Wikipedia:Reference desk/Archives/Science/2007 November 10

= November 10 =

What is the oxidation number of carbon in methanol?
What is the oxidation number of carbon in methanol? According to the article oxidation number, I get 4 (four shared pairs of electrons, in which one of each pair belongs to carbon, results in a charge of +4 upon removal). --212.204.150.105 21:14, 7 November 2007 (UTC)
 * Well, the oxidation state is the charge that carbon would have if the bonds were all ionic, so you need to know, for each bond, which atom is more electronegative. Given that this could be a homework question, I'm not going to tell you the answer.  Someguy1221 21:29, 7 November 2007 (UTC)
 * I know that the oxidation state is -2 (oxygen more electronegative and three hydrogens each less electronegative), however my question pertains to the oxidation number which according to the WP article can sometimes be different. I expect, that the oxidation number should also be two, but when I follow the instruction given in the first sentence of the WP article, I get 4, as described above. Is the article wrong or is my interpretation wrong (and if so, how)? --212.204.150.105 21:44, 7 November 2007 (UTC)
 * I belive that's only in reference to coordinate bonds. Oxidation numbers are usually only used for metals.  Someguy1221 21:55, 7 November 2007 (UTC)
 * The article on oxidation state says Redox (shorthand for reduction/oxidation reaction) describes all chemical reactions in which atoms have their oxidation number (oxidation state) changed - firstly, I don't think it should say oxidation number (especially not preferentially to oxidation state) and thirdly, I read elsewhere that it should take into account the electronegativity - I think it's possible for an atom to be reduced even without a change in oxidation number, so long as the ligands of the product are less electronegative than the ligands before. Thus I think the article is at best misleading, if not wrong. --212.204.150.105 22:31, 7 November 2007 (UTC)

Slowing time
Can you make time for a particular area seem to go slower by moving away from it? That is, lets say, the TV is on, and I start moving away from it at half the speed of light, would the video on the TV appear to be going half its normal rate because it is taking longer and longer for the light to reach your eyes? What if it moved away from a given spot and I move away in the opposite direction at half the speed of light, would it appear to freeze. Of course, a TV that big that my eyes would be able to resolve it is impossible, but just asking... 208.63.180.160 00:01, 10 November 2007 (UTC)


 * If you moved away from the TV, in addition to something similar to the doppler effect causing it to appear in slow motion, and the actual doppler effect making everything have a lower wavelength (redshift), the relatively slower speed of light through you would make time pass slower for you. It wouldn't counter it out perfectly. Two objects moving at half the speed of light away from a certain point aren't moving the speed of light away from each other, due to the time shift. This is all special relativity stuff. If it's too advanced for you, you may prefer Introduction to special relativity or the simple english article for it. — Daniel 01:42, 10 November 2007 (UTC)


 * Moving either towards or away from the TV (or the TV moving towards or away from you) has the same effect, the speed of the show on the TV would change by a rate determined by the square root of 1-v2/c2. If your speed is half the speed of light then you have v=0.5c - so the time distortion would be sqrt(1-0.25) or about 87% of normal speed.  —Preceding unsigned comment added by SteveBaker (talk • contribs) 18:36, 10 November 2007 (UTC)


 * Hold on... I obviously have some sort of misunderstanding here. If the TV is moving towards me, will the photons not hit me a a higher rate than if it is moving away? If a tennis ball launcher throws a ball every second, and it is moving towards me, I will be hit by balls at a rate of greater than one per second; isn't the concept the same, so I will see the events on the TV speed up by moving towards it because the light has to travel a progressively shorter distance and thus I see "newer" light quicker than I would standing still? (I have no understanding of special relativity, and I'm not understanding the articles). 208.63.180.160 01:39, 11 November 2007 (UTC)


 * Your view of the way light (photons) work is incorrect - and that one single fact is the entire reason we have relativity and everything that goes with it. Unlike tennis balls, the speed of light is a constant - irrespective of the motion of the source or the viewer.  This was shown most clearly by the Michelson–Morley experiment.  Hence, the photons coming from the TV will hit your eyes at exactly the same speed regardless of whether you are moving closer to the TV or further away from it.  You'll see 'red-shift' (if you are moving away from the TV) or 'blue-shift' if you are moving towards it - but those are changes don't alter the speed.  What's happening here is MUCH weirder than the 'classical' physics view that you are taking here.  When you move at half the speed of light relative to the TV, the passage of time, distances and masses all change by that 87% factor - even after you take into account the fact that you are getting closer to it so that the light waves have a shorter distance to travel. SteveBaker 02:18, 11 November 2007 (UTC)


 * Actually there is a speed-up effect when the TV is coming towards you at a good portion the speed of light, Steve. Imagine this isn't a TV you're looking at, but just a strobe light that pulses once a second in its own reference frame.  Yes, if it's approaching you at 50% the speed of light, it only strobes once every 1.155 seconds from your perspective.  BUT, since the strobe is coming towards you, the pulses will actually arrive faster than you'd expect.  This isn't exactly the doppler effect, but it's very similar.  I don't feel like precisely calculating which is more important, but its quite obvious if you look at extreme situations.  Imagine the strobe begins transmitting from 1 light year away, but is travelling towards you at a Lorentz factor of one million.  The strobe in this case is trailing so close to its own transmission that the entire 1 year lifetime of its transmission suddenly arrives at you in a 32 second spurt just a year after its light first reaches you.  Now, it so happens that due to relativistic slowdown, the strobe only actually managed 32 strobes in this time, so the visual speed-up effect cancels out the relativistic slow-down effect (the former is entirely a deficiency of observation, the latter is physically happening).  And so at best this can simply cancel out the appearance of relativistic slowdown, though certainly it would exacerbate such if it were traveling away from you.  An effect that is not cancelled out by relativity, however, is the appearance that the strobe was traveling at a velocity equal to its Lorentz factor, in units of the speed of light.  This actually has to be taken into account in observations of the plasma ejections of quasars, some of which have an "apparent velocity" greater than that of light, before this effect is accounted for.  Someguy1221 02:50, 11 November 2007 (UTC)


 * What you are describing is exactly the Doppler effect. Its magnitude is given by the Doppler formula, the longitudinal case of which is quoted in MrRedact's reply below. -- BenRG 00:17, 12 November 2007 (UTC)

Steve made a mistake. Your (the original poster’s) intuition about the case where the TV is moving away from you is almost correct. If you’re watching a TV moving away from you at half the speed of light, it will look like the show is being shown at 0.577 times as fast as its normal rate. In the limit as your speed away from the TV approaches the speed of light, the rate at which the show looks like it’s being shown approaches 0, i.e., it approaches looking like the TV show is stuck on one frame.

