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March 18

Wild bird with a foot band

Weird. I was sitting in my backyard and I saw a wild sparrow with a metal foot band. I'm guessing it didn't put it on its self. How did it get on there? 198.188.150.134 (talk) 01:29, 18 March 2010 (UTC)

I imagine it was Bird ringing. Never knew it was called that before--Jac16888^Talk 01:33, 18 March 2010 (UTC)

That was a very interesting read, thanks for linking to that article. But I fail to see why researchers would add this tag onto a mere sparrow. I'm pretty sure they already know the migration patters of sparrows, and I don't see how it would be practical to count the population of sparrows, I mean , there must be hundreds of thousands of sparrows, there's no way you can catch that many and be sure you've got them all. 198.188.150.134 (talk) 03:25, 18 March 2010 (UTC)

I found it pretty interesting too. Maybe keep an eye out for it, it might still be around, and see if you can get a close up picture of its band and find out who tagged it. Also, found this link [1] which says its for determining survival rates and populations--Jac16888^Talk 03:34, 18 March 2010 (UTC)

To 198...: You'd be surprised. They tag all sorts of animals, even those that aren't endangered, and keep track of them for the very good reason that they can't know if there is or isn't a problem unless they keep track. Yes, there are lots of sparrows, but they need to know exactly how many is "lots" and if there start to be less, they can stay ahead of the problem. They need baseline numbers of populations to know when something does go wrong. And they aren't trying to tag them all. They tag a representative sample each year, and then see how many tagged animals they catch from year to year. If they tag a constant number every year, and the number they get back is steadily declining, then they can get info on the whole population. --Jayron32 03:51, 18 March 2010 (UTC)

To clarify, that: They determine how many there are by seeing what proportion of sparrows they catch have already been tagged. If, say, they have tagged 100 birds and they find that 10% of the birds they catch are tagged, they can conclude there are about 1000 birds. --Tango (talk) 09:02, 18 March 2010 (UTC)

Wow, this is most interesting, thanks Wikipedia. 198.188.150.134 (talk) 07:13, 19 March 2010 (UTC)

has anyone ever been born completely reversed?

I don't mean dextrocardia, but a more complete reversal, including heart, liver, etc. Has it happened? If not, could it? Why or why not? This is not homework. 82.113.121.93 (talk) 01:40, 18 March 2010 (UTC)

Evidently, as the dextrocardia article links to Situs inversus, affecting "less than 1 in 10,000 people". Apparently, Randy Foye has it, and gets along just fine. Buddy431 (talk) 01:59, 18 March 2010 (UTC)

Indeed. A complete reversal shouldn't cause any problems at all (other than doctors misdiagnosing you because, say, the pain from your appendicitis is on the wrong side, or transplanted organs not fitting properly). Partial reversals cause all kinds of problems because things don't fit together correctly. --Tango (talk) 09:04, 18 March 2010 (UTC)

Further to the OP's interesting question, has anyone ever been born inside out, like the monkey in "The Fly"?[Trevor Loughlin]80.1.80.17 (talk) 11:03, 18 March 2010 (UTC)

I'm not familiar with "They Fly", but I don't see how anyone could be born inside out. If the "internal" organs are outside the skin, what would hold them together? --Tango (talk) 11:34, 18 March 2010 (UTC)

its easy to say situs inversus has no problems apart from the appendix being on the left side. however there are innumerable diseases that go hand in hand with situs inversus - many congenital cardiac anomalies, respiratory problems due to ciliary dysfunction and infertility to name a few... —Preceding unsigned comment added by 213.130.123.30 (talk) 11:25, 18 March 2010 (UTC)

I'm not familiar with "They Fly", but I don't see how anyone could be born inside out. If the "internal" organs are outside the skin, what would hold them together? --Tango (talk) 11:34, 18 March 2010 (UTC)

I believe there is a condition where some of the internal organs, like the intestines, are outside of the skin. As you can imagine, this is quite serious and surgery must be performed to place those organs back inside and sew up the skin. StuRat (talk) 14:13, 18 March 2010 (UTC)

Yes, I have heard of people being born with various kinds of hernia. I can't find any mention of it in our article, though (it mentions congenital diaphragmatic hernia, but that involves things from the abdomen protruding into the chest cavity, rather than outside the body). That is far from being inside-out, though. I'm not really sure what an inside-out person would be like. The concept of being inside-out only really makes sense for hollow objects. --Tango (talk) 14:39, 18 March 2010 (UTC)

Gastroschisis. [2] --TammyMoet (talk) 15:40, 18 March 2010 (UTC)

Well, it would mean the ectoderm develops inside and the endoderm develops outside the body. I suspect such a severe deformity would result in a spontaneous abortion early in the pregnancy. (And people/animals can be modeled as a hollow tube.) StuRat (talk) 15:43, 18 March 2010 (UTC)

Do your feet smell? Does your nose run? You're built upside down! --jpgordon^{::==( o )} 23:47, 20 March 2010 (UTC)

Inflation (cosmology)

Can Inflation (cosmology) be thought of as Refractive index just as various materials like water, air and glass are thought of in terms of their effect on the speed of light since such materials in effect impose the effect of inflation (cosmology) upon the speed of light due to their refractive index? -- 71.100.11.118 (talk) 01:42, 18 March 2010 (UTC)

No, it can't. Dauto (talk) 03:29, 18 March 2010 (UTC)

The how do you explain that in the absence of inflation the furthest source of light would be 13.7 billion light years away whereas with inflation it appears to to have traveled for 43 billion years just like light through glass appears to take longer than light through a vacuum? 71.100.11.118 (talk) 05:10, 18 March 2010 (UTC)

During inflation space/time expanded at far greater than the speed of light. But I'm pretty sure that's not all of it. According to observable universe we can see more than 13.7G light years because of the expansion of space. I take it as this (but I'm not a cosmologist, astrophysicist or astronomer): Star 10G LY away throws out some light, at some point in the 10 billion years the light spends travelling towards us, space expands putting the star an extra 10G LY away, so we see a star 20G LY away (also the wavelength of the light has doubled in that time). Perhaps inflation gave it a head start, too, with ultra-super-fast expansion of space. --Polysylabic Pseudonym (talk) 06:24, 18 March 2010 (UTC)

Dauto's brief answer was correct. Refraction follows certain quantitifiable laws. Cosmological inflation follows a different set of laws, including general relativity. If you want to have a very superficial explanation, then "yes, both effects relate to the way light travels." But they are totally different effects caused by different things. Nimur (talk) 14:47, 18 March 2010 (UTC)

Even without inflation, we could receive light from an object more than 13.7 comoving Glyr away, because some of the Hubble expansion (not inflation) of the universe occurred "behind" the photon as it traveled here. -- Coneslayer (talk) 14:52, 18 March 2010 (UTC)

Trains

Are there any electric freight trains?

Are there any monorail freight trains?

Are there any maglev freight trains?

Bowei Huang 2 (talk) 01:56, 18 March 2010 (UTC)

Yes, no and no as far as I can tell. The only proof of any of these methods that I can find is the electric freight train through Google. Regards, --—Cyclonenim | Chat  02:16, 18 March 2010 (UTC)

What route does the electric freight train take on its journey through Google? Edison (talk) 03:41, 18 March 2010 (UTC)

^{Left at I'm feeling lucky station. Regards, --—Cyclonenim | Chat  03:44, 18 March 2010 (UTC)}

Please see electric locomotive, monorail, and Maglev.--Shantavira|^{feed me} 08:29, 18 March 2010 (UTC)

The thing that distinguishes freight from people is speed. There are very few - if any - items of rail freight that needs to get there as fast as people do. Electric trains are about efficiency and such - and hence are useful for both people and freight. But monorails and maglevs are all about speed - and they are a costly way to carry freight. Hence, I agree with Cyclonenim - "Yes, No, No". SteveBaker (talk) 00:32, 19 March 2010 (UTC)

Maglevs are about speed, but most monorails are short-distance, low-speed services in an urban area or a still more confined location like an airport or fairground, and what they're about is having an elevated route without the wide structure needed to support a conventional track. However, such systems would not normally have freight traffic either.

However, at least one monorail did historically exist in a rural location and did carry some freight: the bizarre Listowel and Ballybunion Railway. --Anonymous, 20:44 UTC, March 19, 2010.

Purpose

Does any action other than those done by conscious beings have a purpose or are they just reacting to the laws of physics? —Preceding unsigned comment added by 99.254.8.208 (talk) 05:16, 18 March 2010 (UTC)

Science really can't answer that. See teleology. --Trovatore (talk) 05:19, 18 March 2010 (UTC)

Thank you. On a tangentially related topic, do single-celled organisms do things on purpose? 99.254.8.208 (talk) 05:40, 18 March 2010 (UTC)

There is no contradiction between having a purpose and reacting to the laws of physics. A "purpose" is a label attached to certain types of activities in order to make them easier to predict and understand, namely activities that can be understood as promoting some goal. Currently the only systems we know that can usefully be described in terms of purpose are organisms that result from natural selection, and devices that those organisms have created. Looie496 (talk) 05:41, 18 March 2010 (UTC)

On purpose? Some single celled organisms can follow scent trails to food; can cooperate with others to form colonies. Some pathogenic single celled organisms can cooridinate attacks through quorum sensing. Is sensing something and reacting to is "purpose?". In fact, are "conscious" humans doing any more? Perhaps the Humanities desk can offer a better answer to this philosophical question. --Polysylabic Pseudonym (talk) 06:34, 18 March 2010 (UTC)

J. Scott Turner has an interesting book that I read a couple years ago. He talks a bit about purpose/intentionality, and concludes that to be "purposeful", an organism has to have a mental representation of the world, and to be able to try to make the world more like the one they imagine. So basically, anything that's programmed to modify its environment in accordance with some inner plan could be said to "have a purpose". Beavers, ants, and quite a lot of things would probably fall under this definition. *shrug* I thought it was interesting, anyway. Indeterminate (talk) 11:01, 18 March 2010 (UTC)

I don't think you can sensibly speak of purposes where there is no motive force involved. Did the Colorado River 'purposely' wear down the Grand Canyon? Not really. Vranak (talk) 07:15, 18 March 2010 (UTC)

Conscious beings obey laws of physics. They just think they have a choice. 67.243.7.245 (talk) 15:27, 18 March 2010 (UTC)

If one-celled organisms do all that, what is the most simple thing that tries to preserve (or reproduce) itself by choosing between two actions, even if it's a simple if-then decision? --99.254.8.208 (talk) 22:36, 18 March 2010 (UTC)

It depends what you mean by "choose". Take something like an amoeba - it can detect when there is too little food around and "chooses" to turn into a Microbial cyst to wait out the lean times. But what causes that is a relatively simple chain of chemical pathways...amoeba don't have a nervous system - they consist of just a single cell after all. So is this a "choice"? If you think it is then we can go smaller - does a plutonium atom "choose" to preserve itself for (on average) around 14 years and not to decay into Americium? Well, that's certainly a bit of a stretch - it is a simple physical matter. But why is the chemical reaction that serves to cause the amoeba to turn into a cyst more or less valid than what the plutonium atom does? So if we say that the amoeba didn't have a "choice" then you have to say that a tree doesn't have a choice about whether it's going to drop its leaves in autumn...and (most scientists would say) a human doesn't have a "choice" about whether to have a ham sandwich or a cheese sandwich for lunch...it's still a matter of physics/chemistry...albeit a very complicated matter. So this word "choice" doesn't have a good definition. We either have to apply it to the smallest objects in the universe - or we have to avoid applying it to the most complex objects that we are aware of. In the end, it's just a word. SteveBaker (talk) 02:30, 19 March 2010 (UTC)

It's one of those "what is the definition of this word?" questions. What do you really mean by "having a purpose"?

Consider a definitely inanimate object - a computer let's say - does a computer that is programmed to calculate the first million digits of PI have a "purpose"?

• If the answer is "yes" - then your definition of "purpose" certainly encompasses even the simplest life forms and inanimate objects. A rock that becomes detached and bounces down the side of a mountain has just as much "purpose" as a computer - and so do viruses and bacteria...by that definition.
• If you say "no" (presumably because you feel that a computer is a deterministic system and is "inanimate") then I'd have to argue that from everything science has determined, higher life forms and even humans are simply complex machines. Under this definition of "purpose", nothing in the entire universe has a "purpose" because everything is simply slavishly following the mundane laws of physics.

We could turn to the dictionary definition of "purpose" - which (according to Wiktionary) is:

1. An object to be reached; a target; an aim; a goal.
2. A result that is desired; an intention.
3. The act of intending to do something; resolution; determination.
4. The subject of discourse; the point at issue.
5. The reason for which something is done, or the reason it is done in a particular way.

Of these definitions, our pi-calculating computer has a purpose under definitions (1),(2),(4) and (5)...but not (I'd say) under definition (3). But this is no longer a matter of science - it's a matter of linguistics. SteveBaker (talk) 00:26, 19 March 2010 (UTC)

We don't have the capacity to ultimately control ourselves. We can observe what transpires in our experience from things we do. Therefore we don't have "purpose." Living is comparable to an art. In this sense a boulder rolling down a hill has more "purpose" than a well-lived life of a human being. Bus stop (talk) 00:35, 19 March 2010 (UTC)

Multiverse

Can you get to another universe by physically traveling in 3D space out of the area resulting from the big bang, or would you have to travel in a different kind of "direction" to get there? —Preceding unsigned comment added by 99.254.8.208 (talk) 05:34, 18 March 2010 (UTC)

I'm not sure actual scientific theories of "Multiverses" work in the way you seem to think they do. You may want to read Multiverse and Many-worlds interpretation and the works of Hugh Everett III and Max Tegmark. Any actual discussion of parallel universes is purely speculative and without experimental verification. Such possible universes are perfectly consistant with current understandings of how our universe works, but there is also nothing in the current theories that requires them, so by Occam's Razor, there generally isn't much use for hypothetical alternate universes in the current theory. --Jayron32 05:45, 18 March 2010 (UTC)

I'm pretty sure you couldn't travel outside the universe in normal 3D space. The edge of the universe is moving away from you at (or faster than?) the speed of light. You can't go that fast. --Polysylabic Pseudonym (talk) 06:37, 18 March 2010 (UTC)

If the universe has an edge. See Shape of the Universe for a discussion on the various proposed shapes the universe has. There are some models called "Open universe" models that have no edge because they are infinite; there are also "closed edgeless" models as well. --Jayron32 06:42, 18 March 2010 (UTC)

The notion of an alternate universe suggests a space that exists but is not in any way accessible. And if we cannot access it, then we cannot know it exists. So... pretty much an empty concept, unless I am missing something. Vranak (talk) 07:05, 18 March 2010 (UTC)

There are some theories of "bubble universes", where there are multiple universes separated only by space, but they are always arranged in such a way that it is impossible to travel between them. If you could travel between them, they would be part of the same universe by any reasonable definition. --Tango (talk) 09:08, 18 March 2010 (UTC)

What if they aren't accessible by normal means, but are via wormholes ? StuRat (talk) 14:03, 18 March 2010 (UTC)

Again, wormholes are allowable by, but not necessary for, current theories of how the universe works. That makes them fun for science fiction, but lacking evidentiary support, there's no need to suppose they actually exist. At some level "magic" is also allowable by, but not necessary for, current theories as well. After all "Another land that obeys different physical laws and can only be reached through a cross-dimensional portal" is still basically magic whether its The Lion, The Witch, and the Wardrobe or the work of a well respected cosmologist. Couching the magic with terminology from quantum theory doesn't make it less magical. --Jayron32 14:18, 18 March 2010 (UTC)

What's the difference? As far as I know, there is no difference between a path through "normal" space and one through a wormhole. Locally, space is space, and globally it all depends on your perspective. --Tango (talk) 15:16, 18 March 2010 (UTC)

blue chemical

What is the name of the blue coloring I get after soaking a nickel-copper alloy in white vinegar for a couple of days? Is it copper acetate? Googlemeister (talk) 13:12, 18 March 2010 (UTC)

Our article on copper acetate would seem to imply that that is probably the case - the history section mentions that traditionally copper acetate was prepared using vinegar and copper sheets. I don't know whether the nickel would also react, potentially giving a mixture.131.111.185.69 (talk) 13:31, 18 March 2010 (UTC)

From what I can find on Google, nickel also reacts with acetic acid, and is also bluish. It is probably a mixture of copper(II) acetate and zinc nickel acetate. --Link ^(t•c•m) 13:40, 18 March 2010 (UTC)

zinc acetate? was that a typo? --Ludwigs2 15:04, 18 March 2010 (UTC)

Yes. Blame it on the relative positions of copper and zinc in the periodic table. --Link ^(t•c•m) 22:00, 18 March 2010 (UTC)

bleach

Will two bleached negro parents have white offsprings? —Preceding unsigned comment added by 117.196.140.248 (talk) 14:04, 18 March 2010 (UTC)

No. --Jayron32 14:12, 18 March 2010 (UTC)

Well unless bleached is a bad translation for Albino, in which case maybe? --BozMo talk 14:41, 18 March 2010 (UTC)

More likely the OP is refering to Lamarckism; genetic inheritance doesn't work that way. --Jayron32 14:46, 18 March 2010 (UTC)

In case the OP is a non-native English speaker, he or she may wish to read our article on the word negro to see why this term is not commonly acceptable in modern speech. At the very least, understand the nuances of your word choice. Nimur (talk) 14:51, 18 March 2010 (UTC)

The United Negro College Fund begs to differ. Comet Tuttle (talk) 17:14, 18 March 2010 (UTC)

They probably wouldn't. Like the National Association for the Advancement of Colored People, just because they maintain a historic name does not mean they necessarily think the antiquated terms should be used in other contexts. --Mr.98 (talk) 23:17, 18 March 2010 (UTC)

There is a quote that begins: "If the white man is but a bleached negro,...". If this is what is referred to, it is a trick question and the answer is yes. Other quotes use "bleached negro" to mean a culturally Anglicized African. Rmhermen (talk) 15:29, 18 March 2010 (UTC)

From Doctor Huguet: "And if the taint of the brute origin adheres to the negro, does it not cling to us all? If the son of a murderer stands disgraced, does not his grandson inherit something of the shame? If the white man is but a bleached negro, what right has he to mock his dark progenitor? The credit is due, not to him, but to the cold and clouds of the stormy north, or the darkness of the troglodytes' caves, during myriads of years" (emphasis mine) That's very poetic, if a bit outdated. Buddy431 (talk) 03:10, 19 March 2010 (UTC)

"Bleached negro" could be a non-politically correct version of the also non-politically correct "mulatto", - a person of mixed white and black ancestry. In this case, I would dare to say that yes, they could have child that could be considered white by some standards. However, according to the one drop rule, they could only have black offspring.ProteanEd (talk) 16:33, 18 March 2010 (UTC)

According to the one drop rule everyone on the planet is almost certainly black - see recent African origin 131.111.185.69 (talk) 16:48, 18 March 2010 (UTC)

You are only right if evolution is right. From my experience, however, I know that many racists don't believe in evolution. ProteanEd (talk) 17:41, 18 March 2010 (UTC)

ProteanEd, have you a source for claiming that "mulatto" is non-politically correct? That would be arguable only if you can supply an alternative that expresses the same meaning. The word "quadroon" also has a specific meaning. Cuddlyable3 (talk) 01:21, 20 March 2010 (UTC)

Back to being scientific: If referring to albinoism, wouldn't two albino "black" parents be most certainly have albino children as well? (Our albinoism article doesn't mention this, maybe it isn't necessary true). From what I remember from highschool biology about recessive gene says this would be the case. --Kvasir (talk) 18:02, 18 March 2010 (UTC)

If both parents were affected because of mutations in the same gene, then yes, their children would all be affected (since they could only inherit a mutated copy of the gene from each parent). However, there are several different genes that can be involved in albinism, meaning that if the parents have albinism due to mutations in different genes, then none of their children will be affected, but will all be carriers for two different heterozygous mutations. --- Medical geneticist (talk) 11:29, 19 March 2010 (UTC)

Interesting, thankyou. --Kvasir (talk) 14:54, 19 March 2010 (UTC)

Not sure if it has been mentioned, but actions taken by an individual to change their appearance (i.e. bleaching skin, body building, etc.) would fall under the topic of phenotipic plasticity; those traits do not become part of the individual's genetic make-up. Of course there is some degree of probabilty involved (perhaps a mutation for that allele at an identical locus for each parent), but in short, no, their children won't inherit that trait.161.165.196.84 (talk) 09:52, 21 March 2010 (UTC)

Narcissus in the aquarium

I'm wondering if I can put my post-bloom Narcissus tazetta in the aquarium. I read that narcissus contains the alkaloid poison lycorine. I suppose it's toxic to humans but is it safe for the aquarium? I hope the bulb can soak up the excess nutrient in the tank and also live beyond its normal short flowering life. Thx. --Kvasir (talk) 14:35, 18 March 2010 (UTC)

In my experience, the way to treat indoor narcissi after flowering, is to give them a hefty dose of plant food and then ignore them for about 9 months, until you see a shoot coming from the bulb. They do come back every year, which is what happens in the wild, but they are not repeat flowering. Why would you want to drown a bulb in an aquarium? --TammyMoet (talk) 15:37, 18 March 2010 (UTC)

