Wikipedia:Reference desk/Archives/Science/2014 May 18

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May 18[edit]

Incandescent light bulb color[edit]

The last paragraph of the intro of Incandescent light bulb says "their light color which is almost identical to the sun's spectrum". To me, a traditional soft-white incandescent bulb is distinctly yellow/orange. So what color is an incandescent soft white bulb, and how close to sunlight is it? Bubba73 You talkin' to me? 01:42, 18 May 2014 (UTC)[reply]

That's a misleading statement. See color temperature. The lower the wattage, the cooler the filament will be and the redder the light will be. The higher the wattage, the hotter the filament, and the whiter the light will be (if it was hot enough it would actually turn blue, but I don't think a filament that hot would last long enough to be practical). StuRat (talk) 01:55, 18 May 2014 (UTC)[reply]
There is a tendency for manufacturers to make lower wattage bulbs run at a lower temperature, but this is just manufacturing convenience, possibly avoiding very thin filaments. There is no logical connection between rated wattage and filament temperature. The filament will be hotter if it is run at greater than the rated wattage by applying a higher voltage. (Not recommended because it reduces the life of the filament.) Dbfirs 12:36, 19 May 2014 (UTC)[reply]
Yes, perhaps I should have added "given the same filament". Of course, a thicker filament would waste expensive tungsten (not sure if that's a significant part of the total incandescent bulb price, though) and a thinner filament would burn out too quickly, as you noted. StuRat (talk) 15:32, 19 May 2014 (UTC)[reply]
I think it's not so much about the cost of the tungsten as it is about the balance between bulb life and color temperature. A thinner filament gives whiter light and higher energy efficiency, but burns out quicker. A thicker one lasts a long time, but is redder and less efficient. --Srleffler (talk) 03:20, 20 May 2014 (UTC)[reply]
I agree the statement is problematic, I've tried to reword it to more closely follow the source but I think the problem is it's trying to summarise different ideas. The actual article more accurately describes the situation. Incandescents eare close to an idealised black body radiators, but their colour temperature is far lower than the sun. They also don't filter through the atmosphere. They therefore have a fairly continous spectrum and a near perfect color rendering index but their spectrum doesn't match the sensitivity characteristics of human vision and they appear far redder than daylight. Many people seem to prefer these redder than daylight colours, particularly for home lighting, and there's some suggestion of psychological reasons for such a preference for dim light sources (like most home lighting) but either way the preference doesn't seem to be universal. LEDs and fluourescents, even ones with a fairly high CRI still have a somewhat discontionous spectrum, although there's some controversy over how well the CRI matches subjective color rendering anyway. Nil Einne (talk) 08:25, 18 May 2014 (UTC)[reply]
My wife also prefers the soft white (yellow/orange) color, but not me. I think that is probably because that is what we all grew up on, for the most part. But to me it makes the colors of things unnatural and it is much harder for me to read my soft white than it is bright white or daylight colored lights. Bubba73 You talkin' to me? 18:24, 18 May 2014 (UTC)[reply]
People certainly look better in soft white. Under harsh fluorescent blues, people look pale and veins tend to stick out. StuRat (talk) 15:35, 19 May 2014 (UTC)[reply]
I removed the statement. The claim that the spectrum is "almost identical" to the sun's is simply false, and is not what the source said.--Srleffler (talk) 03:30, 20 May 2014 (UTC)[reply]
The brain automatically corrects the white balance, so we don't notice a big difference. To see the real difference between light from the Sun and light from an incandescent light bulb, you can take a picture with a digital camera of white objects in a room lit by incendescent light with the white balance set to sunlight. Count Iblis (talk) 15:10, 20 May 2014 (UTC)[reply]
The brain tries to correct, yes, but red objects appear brighter under a red light and blue objects brighter under a blue light, and the brain doesn't go so far as to lower the brightness of those objects, to bring everything back into balance. StuRat (talk) 15:13, 20 May 2014 (UTC)[reply]

How long would the TGV take to stop?[edit]