If you’re watching a TV coming at you at half the speed of light, it will look like the show is being shown at 1.732 times as fast as its normal rate. In the limit as your speed toward the TV approaches the speed of light, the rate at which the show looks like it’s being shown approaches infinity.

In general, if you’re approaching the TV at speed v, the apparent frame rate of the TV show is proportional to (1+v/c)/sqrt(1-v2/c2), where c is the speed of light. The same expression also works for the case that you’re moving away from the TV, in which case v is negative.

In reality, the colors of what's on the TV are affected if you're moving relative to the TV, so if you're moving too quickly, the colors would change so much that your eyes wouldn't be able to see them. MrRedact 08:18, 11 November 2007 (UTC)
 * Has any tried watching a TV while moving at a fraction of the speed of light? Most of the time the screen will be too tiny to see!  Even at 0.000001% of the speed of light you will only get a few seconds viewing out of it! (joking here). Graeme Bartlett 20:23, 11 November 2007 (UTC)


 * Those who claim I made a mistake didn't read my reply - I specifically pointed out that you get the time dilation effect even after you take into account the fact that you are getting closer to it so that the light waves have a shorter distance to travel. SteveBaker 02:20, 12 November 2007 (UTC)
 * Not sure what you mean by "even after" -- the answer is that the frames will appear to move faster, not slower. By the way MrRedact's expression (1+v/c)/sqrt(1-v2/c2 simplifies to the nicely symmetrical-looking $$\sqrt{\dfrac{1+v}{1-v}}$$ (where of course I'm using c=1, as is generally a good idea when talking about relativity).
 * Now in practice, of course, if you're going fast enough to notice any effect at all, you'll be going far too fast to process even a single frame from the TV. --Trovatore 02:28, 12 November 2007 (UTC)


 * Your confusion is consistent with misremembering what an "observer" is in relativity. In relativity, an "observer" isn’t one person at one location, but effectively actually a whole system of people spread throughout space, at rest relative to each other, who have a set of synchronized clocks.  The time at which at event occurs is measured at the location at which the event occurs.  So the difference in time between two events, which is what time dilation measures, doesn’t account for any time it takes for light to travel anywhere after either of the two events.


 * Suppose the TV is traveling at half the speed of light in the x direction relative to the lab frame, and suppose the TV has a mirror or two attached to it such that the screen can be seen by someone either in front of it or behind it. We won’t even need a Lorentz transformation for this; in the following, all coordinates are lab coordinates.  Pick units such that c=1, and the period between frames of the show (as measured in lab coordinates) is 1.  Pick the origin such that the TV starts to show frame number 0 when the TV is at x=0, t=0.  At time t=1, when the TV starts to show frame number 1, the TV will be at x=0.5.


 * To someone at rest in the lab frame sitting at the origin, the light from the start of frame 0 will reach them at t=0, and the light from the start of frame 1 will reach them at t=1.5. To someone at rest in the lab frame sitting at x=0.5, the light from the start of frame 0 will reach them at t=0.5, and the light from the start of frame 1 will reach them at t=1.  So the apparent frame rate as seen by the person in front of the TV is 3 times faster than the apparent frame rate as seen by the person behind the TV.  This is consistent with the ratio 1.732/0.577 of the numbers given in my post above.  MrRedact 07:29, 12 November 2007 (UTC)

Greenhouse Gases
Are Greenhouse Gases good for the environment or bad? —Preceding unsigned comment added by 70.171.192.2 (talk) 02:24, 10 November 2007 (UTC)


 * Have you consulted the article greenhouse gas? By themselves in moderate amounts they are not "bad" but in excess, at the rates that they are currently produced by human activities, they increase the greenhouse effect which leads to global warming. Which is bad. --24.147.86.187 02:31, 10 November 2007 (UTC)


 * We need to have some, otherwise the earth would be freezing cold all over. Carbon dioxide is essential for the life of photosynthetic plants.  Graeme Bartlett 12:09, 10 November 2007 (UTC)


 * The 'normal' amount (about 0.038%) of CO2 is essential - less than that would be bad because the earth would freeze, more than that would be bad because of global warming. Right now, we have all the CO2 we need - and it's going up - so adding more greenhouses gas is definitely bad.  Even if humans ceased to produce greenhouse gasses at all, the natural amount from animal respiration and volcanos would be plenty to keep the earth running OK.  SteveBaker 18:31, 10 November 2007 (UTC)


 * I have no idea where the idea that global warming is bad came from. An increase in global temperature has a positive net effect on humanity. Look at climatic changes of the middle ages in europe and you will see that the warm periods were better recieved by the population than the cold ones. The main cause are reduced heating requirements and increased agricultural yields. Putting this to a global perspective: there is not a single area on earth to hot to live in. Too dry, yes, but not too hot. However there is a whole uninhabitable continent at the south pole, covered in mountains of ice.
 * So its good. The real problem is, at some point it will be too much. We don't know how the ecosystem reacts to really dramatic increases in temperature. It might be fine with 8° and collapse at 9°. Who knows? So the reason for reducing CO2 emissions is: we do not want to find this out the hard way. —Preceding unsigned comment added by 84.187.90.130 (talk) 00:49, 11 November 2007 (UTC)