Mostly to soak up nutrients. I also read you can bury the bulb in a pot. Most Narcissus tazetta were sold submerged in water with pebbles for Chinese new year, not in a pot with soil. Maybe that's why i seem to think the species is aquatic or come from swamp like environment. I havne't found any info to that fact. --Kvasir (talk) 18:10, 18 March 2010 (UTC)

I think it's a propensity of plants which come from bulbs not to need soil. For example, I have a hyacinth in a hyacinth vase. Its roots are in the water in the bottom of the glass. All the nutrients it needs are contained in the bulb - hence the advice to give the bulb a good feed before letting it die down and dry off naturally. All it needs in order to fulfil its destiny and produce a hyacinth is water and light. If I'd put the whole bulb in water it would rot: it's just the roots that need to be in water. I bet you'd find the water level only came up to the top of the pebbles in the gifts you mention. You might find this site useful: [3]--TammyMoet (talk) 16:22, 19 March 2010 (UTC)

Interesting, never seen these vases before. --Kvasir (talk) 15:48, 22 March 2010 (UTC)

Calculation of lifespan of the sun

Here's my solution, but it's about 5 magnitudes off the accepted value-
solar luminosity(rate at which the sun gives out energy) $=3.839x10^{26}$
let the lifespan of the sun= energy nuclear fusion /solar luminosity
$e=mc^{2}$.
consider only the mass of the solar core, since nuclear fusion only takes place there.
therefore
$m={\text{volume of the core}}\times {\text{density of the core}}$ $=0.24^{3}\times {\text{volume of the sun}}\times 1.5\times 10^{5}$ (Solar core)
$e=2.12841x10^{49}$
${\text{lifespan}}={\frac {2.12841\times 10^{49}\div (3.839\times 10^{26}}{(60\times 60\times 24\times 365)}}=1759{\text{trillion years}}$

this is obviously off, someone pls help!! —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 15:43, 18 March 2010 (UTC)

Did you use the density of the suns core for the volume of the entire sun, or just for the core of the sun? Googlemeister (talk) 16:01, 18 March 2010 (UTC)

First: e=mc², yes -- but the proton-proton chain is not a full matter-to-energy conversion. Rather, about 0.7% of the initial mass becomes energy, with the remaining 99.3% becoming helium. Additionally, the accumulation of helium in the core accelerates the rate of fusion (see stellar evolution), so a constant rate estimation will be high. Finally, I'd guess that the numbers you're using for "core" may not exactly match the region where fusion occurs. Perhaps the fusing region is smaller, or perhaps hydrogen near the edge of the core doesn't fully fuse. — Lomn 16:01, 18 March 2010 (UTC)

Oops, updating: the bulk of the hydrogen in the sun (that which isn't in the core) will never fuse. There's simply not enough mixing. At some point the helium in the core overwhelms the hydrogen (and being denser, it settles there) and the sun kicks into helium fusion even while still predominantly composed of hydrogen. That's your other missing bit. — Lomn 16:05, 18 March 2010 (UTC)

I have fixed the formatting of your TeX - you need to put \text{} around text or it is treated as lots of variables ("volume" is v*o*l*u*m*e, etc.). The \text command gets rid of the italics and keeps the spaces in. --Tango (talk) 17:35, 18 March 2010 (UTC)

Also, the mass you calculate for the core is already larger than the total mass of the sun! Icek (talk) 03:58, 19 March 2010 (UTC)

expansion exceeding speed of light

If the universe was static and not expanding then it would make sense that the furthest distance one could see would be the age of the universe or at least to the age of the first photon. If the Universe is expanding then at a steady rate of say twice that distance in 13.7 billion light years then it makes sense that we could see light from 27.4 light years away since that light started coming at us only 13.7 billion light years ago. So what if the expansion of the Universe reaches the speed of light, would not this be the point at which we could only see darkness and nothing beyond just like looking into a Black Hole? 71.100.11.118 (talk) 15:51, 18 March 2010 (UTC)

The fault in your explanation is that you are using the universe itself as a frame of reference for measuring the rate of expansion of the universe. The speed of light is based on the current rate of change of the universe - considering the universe to be static. So, if the universe expands or contracts, it is the static frame of reference. The speed of light is measured as though the expansion/contraction of the universe is static. Therefore, light is always light speed in the universe no matter how fast or slow it expands. -- kainaw™ 15:56, 18 March 2010 (UTC)

Well not exactly. Expansion and contraction of the Universe effects the rate at which bodies that emit light are comin toward or going away from us and since some of these bodies emit light if they are ever traveling with the expansion at a rate away from use that exceeds that speed of light then that light being emitted would have to travel faster than the speed of light in order for us to see it. (I am of course assuming that an object traveling away from us faster than the speed of sound can also not be heard.) 71.100.11.118 (talk) 16:49, 18 March 2010 (UTC)

I rarely even belong on the Science desk but I can say that your latter statement isn't true. If there's an airplane 1 mile away from you, travelling at Mach 6 away from you, its sonic boom and the sound of its engines radiates in all directions. The sound will reach you in about 5 seconds, just like any other sound that was emitted from 1 mile away. The sound isn't carried along in the path of the airplane in a "wake" of some sort. (You may have been claiming that the sound couldn't be heard because a quickly expanding universe would keep the sound away from you, but that wasn't how I read your statement.) Comet Tuttle (talk) 16:56, 18 March 2010 (UTC)

Actually a sonic boom and the crack of a whip or lightning occur at a static point in air where the plane or end of the whip or super heated air are compressed and the uncompressed rather than coming out of the plane such a the music from a mp3 player coming out of a speaker that is traveling along at the same speed above the speed of sound which the plane is traveling at. Even if the speaker is pointed in your direction you will not hear the sound coming from it. (I'm pretty sure.) 71.100.11.118 (talk) 17:23, 18 March 2010 (UTC)

You are half right here. There's nothing fundamentally different about a sonic boom from any other type of sound (it's all just compression and decompression of air) so if the plane is playing music you'll still hear it on the ground. But it's also true that sound travels through a medium, which is in this case nearly stationary with respect to the observer, so the fact that the plane is going fast doesn't matter. If on the other hand you're making noise on the ground, the plane can't hear you because it's outrunning the sound. It's also true that using sound as an analogy for light is usually a bad idea. The whole point of relativity is that they can't be treated the same. Rckrone (talk) 18:03, 18 March 2010 (UTC)

I suppose that in the case of the speaker moving with the plane and emitting sound that the frequency reaching the listener's ear would fall below the threshold of human hearing. 71.100.11.118 (talk) 03:00, 20 March 2010 (UTC)

The expansion speed depends on distance. The further something is away, the faster it recedes from us (this is Hubble's law). That means that, if something is far enough away, it will be receding from us faster than the speed of light. This is actually not at all strange. The rule about not going faster than light speed applies to objects moving through space, not the expansion of space. We can still see objects that are receding faster than the speed of light, it just takes longer and longer for the light to reach us. The point where we can't see something because the light hasn't had time to reach us yet is the cosmological horizon (it is very similar to the event horizon of a black hole). --Tango (talk) 17:44, 18 March 2010 (UTC)

In answer to the OP's question, future of an expanding universe says that eventually galaxies and galaxy clusters will be moving away from each other at the speed of light. This will not only make intergalactic travel impossible, but there will basically be no evidence for the big bang for new civilizations to discover. It won't look like looking into a black hole from within any of these galaxies, since you'd still have stars in your own galaxy, but beyond this it would be pitch black. — Ƶ§œš¹ _{[aɪm ˈfɹ̠ˤʷɛ̃ɾ̃ˡi]} 22:51, 18 March 2010 (UTC)

Strange... That is not what I remember the article referenced says. I'm going to see if our library has it so I can check it again. My memory is that it states that increased acceleration will increase the red shift until gamma rays are longer in wavelength than the universe, so it will not be possible to detect electromagnetic waves as waves anymore. It does not imply that the galaxies will be travelling faster than the speed of light or that there will be no evidence of the Big Bang since everything inside the galaxy will still be visible and known in the same way. Interstellar radiation will be undetectable with current methods. A future civilization will require someone to assume that there is another galaxy that must be red-shifted and figure out some method for proving it to be true. -- kainaw™ 00:13, 19 March 2010 (UTC)

Found the article online. Whomever cited it apparently confused reaching the de Sitter horizon with expanding faster than the speed of light. I'll change the article here to properly represent what the paper being cited says. If another paper cites it differently, that should be used instead of the article currently being used. -- kainaw™ 00:28, 19 March 2010 (UTC)

Hmmm, although the article doesn't back up what I said, this video (at the 51-minute mark) does (which is where I learned this idea in the first place). Remember that it's not that the galaxies are moving away from each other; rather the space between them is expanding, which causes the redshifting (the 10 minute mark of the video does a good job of explaining it). At a certain point, that redshifting will be so extreme that no known method can be used to detect other galaxies or the CMB. — Ƶ§œš¹ _{[aɪm ˈfɹ̠ˤʷɛ̃ɾ̃ˡi]} 09:07, 19 March 2010 (UTC)

My main issue isn't that redshift will become extreme - it is the claim that galaxies will be independently travelling faster than the speed of light - which is what the questioner apparently believes. You are very accurate in stating that the distance between the galaxies is expanding. The galaxies are stationary - not moving. So, they are not travelling faster than the speed of light. The way I normally hear it explained is that the galaxy is like a balloon. The galaxies are dots on the outside of the balloon. As the balloon expands, the distance between the dots expands, but the galaxies are still on the same spot on the balloon. So, from the point of view of a galaxy in space, it appears that other galaxies are moving away at incredible speeds, but they aren't actually moving at all. Space is expanding. -- kainaw™ 13:38, 19 March 2010 (UTC)

Kainaw, within the framework of relative your statement "it appears that other galaxies are moving away at incredible speeds, but they aren't actually moving at all" makes no sense since movement is relative. There is nothing wrong with thinking of the movement of the galaxies away from us as real movement. Dauto (talk) 16:17, 20 March 2010 (UTC)

It's an important distinction to make since, according to special relativity, galaxies actually moving at the speed of light is impossible. — Ƶ§œš¹ _{[aɪm ˈfɹ̠ˤʷɛ̃ɾ̃ˡi]} 03:45, 21 March 2010 (UTC)

The question remains unanswered then or in disagreement since blowing up a balloon still puts distance between the dots on its surface. By adding space fast enough eventually this distance will force the dots to separate at the speed of light and possibly beyond even though light will retain its speed limit meaning that the dots will become invisible to each other. 71.100.11.118 (talk) 13:09, 22 March 2010 (UTC)

There is no disagreement. Galaxies won't be able to see each other. It'll be pitch black looking beyond one's galaxy/cluster. — Ƶ§œš¹ _{[aɪm ˈfɹ̠ˤʷɛ̃ɾ̃ˡi]} 22:36, 22 March 2010 (UTC)

Which converter?

If you have a device (a Set-top box) with the specifications of 12V 15W, which voltage converter should you buy? I saw one 12V 1A, but will it fit this device?--ProteanEd (talk) 16:00, 18 March 2010 (UTC)

Watts = Volts * Amps. Your device potentially draws 1.25 A, so the converter is probably not sufficient. — Lomn 16:02, 18 March 2010 (UTC)

And what if I have the choice of a 1.8A, 12V converter? Can this converter fry the device? Or will the device just use enough power as needed? ProteanEd (talk) 16:07, 18 March 2010 (UTC)

A power supply is usually rated by the volts it delivers and the maximum amps it is capable of supplying. Devices draw however much they draw of that. Consider that your 15-amp fuse/circuit-breaker doesn't blow when you are only driving a small lightbulb. DMacks (talk) 16:12, 18 March 2010 (UTC)

It depends on the type of converter. Simple transformer-based ones are designed to be paired with something drawing close to the rated current (if the current draw is too low, the voltage provided will be higher than rated), while switch-mode and other regulated converters will provide the rated voltage for any current draw. --Carnildo (talk) 00:10, 19 March 2010 (UTC)

It's been a while since I did radio electronics, but I seem to recall that a wire transformer rated at (say) 12V, 5A will deliver 12V at 0.05A, will deliver 12V at 1A, still at 5A, but will deliver less than 12V at 6A (or blow its fuse, or fuse its wire windings). A switch mode transformer (which pretty much any transformer you find will be) should deliver close to its rated voltage anywhere under the rated current.

In answer to the original question, you can safely use a voltage converter (which I've been calling a transformer above) rated at 12V and any amperage (A) above 1.25. So you are safe to buy a 12V 2A voltage converter. --Polysylabic Pseudonym (talk) 02:03, 19 March 2010 (UTC)

I would recommend getting something that can produce at least 50% more amps than you strictly need. Power supplies that are being run close to their maximum amperage tend to get hot - and that shortens their lives considerably. The 1.8A/12V gizmo sounds like just the thing. Carnildo's right about not going crazy, you certainly wouldn't want something that could produce (say) 10 times the current you need - but 50% of excess power is perfectly reasonable. The 1.8A/12V gizmo sounds like just the thing. SteveBaker (talk) 02:06, 19 March 2010 (UTC)

I may be old but there is nothing wrong with my short term memory nor is there anything wrong with my short term memory. The 1.8A/12V gizmo sounds like just the thing. Cuddlyable3 (talk) 01:10, 20 March 2010 (UTC)

auxins - geotropism and photo-tropism

auxins are there in growing tips of plants, at shoot tips. are they at root tips too? nextly what about the growing tips of leaf? Next question what's there in roots that makes them response to gravity(mainly growing towards gravity). and whats there in shoot tip that makes to grow towards light? both of them have auxins(main growth promoting harmone)then what makes the above difference....thanx.--Myownid420 (talk) 16:28, 18 March 2010 (UTC)

Don't roots tend to grow toward water more than down ? As for leaf tips, the amount of photosynthesis would seem to be the proper way to guage which way to grow. StuRat (talk) 16:48, 18 March 2010 (UTC)

Auxin, gravitropism, and phototropism cover this issue with varying quality. I don't know how gravitropism works, but with phototropism the auxins are sent to the shaded side of the plant, which then grows faster, which tends to push the plant away from the shaded side, which sends the plant towards the light. It'll be the same with gravitropism, but I don't know what the mechanism is for the plant to detect gravity. --Sean 19:43, 18 March 2010 (UTC)

I'm sure you can get data from the auxin and gravitropism/phototropism articles, but to put it short: a chemical necessary for growth on the side of the stem that is near the light dissolves and/or becomes deactivated, and so growth is greater on the side of the stem farther from the light and the plant 'tips' towards the light. As for gravitropism, intracellular vesicles were determined to control cellular position -- cells have an affinity for them on their floor (thinking of cells as oversimplified cubes). So the cells grow in order to orient themselves so that these vesicles lay on the bottom of the cell -- roots therefor 'know' which side is down and so do stems. I can't seem to remember the name of these vesicles and can't find it online, but they start with the letter 'a'. DRosenbach ^{(Talk | Contribs)} 19:46, 18 March 2010 (UTC)

Got it -- amyloplast. DRosenbach ^{(Talk | Contribs)} 20:16, 18 March 2010 (UTC)

Fascinating!, thanks. --Sean 13:17, 19 March 2010 (UTC)

Phototropism says auxins on the dark side cause softening of the cell wall which allows the cells to expand and thereby tip towards the light. The light-side-growth-deactivation process you describe seems to be something totally different. Citation? --Sean 13:17, 19 March 2010 (UTC)

string theory and space

Some string theories suggests that they are closed rings. And a sting can have enough energy that it will get stretched and form a membrane i.e. a universe. One more thing to note is that space cant be torn, it can be stretched, twisted, bended but not torn. So the hole in between a string will not get filled as it stretches, where will it go? (i got this doubt when i was watching "The_Elegant_Universe-PartIII-Welcome_to_the_11th_Dimension" from NOVA)

Maybe it will stay the same size, and nobody will ever notice it ? StuRat (talk) 16:51, 18 March 2010 (UTC)

See Membrane (M-Theory) and Brane cosmology for discussions over these theories of the universe. --Jayron32 16:52, 18 March 2010 (UTC)

Freshly-made Bradford Reagent

[/i] Has anyone here made their own Bradford Reagent?

I'm trying to make sure the color's right - it's supposed to be a brownish-red.

So far I have:

1. Added 100 mg Coomassie Brilliant Blue G-250 to 50 ml of 95% Ethanol

2. Added 100 ml of 85% Phosphoric Acid to the above

3. Slowly (dripwise) added DI water to volume of 1000 ml

The results of step 3 are a dark purplish blue color, while the results of step 2 are the dark brownish red. So the next step:

4. Filter through Whatman #1 filter paper

...and it's still blue. I'm trying the filtration step again, in the hopes that a second filtration will help, but I'm wondering if anyone has done this before and might have a tip or two.

Thanks! Wevets (talk) 16:46, 18 March 2010 (UTC)

Have you checked the pH of your Bradford reagent after step 3? If higher than 2 that's a problem (how sure are you of your DI water source?). How long did you stir before step 3? It's possible the CBBG did not have enough time (e.g. overnight) to go into solution. -- Scray (talk) 23:35, 18 March 2010 (UTC)

Thanks! Haven't checked the pH after step 3 - I'll try that next - it should be less than 2 you say? The DI water source should be good, and I've tried it with two independent sources... even trying millipore water in a dilution going on now. I took a lot of time with step 3, since I heard it may be a problem (added 500 ml over 14 hours,) but I haven't tried to take much time with step 2... haven't seen any mention of it not going into solution - is that likely to be a problem? Should I mix the phosphoric acid and ethanol and CBBG overnight, then dilute overnight?

Incidentally, a second filtration didn't help. Still blue. Starting to wonder if the dye or the Phosphoric Acid could be too old - I'm not a chemist so I don't know how long these things last. Dye is 5 yrs old according to the bottle. Phosphoric Acid looks very old - can't find a date on the bottle, and the style of the label looks very old.

Wevets (talk) 00:49, 19 March 2010 (UTC)

It sounds like the dye is too dilute. I run this assay (you're doing protein quantitation, right?) at a 1:5 dilution in DI water. Just checking this now in the lab, diluting it too much does indeed alter the colour. I'm not a chemist either, but it seems the pH is raised too much and you shift the dye to it's anionic state (which is blue). Maybe try lowering the volume of DI water? Not sure that it needs to be added dropwise though. Also, the filtration step is just to remove particulate matter, it shouldn't have any bearing on the dissolution of the dye.

Incidentally, we use a commercially prepared solution, and they use methanol instead of ethanol- again, I don't know if that makes much difference chemically. And as for phosphoric acid, that does definitely seem to be somewhat perishable.Wrex-beater (talk) 08:45, 19 March 2010 (UTC)

Thanks - how dilute should the dye be? Maybe 0.1 mg/ml as a final dilution isn't so hot after all. Checking the pH is the first thing I'll do on Monday, so we'll see about that.