Hi, I'm asking this question because I'm curious about using it as a hypothetical to get students talking. How long would a train at top speed take to come to a stop on the Beijing–Shanghai High-Speed Railway, if the brakes failed, and it just rolled to a stop? How long for this one? P.S. I'm editing on a quick break, so if this question is actually answered in the articles, please just abuse the c**p out of me, and we'll all feel a lot better, right? P.P.S. Is it "in" a quick break or "on" a quick break? IBE (talk) 09:51, 18 May 2014 (UTC)[reply]

I don't think we can accurately answer using first-principles of physics. But we can inaccurately answer: you can estimate the coefficient of friction by observing:
  • at maximum speed, all engine power is expended to overcome frictional losses
  • P = F v
  • The engine power, speed, and mass for each train can be read out of our article
  • Assume that full engine power is used at maximum speed. Assume that frictional losses are identical at all speeds. These are flakey assumptions, but if you understand their limitations, you'll see exactly why the problem can only really be answered using empirical data.
With the above assumptions and about two lines of algebra, the braking distance can be solved using Newtonian kinematics. Nimur (talk) 13:36, 18 May 2014 (UTC)[reply]
(EC) Three points. One, I'm pretty sure most high speed, trains, heck most trains have multiple independent braking systems (which are frequently designed to be as fail safe as possible) making complete brake failure (meaning with absolutely nothing in the train trying to slow it down, even if it isn't enough to stop it completely under whatever conditions) difficult. See e.g. [1] [2] [3] SNCF TGV Duplex or even Railway brake#Accidents with brakes.
Two; the time taken will surely depend significantly on the conditions. For example if the train is going a slight incline going upwards it will obviously slow faster than if it's completely level which will likewise take longer than if it's going downhill. (If it's going downhill potentially it may never stop until it either derails or reaches the end of the incline. Of course if it's going uphill it may eventually start going in the other direction, but let's not worry about that.) Similarly a train going at 200 km/h is going to stop fasrer than one going at 300 km/h everything else being equal. I suspect at that level even things like if it's going in to a headwind or tail wind may make a noticable differences. Similarly while high speed trains have gentle curves (as with their inclines) I suspect turning a curve will slow the train more than if it's travelling completely straight.
Three; I'm fairly sure any maglev system will be even more complicated. In many systems I'm fairly sure braking will at least partially come from the propulsion system, but loss of propulsion will often likewise mean a loss of levitation. They are of course designed to fail safe (and some designs may maintain levitation under low speeds) so I guess may have wheels or whatever and so should keep going if this happens (and I guess may have some sort of emergency brakes for when this happens), but it's clearly far from a simple case of 'complete brake failure'.
Nil Einne (talk) 14:09, 18 May 2014 (UTC)[reply]
And the fourth point: At high speeds (80 mph or above), air resistance becomes a significant part of the "frictional losses", and that force varies directly with the square of the velocity -- so the assumption that frictional losses are identical at all speeds is completely inaccurate. I've taken the liberty of deleting the first two paragraphs of Nil Einne's response, because they duplicate the following two paragraphs word-for-word. 24.5.122.13 (talk) 00:05, 19 May 2014 (UTC)[reply]
(Aside: I was going to joke with Nil, yes, I thought you tended to repeat yourself a bit ;) but I was worried it might come out wrong. I was also assuming someone (probably Nil) would take care of it.) Yes, I do remember reading that about friction - at high speeds it is the dominant factor; at low speeds, rolling friction predominates. Interested people can read about it in Energy - a Guidebook by Janet Ramage [4]. I think rolling friction is proportional to the velocity. However, the square of the velocity (for the wind drag) arises because the vehicle has to get the whole column of air moving - the more streamlined the vehicle, the less the wind matters, as it can just slip through the air. Also, for a long vehicle, the head-on air column matters less proportionally. There will be wind drag along the sides of the train, but that again would only be in simple proportion to the velocity. IBE (talk) 00:40, 19 May 2014 (UTC)[reply]
Extra comment: no, I was just reading about friction, and it seems it is constant, and has nothing to do with velocity. IBE (talk) 01:00, 19 May 2014 (UTC)[reply]

Propagation of extremely high frequency sound in air[edit]