 * WHAT! WHAAAT! Where have you been the last decade?  This is quite the most ill-informed, unthinking response I've ever read on this topic!  Firstly, the change in the middle ages was small compared to what we're talking about - also it was over fairly quickly and this time around it's going to be permenant.  Secondly, while crop yields increase in the extreme latitudes, they decline sharply in the equatorial regions - the net effect will be disasterous.  Reduced heating requirements are replaced by increased air-conditioning requirements (you've never lived in Texas have you?!).  The antarctic region might become free of ice - but whats uncovered will be bare rock - no soil.  The sharply increasing sea levels will drown all of the nice flat primo agricultural plains around the coastlines.  You suggest that an 8 degree rise "might be OK" - but I can tell you that a 4 degree increase will be plenty disasterous.  Please - do some reading of the proper scientific literature on this subject...or if (as it seems) you are unable to learn from these sources - rent a copy of "An Inconvenient Truth" - it's very comprehensible. SteveBaker 02:08, 11 November 2007 (UTC)
 * I don't think anyone is talking about an 8C rise. I think a 4C rise was on the tall end of the IPCC report for a 100 years.  Sea level rise will probably affect the salinity of river deltas and a lot of food production areas but to say it will drown all the agricultural plains around the coastline is a little too pessimistic.  As for AIC, any film that links smokestacks to hurricanes as AIC does on it's cover and in the material is not science.  Stick to IPCC.   Summary reports if you don't have time.  Read about the discrepancies and conflicts if you do.  Cloud cover and carbon sinks in the ocean are too big ones that have a lot of research and understanding yet to be done.  --DHeyward 06:36, 13 November 2007 (UTC)


 * 0.038% is the current CO2 concentration, the pre-industrial level was about 0.027%. A question to the experts: How much higher would the equilibrium (glaciers need time to melt...) temperature be at 0.038%, compared to 0.027%? Icek 04:04, 11 November 2007 (UTC)
 * What do you mean? I thought I understood your question except for the glacier part.  Glaciers have their own cycles for formation and melting.  They've been melting on average for 10,000 years.  Melting has been accelerated due to global warming but as I understand it, there was no equilibrium before.  Glaciers can grow and shrink in both warming and cooling periods.  --DHeyward 06:41, 13 November 2007 (UTC)
 * Well, I thought that the melting of the glaciers would be a heat sink for some time, keeping temperature lower as long as there are large glaciers. As the glaciers get smaller, the heat sink would get smaller, and the temperature would rise. Therefore I thought we have to think about the glaciers if we want to compute an equilibrium temperature. I see that the time for reaching such an equilibrium is quite long, maybe longer than climate cycles due to changes in Earth's orbit, obliquity, and equinoxes.
 * Restating my question: Averaged over the long-term climate cycles of the next few 100,000 years, how much higher would the temperature be if we stopped burning fossil fuel now? And how would the CO2 concentration develop? Icek 07:45, 13 November 2007 (UTC)

I'm not an eco scientist but I do Know that the UK's climate relies heavily on the Gulf slipstream. If the polar caps melt, the sea temperature drops by a degree or so and the slipstream stops. That would be the UK buggered! 88.144.64.61 08:40, 16 November 2007 (UTC)

Battery charge
59.93.9.23 05:35, 10 November 2007 (UTC)
 * 1) How do rechargeable devices like mobile phones and mp3 players measure the amount of charge/energy left in the battery? Is it by measuring the terminal voltage? (which would be inaccurate due to polarization effects)? Or do they have ampere hour meters inside them?
 * 2) When we operate them continuously the meter suddenly drops low and after switching off for some time the bar goes back up a little; which would explain the drop in voltage due to polarisation. Am I correct in assuming that?
 * Last I heard, the battery life indicator uses the voltage at the terminals. It is not accurate when it comes to predicting how long until the device quits. Each battery will be in a different stage of its life, and an older one will usually die sooner at, say, two bars. The graph is not linear even for any one battery, the change in voltage is very small, and current demand is unpredictable. You would need laboratory-standard equipment to make the thing at all accurate even under controlled conditions. I'm sure the phenomenon you mention in 2 is due to what is called polarisation, though the battery will drop some voltage internally under load anyhow. If it's right on the line between bars, you'll see a change upon power off just from that. The indicator will have some kind of anti-hunt designed in to keep the display from jumping around, so it will itself be slow to react. --Milkbreath 16:43, 10 November 2007 (UTC)


 * This is a good question, which I wish I had some more definitive answers to. (But since when has lack of definitive information stopped an armchair RD reader from speculating?)
 * There are at least four things a battery-charge indicator could look at in trying to make a determination of how much life the battery might have left:
 * Terminal voltage. A battery is, of course, a two-terminal device, so fundamentally, this is all you've got access to.  Unfortunately, by definition, a battery is supposed to be a constant-voltage device, so its voltage shouldn't (and doesn't) change much over its lifetime.  A theoretically ideal 1.5-volt battery would give 1.500000000 volts for its entire lifetime, then crash precipitously to 0 -- so a charge indicator looking only at voltage would have nothing to go on!
 * Of course, we don't have to limit ourselves to looking at instantaneous voltage; we can also look at rate of change (dV/dt). I'm pretty sure the voltage drops at different rates at various points during a typical (non-ideal) battery's discharge curve.
 * History. Top-end, full-featured batteries (such as the ones used in modern laptops) contain their own microprocessors.  These can learn what that particular battery's discharge curve looks like, and use that knowledge to make a much better estimation of how much life is left based on where in the (now known) discharge curve it looks like we are.
 * Current draw. If the device has a built-in ammeter so that it can measure how much current is being drawn, and if it knows (perhaps based on historical information discovered by #3) what the battery's capacity in amp-hours is, it can make a very accurate estimation of how much life there is left.  Of course, that estimate can and will vary if the current draw changes.  I believe that's one reason why the bar can go back up.  For example, I regularly notice my laptop's expected lifetime jump back up just after I stop doing something CPU-intensive.
 * Of course, this is all complicated by real-world considerations. Rechargeable batteries have a limited number of charge-discharge cycles, and can't hold a charge for as long the more cycles they've gone through, and are also prone to notorious "memory" effects.  (I don't know how hard microprocessor-based batteries work to assess these effects; though they certainly could.)  Also, most batteries show a sort of "rejuvenation" effect when they've been given a chance to rest after working hard (i.e., just like you or me).  That's the other possible explanation for the bar-going-back-up phenomenon you noted. —Steve Summit (talk) 23:04, 10 November 2007 (UTC)

