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Gauge Theory

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When editing a page, as I did on gauge theory, how do you save the change? Denis Lieberman (email address removed)

There should be three grey buttons below the text box labaled "Save Page", "Show Preview", and "Show Changes". Click on "Save Page". --Kainaw (talk) 00:27, 3 July 2006 (UTC)[reply]
Hmmm how did they manage to save this page? Iolakana|T 12:43, 3 July 2006 (UTC)[reply]

Nice Looking Carnivorous Pants

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The above question got me thinking, what sort of trousers are carnivorous?--205.188.116.74 01:30, 3 July 2006 (UTC)[reply]

  • To suggest that trousers can be carnivorous leads into the realm of fantasy and science fiction. There is currently no kind of fabric, that we know of, that has the ability to "eat" (in a very broad sense of the word) meat, let alone humans. However, with the addition of a hot liquid, a fabric can become dangerous to the skin, causing deep burns (and, in a sense, "eating" it). A medical professional is a better person to see concerning such a matter. --JB Adder | Talk 06:41, 3 July 2006 (UTC)[reply]
Hmm, beaten to it. Well, I propose that if one's legs are being simultaneously swallowed by a pair of snakes, the snakes might collectively be referred to as Carnivorous Pants, with the added bonus that the "carnivorous" bit is being demonstrated. Depending on the kind of snake, they might even be Nice Looking! Downside: the wearer is probably already dead, and the snakes aren't going to be happy about the geometric and topological difficulties in their future.
Speaking of which, how is it that we don't have an article on snakeskin? Melchoir 06:56, 3 July 2006 (UTC)[reply]
We don't have one on roof rack either. Wikipedia is crazy like that... --Username132 (talk), UK or Netherlands 19:32, 3 July 2006 (UTC)[reply]
We don't have an article on eelskin, either, but we do have one on moleskin. User:Zoe|(talk) 01:56, 4 July 2006 (UTC)[reply]
What if they're trouser snakes? EdC 14:53, 3 July 2006 (UTC)[reply]
I guess I walked right into that one. Melchoir 18:52, 3 July 2006 (UTC)[reply]

RGB conversion to wavelength

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Is the a formula or simple computer code for converting an RGB value such as #F0F0F0 to wavelength? ...IMHO (Talk) 02:22, 3 July 2006 (UTC)[reply]

I'm looking for a conversion formula or computer code that will do the conversion or generate the table rather than for the conversion table itself. ...IMHO (Talk) 02:48, 3 July 2006 (UTC)[reply]
No such thing, using your example, F0F0F0 is 3 different colors, red, green, and blue, the resulting color isn't any one wavelength, just an overlap between the big 3--205.188.116.74 02:53, 3 July 2006 (UTC)[reply]
(edit conflict) I wouldn't call the task a "conversion", though, since turning an RGB value into a wavelength loses saturation and brightness information, and for #F0F0F0 (light gray) it fails altogether. Also, it depends on which RGB color space one uses, so there won't be a universally correct formula. Melchoir 02:53, 3 July 2006 (UTC)[reply]
I've found several programs that will convert wavelength to RGB I'm just looking for one that does the conversion the other way around. ...IMHO (Talk) 04:56, 3 July 2006 (UTC)[reply]
In general, your request is not possible. Many of the colors that humans can percieve are actually the result of combining multiple wavelengths of light, and consequently can't be represented by a single wavelength. Dragons flight 05:00, 3 July 2006 (UTC)[reply]
You must have a keyboard to press any keys. ...IMHO (Talk) 05:38, 3 July 2006 (UTC)[reply]
In fact, the simplest example of that is white. What's the wavelength of white? --cesarb 05:53, 3 July 2006 (UTC)[reply]
Even if you press all of the keys at one time you still need the keys. ...IMHO (Talk) 05:58, 3 July 2006 (UTC)[reply]
Most perceptual colors don't correspond to a unique temperature either. Also, colors that occur as combinations of wavelengths don't do so uniquely as the same percieved color can usually be constructed from many different wavelength combinations. If you haven't done so, might I suggest you look at color and color vision. Dragons flight 06:16, 3 July 2006 (UTC)[reply]
  • Yes I have done that previously. If you look closely at the image you will see that there is a one to one relationship between RGB values and the combined RGB color vales but that they are out of sync with the visible spectrum above and require a conversion formula to be synchronized with the wavelength spectrum. A synchronization formula if you will. ...IMHO (Talk) 06:43, 3 July 2006 (UTC)[reply]
Ummm, you do know that when you monitor / TV is given an RGB value (x,y,z) it basically just takes x amount of some red wavelength, y amount of some green and z amount of some blue. The actual wavelengths used depend on the device. However, if you percieve the same color in any medium other than a monitor / tv, you will almost never find that it is composed of the same wavelengths. Dragons flight 07:02, 3 July 2006 (UTC)[reply]
Perhaps you could start by first converting the color to another color space. I'd suggest HSV, so you could easily discard the saturation (S) and brightness (V) values that can't be described as a wavelength. Then the only problem is to convert hue into wavelength. Actually, the hue wheel looks quite spectrumish to me. –Mysid(t) 06:56, 3 July 2006 (UTC)[reply]
As a comment on your spectrum image: The "spectrum" in the image is not exactly the spectrum of wavelengths from low to high, but seems to contain changes in brightness as well. See visible spectrum. –Mysid(t) 07:00, 3 July 2006 (UTC)[reply]
Actually you are right. I was thinking "spectrum" where I should have been thinking "combined RGB." However, what i am looking for is a formula to convert "combined RGB" to wavelength or Kelvin. ...IMHO (Talk) 07:13, 3 July 2006 (UTC)[reply]
Perhaps we're beating dead horses at this point, but see also Magenta. Melchoir 07:05, 3 July 2006 (UTC)[reply]
Not a dead horse. I've seen a comparison color band chart (although the above chart is admittedly not it) that does a similar compare for both Kelvin and wavelength. I know the conversion formula is out there somewhere. ...IMHO (Talk) 07:13, 3 July 2006 (UTC)[reply]
Have you already googled? I found a PDF called An RGB to Spectrum Conversion for Reflectances, Dan Bruton's Color Science site, and many others. I also found out that the CIE chromaticity diagram could be useful in these conversions. –Mysid(t) 07:38, 3 July 2006 (UTC)[reply]
Yes I've seen the Fortran program but I sold my Fortran compiler well over a year ago. I've got C++, (and therefore .asm) and Visual Basic capability only. ...IMHO (Talk) 07:46, 3 July 2006 (UTC)[reply]
You might want to try gfortran or g95, which are both free. --cesarb 15:35, 3 July 2006 (UTC)[reply]
  • I've used the color diagram in the CIE 1931 color space in a VB picture to get the RGB color values by mousing over but the program is hanging for some reason and stopped working. Perhaps if I can get it working again I can create a table to do use for the conversion. ...IMHO (Talk) 07:50, 3 July 2006 (UTC)[reply]
That's the purpose of doing the conversion. Suppose for instance I have an International Color Consortium reference color diagram and a camera hooked up to my computer but with the only option of conversion to RGB by the camera. Once I have an image file I can then get the operating system's designation for each RGB value and then sync the values with those on the chart I have read in. Ideally the chart would have a companion floppy disk with all of the values and a program to do the conversion for me by scanning the image file since it knows what the values on the chart are supposed to be and can convert the ones assigned by the operating system. But I don't have such a chart or program yet and so that is what I am looking for. In the mean time I'm trying to do this the hard way by converting the RGB values my operating system assigns to images for which absolute RGB values have been assigned. ...IMHO (Talk) 08:11, 3 July 2006 (UTC)[reply]
I don't think you'll find many decent cameras now that just return "RGB". You are much more likely to find a camera that returns a particular, and documented, color space like Adobe RGB or sRGB (mine has a switch to choose which one of these two). Converting to another color space is then just a matter of using the ICC profile corresponding to the camera. Notinasnaid 08:19, 3 July 2006 (UTC)[reply]