I'm involved in an off-wiki argument with a friend about the "detectability" of sound in air at 11.5GHz. I argue that the attenuation is so rapid that the sound is practically undetectable over a distance of several metres - unless the power level is ridiculously high - enough to obliterate the stones! My friend maintains that such sounds are found in "stone circles" - yes there are other pseudoscience issues that I'm dealing with but for this one I need some help with specific facts and figures. Roger (Dodger67) (talk) 12:13, 18 May 2014 (UTC)[reply]

Let me see if I understand this correctly - you're asking if acoustic pressure waves - i.e., longitudinal compression waves - in air - on Earth, with normal atmospheric conditions - carry energy in the gigahertz range?
No. Vibrations at those frequencies are uncommon in air. Even if we include vibrational modes that are uncommonly described as "acoustic waves", we still don't find energy at 11.5 GHz. For example, atmospheric oxygen has molecular vibrations - the diatomic molecule "spring" stretches and wobbles - in the infrared range. We would consider those vibrations as part of the optical spectrum - still not at 11.5 GHz. And, no microphone in the world exists that could detect such motion as a "sound." And these vibrations do not propagate energy between molecules, through the bulk of the air, in the form of a pressure-wave, so these vibrations are very much not sound-waves.
When we consider Earth atmosphere temperatures and densities, we find that the attenuation won't just take place over several meters - it'll take place within less than one wavelength. The wave will attenuate to "negligible strength" within a distance that we can measure in units of inter-molecular distance: i.e., a few microns. (Millionths of a meter!) The energy doesn't propagate.
Let me see if I can dig up a proper but simplified equation to describe attentuation of acoustic waves as a function of frequency. My first instinct is to pull out Bittencourt, Fundamentals of Plasma Physics, which generalizes the wave equation for a fluid to incorporate electromagnetic forces - because without a superheated ionized plasma - the sort of soup-y mess of dense hydrogen you'd find inside the Sun - there's essentially no way to convey acoustic energy at those frequencies. Even plasmas in laboratories or in near-Earth space resonate at much lower frequencies. Even plasmas near Jupiter resonate at lower frequencies.
At reasonable normal atmospheric temperature and pressure, and atmospheric gases, if a compression wave existed at 11.5 GHz, individual atoms would be moving so fast that the electrons would separate from the nuclei. The wave would devolve into a superposition of many oscillations - the atomic nuclei, and the electrons that cannot move coherently at this frequency. (To do so would imply that the atomic nucleus and the electrons have different temperatures). This would cause a separation of charge and would induce an electromagnetic wave. The wave will disperse. We can write an equation to give you the scale length for that dispersion relation, but it appears to require nine pages in my textbook. Bittencourt writes the spatial frequency (wave number) for a longitudinal wave in terms of electron plasma frequency, which is a long equation and would take some time to define.
Long story made short: your friend is either pursuing counterfactual science, or they've mixed up their S.I. prefixes. Acoustic waves do not propagate in the gigahertz range, not at any condition we find anywhere in our solar system outside of the interior of the sun. Nimur (talk) 12:50, 18 May 2014 (UTC)[reply]
Not disagreeing with any of that, but perhaps a simpler way of putting it is that the gigahertz range is where microwaves live, and if something did vibrate at that frequency, it would generate microwaves, which could easily be detected. Looie496 (talk) 13:55, 18 May 2014 (UTC)[reply]
Thanks Nimur & Looie496. This is exactly what I need. I'm far more familiar with calculations involving electromagnetic energy - I'll do a path loss calculation for a radar without breaking a sweat, but mechanical vibration waves are not my thing at all. I gather that for sound the attenuation v. frequency graph (in air at STP) crosses the fuhgeddaboudit! line well below 11.5GHz. As a radio frequency 11.5GHz is in a band widely used for satellite communication and radars - it's just a bit below the Ku band that most people would come across as it is commonly used for satellite tv transmission. The irony of the whole situation is that both myself and my "opponent" are radio hams! BTW the source of the argument is this video. Watch it at your own risk, I will not accept any responsibility if someone following the link does themself a serious injury due to uncontrollable laughter. Roger (Dodger67) (talk) 14:12, 18 May 2014 (UTC)[reply]
Stokes' law (even without bulk viscosity) gives an attenuation of 31 e-foldings (or nepers) per nanometer in air. --Tardis (talk) 05:01, 27 May 2014 (UTC)[reply]
I agree with everything said above - but I would just mention that in Ultrasound#Imaging it mentions that: "The potential for ultrasonic imaging of objects, with a 3 GHZ sound wave producing resolution comparable to an optical image, was recognized by Sokolov in 1939 but techniques of the time produced relatively low-contrast images with poor sensitivity." - which suggests that GigaHertz-range audio is possible, although perhaps not through air. Our ultrasound article also says that ultrasound reaches up to "several gigahertz".
The obvious way to debunk your friends' argument is to follow the trail of knowledge back to the source. Where did he find out about this? Go there and figure out how the person who told him knew about it...and so forth. One will inevitably find that either the trail goes cold because some idiot just made this up - or a mistake was made and MHz got turned into GHz somewhere - or that you find how this "fact" was ascertained in some kind of experiment or other. My bet it that it's the first of those things. This is obviously complete B.S. pseudoscience. Very likely someone thought about the reputed "energy" found by people in and around stone circles - and when they thought about how it might be transmitted without anyone being able to measure it, they guessed "ultrasound" and made up a plausible-sounding frequency without thinking about whether it's possible or not.
