rotate the whole macroscopic mirror

Gem fr (talk) 21:45, 11 July 2017 (UTC)[reply]

@Gem fr: I don't think physics lets you get away with "small" accounting errors. ;) Wnt (talk) 00:04, 12 July 2017 (UTC)[reply]

statistical physics. But actually the "small" accounting error would be to consider the photon, as if it mattered. it doesn't. Mirrors do not exist at the particle level, it is a wave/macroscopic conceptGem fr (talk) 07:30, 12 July 2017 (UTC)[reply]

There is some truth to what you say, in the sense that AFAIK a line of particles in the mirror bounce photons off them in random directions, and constructive interference from a large number of particles is needed to convert this from being more like a diffraction grating to being something that sends all the light out in one single direction. That said, a single photon with a poorly known position might reflect off many particles and interfere with itself in this same way, much like a double-slit experiment. Alternatively, I assume that a QM analysis of the mirror with many photons whose localization is well known would find that the interference between them ends up working out that the individual photon has a high chance of leaving on the route it is supposed to leave from a mirror, with its angular momentum still being meaningful. Wnt (talk) 12:25, 12 July 2017 (UTC)[reply]

OK, I'm convinced the math works out for linear momentum of a photon bounce. The way it works out is that the mirror yields more when there is low mass, causing more of a change in the photon energy. But ... how does the mirror yield more because it has a low moment of inertia, when a quantum of angular momentum is transferred? And there are still a bunch of leftover questions above... Wnt (talk)

Energy consumption in different contexts

You walk for an hour on a certain piece of land in temperatures that require little thermoregulation by your body. Some days later, when you're in the same physical condition, you perform an identical walk: same route, same duration, etc., but it's significantly warmer, and you're sweating a good deal. Are you likely to consume the same amount of energy, or does the extra thermoregulation require additional energy?

Related question: you walk for a much longer period of time on the cooler day, doing that same route ten times without significant interruption. Of course, by the end, you're more tired and haven't stopped for more than bathroom breaks and drinks. Since you're more exhausted, does it require a greater expenditure of energy to move your body the same distance, especially if you push yourself to walk the same speed? It feels much more difficult, but difficulty is fatigue-related; it doesn't have a 1:1 correlation with energy expenditure. Nyttend backup (talk) 13:54, 11 July 2017 (UTC)[reply]

human power is about ~100 W ;

osmotic power, the one used in drinking and sweating, is about ~1 kWh/m³, that is, ~1 W if you sweat 1 liter in a hour.

so thermoregulation does require some energy, but not an amount significant enough to be felt.

What you will feel, however, is muscle fatigue, lack of water if you run out of water, overheat if thermoregulation is overrun, and other health risks

Gem fr (talk) 14:44, 11 July 2017 (UTC)[reply]

However, generating heat burns lots of energy, with 10 minutes of shivering burning as much as an hour's exercise: [20]. Also note that there's the myth that high temps burn lots of calories, based on the immediate weight loss due to dehydration. But that's absolutely the wrong way to lose weight. StuRat (talk) 15:40, 11 July 2017 (UTC)[reply]

Feynman Lectures. Exercises PDF. Exercise 8-4 JPG. Lecture 8 & 9

. .

Here is my solution png. But I do not fully understand the resolution into components. Feynman speaks about this after eq. 8.11. At a moment of Lecture 8 we do not know about vectors. According to Feynman we can get directly only the components for $ds$ . I have shown in solution that the velocities obey an analogous resolution with cosines of the same angle. But where is the guarantee that accelerations do obey the resolution? We should prove the possibility of acceleration resolution and the formula $dv_{x}=a_{x}dt$ . How to do that? Username160611000000 (talk) 15:10, 11 July 2017 (UTC)[reply]

hell, you make it so complicated... it is not, since you have no horizontal acceleration (

dv_{x}=a_{x}dt=0

; BTW this is just definition of acceleration, no need to "prove" it), and constant (g) vertical acceleration.

Your answer is obviously not stupid since you find 0 distance traveled for both horizontal and vertical shot, and distance traveled proportional to V² and inverse of g, as dimensional analysis suggest.

Indeed it is correct (but you knew it)

Gem fr (talk) 15:57, 11 July 2017 (UTC)[reply]

In the ex. 8-4 we need only components for velocity (so there is no problem yet, but the problem will appear in Lect. 9 with components for acceleration). In the ex. 8-4 the angle Θ is given. Between what is this angle measured ? I think between path (i.e. ds) and X-axis, so we should prove that it is same angle between v and its components.