Thanks for the replies. Googling gives (what seems to be the terminology for this sort of thing) State_of_charge. Some websites and the wiki article say that in laboratory conditions they measure the concentration of the electrodes/electrolytes The results also include the microprocessor thingi. It seems to be called Charge_controller device. 59.93.9.69 04:35, 11 November 2007 (UTC)

Freezing Water Question
What would happen if you tried to freeze water it it were confined so it could not expand? For instance if a quantity of water were enclosed in a solid block of steel and the whole thing were subjected to low temperature and the water could not expand as it would when freezing, what would happen to the water? Would it stay liquid or freeze solid without expanding???? —Preceding unsigned comment added by 207.69.137.23 (talk) 05:57, 10 November 2007 (UTC)


 * Depends on where it is on the phase diagram. Freezing water in a confined space creates pressure on the order of several atmospheres (you can drive a go-cart with it) which may push it back into liquid phase, but again, it depends on both pressure and temperature. The way the ice crystals form is also dependent on this diagram, so you should really just check out the phase diagram article. SamuelRiv 06:11, 10 November 2007 (UTC)


 * Better yet, check out the responses from the last time we had this question. In brief, as the temperature is lowered, at first the water will remain liquid and generate an increasing pressure.  If the temperature continues to be lowered, eventually it will freeze into a form (phase) of ice denser than the everyday kind.  --Anonymous, 06:16 UTC, November 10, 2007.

anti inflammatory
do anti inflammatory medicine (gen-naproxen to be specific) affect your mood? —Preceding unsigned comment added by Morvarid rohani (talk • contribs) 08:08, 10 November 2007 (UTC)


 * In the list of side effects of naproxen (see here), no mood disorders are listed as frequently reported. However, depression has been reported in 1% to 10% of patients, and anxiety has been reported in <1%. Reports do not necessarily indicate causation.  The fact that a drug causes a side effect in some patients does not mean that it is the cause in a specific instance; any question of whether a drug is responsible for a specific clinical condition should be discussed with a physician. - Nunh-huh 08:45, 10 November 2007 (UTC)

Bullets
Hi. How exactly does a bullet to the brain kill? Why do some people shoot the mouth, yet others shoot the temple? At what point does life cease? 203.124.2.43 11:44, 10 November 2007 (UTC) Adam


 * I used to have a great animation of a bullet going through the brain, bouncing off the other side of the skull and basically making a soup of the brain matter after bouncing around several more times, but I can't find it. When a bullet enters one side and exits the other, the entry and exit take a lot of compressed gases and brain matter with them making a small explosion in those areas, which can greatly increase trauma (that mostly depends on the bullet head shape). Sometimes bullet injuries leave people alive, like Manfred von Richthofen (the Red Baron) and Phineas Gage (an iron rod, not a bullet). In Phineas's case, the spike damaged mostly one area of the frontal lobes, which govern a lot of higher-order reasoning and personality, but not so much in terms of low-order processing.


 * So let's get to the meat of your question: assuming a pointed bullet with enough speed to not bounce around the brain so that extra trauma is minimized, how do you kill someone? Shooting through the temple kills mainly by hitting the limbic lobe, which contains a lot of mid-order processing (thalamus), memory, and important regulatory glands. It is a guaranteed kill in some sense because you can destroy everything a person perceives about the outside world, whether or not their heart actually stops immediately. Shooting through the mouth is more appropriate because on the other side lies the brainstem and cerebellum, both of which control low-order function like breathing and heartbeat, with the brainstem being the connection of the brain to the body. So the kill in this case would be roughly instantaneous, though actual brain death would occur a bit later, because there would be some latent blood flow.


 * One more thing - severe brain trauma results in a couple of defense mechanisms by the brain. One is the coma, which for most bullet injuries would set in quickly, so the person would be incapacitated in any case. The other is a release of something (I forget what - I believe glucose) into the cell ether which results in a massive killing of brain cells. My neuroanatomy professor referred to it as a "self-destruct mechanism", but we don't know yet why it exists. Regardless, that can easily make brain trauma much more damaging than that of the actual impact. External links: and  SamuelRiv 14:07, 10 November 2007 (UTC)


 * 1. A brain can't absorp the shock of the bullet's entry without sustaining damage. 2. If that doesn't kill, bleeding will. - 87.211.75.45 19:32, 11 November 2007 (UTC)

Gold-labelled antibody = "fusion protein"
Is it appropriate to cover protein-non-protein ligations in the fusion protein article? I was just starting an article called conjugation (biochemical) and want to determine whether its warranted or not. Conceivably, one article should cover conjugation of proteins with other proteins, non-protein molecules, and possibly even non-protein-non-protein conjugations (none spring to mind). Perhaps the article protein engineering is more suitable for this? In which case, I can add a link at the conjugation disambiguation page? --Seans Potato Business 13:17, 10 November 2007 (UTC)
 * Wikipedia probably has many stubs such as Radioiodinated serum albumin and Biotinylation that could be collected into Protein labeling. I suggest not placing most "protein labeling" methods in the fusion protein article, but some fusion proteins are used to attach a label to target proteins. Sadly, Green fluorescent protein only seems to have an external link for the important topic of GFP fusion proteins. Maybe Conjugated protein could be expanded to include both natural and artificially conjugated proteins. I'd leave protein engineering to itself. --JWSchmidt 14:52, 10 November 2007 (UTC)