Do you use the Bio-Rad Protein Assay #500-0006, I think? That's similar to Bradford and has got Methanol in it. I tried diluting some of that with our DI water for comparison, and similarly, it turns a bit blue (though not the same bright blue/purple I'm seeing in the homemade.) Of course, someone hid the instructions that come with the bottle, so I'm going to download those to see if it recommends anything different. Wevets (talk) 14:31, 19 March 2010 (UTC)

3D TV and Animals

Would dogs or other animals enjoy 3D TV? TheFutureAwaits (talk) 17:38, 18 March 2010 (UTC)

I shouldn't think that they would take too kindly to having the requisite glasses put on them. Vranak (talk) 17:54, 18 March 2010 (UTC)

Koko (gorilla) enjoyed 3D Viewmaster views, so similarly tame gorillas might enjoy 3D TV as well. Edison (talk) 19:02, 18 March 2010 (UTC)

Some dogs will allow glasses to be put on them, if only for a little while. —Pie4all88 ^{T C} 19:05, 18 March 2010 (UTC)

Do dogs see at the same framerate people do? If a dog can see at a faster framerate, the TV picture would seem choppy. Googlemeister (talk) 19:24, 18 March 2010 (UTC)

People don't have a frame rate. Your eyes are being lit up as fast as photons can plink your photoreceptors, and some of those plinks get passed on to higher-level vision equipment, but your brain is definitely not getting a full-frame cinematic update every N microseconds. --Sean 19:48, 18 March 2010 (UTC)

Frame rate explains the 'frame rate of your eyes' (sort of), but it looks like that part of the article needs to be fixed. -Pete5x5 (talk) 19:59, 18 March 2010 (UTC)

Your eyes do not have a Frame rate, they have Persistence of vision. At the article Animation the top illustration of a bouncing ball has about the minumum frame rate 10 fps that can give an illusion of motion. The first surviving movie was shot at 12 fps. Cuddlyable3 (talk) 21:15, 18 March 2010 (UTC)

Even the concept of persistence of vision is kinda archaic - it implies the kind of situation where some light hits your eye and the signal that results from it slowly decays away unless more light hits it within a tenth of a second or so - such that if an image flickers fast enough, it'll seem to be a "continuous" stream of light. But that's not how it works at all. Your eyes are perfectly capable of seeing the light/dark/light/dark of a very high frame rate image - certainly into the 100 to 120Hz range. But it's now understood that the visual system can interpolate the missing data of a flickering image to produce the impression of a continuous image to higher brain functions. This is a useful adaptation when (for example) you're trying to hunt a small, fast moving animal in long grass where the blades of grass chop up the image into vertical strips that come and go as they are hidden and revealed between the blades of grass. There are some elegant experiments that demonstrate this - but the simplest is that even in a system that's updating a moving image 30 times a second on a 60 frames per second display (like many computer games do), most people see double-imaging of fast moving objects. If it were only necessary to update the image 10 times a second to get continuous motion (which is somewhat true) then there wouldn't be a problem. But because each image appears twice at the same position on the screen - then jumps forward a bit and appears twice in a row at the next position, our eyes/brain can't do simple linear interpolation and get the impression of uniform motion. Instead, what happens is that our visual system is forced to deduce that there are two objects moving along close together at uniform speed and each is flickering on and off 30 times a second. The fact that we can see this kind of double-imaging artifact at amazingly high frame rates is proof that we're able to see very short flashes of data and interpolate between them. Further evidence for this is due to interpolation and not something weird to do with having two eyeballs is that if you update the image only 20 times a second and display the video at 60Hz, many people will see triple-imaging. The term "persistence of vision" has come to be a handy short-hand for that - but it's not really the right concept for what's really going on. SteveBaker (talk) 23:59, 18 March 2010 (UTC)

That describes temporal aliasing that is avoidable by reducing the resolution (blurring) of the object in the direction of its movement. I can tell the difference between my 100Hz TV display and a continuous illumination by waggling my finger in front of my view but that does not mean I can see light/dark/light/dark flicker at 100Hz. The articles Flicker fusion threshold and this in Scholarpedia may be of interest.Cuddlyable3 (talk) 01:06, 20 March 2010 (UTC)

Nope - temporal aliassing is a completely different phenomenon. Temporal aliassing is a physical phenomenon related to the nyquist limit. The double-imaging effect happens entirely inside your eye/brain. Appropriate motion blur will indeed eliminate temporal aliassing - but not double-imaging. I didn't say that waggling your finger in front of the TV proved or disproved anything - but since you mention it, if you couldn't "see" 100Hz flicker, then it would appear continuous and the finger waggling trick wouldn't work. But we have to be super-cautious about the word "see". What our eyes can detect and what our consciousness is informed of are two very different things. SteveBaker (talk) 15:22, 20 March 2010 (UTC)

There are lots of reasons why most animals don't really like TV. With some, it's because they live in a world dominated by scent or that they have very precise hearing that isn't fooled even by stereo sound. Stereo vision is mostly important to predatory or arboreal animals like dogs and humans - but of very little interest to herd animals and herbivores who are more concerned with having a wide field of view. Those kinds of animals probably wouldn't even realise that the TV was in 3D. SteveBaker (talk) 00:08, 19 March 2010 (UTC)

Current over large distances

You know how electricity will not travel down a path that leads to a dead end, and it generally takes the path of least resistance? Does electricity decide which path it will take instantly, or is the information conveyed to it at the speed of light? For example, say some current suddenly had the option to travel in one or more of three paths: one that lead to effectively a dead end, one that had a large amount of resistance at the end of it, and one that did not. These paths are several light-years long and are essentially straight. Would the light mostly travel down the path without the large amount of resistance? Sorry if I'm not explaining this well, and thanks in advance for your response. —Pie4all88 ^{T C} 19:05, 18 March 2010 (UTC)

The information is limited by the speed of light. If we think about how the electrons "decide" where to flow, what's happening is that each electron is pushed locally by the electric field of the particles around it. A path that leads to a positive charge means that electrons at the end get pulled toward the positive charge, and leave a positive "gap" behind, which pulls other electrons along and so on. Electric fields can't travel faster than the speed of light. If we cut the wire at one end, electrons would still flow for a bit at the other end until the changes caused in the electric fields could propagate down to them. Rckrone (talk) 19:21, 18 March 2010 (UTC)

If the length of the wire (or other pathway for current) is large compared to the wavelength of the electromagnetic wave (or equivalently, if the switching happens "fast" relative to size of the circuit), then a lumped element model is inappropriate, and a transmission line model is more accurate. This can be approximated with the Telegrapher's equations. Using these models, or a full-blown solution of Maxwell's equations, results in a more physically accurate, conceptually consistent result and shows that changes in the circuit (such as opening or closing a switch, creating the "dead end" the OP is referring to) will propagate at the speed of electromagnetic waves in the medium. Or, to put it very simply, if there were a wire that were extraordinarily long, even if it terminated in an open circuit, electricity would flow down the wire, because that wire would have a capacitance. This current would represent the transient response of the system; clearly, once the current reaches the end of the wire, it has nowhere to go, so "something has to change". This would result in a reflection (of voltage, current, or both, depending on the type of "dead end"). After some characteristic period of time, steady state would be reached (depending on the type of input current, this could be a standing wave at the input frequency, or a DC valued voltage and zero current, etc.) Nimur (talk) 20:12, 18 March 2010 (UTC)

I do not "know how electricity will not travel down a path that leads to a dead end" because that is not the case. Assuming a conducting line that is long, though not necessarily as absurdly long as several light-years, an electric voltage step (or pulse) entering the line will travel along the line. See the article Transmission line. What you may call a "dead end" is a discontinuity on the line, such as an open circuit or a short circuit. When the voltage step reaches the discontinuity it is reflected. The reflection is seen at the input after the round trip of twice the line length. That is the principle of the Time-domain reflectometer instrument for testing electric paths.

Where there are several paths it is not true that electricity has to decide to take only the path of least resistance. Electric current is shared across parallel paths in inverse proportion to the resistances of the paths. See Ohm's law. An Electric arc (such as Lightning) might seem to be an exception but it is not. The arc appears when a path of Plasma (physics) forms that has much less resistance than any other path. Cuddlyable3 (talk) 20:21, 18 March 2010 (UTC)

Also, to the questuon of electrons "deciding" which path to travel down; it happens the same way that water molecules will decide which pipe to go down when approaching a fork in a pipe. The information comes from the electrons already in those wires. Electricity isn't one electron flying down a tunnel. Its a chain of electrons sliding down the wire, so when a lone electron is faced with a junction, it will "choose" the path that at that moment has an opening for it. --Jayron32 21:01, 18 March 2010 (UTC)

Ok, I understand this much better now. Thanks, everyone, for your very informative answers. —Pie4all88 ^{T C} 22:48, 18 March 2010 (UTC)

One last point - the electrons themself do not travel down the wire. Their "push" does. Each electron pushes its neighbor and so on. The push happens at the speed of light (or pretty close). There is also a pull, which works the same way, but in reverse. In a normal circuit both happen at the same time. If the circuit is long enough there is a slight delay, where you have just one but not both. I think measuring effects of this helped figure out which direction electricity travels. Ariel. (talk) 21:08, 19 March 2010 (UTC)

That's not quite true - the electrons do travel down the wire - but very VERY slowly. The best analogy is to imagine a 20 foot-long hose pipe (the wire) that's completely full of half-inch ball-bearings (the electrons). If you push another ball-bearing into the end of the pipe, it'll force one out of the other end. You could imagine that this process happens pretty much instantly - much faster than a ball could roll down an empty pipe an pop out of the other end. But you paint one of your ball bearings red and push it into the pipe, you have to spend an hour pushing more ball bearings into the pipe before the red one pops out of the other end. So while the 'flow' of ball bearings travels really fast - an individual ball bearing takes a heck of a long time to make the journey. Hence, when you turn on a light switch, the delay before the electricity reaches the other end of the wire is as fast as the speed of light - but it could take a very long time for a particular electron to travel from one end to the other.

This analogy also answers the OP's original question. If you have a 'Y' joint in your pipe and cap off one arm of the Y - then how do the ball bearings "know" to travel down the other arm? It's obviously because the capped off arm is already full of balls. Same kinda deal with electrons and a wire that doesn't go anywhere.

SteveBaker (talk) 02:33, 21 March 2010 (UTC)

How long does a snake need to be in order to constrict an average-sized man?

Just thinking back to someone I used to know who had a python (poss. boa constrictor) that tried to constrict him when he was playing with it, whilst drunk. Apparently, the snake wasn't long enough to get into a position where it had enough coils around him at once to make it impossible to unwind, or something. --Kurt Shaped Box (talk) 20:56, 18 March 2010 (UTC)

Indeed, a small sized snake needs about 38 years on the average to constrict an average-sized man. A very big anaconda needs about one minute on the average. So if we tell you that the average snake would need about 19 years and 30 seconds on the average, would that be a helpful answer? DVdm (talk) 21:06, 18 March 2010 (UTC)

I think he's asking for snake length, not constriction duration. Vimescarrot (talk) 21:08, 18 March 2010 (UTC)

Oops. Sorry. Struck. DVdm (talk) 21:11, 18 March 2010 (UTC)

No worries. None at all. --Kurt Shaped Box (talk) 21:15, 18 March 2010 (UTC)

you write "A very big anaconda needs about one minute on the average [to constrict a man]". is it possible to escape during that one minute in typical amaconda habitat settings and how should you do it?. Thanks. 80.187.107.89 (talk) 21:50, 19 March 2010 (UTC)

I think it only needs to be long enough to wrap around the person's neck. But I don't think snakes are smart enough to target that. Well, except for this snake. --Kvasir (talk) 21:26, 18 March 2010 (UTC)

It's not just length though. A Garter snake is long enough to wrap around someone's neck, but just doesn't have the strength needed to kill someone. --The High Fin Sperm Whale 21:52, 18 March 2010 (UTC)

I have watched one video this week of a man versus a python, and the python did indeed go for the neck. Vranak (talk) 01:34, 19 March 2010 (UTC)

most constrictors work by immobilizing the chest cavity, not by actual strangulation. assuming 3 to 4 wraps around the chest, and given a chest size of 42 inches, that would imply a minimum of 10 ft (for 3 wraps) or 14 feet (for 4 wraps) of main body length - add a few feet for tail and head/neck and it seems you'd need a minimum 14 to 18 foot snake to kill an adult male human. Of course, I'm not certain that a snake of that size could capture and pin an adult male - humans aren't strong as animals go, but hands allow for some fairly unique modes of defense (even without weapons) - but it would have to be at least that large to have the possibility of immobilizing the chest and lungs. --Ludwigs2 21:55, 18 March 2010 (UTC)

Going for the chest rather than the neck is actually a rather clever trick that the snake is using quite deliberately. What it does is to wrap initially rather loosly around your chest - but when you exhale, the pressure on its coils reduces and it is able to tighten up just a little bit more. Thus it can use friction between its coils to avoid needing so much muscle power to crush your chest. Some species have interlocking scales to help that process. So with each breath you take, you have to breathe more and more shallowly...not a nice way to die. However, since the snakes muscles aren't that strong - it's possible to unwind it one coil at a time if your hands are free. Most of the prey that constrictors go for don't have hands! SteveBaker (talk) 03:38, 19 March 2010 (UTC)

That said, snakes are apparently a *lot* faster and stronger than they would first appear. I once remember reading that it took five guys to unwind a python than had attached itself to its owner and started to squeeze him until he creaked. I believe that the near-fatal mistake he had made was to handle live rabbits (i.e. the snake's dinner) directly before interacting with the snake - which then smelled rabbit and jumped to conclusions, due to its poor eyesight. --Kurt Shaped Box (talk) 23:03, 18 March 2010 (UTC)

Isn't that a lesson that men should not play with their pythons whilst alone?--79.76.137.66 (talk) 23:49, 18 March 2010 (UTC)

I've read of smallish constrictors, like 10 feet, killing their owners by wrapping around the throat. I would recommend they not kill their owner unless they are prepared to eat him.Edison (talk) 00:01, 19 March 2010 (UTC)

The 'never attempt to swallow anything larger than your own head' rule doesn't apply to snakes, or so I hear. On account of them being able to pop their jawbones out of their sockets... --Kurt Shaped Box (talk) 00:13, 19 March 2010 (UTC)

I misinterpreted your first sentence...I was trying to imagine how it was safe for me to attempt to swallow a snake that was bigger than my head...Mmm'k - never mind! SteveBaker (talk) 03:29, 19 March 2010 (UTC)

OR: I had a corn snake as a kid that I managed to kill by giving it a too-big mouse to eat. It got the thing into its belly, but was dead the next day. :( --Sean 13:26, 19 March 2010 (UTC)

My recollection of snake-handling rules is that you need one handler for snakes up to ten feet in length, and an additional handler for each further five feet of snake. Based on that, I'd guess that ten feet is about the minimum to successfully constrict a person. --Carnildo (talk) 00:23, 19 March 2010 (UTC)

Isn't it easier just to kill the damn snake? --Kvasir (talk) 04:49, 19 March 2010 (UTC)

At the risk of sounding obvious...not if you want to keep it alive. You mean while it's trying to constrict you? Well, either your hands are free, in which case you can just unwind it (sounds easy enough to me, since apparently they aren't all that strong), or they aren't free, in which case - how do you propose killing it? Vimescarrot (talk) 06:38, 19 March 2010 (UTC)

I was referring to having one handler per 5 feet of snake. One spotter would be enough to kill one when it's out of control? Decapitate it or one stab in the heart. Or have I watched too many movies. :) I'm this close to quoting Snakes on a Plane. --Kvasir (talk) 06:54, 19 March 2010 (UTC)

Snakes that long -- and the business disruption of killing one -- are likely to be much more expensive than simply hiring another handler. --Sean 13:26, 19 March 2010 (UTC)

You guys are overlooking another obvious reason not to try and kill the snake - if it's a pet, the owner likely loves the thing to bits and doesn't want it to die... --Kurt Shaped Box (talk) 01:07, 20 March 2010 (UTC)

Decay of large dead animal in water

If a large animal(let's say elephant-sized) drowns in a lake where there are bacteria but no scavenging animals, how long does it take for the carcass to become skeletonized? Assuming that scavengers were introduced to the region, how early would they have to arrive before the corpse no longer had edible flesh on it? 137.151.174.128 (talk) 22:02, 18 March 2010 (UTC)

For starters, our decomposition article says: A basic guide for the effect of environment on decomposition is given as Casper's Law (or Ratio): if all other factors are equal, then, when there is free access of air a body decomposes twice as fast than if immersed in water and eight times faster than if buried in earth. Comet Tuttle (talk) 23:10, 18 March 2010 (UTC)

Anatomists and anthropologists may use "maceration" to clean flesh from bones, by simply placing them in water and pouring off the liquid with the decayed flesh periodically.[4]. They do not give a timetable for large versus small animals. Some 19th century anatomists would place tadpoles in water with the dead animal to remove the flesh. Other 19th century anatomists would place an animal or human in a metal cage and secure it to the bottom of a pond or lake for a year, by which time natural processes and small creatures would have cleaned the bones. The mesh of the cage should be small enough to contain the smaller bones. Could not find schedules or timetables or recipes more precise. Edison (talk) 23:59, 18 March 2010 (UTC)

We have a forensic entomology article, but no forensic ichthyology article. Comet Tuttle (talk) 00:22, 19 March 2010 (UTC)

February 2010 Scientific American magazine has an article called "The Prolific Afterlife of Whales." It is a very good article. Excerpt: "On the deep seafloor, the carcasses of the largest mammals give life to unique ecosystems…" Bus stop (talk) 03:34, 19 March 2010 (UTC)

Our whale fall article covers this topic. --Sean 13:33, 19 March 2010 (UTC)

Aquarium bugs above the waterline

There are hundreds of these specs of brownish, needle head size bugs with antennae 1/2 to 1/3 of their body length crawling about above the waterline on the wet glass of my aquarium. Adult and young, i assume they are all the same. Some might fall onto the water but soon return even after I try to wash them down by sloshing water on them. They seem to be very good at staying above water (probably due to surface tension) and hopping back onto the glass. What are they? I hope my fish will find them as snack if they are quick enough. I guess once I figure out what they are I'll be able to see if it's any indication of my water quality. thx. --Kvasir (talk) 22:03, 18 March 2010 (UTC)

March 19

360 degree lightmeter

Is there any device or instrument that can record the intensity of light in lux falling on itself, not just from one direction like a conventional lightmeter, but all 360 degrees around both horizontally and vertically? And which, rather than merely averaging the light, can tell you the intensity with a resolution of say a few degrees? Thanks 78.151.108.166 (talk) 01:30, 19 March 2010 (UTC)

A simple trick is sometimes used in the movie/ computer graphics business to capture the lighting on a scene in order that graphics may be produced with matching lighting and thereby seamlessly blended into the live action. The trick is to take a mirrored/chromed sphere, place it into the scene where you wish to sample the light and then photograph it with the same camera and camera settings that you're filming the live action with.

You might at first guess that this would only capture half of the incoming light - but if you think about it carefully, you'll realise that all 360 degrees both vertical and horizontal are represented by some point on the sphere. (Well - almost - this assumes that the sphere is small compared to the distances to the light sources). The problem is that the angular resolution of the measurement of light shining from the 'back' hemisphere is very poor compared to the front hemisphere of the scene. Fortunately, when you're doing this for a movie, you don't care as much about light coming from behind the object as in front...so it kinda works out. If you really care - you can photograph the sphere from several directions at once.

So that gave us a way to grab all 360 degrees in a single image - converting that into lux is a problem that I (inadequately) answered in your next question (below). SteveBaker (talk) 03:27, 19 March 2010 (UTC)

We use a Li-Cor LI-193 spherical sensor (link: http://www.licor.com/env/Products/Sensors/193UW/li193_description.jsp) for underwater photosynthetically active radiation measurements. The 360-degree coverage is pretty good and I suspect by changing the calibration constant you might be able to get somewhat accurate readings in air. It is still somewhat directional, there's no real way around that, but it seems to do a pretty good job and it may be the closest thing you'll find.

You ask about getting the intensity to within a few degrees - since you ask for degrees, do you mean getting the direction of the light to within a few degrees of the circle or getting the light intensity to within a few percent or lux? Light intensity measurements in the field (i.e. anywhere but carefully controlled conditions) are going to fluctuate quite a bit. You want an average over time to smooth out those fluctuations. Wevets (talk) 14:41, 19 March 2010 (UTC)

Thanks, but I do not think I could use the LI193 as I assume it averages the light intensity from all around. I said degrees, I meant degrees. 78.147.2.253 (talk) 16:50, 19 March 2010 (UTC)

Perhaps the OP wants a spherical viewing intensity-calibrated camera. Two digital cameras mounted back-to-back, both with Fisheye lens and fixed exposure settings could suffice. Cuddlyable3 (talk) 00:16, 20 March 2010 (UTC)

The LI-193 doesn't average light intensity from different sources, but it does measure the *sum* of light sources from different directions (i.e. a source of 1000 μmol/m² sec to the right of the sensor and a source of 500 μmol/m² sec to the left of the sensor will produce a reading of 1500 μmol/m² sec rather than an average of 750 μmol/m² sec.) Don't know if that helps you at all though. Wevets (talk) 06:23, 20 March 2010 (UTC)

Using a digital camera as a light meter

If I took a digital photo and wanted to estimate the intensity of light in lux entering the camera lens from various parts of the scene photographed, then what calculations would I need to do? Thanks 78.151.108.166 (talk) 01:44, 19 March 2010 (UTC)

You really can't do it in general. The camera adjusts the exposure of the sensor depending on the brightness of the scene so a wide range of different light levels could generate the exact same number in the resulting image. However, if you can get a hold of a "Raw image format" image from the camera, it might contain a header that describes the exposure information...or not...depending on the type of camera...but even then, you'd probably need to know a lot about the exact nature of the sensor that the camera uses. This is either a tough problem - or an impossible one - depending on the exact make and model of camera that you're using. SteveBaker (talk) 03:15, 19 March 2010 (UTC)

If you know the ISO speed, the exposure value, and the average brightness of the image, I don't see why you'd need to know so much about the sensor. Every camera set at ISO 100, f/2.8, and 1/40s shutter will give approximately the same exposure. Using that information, you may be able to make a conversion to lux. Thegreenj 04:06, 19 March 2010 (UTC)

That's still overcomplicating things. Just set the exposure settings manually, calibrate using a surface of known brightness, take a picture of the scene you want to measure, and use one of the many photometry software out there to measure the brightness of the target surface. --09:33, 19 March 2010 (UTC) —Preceding unsigned comment added by 99.237.234.104 (talk)

You'd need to be careful, though. The scale a camera uses to convert what the light levels that it detects to an image close to what humans perceive is nonlinear and does vary from camera to camera; the scale is adjustable, for example, in some Raw conversion software. Edison's formula would likely produce more consistent results unless you had more information. Thegreenj 01:36, 20 March 2010 (UTC)

Most modern digital cameras will include all the data in the EXIF section of JPG file - see the bottom of my File:Carnforth_Canal.jpg for an example, the data has even been kept, although I've reduced the size in Photoshop. Software (like Photoshop) can display the EXIF data if asked. (Note:If I had cut out a section and made a new picture from it, the EXIF data would have been lost)  Ronhjones  ^(Talk) 19:45, 19 March 2010 (UTC)