here, for example is one such site - it mentions stones that it's claimed produce ultrasound - and also block ultrasound (weird!) - and it references a book "Circles of Silence" by Don Robins (Souvenir Press, London, 1985). From what little I could find about it in Google Books, it does make mention of ultrasound - but I couldn't find "GHz" or "gigahertz" in a search of the book. Don Robins also wrote "The Secret Language of Stone: A New Theory Linking Stones and Crystals With Psychic Phenomena" and has made claims that ultrasound is the cause of crop circles. (<sigh> - this isn't looking good is it?!).
Anyway, those claims for stone circles come from something called "The Dragon Project". Digging into that a bit turned up "The Dragon Project and the talking stones." an article in the "New Scientist" journal(!) - which has graphs and stuff in it! Now we might be on to something! That article says that these ultrasonic waves were detected using a bat detector...well, our bat detector article says that these work in the 15 kHz to 125 kHz range. So right there - we're not talking Gigahertz - or even Megahertz - we're in the range only just above human hearing...and far from being hard to detect, a $85 (Amazon.com) detector (or this $26 kit) is sufficient to pick them up.
Now, those graphs of the ultrasonic energy of the stone circles says that the energy is highest in March and November...and I'd bet that March is breeding season for bats. They talk about building a whole series of increasingly sophisticated ultrasound detectors - but there are no details about what frequency range they detected or what intensity levels. The article is full of self-contradictions (eg he says that February is one of the high points of ultrasound levels - when his graph says March).
So I'd have to say, that this was a fairly amateur effort and the results are not accurately reported. There is certainly no evidence that I could find for frequencies up into the Megahertz range - let alone gigahertz...and far from being hard to detect, you can pick them up with a sub-$100 piece of equipment.
SteveBaker (talk) 15:32, 18 May 2014 (UTC)[reply]
Hi SteveBaker - The "trail of knowledge" starts with this video - I also linked it in my post just above yours - see my "disclaimer" above. Roger (Dodger67) (talk) 15:43, 18 May 2014 (UTC)[reply]
Yes, the guy who posted that is Michael Tellinger - who is a politician, song writer and trained as a pharmacist. I wouldn't put much store by his science credentials, and from all I can find, he's basically just repeating the claims of Zecharia Sitchin. That guy is into the whole "aliens came to earth and played god" thing...and our article about him says "Sitchin's ideas have been rejected by scientists and academics, who dismiss his work as pseudoscience and pseudohistory. His work has been criticized for flawed methodology and mistranslations of ancient texts as well as for incorrect astronomical and scientific claims.". So that just about wraps it up for that trail of evidence! At least the Dragon Project made an effort to some up with something. SteveBaker (talk) 16:59, 18 May 2014 (UTC)[reply]
Just a quick comment: any physicist could recognize that a 3 GHz sound-wave would permit acoustic images with high resolution. In the same way, I can recognize that a glass lens whose refractive index were an imaginary number would allow me to build an infinitely-thin camera. But I still can't make one! That's the difference between theoretical physics - in which we crunch the equations just to see what should happen - and experimental physics, where we see what actually happens, then write a better equation to describe it! Nimur (talk) 18:33, 18 May 2014 (UTC)[reply]
  • There are some great arguments here for why gigahertz frequency sound is impossible. The only nagging problem I have is that our article on ultrasound pointed me to acoustic microscopy, which uses gigahertz frequencies routinely. There are lots of articles that come up in a search for "acoustic microscopy" and "gigahertz"; unfortunately, sampling these, it looks like this is one of those fields you're not allowed to know much about unless you pay or your owner did. Wnt (talk) 15:51, 18 May 2014 (UTC)[reply]
There is no air involved in acoustic microscopy - it's all about solids and liquids. See Acoustic microscopy#Sample types and preparation. The highest frequency mentioned in that article is 400MHz - a very long way from 11.5GHz! Also take a look at Medical ultrasonography where "microscopy" is discussed as an experimental technique to examine structures in the eye - the highest frequency mentioned there is 100MHz. The medical article also briefly discusses attenuation in (mostly liquid) human tissues. Roger (Dodger67) (talk) 16:21, 18 May 2014 (UTC)[reply]
I invite you to run the search - many snippets mention 1 GHz or 3 GHz. I don't deny that ultrasound attenuates rapidly in air, but the argument about the frequency turning into microwaves should apply in water too, shouldn't it? Actually, this reminds me of an old question that never got an answer about whether materials where slow light travels at the speed of sound are able to convert sound to light or vice versa. It'd be interesting to hear more. Wnt (talk) 19:38, 18 May 2014 (UTC)[reply]
I think the suggestion that they would turn to microwaves is simply fishing something out of the air. Given the massively different propagation velocities, even if there was a radiative mechanism (e.g. relative displacement of nuclear orbital charge), it would not be self-reinforcing. This speculation simply does not belong here. —Quondum 20:56, 18 May 2014 (UTC)[reply]

