Wikipedia:Reference desk/Archives/Science/2007 November 3

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November 3[edit]

How to disappear completely[edit]

Teleportation: "Until recently, scientists had been able to transport only light or single atoms over short distances (millimeters)."  !?! In teleportation our original copies don't have to die after all: there will be no pesky original copies of us to trouble over. And here I was worrying about a convenient future that might involve the existential murder trial for killing myself just to make it to Sydney in the blink of an eye. Joking aside, how does any sort of teleportation of matter not tear to shreds the Heisenberg uncertainty principle? —Preceding unsigned comment added by Sappysap (talkcontribs) 01:39, 3 November 2007 (UTC)

There's something called quantum teleportation that's specifically about getting around the uncertainty principle in situations like this. But that's kind of a misnomer, since you can't use it to help you teleport something until you've solved all the problems attendant on classical teleportation, like the whole determining-the-complete-structure-of-the-object-and-then-reassembling-it-on-the-other-side part. I see no reason to believe that this kind of technology will ever exist. -- BenRG 02:23, 3 November 2007 (UTC)
I love that song.--The Fat Man Who Never Came Back 02:28, 3 November 2007 (UTC)
  • The movie The Prestige addresses this existential clone murder in fascinating fashion. --Sean 19:15, 3 November 2007 (UTC)

Artery defurring[edit]

Is it possible over a period of time to defurr your arteries by eating zero fat? —Preceding unsigned comment added by (talk) 01:51, 3 November 2007 (UTC)

Defurr? Dureo 08:44, 3 November 2007 (UTC)
The technical name for "furring" of the arteries, also known colloquially as "hardening" of the arteries, is atherosclerosis.
The research I am aware of pertaining to your question is the Lifestyle Heart Trial of Dr. Dean Ornish. His research was widely popularized in his best-selling book Dr. Dean Ornish's Program for Reversing Heart Disease.
We can’t give any advice pertaining to using diet and lifestyle to prevent or treat atherosclerosis, so consult with your doctor as to whether a change in diet and lifestyle may be appropriate for you. You may also find it valuable to consult with a registered dietitian. MrRedact 17:30, 3 November 2007 (UTC)


What causes it? — Daniel 03:39, 3 November 2007 (UTC)

I'm assuming it's because the products will be at a more favorable thermodynamic state than the reactants, but I could be wrong. Specifically what substance(s) are you wondering about? shoy (words words) 03:52, 3 November 2007 (UTC)
Thermodynamically, you're simply looking for the state with the lowest free energy, and a substance that dissociates will have a lower free energy of its separated components than if those components were together. This depends on the nature of the solvent, and I'm gonna be careful here from the mistake I made before and say that the textbook statistical mechanics analysis of solutions won't necessarily apply if tthe solute dissociates. From chemistry class though, I remember that conceptually when a compound dissolves in water, it will dissociate if the pull of the polarized water molecules is greater than that binding polarized components of the compound. So in an ionic compound like NaCl, the Na+ ions are attracted to the O-2 side of the water molecules while the Cl- ions are attracted to the H+ side of the water molecules, and this attraction breaks apart the bonds of the salt crystal. SamuelRiv 06:34, 3 November 2007 (UTC)
DOn't forget Entropy! Dissociating anything results in two or more molecules/ions from one --> more disorder --
This isn't strictly true. While dissociation might seem to imply the creation of more molecules, one needs to be careful here. For example, the dissociation of water is often represented at H2O <-> H+ + OH-. But what is really going on is 2 H2O <-> H3O+ + OH-, so no new molecule is actually created. shoy (words words) 18:03, 3 November 2007 (UTC)
Entropy factors into the rate equations. That's why rate is proportional to reactant concentration raised to the power of the number of molecules needed per unit of product. SamuelRiv 14:36, 3 November 2007 (UTC)

use of superbugs in oil slicks[edit]

hello i would like to know about the process involved in use of superbugs in oil slicks. ie how exactly the superbugs help in getting rid of oil slicks & the pros & cons of it.i would also like to know a few strain names of the superbugs as examples. i'm aware of the diamond vs. chakraborty case. i'm asking this ques as r dear teacher hasnt taught this topic properly & cudnt find enough info to write a 10 mk. ans in exams !!! so pl. help me out. 07:05, 3 November 2007 (UTC)

I believe the Bioremediation, Biodegradation, Microbial_biodegradation, and Biotransformation articles will help you. Dureo 08:42, 3 November 2007 (UTC)
The basic concept is that they would digest the toxic components into waste products which aren't toxic. StuRat 14:20, 3 November 2007 (UTC)

The night sky, stars, and galaxies[edit]

Hi, I have sort of a random question just out of curiosity. It may be a dumb question but I have never heard it explained. When one simply looks up into the night sky and sees all of the different stars, are they individual stars like our Sun? Or are some of them that we observe stars and others actually galaxies? And if they are mostly individual stars, are they all located within our own Milky Way galaxy? Again I apologize if this seems elementary, but I looked in both the Star, Galaxy, and Night Sky articles and was not quite sure of what the answer is. Thanks in advance. Zenislev 08:19, 3 November 2007 (UTC)

Some are individual like our sun, and some are further away and appear individual, but are actually galaxies and star clusters, the Universe article may help also. Dureo 08:29, 3 November 2007 (UTC)
Most are stars as in the sun, but actually most are much brighter than our sun. Few Galaxies are visible to the eye, and they appear as fuzzy patches. Quite a few stars are double stars. LMC. Graeme Bartlett 08:36, 3 November 2007 (UTC)
You also see the occasional planet.--Shantavira|feed me 09:00, 3 November 2007 (UTC)
I've heard a number quoted a lot that approximately 2/3 of the stars that can be seen in the night sky are actually binary (or higher) stars --Shniken1 10:14, 3 November 2007 (UTC)

Besides stars, clusters, and planets, there's one more possibility: while a comet usually has a tail or at least a fuzzy appearance, it can occasionally look like a star to the naked eye. I saw one like that a few days ago.