Camera light sensors detect reflected light rather than incident light, which is more closely related to Lux, which is one lumen incident per square meter. It may not be straightforward to define "incident light" from "various parts of the scene." If you want a spot meter, then use one rather than wanting "incident light" from a narrow spot. Some meters like an old Gossen Luna-Pro have a little white plastic dome which slides over the light sensor to directly measure incident light in "EV" or exposure value. Then various websites give formulas for converting that EV (for a specified ISO, typically 100) to Lux. One is Lux=2.5*2^EV at [5] for ASA 100. (ASA was the predecessor term for ISO to indicate film speed). See also Sekonic support which has a conversion table from (incident?) light measured for ISO 100 with their meter, in EV, to Lux. See also photonet. Edison (talk) 19:27, 19 March 2010 (UTC)

The idea behind the question is to measure incident light coming from many different points rather than just averaging adding them all together with an invacone. Camera light sensors will detect whatever light hits them, they have zero brain-cells and do not understand the difference between incident or reflected, which is about the conventions of positioning camera and light meter and not the physics of photons. 89.242.46.75 (talk) 23:56, 19 March 2010 (UTC)

1/2 degree spot meters have been in common use for decades, and detect the light reflected from a tiny part of the scene. They are by definition very directional, and with suitable formula juggling should be able to provide the desired information. Lumens per steradian and all than. I just wish I had taken coursework and done problem sets and labs in photometry so I were as confident in it as with some aspects of electricity. Edison (talk) 01:22, 21 March 2010 (UTC)

The "suitable formula" is what I'm interested in finding. 84.13.41.17 (talk) 13:46, 21 March 2010 (UTC)

The spot meter reading from a bright or dark portion of a scene may vary by 4 fstops or so. Each reading, for a given ISO setting such as 100, will generate an exposure value, which could be plugged into the formula above for an incident meter. But I suspect that there may be a definitional problem with this approach. Edison (talk) 19:05, 22 March 2010 (UTC)

psychology

Is there a term or terminology in psychology to describe a situation in which an individual has both an undesirable behavior (like dependency on alcohol) and a desirable goal (like marrying into wealth) where such a goal can not be fulfilled so long as the individual posses the undesirable behavior under circumstance where the individual is permanently incapable of acknowledging they posses the undesirable behavior or comprehending that it is the obstacle keeping their goal from being reached? 71.100.11.118 (talk) 01:53, 19 March 2010 (UTC)

Did you choose the 'desirable goal' randomly, or did you need something specific to that situation, because I think we have a term for that. Beach drifter (talk) 01:57, 19 March 2010 (UTC)

"Desirable goal" could be writing a best selling novel or managing of owning a successful and growing business or seeing your children become wealthy or at least succeed. "desirable goal" is intended more as a general term related to financial and general success in life than to any specific or narrower goal. 71.100.11.118 (talk) 02:37, 19 March 2010 (UTC)

Bad habits are getting in the way of goals? A blind spot to one's own bad habits? Bus stop (talk) 03:03, 19 March 2010 (UTC)

It's not a psychological term per se, but I think character flaw pretty much covers it. Vranak (talk) 03:22, 19 March 2010 (UTC)

A person's character refers to the thing they would normally do. For instance, a person might be described as the type of person who never tells a lie. A character flaw then is indicated by something that a person normally does which might break some social rule such as cursing in public or entering someone's house without knocking. The flaw in their character is that as a rule they are not able to see the need to comply with social expectation or norm. 71.100.11.118 (talk) 05:12, 19 March 2010 (UTC)

I think the term you are looking for is denial. Looie496 (talk) 02:56, 20 March 2010 (UTC)

I'm quite curious how these tricks are done:

http://www.youtube.com/watch?v=hwVy_2eOfsE I've studied it a bit and just can't work it out, which is of course what makes a magic trick great but I am have always been one who wants to look behind the curtain.--Fuhghettaboutit (talk) 03:48, 19 March 2010 (UTC)

The first few tricks seem to be based on trick coins (there are more coins than there appear, but they can stick together to make one or more disappear when needed). Watch in slow motion (Youtube doesn't make freeze-framing very user-friendly, but give it a shot). You'll see that, for example at 3:15, the magician has a fourth coin in his hand before he even slams the table from the bottom (screengrab!); and after the slam, the coin-flip on top causes two coins to stick together. (They're probably magnetized, so that if shaken up or thrown, the coins will automatically go for each other). In the 1080p high quality video, if you freeze-frame on the stuck-together coin, you can actually see both coin-rims, not 100% perfectly aligned - the magician should know better than to try this trick in HD video!

When you notice this, you can start looking back at the other tricks (at 0:29, for example, the magician flips the coin in the accomplice's hand, but never actually grabs it); three coins "become" two with the stick-together-trick; and we don't really know what he's holding underneath the velvet (hint: it's the fourth coin). You can see the same sort of trick applied repeatedly through the rest of the video.

It seems that the final trick is accomplished with a trick table; there is a hole in the plexiglass off the the right (probably normally hidden under the teapot, which is on an upper, non-moving surface), and the table surface can be rotated in so that the hole slides into the work area. I suspect at least one of the audience members is a cohort of the performer. If you look closely, 6:18, you can see the performer's hand wobbling as the table surface is rapidly moved (the camera angle is intentionally obscuring this with a "face-on" shot instead of an overhead shot). As with all sleight of hand, misdirection is the magician's friend - he plays all kinds of games wiping the table and wiggling his hand, while the actual work is being done by an accomplice (who is rotating the trick surface for him). Nimur (talk) 04:52, 19 March 2010 (UTC)

Nice analysis. One clue is that there seem to be extraneous people around the table with no apparent purpose. Their real purpose must be to help rotate it into position. Also, anything on video could always use stop-action, where they stop filming, bring out a table with a hole in it, then resume. But, in this case, this doesn't seem to be necessary. StuRat (talk) 05:27, 19 March 2010 (UTC)

Learning Fourier Transform in 2 days

I learned it in advanced biochem protein course. I have a test on this next Monday. I'm really scared I don't understand anything. The instructor assigned readings but I don't understand them at all. I feel those readings assume certain background that I lack (Eg. the way that NMR is talked about in those papers is not like the NMR that I know.) I really need to understand this. Any great resources that don't assume advanced background in math, spectroscopy, etc. would be appreciated. —Preceding unsigned comment added by 70.68.120.162 (talk) 04:35, 19 March 2010 (UTC)

Sorry to start with the obvious resource, but have you read our Fourier transform article? Comet Tuttle (talk) 05:25, 19 March 2010 (UTC)

Before we can help you, it's important for us to know: do you need to know how to take a Fourier transform, or do you just need to understand the conceptual ideas behind it? Unfortunately, if you need to compute the Fourier transform of a signal, you need to be fairly proficient at integral calculus, and 2 days is a bit short to ramp up on that. (Alternatively, we can recommend some software which can do this for you - but that's probably not much help on a written test). If you just need conceptual help, you might want to start with the frequency spectrum article, which is a little more user-friendly than the Fourier transform article. Just keep in mind that a Fourier transform is simply the mathematical technique used to convert back and forth from time-domain to frequency domain. Nimur (talk) 05:28, 19 March 2010 (UTC)

This explanatory video takes less than 8 minutes. Cuddlyable3 (talk) 23:45, 19 March 2010 (UTC)

And here's EE261 "The Fourier Transform and its Applications", a ten-week/30-lecture course from Stanford, officially available online for free as part of an OpenCourseWare system. (Supposedly they charge something like $15,000 if you want to sit in on this class through SCPD - so grab it while it's free and free! Nimur (talk) 10:53, 20 March 2010 (UTC)

GREAT link to high-quality content - thanks for that, Nimur! -- Scray (talk) 16:03, 20 March 2010 (UTC)

Fourier transform 2

Like the poster above, I'd also like to learn Fourier transforms within a few days. To answer Nimur's suggestions to the poster above:

I'd like to know both the concepts behind it and how to actually perform it. I'm pretty good at integral calculus, so that's not a problem. Also, the Fourier transform article is very hard to understand by non-mathematicians and has no explanation of the principles behind Fourier transforms and why such transforms are so useful. --99.237.234.104 (talk) 08:46, 19 March 2010 (UTC)

The set of all functions is a vector space (mostly because you can multiply functions with numbers and because you can add two functions). See http://en.wikipedia.org/wiki/Vector_space . Vectors (and functions) can be expressed as coordinates relative to a certain basis ( http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29 ) . This basis is not unique : you can express vectors (and functions) relative to many different basisses (that's not the correct plural of basis, is it? ). In particular, you can choose a "position basis" : $f(x)=\sum _{i}a_{i}R_{i}(x)$ where $a_{i}$ is the value at $x=i$ and $R_{i}(x)$ is a rectangular function centered at $x=i$ (this works for continuous functions too, only using integrals and Dirac functions rather than sums and rectangular functions). $a_{i}$ are the "coordinates", $R_{i}(x)$ is the "basis". Unsurprisingly, you can use another basis, the so-called fourier basis, consisting of sines and cosines of different frequencies (or complex exponentials, if you like that better). The formulas given in http://en.wikipedia.org/wiki/Fourier_transform#Definition are projection operators ( http://en.wikipedia.org/wiki/Projection_%28linear_algebra%29 ) from the position basis to the fourier basis (and vice versa) . They give you coordinates in frequency space, so to speak. Knowing the function in the position basis allowed you to see instantly what the value of the function was at a given point. Knowing the function in the fourier basis allows you to see how much of a certain frequency the function contains. It allows you to find superimposed periodicities in the function. You could for example fourier transform a calendar and you would find the periodicity for months, for years and even for leap years.

Now to actually calculate the fourier transform, first know what you are transforming: a continuous function R->R ? Use http://en.wikipedia.org/wiki/Fourier_transform#Definition . a function on a continuous interval [a,b]->R ? See http://en.wikipedia.org/wiki/Fourier_series#Fourier.27s_formula_for_2.CF.80-periodic_functions_using_sines_and_cosines (and read the rest of that article). A function on a discrete interval? See http://en.wikipedia.org/wiki/Discrete_fourier_transform#Definition 83.134.168.74 (talk) 09:37, 19 March 2010 (UTC)

Basis plural - Bases but pronounced "basees". Graeme Bartlett (talk) 22:01, 19 March 2010 (UTC)

CFL Afterglow

I have several compact fluorescent lamps at home. I noticed that they continue to glow for a few minutes after I turn them off. Big helix types glow the best. Since I cut the power completely, where does the energy come from? They get warm after working, is it the heat? I thought about heating the lamps with a hair dryer to see if they glow or not, then gave up fearing they'll break and spill mercury. 88.242.232.209 (talk) 10:17, 19 March 2010 (UTC)

A Fluorescent lamp "is a gas-discharge lamp that uses electricity to excite mercury vapor. The excited mercury atoms produce short-wave ultraviolet light that then causes a phosphor to fluoresce, producing visible light." Both of these continue to some extent but I believe the phosphors in particularly usually continue to flouresce for a while (by both I meant the the mercury vapour remains at an excited state for a short time after you switch off the electricity). As you may guess heating the lamp won't do much useful Nil Einne (talk) 10:37, 19 March 2010 (UTC)

You'll notice that the phosphors in your (CRT-based) TV and computer monitor do the same thing. --Sean 13:39, 19 March 2010 (UTC)

It helps, though it's not necessary, to see this if all the other lights in the room are off when you turn the TV off. Dismas|^(talk) 11:33, 20 March 2010 (UTC)

Reusable chemical, cyronic or organic movie camera?

Since my study of the issue seems to show that even next to the best UHDTV digital movie cameras, chemical film still has the edge in terms of dynamic range, would it be possible to build a system that continuously recycled and reused all the "consumables" in film in real time, transferring it to digital format in-between? Alternatively, could a living eye (either tissue engineered or from a donor animal) be supported by a compact artificial life support system with the optic nerve wired in to a digital processor (with nerve growth factors on the chip) and if so, what would the dynamic range and other specifications be in comparison to high end digital movie cameras such as Red or conventional 35 mm movie cameras? Also, would a digital movie camera with its sensor chilled to cyronic temperatures have superior characteristics?80.1.88.11 (talk) 10:21, 19 March 2010 (UTC)[Trevor Loughlin]

It would just be a whole lot easier to build a digital camera with higher dynamic range. That's a relatively simple thing to do - even if there were limitations in the sensors (which I don't think is the case) - you could have (say) three or four sensors - each with different aperture and exposure and capture more dynamic range that way. However, astronomers long ago switched from film to digital imaging arrays precisely because the digital stuff is better - so I'm fairly sure this isn't a limitation in principle. We don't do it normally because it's expensive and TV displays have an even more limited dynamic range than digital movie cameras - so there is not much point. The idea of transferring film to digital "in real time" simply pushes the problem back one step to making you ask how you're going to go from film to digital without losing dynamic range. Biological eyes don't work like cameras - they don't deliver a simple grid of pixel values - instead there is a processing layer at the back of the eye that converts the image into data like what the orientation of edges are in the image and what direction they are moving in. Turning that back into a video signal would be an incredibly difficult process. SteveBaker (talk) 13:47, 19 March 2010 (UTC)

Some potential problems with using multiple digital cameras like that:

1) Having a longer exposure would tend to blur moving images.

2) You could potentially damage the most sensitive camera if pointed at a bright object.

For these reasons, I believe those using digital cameras to film TV shows instead concentrate on better lighting, so there aren't any dark corners that don't record on digital media. However, one variation on the multiple digital camera approach might be to use a normal camera and an infrared camera. The infrared would work to fill in the dark spots, since there isn't nearly as much variation in IR as visible light (unless you happen to be filming a fire). Something like green screen replacement could then be used to put in the IR image wherever the visible image is black. Note that this method wouldn't provide color for the dark areas, but I'd think a black and white area would be better than nothing, and it also simulates how our eyes see black and white in the dark. StuRat (talk) 14:23, 19 March 2010 (UTC)

In most cases the use for higher dynamic range is you have more room for "adjustment" before you output it to it's final form. This is done either in the dark-room or with high bit-depth digital files in photoshop. So for most uses converting to a lower dynamic range format on-the-fly would defeat most of the purpose.

If your on-the-fly film scanner is also operating in high dynamic range, then you've now go an HDR imager that takes one image per frame, why not just put a lens on it and use it directly?

Not that a perpetual film camera wouldn't be a cool hack, of course. APL (talk) 15:25, 19 March 2010 (UTC)

Hi 80.1.88.11

Improvements in sensor technology is I think the way to go. UHDTV will need to be viewed via a projector to appreciate the dynamic range that it can handle. As current projectors have a DR of 2000:1 UHDTV has 10 bits per channel and four sensors so that gives (I think) about four billion colours but it is limited in DR by the sensor signal over thermal noise and so is about 1000:1 or roughly half the projector performance, (not that this wont be amazing in itself!). Is this similar to how you see UHDTV? If so, back to your question. Cooling might help (as in space applications) but there is the problem of condensation to over come. However, solid-state heat cooling chips should make this as simple as in current military thermal IR scopes. And there is always the possibility that better sensor materials may get developed allowing for larger signal to noise ratios. Increasing the colour depth to a higher figure might also be possible but that might need a more efficient compression algorithm to get it all into the same transmission bandwidth. Leave the bionic eyes to Hannibal Chew (James Hong) of the Tyrell Corporation.--Aspro (talk) 17:20, 19 March 2010 (UTC)

Remembered ! The things that are (or were, maybe still is, perhaps, etc) secret in military thermal IR scopes are “Peltier elements” This section has a short bit about the CCD application.Thermoelectric_cooling#Uses--Aspro (talk) 21:52, 19 March 2010 (UTC)

PROJECT WORK

ARTICLE ON ELECTRONIC VOTING SYSTEM —Preceding unsigned comment added by ETE0362 (talk • contribs) 14:37, 19 March 2010 (UTC)

Electronic voting machine? DMacks (talk) 14:43, 19 March 2010 (UTC)

Electronic voting? Regards, --—Cyclonenim | Chat  14:44, 19 March 2010 (UTC)

ignitor and voltage

I was trying to make an ignitor, but the problem arose was that i used 18 volts(two 9 volts battery in series) still there was not even a speck of spark. I cant figure out why? I even tried thick aluminum wires and also tried simple copper wires. will spark produce at a particular voltage?

one more question - spark of what voltage can be easily felt by a human. i dont think this question very correct, but i have seen shock giving chewing gum packets and also shock giving tic toc pens, pens use 2 1.5 button cells ..........thanx--Myownid420 (talk) 17:15, 19 March 2010 (UTC)

You might want to try a lantern battery for a visible spark. As for what a person feels, that depends. A weak battery placed with it's contacts on the tongue can still cause a tingle. StuRat (talk) 17:42, 19 March 2010 (UTC)

Be aware that electricity can kill people. Someone with a pacemaker might be especially vulnerable, so I mention in passing (making every assumption of the questioner's good faith) it is not a good idea to build a gadget which gives unsuspecting people a shock. If you don't know much about electricity, it is very easy to make a fatal mistake, especially if experimenting with high voltages or things which can supply high current. A lantern battery produces 6 volts, the same as 4 AA batteries in series, but can produce much more current for a longer time. For information about a commonly used low current, high voltage spark source/ignitor see Piezoelectricity#High voltage and power sources, about crystals which produce a brief high voltage when they are mechanically struck. A circuit with inductance in it produces a spark at the opening of the circuit, as the inductor resists the decrease in current by generating a counter-EMF, so you might be interested in Inductor. A Transformer can produce a high voltage as well. A magnet moving rapidly near a coil can produce a high voltage, and is the basis of some toy shock trick devices, as well as the magneto on gasoline engines which produces the spark in the spark plug. It is hard to accurately define the minimum "spark" that can be perceived, since a tiny spark could produce enough heat to be felt. You could check Google Book Search for experiments where the minimum perceptible voltage for feeling a shock was measured, but it will depend on the surface area and the resistance of the skin at the site. Under the same conditions of walking across a wool rug in winter (which builds up a static charge on the person) and touching a doorknob (which discharges the electricity), I have noticed that touching the knob with the extended finger produces a painful shock, but touching it with a key held by the fingers produces no discernible shock, because of the large surface area and low resistance through which the electricity passes from the key to the hand. The threshold for perceiving electricity is usually measured in term of current, and is under one milliampere (varying for frequency). I will not state a "safe level" of current, since individuals vary. I expect that if a shock gadget operates on two 1.5 volt batteries, there is circuitry to step up the voltage many times. Edison (talk) 18:52, 19 March 2010 (UTC)

To give a sense of perspective, a spark plug uses around 20 000 V. The dielectric strength of air is 310 V/mm, which means you need 310 volts per mm of separation to produce a spark. --99.237.234.104 (talk) 19:22, 19 March 2010 (UTC)

One method is to put a large value inductor, may be 1 Henry, with a low resistance inline with your sparker and battery. Yo need a battery that can supply the highest current, and then break (open) the switch, you will get a high voltage and a spark. Graeme Bartlett (talk) 21:53, 19 March 2010 (UTC)

cant a spark can be produced using just batteries and wire? and i didane clearly understood inductors -- it says 'Typically an inductor is a conducting wire shaped as a coil, the loops helping to create a strong magnetic field inside the coil' its the loops we could see in the picture or it is inside the thing it is wound up? Has the core to be there to let it work..thnx--Myownid420 (talk) 07:18, 20 March 2010 (UTC)

Batteries can produce a spark, but typically need to have a voltage of a hundred volts or so - so you would need 10 or more of your nine volt batteries to produce much of a spark. The length of a spark is a function of the voltage - the higher the voltage, the longer the spark. Please do be very careful - once you get a voltage high enough to create sparks, it's possibly high enough to electrocute you and possbily kill you.--Phil Holmes (talk) 13:15, 20 March 2010 (UTC)

You can get a spark from a 12 volt battery or a 6 volt battery, just not a sustained spark across much of a distance, which we call an "arc." A coil of wire generally needs a core of iron or ferromagnetic material to have enough inductance to make much of a spark. A Model T ignition coil could make an impressive spark when DC current was started and stopped in the primary winging ( I expect it was in fact a transformer, like the Induction coils used in radio for "spark transmissions" before vacuum tubes were introduced. Here is the full text of a 1908 book by radio pioneer A. Frederick Collins telling all about induction coils and how to make them. (A great deal of care and skill is needed). Tesla coils also make impressive and dangerous sparks, and take a fair amount of money and skill to make. Van de Graaf Generators are available from science supply companies or EBAY and make impressive sparks. Remember that electric shocks can be painful and dangerous. Edison (talk) 01:15, 21 March 2010 (UTC)

NMR

I previously learned NMR in organic chemistry 1 & 2 and I'm taking an advanced biochem course that covers NMR. Strangely, the NMR covered in this biochem course seems like a different NMR from the one I know. Here's what I understand about 1Hydrogen NMR:

1. Expose sample to an external magnetic field Bo, which sets up the energy difference between higher and lower energy states of two different orientations 1/2 and -1/2 for 1Hygrogen isotopes in the sample. 2. In the presence of this Bo, treat the sample with photons of different wavelengths in radiowave region. The specific wavelength which corresponds to the energy difference between the two orientations is the resonance wavelength. 3. These photons with the resonance wavelength will be absorbed by and excite whichever oritentation with the lower energy (1/2 for 1Hydrogen), and this will put those lower energy orientation isotopes to higher energy level (-1/2). 4. Now I'm unsure about exactly what happens after this point. Do the excited states spontaneously fall back to the ground state orientation, emitting resonance wavelength which we can detect and use to measure the energy difference?