My Camera only has vividness options of standard, vivid, black & white, Sepia and Cyanotype with no Adobe or sRGB options. Therefore I actually have to use a standard color swath by holding it up to the monitor and trying to read embeded test under a variety of conditions until what I see on the monitor is what I see on the swath. What is actually being calibrated are my own eys rather than the equipment so that when I say I see ICC red you know exactly what color I am seeing so long as you have done the same thing. ...IMHO (Talk) 08:55, 3 July 2006 (UTC)[reply]

What is the camera? If it has fixed sRGB or Adobe RGB you won't see options, but that doesn't mean it isn't using a standard color space. Notinasnaid 08:57, 3 July 2006 (UTC)[reply]
The camera I am using for macro shots right now is a Nikon Coolpix 4800. However, I have other reasons for wanting to do a conversion from RGB whether in a standard color space or not. ...IMHO (Talk) 09:09, 3 July 2006 (UTC)[reply]
Ok... I hope it's already clear that while you can usually convert a wavelength to a particular RGB space, the reverse is not true. What might help is a set of RGB profiles, one for each image setting, if the camera isn't a fixed space. Anyway, you might find [3] interesting. Notinasnaid 09:22, 3 July 2006 (UTC)[reply]
ummm... Seems like it would be possible to convert each of the primary colors independently of the other primary colors to a wavelength which is all I need to do. ...IMHO (Talk) 10:50, 3 July 2006 (UTC)[reply]
Certainly, any color could be converted to three separate wavelengths and intensities. But that's really implicit, it's the same as discovering the color space. The color space defines the primaries (R,G,B) and each of these has a fixed wavelength. The color itself is a mixture of primaries with intensities; so every color will have the same three wavelengths. Maybe we're talking at cross purposes; it seems to me you want to match colors, so what you need is the ICC profile for the color space of the camera data. It seems to me that finding wavelengths is a red herring in this process, though I don't know the internals of the analytical process of calibrating a set of samples into a profile. Notinasnaid 11:20, 3 July 2006 (UTC)[reply]

Further discussion

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Well lets just say that knowing the wavelengths of each primary color that is displayed on my monitor and how far it deviates from the standard wavelength for that primary color will satisfy my curiousity as to how far off my monitor (and system and camera and scanner and the files created by others) is from the actual primary colors that nature provides for me to see. ...IMHO (Talk) 16:17, 3 July 2006 (UTC)[reply]

Well, you aren't going to be able to use an algorithm to determine what wavelengths your monitor produces. Perhaps you should look at your monitor through a spectroscope. —Bkell (talk) 02:02, 4 July 2006 (UTC)[reply]
There seem to be multiple confusions going on here. There are no primary colors in nature. "Primary colors" are entirely a human invention, and there isn't even a single well-defined set of primaries. One can choose different sets of three "primary" wavelengths and use them to mix a full gamut of colors. Red green and blue are convenient choices that produce a broad gamut, but exactly which wavelength of red, green, and blue is used can vary. Additionally, many monitors do not use spectrally pure primaries—the primaries themselves are mixtures of multiple wavelengths.--Srleffler 18:44, 2 September 2006 (UTC)[reply]
If you want to mix colors and you know the wavelength of the primary colors then the resulting color will be the result of a mixture of the primary colors dependent upon the intensity of each right? Thus the wavelength of the resulting color can be determined if the wavelength of the primary colors are known and the intensity of each primary color that makes up the resulting color. Or is this just bunk and I need to get a refund from school for the cost of my textbooks? ...IMHO (Talk) 11:17, 4 July 2006 (UTC)[reply]
Here is the crux of your confusion. Suppose you start with three spectrally pure "primary" colors of known wavelength, and you mix them to obtain another color. That resulting color does not have "a wavelength". The resulting color is still a mixture of three different wavelengths. It's not a spectral color at all. It just doesn't appear on the spectrum. The human visual system, though, processes mixtures of wavelengths in such a way that we perceived them the same as pure spectral light. The mapping is not one to one. There are some spectral colors that can not be mixed from red, green, and blue primaries. Pure spectral violet is an example. The best one can do with RGB is only an approximation to true spectral violet. About the only way you can ever see true violet is with a prism or other dispersive element. In the other direction, there are colors that humans can perceive, and which can be generated from RGB, which do not appear in the spectrum at all. There is no magenta in the spectrum, nor is there brown or pink. These colors do not have a wavelength. They only occur when there is a mixture of wavelengths.--Srleffler 18:44, 2 September 2006 (UTC)[reply]
That information is implicit in the ICC profile for your monitor. You seem to be intent on recreating a very, very complicated wheel, whose full details I don't pretend to understand. But there are well established practical tools for people who want to match color between their devices. To start with, look for a profile for your monitor from the manufacturer. Notinasnaid
I've heard that there a color transparencies you can get through which you can view the output of your monitor which will then tell you the offset in wavelength or whatever. Is this not true? ...IMHO (Talk) 11:17, 4 July 2006 (UTC)[reply]
That may be possible. Done one color at a time it sounds plausible. But what would you do with the information? There is no "standard" set of primaries for monitors, so what would you compare against? That's why getting an ICC profile seems to me to be the way to go, since that defines what the primaries are, and color management software can then (given accurately profiled color data) display it accurately. Notinasnaid 20:04, 4 July 2006 (UTC)[reply]
By the way, I don't present working with profiles as an easy alternative. You may find plenty of challenges there. I just think you are more likely to end up where you want to be. Notinasnaid 07:17, 5 July 2006 (UTC)[reply]
This article and program may be of interest to you. http://www.fourmilab.ch/documents/specrend/ "Colour Rendering of Spectra"

