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August 13

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Insane Plants

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I am researching plants. Specifically plants with some exciting or dreadful, indeed, insane aspects to them. I have covered the corpse flower and things like henbane and datura and am looking to see if anyone can point me to some more. 216.6.159.83 (talk) 00:05, 13 August 2009 (UTC)[reply]

Well I am very fond of all carnivorous plants. Plants that eat animals... insane! --98.217.14.211 (talk) 00:14, 13 August 2009 (UTC)[reply]
"Mad cucumber" Ecballium elaterium is fun :) --Dr Dima (talk) 00:16, 13 August 2009 (UTC)[reply]
Those are much more interesting than their stub article would suggest. There are a lot of videos of them if you look on YouTube for "exploding cucumber" and things like that. They don't just release their seeds in a way that can be "seen by the naked eye," they apparently shoot them out quite violently... [1] In general, the Rapid plant movement plants are pretty neat. --98.217.14.211 (talk) 00:36, 13 August 2009 (UTC)[reply]
How about the mandrake, 'sprouted from the death-ejected sperm of a hanged man'? Adambrowne666 (talk) 00:44, 13 August 2009 (UTC)[reply]
Mandrake (plant) aka Jihn's eggs = fun indeed. --Dr Dima (talk) 00:47, 13 August 2009 (UTC)[reply]
In the "not so fun" category you can try Castor oil plant, curare vines (Abuta, Chondrodendron, also Strychnos) - all very poisonous. Try also cholla cactus, it's just nasty :) --Dr Dima (talk) 00:51, 13 August 2009 (UTC)[reply]
There is also sensitive plant, which closes its leaflets when disturbed.CalamusFortis 03:18, 13 August 2009 (UTC)[reply]
You might consider the plants depicted and described by Leo Lionni in his book Parallel Botany, the contemplation of which might well cause the contemplatee some mental distress. 87.81.230.195 (talk) 03:29, 13 August 2009 (UTC)[reply]
Weird: Lithops, Resurrection fern, Wollemia, Welwitschia, Rafflesia arnoldii, Wolffia. --Sean 17:43, 13 August 2009 (UTC)[reply]
Catnip? It has mind-control over cats :) --131.188.3.21 (talk) 16:21, 14 August 2009 (UTC)[reply]
Dunno - "control" isn't the word that comes immediately to mind when you see a cat tripping-out on catnip. SteveBaker (talk) 02:34, 15 August 2009 (UTC)[reply]

Why are some dishes so greasy?

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Most oily dishes can be washed easily with soap, but some require dishwashing fluids. What's the diff? 67.243.6.13 (talk) 00:20, 13 August 2009 (UTC)[reply]

Probably the composition of fat. Some fats solidify at room temperature, some don't. See Fat and saturated fatty acids. --Dr Dima (talk) 00:27, 13 August 2009 (UTC)[reply]
You wash dishes with soap? Unless you guys use the word soap to mean some sort of dish soap, but I only use soap for my hands and in the shower. Everyone I know uses some sort of dishwashing liquid in the kitchen sink. Vespine (talk) 02:03, 13 August 2009 (UTC)[reply]
In the USA, where the IP is from, we call dish soap "soap." A different kind of soap, but as you say, soap nonetheless. There are even kinds sold that claim to function as both dish and hand soap. ~ Amory (usertalkcontribs) 04:48, 13 August 2009 (UTC)[reply]
There is also no good reason why you couldn't use dish soap in the shower in a pinch. It just takes a lot more water then normal soap for it to stop making suds. Googlemeister (talk) 13:45, 13 August 2009 (UTC)[reply]
I imagine some use mere surfactant effects, emulsifying oils and water, while stronger liquids might directly attack lipids. John Riemann Soong (talk) 08:46, 13 August 2009 (UTC)[reply]
To the best of my knowledge, all dishwashing liquids use surfactant effect and do not "directly attack lipids". You want something to "directly attack lipids", try baking soda or sodium carbonate (powdered "detergent" for dishwashing machines), or even Drano (though this last option could be dangerous for washing dishes). 98.234.126.251 (talk) 04:27, 16 August 2009 (UTC)[reply]

Hideous mermaid image

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Ages ago, I found on the internet a black and white illustration, possibly an engraving, of a skrinkled up little corpse of a purported mermaid - not the Feejee Mermaid, but one less human, more ghastly. I wanted to use it on the cover page of something I'm writing, but foolishly I seem to have deleted it from my files. Can anyone find it for me? (I'm thinking now it may not have been a mermaid after all, but just some random scientific curiosity from a wunderkammer somewhere).

Thanks in advance.

Adambrowne666 (talk) 00:41, 13 August 2009 (UTC)[reply]

A shrinkled up little corpse that’s more ghastly than the Fiji mermaid? That looks like a pretty ghastly shrinkled up little corpse to me. Red Act (talk) 00:56, 13 August 2009 (UTC)[reply]
It's plenty skrinkled up, let me tell you. Adambrowne666 (talk) 01:01, 13 August 2009 (UTC)[reply]
There is [this http://www.thegatherer.co.uk/html/The_Mermaid_The_Mirror_Saturday_November_9_1822.html], but it's probably not what you're looking for. I'm pretty sure that this is the FeeJee Mermaid before it became known by that name. It's uglier than the usual drawing, though. APL (talk) 01:39, 13 August 2009 (UTC)[reply]
Thanks, APL, but yeah, not what I'm looking for - my uncertain memory tells me there wasn't much of a face at all, and if it was engraved, it was finer work, darker in hue. I think it predates the Feejee Mermaid by at least a century. Sorry, I realise how unlikely it is for this question to be answered. Adambrowne666 (talk) 02:25, 13 August 2009 (UTC)[reply]
What about this little lady? LANTZYTALK 06:52, 13 August 2009 (UTC)[reply]
I'm starting to feel like an ingrate - everyone's been v generous, and here I am just saying no no no; so, sorry, Antzy, but no; that's another incarnation of the Feejee Mermaid Adambrowne666 (talk) 08:54, 13 August 2009 (UTC)[reply]
It's frustrating. I'm sure I've seen woodcuts of dead alleged mermaids before that weren't the FeeJee mermaid, but I have no idea where I've seen them. APL (talk) 13:41, 13 August 2009 (UTC)[reply]
Jenny Haniver

May I introfduce you to Jenny Haniver? Perhaps not in thr Littlr Mermaid class nor yet in that of Wonder Woman, but let's not ne picky. B00P (talk) 07:35, 14 August 2009 (UTC)[reply]

This really doesn't match your description exactly, but it is what I immediately thought of. I remembered reading about a "hoax mermaid" that had purportedly washed up on a beach: http://urbanlegends.about.com/library/bl_merman_carcass3.htm. The link is on page three of four, the first page is an article discussing the hoax itself. Maedin\talk 19:39, 14 August 2009 (UTC)[reply]

Thanks heaps, all, for your tireless attempts - I'll likely use one of the better Feejee Mermaids if I can't find the one I'm thinking of. In the meantime, though, I'm gonna try the Humanities Desk. Thanks again Adambrowne666 (talk) 02:30, 16 August 2009 (UTC)[reply]

Ice Age

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Does ice Age happen once every certain amount of time or ice age is non-stoppable matter like Greenhouse effect. Does Ice Age go away after certain time, or it is accumulative like global warming?--69.229.108.245 (talk) 02:55, 13 August 2009 (UTC)[reply]

Have you read Ice age? You may also find Timeline of glaciation interesting. Global warming will go away after a certain time if we stop pumping extra greenhouse gases into the atmosphere, it might just take a long time and do a lot of damage along the way (especially if we trigger a runaway greenhouse effect as some people fear we are going to). --Tango (talk) 03:01, 13 August 2009 (UTC)[reply]
How do you know if "global warming" "will go away... if we stop pumping extra greenhouse gases into the atmosphere" -- will doing that change the amount of solar radiation reaching the Earth? And besides, what do you mean "we" will "trigger a runaway greenhouse effect"? In other words, what evidence do you have that "global warming" is caused by some so-called "greenhouse effect" and not by an increase in the amount of solar radiation? 98.234.126.251 (talk) 04:34, 16 August 2009 (UTC)[reply]

neural optimization.. how to,..