The answer to "And if they are mostly individual stars, are they all located within our own Milky Way galaxy?" is normally yes. The only way that a star located outside the Milky Way could be bright enough to see would be if it was going supernova, which happens very rarely indeed. --Anonymous, 10:37 UTC, November 3.

The most distant object visible with the naked eye (and only in the most favourable conditions) is the Triangulum Galaxy. Gandalf61 11:23, 3 November 2007 (UTC)

What's even more interesting is that if there are trillions upon trillions of stars in the universe, why is the sky is so dark at night? -- MacAddct  1984 (talk &#149; contribs) 13:55, 3 November 2007 (UTC)

Two reasons:
  • There aren't an infinite number of stars.
  • The light from distant stars is absorbed by dust or spread out so thinly that not even one photon hits our eyes or telescopes.
The sky does appear much brighter through a good telescope, however. StuRat 14:17, 3 November 2007 (UTC)
Well, yes - but that's kinda meaningless in this context. It's because you're gathering light through a large aperture and squishing it down into a little eyepiece - it's not because you are somehow seeing more photons per steradian of sky or anything of that sort. The deal is that the amount of light from a distant source decreases as the square of the distance - but the number of stars in a spherical volume of space averages out to the cube of the radius. So, were it not for the interstellar dust clouds, the amount of light coming from stars at greater and greater ranges would be greater and greater and the night sky would be insanely bright instead of mostly black (and we'd be cooked by the infrared, hard X-rays, etc. However, the amount of dust obscuring the light from those stars is also proportional to the distance - but the rate of light absorbtion is more than linear. eg If one lightyear of dust absorbed half the light from a star, then the next light year of dust would absorb half of what came through the first lightyear of dust, and so on. Hence the amount of light you see through some amount of dust is an exponential thing. When you work it all out, it's clear that even if there were infinite numbers of stars, the amount of light we'd see would be what we see now. SteveBaker 16:16, 3 November 2007 (UTC)
Those are not the reasons the sky is dark at night. The reasons are (as the Olbers' paradox article says) cosmological redshift and the finite age of the universe. Absorption by dust has no net effect because all of the absorbed energy is reradiated. -- BenRG 16:44, 3 November 2007 (UTC)
It's also possible to see satellites and (especially) the International Space Station - but they move pretty fast so you can generally tell that they aren't stars. SteveBaker 16:16, 3 November 2007 (UTC)
Yes, and the observable universe is only about 15 light-years in radius, so eventually you'll run out of bright objects altogether as you go past. It's much easier to see a star than another galaxy if you look anywhere in the sky. However, the central disk of the milky way is made of millions of stars, which is why, even being about 10,000 light-years away, it is still possible to see it in a dark area. I've seen the milky way through binoculars, although not very clearly, from a light-polluted location. Comet 17P/Holmes, by the wat, is currently just west of the Milky way. Hope this helps. Thanks. ~AH1(TCU) 17:08, 3 November 2007 (UTC)
Huh? The observable universe has comoving radius about 46.5 billion light years. Algebraist 17:41, 3 November 2007 (UTC)
Oops, minor error, I meant 15 billion light-years. That's assuming you can measure the distance from Earth to the edge of the universe without the universe expanding while you measure (eg. the time for the measuring "ruler" to reach the observable edge of the universe from the Earth is 0). This is based on the fact that the universe is about that old and that the farthest observed galaxies are about 17 billion light-years away. Hope this helps. Thanks. ~AH1(TCU) 18:02, 3 November 2007 (UTC)
You're wrong here: it's not the distance (which is closer to 14 billion (more precisely 13.7) light years according to the latest data) measured by an instantaneous ruler but the light travel time times the speed of light. The comoving distance is more like what you mean, and it's larger, as Algebraist pointed out. Icek 01:55, 4 November 2007 (UTC)

Thanks for all of the helpful responses! That sure is enlightening (I'm sort of unlearned in the field of cosmology). It's very interesting to think that all of those stars we see are mostly just from our own galaxy and that there are billions of other galaxies out there. Well have a good one guys and thanks again ;) Zenislev 19:50, 3 November 2007 (UTC)

If you live in an urban environment, you might only be able to see a thousand stars with the naked eye, all of which are in the nearby parts of the Milky Way galaxy. Fainter, or more distant stars, and the other galaxies would require at least binoculars or the very dark skies you get far away from cities. When I first saw the nearest bright galaxy (Andromeda, M31) through a pair of binoculars, I was surprised how faint it really was - a small fuzzy patch, especially compared to those great photos you see. Astronaut 03:41, 4 November 2007 (UTC)
Zenislev, if at all possible, grab a pair of binoculars and look at the Andromeda galaxy! It will look like a luminous smudge, but it's quite cool to be able to see another galaxy. And in fact, the Andromeda galaxy is *huge* on our night sky... the problem is that most of it is relatively dim. You can only see the brightest, central part. If I remember right, the whole thing is as big as about 7 full moons! So the difficulty is not magnifying something that appears to small - it's enormous! Instead, the hard part is collecting enough light to see something that's relatively dim. Then point your binoculars at Jupiter and look for the Galilean moons. Even with small binoculars you might see 2 or 3 of them! --Reuben 04:36, 4 November 2007 (UTC)
I do indeed live in the city, so I know what you mean there. I did however bring binoculars with me on a camping trip a few months ago and was amazed at how much I could see; I really didn't know what to make of any of it though. That does seem interesting to be able to see Andromeda, I never knew you could. I'll have to try that when I get a chance. It's hard to comprehend the scale of things in our universe though; to think that this blob in the sky really spans thousands of light-years is really thought-provoking. Zenislev 05:30, 4 November 2007 (UTC) —Preceding unsigned comment added by Zenislev (talkcontribs)

motor tracts from the cortex[edit]

Is the diagram below correct?

motor tracts from the cortex

lateral system..................................medial system

lateral corticospinal tract.....................anterior/medial corticospinal tract
.........................................................corticobulbar tract

rubrospinal tract..................................tectospinal tract
lateral reticulospinal tract.....................medial reticulospinal tract
lateral vestibulospinal tract...................medial vestibulospinal tract

(Please don't be bothered with the "...", they're just there because spaces don't work.)