Also, in the biochem course, it's said that a magnetic dipole precesses around a Bo, and then absorption of resonance frequency makes the magnetic dipole to lie on x-y plane while still precessing about z-axis. Then as time goes, this magnetic dipole is restored to its previous position (precessing about z-axis and not lying on x-y plane), and the amount of time it takes for that to happen is called free induction decay (FID). I totally don't get this. How does this precessing magnetic dipole relate to the emission resonance wavelength we see? Why does FID happen? —Preceding unsigned comment added by 142.58.129.94 (talk) 17:32, 19 March 2010 (UTC)

The idea you originally described, where you gradually scan across a range of energies looking for resonant frequencies, is the way a continuous-wave instrument works. The idea of a single pulse that polarizes everything and then monitoring its decay/precession is a Fourier-transform instrument. By mathematically processing the precession, one can determine the resonant frequencies. Our NMR article talks about these different ideas.DMacks (talk) 17:44, 19 March 2010 (UTC)

Extractor fan overrun timer

Resolved

Often in a bathroom, there will be an extractor fan which is triggered when turning on the light or (power to the) shower and which continues to run for a configurable period after the switch has been turned off. Typically, how does such an overrun timer work? (I mean at one extreme it could be clockwork and at the other extreme might involve configurable software but I imagine it's somewhere in between.)--Frumpo (talk) 18:57, 19 March 2010 (UTC)

It's a circuit similar to this (fig 2):[6]. However, instead of S1 you have a current sensor switch to let the timer know that the light has been switched on. RV 1 adjusts the delay period. All these devices are all bound to be based on the very cheap 555 timer IC--Aspro (talk) 19:37, 19 March 2010 (UTC)

re-reading this I don't think I have fully answered your question so in short: The adjustment control ( inside the fan unit) is just a variable resistor! By tweaking this, one can change the rate that a capacitor discharges through the 555 timer. At a certain point, a comparator circuit within the 555 tells a flip-flop circuit (which 'flipped' on when the light came on), to now flop off. This time to discharge to the 'flop off' state can be stretched out to several minutes by adjusting the resistance (or capacitance).And is is it in graph form with the 'out put pulse' representing the fan current.

So it is all done with transistors, resistors, and capacitors and a screw-driver. You are likly to have several 555's in your home all ready. They are very handy. --Aspro (talk) 21:01, 19 March 2010 (UTC)

Perfect. I thought it must be something along those lines but wasn't sure how you'd get a switch to toggle when a capacitor had discharged by a certain amount. It's hard to search for this stuff unless you know the already know the terminology. Thanks very much. --Frumpo (talk) 16:16, 20 March 2010 (UTC)

effects of alkaline on carbohydrates

on the moore's test why sodium hydroxide doesnt react with starch and sucrose unlike glucose and fructose? —Preceding unsigned comment added by 88.242.239.194 (talk) 22:09, 19 March 2010 (UTC)

This sounds suspiciously like a homework question, and it should be noted that we have a policy of "please do your own homework". (We're more than happy to assist, but you need to demonstrate that you've at least attempted to answer it yourself, and point out what, specifically, you are confused about.) - That said, I'm not familiar with Moore's test, although I imagine your answer may have to do with the most noticeable difference between glucose/fructose and starch/sucrose (hint). -- 174.21.243.94 (talk) 21:06, 21 March 2010 (UTC)

Anti-plasmodial

I'm trying to add a link to Piper sarmentosum to the word "anti-plasmodial" (in the "In medicine") section, but I'm having a hard time figuring out what "plasmodial" means in this context. Any ideas? Thanks. howcheng {chat} 22:13, 19 March 2010 (UTC)

An anti-plasmodial is an agent which acts against parasites of genus Plasmodium — in other words, they're antimalarials. TenOfAllTrades(talk) 22:43, 19 March 2010 (UTC)

Excellent. Thanks. howcheng {chat} 00:01, 20 March 2010 (UTC)

March 20

Location of herpes hideout

I was reading this article about Swedes working on a herpes vaccine: [7] and it gives me the surprising impression that the virus travels up the nerve fiber from the site of infection and hides out in the spine. Is that correct? I know it resides in sensory nerves, and I always assumed that meant the viruses would stay at the site of infection, but the article says "We found the genome of the virus in the spinal cord" and "somewhere between the mucous membrane and the spinal ganglion the infection lost its virility". 213.122.1.147 (talk) 00:57, 20 March 2010 (UTC)

Our article on herpes simplex virus notes that sites of latent infection do indeed include the sacral ganglia (particularly for HSV-2). TenOfAllTrades(talk) 01:52, 20 March 2010 (UTC)

Symbol for angular impulse

What's the standard or semi-standard symbol for angular impulse in physics? A few places on the internet I saw it as H, but they also seemed to denote torque or what they called moment with an M, whereas I was taught to use tau. Is H the correct symbol to use? 68.160.248.223 (talk) 03:09, 20 March 2010 (UTC)

Angular friction

I made up this expression which is supposed to calculate the amount of torque friction causes on an initially spinning object (like spinning something on a table that eventually comes to a stop). I'm not very experienced with calculus, but I felt like experimenting to see what I could come up with. Starting with $F_{f}=\mu _{k}F_{n}$, where $\mu _{k}$ is the coefficient of kinetic friction and $F_{n}$ is the normal force, I got $d\tau _{f}=r\mu _{k}gdm$, where $d\tau _{f}$ is the frictional torque, $g$ is gravity, and $r$ is the radius of mass element $dm$ from the center of mass. Integrating both sides gives $\tau _{f}=\mu _{k}g\int rdm$. Is this expression correct?

If this expression is in fact correct, could someone try checking my answer to the following? I tried evaluating this expression for a disk of mass $M$ and radius $R$ and got $\tau _{f}={\frac {2}{3}}\mu _{k}MgR$. Does this seem right?

If all this stuff is correct, I will be very happy with myself. :P

Thank you for all your help!!! 68.160.248.223 (talk) 03:40, 20 March 2010 (UTC)

It looks right to me. Dauto (talk) 04:44, 20 March 2010 (UTC)

I don't see how the radius matters. First, are we talking about a round object spinning on edge, like a coin, or on it's face, like a puck ? However, in any case, friction isn't theoretically dependent on the area of contact, but only on the coefficient of friction and the normal force (which is mass times g in both cases here). StuRat (talk) 17:52, 20 March 2010 (UTC)

I was referring to it being on its face, like a puck. The radius here does matter because we are talking about a frictional torque now and not a frictional force. That's why I set up my equation to integrate the radius with respect to each infinitesimal element of mass. I guess more generally, it should be $\tau _{f}=\mu _{k}\int rdF_{n}$ for situations where the normal force for the whole object is not necessarily $mg$ (i.e. - a puck coming to a stop from spinning on a ramp or in an accelerating elevator or something like that). 68.160.248.223 (talk) 18:17, 20 March 2010 (UTC)

<<<Static Electricity Question>>> Is it possible to excite BOTH light particles (in a small area) and the air surrounding them, with the SAME charge?

Greetings.

I have a question that has been bothering me for quite some time, and I hope somebody can answer it.

Imagine a cubic box (about 15cm X 15cm X 15cm) that is closed on 5 sides and open on 1 side.

Now, imagine that the air particles INSIDE THE BOX --to wit, the Nitrogen, Oxygen, and Argon atoms-- are all excited. And also, imagine that all light particles (photons, etc.) ENTERING THE BOX VIA THE OPEN END are also excited WITH THE SAME CHARGE as the air particles in the box (both positive or both negative).

My question is:

A.) Would this result in reflection of the light away from the box, total darkness in the box, or some other visual effect?

and B.) Can such a phenomenon be created using extant machinery and documented, scientific processes?

--Thank you for reading this. I know it's a big question, but it's also one that is driving me crazy! Pine (talk) 07:28, 20 March 2010 (UTC)

Photons do not carry electric charge. When an atom is in an excited state, it usually does not have an electric charge either - unless enough energy has been added to ionize it. I think you have some confusion between energy and electric charge. If, however, the photons have the correct frequency, they will interact with the atoms in the box, possibly resonantly. Note that when a photon excites an atom, the atom is briefly excited, but later decays to a lower energy level and re-emits a photon in an unpredictable direction ("spontaneous emission"). This means that there will be light in the box. Some technical terms for related processes include Atomic absorption spectroscopy, Compton scattering, spontaneous emission, stimulated emission. If you carefully control the "box" and its resonance, you will have created a laser. By injecting photons (energy) into the chamber, and controlling resonant frequencies, you can produce a coherent beam of light. This is nearly impossible to do with regular atmospheric air mixtures, but works great with mixes of helium and neon, or other setups. Nimur (talk) 09:27, 20 March 2010 (UTC)

Confused about Fourier Transform

How is a time-domain function represented in a frequency-domain function? Say we have f(t)=sin(t), a time-domain function. If we do Fourier Transform to convert it to f(v), then what do the range and domain of f(v) mean, and what do they look like?

In the article http://en.wikipedia.org/wiki/Frequency_spectrum, it's said, "A source of light can have many colors mixed together and in different amounts (intensities). A rainbow, or prism, sends the different frequencies in different directions, making them individually visible at different angles. A graph of the intensity plotted against the frequency (showing the amount of each color) is the frequency spectrum of the light. When all the visible frequencies are present in equal amounts, the effect is the "color" white, and the spectrum is a flat line. Therefore, flat-line spectrums in general are often referred to as white, whether they represent light or something else."

How does the frequency-domain graph (i.e. frequency spectrum) of a visible light look like and what does it actually mean?

Also, it's said that the values that Ff(u) (the FT function) take on are usually amplitudes and phase, both plotted against frequency. Then what is the y-value for the time-domain function?

—Preceding unsigned comment added by 70.68.120.162 (talk) 07:42, 20 March 2010 (UTC)

The map from time domain f(t) to fourier domain F(ω) creates a set of sinusoidal functions. F(ω) is a new function which represents the defining parameters for that set of sinusoidal functions: F(ω) gives you an amplitude and a phase for the sine wave at frequency ω. Adding up all those sine-waves, with the correct amplitude and phase, recreates the original time domain signal. So, the Fourier transform contains the same information of the original signal, but in a different format. A fourier transform can transform any type of "y-value". Scientists take frequency spectra of radio signals, light signals, height of ocean waves, comet orbital radius as a function of year (R_comet(t)), stock price as a function of day ($(t)), and so on.

An NMR machine measures an RF or microwave signal as a function of time - this is the "y-value" of its signal in the time domain. That RF signal is probably around a 30 MHz-ish intermediate frequency downmixed by the machine from the actual NMR physical frequency. This is an engineering detail you probably don't care about.

Light is a sort of pathological example, because we almost never think about light signals in terms of their time domain representation. (This is because they are extremely high frequency - 10¹³ to 10¹⁴ Hz. In other words, the oscillations are too rapid for most of our technology to work with in the time domain). But, if you wanted, you can describe the "y-value" as the time variation of the electric field for a light signal - there really is an electric field and a magnetic field, and it really is producing an effective voltage in free space, and so you can think of the time-domain signal as a function of time and space, where E is electric field strength: E(x,y,z,t). At a single point (x,y,z), you can simply consider E(t), the electric field fluctuation with time. Represented this way, light is exactly like a radio signal (except at higher frequency); so just like picking up an AM broadcast with a radio antenna, you could pick up the electric fluctuations of the light. (Technical details about building an antenna and amplifier are non-trivial - but let's ignore that issue for now).

So now you have a well-defined signal: E(t). It has values for all time t. Just as the mathematics describes, you can perform the time-integral of this signal, and calculate its fourier transform (or approximate its fourier transform for a short sampling window, since it is impossible to measure E(t) for all t out to ± ∞). Let's define E(ω), where ω is the frequency: E(ω) ≡ FOURIER{ E(t) } = α ∫ ( E(t) e^-2πiωtdt . You know E(t); and let's assume you're an expert at calculus and/or Mathematica software - so you can calculate the integral.

You need to determine the value of α based on your sampling window (all this does is normalize the values to the correct amplitude; in other words, "calibrate" for the units of energy, based on how long you collected energy for E(t)). There are some standard values, standard formulae, based on "standard" integral limits. This is common if you know the frequency content you are interested in, and can easily define a sampling time based on that - see Nyquist sampling theorem. For now, we'll just assume α doesn't matter. (It really doesn't make a huge difference; if you're an engineer or applied scientist, you probably read this value off of the instrument calibration spec anyway, which accounts for other practical details we don't care about here). There are also issues about whether you are using a digital system with discrete sampling (signal processing); this forces you to use a discrete fourier transform, which is an approximation of the real, full-blown continuous version. Most of the time, you won't care about the difference, because the NMR machine has been designed to do this approximation very accurately (by sampling fast enough for its needs, and computing the answer very quickly, using the FFT algorithm).

So now you have E(ω) - the amplitude of the electric field as a function of frequency, and you can describe the entire light signal in terms of its frequency content. (You can easily convert frequency to wavelength, if you prefer - that's a fairly trivial unit change, and it scales and flips the axis, because wavelength λ = 1/ω). You asked about what white light looks like in this domain. Well, it depends on how precise we're being in the definition of "white". Sunlight is usually the "golden standard" of white light - but its spectrum is actually blackbody radiation, plus a few resonances from nuclear fusion, noise, and other processes - technically, not really white noise: see a standard solar frequency spectrum). But in a small range around the visible spectrum, the amplitudes are reasonably flat - "white enough." And if you're looking at a real (non-theoretical) source, like a measurement off a spectrophotometer, even a really good white light source with a perfectly flat spectrum has a noise component - see Welch's method and spectral density estimation for some conceptual and mathematical analysis about estimating noise of a noise-spectrum. (This can be a little abstract).

So what does it mean when we have a frequency spectrum? Well, it means that we have represented the same signal (the light we measured, with our antenna apparatus), but we represented it in a different way. Now we can analyze the signal in terms of the frequencies that make it up. A lot of times, like if we are looking at a light source from a chemical reaction, or if we are looking at terahertz or IR spectra from an Nuclear Magnetic Resonance machine, we will see specific frequencies popping out of the noise. Since physicists know a lot about simple harmonic oscillation, if you tell them a frequency, they can estimate the mass (or some other parameter) that would correspond to that frequency. This really helps, if you care about mass (and in NMR, you usually are using mass as a proxy for chemical or functional-group identification). So, what we did is convert the signal into a different format (in fact, a different domain, the frequency domain), where the physical process that created the signal shows up very cleanly. Nimur (talk) 08:32, 20 March 2010 (UTC)

Your time-domain function f(t)=sin(t) is a sinusoidal waveform (with no phase shift i.e. it goes through the origin f(0)=sin(0)=0). Note that it is periodic i.e. it repeats forever in both positive and negative directions of t. Its FT (Fourier transform) is simply a spike at the real part of frequency 2*pi/t. Now some variations: Change the phase of the sine wave i.e. f(t)=sin(t+P) and the FT spike spreads to both real and imaginary parts of the frequency. However the power Re²+Im² is unchanged. The phase shift just rotates the vector in the complex plane without changing its length (magnitude). But real waveforms don't go on forever. Taking the FT of the sine wave over a limited time causes the spike to spread and look like the Sinc function. That has distracting side lobes but they can be reduced by applying a window function explained in the article Spectral leakage with examples here. Real waveforms need not be sine waves. Often the time domain samples are treated as real values by an FT because we ignore (set to zero) their imaginary components. Note however that both time-domain samples into the FT and frequency samples out of the FT are complex i.e. numbers with two variables. Therefore the same number of variables enters and leaves the FT. All the information is preserved and the time-domain wave can be recovered by doing an inverse FT on the complex frequency samples. For light a glass prism can act as a simple analog spectrometer doing what an FT does digitally. Cuddlyable3 (talk) 15:45, 20 March 2010 (UTC)

Phase

The Wiki's Phase article says, "The phase of an oscillation or wave is the fraction of a complete cycle corresponding to an offset in the displacement from a specified reference point at time t = 0."

The definition makes no sense at all. I'm learning Fourier Transform and NMR and I'm trying to understand how phase is related to FT. Any resources that don't assume high-level math/spectroscopy would be appreciated. —Preceding unsigned comment added by 70.68.120.162 (talk) 08:23, 20 March 2010 (UTC)

Remember that a Fourier transform is describing a signal as an equivalent set of sinusoidal functions. You already know from basic math (or you can trust me here) that a sine wave can be totally defined by three parameters:

A*sin(ωt+φ), where A is amplitude, ω is frequency, and φ is phase

So, when you represent an arbitrary function f(t) in Fourier domain, you generate those three parameters. For every frequency ω, there is an amplitude A and a phase φ. Many applications of spectrum analysis ignore the phase; but it contains information which is required to totally describe the original signal f(t). In conventional NMR, you probably care more about amplitude, because your source is incoherent (probably), so your phase signal ends up being mostly stochastic noise. Detailed analysis can extract information from that phase signal, but most physical chemists who use NMR only care about the amplitude spectrum because that can be used to easily identify chemical function groups or individual chemical elements. Nimur (talk) 08:41, 20 March 2010 (UTC)

The definition is poorly worded. Here is another shot: The phase of an oscillation at a given moment is the fraction of a cycle that has been traversed at that moment, after taking away all complete cycles that have been traversed. The concept of phase requires that some point in the cycle be designated as the beginning of the cycle. That's still a bit awkward -- really it's hard to be both clear and precise without resorting to equations. Looie496 (talk) 17:29, 20 March 2010 (UTC)

Essential oils

In extracting the essential oils from a plant, how much oil, roughly, could you expect for a given weight? Ballpark figures are fine. Thanks. 173.179.59.66 (talk) 08:31, 20 March 2010 (UTC)

Juniper Oil Yield, Terpenoid Concentration, and Antimicrobial Effects on Deer (1980), indicates the process. Depending on what you consider "raw material", (i.e., full plant; just the leaves; or a prepared substrate mash), your answer will vary wildly. Once you get a mulched-up leaf pulp, fractional distillation yields about 1-2 microliters of juniper oil per milliliter of plant mash. Nimur (talk) 09:10, 20 March 2010 (UTC)

And it seems that if you use a captive deer to digest the plant, (instead of a laboratory and a distilling machine), the yield can double or triple. Unfortunately, this requires "stomach pumping a captive deer"[8] to extract "rumen inoculum". Before Wikipedia, I suspect that insomniac scientists had far too few options for intellectual stimulation... Nimur (talk) 09:14, 20 March 2010 (UTC)

It really is impossible to say, depending on the plant itself and the method of extraction. This site [9] claims that "200 kg (440 lb) of fresh lavender flowers, between 2 and 5 metric tonnes of rose petals and 3,000 lemons are needed to produce 1 kg (2 1/4 lb) of essential oils of lavender, rose and lemon respectively". --TammyMoet (talk) 09:16, 20 March 2010 (UTC)

That's a bit unfortunate, but thanks. 173.179.59.66 (talk) 22:59, 20 March 2010 (UTC)

As an aside, it's also possible to produce rose oil from benzene and ethylene, and oil of geranium from isoamyl alcohol or even isopentane (and you could get 80-90% or better yield in both cases). FWiW 24.23.197.43 (talk) 05:32, 22 March 2010 (UTC)

Relationship between Bragg planes, atoms, unit cells in crystallography

How are Bragg planes related to individual atoms in a lattice, for the purpose of X-ray crystallography? Is each Bragg plane horizontally lined by atoms? —Preceding unsigned comment added by 70.68.120.162 (talk) 08:57, 20 March 2010 (UTC)

Miller indices uniquely identify the atomic arrangement with respect to a particular crystal face. Shown here are alignments for (111) and (221) planes.

.