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hey friends,


i have obtained 10 straight lines as solution of a 10 differential equations say u=f(t)(t=time). i need to optimize u by varying t.how to go for it in matlab tool box. 218.248.11.212 (talk) —Preceding undated comment added 07:37, 13 August 2009 (UTC).[reply]

CVX: Matlab Software for Disciplined Convex Programming is the standard around these parts. You might also find LSQR useful. I believe LSQR is now included with MATLAB. Nimur (talk) 19:49, 13 August 2009 (UTC)[reply]

Moon at 100 km

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What would happen if tomorrow the moon just appeared in a new, circular orbit, where it skirted the atmosphere through the length of the journey? (I know the moon is actually retreating and this couldn't really happen but I'm curious about the possible consequences). How long would it take for atmospheric drag to bring its path into contact with the surface? TheFutureAwaits (talk) 09:53, 13 August 2009 (UTC)[reply]

Tidal forces would likely disrupt the moon into a ring. The earth also would be seriously damaged by tidal forces. Graeme Bartlett (talk) 09:57, 13 August 2009 (UTC)[reply]
A ring like a planetary ring? —Akrabbimtalk 12:47, 13 August 2009 (UTC)[reply]
Yes, as a planetary ring. 100 km is well within Earth's Roche limit. Interestingly, the Earth would also lie within the Moon's Roche limit (which is about 120 km above the lunar surface) -- so yes, Graeme Bartlett is entirely correct that damage to the Earth, from tidal forces alone, would be incredibly severe. I would guess you're talking damage on the order of liquifying most of the surface, even before impact. A wild-ass guess puts time to collisions starting within a week, but not finishing for years, due to the Moon being ripped to shreds (some portions of the moon would be 1800 km up and well above most of the atmosphere). — Lomn 12:57, 13 August 2009 (UTC)[reply]
Whaaat, you mean The Distance of the Moon didn't really happen...? 88.114.222.252 (talk) 13:19, 13 August 2009 (UTC)[reply]
Very low earth orbits, like a 100 km altitude, usually suffer from air drag, and decay rapidly (within weeks or months, depending on characteristics of the object. Nimur (talk) 15:59, 13 August 2009 (UTC)[reply]
Yes ... but bear in mind that the Moon is much larger than your average (artificial) satellite, most of it is outside of the atmosphere, and it has an awful lot of momentum. As far as atmospheric drag is concerned, this would dump so much energy into the atmosphere that it would probably strip off most of the atmosphere before it had any appreciable effect on the Moon. But as other contributors have pointed out, tidal forces will rip the Moon apart long before that happens. Gandalf61 (talk) 16:28, 13 August 2009 (UTC)[reply]
Yes, I don't think the atmosphere would last more than a few hours, a couple of days at most, although some of it might be replaced by the water and even rock on both the Earth and Moon vaporising. The Roche limit doesn't really apply to objects as rigid as the Earth and Moon, but it gives a good idea of how close you can get before you start to have problems. Anything not tied down is in trouble! --Tango (talk) 16:40, 13 August 2009 (UTC)[reply]
Why doesn't the Roche limit apply? I thought it applies exactly for such a situation - the tidal forces are so overwhelmingly strong that they outweigh the gravitational self-attraction of the constituent materials of the moon. The only reason the moon is "rigid" is because gravitational energy was sufficiently strong to fuse the rocks together (i.e. by melting) - but the Roche limit defines the point where that effect is smaller than the tidal shear force. Needless to say, we don't have a lot of empirical examples for comparison; bt I think the principle applies, and the moon would fracture. Just figure that the orbital period at 100 km should be about one or two hours; while at the far side of the moon (100km + moon diameter), the stable orbital period would be some factor slower (10x? I'm too lazy to do a back of the envelope calculation). Whatever that differential in speed is would be shearing the moon apart. Nimur (talk) 17:12, 13 August 2009 (UTC)[reply]
Now those rocks have fused they would continue to be held together even if gravity disappeared. The moon (and much of the Earth) would indeed be ripped apart, but you can't simply look at the Roche limit. Read the second paragraph of Roche limit. --Tango (talk) 17:18, 13 August 2009 (UTC)[reply]
Frankly, I think our second para of Roche limit looks pretty weak. There's no cite on its claims (of the anomalous moons being held together by non-grav forces), and Metis (moon) (one of the supposed Roche-defying bodies) cites the moon's continued integrity as evidence that it's outside the Roche limit. Since the RL is density-dependent, an orbital radius that turned one moon into a ring system may leave another intact. Additionally the selected examples subsection of RL notes both anomalous moons as orbiting outside the rigid Roche limits, and a great many moons orbiting inside the fluid limits. That said, I'm no expert in this field. — Lomn 18:34, 13 August 2009 (UTC)[reply]
(outdent) - I disagree that melting a rock gives it a stronger bond than the gravity which provided the source energy for the melting and physical bonding to form. I'll try to find some sources and edit Roche limit - but I'm pretty sure that conservation of energy is at play here - the rock can not be held together by a binding energy stronger than the gravity which initially bound it (unless it started out molten - but this is not widely supported by the accretion theory - the rock should have started off cool, and only melts due to gravitational compression heating). Nimur (talk) 19:16, 13 August 2009 (UTC)[reply]
I think somebody owes Steve a dollar. We're not talking about energy, we are talking about force, they are very different (albeit related) things. When physical bonding has taken place that is going to mean it can survive some force pulling it apart, that means the tidal forces will have to cancel out gravity and then pull apart the bonds, that is going to take greater force that just cancelling out gravity, which is all that would be required to pull apart a pile of rubble (a lot of these kind of calculations are based on the moons/asteroids being piles of rubble, as many smaller ones are). --Tango (talk) 21:06, 13 August 2009 (UTC)[reply]
I don't think I owe Steve a dollar. I'm simply stating that a force can't be stronger than the energy gradient which creates it. That's the definition of a force, , where U is a potential energy field. In this case, U is gravitational energy. The rock can only melt and "fuse together" if the force due to gravity is sufficiently strong to melt it - there's no other source of energy (or force, or anything). Nimur (talk) 21:38, 13 August 2009 (UTC)[reply]
I never said the force was stronger than gravity, it is in addition to gravity so the total is, of course, greater. --Tango (talk) 22:20, 13 August 2009 (UTC)[reply]

Interesting answers guys! I'm currently writing a sci-fiction story about this situation and appreciate all your insight! (Essentially aliens slow the orbit of the moon to force it into an extremely low orbit)So what would this all be like for a person on the surface? Would they feel themselves being pulled in the direction of the moon? Could they jump higher? How big would it look? What would happen if you woke up one day, walked outside and the moon was 100 km away? TheFutureAwaits (talk) 19:13, 13 August 2009 (UTC)[reply]