by John Walker --GangofOne 06:19, 8 July 2006 (UTC)[reply]

Evolution of intelligent species

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I heard somewhere that a major factor in the evolution of intelligent species is the use of tools, which requires dextrous hands or limbs of some kind. This is why humans evolved from tree-dwelling apes (they had dextrous hands, because they needed to grab branches).

Does this mean that, among intelligent alien species (assuming they exist), they would all be descended from animals that had dextrous limbs (for tree-climbing or worm-digging or whatever)? Would it be unlikely, or even impossible, to find a sentient bird-like species or sentient steppe grazers? Battle Ape 06:01, 3 July 2006 (UTC)[reply]

  • Despite our own (very limited) experience here on Earth, which suggests that "intelligent" species would be very likely to have evolved some means by which they may manipulate tools, to speculate beyond that would be completely meaningless. We have no idea what may or may not be out there in the universe, and evolution has an uncanny knack for surprising us. – ClockworkSoul 06:14, 3 July 2006 (UTC)[reply]
    • Let's face it, the most "intelligent" species we know of on Earth aren't all equipped with dextrous hands. Dolphins certainly aren't, and parrots such as kea use their beaks. Octopodes/octopuses/octopi are also pretty smart - though not up to human standards (he says egotistically) - and their limbs are far more dextrous than ours. Manual dexterity is highly correlated with intelligence, but it isn't an absolute 1:1 correlation. Grutness...wha? 07:38, 3 July 2006 (UTC)[reply]
they may have developed from creatures that learned to manipulate objects using only the power of their minds...
they may have developed from creatures that have no requirement of physicality
it's all speculation... --Dweller 11:17, 3 July 2006 (UTC)[reply]
Intelligence needs something to work on. So it needs input. It would also need to experiment (what happens if I do this?) so it'll need actuators as well as sensors. Which ones it has determines how it perceives reality and what kind of intelligence it has. All life has sensors an actuators (else it wouldn't be alive), so all life is intelligent. It's just that we value our type of intelligence the most. So you're really asking what it takes to develop an intelligence like ours. The same type of sensors and actuators, I assume. So the more something is intelligent according to our standards, the more it will be like us, I suppose. DirkvdM 11:37, 3 July 2006 (UTC)[reply]
The simple answer is we really don't know how intelligence evolved, or even what intelligence is. But one theory I thought was intriguing was that it was due to the development of the language faculty as a mechanism for modelling reality, allowing humans to think about things independently of stimulus. Peter Grey 01:47, 4 July 2006 (UTC)[reply]
An extremely common misconception, that language evolved before intellignece. How can you talk if you have nothing to talk about? DirkvdM 10:47, 4 July 2006 (UTC)[reply]
Language is thinking - talking came later. Peter Grey 13:32, 4 July 2006 (UTC)[reply]
How can you talk if you have nothing to talk about? You clearly don't listen to much talkback radio, do you Dirk? :) Grutness...wha? 14:01, 4 July 2006 (UTC)[reply]
I can proudly say I never even heard of the term. Anyway, I don't even have a radio. DirkvdM 15:35, 4 July 2006 (UTC)[reply]
"Language is thinking". An interresting concept. Do you mean people talked to themselves before they talked to others? :) DirkvdM 15:38, 4 July 2006 (UTC)[reply]
Language is a mechanism for modelling reality. (According to some; like intelligence, we're not completely sure what exactly it is.) People had to be thinking the same idea before one could communicate it to another. Once language also became used for communication, Homo sapiens could share intelligence, which is really what set us apart from the other animals. Peter Grey 19:41, 4 July 2006 (UTC)[reply]

It's all very interesting, but the questioner asked about an alien species. An alien species may have developed in quite different ways than those on this planet. But then again, it might not. Btw, several species of bird can "talk". Whether they are "intelligent" or not, I have no idea, but DirkvdM, I believe the point Grutness was trying to make is that the average parrot is probably capable of conversation several notches more intelligent than callers to some talk radio shows. --Dweller 19:49, 4 July 2006 (UTC)[reply]

An alien species and tools. Again, we're not really sure how intelligence came about, so it's hard to say. Maybe intelligence was a desperate last resort for a species that didn't fit its niche very well. Peter Grey 23:57, 4 July 2006 (UTC)[reply]
Perhaps when we humans are able to get artificial intelligence AKA robots to work effectively, we will have a better understanding of what it takes to get self aware functioning. User:AlMac|(talk) 18:17, 12 July 2006 (UTC)[reply]

Antimatter

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How do Antimatter and Matter Annihilate in a 100 percent perfect mass to energy conversion?