Breast shrinkage[edit]

Why do not womens breasts shrink back after finishing baby feeding?--86.160.193.27 (talk) 12:17, 18 May 2014 (UTC)[reply]

Who says they don't? ←Baseball Bugs What's up, Doc? carrots→ 13:34, 18 May 2014 (UTC)[reply]
Breast#Hormonal_change covers this pretty well. SteveBaker (talk) 14:26, 18 May 2014 (UTC)[reply]
Also, they aren't filled with milk the way a water balloon is filled with water. It not just a bag to hold the milk, it's still fat, tissue and all the other stuff. Mingmingla (talk) 17:32, 18 May 2014 (UTC)[reply]
The question is a little unclear. In humans and a few other species, the post-pubescent females have noticable breasts. In most other species, the swollen breast only forms during lactation, and sometimes not even then. There have been a number of reasons proposed for why this arrangement is the way it is (Desmond Morris proposed that the cleavage of breasts was meant to evoke the cleft of the buttocks, but I don't know how popular that hypothesis has been). Whatever the original cause, female breasts are a classic secondary sex characteristic. When a woman is breastfeeding, the size and shape of the breasts will change depending on how much milk has been expressed and other factors. But once that's all done and the child has been weaned, women retain their breasts because they still fulfill their original role. Matt Deres (talk) 16:46, 21 May 2014 (UTC)[reply]

Sunset time and darkness[edit]

How long does it take to get completely dark in London after sunset in May? Can you provide reliable evidence for this? 82.40.46.182 (talk) 13:46, 18 May 2014 (UTC)[reply]