Have you seen Miller index? This is the quantitative way to identify how a crystal plane relates to the actual atomic positions. The diagram at right shows two variants. The Bragg diffraction always occurs with respect to some planar alignment of atoms; depending on the geometry of the atoms in the lattice, that plane may be oriented in any particular way. Nimur (talk) 09:02, 20 March 2010 (UTC)

determining a person's age

I've heard one can tell how old a horse is by looking at its teeth. For a tree, you can count the rings, though you have to cut it down. Is there a medical way to tell how old a human is, if their age is disputed? Some invasiveness (e.g. an X-ray) is ok, but you can't decapitate the person or anything like that. Inspiration for asking: Abduwali Abdukhadir Muse. 66.127.52.47 (talk) 19:22, 20 March 2010 (UTC)

Not precisely. As with horses, the younger they are, the more accurately you can tell their age. I don't think anyone could accurately determine if someone is less than a year younger than 18 or less than a year older. There are all kinds of changes happening to the body at that time, but they happen at slightly different times in different people. --Tango (talk) 19:40, 20 March 2010 (UTC)

You might find the Wikipedia article Bone age informative. Also [10]. Disorders can often be diagnosed by retardation in development: [11]Osteo- archaeologist can also disern years of famine from bone growth rings. But I don’t have a reference for this.--Aspro (talk) 21:05, 20 March 2010 (UTC)

In forensic science the most common are what they call morphological techniques—estimating based on tooth or bone growth. The problem is the margin of error with adults in particular is quite large, some plus or minus 10 years. There are some new methods being investigated (but not yet widely used, I don't think) that show promise at getting more precise estimates. One is by using teleomeres (little bits on the end of the DNA in your cells that get smaller every time the cell duplicates); another is by radiocarbon dating of teeth (which only works if you were born after nuclear testing took place, and you have to know which hemisphere the person was born in; another is to look at the amount of aspartic acid in human dentin, which apparently correlates with age. The latter two are somewhat destructive—they require teeth and you can't get them back, I don't think. --Mr.98 (talk) 21:40, 20 March 2010 (UTC)

Well, you can tell a lot about a person by having a conversation with them, or reading their online profile, etc. I bet an accurate scheme could be created to estimate age through natural language processing or metadata analysis of a personal website. In this controversial experiment from MIT, a computer system accurately estimated a person's sexual orientation based on metadata in their Facebook profile, using a statistical database of keywords, friend/social-network patterns, and other public data. If sexual orientation can be uniquely identified, I wouldn't be surprised if an equivalent system could be trained to accurately estimate a person's age based on publically available social-network metadata - keywords, preferences for TV shows, etc. But the OP was probably more interested in biological tests.... Nimur (talk) 00:20, 21 March 2010 (UTC)

That's all assuming they put in accurate data, though. The example that inspired the original poster is someone whose age is under dispute, who is not on Facebook, and has apparently either lied about or does not their own age. (I have to say, I'm not impressed by the MIT experiment. Half of my gay friends on Facebook have pictures of them running around Fire Island-like environments in speedos, and the other half are heavily involved in groups like "Repeal Don't Ask Don't Tell" and have "The Birdcage" as their favorite movie. This isn't rocket science...) --Mr.98 (talk) 01:28, 21 March 2010 (UTC)

Sure, but you could probably use "favorite TV show" as a highly correlated indicator of age-group; and narrow down based on other details... and as far as "fake" data - there's no such thing! Provided that your system trains on a good set, "inaccurate" data is a statistical (albeit, noisy) signal which can be observed and used. Nimur (talk) 18:43, 21 March 2010 (UTC)

I am just not sure your system could distinguish between a heterosexual 16 year old girl and a homosexual 50 year old man pretending to be a heterosexual 16 year old girl. The latter case definitely does exist on these here internets! --Mr.98 (talk) 19:23, 21 March 2010 (UTC)

I don't see evidence he either lied about or does not know his own age. According to the article, his mother gave one date, his father another and the US government yet another. We don't have any evidence of what he stated his age was, if anything nor if what he stated was wrong Nil Einne (talk) 15:19, 21 March 2010 (UTC)

I was referring to this part: "Assistant United States Attorney Brendan McGuire informed U.S. District Court Judge Andrew J. Peck at a hearing to determine Wali Muse's age that he had told Americans he was 16, 18, 19, and 26 years old." --Mr.98 (talk) 15:27, 21 March 2010 (UTC)

Apologies, I missed that part Nil Einne (talk) 16:22, 21 March 2010 (UTC)

And you don't have to cut trees down in order to count their rings (we do not have an article on tree coring, but we do on ice cores and core drilling, which work in a similar manner. DRosenbach ^{(Talk | Contribs)} 18:09, 21 March 2010 (UTC)

Solar power satellite

can solar power satellite affect ozone layer? --Siddharth9936 (talk) 20:03, 20 March 2010 (UTC)

Don't think so. I can't imagine any mechanism by which one would. --Jayron32 20:05, 20 March 2010 (UTC)

No, I don't see how they could. Satellites orbit high above the ozone layer and they don't generally emit much pollutants once they are up there so they don't have a direct effect. The amount of sunlight blocked by their solar panels is utterly negligable - so there is no effect there. Probably the only possible effect would be from the exhaust gasses produced by the rocket while getting it up there - and that's the same no matter whether the satellite is solar powered, runs on batteries or something else. SteveBaker (talk) 00:45, 21 March 2010 (UTC)

I think they are asking about the microwave beam from a theoretical solar power satellite and whether the beam could damage the ozone layer. 75.41.110.200 (talk) 01:41, 21 March 2010 (UTC)

I agree. And the answer is still no. Dauto (talk) 02:34, 21 March 2010 (UTC)

If the silicon chips for the photovoltaics are washed in chlorofluorocarbons, then that chemical can cause ozone depletion. Nowadays I think they use other fluorocarbons for washing which are greenhouse gases instead.

raw data on smoking deaths

I would like to know where I can find raw data, or something closer to it, on "deaths caused by smoking". I am particularly interested in how deaths by cancer are separated into cancer caused by smoking and cancer caused by something else, since I don't think we know a lot about the specific mechanisms that cause cancer in general.

I am not interested in debating whether smoking is harmful to health, or whether it causes cancer, etc. I believe that it is, and that it does. I don't smoke, I've never smoked, I don't want to smoke, and being around cigarette smoke irritates me (though not as much as some people, evidently). But I do wonder whether the stats are "puffed" a bit (if you'll pardon the expression), since there is so much moral overtone in a lot of anti-smoking material.

I did a Google search and went through several pages of references to stats; they are pretty uniform in what they report, but nowhere did I find a reference to, say, a literature review on the science behind the summary numbers. I am always hearing things like "450,000 deaths in the US each year caused by smoking." So from what scientific evidence do they get this number?

Ralphcook (talk) 22:24, 20 March 2010 (UTC)

You will find more than you probably want to know in this survey paper. It contains a lot of different sources of data and different methods of analysis. -- kainaw™ 22:51, 20 March 2010 (UTC)

Thanks, that is the kind of information I'm looking for, but only a small part of it. It is an 11-page paper from 1956, summarizing data from a survey of 40,000 British doctors sent out in 1951. It compares the number of deaths in broad groups of smokers -- heavy, light, non -- and draws the not-unreasonable conclusion that people who smoke, especially cigarettes, have to expect a greater chance of lung cancer.

But it's a long jump from there to the quoted figures of "numbers of deaths caused by smoking", especially in a different country and culture and outside the population of this study. Does anyone have anything on the source of those figures, especially anything more recent than 45-50 years ago?

Ralphcook (talk) 15:52, 21 March 2010 (UTC)

Testosterone and intelligence

I heard someone make the claim that high levels of testosterone correlate with generally lower intelligence. Is this true? ScienceApe (talk) 23:03, 20 March 2010 (UTC)

It's not a simple matter - I suggest you read Testosterone#Brain. SteveBaker (talk) 00:39, 21 March 2010 (UTC)

Reading the article linked to by Steve Baker, it seems that it would be truer to say the reverse. I hate these stereotypes of masculinity. 92.29.149.119 (talk) 17:11, 21 March 2010 (UTC)

Part of the problem with that is the (belated) understanding that male and female brains are different. As that section of the article says, men have bigger brains but women's are better interconnected - as a direct consequence of testosterone levels. In a computer system, both size and interconnectedness produce higher processing speeds - but some ways of implementing algorithms are better for a bigger computer while others are better for a more interconnected computer. Hence, we find that women are better at multitasking - while men are better at 3D visualisation. The difficulty then is how we measure intelligence and (at the heart of the problem) what that word actually means. SteveBaker (talk) 19:26, 21 March 2010 (UTC)

Woman being better at multitasking is a prime example of a stereotype. I do multitasking all the time - I'm a male. Although studies have shown men to be better at spatial stuff, this may be just the effect of a lifetime of practice. Nobody can really say what's innate and what is due to a lifetime of conditioning from gender roles. 92.29.149.119 (talk) 20:00, 21 March 2010 (UTC)

I'm Chinese, yet I'm living in Canada. I suppose that must mean most Chinese people live in Canada, right? --99.237.234.104 (talk) 00:51, 22 March 2010 (UTC)

Indeed - and nobody said that men can't multitask - only that women are better at it. SteveBaker (talk) 02:06, 22 March 2010 (UTC)

I noticed that the claim in our article that women's brains are better interconnected is unsourced. I couldn't find a source that claims testosterone is responsible for a decreased number of dendritic connections. --99.237.234.104 (talk) 01:03, 22 March 2010 (UTC)

Although no differences in performance at certain tasks can be proven outright; it would be foolish to suggest that the physiological differences in the brains of the genders would have no impact on their ability at tasks.

Not all the differences between male and female brains are simply the result of presence or absence of T. However, back to the original question: has anyone ever done a research study in which people took an IQ test and had their T levels measured? I suspect it has been done, and it would not shock me if they found a mild negative correlation, but if so, all further interpretations founder on all the factors mentioned by the previous respondents plus a few others: the questionable ability of an IQ test to measure all forms of intelligence, and the relatively poor ability of T assays to accurately measure T production or effect. alteripse (talk) 19:38, 21 March 2010 (UTC)

We have no solid definition of intelligence beyond "the ability to score well in an IQ test" - and no solid grounds for saying that IQ tests measure intelligence since that is a circular argument. How could we be sure that we were measuring the right thing? We know that male and female brains are qualitatively different - and it seems quite likely that differences in IQ test results could be more to do with the test itself measuring properties that are different in male and female brains that have little to do with whatever we regard as "intelligence" on a day-to-day, practical basis. You simply can't make a proper scientific study of intelligence versus testosterone levels until/unless you have a proper scientific definition of what we mean by "intelligence"...and we don't. So any results you might come up with would be debatable to say the least. SteveBaker (talk) 02:06, 22 March 2010 (UTC)

Let me address the perception that testosterone lowers intelligence (rather than if it actually does). If two men have a disagreement, it's probably true that those with high testosterone levels are more likely to end up fighting, while those with a moderate level might reason out a compromise. Thus, the "more intelligent solution" is associated with the lower levels. However, this doesn't mean the high testosterone subjects were actually less intelligent, just that they didn't use their intelligence in this situation. StuRat (talk) 02:37, 22 March 2010 (UTC)

March 21

Intermittent blood supply

Hi. While all organs of the body are intact and connected by the nervous and lymphatic system, are there any organs that would either not survive, suffer permanent damage or organ failure if blood supply to the organ was cut off or significantly reduced for say, one minute? This is not a request for medical advice. Of course, this question does not include the heart, which transmits the blood supply, and "organ" in this case could mean intermittent or interrupted supply to a portion of the organ rather than its whole. Thanks. ~AH1^(TCU) 02:12, 21 March 2010 (UTC)

I took the liberty of fixing your red link. Dauto (talk) 02:22, 21 March 2010 (UTC)

So did I. - Nunh-huh 02:28, 21 March 2010 (UTC)

But history says that Dauto didn't fix any red links...can my question be answered? ~AH1^(TCU) 02:44, 21 March 2010 (UTC)

Of course, we have an article that deals with this. As that section states, highly aerobic organs like heart and brain sustain irreversible damage after 3-4 minutes. Our Cardiac arrest article is also relevant, and highlights the importance of body temperature and other variables. So, I don't think one minute of ischemia would predictably cause irreversible damage to any human organ under normal circumstances. Of course, I'm not recommending that anyone try this at home... -- Scray (talk) 04:06, 21 March 2010 (UTC)

Indeed not! Brain ischemia says that 10 seconds without blood supply to the brain will cause unconsciousness and 20 seconds shuts down all electrical activity. [12] says "Interruption of blood supply to the brain tissue for 2-5 minutes may result in permanent damage." - so in theory, a one minute interruption might not cause lasting damage. SteveBaker (talk) 04:19, 21 March 2010 (UTC)

To what is "Indeed not!" directed? My answer addressed irreversible damage, and nothing in your response refuted what I said. -- Scray (talk) 15:06, 21 March 2010 (UTC)

I presume the disclaimer (don't try this at home) hence why SB first pointed out the potential negatives before agreeing that it might not cause lasting damage Nil Einne (talk) 15:12, 21 March 2010 (UTC)

Thanks for clarifying - makes sense! One also has to worry whether our ability to detect irreversible damage might be too insensitive to detect microscopic foci of damage that, cumulatively over the years, result in changes known as "aging" or "dementia". -- Scray (talk) 21:04, 21 March 2010 (UTC)

Lorentz invariant

I have seen the Lorentz transformations used to derive the Lorentz invariant, but can the Lorentz invariant be used as a postulate to derive the transformations? 173.179.59.66 (talk) 06:36, 21 March 2010 (UTC)

Yes, in fact that's the way it's uasually done. You star with the relativistic interval which is a lorentz invariant. Assuming $dy=dz=0\,$ for simplicity,

$dS^{2}=(cdt)^{2}-dx^{2}$, and

$dS'^{2}=(cdt')^{2}-dx'^{2}$.

Now chose an interval for which $dS=dS'=0\,$ and you get

${\frac {dx}{dt}}={\frac {dx'}{dt'}}=c$

In other words, if an object moves at the speed of light in a coordinate system ${\frac {dx}{dt}}=c$, then it is also moving at the speed of light in the $S'\,$ system ${\frac {dx'}{dt'}}=c$. But $S'\,$ is completely arbitrary so the speed of the object must be the same in all reference systems. which is the usual starting point of the theory of relativity

Dauto (talk) 15:41, 21 March 2010 (UTC)

But it seems that the Lorentz invariant is justified on the grounds of c being the same in all reference frames. If we consider a beam of light emitted at the origin in two reference frames, x^2 + y^2 + z^2 = (ct)^2 and x'^2 + y'^2 + z'^2 = (ct')^2. Then x^2 + y^2 + z^2 - (ct)^2 = x'^2 + y'^2 + z'^2 - (ct')^2 = 0. So for any arbitrary displacement (x, y, z), we should have x^2 + y^2 + z^2 - (ct)^2 = A[x'^2 + y'^2 + z'^2 - (ct')^2] (there's no exponent to preserve linearity). But because y' = y and z' = z (justified under the grounds of symmetry), x^2 + y^2 + z^2 - (ct)^2 = x'^2 + y'^2 + z'^2 - (ct')^2 for any (x, y, z). My question is how we can go from this, to the equations x' = (x - vt)/sqrt(1 - (v/c)^2) and so on. 173.179.59.66 (talk) 00:58, 22 March 2010 (UTC)

I see that I missunderstood your question. Let me have a second stab at it.

Due to spacetime homogeneity we can assume that the transformation is linear so we can write (using Einstein's summation convention).

$x'^{\mu }=\Lambda _{\lambda }^{\mu }x^{\lambda }$

Where the $\Lambda _{\lambda }^{\mu }$ are to be considered as yet unknown.

Now we can use the metric tensor $g_{\mu \sigma }$

$g={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}\$

And build invariants (note that I will be working with 2-dimentional spacetime for simplicity)

$dS^{2}=g_{\lambda \rho }dx^{\lambda }dx^{\rho }$

and

$dS'^{2}=g_{\mu \sigma }dx'^{\mu }dx'^{\sigma }$

$dS\,$ is an invariant so we have

$dS'^{2}=dS^{2}$

$g_{\mu \sigma }dx'^{\mu }dx'^{\sigma }=g_{\lambda \rho }dx^{\lambda }dx^{\rho }$

$g_{\mu \sigma }\Lambda _{\lambda }^{\mu }dx^{\lambda }\Lambda _{\rho }^{\sigma }dx^{\rho }=g_{\lambda \rho }dx^{\lambda }dx^{\rho }$

$g_{\mu \sigma }\Lambda _{\lambda }^{\mu }\Lambda _{\rho }^{\sigma }=g_{\lambda \rho }$

which gives us four equations, but only three are independent, namely

$\Lambda _{0}^{0}\Lambda _{0}^{0}-\Lambda _{0}^{1}\Lambda _{0}^{1}=1$

$\Lambda _{0}^{0}\Lambda _{1}^{0}-\Lambda _{0}^{1}\Lambda _{1}^{1}=0$

$\Lambda _{1}^{0}\Lambda _{1}^{0}-\Lambda _{1}^{1}\Lambda _{1}^{1}=-1$

Now we square the middle equation and make a couple of substitutions from the other two

$\left[\Lambda _{0}^{0}\Lambda _{1}^{0}\right]^{2}=\left[\Lambda _{0}^{1}\Lambda _{1}^{1}\right]^{2}$

$\left[1+\left(\Lambda _{0}^{1}\right)^{2}\right]\left(\Lambda _{1}^{0}\right)^{2}=\left[1+\left(\Lambda _{1}^{0}\right)^{2}\right]\left(\Lambda _{0}^{1}\right)^{2}$

$\left(\Lambda _{1}^{0}\right)^{2}=\left(\Lambda _{0}^{1}\right)^{2}$

$\Lambda _{1}^{0}=\Lambda _{0}^{1}$

We take the positive square root here because the negative one leads to either time reversals or space reflexion which we are not interestead in.

Plugging that back into the equations above (after the word "namely"), we get

$\left(\Lambda _{0}^{0}\right)^{2}=\left(\Lambda _{1}^{1}\right)^{2}=1+\left(\Lambda _{1}^{0}\right)^{2}$

Which allows us the parametrizations

$\Lambda _{1}^{0}=\Lambda _{0}^{1}=sinh\phi$

$\Lambda _{0}^{0}=\Lambda _{1}^{1}=cosh\phi$

$\phi \,$ is called the rapidity.

Another (more familiar) possible parametrization is

$\Lambda _{1}^{0}=\Lambda _{0}^{1}=\beta \gamma$

$\Lambda _{0}^{0}=\Lambda _{1}^{1}=\gamma$

in which case $\beta ={\frac {\Lambda _{1}^{0}}{\Lambda _{0}^{0}}}=tanh\phi$

Also

$\gamma ^{2}=cosh^{2}\phi ={\frac {1}{1-tanh^{2}\phi }}$

Which finally leads to

$\gamma ={\frac {1}{\sqrt {1-\beta ^{2}}}}$

Short-pulse NMR

I'm trying to understand the principle/concept behind short pulse-NMR (not the original continuous-wave (CW) method). I read the NMR article's relevant section but there are several things in the article that don't make sense. Please clarify:

"A short square pulse of a given "carrier" frequency "contains" a range of frequencies centered about the carrier frequency, with the range of excitation (bandwidth) being inversely proportional to the pulse duration" >>>Is this because period is just inverse of frequency? Is the pulse duration the same as period of the pulse? Also, how can a group of frequencies give a single, unified period? I thought frequency and period of a wave were in a 1-to-1 relationship.

"Applying such a pulse to a set of nuclear spins simultaneously excites all the single-quantum NMR transitions." >>>After resonance happens, how do we actually know which frequency among many others contained in the pulse was the resonance frequency? Say we treated a sample with a pulse containing waves of 3 frequencies: 1nm, 2nm, or 3nm. If 2nm were the resonance frequency, how do we know if it's absorbed and then emitted?

"In terms of the net magnetization vector, this corresponds to tilting the magnetization vector away from its equilibrium position (aligned along the external magnetic field)." >>>What is this magnetization vector? I have no intuition about it and I've never seen it mentioned when I learned the CW method NMR

"The out-of-equilibrium magnetization vector precesses about the external magnetic field vector at the NMR frequency of the spins. This oscillating magnetization vector induces a current in a nearby pickup coil, creating an electrical signal oscillating at the NMR frequency. This signal is known as the free induction decay (FID) and contains the vector-sum of the NMR responses from all the excited spins. In order to obtain the frequency-domain NMR spectrum (NMR absorption intensity vs. NMR frequency) this time-domain signal (intensity vs. time) is FTed." >>>My understanding is that the pulse that has the resonance frequency among its many other non-resonance frequencies will cause the magnetization vector to move out of the equilibrium and the out-of-equilibrium vector will precess in a different space, sending signal to a coil. The coil measures the signal and plots its intensity VS time. Then we FT this to turn this into the typical NMR spectrum we know. The intensity VS time graph only has one frequency so if we FT this, then the intensity VS resonance frequency (chemical shift) graph will have only one peak, at the resonance frequency (x-axis). Please correct me if anything I just said is wrong. —Preceding unsigned comment added by 70.68.120.162 (talk) 07:15, 21 March 2010 (UTC)

This excellent blog entry answers many of your questions, with (sound) examples : http://scienceblogs.com/builtonfacts/2010/03/hearing_the_uncertainty_princi.php 83.134.177.217 (talk) 08:36, 21 March 2010 (UTC)

You might find impulse response a very useful concept also. Note that a short pulse has a large bandwidth, so it captures a lot of information in the frequency domain. Nimur (talk) 18:46, 21 March 2010 (UTC)

This image shows how a square wave consists of the sum of many frequencies, as can be demonstrated by taking its Fourier transform. In pulsed NMR a single-frequency carrier is gated by a lower-frequency square wave, effectively multiplying the carrier frequency by all the component frequencies of the square wave. Cuddlyable3 (talk) 23:02, 21 March 2010 (UTC)

speaker and receivers

what is the main difference in structure of receivers and speakers. can speakers be changed into receivers by just altering some connections?