 * (EC) As a biochemist, I've never seen the term 'fusion protein' used to describe a non-protein combination. 'Fusion protein' is a bit more specific; it implies that a change has been made (insertion, replacement, and/or concatenation) to the primary amino acid sequence of the protein.  This runs the gamut from adding a little tiny His tag through humanizing an antibody to attaching a whole additional protein (like GFP or a second part of an enzymatic complex).  I would include under the 'fusion protein' definition the addition of domains that will dock particular prosthetic groups (a domain that binds a single metal atom, for instance).
 * For something like a gold-labelled antibody, you could use exactly that term. 'Immunogold', 'gold-tagged antibody', and even 'gold-conjugated antibody' come up a fair bit, too.  For a general term to describe 'sticking something interesting to a biomolecule', 'conjugation' is probably as good a word as any.
 * As an aside on the topic of 'non-protein-non-protein conjugations' you need to remember the other important classes of biomacromolecules: DNA, RNA, and polysaccharides. All of them can be (and often are) modified with various sorts of labels (fluorescent, radioactive, immunogenic) to allow them to be studied.  Conjugates that modify their function are also used sometimes, though this is perhaps less common.  On a terminology point, the DNA equivalent to a 'fusion protein' would probably be 'recombinant DNA'. TenOfAllTrades(talk) 15:11, 10 November 2007 (UTC)


 * So perhaps conjugation (biochemical) would be a suitable umbrella for all conjugations, protein and otherwise. I don't have time for it now, but eventually... --Seans Potato Business 23:36, 10 November 2007 (UTC)


 * I think that should read Conjugation (biochemistry) (so it avoids adjectives in the brackets). - 87.211.75.45 19:29, 11 November 2007 (UTC)

Neurotransmitters
In "Talk:Neurotransmitter", I have described a recent experience involving dopamine, norepinephrine, and serotonin, attempting to link symptom clusters with specific neurotransmitter changes and extremes. Does my interpretation appear correct? Which receptors appear to have been preferentially overstimulated or understimulated? 66.218.55.142 15:13, 10 November 2007 (UTC)

DNA sequence around integrated HIV viral DNA attachment site?
Can anyone point me to books, published articles or other currently available research data which describe exactly what is the base pair sequence of integrated HIV viral DNA - I mean after the integrase process is completed, what is the DNA base pair sequence around those "attachment" sites? For example, what is the sequence when DNA in cell is cut in order to integrate viral DNA segment:

--TG...and_here_comes_HIV_DNA...CA--

--AC............................GT--

(This above is just example of what I'm looking for, those TGAC.. are just for example!) And then once the whole thing is integrated, what are those first few base pairs around both attachment sites? For example:

--TG...(U3RU5---HIV-DNA---U3RU5)...CA--

--AC...............................GT--

So, what are U3 and U5 of linear terminal repeat (LTR) attached to once the segment is integrated (the sequence of first few base pairs on both sides) and what does it look like together with U3 and U5 on both sides?? MANY THANKS to anyone who can point me to literature which covers this. --80.95.231.124 16:13, 10 November 2007 (UTC)


 * Have you seen HIV structure and genome? The literature at the end might be useful. SamuelRiv 17:34, 10 November 2007 (UTC)

Rocket speed
After how long does a space rocket reach 500 km/h and 1000 km/h ? Is there a graph of speed/time for rockets somewhere? I'm not very good with equations and I couldn't find the ones I thought necessary in the article rocket. Keria 17:54, 10 November 2007 (UTC)


 * Well, obviously, the time dramatically varies depending on the type of rocket. Something like the AIM-9 air-to-air missile accellerates at over 20g's.  The space shuttle accellerates at a more pedestrian 3g's - so it piles on speed at roughly 30 meters per second every second.  So we convert 500km/h into meters per second (500,000m/3600seconds = 138 meters/second) - takes roughly 138/30 seconds...4.6 seconds!  To get to 1000km/h takes only 9.2 seconds.  Well, in reality it's a lot more complicated than that - the various rockets start off accellerating fairly slowly because of the weight of all of the fuel - as the fuel burns off, the accelleration builds up - but at some point before the shuttle leaves earth's atmosphere, the speed becomes too high and the engines have to be throttled back to relieve the pressure on the spacecraft.  Then as they get higher up and the air is thinner, they can go faster.  When the SRB's detach, the spacecraft gradually slows down until (as yet more fuel is consumed), the engines can push the accelleration up again.  Finally, the g forces start to get too large and again the engines have to be throttled back.  Hence, in reality, the total time is going to be longer than 9.2 seconds - but still, it's piling on speed pretty amazingly fast.  However, that AIM-9 rocket adds 200 meters per second every second - so it hits 500km/h within about two thirds of a second and 1000km/h in one and a half second!  SteveBaker 18:23, 10 November 2007 (UTC)


 * You're a hell of a pedestrian. Hope I'll never run into you. —Preceding unsigned comment added by 84.187.90.130 (talk) 00:35, 11 November 2007 (UTC)

Robot Cars
What exactly is an autonomous robot car and how does it work? Could it work for a smaller or toy car? —Preceding unsigned comment added by 68.120.224.217 (talk) 19:27, 10 November 2007 (UTC)
 * See DARPA Grand Challenge and links. --Sean 20:35, 10 November 2007 (UTC)


 * Currently, they take a standard car and add levers to push on the pedals and shift the shifter - also something to turn the steering wheel - those are hooked up to a computer that can therefore drive the car just like a human would. The tough part is that the computer has to be able to see where it's going and have enough intelligence not to do anything stupid like ramming the car into a brick wall.  For this they use a combination of digital video cameras, and laser range finders (see Lidar - like Radar but with light instead of radio waves).  Each individual technology is quite well known and understood - the tricky part is getting them all to work well together.  We could make toys that did this - but the lidar sensors and the amount of computer power needed would be quite significant.  But toys like the Aibo robotic dog are fairly sophisticated and can do some quite impressive things.  It'll happen sooner or later.  SteveBaker 00:24, 11 November 2007 (UTC)


 * You can do a lot with Lego Mindstorms, too. (I think; I still haven't given in and gotten a set for myself yet.) —Steve Summit (talk) 02:36, 11 November 2007 (UTC)