I have a general question , not directly connected with ex. 8.4: If we have acceleration direction and magnitude and XYZ rectangular coordinate system, then we can project magnitude on axes. Do these 3 projections coincide with the acceleration 3 components received from table of x, y positions? Username160611000000 (talk) 16:48, 11 July 2017 (UTC)[reply]

\definecolor {myorange}{rgb}{0,0.52,0.37}\color {myorange}dv_{x}=a_{x}dt=0

; BTW this is just definition of acceleration, no need to "prove" it -- a = dv/dt is the definition, but Feynman have said nothing about components. E.g. we have a table with columns t, x, y and infinite number of rows for each particle position. Then from this table we know dx and dy, therefore we can find v_x, v_y (organize 2 more columns). Then we can find dv_x, dv_y and therefore a_x, a_y. We know only

ds={\sqrt {dx^{2}+dy^{2}}}

from assumption that trajectory can't be very curved over small

ds

and

v={\sqrt {v_{x}^{2}+v_{y}^{2}}}

as proved, but we know nothing about dv, dv_x, dv_y. How can we calculate resultant

a

and angle

\angle (a,a_{x})

and check these values with directly measured ones by the accelerometer? If such formula as

{\text{magnitude}}={\sqrt {{\text{x-component}}^{2}+{\text{y-component}}^{2}}}

is axiom, then why does Feynman prove it for

v

?Username160611000000 (talk) 20:34, 11 July 2017 (UTC)[reply]

Again, you turn very simple things into horrible. dv = dv_x + dv_y, that's a definition, too. Not a definition of v, but of a coordinate system. This answers your "general question" above.

there is no such thing as an assumption that "trajectory can't be very curved over small

ds

"; physicists are simple-looking, pragmatic, persons, that first try simple things and are happy with them as long as they works. Here, the idea is just to try to find a solution in the realm of the simplest, that is, "not very curved". If it works (and it does, it eventually appears), why look for complicated solution in the realm of erratic, brownian, fractal bang-around kind of motion, where the word "trajectory" lose relevance? If, however, it appeared that so such solution could be found, or if it didn't match the observations, well, some other, more complucated solution would be looked for. happily this is not needed here.

An accelerometer is not a mean to measure angle. It will just give you g, not the current angle of the trajectory relative to the flat ground. This you'll have to calculate with usual trigonometry.

Gem fr (talk) 22:35, 11 July 2017 (UTC)[reply]

dv = dv_x + dv_y, that's a definition, too. -- it is a vector form, which we officially do not know. If v = (v_x² + v_y²)^0.5 is definition, then why does Feynman prove it by formulae 8.11 - 8.15?Username160611000000 (talk) 03:56, 12 July 2017 (UTC)[reply]

facepalm. Feynman is teaching, for god sake. Teaching implies saying things so that even the dumbest get it. That's why. Gem fr (talk) 07:09, 12 July 2017 (UTC)[reply]

First, please use dry scientific voice. Second, such answer does not agree with Feynman's approach. Feynman gives some mathematics subjects but not in a style of writing down redundant data.

The Eq. (9.4) and the same in Lect.8 are also redundant information for poor students, are not they? Username160611000000 (talk) 09:36, 12 July 2017 (UTC)[reply]

IMHO, saying that v_x is the horizontal component of v is a definition. Coming up with a Pythagorean Theorem solution for the relationship of its magnitude with v is a calculation. Wnt (talk) 12:44, 12 July 2017 (UTC)[reply]

Point of order: the link above is actually to "The Swiss Bay", which apparently is a The Pirate Bay knockoff of some sort (I recognize the ship). I don't think there's any need to resort to this when the text is readily available at Caltech, and those links might be more stable. In any case, there is no earthly reason not to simply reprint Question 8-4 here in its entirety for the reader's convenience:

"8-4. A projectile is fired over level terrain at initial speed V, at an angle

\theta

with the horizontal. (Neglect air resistance) Find the maximum height attained and the range."