1 MORE THING

i have a receiver with 2 wires coming out, and a speaker with 2 wires coming out, both will work with 2 pencil cells (i suppose so), how should i connect them to make a mic {receive my voice and give it out too}.... thanx--Myownid420 (talk) 10:16, 21 March 2010 (UTC)

I suspect there may be a language problem here. A receiver normally means a device that receives radio wave transmissions and passes them on to the amplifier and speakers. However, I think you are using "receiver" to mean microphone. Is that right ? StuRat (talk) 14:26, 21 March 2010 (UTC)

As for whether a single device can work as both a microphone and speaker, I think it's possible, but the sound quality is quite low, so it isn't normally done that way. For your 2nd Q, the wires are inputs in the case of the speaker and at least one is an output for the microphone, so it would require switching where they plug into the recording device. StuRat (talk) 14:33, 21 March 2010 (UTC)

oh! yes i mean microphone no receivers.. sorry--Myownid420 (talk) 16:45, 21 March 2010 (UTC)

Now, as to why you get worse sound quality when using a combination microphone/speaker, I don't know the details, but you do need larger speakers to get good bass, while this isn't as much a factor for microphones. The speaker and microphone on a telephone are of similar size, but they pretty much just cover the treble frequencies. So, if you don't care about bass, it's more doable. StuRat (talk) 17:51, 21 March 2010 (UTC)

Loudspeakers were used for decades with great success as both microphone and reproducer in intercoms. An electrodynamic microphone is basically a loudspeaker. Alexander Graham Bell's primitive telephone used them interchangeably (without any amplification) before the carbon microphone was invented by others. Edison (talk) 03:21, 22 March 2010 (UTC)

thanx for answers but how could i change a speaker in to a microphone —Preceding unsigned comment added by Myownid420 (talk • contribs) 08:17, 22 March 2010 (UTC)

guns in video games

I remember those days when i would play a bird shooting video game. a simple gun having one lens and a chip and a led and a tree type thing made of glass in it {i figured it out when i broke it}. it was connected to the video game keyboard by a fifteen pin socket, what you have to do is to just aim the flying bird {in tv screen} and shoot, and the bird is gone. that wasa a simple tv and even if precessor of video game come to know that the player had shot the shot, how would it come to know the aim was correct?, tv dosent had any sensors.

also note that it would not work if i touch the barrel of the gun and shoot

please explain how it worked..............thanx--Myownid420 (talk) 10:36, 21 March 2010 (UTC)

There are a few different ways to make this work- see Light gun. Staecker (talk) 11:06, 21 March 2010 (UTC)

You probably were using a NES Zapper. Basically whenever you press the trigger the screen would turn black except the targets (ducks) which would turn white. The zapper had a little chip in it that could detect if it was aimed at something white (of the right intensity of light). That way the gun would know if it was pointed in the right direction or not. Then the screen would go back to being in color again. Not the most accurate or foolproof method but it worked pretty well for children's games. You could notice the quick change if you looked at it out of the corner of your eye, I recall. --Mr.98 (talk) 15:05, 21 March 2010 (UTC)

oh! ya i remember it flashed whenever i shoot. but why it dosent work from snall range(very near) why?--Myownid420 (talk) 16:48, 21 March 2010 (UTC)

This is rather old tecnology. Our Sears Telegames Pong system had a gun for shooting a white dot on a black screen. This would have been in about 1978 or so. 75.41.110.200 (talk) 18:22, 21 March 2010 (UTC)

There are two ways these things could work - the simplest (as described above) just had a photocell in the barrel of the 'gun' which determined if you were pointing at something white when you pressed the trigger. That's really crude though because you only know that the gun was pointed at something bright - you don't know WHICH something. Also, you have to have your targets be white against a dark background - you can't draw clouds and trees and such. A more sophisticated version displays a light color on the screen all the time (grey, probably) and measures the amount of time between the vertical refresh signal going to the TV and light entering the barrel of the gun. Knowing that time lets you know where the gun is pointing because the progress of the raster scan is known. That approach allows the computer to show misses graphically - and if you hit a bird, then you know which one. There can also be other things on the screen, allowing you to draw anything you like as a 'background'. SteveBaker (talk) 19:17, 21 March 2010 (UTC)

Note that you could have clouds and trees with games like Duck Hunt, because of the aforementioned "flash" to the black/white mode. Which could have multiple targets, too? I wonder if they had two separate black/white flash modes. That would probably put an upper limit on the number of targets, though. --Mr.98 (talk) 19:20, 21 March 2010 (UTC)

The NES Zapper article suggests that different targets flashed separately, so the computer only has to match the gun-detector to the frame display. The idea of watching the raster-scan to know directly "where is the thing pointing" on an arbitrary screen is the basis for light pen devices. DMacks (talk) 19:25, 21 March 2010 (UTC)

Copperhead lookalike?

(I put this on the herp project page this morning, but then noted that the project has low activity)

I'm having a few worries about a species of snake that keeps popping up near our springhouse/smokehouse (the smokehouse has been re-purposed as a playhouse for my daughter). It certainly resembles a copperhead, but the markings are slightly more muted and they don's seem to get quite as large. I shot one of them last year, but another has moved in, and I don't want to kill it if it's not venomous (we have black northern watersnakes and garter snakes living in the same pond, and the garters actually "snuggle" with this snake on warm days). However, if it is venomous, it can't stay!

I didn't see any list of lookalikes on the article, but I've been told that there's a snake called a rat snake or king snake that is similar. Anyone know what they're talking about? I'm in eastern Pennsylvania. --SB_Johnny | ^talk 09:46, 21 March 2010 (UTC)

You might find this site [13] interesting, it actually discusses copperhead lookalikes. Richard Avery (talk) 13:36, 21 March 2010 (UTC)

Awesome, thanks! The one by the house is not a copperhead, thankfully. (We do have them in the woods, but that's not as alarming). --SB_Johnny | ^talk 14:12, 21 March 2010 (UTC)

hairdryers

Hypothetically were I to tape over most of the bit of a hairdryer where the air goes in, leaving only a little hole, would the air be sucked through there much faster or would it blow out the other end slower, or would the whole thing just explode? And would that make it louder or quieter? 148.197.114.158 (talk) 15:35, 21 March 2010 (UTC)

The fans are not fixed displacement types, so it would...

• Blow out slower
• Blow out much hotter air
• (as one did at work last week, when the fan fell off the shaft) maybe catch fire!

 Ronhjones  ^(Talk) 15:45, 21 March 2010 (UTC)

I'd say it would catch fire. The lubricating oil for the fan, in particular, as well as dust inside from previous usage. StuRat (talk) 17:37, 21 March 2010 (UTC)

Should have a high temp cut out, really. --BozMo talk 20:24, 21 March 2010 (UTC)

Eventually you would burn out or wear out the motor. It would be quite noisy and not blow out much air until then. Googlemeister (talk) 13:24, 22 March 2010 (UTC)

Why would it catch on fire? Just from that lack of air flowing through it? also, is there anything with a different sort of fan then, that might be affected differently? 148.197.114.158 (talk) 15:27, 22 March 2010 (UTC)

The heating elements must be cooled by blowing air or they get hot enough to ignite a fire. As BozMo pointed out, many countries require a temperature sensor to shut it down if it gets too hot, to prevent this. Some hair dryers also have a "fan only" setting, where no heat is supplied. Then the only heat would be that generated by friction in the fan motor. StuRat (talk) 15:34, 22 March 2010 (UTC)

Would a rediculously cheep hairdryer have only the fan and no extra heat?148.197.114.158 (talk) 20:30, 22 March 2010 (UTC)

Natural Gas

What is the chemical compund for Natural gas which is supplied by utility companies to businesses and residences? —Preceding unsigned comment added by 71.84.229.190 (talk) 17:28, 21 March 2010 (UTC)

It's mostly methane with some ethane, propane, and butane. Dauto (talk) 17:31, 21 March 2010 (UTC)

Odorants like t-butyl mercaptan are usually added too so that you can notice leaks. 75.41.110.200 (talk) 17:57, 21 March 2010 (UTC)

Fixed redlink 94.168.184.16 (talk) 18:21, 21 March 2010 (UTC)

And a possibly confusing typo. --Anon, 20:30 UTC, March 21, 2010.

For further information, we have a whole article with the obvious title of natural gas about the contents, odorants, etc. DMacks (talk) 22:54, 21 March 2010 (UTC)

Hmm, I wonder why the odorant in natural gas smells similar to a durian? ~AH1^(TCU) 23:34, 21 March 2010 (UTC)

Durian#Flavour and odour says there are sulfur compounds in it. I've never smelled one (and from that article, doesn't sound like I want to) so no idea myself. The talk-page mentions some disagreement about whether H2S in particular is involved, and the article relies heavily on a wide-ranging secondary/review article that doesn't exist (!) so can't find the underlying actual supporting refs. DMacks (talk) 09:30, 22 March 2010 (UTC)

The smell is mildly reminiscent of it, but not to the point where you'd confuse one with the other or anything. The durian's scent has a fruity, sweet, kind of undercoat to it. It's not so much that it smells inedible, so much as it smells like something very edible that's already been partly digested... Difficult to describe, really, but you can almost taste a sweet sliminess in the air. If you're curious, many Thai and Vietnamese restaurants will have, if not the fruit itself, at least a frozen milkshake version which will give you a feel for it. Matt Deres (talk) 16:42, 22 March 2010 (UTC)

Medical Advice

Is there somewhere on the internet where I can ask medical advice questions? I think I have a medical problem, but I cannot just go to a doctor easily and it'd be such a wase of time to travel weeks to see one if my problem isn't serious. Thank you —Preceding unsigned comment added by 94.4.255.173 (talk) 17:41, 21 March 2010 (UTC)

You should be able to phone a doctor, even if it isn't practical to visit them. There are lots of places on the internet where you can ask for medical advice, but I know of none where you can have any confidence in the answers. You need to talk to a doctor. --Tango (talk) 17:47, 21 March 2010 (UTC)

The answer to this is highly dependent on the person's location and resources. The IP address is London, UK, unless he/she is hiding his location. That does not match the "travel weeks to see one" phrase. Almost no doctors will give free advice by telephone without an institutional or already established relationship. There are for-profit internet medical consultation sites (just google "online medical consulting" and you will get dozens of choices), so if the questioner is willing to pay for the consultation, he can get answers from a doctor without a face-to-face visit. Obviously this would work better for some types of problems than others. alteripse (talk) 18:29, 21 March 2010 (UTC)

In the UK, almost all doctors will give free advice over the phone (to patients registered with them, anyway, and I'm assuming the OP has a specific doctor weeks away that they could call). An alternative is phoning NHS Direct (if the OP really is in the UK, this would be my recommendation - NHS Direct is really good. You'll get to talk to a nurse and they'll get a doctor to call you back if necessary). --Tango (talk) 18:34, 21 March 2010 (UTC)

I had no idea medical practice in the UK is that different from North America. Do British doctors have telephone hours, as I assume the volume could be horrific? Will they talk to patients not assigned to them by the NHS? alteripse (talk) 18:51, 21 March 2010 (UTC)

Exact procedures vary from GPs surgery to GPs surgery. Talking to a doctor on the phone would normally be used in emergencies, rather than for routine stuff. You would either phone the surgery and talk to the receptionist, who would choose to give you an emergency appointment or get a doctor to call you back, or you can go through NHS direct as I mentioned above, and a nurse will either advise you themselves or get a doctor to call you back. Patients aren't really assigned by the NHS. You register with a GP of your choice (assuming they have space on their books, anyway). For routine stuff, you have to go through your GP (who may refer you to a specialist), but for emergency stuff you can just call the nearest surgery (eg. if you fall ill while on away from home). --Tango (talk) 19:02, 21 March 2010 (UTC)

Thanks. Perhaps the reality is not as different as it seemed. In the US, if you are an established patient of a practice, you can usually call and talk to a nurse who will decide if you need an appt or might get an answer from the doctor or in some cases might have the doctor call you back. No US doctors will simply provide telephone consultations to new patients who may not be coming in. It seems unlikely to me that even a UK doctor would provide a telephone consultation to a patient who has not been seen and will not be seen in his practice. What else would they have time to do? alteripse (talk) 19:23, 21 March 2010 (UTC)

Tango represents things correctly but in my experience if you are away on holiday in a different part of the UK a local GP will generally return a call and try to be helpful, without requiring a temporary registration (which they will organise if they actually have to see you). A typical rental cottage with list the local GPs phone number. I think most GPs in the UK would return a call but I guess they would be circumspect about giving some types of advice. Most of such advice is probably about children anyway. Lots of things though can be answered on the phone. --BozMo talk 20:20, 21 March 2010 (UTC)

There are many docs in the U.S. who provide a call-in number to patients, along with instructions regarding when to use it. For some of us, that means under certain circumstances, for others that means at certain times. This is entirely up to the individual physician, but enhanced access can be very reassuring to the patient, and most patients respect the privilege. This is not just "boutique" medicine. -- Scray (talk) 21:00, 21 March 2010 (UTC)

Not for people who have not been their patients because there is no mechanism to charge, unless you are talking about an online or telephone only practice for profit. You ignored a key part of my assertion. alteripse (talk) 11:32, 22 March 2010 (UTC)

Are you trying to treat or diagnose? If you are trying to diagnose then I'd say definitely go to a doctor. No one over the internet will be able to diagnose you as well as a doctor in person and it is too easy to misdiagnose yourself just from what you think your symptoms are, confirmation bias and all that. If you are trying to treat then we have articles for a lot of illnesses which include typical treatment regiments and there are support groups and forums for a whole range of illnesses online. Vespine (talk) 23:44, 21 March 2010 (UTC)

Judging by the OP's wording I'm guessing they do not live in the US/UK since it would take weeks to get to a doctor? I have no idea in which countries this would be the case. Regards, --—Cyclonenim | Chat  09:21, 22 March 2010 (UTC)

Meaning of retrolateral and prolateral

I see the terms retrolateral and prolateral used all the time in entomology papers, but I haven't been able find out what these terms actually mean. I've already checked anatomical terms of location, but they aren't mentioned there. Anyone know the answer? Kaldari (talk) 19:08, 21 March 2010 (UTC)

Since retro means behind and pro means forward and lateral means side, I would assume that retrolateral means along the side toward the rear of the body and prolateral means along the side toward the front of the body. However, IANAE. alteripse (talk) 19:30, 21 March 2010 (UTC)

Hmm, I think that's close, but not quite right. The terms are usually used in reference to drawings of pedipalps, which are basically the equivalent of a "hand" or "arm". I'm thinking that one refers to "side view from the side closest to the body" and the other refers to "side view from the side away from the body". Perhaps retrolateral is the side towards the body (behind), and prolateral is the side away from the body (forward). Kaldari (talk) 21:00, 21 March 2010 (UTC)

In this article about scorpion'a teeth (behind a subscription wall) the author explains "I use the terms prolateral and retrolateral respectively". Cuddlyable3 (talk) 22:19, 21 March 2010 (UTC)

After some more reading, I think alteripse might be correct as the term is also applied to arachnid legs, not just pedipalps. Kaldari (talk) 22:52, 21 March 2010 (UTC)

Reaction quotient, dependent on standard state?

I am trying to understand some quantitative chemistry, but continue to be stumped by quantities that are ostensibly dimensionless but whose values nevertheless depend on essentially arbitrary choices of "standard states". One of these is "activity"; another is "reaction quotient". But the latter seems to be used in contexts where it cannot be allowed to depend on our choice of standard states.

For example, imagine that we fill a vessel with an equal (by volume) mixture of diatomic ozygen and ozone at standard temperature and pressure, and consider the equilibrium $3\,O_{2}\leftrightarrow 2\,O_{3}$. In gases, the activity (chemistry) of a species is roughly equal to its partial pressure divided by the standard pressure, so we have $\{O_{2}\}=\{O_{3}\}=0.5$, and the reaction quotient $Q_{r}$ becomes $(0.5)^{2}/(0.5)^{3}=2$.

Now, if "standard pressure" had been defined as 25 kPa instead of 100 kPa, the numbers would be different. Namely, then for the same situation as above, $\{O_{2}\}=\{O_{3}\}=2$ and $Q_{r}=0.5$.

The article Gibbs free energy tells us $\Delta _{r}G=\Delta _{r}G^{\circ }+RT\ln Q_{r}$, so the sign of $\ln Q_{r}$ (i.e. whether $Q_{r}$ is less than or greater than unity) determines whether $\Delta _{r}G$ will increase or decrease if we change the temperature a small bit. But how $\Delta _{r}G$ varies with temperature is something that should be discernible by an appropriate experiment and cannot be allowed to depend on which standard state we use for our computations!

Clearly I am missing something here – but what? Does one of the articles have it wrong? Is my math wrong? Is $\Delta G$ not actually a physical quantity (even though the dimension $[\Delta G]=\mathrm {J} /\mathrm {mol}$ looks respectable enough)? Or does $\Delta _{r}G^{\circ }$ secretly vary with the temperature (even though the "$X^{\circ }$" notation is supposed to mean, among other things, "at standard temperature")? –Henning Makholm (talk) 21:55, 21 March 2010 (UTC)

$\Delta _{r}G^{\circ }$ changes depending on what the standard states are defined to be: It's the standard state Gibbs energy change, and does of course depend on what you consider to be standard. Typically, we choose standard states to be 1 atm (or 1 bar, I guess; they're nearly the same]] for gasses, 1 M for solutes in solution, and 298 K (?) for temperature, but those are pretty arbitrary. $\Delta _{r}G$, on the other hand, does represent a constant physical value that can be measured, and is the same under given conditions no matter what you define your standard states at. Buddy431 (talk) 14:51, 22 March 2010 (UTC)

OK, so $\Delta _{r}G^{\circ }$ is really a constant once I have chosen standard states. That elminiates one possibility, but I still haven't found my misunderstanding. If I can measure $\Delta _{r}G$ for my sample gas mixture at 298 K and 299 K, the difference between those measurements is independent of $\Delta _{r}G^{\circ }$. Of my two calculations, one says that $\Delta _{r}G$ must be higher at 299 K, and the other says that it must be higher at 298 K. But those can't both be the case.

Doing the math in more detail:

${\frac {\partial \Delta _{r}G}{\partial T}}={\frac {\partial }{\partial T}}(\Delta _{r}G^{\circ }+RT\ln Q_{r})=R\ln Q_{r}+{\frac {RT}{Q_{r}}}\,{\frac {\partial Q_{r}}{\partial T}}$

ought to be a measurable physically quantity (and so not depend on our arbitrary choice of standard state). The final term is the same under all choices of standard states, because another choice just shows up as a linear factor in the activity of each species, which applies to the same power in the $Q_{r}$ denominator and the $\partial Q_{r}$ numerator. Therefore, by choosing sufficiently extreme standard states, we can make $R\ln Q_{r}$ so large (of whatever sign) that it dwarfs the final term, and thus both sign and magnitude of $\partial \Delta _{r}G/\partial T$ seem to depend entirely on which standard state I choose for my calculations. Which is absurd.

Again, what am I doing wrong here? –Henning Makholm (talk) 17:33, 22 March 2010 (UTC)

Teleportation

This video with physicist Michio Kaku, http://www.youtube.com/watch?v=-FqLCLooayM&feature=PlayList&p=1B829DF36754F91C&playnext=1&playnext_from=PL&index=2, talks about teleportation. He mentions that we were able to teleport photons of light some distances. He also states that we are working on teleporting particles of matter some distances as well in the future. How is this possible? Our article on teleportation doesn't explain anything at all. Wouldn't teleportation constitute as a FTL (a circumvention of it)? I guess my confusion is, my understanding is that we have never solved the way to circumvent the light speed barrier. All of our ideas were merely theoretical. If Michio Kaku is telling the truth, didn't we already solve it? ScienceApe (talk) 22:09, 21 March 2010 (UTC)

Extraordinary claims need extraordinary evidence. Saying "We have teleported photons 600 meters across the Danube" doesn't do it. Wikipedia has an article about Michio Kaku. Cuddlyable3 (talk) 22:30, 21 March 2010 (UTC)

He is almost certainly talking about quantum teleportation (see that article for an explanation). As the first paragraph of that article says, it doesn't allow FTL travel or information transfer. --Tango (talk) 22:41, 21 March 2010 (UTC)

(EC) The obvious question is did you search, the answer I presume is no since two simple searches [14]/[15] [16]/[17] would have lead you to [18] and [19] or similar articles and eventually Quantum teleportation which should get you started Nil Einne (talk) 22:45, 21 March 2010 (UTC)

Oh no, you're wrong. I did search the answer. ScienceApe (talk) 23:05, 21 March 2010 (UTC)

What terms? In this specific case, I can find the answer very easily, e.g. even 'teleport photons of light some distances' finds one of the earlier links as does 'teleportation light & 'teleportation particles'; meanwhile and 'teleporting particles of matter some distances' finds [20], 'Michu Kaku teleportation' finds [21], 'teleportation ftl' leads to [22]. In fact even 'teleportation light speed barrier' in Google Bing show as the first and fourth result respectively the wikipedia article Faster-than-light both of which highlight this quote "Quantum teleportation transmits quantum information at whatever speed is used". So yeah, I'm having great difficulty finding any likely search terms that doesn't fairly quickly lead you to the answer, i.e. it'll be more helpful for you to learn where you went wrong then anything else Nil Einne (talk) 23:15, 21 March 2010 (UTC)

This is relevant. —Akrabbim^talk 23:15, 21 March 2010 (UTC)

The real question you should be asking is not that, but why did I ask the question, when I already knew the answer. :) ScienceApe (talk) 23:22, 21 March 2010 (UTC)

Fourier Transform

When we FT a time-domain function made up of only one frequency, why do we see two peaks (at + and -)? (Eg. FT of a sine function has a peak at freq=1 and -1.)