You might find something by reading the articles annihilation and antiparticle#Particle-antiparticle annihilation. –Mysid(t) 06:49, 3 July 2006 (UTC)[reply]

Time cause of motion?

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Hi everyone,

Here is my question: can time be considered as the cause of motion?

This should probably be kept at Wikipedia:Reference desk/Humanities; please don't double-post. Melchoir 07:56, 3 July 2006 (UTC)[reply]
According to the theory of relativity, time is just a component of 4-dimensional vector of spacetime. See also, Metric tensor (general relativity). 62.63.84.118 09:51, 3 July 2006 (UTC)[reply]
The answer to your question, or more fundamentally, whether there is an answer to your question at all, depends critically on what you mean by "the cause of motion". Exactly what does it mean for something to be "the cause of motion"?--72.78.101.61 04:17, 4 July 2006 (UTC)[reply]

By motion I mean motion of macro and micro-elements, that is, objects, atoms and particles... By cause, I mean the cause of all causes, the cause that allows causality itself... In other words, can time be considered as a fundamental force ?

Am I clear?

Consider a very simple processor receiving a tiny amount of data, doing a couple of operations on it, and writing it back out. Over the course of these operations it takes on 8 states. The final state includes that it is ready to write out the answer. Now, you could represent the eight states on eight sheets of paper and flip through them. Each state also has a causative effect on the next one: the next one "follows" from it based on the laws of physics and the design of the processor. Now time is "what keeps everything from happening all at once". If you didn't have time, you could still represent all eight states, and the causative effect each has on the next one, in much the same way as I can represent the fibonacci sequence in a single line:
1, 1, 2, 3, 5, 8, 13, 21, 34...
Now, I could have written a javascript program to flash the successive values over time: instead, I wrote them on a single line. Does the fact that as you're looking at them they're "all there" instead of appearing one after another mean that I'm not following the definition of the series? Of course not. Likewise, time is not necessary for a causative relationship between two things. Indeed, mathematically, there's no reason that anything "caused" by something else must happen later, and not at the same time. So, I do not think that conflating time with causality is useful or productive. 82.131.187.36 09:09, 4 July 2006 (UTC).[reply]

Quantify heat effect

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At my work because it is so hot (32°C and above) we've got the message that we're allowed to remove our ties (lucky us). Does anyone have any idea how much difference this will make in cooling us down/reducing effects of the heat? I don't think it'll have a great deal of effect but does anyone know if there is are any actual numbers or data on this? AllanHainey 07:47, 3 July 2006 (UTC)[reply]

Last summer in Japan there was a campaign encouraging people to go to work without a tie, so the airconditioning could be set lower. This apparently resulted in a 0,08% saving on energy. Not a lot, but every bit helps and if it's so simple and comfortable why not do it? It may also need some startup time, letting people get used to the idea of turning the airco down. Also relevant here is how much of the energy consumption in Japan is gobbled up by ariconditioning. This was to be followed up with a campaign encouraging people to go to work in a sweater last winter. Don't know about the result of that.
I prefer to wear nothing but a Speedo to work myself. :-) StuRat 20:55, 9 July 2006 (UTC)[reply]
Also, when my father worked in Curaçao, office workers were allowed to not put on a tie if they wore a specific type of shirt with some frills down the front. That was in the fifties, so I don't know if they still have that rule. DirkvdM 11:49, 3 July 2006 (UTC)[reply]
Frills down the front ? So how long was your dad a bartender in a gay night club, anyway ? :-) StuRat 20:55, 9 July 2006 (UTC)[reply]
Wouldn't it be a little hard to quantify? There would be personal characteristics to take into consideration, such as the size and weight of the person, thickness of their shirt, chest hair (?) etc. At least you could pull the top of your shirt in and out to get the air circulating; I suppose this would change the results also. BenC7 03:10, 5 July 2006 (UTC)[reply]
What about a motorised necktie that does that for you? Personalised airco! DirkvdM 05:36, 5 July 2006 (UTC)[reply]
You might want to check with your workplace health and safety representative about that. In many jurisdictions, offices at 32°C is a health issue and a breach of regulations. You may even have a legal right not to go to work under these conditions.--JLdesAlpins 12:45, 5 July 2006 (UTC)[reply]
"many jurisdictions"? Since this happens a lot here, I have to ask - are you talking specifically about the US? I can't imagine such a rule existing here in the Netherlands. Would be nice, though because social security is making me do forced labour for no pay and the temperature there might very well be over 32 C. It's just that I would feel like a wimp complaining about that. DirkvdM 18:35, 5 July 2006 (UTC)[reply]
High heat is a well known health hazard, especially for individuals with medical conditions such as heart or respiratory issues. Now, how legally "high" is too high is a matter of local regulations. There has to be provisions for workers' protection against heat-related hazards in the Arbowet[4]. If you feel that your workplace is unsafe (I mean "not safe", not "not comfortable"), then you have to have recourse... I hope.--JLdesAlpins 22:36, 6 July 2006 (UTC)[reply]
That temp is high enough to be quite miserable, and lower productivity dramatically, but probably not enough to be dangerous (unless the relative humidity is near 100%). I suggest removing that tie (good for a degree or two), wearing a short sleeved shirt, pointing plenty of fans directly at yourself, and drinking lots of ice cold drinks. I would probably go even further and bring dry ice in with me in a cooler, but then I'm a nut. StuRat 20:51, 9 July 2006 (UTC)[reply]

bilogy

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how does cockcoach survive even in the presence of nuclear rays?