Given that we've been discussing how quickly the Moon would render the Earth entirely uninhabitable at that distance, and how the Earth would rip the Moon entirely to shreds, this is all moot. You can easily calculate the Moon's angular diameter at 100 km altitude and subtract the relevant surface gravities if you like, but I won't imply that any of this is what it "would be like for a person on the surface". By this point, either the Moon would already be gone (per the Earth's Roche limit, if aliens move the moon), or the Earth would be uninhabitable (per the Moon's Roche limit, if aliens teleport the moon). Either way, rocks fall, everyone dies. — Lomn 19:54, 13 August 2009 (UTC)[reply]
Yeah, but you have about three days to prevent complete disaster at that point. —Akrabbimtalk 20:26, 13 August 2009 (UTC)[reply]
If aliens want to kill everyone, there are a lot better ways then to move an object the size of the moon. Heck a couple of well placed nukes and we would more likely do it ourselves with our own equipment. Googlemeister (talk) 20:29, 13 August 2009 (UTC)[reply]
A "couple of well-placed nukes" could kill maybe about twenty million people, but won't "kill everyone" -- you'd have to start an all-out nuclear war between Russia and the US for that to happen. 98.234.126.251 (talk) 05:20, 16 August 2009 (UTC)[reply]
A couple of well placed nukes WOULD start an all out nuclear exchange between Russia and the US in all likelyhood. Googlemeister (talk) 15:10, 17 August 2009 (UTC)[reply]
Not if they're traced to someone else, like Iran or Pakistan (yes, there are ways to trace a nuke back to its manufacturer even after it's gone off). It would have to be done very soon after the explosion, but it can be done. (See the novel The Sum of All Fears by Tom Clancy.) 98.234.126.251 (talk) 22:53, 17 August 2009 (UTC)[reply]
I think you need to understand the concept of tidal forces a little better. If the moon were to get that close - the force of gravity due to the earth on the side of the moon nearest to the earth would be enormously larger than on the side furthest from the earth. Moreover, the side of the moon furthest from the earth is travelling around in a circle with a much greater diameter than the side closest to the earth - and that adds to the centrifugal force that it experiences. The result is that the moon is being pulled apart by gravity on the one side and centrifugal force on the other. At some point, (The "Roche limit") the pulling and tearing forces become greater than the structural strength of the moon - and it simply flies apart. When that happens, each individual little rock flies off into it's own orbit - but their mutual gravitation flattens that orbit into a disk. So long before the moon got as low as you are suggesting, it would have disintegrated - turning Earth into a mini-Saturn. As others have pointed out, the reverse is also the case - the Earth is obviously affected by tidal forces of the moon (if you leave on a beach, you'll notice this!) - and at the altitude you specify, the earth would also be beyond the roche limit of the moon - so both bodies would simply fly apart at that separation. Eventually, all of the pieces would fall back together making a single new planet...which would likely retain a ring system for a few centuries before all of the debris either collected to make a new moon - or fell back onto the newly formed planet. But if the moon were nudged gradually closer rather than magically teleported to that distance, it would simply start to disintegrate long before it got as close as you suggest.
From the point of view of science fiction - you can go two ways with it. Either the universe doesn't work the way we think - and (essentially) the aliens can use magic to slow down the moon and stop it from breaking up - and give you the story you want regardless of mere scientific niceties...or you're trying to write a reasonably scientifically plausible story. If it's the former - then just say that the aliens use a force field to hold the moon together and ignore what everyone here is saying. If it's the latter than ask yourself about aliens with the power at hand to slow down 7x1022kg of moon by something close to a kilometer per second - that's an INSANE amount of energy. If they have that much energy at hand, wouldn't it be simpler just to pick up a couple of mountain-sized asteroids and slam them into the earth? The effect on life would be about the same...and the energy requirements would be microscopic in comparison. It just doesn't make sense that they'd do that. SteveBaker (talk) 22:48, 13 August 2009 (UTC)[reply]
I doubt the Earth would be completely ripped apart. It would be only just inside the Roche limit and has significant tensile strength, so I'm fairly sure it would survive, but the damage would be enormous. The atmosphere and oceans would be in trouble - they have essentially no tensile strength and are right on the edge of the Earth, so would be ripped away, the Earth's core, on the other hand, would be far enough away from the Moon and it is held together quite strongly so would remain intact, as would most of the mantle, probably all of it and most of the crust. --Tango (talk) 23:25, 13 August 2009 (UTC)[reply]
Oh - just all of the oceans, atmosphere and the first few kilometers of dirt and rock....well, that's OK then! I can't imagine what I was worried about! :-) SteveBaker (talk) 00:15, 14 August 2009 (UTC)[reply]

First, remember than an object in a lower orbit moves faster than one in a higher orbit. So to move an object from a high orbit into a lower one, you don't just "slow it down". If a force is applied over an extended time in a direction that slows the object's motion, it will gradually move into a lower orbit but it will actually be moving faster all the time, as its potential energy gets converted to kinetic energy. This is the way satellites behave when their orbit decays due to atmospheric drag, for instance. (For simplicity I assume we are talking about orbits that are circular or pretty nearly so.) With a suitable power source, the Moon could be lowered into a low orbit this way over a period of months or years.

The other way to move an object into a lower orbit -- the way it would be done with today's rockets -- is to start by applying a stronger force in the direction that slows its motion. Its speed will become lower, but it will now be at the high end (apogee) of an elliptical transfer orbit. As the satellite falls closer to the primary (the Earth), its speed increases. When it reaches the low end of the orbit (perigee), you again apply a force to slow it down, and this converts the orbit to a circular one. (The important thing to remember is that when no external force is being applied, the orbit always passes through the last place where a force was applied. So to change to an orbit that does not intersect the original one, you need to apply force twice.) By this method the Moon could be lowered into a low orbit over a period of a couple of days.

With a still stronger force, it could be done even faster, but then the transfer orbit would have to be hyperbolic -- without the second application of force to put the Moon back into orbit, it would escape the Earth's gravity altogether.

If you're talking about the Moon being teleported rather than accelerated into the new orbit, then it's not being slowed down. It's being placed in a new position and given a suitable higher speed that will put it in circular orbit, although it will have less potential energy than before.

At 100 km above the Earth's surface, the center-to-center distance is 1,738 + 6,378 - 100 = 8,016 km. The normal distance varies from 363,100 to 405,700 km, so let us say it would be 1/48 as much. So the first thing is that the Moon would look 48 times as wide as it does now.

Next, the period of the Moon's orbit would be greatly reduced according to Kepler's Third Law. It would be to sqrt(1/48³) power of what it is now, or about 2 hours. As seen from a given location, there would be a Moonrise or Moonset every hour or so. In latitudes near enough to the equator, the much larger Moon would eclipse the Sun 6 times a day, but they would not be like the eclipses we know now where the Sun is just barely covered and the Moon's motion in front of it is slow. The Moon would blank out the Sun in less than a minute, and would in many cases cover the whole corona, but within a few minutes it would equally rapidly move clear of the whole sun again, returning daylight to normal.

The Moon would also be in total eclipse most of the time it was above the ground at night; it would only shine brightly when close to the horizon.

Now the tides. Tidal force varies as the inverse cube of the distance, so it would be about 48³ = about 110,000 times normal! This is still not enough to disrupt the solid Earth, but it's plenty to tickle some seismic faults and trigger some of those earthquakes that are expected "sometime in the next hundred years". More important, effects on the sea would be huge -- coastal lowlands throughout the world might become wholly or partly uninhabitable due to high tides, and that includes major cities such as Los Angeles, New York, Rio de Janeiro, London, Mumbai, Tokyo, Sydney, and many, many more throughout the world's seacosts. It's hard to estimate how high the tides would be since they would come once an hour instead of every 12+ hours, and the patterns of motion would be different than today. But they would certainly be big.

And the atmosphere would be similarly disrupted -- it feels tides too -- although it's hard to say what the effect would be. Violent weather seems a likely bet, though.

All this is true as long as the Moon stays more or less in one piece. The Earth's tidal force on the Moon, which is much stronger than the Moon's tidal force on the Earth, would also be 110,000 times stronger than normal, and stronger than the Moon's own gravity -- that's what "being inside the Roche limit" means. So maybe the Moon would break up under the stress and form a ring around the Earth. That depends on its tensile strength and I won't attempt to calculate whether it's likely. Dust and loose objects on the Moon's surface, like the Apollo landers, would float free of it, in any case, and enter orbit independently.

Finally, note that the Moon rotates once per month and unless something was deliberately done to change this, it would continue to be true in its new orbit. Consequently it would no longer always have the same face turned towards the Earth, but would look noticeably different from one Moonrise to the next.

--Anonymous, 22:01 UTC, August 13, 2009.