Wnt (talk) 12:35, 12 July 2017 (UTC)[reply]

Energy loss, planes and ships

How much energy planes and ships lose by generating a vortex behind them? — Preceding unsigned comment added by Tyief (talk • contribs) 17:24, 11 July 2017 (UTC)[reply]

See lift-induced drag. It isn't a simple "how much" question. There are a lot of variables. 209.149.113.5 (talk) 17:49, 11 July 2017 (UTC)[reply]

Relative calorie burn in air and water

Hello. How many calories would a person standing in water burn per minute, compared with the same person standing on dry land at the same air temperature? Twice as many? Thrice as many? Thanks.--90.69.12.160 (talk) 18:00, 11 July 2017 (UTC)[reply]

Assuming that the air and water are the exact same temperature and there is absolutely no air or water motion (impossible), then the only difference would be weight change due to buoyancy. You weigh less in water (mass is the same, weight changes) because you float a little. Therefore, your muscles have to hold up less weight. You should burn less calories, but not much less. How much less is dependent on the person. Some people are more buoyant than others and body mass is highly variable. 209.149.113.5 (talk) 18:07, 11 July 2017 (UTC)[reply]

Assuming that the temperature of the environment (water or air) is less than body temperature, the body loses heat. The loss of heat is larger in water, as the heat transfer coefficient from skin to water is larger than from skin to air, and the thermal conductivity of water is larger. The heat has to be produced by the body, so more energy is converted to heat in the case of water (the picture is made more complex by many factors, e.g. the constriction of blood vessels close to the skin as the skin is colder in the case of water). Icek~enwiki (talk) 20:32, 11 July 2017 (UTC)[reply]

you need to consider not just temperature, but also speed of the air/water, and Thermoregulation in humans.

I remember (no ref at hand, sorry) that confort temperature (that is, when thermoregulation is mostly in idle state, the loss being more or less equal to basal production, around 100W) is around 20-22°C in the air, but 26-28°C in water

also, you can easily stand -40°C in the air (with no wind) for as long as enough food is available , while a mere 2°C in water will kill you in few minutes (hypothermia).

Gem fr (talk) 23:23, 11 July 2017 (UTC)[reply]

I think that being correctly dressed would also assist in surviving at -40C. I don't think -40 in a t-shirt and jeans is going to do it. I would like to see a source for the idea that you could survive at -40 provided you had food and were correctly dressed. CambridgeBayWeather, Uqaqtuq (talk), Sunasuttuq 00:04, 12 July 2017 (UTC)[reply]

I seldom if ever try to argue from experience/authority. But User:CambridgeBayWeather I think hails from Cambridge Bay, and so I respect their thoughts on human survival in cold weather and climates ;) SemanticMantis (talk) 00:38, 12 July 2017 (UTC)[reply]

Turns out this is hard to search for. I suspect because nobody has tried it. I just realised that without water you aren't going to live too long. Melting snow in your mouth will just cause hypothermia to set in quicker. This also means that you are putting frozen food in your mouth and that's going to lower your body temperature. By the way it is possible to survive at least 18 minutes in −1.7 °C (28.9 °F) water, see Lewis Pugh#North Pole. I uploaded a couple of pictures of people swimming in the Arctic Ocean, File:Swimming in the Arctic Ocean 01.JPG and File:Swimming in the Arctic Ocean 02.JPG. CambridgeBayWeather, Uqaqtuq (talk), Sunasuttuq 04:22, 12 July 2017 (UTC)[reply]

I remember a documentary about these things, picturing a man swimming in cold water (without swimsuit). But the man was trained, and the usual "don't do this, you may kill yourself" message was clear, as the body of the man had adapted in very interresting ways (hence the point of the documentary). He could swim and hence generate heat in conditions the normal person just dies. Wasn't this one [21]; may be this one [22] (i didn't check)

Henri Guillaumet was not trained, but he nonetheless survived 1 week in harsh climate above 3000m

Gem fr (talk) 10:42, 12 July 2017 (UTC)[reply]

July 12

Intelligence difference between animals and humans

So... Why is it that I notice that many animals don't even really want to try to do what we do? For instance, why will a cat not see a human pick up a pencil to write, and see that same thing often, and then try to pick up the pencil themselves. (though I don't believe it'd work, I've never seen a cat try) What sort of lack of intelligence is this really? To make the decision "I will try to pick up a pencil like that human is" doesn't sound like it takes that much thought, especially if you can catch a mouse so easily. What field or term in neurology or psychology refers to this? Something that an animal could possibly think of, but that they won't?

To expand my question, in a similar manner, what is one of the things that a human would pretty much never decide to do that we probably could at least think of trying to do? I really can't think of any examples, because whenever I think of an example of one of these things I can almost definitely imagine some human doing that thing, no matter how crazy or stupid it sounds; like trying to jump to the moon with his or her own two feet. Yeah I think some human would be crazy enough to try that.