The usual Fourier transform does not use sines and cosines as base functions, but $e^{ikx}$, and you need two of those to make up a real sine. See Euler's formula#Relationship to trigonometry. –Henning Makholm (talk) 23:18, 21 March 2010 (UTC)

If we FT a sine function, why do we see a peak at frequency=1 instead of at 2pie which is the period of a sine function?

Because the k axis of the transformed function is actually angular velocity, not frequency. –Henning Makholm (talk) 23:18, 21 March 2010 (UTC)

If we FT a sine function that is shifted up, why does its freq-domain graph have a peak at frequency=0 in addition to at -1 and 1?

Because the way to get a constant term is $c\cdot e^{i\cdot 0\cdot x}$. –Henning Makholm (talk) 23:18, 21 March 2010 (UTC)

Also, where can I get some practice questions on guessing the result of doing an FT on a function (whether time-domain or freq-domain)? —Preceding unsigned comment added by 70.68.120.162 (talk) 22:50, 21 March 2010 (UTC)

• 1) Think about the identity $cos(\theta )={\frac {e^{i\theta }+e^{-i\theta }}{2}}$. The negative frequency is there
• 2) $\omega =2\pi f\,$ there is your $2\pi \,$
• 3) Shifting up is adding a constant which has zedro frequency. Think $e^{0}=1=const.$

Dauto (talk) 23:23, 21 March 2010 (UTC)

For practice working fourier transform problems, I recommend Signals and Systems, by Simon Haykin and Barry Van Veen, (available at Amazon, $52). There are hundreds of Fourier transform problems, with answers, and many worked examples, ranging from theoretical to applied to numerical/computational problems. There is also a great MATLAB tutorial for FFTs. This is a great textbook if you have never seen Fourier theory before, but want a complete mathematical and conceptual introduction to it. Nimur (talk) 00:21, 22 March 2010 (UTC)

are thermal baths bad for the skin

are thermal baths and mineral baths like salt baths bad for the skin, and is the sauna —Preceding unsigned comment added by 82.113.121.95 (talk) 23:26, 21 March 2010 (UTC)

Take a look at sauna#Health Risks & Benefits and Epsom salt#Applications. ~AH1^(TCU) 23:37, 21 March 2010 (UTC)

CO2 in Antarctic Ice Sheet

Hi. How many tons of CO₂ is stored in the Antarctic Ice Sheet, including gas frozen in the ice within bubbles of air, carbonic acid in precipitation that piled up into ice, and the carbon dioxide frozen onto the surface of the ice over time due to the cold temperatures and katabatic winds? Also, if all of this carbon dioxide was to be released into the atmosphere, how much of an increase in atmospheric CO₂ concentrations in ppm would result? In addition, could the cold temperatures over East Antarctica be freezing the anthropogenic additions of CO₂ out of the immediate troposphere, reducing the greenhouse effect and resulting in the observed slight cooling in the region? Thanks. ~AH1^(TCU) 23:42, 21 March 2010 (UTC)

Also, could there be deposits of carbon dioxide or methane clathrate hydrates frozen into the ice, and could there be any carbon-based soot in the ice that could either enter the ocean or speed up melting on the continent as the ice begins to melt? Thanks. ~AH1^(TCU) 23:43, 21 March 2010 (UTC)

Yes there could be (I don't see any reason why not); the real question is, are there? 24.23.197.43 (talk) 05:18, 22 March 2010 (UTC)

According to File:Vostok-ice-core-petit.png, the average CO₂ Antarctic ice core content is about 240 ppmv. (I don't have any immediate data on ice core porosity, which would be needed to calculate bulk CO₂ content.) Bear in mind that the ice has trapped bubbles or atmospheric air (and not just carbon dioxide), so the release of gasses due to the melting of ice sheets may not change the overall percentage of CO₂ in the atmosphere. caknuck _° needs to be running more often 07:41, 22 March 2010 (UTC)

According to [23] the volume of the antarctic ice cap is around 30 million cubic kilometers. At 240 parts per million by volume of CO2 - that could contain 7200 cubic kilometers of CO2. Presuming the bubbles are at normal air pressure (because that's the pressure they were at when they were formed) - then that's something like 10¹⁰ tonnes. That sounds like a lot - but it's about the same as the total CO2 output by US coal fired power plants in a couple of months. But as Caknuck points out - it won't make much difference to the percentage of CO2 in the atmosphere - and that's what matters from a global warming perspective. The amount of CO2 in the air has varied over history - but the long term average probably isn't much different from today.

As for the other things - I seriously doubt that much CO2 is freezing into the antarctic ice in the long term - that requires temperatures down at -78C which only happens for very short periods of time in the center of the continent during the depths of winter. As soon as the temperature warms up again, it would sublime back into a gas before it could get too deeply buried and form long-term deposits. Katabatic winds and the kinds of deep snowfalls needed to bury frozen CO2 only happen near the coast of antarctica - and it doesn't get cold enough there for CO2 to be a solid.

Black, sooty deposits in the ice are likely to be one reason for the melting we're seeing - but that stuff is coming from forest fires and things like diesel engines that are swept inland in modern times. It's possible that some major earthquake or forest fires in the distant past laid down layers of dark material that could eventually become exposed - but I've not seen any mention of that from the ice-core studies. SteveBaker (talk) 10:54, 22 March 2010 (UTC)

March 22

Tides

I was reading my textbook which tried to evaluate the height of the tides. To do this, they borrowed what was apparently a method devised by Newton, and "pretend[ed] that two wells full of water run from the surface of the Earth to the center, where they join. One is along the earth-sun axis, and the other is perpendicular." I'm having trouble seeing how this can in any way help in determining the equilibrium height of tides. After all, the ocean runs along the surface of the earth, not through it. 173.179.59.66 (talk) 00:50, 22 March 2010 (UTC)

Yes, I agree that this sounds like an unreliable model. While tides do affect solids as well as fluids, they deform far less than the fluids do. StuRat (talk) 02:40, 22 March 2010 (UTC)

Stu is misunderstanding the thought-experiment. There's no assumption that the solids will deform.

Suppose you have two lakes that are at somewhat different levels, and you dig a canal connecting them so that water can flow freely between them. If the canal and the lakes are deep enough that nothing empties out, then the water will flow from the higher-surface lake to the lower until the level is the same in both lakes, right? Okay, now instead of the canal, dig an underground tunnel connecting the two lakes. The water will still flow in the same way, leaving the lakes at the same level, right?

Now imagine that you have two lakes, one at the South Pole (which thanks to global warming is now above the freezing point) and one at the Equator. Join them with a tunnel through the Earth and the water will still try to flow to the same level, right? That's true even if the tunnel down goes all the way straight down to the center of the Earth and makes a right-angle turn to rise back to the surface. (Pretending that you could actually build such a tunnel and that water would remain liquid in it.)

Well, that's the layout that the thought-experiment is about. Two deep wells meeting at the center of the Earth. And the point is that when we say the water reaches the "same level", the idea of what is the same level is affected by the tidal force. Because the water at the equator is pulled toward the Sun if the well is pointing toward the Sun, then it will rise higher to reach what "feels" to it like the same level.

And similarly with the ocean -- it tries to reach the same level everywhere, but its idea of what is the "same" level is affected but the tidal force. And what's more, it's affected in exactly the same way as in the thought-experiment configuration with the wells or tunnels. So the simple configuration described in the experiment tells you how high you could expect the ocean tide to rise, if things like friction, and the shape of the seafloor in coastal areas, did not have a big effect on it.

--Anonymous, 05:51 UTC, edited 18:55, March 22, 2010.

How did the Atlas ICBM navigate?

How did the SM-65 Atlas ICBM navigate, i.e. how did it know where it was and where to go? Was it a) remote-controlled, or b) steering autonomously, or did it c) simply follow a fixed pre-programmed flight schedule without any adjustments? In case of a) or b), how did it know its location in the absence of GPS? Did it have a camera built in, maybe to observe the stars? Or did the ground stations follow it by radar and use option a)? AxelBoldt (talk) 03:16, 22 March 2010 (UTC)

Our article describes the guidance of the Atlas D variant as radio-based (an inertial system with ground-supplied course correction information), and the E and F variants as having true inertial guidance systems. This page (see 'System Operation') has more detailed information about the specific navigation equipment. TenOfAllTrades(talk) 03:34, 22 March 2010 (UTC)

Just a note that celestial guidance was not around until the 1970s (used first in the UGM-73 Poseidon). Missile guidance has more general information on technologies developed to get the missile from launch to target. Note that the Atlas D had a CEP of 1.8 nautical miles and Atlas E-F had 1.0. So their accuracy was not exactly pin-point (though still impressive for that generation of missile), but when you have a 4 Mt warhead on the tip of it, it doesn't have to be. --Mr.98 (talk) 14:57, 22 March 2010 (UTC)

Cloud base measurement

Here's a question for all the weathermen in here: Suppose that you have to measure the cloud base at your location in support of an important aerial mission. Suppose also that you don't have an operating ceilometer (what the hell, no article?!) at your disposal, so you have to find another way to do this. Further suppose that it's nightttime, so you can't use any method that relies on visual observations by ambient light alone. Given these conditions, how would you do this? Thanks in advance! 24.23.197.43 (talk) 03:37, 22 March 2010 (UTC)

I before E fixed your link :). Well our Cloud base article says The height of the cloud base can be estimated from surface measurements of air temperature and humidity. . But stuffed if I know how. At least it sounds possible. Vespine (talk) 03:56, 22 March 2010 (UTC)

Unless this is what a celiometer is, you could shine a laser at the cloud base and time how long the light takes to get back. --The High Fin Sperm Whale 03:57, 22 March 2010 (UTC)

Unfortunately that's exactly how a ceilometer works, so this method is not available in my case. Also, what is a ceiling projector and how is it used? I've got a vague feeling that it might be one of the solutions to this problem. 24.23.197.43 (talk) 05:06, 22 March 2010 (UTC)

Actually just google Cloud Base Calculator, i've just found a bunch. Vespine (talk) 04:23, 22 March 2010 (UTC)

Oh, that's how it's done -- you shine a ceiling projector at the cloud base and measure the bright spot's angle of elevation with an alidade or theodolite, and then take the tangent of that angle to get the cloud base height. Well, thanks anyway :-) 24.23.197.43 (talk) 05:13, 22 March 2010 (UTC)

Sort of like the way the Dambusters measured the altitude of the bombers above the lake water during Operation Chastise, only pointing the other way. --Anonymous, 05:54 UTC, March 22, 2010.—Preceding unsigned comment added by 70.48.228.240 (talk) 05:56, 22 March 2010

Exactly. 24.23.197.43 (talk) 06:03, 22 March 2010 (UTC)

Resolved

-Pete5x5 (talk) 15:54, 22 March 2010 (UTC)

Resolved? Nobody mentioned base reflectivity - which is a standard data product output from a NEXRAD or a WSR RADAR! We have Composite reflectivity, which explains more detail. Note that there may be a cone of silence (i.e., the airfield may be so close to the RADAR that the RADAR can't image the cloud base at such short range), so in practice, a combination of observations are used. Nimur (talk) 16:23, 22 March 2010 (UTC)

Number of conjugated double bonds and photon energy absorbed

The wiki page says, "With every double bond added, the system absorbs photons of longer wavelength (and lower energy), and the compound ranges from yellow to red in color. Compounds that are blue or green typically do not rely on conjugated double bonds alone."

I thought it should be the opposite, because the more double bonds we have in a conjugated system, the more overlapping p-orbitals there are. So the system can absorb or bare with a greater amount of energy (The electrons have much more space to travel around when they are excited, so the system can take in more energy). It doesn't make an intuitive sense as to why a conjugated system with more double bonds absorbs photon of lower energy. —Preceding unsigned comment added by 70.68.120.162 (talk) 04:34, 22 March 2010 (UTC)

Because it takes less energy to get those conjugated electrons racing around back and forth? 24.23.197.43 (talk) 05:16, 22 March 2010 (UTC)

More space for the electrons means higher uncertainty about their positions which allows for lower uncertainty about their momentums momenta which permits lower excited states which allows absorptions of lower energy photons. Dauto (talk) 05:40, 22 March 2010 (UTC)

Right. Personally, I prefer to visualize this a bit differently, though. Imagine you have a potential well with some discrete energy levels. If you connect two wells like that, each level will split into sub-levels. The more wells you connect, the more components the original levels split into, and the broader the absorption line (band) becomes. Benzene is transparent (absorbs UV only), tetracene is orange (absorbs UV and green/blue), and graphite & amorphous carbon absorb UV and any wavelength in the visible range. The two explanations (Dauto's and this one) are not contradictory, of course. Hope this helps. --Dr Dima (talk) 07:34, 22 March 2010 (UTC)

I'm really confused. "More space for the electrons means higher uncertainty about their positions" Okay. "...which allows for lower uncertainty about their momentums" I don't understand this. "...which permits lower excited states which allows absorptions of lower energy photons." Oh my goodness. It doesn't make any sense at all.

"Imagine you have a potential well with some discrete energy levels. If you connect two wells like that, each level will split into sub-levels." Why? "The more wells you connect, the more components the original levels split into, and the broader the absorption line (band) becomes." I just don't see it intuitively. "tetracene is orange (absorbs UV and green/blue)" Why is tetracene orange if it absorbs green/blue? Shouldn't its color be some mixture of colors that are not green/blue, since objects' color is determined by the colors they don't absorb? For example, if something does not absorb only green, then its color will be green".142.58.129.94 (talk) 17:25, 22 March 2010 (UTC)

From the point where your confusion sets in, it seems that you have too little knowledge of quantum mechanics to understand the full explanation (which is no shame, but a bit more than can be remedied in a refdesk exchange). So here is a non-quantum half-story: think of the sequence of conjugated double bonds as a conductor that works like an antenna to pick up electomagnetic radiation. Now, the longer an antenna is, the longer is the wavelength it is tuned to. So more bonds means a longer antenna, which means that the antenna can pick up longer waves. (In neither case does the antenna reach an entire half-wave of visible light, but that is where quantum stuff comes in).

And yes, orange is "a mixture of colors that are not green/blue/UV". –Henning Makholm (talk) 17:45, 22 March 2010 (UTC)

It still doesn't make intuitive sense, but I'll just take what you said for granted. I'm concerned that I have "too little knowledge of quantum mechanics". I've taken college intro physics, chem, calculus, bio and organic chem. Are intro physics/chem supposed to equip you with enough knowledge of quantum mechanics? If not, what can I do now to get such knowledge? Taking an upper level course is not an option since I don't want to risk my GPA. —Preceding unsigned comment added by 142.58.129.94 (talk) 18:42, 22 March 2010 (UTC)

Conjugation/resonance should have been covered in orgo (electrons delocalizing along the whole system rather than being in specific alkene-locations). The specifics of UV/vis is often not. But the relationship of wavelength to energy should have been covered in physics (longer wavelength is lower energy). So now you have electrons resonating in a longer space, which is analogous to a wave, and therefore lower energy. DMacks (talk) 18:50, 22 March 2010 (UTC)

Sure you've heard of Heisenberg's uncertainty principle that states that the higher the uncertainty in the position is the lower the uncertaity in is momentum can be, haven't you? Dauto (talk) 18:52, 22 March 2010 (UTC)

If you really want to understand thaose things you really need to take those advanced courses. There is no royal road to science. Dauto (talk) 18:52, 22 March 2010 (UTC)

Radiocarbon dating

Near my home in California a fair number of old pictograms are to be found. One set of them is rather controversial at the moment, with one faction claiming the drawings are of considerable antiquity (one to two thousand years), while another faction claims they were drawn by local kids in the 1930s. Radiocarbon dating has been discussed; how would that work with stuff painted on the boulders? What would they test for decay? --jpgordon^{::==( o )} 05:46, 22 March 2010 (UTC)

They would test the material used to make the pictograms. With powdered plant pigments or charcoal paintings, radiocarbon dating may work (depending on human/environmental contamination). If they are chalk scrawlings, then radiocarbon would be next to useless. caknuck _° needs to be running more often 07:27, 22 March 2010 (UTC)

Looks like red ochre to me. --jpgordon^{::==( o )} 17:17, 22 March 2010 (UTC)

Sabbath feeling

Is there a name for it when you somehow got the feeling that you shouldn't do any (specific kind of?) work? Quite unreligiously I get this feeling whenever I have to do my tax declaration, even when I'm sure I'll get money back. I can see it sometimes in other people, too. In some fantasy/fiction stories there are alien races with a limited supply of "vital energy" and every time they make a decision, it decreases.

I'm sure there is a name for that, perhaps an article. 95.115.136.225 (talk) 09:33, 22 March 2010 (UTC)

Procrastination? (This really belongs on the language desk - it's hardly a scientific matter - maybe I'll move it...tomorrow). SteveBaker (talk) 10:28, 22 March 2010 (UTC)

Religious guilt? 67.243.7.245 (talk) 13:14, 22 March 2010 (UTC)

I used to get that, but it stopped when I started dressing more apropriately for the cold weather, that lack of energy and feeling it gradually draining away. Or is it more like the way I feel when I've been working almost non stop all day for a week or two. I call that work tireness, though I am sure that isn't the official name, if there is one.148.197.114.158 (talk) 15:25, 22 March 2010 (UTC)

Work aversion ? Gandalf61 (talk) 15:53, 22 March 2010 (UTC)

Theory of relativity

Would somebody kindly explain Theory of Relativity (and related concepts) in simple layman's terms? I have been struggling to grasp the concepts for some time now. Everything I can find is written so complexly that I understand next to nothing. Also, something about String theory would be useful, I have had mentioned that in conversation few times, and it's been sufficiently awkward. 203.206.49.48 (talk) 15:59, 22 March 2010 (UTC)

Have you looked at Theory of relativity? It is hard to explain this physical theory in simpler terms than the article uses. What specific concept(s) are you struggling with? Nimur (talk) 16:26, 22 March 2010 (UTC)

You should take one step at a time. If you're struggling with relativity, than you should concentrate on that and leave string theory for later. Relativity is vast subject. You probabily should start by reading time dilation, length contraction, and relativity of simultaneity. Let us know if you find those articles hard to read. Dauto (talk) 16:29, 22 March 2010 (UTC)

If you're interested in the whole gamut of relativity and string theory, I'd recommend getting one of the many popular books about it. Brian Greene's The Elegant Universe (1999) is well-known for its accessibility and readability in explaining relativity, quantum theory, and string theory. It's not totally up to date anymore, but at the level of your interest it is probably one of the better launching points. I find Wikipedia is a little spotty on things like this—its topical coverage swings between the exceedingly general and the exceedingly technical. --Mr.98 (talk) 17:05, 22 March 2010 (UTC)

Out of the articles cited above you probabily should start by making sure you understan the section time dilation#Simple inference of time dilation due to relative velocity. Dauto (talk) 18:33, 22 March 2010 (UTC)

Angular frequency in x-ray crystallography

I've already read over the wiki page for angular frequency, but how is angular frequency different from normal frequency and what does it mean in terms of photons when used in x-ray crystallography? How is sin(2pie*x) different from sin(x)? —Preceding unsigned comment added by 142.58.129.94 (talk) 16:25, 22 March 2010 (UTC)

It is merely a matter of units - are the angles being measured in terms of radians or cycles? The conversion is simply a multiplication by a constant scale factor; the difference can be summarized by saying that angular frequency yields fewer messy terms in the algebra; while frequency in cycles (or hertz) is more intuitive and works better for engineers or scientists who are reading off results from a machine calibrated that way. Nimur (talk) 16:29, 22 March 2010 (UTC)

Emission (Fluorescence) in Polar and hydrophobic environments

Emission energy decreases (i.e. emission wavelength increases) in polar environment ("red-shift"), and it's the opposite in hydrophobic environment ("blue-shift"). Is that because, in polar environment, the molecules stabilize the excited molecule and help dissipate some of its excited energy through non-radiative means, thus decreasing the amount of energy emitted? —Preceding unsigned comment added by 142.58.129.94 (talk) 16:26, 22 March 2010 (UTC)

I haven’t had time to read these two through but they may have some answers.Vibrational transitionMolecular electronic transition. Just curious: Why are you asking?--Aspro (talk) 18:50, 22 March 2010 (UTC)

Note: This is presumably a continuation of Wikipedia:Reference desk/Archives/Science/2010 March 11#Fluorescence in polar and nonpolar environments. DMacks (talk) 18:52, 22 March 2010 (UTC)

Thanks for that Dmacks. I am out of my depth on this. However, page 21 of this pdf [24] talks of “Energy loss and quenching.”The Wikipedia articles to look at, are I suppose: Quenching (fluorescence) and a radiationless mechanism known as Förster resonance energy transfer & Vibrational transition & Molecular electronic transition. However, there must be a simpler way of explaining it but I suppose the answer looks if its going to be ‘yes’ and the pdf may make the bumbling prof’s mumblings more intelligible. . . or on the other-hand it might not.--Aspro (talk) 19:59, 22 March 2010 (UTC)

Is E2 (17beta-estradiol) a dioxin?

Is 17 beta-estradiol a form of dioxin? —Preceding unsigned comment added by ValkyrieKnight (talk • contribs) 20:29, 22 March 2010 (UTC)