This has been discussed before, see Cockroaches surviving nuclear explosions? in the archive. This link proved to be useful: Great Moments in Science: Cockroaches & Radiation. –Mysid(t) 12:24, 3 July 2006 (UTC)[reply]
We do too. Radioactivity is all around us. It's just that, as with everything, too much of it is unhealthy. And what is too much depends on your fysiology. And for insects that is different. DirkvdM 10:54, 4 July 2006 (UTC)[reply]
Physiology, Dirk? ;-) — QuantumEleven 13:30, 4 July 2006 (UTC)[reply]
Oops. Well, at least I didn't ask a question about the science of bile (see header). :) DirkvdM 14:02, 4 July 2006 (UTC)[reply]

Physics, Maths and Mathematical Physics

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Dumb question from a non-scientist Alert Are "Physics", "Mathematics" and "Mathematical Physics" 1, 2 or 3 different disciplines? Thanks in advance for your tolerance. --Dweller 11:52, 3 July 2006 (UTC)[reply]

Physics and mathematics are clearly two different disciplines, though physics uses mathematics extensively. Mathematical physics is a subfield of theoretical physics, but a person with affinity to applied physics will probably tell you it has more to do more with mathematics than with physics. Conscious 13:42, 3 July 2006 (UTC)[reply]
Interesting. At Talk:Polymath#Einstein - not a polymath it's been suggested that discounting any musical skills Einstein had, he does not qualify as a polymath because either Physics and Maths are one subject or, putting the same argument into a different terminology, because he was "only" (!) a genius in the field of "Mathematical physics". So you might go along with the latter argument, that he was outstanding in a single field that was a subfield of theoretical physics, or would you shoot that particular argument down in flames? --Dweller 15:00, 3 July 2006 (UTC)[reply]
Physics and Math are certainly considered two separate subjects (physicists are generally not mathematicians and vice versa, even if physics use mathematics and mathematicians can use physics). As for whether doing both makes you a polymath or not, it seems like a pretty strange and subjective question to me, more about disiciplinary infighting than any hard criteria. Einstein was not, however, a mathematician by any stretch of the imagination, which would make him solidly a physicist in my book, and so by this strange definition of polymath, he would not be one (assuming you discount the musical skills, which seems a rather arbitrary move to me). --Fastfission 02:35, 4 July 2006 (UTC)[reply]
Excellent answer. I agree that dismissing musical skills is a nonsense, but I wanted to get a scientist's view on the Physics/Maths issue, without music blurring the argument. --Dweller 19:38, 4 July 2006 (UTC)[reply]

TEREPHTALYLIDENEDICAMPHOR SULFONIC ACID

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Hello dear people I would like to know what this compound is and for what it is used

furthermore I would like to know which companies produce this product

kindest regards

RT

See Mexoryl. An UV-absorbing chemical in sunscreens, patent held by L'Oréal. Femto 12:15, 3 July 2006 (UTC)[reply]

RT,it's a good way to attract one's attention, but, you may not be so lucky next time.PLEASE, be POLITE to your fellow-users(this is a request,not an advice).Thanks,Pupunwiki 15:25, 4 July 2006 (UTC) Being polite, I mean, don't write all in CAPS, unless absolutely necessary.Thanks again,Pupunwiki 15:28, 4 July 2006 (UTC)[reply]

human science

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what causes discomfort when our buttocks sweat excessively when we sit ? --59.93.0.78 12:20, 3 July 2006 (UTC)[reply]

The articles on buttocks and sweat may help you. Of the top of my head I wouldn't know as I have air conditioning.--Frenchman113 on wheels! 13:04, 3 July 2006 (UTC)[reply]
I guess you're talking about your buttocks when you sit (rather than include the rest of us in your problem). The discomfort is caused by the sweat not evaporating in order to cool you down, which is what it's for. It then irritates the skin because the skin cannot breathe. On another level, the discomfort is caused by such things as plastic chairs and nylon pants. See sweating.--Shantavira 13:12, 3 July 2006 (UTC)[reply]

Open proxy

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How do you determine from an IP address (or whatever) if someone is coming from an open proxy? Thanks, Iolakana|T 13:29, 3 July 2006 (UTC)[reply]

Contact the owners of the proxy and see if they will tell you the address of the person using it. The whole point of the proxy is to keep information like that from passing through. --Kainaw (talk) 13:49, 3 July 2006 (UTC)[reply]
Proxycheck is a tool that can do it. [5] My guess is that you just try to connect to a proxy. If you succeed, it's open. Conscious 13:58, 3 July 2006 (UTC)[reply]
I believe that either I'm confused about the question or this answer. I read the question as: "I want the IP address of a person connecting to my computer|server through a proxy." This answer appears to be "Proxycheck will tell you if an IP address has an open email proxy on it." Perhaps I just need to take more Excedrin. My headache is so bad now that my eye's are twitching and the words are getting all fuzzy. --Kainaw (talk) 14:03, 3 July 2006 (UTC)[reply]
I interpret it as: "Given an IP address, how do I determine if it's the address of an open proxy?" The answer, in general, is that you can't; however, there are several ways that can work in some cases:
  • Check if it's listed on public lists of open proxies, including DNSBLs, or is listed as part of a proxy network such as Tor.
  • Try to use it as a proxy. Note that there are many different kinds of proxy servers, including HTTP, SOCKS and CGI proxies. You will also need to guess the port (or, for CGI proxies, the URL), which may require running a port scan.
  • Check for telltale signs in the connection coming from the address. Well-behaved HTTP proxies often add certain headers, such as "Via" and "X-Forwarded-For", to requests. While the contents of such headers may not always be trustworthy, their presence is a good indicator that the IP may be a proxy. Also, some common CGI proxies have a known bug where every apostrophe and backslash in form submissions is prefixed with a backslash.
Ilmari Karonen (talk) 14:51, 3 July 2006 (UTC)[reply]

Chiropractors (title added)

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Is there objective evidence that chiropractors work? If so, what is it? —The preceding unsigned comment was added by 206.162.181.34 (talkcontribs) .