But this seems to be because humans are so diverse in thought, and really maybe a human could think of or perceive or try anything? Philmonte101 😊😄😞 (talk) 01:30, 12 July 2017 (UTC)[reply]

The term is observational learning. Spitballing, curiosity is also probably important here (sadly we have no article on animal curiosity that I can find). What if the cat totally understands what you're doing, but just doesn't want to copy you? Someguy1221 (talk) 03:05, 12 July 2017 (UTC)[reply]

It has puzzled me for some time how the dog, despite being in close proximity to humans for thousands of years, has never learned to raise one leg and point to something, such as its food bowl when it wants food. Instead, all it can do is bark and run back and forth from the object. Akld guy (talk) 07:44, 12 July 2017 (UTC)[reply]

Akld guy based on the excellent post by SteveBaker here, the answer may be that dogs equate humans' hands with extra mouths, rather than with paws. Adrian J. Hunter^{(talk•contribs)} 08:26, 12 July 2017 (UTC)[reply]

It may also puzzle dogs how the human, despite being in close proximity to dogs for thousands of years, has never learned to walk on all fours, raise a leg and.... Proximity learning only works between animals of the same species; the dog or cat does not identify with the human or their wish to pick up a pencil. Similarly you don't train a dog to do tricks by doing them yourself.--Shantavira|^{feed me} 08:02, 12 July 2017 (UTC)[reply]

As anyone who has interacted much with dogs knows, dogs 'point' by attracting a human's attention and then directing their gaze, which any dog-familiar human readily understands. This natural, mutually understood interaction has been refined by artificial selection in the particular dog breeds known as 'pointers', but the unaugmented version works well because over the thousands of years that dogs became self-domesticated by their associations with humans, humans also Coevolved to better interact with dogs, in what might be termed "co-domestication" (see Section 3.6 Convergent evolution between dogs and humans). {The poster formerly known as 87.81.230.195} 90.204.181.91 (talk) 08:38, 12 July 2017 (UTC)[reply]

Dogs can learn to point, it's just they normally they do so with their face and eyes rather than their paw (after all, dog anatomy is different to ours, and their "wrist" is effectively just a second knee - they lack the ability to point as precisely as we can.) Smurrayinchester 09:28, 12 July 2017 (UTC)[reply]

Pointing requires, apart from a faculty for cooperative behavior, a certain recursive way of thinking that may well be unique to humans. Chimps are generally bad at understanding finger pointing (apparently, worse than dogs.) Some actually believe the neurocircuitry for pointing, creating joint attention, shared intentionality etc had to be already in place for speech to evolve. I once worked in a manual job with people whose language I didn't speak. You won't believe how well one can get along by just pointing at stuff Asmrulz (talk) 13:03, 12 July 2017 (UTC)[reply]

Our cat does like to grab pens, but not to write

. Interestingly they have their own limited language, which we can learn and exploit to our advantage. I'm not a cat or a dog, but I'm pretty sure that they simply find pointless that we mysteriously gaze at paper and computers, except for the fact that those objects steal our attention from them... But it can be fun to watch their amazed reaction and wondering when we do peculiar things or manipulate new objects. They probably are trying to understand what occurs for a moment (the aforementioned curiosity is also a factor here) and to evaluate threat and safety, if it can be used as a toy or food, etc. Their needs are simple and they are not very interested in anything beyond those, we can trick them into doing things they would not normally do, but they don't do them for the same reasons we do (except perhaps for the rewarding factor)... Animal cognition is probably a good starting point to read more about the topic. —Paleo Neonate - 12:56, 12 July 2017 (UTC)[reply]

How does polysiloxanes effect scar development?

How does polysiloxanes, common in the scar-treatment gel "Kelo Cote" effect scar development? Are these polysiloxanes become a permanent part of the scar tissue, or what? 109.66.156.177 (talk) 01:54, 12 July 2017 (UTC)[reply]

According to the best sources I can find, "the mechanism of action is unknown". That said, scar treatment gels were preceded by solid scar treatment sheets that most likely were not diffusing into the skin. There are a number of theories as to how scar treatments work that you can read about here, if that's not behind a paywall for you. Someguy1221 (talk) 03:02, 12 July 2017 (UTC)[reply]

My plastic surgeon says that it keeps the scar tissue hydrated. That is true of the silicon sheets, as the skin underneath gets distinctly clammy. Whether that really improves scarring is a whole nuther thing. Greglocock (talk) 06:53, 12 July 2017 (UTC)[reply]