The Moon would be so far inside the Roche limit that I would be very surprised if the tensile strength was enough to save it. You also having taken atmospheric drag into account - there is still quite a lot of atmosphere at 100km, maybe not enough to slow the Moon down considerably, but plenty to cause some considerable heating. --Tango (talk) 22:42, 13 August 2009 (UTC)[reply]

I just wan't to point out that the RL calculation assumes that the orbiting body is much smaller than the other one. Because of that, all statements above mentioning the fact that the earth would be within the moon's RL are nonsense. Pieces of the earth would not fly upwards. The moon, on the other hand, would be destroyed as pointed out. Dauto (talk) 23:51, 13 August 2009 (UTC)[reply]

I'm just going to go out on a limb and suggest that Roche-limit-related fracturing would be most likely moot, because at 100 km, I suspect the moon would simply collide with the earth due to orbital instability and amplifying perturbations from circular orbit; after collision, we've got a whole new mess of non-ideal, non-rigid-body mechanics to deal with. I think we're all in consensus about the catastrophic nature of this scenario, one way or the other. Nimur (talk) 01:07, 14 August 2009 (UTC)[reply]
Just one thing Nimur. A quick search of Wolframreveals... The Roche Limit is the point where a body with no tensile strength would be torn apart. As people say this wouldn't tear apart the Earth because it has tensile strenght. A real body can pass the Roche limit of the planet its orbiting without being torn apart. From what76.94.88.196 (talk) 06:36, 14 August 2009 (UTC)[reply]
Ok, I suppose I might have been somewhat incorrect in the definition (by assuming that tensile strength was accounted for); apologies to Tango and others for my adamant statements earlier. Nimur (talk) 14:40, 14 August 2009 (UTC)[reply]

While we're hypothesizing...

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One question that came up in my mind hasn't been addressed. (I haven't read any of the referenced pages, by the way.) As a moon-sized object in a decaying orbit starts to hit upper atmosphere, would the friction on one side of said moon be enough to start it revolving on an axis again?

Think of the tires on a landing aircraft. They get up to speed pretty quickly from friction with the runway, which is only on one side of the tire. Is there enough friction present in the low-orbiting-moon scenario to cause the same effect, or to override whatever force has stopped the moon from revolving in the first place?

Further unrealistic assumptions about breaking up in orbit may be necessary :-) --DaHorsesMouth (talk) 22:35, 14 August 2009 (UTC)[reply]

I suspect the answer is "yes", but I should point out that the Moon is revolving on its axis now, it's just revolving at the same rate it is orbiting, so we always see the same face. Its orbit and revolution become locked like that due to tidal forces, it's called tidal locking. --Tango (talk) 00:40, 15 August 2009 (UTC)[reply]

CPU gold content

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Who can tell me an approximate quantity of gold and other noble metals in various CPU? Renaldas Kanarskas (talk) 10:57, 13 August 2009 (UTC)[reply]

Googling "how much gold in a computer" comes up with a number of links to suggest that there are very small amounts of gold in a computer, in the form of gold plating, which is not very much gold. One thread suggests that "One ton (2000 lbs) of "average" circuit board from modern computers and electronics generally yields (in a very good system) between 8 and 11 troy ounces of 24k gold." There is much more silver, and minute amounts of everything else. [2] --98.217.14.211 (talk) 13:47, 13 August 2009 (UTC)[reply]
It really comes in two places - some of the pins of some of the connectors are coated with a microscopically thin layer of gold - and inside the actual chips themselves, incredibly thin gold wires are sometimes used to connect the edges of the silicon wafer to the pins of the chip that lead out onto the circuit board. In both cases, the amount of gold is incredibly small. It's also exceedingly difficult to extract on any commercial scale because it's mixed up with all manner of other weird and wonderful metals. But look at it like this: We can now make low-cost net-books for $100 to $200. With gold costing $950 an ounce - if there was any significant amount of gold in there - they'd cost a lot more! Manufacturers are not obsessing about the amount that the gold in these machines is costing them - so we may conclude that it's negligable. SteveBaker (talk) 14:03, 13 August 2009 (UTC)[reply]
AFAIK there will be no gold actually in the CPU - i.e. the inside the chip compartment. In the "olden" days gold was indeed used for bond wires, and occasionally even for "wiring" on the chip. However, this is no longer the case and aluminium based bond wires and chip wiring have been the norm for a while. Gold is generally a bad thing for silicon chips, since it causes recombination centres which shorten minority carrier lifetimes, so it's pretty common to avoid it at all costs. --Phil Holmes (talk) 14:07, 13 August 2009 (UTC)[reply]
On the other hand, older computers and electronics (1970's) had quite a bit of gold, mostly because the technology to apply very thin conisitent coatting was not available, so the coating on the connectors was much thicker. Also, since the individual chips were much less integegrated, there were proportionally more wires and connectors. -Arch dude (talk) 17:24, 13 August 2009 (UTC)[reply]

Are cats really this clever?

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http://news.bbc.co.uk/1/hi/england/berkshire/8195156.stm Was this just coincidence or are cats really this clever? The cat would have needed to have gone through quite a long chain of thought, and be capable of altruism, to have purposedly done this. Is it more like to be merely coincidence, since many fires occur in which cats do nothing. 78.147.132.206 (talk) 12:09, 13 August 2009 (UTC)[reply]

Cats are clever enough routinely to hassle humans when they want some change in the environment: the production of food, the opening of a door, &c. It's not a very great stretch between such commonplaces and the event described. --Tagishsimon (talk) 12:14, 13 August 2009 (UTC)[reply]
If the story happened exactly as presented it would be a remarkable story. But the detail that makes it remarkable seems completely unverifiable. How does the author know that the cat entered a burning building? (Did a human observe the cat enter the burning building? If so, why didn't the human raise the alarm?) If the cat was already sleeping in the building when it caught fire it's not hard to believe that it would seek out a human to deal with the issue.
There have been other cases[3][4] of cats waking their owners in fires. All of these involve cats that were already in the house. If Hugo really entered the building after it had caught on fire as the article suggests, that would be really impressive.
While the cat was certainly a hero, I'd be surprised if that was its primary intention. Too me, it seems far more likely that the cat perceived that the house had a serious problem and he wanted it fixed immediately for his own sake. APL (talk) 13:29, 13 August 2009 (UTC)[reply]
There are stories of dogs doing this kind of thing too - but in their case, it's more likely because they are pack animals. But I agree, the only remarkable thing is that the cat supposedly re-entered the building after it caught fire - and that's a completely unverifiable detail. However, if you Google "cat enters burning building" - you get 100,000 hits - it's a common meme that a mother cat will re-enter a burning building multiple times to bring out her kittens, so there may be basis in fact. SteveBaker (talk) 13:52, 13 August 2009 (UTC)[reply]
I've certainly seen a mother cat making multiple trips to move kittens away from something she considered dangerous (me and my friend who were playing with the kittens on top of the bed they were living under!), so I can't see why she wouldn't have done the same in the case of fire. --Tango (talk) 16:45, 13 August 2009 (UTC)[reply]

non-therapy ancillary services

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In medicine, what does the term "non-therapy ancillary services" refer to? This is included in the House healthcare bill (HR 3200) page 230 and other places. 68.54.198.247 (talk) 12:29, 13 August 2009 (UTC)[reply]

Respite care, financial aid, home help, assistance with transportation etc. Caring for the non-medical needs of ill patients. Fribbler (talk) 12:34, 13 August 2009 (UTC)[reply]

Quantum mechanics and determinism

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Please excuse the probable ignorance displayed in my asking this question, as I have only a poorly informed lay understanding of how quantum stuff works- but I guess if I knew it all I wouldn't have to ask questions about it!

I was wondering about whether quantum physics and the uncertainty principle actually preclude the possibility of a deterministic classical Universe even at the smallest scale. I was reading A Brief History of Time, and Hawking commented that if anything did happen before the Big Bang (like a previous Universe or something) it could have no observational consequences for this Universe so science could not speculate about it even if it wanted to. I was wondering if this was parallelled in the uncertainty principle- is there a possibility of particles having an exact speed and position, but because they could never be observed science cannot speculate about them?

Basically my question is: is quantum mechanics a definition of the boundaries of what is knowable about the Universe, or does it actually preclude the existence of anything certain? And if so, how?