 * I have a few of the older sets - the newer ones are simultaneously better and less good for complicated reasons. But yes - you can easily build a "blind" robotic car using Mindstorms - but one that senses it's world and reacts accordingly (such as the ones in the Darpa Challenge) is MUCH harder and isn't really possible with the limited sensors that Mindstorms provides.   It's relatively easy to do things like building a car-like robot and having it follow a black line made with electrical tape - or having it seek out light and park itself in a puddle of sunlight on your livingroom floor.  But driving around in a complex environment without colliding with things and mowing down pedestrians...there is no way that a Lego Mindstorms machine could do that.   It's a lot of fun though - and it's about on the limit of how complex something can be and still be considered a "toy" that kids could build and program.   The most fun I ever had with it was making a pair of identical robots that could play "tag" in a darkened room.  Each one had a bright light on the top and a 'bump' sensor that told it if it had been run into.  One robot was programmed to seek light and the other to avoid it.  When the 'bump' sensor detected that it had collided with something, the robot that was seeking light would send an infra-red message to the other robot  saying (essentially) "You're It!" and switched from seeking light to avoiding it.  When the other robot received a "You're It!" message, it would stop still, beep ten times at one second intervals and then switch from avoiding light to seeking it.   The result of two of these little guys scurrying around the room was just hilarious to watch - but (and this is the ENTIRE problem here) the system only worked if neither of them went behind the sofa (killing the light seeking/avoiding behavior) or if they bumped into something else other than the other robot.  I was able to add some sophistication to kinda/sorta fix those problems - but it rapidly becomes obvious that you need something better than a simple directional light sensor.  Having a camera and lidar would help a lot!  Some enterprising people were able to make a very crude lidar-like system using the infrared messaging system to send a message and looking for IR reflections in the light sensor.  The strength of the returned reflection enabled some limited range measurement - but it was very crude.  SteveBaker 02:51, 11 November 2007 (UTC)

Jupiter (the planet)
How was Jupiter formed and how old is it? —Preceding unsigned comment added by 72.38.227.206 (talk) 22:07, 10 November 2007 (UTC)


 * You can read about it at Jupiter. —Steve Summit (talk) 22:21, 10 November 2007 (UTC)


 * ...which, I'm sorry, doesn't really answer your question. Anybody have a better reference?  (There's a bit of information at Solar System.) —Steve Summit (talk) 22:27, 10 November 2007 (UTC)


 * Planetary formation (which, for some reason, is currently a redirect to "Nebular hypothesis") may answer this and the following question in more (perhaps too much) detail. In any case, the best one can say about the ages of the planets is that, as we currently understand planet formation, they're all about equally old, and slightly younger than the Sun itself.  The reason it's hard to even define an exact age for a planet is that, after the initial aggregation of small planetesimals from the protoplanetary disk, they are believed to have undergone an "oligarchic growth" phase where the proto-planets of various sizes grew by colliding and merging at random.  Thus, it's hard to say which collision was the one in which the resulting planet became the one we know today.  There's no specific start or end to the oligarchic growth phase either; while the biggest planets eventually settled into more or less stable and non-colliding orbits, the smaller, more numerous bodies continued to collide with them and each other at gradually decreasing frequencies, as they still do to this day.


 * Also, it's worth noting that, for gas giant planets like Jupiter, there are two competing theories of their formation. The one that, as far as I can tell, seems to be enjoying the greatest popularity at the moment is that they started out like the smaller planets, as solid bodies growing by collisions, but eventually grew big enough that their gravitational pull could directly capture and hold down gas from the surrounding protoplanetary disk, leading to runaway growth as more and more gas fell down upon the initial rocky core.  This didn't happen with the inner planets because, by the time they grew big enough, the solar wind had already blown most of the gas away from that part of the solar system.  The alternative hypothesis, however, is that the gas giant planets formed directly from the gas disk without any solid "seed"; according to that theory, an eddy in the protoplanetary disk simply pulled enough gas together that its mutual gravitational pull caused it to collapse, much as stars are believed to form.  Neither hypothesis is by no means disproved yet — it's even possible that gas giant planets can and do form by both mechanisms.  —Ilmari Karonen (talk) 00:07, 11 November 2007 (UTC)

The planet Mars
How old is Mars? —Preceding unsigned comment added by 72.38.227.206 (talk) 22:15, 10 November 2007 (UTC)


 * You can read about it at Mars. —Steve Summit (talk) 22:21, 10 November 2007 (UTC)


 * ...which, I'm sorry, doesn't really answer your question. Anybody have a better reference?  (There's a bit of information at Solar System.) —Steve Summit (talk) 22:27, 10 November 2007 (UTC)


 * I presume Mars must have been formed at roughly the same time as the Earth - roughly 4.5 billion years ago. (See Formation and evolution of the Solar System and Age of the Earth.) We haven't studied the geology of Mars well enough to know for sure. But the age of the Earth is something of a compromise between the age of the oldest rocks we can find (3.9 billion years) and the age of the solar system (4.6 billion years).  It's very likely that even after we have crawled over every inch of Mars looking for old rocks, our answer will be about the same.  SteveBaker 00:17, 11 November 2007 (UTC)


 * Part of the problem with rock ages is that rocks form and reform due to subduction of the Earth's crust and the much different environment of early Earth, so today's rocks are going to be younger than Earth as a whole. Also note that most of our aging techniques won't work on liquids like magma. SamuelRiv 01:27, 11 November 2007 (UTC) Addendum: note there is no plate tectonics on Mars, though there is volcanism. Thus somewhere deep under the surface may be a rock as old as the solar system. SamuelRiv 01:28, 11 November 2007 (UTC)


 * I have some quibbles with that. Firstly, we know for sure that the young Mars had volcanoes.  Olympus Mons - for example - is the largest extinct volcano in the known solar system!  If Mars once had a liquid core sufficiently close to the surface to allow volcanism then it seems entirely possible that it once had plate tectonics too.  Secondly, despite the ravages of plate tectonics, the date of 3.9 billion years for the minimum age of the earth comes from dating zircon crystals from Western Australia - so despite subduction and volcanism - we still have some pretty old rocks lying around at the surface where we can find them. SteveBaker 02:30, 11 November 2007 (UTC)


 * I addressed the issue of volcanism, and note the volcanoes on Mars are all shield volcanoes (like Hawaii), and thus are not dependent on plate tectonics. I'm aware that very old rocks have been found on Earth - I'm just addressing why they might be hard to find, and indeed why one as old as the Earth may be impossible to find. SamuelRiv 16:23, 11 November 2007 (UTC)