I know a chiropractor and he works. I'm not sure how hard though. --Dweller 14:55, 3 July 2006 (UTC)[reply]

There are many people who feel better after chiropractors have manipulated their spines. The reasons chiropractic is only rarely integrated into standard care is that (1) their principal theories of disease and health have been inconsistent with scientific medicine and (2) many of them have rejected many aspects of scientific medicine. There are exceptions to both generalizations but they are exceptions. alteripse 15:31, 3 July 2006 (UTC)[reply]

The use of the unit 'mole'

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The eternal question... What is the use of the mole? One hydrogen atom weighs 1.67x10^(-24) grams, one gram of hydrogen contains 6.022x10^23 atoms, as does x grams of the xth atom. This amount of atoms weighs 1.67x10^(-24) x 6.022x10^23 = 1.005674 grams. Why complicate things by introducing the mole? Isn't the mole the same as Avogadro's number? Jack Daw 14:31, 3 July 2006 (UTC)[reply]

A mole is the same as Avogadro's number, but "a mole of hydrogen" is easier to say than "Avogadro's number of hydrogen molecules". —Keenan Pepper 14:32, 3 July 2006 (UTC)[reply]
Yep - just for convenience.
Plus it allows you to do all sorts of geeky puns like guaca-mole. --Fastfission 16:39, 3 July 2006 (UTC)[reply]
Well there you go... Is a mole of hydrogen the same quantity as a mole of carbon, or is a mole of carbon 12x Avogadro? Jack Daw 17:12, 3 July 2006 (UTC)[reply]
A mole of anything is Avogardo's number of whatever it happens to be. So a mole of hydrogen contains the same amount of molecules as a mole of carbon. –Mysid(t) 17:14, 3 July 2006 (UTC)[reply]
A mole is a certain number of items (6.022 * 10^23). A mole of hydrogen is a certain number of hydrogen atoms (or molecules) and a mole of carbon is a certain number of carbon molecules. It has nothing to do with weight, size, volume, or any other quantities. Just a number. Exactly like the unit "dozen". You could have a dozen hydrogen atoms, or a mole of hydrogen atoms. Just instead of 12, 6 * 10^23.
Ugh. Please, number of molecules, not amount of molecules. That really gets on my nerves, especially in cases like this where the difference is crucial. —Keenan Pepper 20:22, 3 July 2006 (UTC)[reply]
A mole is a number, Avogadro's number is the name of that number. --Yanwen 17:21, 3 July 2006 (UTC)[reply]
So then x moles of any given molcule is simply Avogadro's number * all atoms in the molecule; for example 5 moles of glucose C6H12O6 is 5 * (6+12+6) * Avogadro = 7.2264x10^25 ?

And what does grams/mole mean? That is, when is it a relevant concept, what calculations result in something becoming "grams per mole"? I am so stupid... Jack Daw 19:37, 3 July 2006 (UTC)[reply]

Not stupid, just new to the idea. The unit "grams per mole" is a measure of how many grams (a unit of mass) are in a single mole of a particular item - it is a property of the item. It's also known as the "molar mass", i.e. the mass of one mole. For instance, one could say, hydrogen atoms are about 1 gram/mol, meaning that a mole of hydrogen atoms (6 * 10^23 of them) weighs 1 gram. Hydrogen molecules (which have two atoms in each molecule), on the other hand, are about 2 gram/mol; since each molecule weighs twice as much as the atom, a mole of molecules weighs twice as much as a mole of atoms. Thus, its "molar mass" is twice as big. It's a similar to the unit "pounds per dozen" - you could say eggs are 1.3 pounds/dozen, meaning 12 eggs weighs 1.3 pounds. Note: mol is the abbreviation for mole, and technically gram is not a unit of weight, but let's not nitpick.
Jack, it seems you might be thinking a mole is (always) a number of atoms. It's not. A mole is just a number of "things". The things could be atoms, or could be molecules.
Saying a mole of "any given molcule is simply Avogadro's number * all atoms in the molecule" seems to be an attempt to convert into atoms. There's no conversion to be done. A mole of a given molecule is simply Avogadro's number of those molecules. There is no need to reference the atoms in the given molecule in order to call the number a mole.
You originally asked: why use it? It's actually for convenience for questions like: "If I have 12 grams of carbon, how many grams of hydrogen will it combine with to form methane?" -R. S. Shaw 05:54, 4 July 2006 (UTC)[reply]
Right, and what is the answer? Could you show me the "chemical math" for calculating that please? Jack Daw 13:18, 4 July 2006 (UTC)[reply]
  • To expand on that say for example you are electrolysing water. 2H20 --> 2H2 + O2. So 2 moles of water will fall apart into 2 moles of hydrogen gas and 1 mole of oxygen gas. Knowing how many molecules are involved in a reaction makes calculating the resulting masses a lot easier. All you have the know is the substances' molar mass. - Mgm|(talk) 07:42, 4 July 2006 (UTC)[reply]
      • No. One molecule of water decomposes to two atoms of hydrogen and one atom of oxygen. To relate that to moles, one mole of water will decompose into two moles of hydrogen (because of the two hydrogens on the molecule) and one mole of oxygen. Scienda 17:44, 4 July 2006 (UTC)[reply]
      • To be more explicit, the formula 2H20 --> 2H2 + O2 says that 2 molecules of water (H20) become 2 molecules of hydrogen (H2) and 2 molecules 1 molecule of oxygen (O2). The subscripts say how many atoms are in each molecule, so there are, for example, 2 hydrogen atoms (H) in each hydrogen molecule (H2).
      • That formula describes what happens at the molecular level. Moles are handy at our more normal scale. The usefulness of the mole is that a formula like the one just given can be directly interpreted in units of moles instead of molecules: 2 moles of water give 2 moles of hydrogen plus 2 moles 1 mole of oxygen. Looking up the weight (mass) of a mole of H2 as 2 grams, and of O2 as 32 grams, you find that for every 4 grams of hydrogen you get, you'll get 64 32 grams of oxygen. Check out the Utility of moles section of Mole. -R. S. Shaw 19:49, 4 July 2006 (UTC)[reply]

I realized today that the amu is the inverse of Avogadro's number, so I thought maybe the mole is useful in this fashion. "One u is the weight of one hydrogen atom, which is 1.66x10^(-24) grams. How many hydrogen atoms, then, are contained in one gram of hydrogen atoms? The answer, is (1.66x10^(-24))^(-1) hydrogen atoms." Is it in this correlation it's useful? Anywho, would I be unable to pass college chemistry without understanding the concept of moles? 81.233.224.235 21:14, 4 July 2006 (UTC) <- Jack Daw 21:15, 4 July 2006 (UTC)[reply]