Thanks much, Dan Hartas (talk) 12:31, 13 August 2009 (UTC)[reply]


Quantum theory certainly precludes perfect determinism at the smallest scales. If you take a single atom of Uranium-232, you have literally no way to determine when it will decay into an atom of Thorium. You can say that there is a 50/50 chance that it'll have decayed within the half-life of U232 (68.9 years) - but if you come back 7 years or 70 years later and discover that it hasn't decayed yet - then ask the same question, the answer doesn't change - it's still a 50/50 chance that it'll decay in the NEXT 68.9 years! The atom doesn't have a 'clock' inside telling it when to decay - and it doesn't get worn down with time and become 'ready' to decay...it's just completely random. But as you work at higher scales, that randomness generally averages out so at the scale of humans, things seem pretty deterministic. If you take a bucketful of Uranium-232 (which is probably a bad idea!), you know with absolute certainty that half of it will have decayed into Thorium after precisely 68.9 years. However, since some things are "chaotic" (in the mathematical sense of being highly sensitive to small things) - it is true to say that quantum level indeterminism does have an effect at the 'macro' scale. To pick the commonest example of the weather - which is "chaotic" - we've all heard that the beat of a butterflies wing could cause a tornado halfway around the planet years later...well, the decay of a single atom of uranium could cause changes on a similar scale...so quantum randomness can indeed drive extremely unstable macro-scale things like weather patterns.
However, the issue of there being no information from before the big bang is a different matter entirely. Because the universe appears to have started as a 'singularity' (an infinitesimal dot - a point with zero size), then there is very little information content possible within it. Particles don't exist, there are no velocities, positions, nothing like that. The point may have mass - but that might be infinite, so there isn't any information there either. In an information-theory sense, there is no place for the information about what happened "before" to be stored. Of course Hawking claimed that there was no "before".
I suppose that since the singularity was so small, it would be affected by quantum theory indeterminism - so for the very first few picoseconds of the big bang, it might also have been ruled by things like the uncertainty principle - meaning that any information that had passed into our universe from whatever came before would be totally scrambled anyway...but that's not a necessary precondition. The mere fact that it was a singularity sharply limits the amount of information it could possibly contain...and that's enough to ensure that we can't directly observe what came before. However, even the existance of a "before" is somewhat contentious. It's arguable that time itself was started by the big bang and even the concept of a "before" is meaningless. That would be a somewhat beautiful conclusion because it cuts off the need to go into this infinite regress of asking what created the big bang...then what created the thing that created the big bang...and so on. If time started at the instant of the big bang then we have the complete "creation story" nailed down already.
SteveBaker (talk) 13:30, 13 August 2009 (UTC)[reply]
So if you take 2 U232 atoms and wait one half life, is one of them going to be Thorium and the other U232 or could they both be one or the other? Googlemeister (talk) 13:40, 13 August 2009 (UTC)[reply]
The configuration would have the same probability distribution as flipping two coins, substituting "heads" with "does not decay / uranium" and "tails" with "decays / thorium" (assuming the thorium hasn't itself decayed). 25% chance of no decay, 25% chance of total decay, and 50% chance of partial decay. — Lomn 13:44, 13 August 2009 (UTC)[reply]

Our article on Hidden variable theory deals directly with the OP's question, although it is written at a level that may present difficulties for somebody without much physics background. Looie496 (talk) 15:32, 13 August 2009 (UTC)[reply]

We need to be clear though that only a minority of physicists really support hidden variable theory - and even then, only one particular flavor of it (Bohm theory) is really acceptable since it's the only one that agrees with conventional indeterministic quantum theory and experimental evidence. This addresses the OP's question - but it's not a solid answer. Since Bohm theory doesn't exhibit the property of 'locality' - this interpretation poses problems that are exactly as bad as indeterminism in practice - they just give you the warm, fuzzy feeling that at it's heart, the universe isn't random - it simply behaves precisely as if it were totally random! SteveBaker (talk) 21:56, 13 August 2009 (UTC)[reply]
The universe described by Quantum Mechanics is completely deterministic, however, from the point of view of an observer within the universe, the outcome of an event can be completely non-deterministic, without even a theoretic possibility of predicting an outcome. To see how this is the case, consider the following thought experiment: I challenge you to predict the outcome of a random event, in this case flipping a coin. It is my contention that I will win our challenge at least some of the time because you are unable to predict the outcome of the random event. What I haven't taken into account is that you have a special power of knowing the current state of every particle in the universe. Since all you have to do is measure the exact position of the coin and all the variables that contribute to the flipping motion (the state of my hand muscles, the coins weight, air resistance, etc.), you will win the challenge every time. The question is, how can I possibly regain my ability to win the challenge given your remarkable super abilities? If I utilize the principles of quantum mechanics, I can do it like so: when flipping the coin, I actually create two copies of it, one that lands on heads and the other that lands on tails. At the same time, I am creating two copies of you, and the two copies will see different results for the coin flip. There is no point in asking which coin landed on which side; since else everything except the coin state is identical, which one is which is completely arbitrary. In addition, I'll add the constraint that you can't communicate with your parallel self copy in any way; once you get copied you go along after that completely independently. Now, in this scenario, you will always see the coin flip as being a completely random event. There is no way, even in principle, to know which side you will see come up, because you cannot "choose" which copy of yourself becomes "you"; "you" are observing both possible scenarios in separated lines of causality, but each copy of you assumes that his version of events is actually what happened. Your complete knowledge of the state of every particle in the universe still stands; you knew with 100% certainty that I was going to make a copy of the coin and of you the moment before I did it, so the situation is completely deterministic (and might still appear completely deterministic from the perspective of a godlike observer who could "see" both outcomes happening), but as an actor who is made up of the same particles in the universe as the coin is made of, the randomness of the event is absolute. Truthforitsownsake (talk) 03:32, 14 August 2009 (UTC)[reply]
But if absolutely everything is deterministic, how do you create two copies that will see different results for the coin flip ? If the two copies start out identical (as they presumably are if they are faithful copies) and everything is deterministic then they evolve identically and so they see identical results for the coin flip. And if you reverse the coin in one copy and not in the other then you have created a non-determinsitic discontinuity in the history of one copy, and I can tell whether I am in the "true" copy or the "reversed" copy. Gandalf61 (talk) 13:03, 14 August 2009 (UTC)[reply]
The fact that I am holding a quantum coin and not a classical one is what enables me to create two copies with different results deterministically. I am not reversing the coin in one copy and leaving it alone in another; before I flip the coin, it is in neither the heads or tails flipped state because I haven't flipped it yet. After I flip the coin, it will always be in both the heads and tails flipped state. From your perspective before I flip the coin, you can determine that it will definitely be in both the heads and tails flipped state after I flip it, because my action of flipping it is already determined. The problem is that each copy of you after I flip the coin will see the opposite result state from the other, and see their result as a single canonical random observation. Of course, one cannot actually do such things with real coins since they are macroscopic; the coin in this example is just standing in for the state of a quantum particle. —Preceding unsigned comment added by Truthforitsownsake (talkcontribs) 14:21, 14 August 2009 (UTC)[reply]
Okay, so its a quantum coin, and you duplicate its wave function, which evolves identically in the two copies. Then in one copy I observe the coin and its wave function collapses into the state "heads" and in the other copy I observe the coin and its (identical) wave function collapses into the state "tails". So ... how exactly is that wave function collapse a deterministic process ? Gandalf61 (talk) 15:49, 14 August 2009 (UTC)[reply]
The wavefunction is never duplicated; there is only ever one. The wavefunction is what evolves deterministically in time, and it contains all the information about which states the coin (and you) will be in in the future. What you're calling a "collapse" is a shorthand way of referring to a particular configuration which will be in a part of the wavefunction at some point in time, but there is a corresponding other portion in which the opposite "collapse" configuration will occur as well. Truthforitsownsake (talk) 18:22, 14 August 2009 (UTC)[reply]
Are you talking about quantum entanglement ? Gandalf61 (talk) 20:04, 14 August 2009 (UTC)[reply]
I'm talking about whether it's possible to predict the outcome of a quantum event, even in principle. Truthforitsownsake (talk) 23:09, 14 August 2009 (UTC)[reply]

insecticide

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I am a retired American living in northeast Brazil where ant and termite invasions are a constant problem. In fact, no matter where I have lived (Michigan and Florida) ants have been an aggrevation. In Brazil, I recently bought an insecticide named "Ecoatta" that claims to be non-toxic and comes highly recommended here. Supposedly, it destroys the "queen" when the "workers" carry the poison back to the nest.