Cause and Effect
Is it right that not one thing in the Universe hasn’t obeyed the Laws that govern it? If EVERYTHING obeys the rules of Cause and Effect is it not true that everything happens for a reason? If this is true it would seem that nothing can be random and everything is predicatbale. Sorry if this is the wrong place to ask such a question. —Preceding unsigned comment added by 71.150.248.92 (talk) 22:34, 10 November 2007 (UTC)


 * This topic has dogged philosophy and religion since time out of mind. There's a bunch of stuff on the topic at Causality, Determinism, and Free will. It seems obvious that events are caused by other events. I wouldn't be typing this right now had you not asked the question. But it seems just as obvious that we are not robot-like automatons. I feel as if I am making choices about the words I type. The conflict between these two obvious things is, I think, at the heart of most religions and much of philosophy. Personally, I think that something must be wrong with the question, since people have been asking it for thousands of years and still end up in paradox. Pfly 22:57, 10 November 2007 (UTC)
 * For many interesting physical systems, predictability is not possible because even small variations in where you are now can result in large changes in where you will be. See Chaos theory. --JWSchmidt 23:29, 10 November 2007 (UTC)


 * Well, let's be really careful here:
 * Is it right that not one thing in the Universe hasn’t obeyed the Laws that govern it?  - Yeah - that's right. If some 'law' is ever disobeyed then it isn't a law...so this is true by definition.  Is it the case that some of the things we humans believe are 'laws' are wrong?  Yes - it's really unlikely that everything we think we know is 100% correct.  Newton's "laws" of motion were proven wrong by Einstein's relativity.
 * If EVERYTHING obeys the rules of Cause and Effect is it not true that everything happens for a reason? - "Cause and Effect" is not a law. To the contrary, many quantum effects do not follow 'cause and effect' and are fundamentally unpredictable.  If you irradiate an atom and stuff some extra neutrons into it - it will eventually decay back down to it's original state by emitting a neutron.  When will that happen?  Well, we don't know, we cannot know - it's truly, utterly random.  Where is the "cause" of that neutron being emitted?  There isn't one.
 * If this is true it would seem that nothing can be random and everything is predicatbale. - To the contrary, at its heart absolutely everything is completely random and unpredictable. The only thing that makes the universe seem stable and follow nice cause-and-effect rules is the effect of statistics.  We don't know when one irradiated atom will decay - but we know with great precision how a kilogram of atoms will decay.  We can't know the exact position of an electron - but we can deduce the position of a planet orbiting a star a hundred light years away by measuring the tiny wobble it induces in that star.  The universe on the large scale obeys rules - but at the small scale, it's truly random.  But the large scale effects are pure statistics.  It is perfectly possible (although exceedingly unlikely) for a grand piano to appear out of nowhere in your living room right now.  It's only statistics that enables us to say that this "Won't ever happen".
 * Look at it like this, if you flip a coin, you have no idea whether it'll come up heads or tails. If you flip 100 coins, you can be pretty sure that between 40 and 60 heads will show up - but predicting that you'll get 50% heads is a bit 'iffy'.  If you flip a million coins, you can be quite sure that between 499,000 and 501,000 heads will show up - so a 50% prediction is a fairly accurate 'law'.  If you flipped as many coins as there are atoms in a grand piano, your prediction of 50% heads would be precise to within one part in a billion billion billion (probably much better than that actually).   In effect, you have a cast iron "law" of nature that says "when you flip coins you absolutely always get exactly 50% heads" - but that's not even close to being true for four coins - and it's POSSIBLE to flip a million coins and for them all to come up heads...it's just so unlikely that on a large scale, it's not going to happen.  That's how our "large scale" laws operate.  They are so accurately true that we can rely on them - even though at their heart, they are relying on completely random events.
 * Sorry if this is the wrong place to ask such a question. - This is the perfect place to ask this question!
 * SteveBaker 23:58, 10 November 2007 (UTC)

Thanks for all the replies. Very interesting answers. —Preceding unsigned comment added by 71.150.248.92 (talk) 02:45, 11 November 2007 (UTC)
 * I have read alot of extremely well-researched, wonderfully articulated replies on the Science Desk, but SteveBaker's reply to this ultimate of all quandries is truly amazing. The idea that science, even at its finest, boils down to statistics, and that those statistics when viewed from an appropriate macroscopic level may be termed as "laws" is an idea I've been aching to come across.  Thank you! Sappysap 04:05, 11 November 2007 (UTC)


 * Er...wow! Well, thanks!  Before we get carried away though - the critical 'take away' point here is that while these macroscopic laws are "only" statistical, the magnifying effect of the sheer quantity of particles on the certainty of the result makes the resulting law quite utterly cast-iron.  You cannot and must not take from my explanation the idea that the macroscopic laws are broken routinely because of this statistical stuff.  On human scales - they absolutely are not.  The probability of anything measurably different from what we expect actually happening is so astronomically small that this makes it impossible for any practical measure whatever.  So "certainty" is still present at our scales.  But when we deliberately make the small scale visible on the large scale, weird stuff can happen.  Listen to the individual clicks of a Geiger counter picking up background radiation (Image:Geiger calm.ogg for example) - each click is the result of the decay of a single atom producing a single neutron.  Guess what?  It's utterly random - you can clearly hear that - there is fundamentally no way to predict when the next click will happen. SteveBaker 05:43, 11 November 2007 (UTC)


 * Well, I'm not listening to a Geiger counter per se, but just now I happen to be listening to "Radio-Activity", by Kraftwerk, which starts out with the sound of one. Weird coincidence, or subtle causality?  (You be the judge. :-) ) —Steve Summit (talk) 06:01, 11 November 2007 (UTC)