Second question: Where in this diagram does Corticobulbar tract go?

Thank you! Lova Falk 11:01, 3 November 2007 (UTC)

It seems correct after I cross-referenced, but it's been a while since neuroanatomy. Try google images on the coritcobulbar tract, and you will see cross-sections of it through the midbrain, pons, and medulla that should give it away. SamuelRiv 15:03, 3 November 2007 (UTC)
According to the textbook "Principles of Physiology" by Berne and Levy there are both lateral and medial vestibulospinal tracts. Other textbooks just list the vestibulospinal tract as medial. Berne and Levy say, "Much of the corticobulbar tract belongs to the medial system". Course and distribution of facial corticobulbar tract fibres in the lower brain stem goes into some of the details. --JWSchmidt 16:25, 3 November 2007 (UTC)
Thank you! I changed the diagram by putting the lateral and the medial vestibulospinal and reticulospinal tracts into the diagram. Am I correct to put the corticobulbar tract with the pyramidal tracts? Lova Falk 18:31, 3 November 2007 (UTC)
Yes, the corticobulbar tract is counted as part of the pyramidal system....functionally it has axons that carry signals for good voluntary motor control.....they just carry signals that control motor neurons located in cranial nerve brain nuclei rather than motor neurons located in the spinal cord. --JWSchmidt 22:25, 3 November 2007 (UTC)

FYI, Lova, the "correct" way to make columns is with a table – hit the rightmost button above the edit window to start one. —Tamfang 09:21, 7 November 2007 (UTC)

lateral system medial system
pyramidal pyramidal
lateral corticospinal tract anterior/medial corticospinal tract
corticobulbar tract
extrapyramidal extrapyramidal
rubrospinal tract tectospinal tract
lateral reticulospinal tract medial reticulospinal tract
lateral vestibulospinal tract medial vestibulospinal tract

Thank you all for answering! I will also have good use of the table. Lova Falk 19:11, 7 November 2007 (UTC)


Is it true that some women take cocaine gynaecologically or is it just a myth? Everyone knows someone called Dave 12:11, 3 November 2007 (UTC)

Can you rephrase the question? I'm not sure if you mean illegally by some type of insertion, or prescribed by an OBGYN, or um something else. Maybe I am not thinking outside the box. Dureo 12:24, 3 November 2007 (UTC)
If you're asking about routes of exposure:
"[Cocaine] in this hydrochloride salt form may be injected; swallowed; applied to oral, vaginal, or even rectal mucous membranes; or mixed with liquor. Coke is most commonly used by snorting or sniffing." [1] (emphasis added.)
Hope that helps. TenOfAllTrades(talk) 13:40, 3 November 2007 (UTC)

Atomic movement[edit]

Can atomic movement of an atom be stopped an if so what would be the result? I am wondering if atomic movement if of a particular fequency, could you apply the same frequency of opposite phase to cancel it out and if so what would the effects of this be? If you can accelerate particles by superheating them, would stopping the movement have a cooling effect? —Preceding unsigned comment added by (talk) 13:50, 3 November 2007 (UTC)

Yes. At absolute zero, there is absolutely zero (har har) movement. Although it's not possible to replicate in a laboratory (although they've gotten very close), some scientists believe that in the far reaches of space the temperature exists. -- MacAddct  1984 (talk &#149; contribs) 13:58, 3 November 2007 (UTC)
I see that in discussing Bose–Einstein condensate, the absolute zero article mentions "pK" (picokelvin). How does someone measure a temperature that low? --JWSchmidt 15:01, 3 November 2007 (UTC)
At absolute zero there is still movement predicted by quantum mechanics and Quantum Field Theory (QFT). A simple way to see this is by noting Heisenberg's Uncertainty Principle which shows that a zero change in position requires infinite momentum, and vice versa. Note that this is a mathematical fact of quantum theory and not some philosophical abstraction. Empty space in the current universe is at about 3K and does not approach zero as one goes to the "far reaches of space" because there simply is no far reach of space as expansion is uniform. As one goes farther into the future, however, the temperature of the universe is falling as the expansion of the universe accelerates.
Extremely low temperatures are achieved first by evaporating liquid helium (liquefaction of helium being achieved by the Joule-Thomson effect) and then by isentropic demagnetization of the subatomic particles (so demagnetizing particles without changing entropy). They are measured by reversing the effect: measuring the magnetization of subatomic particles and measuring the vapor pressure of liquid helium. This is only effective to a scale of millikelvins, below which I'm not sure what they use. Note the validity of these methods of course depends on having a very good theory of the behavior of particles at low temperatures, which we do (QFT is the most numerically accurate theory known).
On a side note, temperature is defined as if the number of particles in the system doesn't change. That is, the inverse of temperature is the change in entropy over the change in energy. This definition, and indeed statistical mechanics itself, sets no restrictions on how high or low temperature can be. So we can have systems with Negative temperature! SamuelRiv 15:31, 3 November 2007 (UTC)
Oh, and physically slowing the motion of atoms does cool them: that's how liquefaction of gases is done. See cryogenics and the Joule-Thomson effect. Basically, gas is expanded through a valve which causes energy to be absorbed against the intermolecular forces of the gas, which results in a net temperature decrease as long as the initial temperature of the gas is below the inversion temperature. Again, the motion of the atoms can never be stopped completely. SamuelRiv 19:34, 3 November 2007 (UTC)