When I spray a small amount on a cluster of ants that are swarming a morsel of food that someone has inadvertently dropped on the floor or left on a table they only become agitated and disperse--perhaps back to their nest. Within a day or so, there are absolutely no signs of ants where before there had been. I am impressed and would like to know more about these products to recommend to family and friends in the U.S. but Google and other searches have not turned up anything useful. Perhaps you can help me find an equivalent product in the U.S. if one exists. If such a product is available, I believe it obsoletes the use of many conventional spray insecticides that typically line the shelves of our stores.

Specifics on the spray bottle:

Name: Eccoatta, non-toxic, "natural ingredients." Composition: Lactic acid 0.1%, Animal protein 0.12% Manufacurer: Empreendimentos Azuleida LTDA Contact: amp2000@bol.com.br, ph 55-71-9987-7711

Thank You189.97.59.48 (talk) 13:20, 13 August 2009 (UTC)[reply]

Does it say anything else on the bottle - ie "active ingredients" ? Unless lactic acid is toxic to ants, which seems unlikely.83.100.250.79 (talk) 21:10, 13 August 2009 (UTC)[reply]
I believe it is actually named EcoAtta (eco for "eco friendly" and atta for "ant"). It is referred to here as being a Brazilian product. There are no other hits on Google for EcoAtta. Using other search engines, I found a bottle of EcoAtta for sale here. -- kainaw 22:24, 13 August 2009 (UTC)[reply]
Well some proteins can be potential toxins -- and they can also interfere with pheromone and hormone systems -- but perhaps the company doesn't want to disclose just what type of proteins it's using (for trade secret purposes). I'd be worried that if the proteins are too specific it'd be easy for colonies adapt via mutation though. John Riemann Soong (talk) 13:12, 14 August 2009 (UTC)[reply]
Besides, a venom that kills ants might not necessarily work against termites. FWiW 98.234.126.251 (talk) 05:16, 16 August 2009 (UTC)[reply]

Influence of the Southern Oscillation on tropospheric temperature

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http://www.agu.org/pubs/crossref/2009/2008JD011637.shtml

Any thoughts on this, and any studies that you've found that you can cite which comment on it in anything other then glowing and fanatical words?

I haven't heard the supporters of global warming comment on it and I'd like to see if some have.

Thanks, Chris M. (talk) 17:18, 13 August 2009 (UTC)[reply]

It's to early for peer-reviewed comments to have made it through the system, but there are comments at RealClimate [5] and Open Mind [6]. In particular, the first author of the paper agrees that their methodology cannot detect a secular underlying trend [7]. Also see [8] for some more comments. --Stephan Schulz (talk) 18:26, 13 August 2009 (UTC)[reply]

Paper coffee cups and cold liquids

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I have noticed that whenever I use a standard paper coffee cup for iced coffee, the cup almost immediately starts to go soggy, whereas with hot coffee, the same effect does not happen. I was curious about the reasons for this and have a suspicion though I might be way off base. Since these are wax impregnated (I believe), is it that the cold contracts the wax and thus causes it to flake or open up lots of seams such that the liquid then penetrates the paper, whereas hot liquid, though it could dissolve some of the wax (I wonder how much wax I eat a year in this manner) would do so uniformly, so it would not cause a fundamental breakdown of the water repelling properties of the wax so quickly? As I said, just hypothesizing.--Fuhghettaboutit (talk) 17:22, 13 August 2009 (UTC)[reply]

I never drink iced coffee from paper cups, but my guess is that the cups are only waxed on the inside, and that iced coffee causes condensation on the outside which soaks into the paper. Looie496 (talk) 18:08, 13 August 2009 (UTC)[reply]
I agree, that's what it looks like to me, too. The same will happen if you take a paper cup of ice-cooled soft drinks to a reasonably humid place (i.e. outside in Miami). --Stephan Schulz (talk) 18:50, 13 August 2009 (UTC)[reply]

What about Jerome Drexler and cosmology?

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I see many references to Jerome Drexler and his theories about dark matter, dark energy, and relativistic particles, etc. What I find is that they are all written by Jerome Drexler himself, and they are often flogging his three books. I was hoping to find some objective discussion of his theories on Wikipedia, but he is not mentioned once. This leads me to believe is a wacko, but it would be helpful to find some objective evaluation somewhere, such as Wikipedia.

I have one theory and that is that Drexler himself has suppressed any discussion here because it would be unfavorable.

I would certainly appreciate your comments, and any pointers to an authoritative evaluation of Drexler and his theories.

JFistere (talk) 17:42, 13 August 2009 (UTC)[reply]

Google scholar doesn’t turn up one single paper by him on astronomy or cosmology in a peer reviewed source, which would be rather astonishing it was really true that he was such a genius that “No one else has plausibly explained these cosmic phenomena”[9] (five separate astronomy and cosmology topics). Sounds like just another nut job. Red Act (talk) 18:16, 13 August 2009 (UTC)[reply]
(edit conflict) Wikipedia articles require reliable sources, and it's hard to find any for him. He has no journal publications, his books are self-published, and using Google I haven't been able to find any reviews of his work in reputable places. So it seems that there isn't really any material available to use for a Wikipedia article. This is a relatively common situation with writers on "fringe" topics, unfortunately. It would be nice for Wikipedia to give the "real story", but we can't do it if there aren't any sources. (PS, I've found that it's better to avoid phrases like "nut job".) Looie496 (talk) 18:19, 13 August 2009 (UTC)[reply]
What term do you prefer? “Crackpot”? “Crank”? At any rate, we certainly don't need an article on a self-published crackpot. Red Act (talk) 18:45, 13 August 2009 (UTC)[reply]
I think it would be very difficult for Drexler to suppress an article about himself here on Wikipedia. If someone had created that article, he could vandalise it - maybe delete all of it's content even - but you'd still find it in a search. It takes admin privileges to actually delete an article - and admins who go around deleting articles without going through the official processes don't remain as admins for very long! It's much more likely that he simply does not have the notability required to justify an article...
  • He fails all the criteria as an academic (Wikipedia:Notability (academics)) because his work isn't cited elsewhere.
  • His books clearly fail Wikipedia:Notability (books) because they have not been given awards, or mentioned much elsewhere.
  • WP:AUTHOR is his best chance...but he fails almost all of those criteria too - and is pretty weak on the ones he doesn't outright fail.
So, no - I don't think he's worth a Wikipedia article - and I'm sure he hasn't bee suppressing any article that may have been written. SteveBaker (talk) 21:05, 13 August 2009 (UTC)[reply]
These sorts of fellows (it is always, isn't it?) are either deluded or charlatans, so perhaps "nut job" is the more charitable phrase. --Sean 15:57, 14 August 2009 (UTC)[reply]
It's better to avoid such terms, because they potentially violate WP:BLP (which yes, does apply to all pages on wikipedia, including the RD) unless perhaps you are just describing what a reliable source says about the person, but it's clear no one is, because we've established there are few or none on him. If his theories are completely unsupported, just say that, similar to the way you would (hopefully anyway) discuss problems concerning a wikipedian. Outside wikipedia you are of course free to say whatever you want, although it often remains best to avoid getting too personal Nil Einne (talk) 16:58, 16 August 2009 (UTC)[reply]

Micro black hole scenario

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I think we can agree that there is no significant black hole threat to the earth from the LHC or any other machine being thought about.

However, that doesn't keep us from conducting a thought experiment: A microscopic black hole will "evaporate" very quickly, before it can collect enough additional matter to stay in existence. But what if were a little more massive? Would it begin to collect particles from nearby atoms? What would be the geometry of such a situation? How close would the nearby atom, or particle have to be?

Let's specify that it begins to collect particles. There are two possibilities: 1) It continues to grow and 2) Due to a random fluctuation, it runs out of nearby particles and evaporates anyway.