You would probably find it very difficult. Don't try. As someone said earlier, it's just a number, like "dozen". BenC7 03:20, 5 July 2006 (UTC)[reply]
Well I just don't get it, could you give me an example of a chemical calculation where the number mole is crucial? Why can't I just use a few grams of this and a couple grams of that? Where's the starting point? If I'm thinking, What would happen if I put these two elements together and add some heat?, where do moles come into the picture? Jack Daw 16:23, 5 July 2006 (UTC)[reply]
Take a look at Mole_(unit)#Example_calculation. In a chemical reaction, you normally want to use just the right amount of the reactants, not a large surplus of one or the other, because the reactants may be expensive, or the remaining surplus reactant may be toxic, or you just have to then isolate the product from the surplus reactant. You might also get further reactions taking place between the surplus reactant and the desired product. Say you have a formula of:
2X + Y -> X2Y
You want to mix two moles of X with one mole of Y to get one mole of the product. This is not the same as mixing 2 g of X with 1 g of Y, because the molar mass of X and Y will be different. To put it in different words, 2 g of X will have moles of X, and you want half as many moles of Y, so you need g of Y to match your 2 g of X.-gadfium 23:44, 5 July 2006 (UTC)[reply]

Could you give me an example of a chemical calculation where the number mole is crucial? OK. Let's say, hypothetically, that we want lead (Pb) to react with elemental oxygen (O) to form lead oxide (PbO). If we had one gram of lead and one gram of oxygen, does that mean that there is the same number of atoms of lead as the number of atoms of oxygen? No. Why? Lead is heavy (big), oxygen is light (small). One gram of lead might contain (to use round figures) 10 trillion lead atoms. But one gram of oxygen might contain, say, 300 trillion oxygen atoms, because oxygen is small and light. In other words, one gram of oxygen contains more atoms than one gram of lead.

Now, instead of saying '10 trillion atoms' and '300 trillion atoms' of a particular substance, it's more conventient to say, '0.9 moles' of this, or '0.005 moles' of that, where one mole is about 602300000000000000000000 atoms (or molecules).

So - if you say, "I will mix 1g of lead with 1g of oxygen", you will have left over oxygen, because there are more oxygen atoms in 1g of oxygen than there are lead atoms in 1g of lead. You have to mix 602300000000000000000000 (one mole) of atoms of lead with 602300000000000000000000 (one mole) of atoms of oxygen. So you might need, say, 1g of lead, but only 0.1g of oxygen.

Still don't get it? Think of atoms/molecules like polystyrene balls. Large ones would be heavier than smaller ones. If I had 602300000000000000000000 large balls, it is going to weigh a lot more than the same number of small balls.

(Before I get chemistry people picking out errors, note that the figures above are deliberately oversimplified - I have skimmed over some details to try to explain the concept as simply as possible.)

Hope this helps! BenC7 01:25, 6 July 2006 (UTC)[reply]

Yeah I got it now, finally... THANK YOU ALL FOR THE HELP! Jack Daw 02:10, 6 July 2006 (UTC)[reply]

Cost of GPS?

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Once you've bought and installed TomTom Mobile for a mobile phone, do you then have to pay further data charges to TomTom or the phone network? --Username132 (talk) 15:27, 3 July 2006 (UTC)[reply]

Using GPS is free, it won't cause any additional costs. –Mysid(t) 16:49, 3 July 2006 (UTC)[reply]
I'm not that sure about those additional services that TomTom might offer, though. –Mysid(t) 16:52, 3 July 2006 (UTC)[reply]

How is polyethene prepared?

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Please tell me how is Poly Ethene prepared and please provide with it the required equations also.

Thank you. S.Nanda

See polyethylene. The article has list of production methods for different polyethylene types. Conscious 17:37, 3 July 2006 (UTC)[reply]

hawkings new theory

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can anyone explain what stephen hawking's new theory states?--219.91.153.254 17:19, 3 July 2006 (UTC)kaushal[reply]

Are you talking about Hawking evaporation or something more recent? Conscious 17:35, 3 July 2006 (UTC)[reply]
I think recent puts anything worked out in the seventies out! Philc TECI 22:51, 3 July 2006 (UTC)[reply]
The anonymous user may be asking about Hawking's recent (2004) reversal on his views about black holes, which is well explained in his article, as well as at Thorne_Hawking_Preskill_bet, and, of course, at Hawking radiation, as Conscious has hinted.  freshofftheufoΓΛĿЌ  04:26, 4 July 2006 (UTC)[reply]

Site archiving software?

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I'd like to be able to point a program at a website and download all of its content, including graphics and HTML and stylesheets and so forth. Googling "site archiving software" or "download website software" and so forth doesn't work well (it gets confused with the "archive" pages that almost all sites have). So I'm hoping someone will know of one off hand. This is a one-time archival job, not anything that needs to run regularly. Something that worked on either Windows XP or Mac OS X would be great. Linux solutions might work if I can figure out how to implement them from OS X's terminal but I'm less comfortable with this option at this point. Many thanks. --Fastfission 17:30, 3 July 2006 (UTC)[reply]

How about HTTrack? --Bowlhover 17:45, 3 July 2006 (UTC)[reply]
HTTrack is good, but it needs careful handling - there really aren't many good fire-and-forget archiving programs AIUI. The best way to do it is talk nicely to the website maintainer... Shimgray | talk | 18:54, 3 July 2006 (UTC)[reply]
Wget is the standard for recursive download. There's GNOME and Windows GUIs, but you can use it from the command line on pretty much any OS. EdC 19:53, 3 July 2006 (UTC)[reply]
Awesome, one of these should work for me. Thanks. There are just a few websites that I find useful/interesting with very finalized content that I'd like to have a personal archive of in case they ever go down (and I don't trust the waybackmachine to necessarily get all of the files). --Fastfission 02:29, 4 July 2006 (UTC)[reply]

Teleport Pro is a very good software which you might find useful - Nikhilthemacho

But it isn't free. DirkvdM 11:02, 4 July 2006 (UTC)[reply]

internet privacy

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Can one's ISP service, such as Comcast, have the ability to see your online activities through the eyes of your monitor without your expressed consent? what I mean, is access your computer and see exactly what you are doing at any given moment and time?