Glue board adhesive composition[edit]

I'd like to make my own glue boards, similar to those for catching mice, although mine will be used for insects. The hard part is the adhesive. What ingredients can I use to make my own ? I found this list of ingredients for a mouse glue board, but the portions aren't given and many terms are vague (which stabilizers or hydrocarbons ?):

  • Styrene copolymer
  • Paraffin oil
  • Stabilizers < 1%
  • Anti-oxidants < 0.5%

I'd like to make this from items I can buy at a grocery store, if possible. Dish washing detergent works, but stays liquid for weeks, until all the water evaporates. Could I make it immediately solid by adding gelatin or something else ? Could I heat it to drive off the excess water ? StuRat 14:07, 3 November 2007 (UTC)

Dishwashing liquid is about 95% water so it'll take a long time to dry. Try dissolving salt in it to make it get thick. Car wash detergent may be more concentrated. You can try the salt trick with that too. Alternatively citric acid can be used as a thickener. You'll notice a lot of shampoos contain a salt, which is just there to make it thick.
Pressure sentitive adhesives contain a polymer for strength and a tackifier to make it sticky. It is a synergistic combination: the mixture is "stickier" than the two components. For your mouse-catcher, the styrene copolymer is probably a styrene-butadiene-styrene block copolymer like Kraton. The hydrocarbon resin is the tackifier and is probably a derivative of pine tree resin. Delmlsfan 14:58, 3 November 2007 (UTC)
I guess you could try boiling the washing up liquid to drive the water out more quickly - but I have no idea on the consequences of doing that! SteveBaker 15:48, 3 November 2007 (UTC)
Just buy some flypaper. -Arch dude 16:54, 3 November 2007 (UTC)
I have tried flypaper. I found it unacceptable because it's ugly and the glue gets all over everything, including my hands. Also, I want to get all insects, not just flying insects. Spiders and centipedes need to die, too. I've tried to lay flypaper out on a flat surface, but the glue got all over everything in the process. But, that is the idea, I want one-sided flypaper laid down on a flat sheet, say the size of a newspaper. I want to do so on a large scale for a low price. StuRat 17:06, 3 November 2007 (UTC)
(Spiders and centipedes are not insects). SteveBaker 22:58, 3 November 2007 (UTC)
Ok, let me be more scientific and say that I want to kill all "creepy-crawlers" which drag their asses across my floor. :-) StuRat 00:14, 6 November 2007 (UTC)

Do a search with "make flypaper" and you can get recipes using sugar, golden syrup etc. The advantage of these edible glues is that they wash off very easily in water. Keep in mind some invertabrates wont walk out in the open. I accidentally caught silverfish with a roll of gaffer tape that was a little unrolled, they had wedged into the sticky crevice. IF you are after a more specific result, let us know. Polypipe Wrangler 22:57, 4 November 2007 (UTC)

I'm a bit worried that if edible glue isn't sticky enough, I will end up feeding bugs instead of killing them. StuRat 23:31, 9 November 2007 (UTC)

X-inactivation and color blindness[edit]

Several forms of color blindness are far more prevalent in male humans because they are caused by a defective gene on the X chromosome of which males only posses 1. If you look at the prevalence for males and females in out article, the numbers seem to agree with this theory (1% vs. 0.01%, 6% vs. 0.4%). However, in females one X chromosome is inactivated; this inactivation happens early in embryonic development, at the 70-100 cell state (at least in the mouse, according to the Wikipedia article) and is apparently random. There are several genes which are nonetheless expressed on an "inactivated" X chromosome, but these are mostly genes which are also found on the Y chromosome (pseudoautosomal region).

I don't know how many of the 70-100 cells which all individually undergo X-inactivation end up as the relevant cells in the retina, but even if it's more than one, one would expect that there are cases of female "patchy" colorblindness, i. e. affecting only part of the visual field. This condition then should be even more prevalent than colorblindness in males: If the fraction of X chromosomes with the color-blind property is r, then (assuming random distribution) the fraction of males carrying such X chromosomes is r, and the number of females carrying at least one such X chromosome is 2*r - r2.

On the other hand, if all relevant cone cells originate from only one embryonic cell at the time of X-inactivation, the probability that a female is colorblind is 1/2 if she carries 1 "color-blind" X-chromosome and 1 if she carries 2 "color-blind" X-chromosomes; together that makes for the rate of color-blind females the same as the male rate.