Continuing with Scenario 1), it will be attracted by gravity, and start falling towards the center of the earth, eating its way through the bottom of the LCH or whatever created it, and continue on down. Would it oscillate from one side of the earth to the other indefinitely, or is there a damping factor to cause it to come to "rest" at the center of the earth?

The creation of this black hole would be hard to detect, since it is happening on a subatomic scale, and it is collecting matter at a rate just barely enough to allow the black hole to survive, a few particles at a time, at first.

Let's assume it came to rest at the center of the earth as a very small black hole, and continued to collect matter. I think there would be no net change in observable gravity for some time period. How long would that period be? Seconds, months, years? Would the earth become a more or less empty shell? If so, how large would the "hollow" region be? A diameter of one cm, one km, 1,000 km? Of course the central region wouldn't be empty, because the black hole would be pulling matter towards itself.

How long would it be before its effects became noticeable? It could be quite a while since in the radial direction at least, gravity would not change. From the surface of the earth, does a point mass look the same gravitationally as the same mass distributed through the earth's volume? Let's change the scenario and allow the black hole to continue oscillating through the earth. Now the direction of gravity will fluctuate. In that case, how long before it is detected?

What other aspects of this thought experiment are interesting?

JFistere (talk) 18:30, 13 August 2009 (UTC)[reply]

Skipping past the "how big should it be?" part, I can answer the rest.
  • It'll eventually come to rest at the center of the Earth, per friction.
  • There will never be a true change in gravity -- no new mass is being created. The Moon's orbit will never be affected.
  • However, the Earth will collapse to fill any hollows, so as to attain hydrostatic equilibrium. This will reduce the Earth's radius, and thus increase gravity at the surface of the Earth (because the surface has moved).
  • How long? Eh, too much theoreticalness. I'd ballpark millions of years, but it's very heavily dependent on starting conditions (the rate of consumption will rise exponentially). — Lomn 18:57, 13 August 2009 (UTC)[reply]
So for some time we would just notice a gradual weight loss(edit) the same weight when we stepped on the scales, and an around the world trip would be shorter? Edison (talk) 19:11, 13 August 2009 (UTC)[reply]
No, why would you feel lighter? Anyways, a black whole that would normally evaporate in one second has a mass of about 20 tons, and a Schwarzschild radius of 3e-23m. That's about 100 million times smaller than the diameter of of a proton. It will be very hard to feed the 20 tons of mass or so per second needed to balance the Hawking radiation into that small a space. --Stephan Schulz (talk) 19:24, 13 August 2009 (UTC)[reply]
I think that was a typo. He said earlier one's weight would increase. JFistere (talk) 20:38, 13 August 2009 (UTC)[reply]
What everyone tends to miss here is that a mini-black hole with a mass of (say) 1kg - still only has the gravitational attraction at (say) 10cm from it's center that a 1kg ball of iron would have at 10cm...which is to say, utterly negligable. It's only when you get very close to the black hole that you notice its gravitational pull at all. That's because the acceleration is proportional to the square of the distance. But in order for a 1kg mass to have enough gravitational pull to make a difference, you have to be insanely close to it...much less than the radius of an atom. That's the only reason we don't get sucked into 1kg iron spheres...we just can't get close enough to them (even when we're touching them). So the probability of even one more atom happening to get close enough to our 1kg black hole to get 'eaten' by it is pretty small...and even if it does eat an atom once in a while, atoms don't weigh much. You won't even start to notice it's gravitational pull at 'reasonable' distances until it's grown enormously - and statistically, that's really unlikely. When you start to factor in the evaporation rates - it's truly impossible for a mini-black hole to have any detectable effect on something as big as a planet. SteveBaker (talk) 20:56, 13 August 2009 (UTC)[reply]
20 tons of matter evaporating in a second is going to be releasing a great deal of energy, so much so that nothing will get close to it due to radiation pressure if not particle pressure. Graeme Bartlett (talk) 12:19, 14 August 2009 (UTC)[reply]
Or we could work backwards. Say we want to start with a black hole with an event horizon with a similar scale to an atomic nucleus - say 10-15 m across. This is about 1020 Planck lengths, so mass is of the order of 1020 Planck masses or about 1012 kg (modulo some factors of 2). I reckon this is about the mass of a modest mountain or small asteroid - so not so very "micro". But we don't have to worry about evaporation any more, as primordial black hole says a black hole with a mass of 1012 kg has a lifetime equal to the age of the universe. Gandalf61 (talk) 12:42, 14 August 2009 (UTC)[reply]
Question: At the scale of an atomic nucleus - would such a micro-black hole produce the strong & weak nuclear forces - or would they be unable to escape the event horizon? If it does, wouldn't it actually repel atoms more strongly than it attracts them? SteveBaker (talk) 12:59, 14 August 2009 (UTC)[reply]
Why would the strong and weak forces repel atoms? The weak force doesn't do much, but the strong force is what holds nucleons and nuclei together. What usually holds atoms apart is electromagnetism and their electron clouds. --Tango (talk) 18:23, 14 August 2009 (UTC)[reply]

Study on tea and stress

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The media in the UK is currently promulgating that "tea reduces stress", referencing a study by a psychologist at City University London named Malcolm Cross. Unfortunately I can't find the study anywhere, including Google, Google Scholar and Pubmed. Can anyone else find it? Here are some media sources: [10] [11] [12] [13]. --Mark PEA (talk) 18:49, 13 August 2009 (UTC)[reply]

I searched a bunch of medicine and psychology databases with no success. Perhaps the paper hasn't been published yet? I've certainly come across previous occasions when the newspapers have gotten all worked up by a pre-publication press release, for example, although I can't find one of those on the university's website or the site of the sponsors, Direct Line, either. A paper isn't listed on Dr. Cross's bio. Nothing at the BBC, either, which is unfortunate, as they tend to link to original documents. While I couldn't find a paper, Direct Line have put up a website and apparently had someone blend some special tea for them. On another note, I never knew we had an article on Potential effects of tea on health! --Kateshortforbob talk 21:17, 13 August 2009 (UTC)[reply]
I had better luck using Google News -- you can download a PDF of the executive summary here. Looie496 (talk) 23:14, 13 August 2009 (UTC)[reply]

Human skin color

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What kind is human skin response to different light frequencies? I'm thinking about projecting the colors of the rainbow onto a face to 3d scan it with just a few images. --194.197.235.26 (talk) 20:28, 13 August 2009 (UTC)[reply]