Seeing exactly what is on your desktop - no, not necessarily. If you are using Windows and have been dumb enough to install tons of spyware on your computer, then they can exploit the spyware to see what is on your desktop, steal your passwords, and use your computer for whatever they want to do. What they can easily see is every packet of information that goes between your computer and the internet, which is why you want to make sure all secret stuff is done with encryption (such as using https for your online banking instead of http). --Kainaw (talk) 18:29, 3 July 2006 (UTC)[reply]
The stuff that is not encrypted they can see. They probably can't access it legally, but I am not completely sure. --Proficient 11:11, 5 July 2006 (UTC)[reply]
ISPs can passively "see" every unencrypted action and transaction you perform on you PC that involved your internet connection. (Passively means here that they do not need to breach the security of your machine to do so). Anything that is performed locally to your machine can only be "seen" by your ISP via a security vulnerability.--JLdesAlpins 12:38, 5 July 2006 (UTC)[reply]
However, any party with TEMPEST capabilities can literally see your desktop from hundreds of yards away without leaving any clue of their surveillance activities.--JLdesAlpins 12:38, 5 July 2006 (UTC)[reply]

how to speed up the download speed using router

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what are fowarding ports & what are the other things that are benefical while using a router. And want to know how configure advancily a router currently i am using linksys WRT54g router, i am using it, i configured it any how but i want know with each menus in it what i can doin configure process.

Believe me that your router manual is a better resource than we are. That's the truth. --mboverload@ 19:38, 3 July 2006 (UTC)[reply]
To my knowledge forwarding ports are only useful if you are using Torrent programs. If that's what you want to set up, do a Google search for "torrent port forwarding linksys" and it should give you some tips. --Fastfission 02:27, 4 July 2006 (UTC)[reply]
You can't arbitrarily increase your connection speed. Your ISP would definitely not like it if that were possible. However, if you're not getting the speed that your ISP advertised, you may want to tweak your TCPIP settings with something like TCP optimizer (google for it). If you're using bittorrent, your ISP may be traffic shaping you, in which case you would want to use encryption. If it's not any of those, then you're pretty much screwed. Forwarding ports is just a method to allow incoming connections on them (if your router has a firewall).--Frenchman113 on wheels! 13:44, 4 July 2006 (UTC)[reply]
It depends what you mean by "increase your connection speed". You can optimize the way your router works with certain types of p2p programs (such as with port forwarding), which won't increase your connection speed, but will result in much faster downloads by better using the bandwidth you already have available. --Fastfission 14:14, 4 July 2006 (UTC)[reply]
Just a note, uncapping is illegal. --Proficient 11:12, 5 July 2006 (UTC)[reply]

claims

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Are claims? A)Nondebatable B)Factual C)Values D)Religious E)All the above. I can seem to find the answer,please HELP!!!

Is this question about the meaning of the word claim? In that case it belongs on the language desk, but I'll answer it anyway. Claims are always debatable; if they were nondebatable they'd be definitions or self-evident facts or something. Anyone can claim whatever they want, so claims don't have to be factual. Claims are not always values, in fact I find it hard to think of a claim that's also a value. Maybe moral claims like "murder is never excusable" are values. Clearly some claims are religious and some are not. So I'd say "none of the above". —Keenan Pepper 21:11, 3 July 2006 (UTC)[reply]
I feel it is important to note that in logic, factual claims are not facts. A fact is something that is true ("Wikipedia is a website"). A factual claim is something that it may somehow be possible to prove true or false ("There is life on Pluto"). It is not opinion ("Kainaw is stupid"). So, a statement that is blatantly false can be a factual claim - just a false one. --Kainaw (talk) 00:54, 4 July 2006 (UTC)[reply]
That's a good point. I was thinking of factual as a synonym for true, but in this case it probably means "of the nature of a fact", so that must the answer. —Keenan Pepper 01:08, 4 July 2006 (UTC)[reply]

Ramjets- how do they get to the operating speed?

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Hi,

I have read several different and rather educational texts concerning the operation of ramjets in aircraft and missles. What I cannot understand, however, is how a 'plane such as the Lockheed SR 71 Blackbird can initially climb to speeds high enough for the operation of the ramjets to become functional.

I am led to understand that the operation of ramjets requires a speed high enough to render the compression of a turbine redundant. That I can follow without much trouble, but if the Blackbird only carries ramjets then how can it attain such speeds in the first place? I have examined articles on the aircraft and the engines, but nothing seems to fill in these gaps. I can make the leap from ramjets to scramjets, and pulse jets to pulse detonation wave engines, but all I can find on early ramjet design suggests that rockets are needed to propel the Blackbird to ramjet speed (or damn good turbofans) and I can't seem to find this information.

As I'm writing a story concerning the advances in this area, any explanation on this would be most gratefully received.

for the sr-71, the engines were Pratt & Whitney J58's, which are a hybrid turbojet/ramjet engine. at low speed the necessary thrust comes from the turbojet section, at higher speeds the ramjet begins to contribute more and more thrust. check the engine page for more details. Xcomradex 23:20, 3 July 2006 (UTC) 23:18, 3 July 2006 (UTC)[reply]
(edit conflict) Some ramjet craft piggyback off of other aeroplanes and drop into a dive after they're released until they gain enough speed for the ramjets to function correctly. Obviously this doesnt apply to the SR-71 though. Philc TECI 23:21, 3 July 2006 (UTC)[reply]
There's an article about the Bomarc missile system: Although it doesn't go into much technical details, the missile had two stages. The first stage was a solid-rocket booster that propelled the second stage up to a velocity where the second stage ramjet engines were able to operate.
You are right - ramjet-powered vehicles always need another way of accelerating to the speeds where their ramjet would work. Either they have another set of engines (or a hybrid solution like the SR-71), or they are attached to another vehicle which accelerates them to the appropriate speed and then detaches. — QuantumEleven 13:16, 4 July 2006 (UTC)[reply]