See Twilight#Duration - Roger (Dodger67) (talk) 14:41, 18 May 2014 (UTC)[reply]
Since it's a big city with lots of artificial lighting, it never gets "completely" dark in London. Our article Dusk discusses the various definitions of twilight and sunset. Basically, the definition depends on whether you're talking about "civil", "nautical" or "astronomical" twilight - and the definition is in terms of the angle of the sun below the horizon rather than how dark it actually is. This site (amongst many others) documents the sunset time and the time that twilight ends for each definition of twilight. So for London on 18th May 2014, the sun officially sets at 20:50:28, Civil twilight ends at 21:32:42, Nautical twilight ends at 22:31:21 and Astronomical twilight ends 00:08:09. Which of those you take to be "completely dark" is entirely a subjective matter since it's never going to be particularly dark in or near such a large city - and for any definition based on actual light levels, it's going to depend a lot on the cloud cover, moon phase and whether you are standing in a dark alleyway or on top of a tall building. I suppose the best generic answer is "about 40 minutes" - which is the duration of civil twilight - but in truth, this is one of those "How long is a piece of string" questions. SteveBaker (talk) 14:42, 18 May 2014 (UTC)[reply]
The rule of thumb that I use is about an hour. Bubba73 You talkin' to me? 18:27, 18 May 2014 (UTC)[reply]

Observant Jewish people need to know these sorts of things and have precise definitions for them, see Zmanim. The times you're talking about would correspond to Shkiyat Hachama and Tzet HaKochavim. According to chabad.org those times for today, May 19, are 20:50 and 21:42 respectively, so 52 minutes, pretty close to Bubba's hour. --Dweller (talk) 17:15, 19 May 2014 (UTC)[reply]

After reading the Zmanim article you linked to, I really wouldn't describe the Jewish reckoning of times pertaining to twilight as involving "precise definitions". The definitions aren't particularly precise given that they involve such vague quantities as the amount of time it takes to walk a mil, the amount of light needed to recognize a person four cubits away, or the amount of light needed to distinguish the blue threads of a tzitzit from the white ones. The Jewish holy books aren't even internally consistent about daylight reckoning, with the Talmud in Pesachim saying there are four mil between sunset and nightfall, and the Talmud in Tractate Shabbat saying there are just three-fourths of a mil between sunset and nightfall. With all that vagueness, it's unsurprising that when exactly nightfall is considered to be depends on which group of rabbis you choose to follow the interpretations of. And even if the Talmud and the rabbis did all agree, it would still just amount to a rather arbitrary set of definitions that only have meaningful significance to the 0.2% of the world's population that follows Judaism. Red Act (talk) 20:36, 19 May 2014 (UTC)[reply]
Believe me, by the time the Acharonim finished with the Talmudic definitions, the calculations have become very precise indeed. I appreciate very few people follow these arcane rules, but it seemed a way to answer the question using references, rather than finger-in-the-air speculation, which is, after all, what we're here for. And "52 minutes" is a nice, specific answer, from a proper reference, for the OP. Whatever religion they may or may not follow. --Dweller (talk) 06:56, 20 May 2014 (UTC)[reply]
The definitions for "civil twilight", "maritime twilight" and "astronomical twilight" are incredibly precise - we have references (indeed, entire Wikipedia articles) for them - you can calculate their times to the nearest millisecond for any day of any year on any point on the Earth if you want to. They are also employed more or less universally by people who care about such things (versus the very small percentage of people who are jewish and the yet smaller percentage of that small percentage who either know or care about these religious pronouncements). So if you need a 'hard' number (like "52 minutes") - then choose "civil twilight" or another one of those entirely arbitrary definitions. But the unassailable point here is that it's still not going to answer the OP's question and tell him/her a damned thing about when it gets dark in London in May 2014 - because that depends on things like "How dark does it have to be for our OP to call it 'dark'?" and "Is it a weekday or not and are the floodlights turned on at Wembley Stadium that night?" and "How cloudy/foggy is it?". Compared to the HUGE differences those kinds of things make to the final determination of when darkness has actually arrived (indeed, *if* it ever arrives!) - specifying a number accurate to within plus or minus an hour would be ridiculously over-precise.
So, the answer is STILL "there is no good answer" - and pulling some nonsense definition from some arcane religious sect surely isn't any better! SteveBaker (talk) 16:22, 21 May 2014 (UTC)[reply]