How do you explain the observed approx. r2 rates in females? Icek 17:53, 3 November 2007 (UTC)

It turns out that not all genes from the so-called 'inactive' X chromosome (Xi) are actually fully silenced (see X-inactivation#Expressed genes on the inactive X chromosome). Of those, a few are transcribed at the same level as on the active X chromosome (Xa); a very few are even transcribed at higher levels from Xi than from Xa. (The obvious example for this last case would be the RNA gene Xist, which effects the X-inactivation.)
I don't know if this is the case for some of the genes related to colour vision, however; I'm still digging in the literature. TenOfAllTrades(talk) 18:52, 3 November 2007 (UTC)
Some further interesting reading from PNAS. A survey of genes on the X chromosome found that as many as twenty to thirty percent(!) of genes on the short arm of the X chromosome (Xp) may escape inactivation at least some of the time. TenOfAllTrades(talk) 19:13, 3 November 2007 (UTC)
Thanks for the answer. From the supplementary data of an Cell article drawing the data from this Nature article it appears (see figure 8, last page of the supplementary data) that the relevant genes for red-green blindness at locus Xq28, OPN1LW (for the long-wavelength (red) receptor) and OPN1MW (medium-wavelength (green) receptor) are not at all expressed from the 'inactivated' X chromosome. Of course, the study was made with fibroblasts and the situation may be different in cone cells. Icek 01:01, 4 November 2007 (UTC)
FYI, here is the OMIM page for OPN1MW and here for OPN1LW. Icek 01:05, 4 November 2007 (UTC)
Thinking again, it's likely that opsin (the product of the aforementioned genes) may not at all be expressed in the studied fibroblasts; maybe the fact that there is no gray bar at all instead of a gray bar extending only below the zero line for some genes in the supplementary data figure I linked to before means that the expression was zero in the activated as well as in the inactivated X chromosomes. Icek 01:23, 4 November 2007 (UTC)

Detecting nuclear blasts[edit]

A recent news article claims that the US used tactical nukes at Operation Orchard. Seems incredible to me. Is it possible nowadays for an above ground nuclear blast to go undetected? --Duk 18:38, 3 November 2007 (UTC)

Undetected - no I doubt it, unreported - perhaps. Lanfear's Bane | t 20:20, 3 November 2007 (UTC)
Fuel-air bombs can get into the neighborhood of a (very) small nuke, as far as TNT equivalent, so there could be some ambiguity in seismic detection, but there is basically *no* way that the radiation would go undetected, especially if people were looking for it. Also, this government can't even keep it secret that they flew some nukes across the country accidentally or that they like to pour water up some poor bastard's nose; there's no way it would stay quiet if they actually used a nuke. --Sean 20:20, 3 November 2007 (UTC)
I haven't yet found a Wikipedia article specifically about detection of nuclear explosions, but in the articles on the Ryanggang explosion and 2006_North_Korean_nuclear_test it is assumed that any nuclear explosion would be detectable by the characteristic isotopes produced. An explosion of such magnitude as to have been a nuke would register on seismographs all over, and, although we can't know where the satellites are looking, I'd expect one or two to be pointed at the Middle East a lot of the time. Any large explosion in Syria would certainly be detected and probably pinpointed, and it would be a simple matter to test the area, or even downwind, for the telltale isotopes. So, no. --Milkbreath 20:45, 3 November 2007 (UTC)
Seismographs can only measure the size of the explosion - so a small nuke and a very large conventional bomb would look identical. Direct photography of the mushroom cloud or the resulting debris field could probably tell the difference - as could eye-witness testamony...but we don't have that kind of evidence. Radiation is another way - but only people with access to the site could measure that without some pretty high-tech equipment. The people with the high tech equipment aren't admitting to it - so unless someone has actually been out there with a geiger counter, we can't know. HOWEVER: If the US have fuel/air bombs with equivelent destructive ability to a small nuke, why on earth would they risk world condemnation following the use of nuclear weapons when they could use a conventional bomb with little or no comment and get much the same result? It just doesn't seem plausible. SteveBaker 22:48, 3 November 2007 (UTC)
A neutrino detector should be able to single out nuclear explosions, in theory, or at the very least determine the exact time that one occurs. Someguy1221 22:54, 3 November 2007 (UTC)
(Seismographs also show roughly where an explosion took place, and, of course, precisely when.) There are plenty of other signs of a nuclear explosion. They get extremely hot, so there would be melting at ground zero that would be hard to fake, and fires all around. There is an electromagnetic pulse that could be expected to at least have some effect on the power grid. Nukes are very, very, very bright. The hard radiation in its vicinity would show in the people and the animals near it. Sounds to me like Comical Ali has come out of retirement. --Milkbreath 01:19, 4 November 2007 (UTC)
If it was above ground it would produce easily measurable fallout, even if it was very far above ground or most of it was buried. Such conditions would produce less fallout, perhaps not enough to hurt anyone at any distance, but it would certainly be measurable by a variety of stations around the world within a week or so. So the fact that none of the many, many stations in the area with radiation detectors (even a simple reactor will have those) reported anything awry is pretty much good evidence that it was not used, in my book. But of course the most compelling answer is the one you've given — it would be tremendously stupid and wholly unnecessary to use nuclear arms in such a situation. -- 00:18, 5 November 2007 (UTC)
If you read the story carefully you'll see that they never say that it was a nuclear weapon that actually detonated at the site. It is poor reporting in any case and is probably meant to be misinterpreted. I doubt the USAF was involved but even if it was, and even if we take as a reliable claim that the escorting planes would have had tactical nukes on board (highly, highly unlikely that they have such things flying around a theatre like that just for the heck of it, much less two of them), the idea that they would non-chalantly use such weapons is pretty absurd, and the article doesn't even really assert that. The whole thing sounds like misinformation to me. -- 00:18, 5 November 2007 (UTC)

Any clues as to this equation?[edit]

Hi all, does anyone have any idea what this represents? [2]

Many Thanks! --Chachu207 talk to me 19:56, 3 November 2007 (UTC)

Maxwell's equations. Clarityfiend 20:04, 3 November 2007 (UTC)
Thanks! --Chachu207 talk to me 20:07, 3 November 2007 (UTC)

Two locomotives[edit]

How can two locomotives pull the same train together? It seems to me that if one pulls even a tiny bit harder it will be dragging the other one. --Milkbreath 21:14, 3 November 2007 (UTC)