Your phrase "response to different light frequencies" is a very complicated way of saying the word "color". In other words, we could rephrase your question as: "What color is human skin?"...which I think you already know the answer to! Using "structured light" to do 3D digitisation is a reasonable one - but it's extremely tricky when you have to rely on the color of the surface. Projecting a grid of white light onto the face - in an otherwise dark environment is really a better approach - which is why that's how it's generally done. Another thing you need to think about is that pretty much all cameras don't respond to all the colours of the spectrum - they respond to red, green and blue. Hence you won't have the separation of information you're expecting to get. SteveBaker (talk) 20:44, 13 August 2009 (UTC)[reply]
Thanks for the answer. When I made the question I thought more about some generic rule for darker vs paler vs whatever skins, but you are probably right hinting this isn't a too good idea. On a second thought high contrast shapes sound a lot better (especially when I don't have access to too fine cameras). --194.197.235.26 (talk) 21:17, 13 August 2009 (UTC)[reply]
Reflectance of the human skin has been studied in detail. You can start with the Color Science textbook by Wyszecki and Styles. In the 1982 edition see page 63, figures 2(1.4.6) and 3(1.4.6). --Dr Dima (talk) 22:21, 13 August 2009 (UTC)[reply]
Actually, Wyszecki & Styles is a really good textbook. It is a much better idea to just read it (rather than to jump straight to the page 63). --Dr Dima (talk) 22:25, 13 August 2009 (UTC)[reply]
This is actually something of a nontrivial question - human perception of colour has only about three degrees of freedom, which is not nearly enough to uniquely describe absorption across the entire visible light spectrum. I think it would be fun to do more research into the "complete colour" of many different everyday objects. Dcoetzee 22:28, 13 August 2009 (UTC)[reply]
Correct, with only three cone types in our retinas we can not perceive the details of the absorption spectra of surfaces under any given natural illuminant. That's why it is such a good idea to read the textbook, and to try to understand what it is that we see perceive as color, and how the color percept changes with the changes in illuminant, absorption spectrum, background, and so on. --Dr Dima (talk) 22:53, 13 August 2009 (UTC)[reply]
However, the question is about digital imaging. I suspect that the rainbow being projected is not a true spectrum, but an image of one, projected with only three primary colors. The camera used to image the face after a rainbow has been projected on it will likely also use three primary colors, hopefully ones very close to the projector's. So the question becomes simply "What color is human skin, as seen by an RGB camera." You could easily get a "baseline" value by switching the projector to show all white, and then compare that image to the rainbow colored image. APL (talk) 23:23, 13 August 2009 (UTC)[reply]
While it's true that our poor, pathetic eyes can't see all of those colours, that's unimportant in this case because all commonly available cameras have the exact same restriction. They are designed to see the same way we do. So what we see is what the camera sees. You could certainly use a monochrome camera and a large selection of colored filters to build a camera that could see in many more frequencies - but the entire point of the OP's proposed technique is to do it all with just one picture. If you can take your time and take lots of pictures then you can either move the projected light source, move the camera or move the subject in order to get enough information to make a 3D image...those techniques being much simpler than messing around with 'pan-spectral-imaging' and completely avoiding the issues of people with different colored skin, what to do about colored hair, eyes, lips and other things that would undoubtedly complicate getting a 'rainbow' image into a 3D form. I agree with the point that actually generating a true spectrum isn't exactly trivial. Using sunlight and a prism is probably the simplest solution - but it would be horribly inconvenient in many applications of 3D scannera. Synthetic "true white" light sources (those with roughly equal quantities of all of the frequencies) are very rare indeed...in most cases, the spectrum of artificial light is such that you don't get a good 'rainbow' at all.
Our OP might like to investigate my personal Wiki - where I present a design that I did for a 3D scanner using a video camera and a laser pointer: [14]. That project has been taken up by a team of OpenSource programmers who have turned my 'quick hack' into something that really works very nicely: [15]. If you have a decent digital video camera already - you can build this in an hour and have working 3D scans in a day.
SteveBaker (talk) 00:08, 14 August 2009 (UTC)[reply]
It's really only two degrees of freedom. While the eye's response has 3 degrees of freedom, only two of those are what we call "colour", the 3rd is brightness. There are other ways of dividing them up, of course, but having one as brightness is the most useful. This diagram shows one way of depicting the two remaining dimensions (in a nice normalised fashion). As you can see, all the colours are shown on a region of a plane, so it is clearly a 2D space. (Note, the colours in the image are approximations, your screen can't produce the whole range.) --Tango (talk) 00:31, 14 August 2009 (UTC)[reply]
I'm a little lost - how do you intend to get depth information from color information? Amplitude is only weakly related to depth; so whatever color light you shine, you still have to do some postprocessing to convert (guess) the true depth. I don't know why different colors would have different mappings between depth and amplitude, so a single image would probably work just as well as several color-channels. Nimur (talk) 01:17, 14 August 2009 (UTC)[reply]
The general idea is to use some kind of 'structured' light to extract 3D information from a 2D photograph.
Put simply - suppose you use a slide projector to shine a regular grid of white lines onto a bumpy surface like someone's face. Then, take a photo of that surface from a little way off to one side of the projector. By noting the position of the grid lines in the photo, you can deduce how the regular grid was distorted by being projected onto that surface. That allows you (in principle - and in rather limited situations) to recover three dimensional position information from a two dimensional photograph.
The problem is that you only get to sample at the resolution of the grid - and it's hard to tell from the photo which grid intersection in the photo is from which intersection from the projector. The finer you make the grid, the harder that determination becomes - and the coarser you make the grid, the more surface detail you miss.
For that reason, more often, instead of using a grid - you use a simple straight line or even just a dot and scan it across the surface - taking lots of pictures, maybe with a video camera. This lets you use laser light - for extra precision. As demonstrated on my Wiki. Using "time" as one of the parameters gives you a way to know precisely where the light was being shone when each photo from the video was taken - and gives you vastly higher precision (my home-made scanner gets better than 0.5mm precision - and it's built out of Lego, a cam-corder and a $5 laser pointer!).
The trouble with that for the particular case of scanning human faces is that people can't sit still long enough - and shining lasers into people's eyes is NOT recommended!
Our OP is thinking to shine a "rainbow" of light onto the surface instead of a grid and to measure the color of each point in the photo to discover where a particular color from the projector ended up on the surface...this works just like the grid - but leaves no ambiguity because each point indicates where from the projection it came from by virtue of it's color - and it's as precise as you have color fidelity in light and camera. It's a rather clever idea actually.
Sadly, it fails for anything other than a perfectly uniform white surface because you can't easily tell whether a color variation at a particular point is due to a 3D displacement of the surface - or whether it's a change in reflectivity of the underlying material. So, the light that falls (for example) on the person's lips might look a little more red than it would if it had landed on the person's cheek (say)...and that would fool the image processing software into guessing that the light from the projector had been displaced by more than it really had. Bottom line is that the OP's idea is quite ingenious - but sadly not very useful for annoying practical reasons.
My rotating scanner is also imperfect - it doesn't work well for very shiney surfaces because they bounce the light off so it hits other parts of the object. None of these kinds of scanners handle parts of the surface that are never lit by the light (or which are never visible from the camera). There is an inherent compromise between this 'shadowing' problem (which forces you to keep the light source and camera very close together) and the desire to get more displacement of the image for any given 3D bump size by moving the two further apart.
However, for things like faces, the technique works really well.
SteveBaker (talk) 02:40, 14 August 2009 (UTC)[reply]
You might make it work with two images, a white light image then the rainbow image. Still seems like it'd be tricky. APL (talk) 05:42, 14 August 2009 (UTC)[reply]
Also, why go with a regular rainbow? You could go with a much more complex pattern. (Something that would probably look like noise) that way pixels that are similar in tone are not normally next to each other. APL (talk) 06:19, 14 August 2009 (UTC)[reply]
Yep - there are many alternative ways to do it - but they all require some kind of "structured light" - and some means to identify which "pixel" from the projected light source you are seeing at which position in the photograph(s). I could certainly imagine some kind of random dot pattern and some kind of image recognition trick that would match the pattern in the photo to the original pattern that was projected...but the distortion of the image and the problems of some parts of the source image not being present at all in the photo conspire to make this a fairly tricky software task. With a simple turntable-based system where the object was rotated under a vertical line of red laser light - I was able to extract a respectable 3D object with just 500 lines of software...very simple indeed. There are other techniques such as "shape from shading" (Photometric Stereo) - and I see we have an article about Structured light and even Structured-light 3D scanner. One cool trick I saw recently entailed taking two photographs, one with a flash and one without. Subtracting one image from the other gets you an approximation for the surface reflectivity - and factoring that out of the image allows you to use the brightness of the resulting monochrome image to estimate the slope of each pixel relative to the light source direction using lamberts law. Then you can integrate the slope information to get height. It's actually rather error-prone - but for many purposes it lets you get what you need. SteveBaker (talk) 12:52, 14 August 2009 (UTC)[reply]

Fun with negative refraction?

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Hi. Are there any specific optical effects with negative refractive materials that one would consider interesting? For example, how would I look in a mirror with a negative refractive index? Or, how would the world look like through a negative refractive window, etc. etc.? Thanks in advance, Kreachure (talk) 23:49, 13 August 2009 (UTC)[reply]

Basically negetive means that it will go the opposite way. For example when you put a a pen in water it appears bigger, negetive will make it appear skinnier, so much so that it might appear to be invisible (so thin you can't see it) but light from the other side still gets through ok. —Preceding unsigned comment added by 155.144.40.31 (talk) 03:36, 14 August 2009 (UTC)[reply]