Excellent question. I've wondered about this, too.
First, it's not just two locomotives. Suppose you and a friend are both trying to push a car -- doesn't the same argument apply?
I've thought of two answers, though there may be more.
  1. If there's any flexibility in the system, it can "even out" the two force providers (locomotives or people) if they're unbalanced. For example, think about the various couplings (gears and such) via which the mechanical energy is transmitted from the locomotive's actual engine down to its wheels. Suppose furthermore that there's some "lash" or "spring" in that power transmission chain (as there just about always is, in practice). Now, if one locomotive pulls a little bit harder, it might seem as if the other one wouldn't have any work to do, because the train is being pulled out from under it, so to speak. But if one engine is doing all the work, its transmission chain will be maximally compressed. The other engine (not doing any work) won't have its transmission chain compressed as much; it will begin to untwist (precisely because the stronger engine is in the process of pulling the train out from under the weaker engine). But after an infinitesimal moment, after the weaker engine's transmission chain untwists just a bit, the weaker engine will have something to push on, after all, so it will do some work -- precisely its share -- and all of a sudden the stronger engine won't be pulling the whole train, after all.
  2. The more accurate explanation, I think, is based on the fact that each locomotive is a constant force device. Each engine is doing whatever it can to keep pulling on that train. If one engine pulls harder, threatening to pull the train out from under the other one, suddenly the other one won't be applying as much force any more. So it will do whatever it has to do -- i.e. speed up a bit (which is easy to do when suddenly it's not meeting as much resistance when it tries to pull) until it's pulling its weight again. And the same explanation applies to you and your friend trying to push that car, even if one of you is pushing harder than the other.
Now, with that said, it's still an at least somewhat tricky problem. As I understand it, there had to be some real breakthroughs in electric motor control before two diesel-electric locomotives could each act like constant-force devices and pull a train in harmony. (And as I think about it, there's a second problem: if one engine is more powerful than the other, it ends up having to apply a higher constant force than the other. I don't know precisely what mechanism arranges that the constant forces that the two locomotives try to apply are in proper proportions to the power ratings of the two locomotives. Perhaps I was wrong when I said the locomotive had to be a constant-force device; perhaps the more accurate description is "constant power".) —Steve Summit (talk) 21:54, 3 November 2007 (UTC)
P.S. See our article on Diesel locomotives, particularly the "Adaptation of the Diesel engine for rail use" and "Diesel-electric control" sections, which talk about these control issues. (In particular, it looks like diesel-electric locomotives are indeed constant-power. Steam locomotives, on the other hand, are described as "constant torque", which I think is equivalent here to constant force, so perhaps the arguments for multiple-unit hauling with steam locomotives are different than for diesel.) —scs 22:06, 3 November 2007 (UTC)
The key is that either engine is capable of pulling faster than the entire trainset is moving. So if the #2 engine is pulling a bit harder than the lead, the lead is still pulling at the front of the #2 engine. Its very much like a group of people pulling together on a rope to haul something. But here's an even funkier configuration for a consist: 2 locomotives at the front, and one at the tail end of the train. :-) ArakunemTalk 22:02, 3 November 2007 (UTC)
A simple mechanical contraption will allow two locomotives to combine variable pull forces - see Differential (mechanical device) for an example of such. As for the actual nature of the machinery, I would assume as said before that the locomotives approximate to a good extent a constant-force machine. SamuelRiv 00:41, 4 November 2007 (UTC)

Ah. OK. Minor duh. The slack I was looking for is in the power. As long as the power delivered by the two engines is kept in an overlap range, the rotational speed of the wheels will vary as needed without further control, right? I was thinking originally about the old steam engines with rigid mechanical linkage without considering the steam itself. I guess you'd still have to be extra-careful about your pressure when running two steam engines together. The push-pull configuration sounds tricky, too. I imagine the rear engine would pile up as many cars at the back as it could. Rough on the couplers, I would guess. --Milkbreath 01:52, 4 November 2007 (UTC)

I think people are making this question harder than it is. There's no need for careful power control. As stated earlier, a locomotive provides an essentially constant force (until moving fast enough that friction becomes an issue, that is). If the front locomotive is pulling with 12,000 pounds force and the second one with 20,000 pounds, then the strain on the coupling between them is 12,000 pounds and on the coupling behind the second one it's 32,000 pounds. The train can't tell whether that force is coming from one locomotive at the front, or two or six. And the second locomotive can't tell whether the train is accelerating faster its 20,000 pounds would produce because there's another locomotive, or because it's running downhill. Same thing if there are engines at the rear, only in that case the strain on the couplings near the rear will be compression instead of tension.
So long as the controls allow all the locomotives to produce the necessary level of power, the only possible concerns with the relative contributions of different locomotives are efficiency and whether it's possible to put a dangerous strain on the couplings. Remember, double-headed trains were common on some railways -- triple-headed trains on some -- in steam days when there was no such thing as multiple-unit control; the crews of different locomotives could communicate only by whistle signals. The trains still moved fine.
--Anonymous, 01:18 UTC, November 5, 2007.
Upon reflection, perhaps the simple answer is just that modern locomotives are not constant-velocity devices.
If they were, then yes, the fast one would drag the slow one -- but they're not. —Steve Summit (talk) 02:34, 5 November 2007 (UTC)
The locomotives all have torque regulators so they all run at the same torque. If one goes over torque, it goes underpower, if it is torque it goes over power. If there isn't enough power it will cut off and go free-wheeling.--Dacium 03:12, 5 November 2007 (UTC)
So they all run at the same torque, or the right torque? Suppose I try to pull a train with two locomotives, one having twice the power of the other. The more powerful one would have to apply twice as much torque, I think. —Steve Summit (talk) 13:02, 5 November 2007 (UTC)

Anonymous has it correct; there's no fancy regulator (at least on early diesel-electric locomotives), each locomotive just does what it can to add to the overall tractive effort. In fact, you can break this down to the level of traction motors; each traction motor and its associated wheelset turns independently and there can be imbalances between them as well with relatively little harm. The key to all of this is that a locomotive doesn't resist motion that is externally applied to it so the worst that would happen is one locomotive would "goof" off while the other does all the work. (That is, it's not like multiple synchronous motors fighting each other.) But if the tractive load required exceeds the capabilities of the individual locomotives, then it's a sure bet that they're sharing the load at least somewhat.

Nowadays, individual locos do have some amount of automated wheel anti-slip control within a single locomotive, but I don't know if multiple locomotives share any data to optimize this across multiple locomotives.

Atlant 17:00, 5 November 2007 (UTC)

Computers and adverse health effects[edit]

Hi. I'm not asking for medical advice. I've heard that computers and cause, at least in part, everything from eye problems to headaches to brain tumours. Is at least some of this true? I know people, myself included, who have felt the effects of over-computer use. What causes this? What could possibly cause these if computers aren't responsible? Thanks. ~AH1(TCU) 21:24, 3 November 2007 (UTC)

The only adverse health effects from computers that I've found to have any basis in fact is toner dust from laser printers getting into the air and causing lung problems. That is the reason that toner no longer comes in a bucket that you pour into the printer. As for the brain cancer and eye prblems - this was believed to be a health problem from the monitor. Some believed that the radiation from the cathode ray tube caused brain cancer and sitting too close to the monitor caused eye problems. Neither is based on scientific observation. The computer itself is merely an electronic device (and a low-voltage one at that). The only real health problem it can cause is the side effect of sitting in front of one all day and not getting any exercise. However, you will find people who claim anything and everything causes health problems. -- kainaw 22:32, 3 November 2007 (UTC)
CRTs do emit X-rays, and X-rays can cause cancer; however, I'm pretty sure the level of radiation is well below the background level unless you use a (powered on) CRT as a pillow every night. Repetitive stress injuries are a genuine health problem. As far as I know it's impossible to permanently damage your eyes by staring at a screen, though you can certainly tire them in the short term. Concerns about cancer from low-frequency radiation are groundless. -- BenRG 23:37, 3 November 2007 (UTC)
I agree. The most important thing is to use good posture, to avoid injuring your neck and back (by keeping your head in the same position for looking at the screen) and your wrists (by typing with bad posture). And take breaks every once in a while. —Keenan Pepper 00:39, 4 November 2007 (UTC)
I agree. Even a trip to the kitchen to make tea every hour is enough when it comes to exercise. --Ouro (blah blah) 08:38, 4 November 2007 (UTC)
I've been using computers for 40+ years now. I've suffered some pretty nasty symptoms over the years - but the idea that radiation is a problem has been completely debunked. Eye strain, bad posture and repetitive strain issues are real - but there is no mysterious death-rays causing them! All three symptoms can be greatly reduced by two things:
  • Firstly, find an expert to set up your work-station. Many companies will employ the services of an ergonomics expert to figure out your optimium seating position and set up your work area correctly for your particular body dimensions. This may require chair adjustments, desk height adjustment, possibly a foot-rest, almost always something to raise the height of your monitor. Sadly, some companies wait until you've sustained an injury before they'll do that for you - but you should definitely ask and if they refuse, get that in writing because you're going to need it when you come to apply for workers injury compensation later in life. If you can get an expert - take their advice seriously - and if you use a computer at home - measure how your work setup is arranged and try to set up your home set up identically.
  • Secondly, take a few minutes break every 20 to 30 minutes - get up walk around - stretch and stare off into the distance. In most places there are health & safety at work laws that require your employer to allow this. I use a kitchen timer to remind me to take a break because it's easy to forget.
If you start to get repetitive strain symptoms in your fingers - stiffness and a burning feeling on the backs of your hands - then you absolutely need to do something about it. If you carry on working, then over the next few weeks it will spread to your wrists, then your elbows then to your shoulders. The longer you leave it - the longer it'll take you to recover. I'm a huge fan of split keyboards - the ones I use are made by 'GoldTouch' and they are literally split in two with a ball and socket joint between the two halves that lets you adjust the split and tilt for each hand and then lock it in position.
SteveBaker 04:33, 5 November 2007 (UTC)

Returning to laser printers, they can emit ozone, especially if the recombination catalyst on their ventilation air exhaust has become old and tired. They can also emit styrene vapors from the toner. The fuser units within them can pose a thermal burn hazard if you attempt to service them improperly.

High voltages used within various computer subsystems can also be an electrocution hazard if you attempt to service your own computers. Certain computers will also pose electrical hazards when not properly earthed/grounded.

There is also the hazard of abrading your fist by punching the screen if you spend too much time at certain websites.

Atlant 17:04, 5 November 2007 (UTC)

General relativity: gravity Doppler?[edit]

This is a rather complex question that I ask for my own understanding as it has been bothering me for some time. It is my current understanding that gravity (or changes in gravity) travel at the speed of light. For example, if the sun were to disappear now, earth would continue to orbit its location for another 8 minutes or so, "using up" the gravity that the sun had "sent out" before disappearing and which is still rippling out before earth finally sets off in a straight line.

Now, the hypothetical observations I mention below is meant after compensation for other known gravity effects. Assuming speeds that are significant compared to the speed of light, my question is whether (and if not, why not?) objects moving towards an observer appear heavier and objects moving away appear lighter because one would think the gravity "wave" traveling towards the observer "builds up" or "stretches/thins out" respectively. What I mean by this can be thought of as a type of gravity Doppler effect. So according to my hypothetical model, in an expanding universe the gravity experienced by an observer from other bodies would generally be reduced. As far as I know no such effect has ever been observed; my question is where the flaw lies is this reasoning given the laws of relativity.

Note that I am not referring to the inverse square law of gravity which has to do with relative position rather than relative motion. I am also not referring to the law of always increased observed mass due to motion/relativity in both away and towards cases. 22:35, 3 November 2007 (UTC) Eon Zuurmond

I think I see what you're getting at, but allow me to give you a simple but complete example. Imagine you have two perfectly elastic (and very heavy) superballs out in the depths of space away from everything else. They can collide off of eachother with no loss of energy. Now, say you start them off hurtling towards eachother. They meet, collide, and take off in the opposite directions that they came, and you would like to say they will continue like this for an eternity, since no energy is lost anywhere. But then recall that changes in a gravitational field are transmitted at the speed of light, not instantaneously. As the balls are approaching eathother, each ball "sees" gravity the other sent out fractions of a second earlier, which is weaker than it should be. So the gravitational force bringing the balls together is weaker than one would calculate classically. There's no crazy physics to deduce this, just the simple fact that gravity travels at the speed of light. So, while the balls approach, they don't convert all of their gravitational potential energy into kinetic, some is just magically (so it would seem at first) lost. And when they collide and take off, each "sees" the other's gravity as stronger than it should be. So not all kinetic energy is converted into potential, again, some is just lost. So it's a bit opposite than you suspected. Approaching objects seem lighter than they really are, and receding objects seem heavier. Now, you might ask, where did all that energy go? Gravitational waves. Someguy1221 23:04, 3 November 2007 (UTC)
I think both of you are confusing gravitational fields with gravitational radiation (waves). Gravitational radiation propagates outward from the source; a gravitational field does not. Say you have an object resting on a table and you push on one end of it, causing it to shift to a new location. The whole object doesn't move immediately; the change in position is transmitted from molecule to molecule outward from the point you pushed at a finite speed (the speed of sound in the solid). Eventually the object stabilizes in its new position. This isn't a very close analogy, but if you think of the gravitational field as the solid and gravitational waves as sound waves in it, you broadly have the right idea. An object's gravitational field is not emitted by it, it's just there. Where did it come from? It didn't come from anywhere; it's always been there. This is why conservation of energy is absolutely necessary for a relativistic field theory of gravity to work at all. If mass could just appear or disappear, it would have to take its field along with it, but that would violate locality.
To answer your (original poster's) question more directly, the gravitational field at a given distance in front of a uniformly moving object has the same strength as the gravitational field at the same distance behind. The headlight effect applies to waves, not fields. Again, the field behaves something like a solid structure attached to its source; when the source is moving, the field is Lorentz contracted. It doesn't lag behind nor get concentrated in front.
I think you're creating confusion for yourself by asking whether objects would "appear heavier" in certain situations. This is a meaningless notion unless you can give it an operational definition. It almost never makes sense to ask about "apparent" properties of objects, unless you're literally referring to what the object looks like, i.e. to the nature of the electromagnetic radiation which is emitted by the object and subsequently detected by your experimental apparatus. -- BenRG 23:11, 3 November 2007 (UTC)
Oh, I wasn't confusing anything, I just neglected to mention that I was talking about something else :-D But yes, the doppler effect applies to waves, not fields in general. And the static field about an object has no wavelike characteristics. Someguy1221 23:34, 3 November 2007 (UTC)
Keep in mind that mass undergoes Lorentz contraction, so as one approaches or recedes from an object one sees a heavier mass. The most definitive answer would be an actual solution to the Einstein Field Equations for a central field at constant velocity, for which all I can say is that there will be off-diagonal terms in the stress-energy tensor, which, given a Schwarzschild metric, means interesting curvature. I say interesting because I have no clue how to interpret the results, other than saying that the off-diagonal terms are not present in the stationary case. Some help would be appreciated. SamuelRiv 05:48, 4 November 2007 (UTC)

Thanks for all the help so far. As for BenRG's comment about my notion of "appearing heavier", a simple explanation would be this: Consider two heavy objects in far space accelerating towards each other purely due to each other's gravity. Both are emitting a single spectral line of red light. The two can measure speed relative to each other by observing light Doppler shift in the color of the observed light from the other (which should shift towards green or blue). What I mean by "appearing heavier" is that the infinitesimal acceleration of object A towards object B due to gravity would (in my hypothetical model) be slightly more than just the combined value of (1) The gravitational effect using the 1/r^2 law and (2) compensating for the fact that B has a heavier apparent mass due to relativity. In both (1) and (2) cases the distance and speed of a short while back are used due to the speed of light delay combined with a nonzero distance between A and B. This "appearing heavier" effect would then be notable in the spectral profile of the observed color. I realize now that this "effect" that I want to give a name might be something trivial that is automatically taken into account with most such calculations when calculus is used. 12:26, 4 November 2007 (UTC) Eon Zuurmond

I would also like to add the following: words like "Doppler", "ripple" and "wave" are apparently bad choices because they all imply I'm assuming something with a frequency or wavelength. I just use these for analogy. 20:19, 4 November 2007 (UTC) Eon Zuurmond

Radiative Forcing[edit]

Why is the amount of radiative forcing greater on a south-facing slope than on a north-facing slope?? 22:48, 3 November 2007 (UTC)

This sounds like it could be a homework question, so I'm not going to give the full answer, but here are a couple hints. A) The reverse is true in the Southern hemisphere, and B) think about the position of the sun. Dragons flight 23:02, 3 November 2007 (UTC)
And it's completely backwards (says me, south of the equator). --Psud 20:28, 5 November 2007 (UTC)