 * Responding here to Steve's claim that "certainty is still present at our scales" except when "we deliberately make the small scale visible on the large scale". I don't believe that's true. The effect of quantum indeterminacy can easily be blown up to a macroscopic scale by processes not requiring our deliberate intent, by any system that chaotically amplifies small differences with positive feedback. Like weather. So I think, for example, that the question "will it rain in Dallas on the afternoon of July 17, 2063?" is truly non-determined; quantum uncertainty now, on a microscopic scale, will have been amplified to different macroscopic answers by then.
 * He's certainly correct that there are questions we can ask where the answers are quite deterministic for practical purposes. But those are the questions where small differences tend to cancel out, rather than being amplified. --Trovatore 22:00, 11 November 2007 (UTC)


 * You're talking about chaos theory. The thing about that is that it produces infinite sensitivity on initial conditions.  Chaotic events would be unpredictable as a practical matter (and perhaps as a theoretical matter too) no matter whether quantum-level randomness existed or not. SteveBaker 01:53, 12 November 2007 (UTC)
 * But the point is that, because of the quantum-level stuff, whether it will rain in Dallas on that afternoon is (I think) unpredictable even in principle, because not enough information exists to determine it. That's different from deterministic chaos, where the only issue is whether you have enough computing power available to run the simulation. --Trovatore 02:00, 12 November 2007 (UTC)


 * NO! That's not true.  It's not about computing power.  I can explain this but we need a simpler concrete example.  The weather is too complicated to discuss - let's talk about a simpler (but still deterministic and still chaotic) system.  This is one of my favorites because it's easy to imagine:


 * First, the equipment: Take a couple of small, strong magnets and place them about six inches apart in the middle of a large sheet of paper your desk. Now hang a magnetic pendulum bob a few inches above the magnets on the end of a nice long string suspended from the ceiling.  You also need a red and a blue crayon.  OK - so here is the experiment:


 * Hold the pendulum over some point on the paper, release it and after it swings around a bit and if the magnets are strong enough, the pendulum will end up hovering over the center of either one or the other magnet. Colour that 'release point' with a small red dot if the pendulum ended up over the right-hand magnet - colour it blue if it ended up over the left-hand magnet.  Repeat this experiment for every point on the paper so it's completely covered in red and blue dots.


 * So what red/blue pattern results?  Well, when you release the pendulum near the right magnet the result is always that the pendulum swings immediately over that magnet - so there is obviously an area around the right magnet that ends up red, and an area around the left magnet that ends up blue.  Now, if you hold the magnet a bit further off to the right - beyond the righthand magnet, the pendulum will fly right over the righthand magnet and swing over to the lefthand one - by then it's lost enough energy due to air resistance that it'll stop over the lefthand magnet.  But you can imagine that from some places the pendulum loops around one magnet then the other crazy swings until it finally loses energy and winds up over one or the other.


 * So you can imagine a fairly complex pattern of red and blue on the paper.


 * Now this magnetic pendulum setup is 'chaotic' (in the same way that the weather is). If you crunch the math on this, the actual mathematical pattern you wind up with is a fractal - something like the Mandelbrot set.  There are regions of our red/blue pattern that are in big solid patches (like immediately around each of the two magnets) - but there are regions where the red and the blue is all mixed up in whorls and patterns of great complexity.  Write a computer program to generate this pattern and you can zoom into these patterns and you see more patterns, you can zoom in deeper and deeper into that map - and in some areas you'll keep on getting more and more red/blue patterns no matter how tightly you zoom.  The image is infinitely complex...fractal...chaotic.


 * What is the physical meaning of this infinite complexity of red and blue dots? It means that if you start the pendulum over one of those chaotic regions and release it, the magnet it will end up over is very sensitive to where you started it from.  Move a millimeter to one side and the answer may be different.  Move a millionth of a millimeter to one side and the result will be different, move the width of a hydrogen atom to one side and you'll get a different answer.  In fact, move an INFINITELY SMALL distance to one side or the other and the pendulum may end up over the other magnet.  The result is that moving the pendulum by (1/infinity) meters can change the answer...but (1/infinity ) is zero (well, kinda - mathematicians might argue it 'approaches' zero - but the result is the same)...so if you move the pendulum by zero distance, the answer can change.   It's deterministic - in that there is no randomness in the equations - but you need infinite precision to calculate it - and even if you had that, you still wouldn't get the right answer because displacing the initial position of the pendulum by 1/infinity meters changes the answer.


 * CONCLUSION: You don't need quantum effects to get a random answer - you don't even need an inaccurate measurement of the initial position because an error of (1/infinity)% is enough to change the answer. The result is independent of computer power or precision.


 * SteveBaker 15:46, 12 November 2007 (UTC)


 * Um. The boundary between the red and blue regions may well not be computable. But I think you'll find, in the deterministic idealization of the pendulum problem, that for any &epsilon; greater than zero, there's a computer program and a finite amount of information about the initial conditions that -- given sufficient time to run and resources -- would give you the right answer for at least a proportion of 1-&epsilon; of the space of possible initial conditions. Which is not true for the quantum version of the same problem.
 * Computability is beside the point here anyway (or at least, beside my point). In the deterministic case, there exists a function, whether you can compute it or not, that takes the initial conditions and returns the final answer. In the quantum case, no such function even exists,computable or not. --Trovatore 19:47, 12 November 2007 (UTC)


 * That's true - but you miss the point. For any given starting point, even if you could compute whether to label it red or blue, the problem is that in some areas of the map, the individual red and blue dots are INFINITELY small.  So the information from some theoretical computation would be utterly useless because nothing can be positioned infinitely accurately (irrespective of quantum theory).  For every red dot in one of these chaotic regions, the distance to the nearest blue dot is zero.  But worse, the answer isn't computable because to arrive at the right answer you need infinite precision - that requires an infinite number of bits of storage and (on a finite computer) infinite computation time.


 * Chaotic systems are deeply weird - even in a 'classical' universe. In a practical quantum universe, there is clearly no way to position anything with anwhere near enough precision and still keep it stationary because of the uncertainty principle.  But Newton and Einsteins idea of a 'clockwork universe' where everything is ultimately predictable is blown away by chaos theory every bit as efficiently as by quantum theory.  Better actually because you can show the correctness of chaos using pure mathematics.  Quantum theory still requires experimental evidence and Einsteins idea that there might be an 'underlying certainty' is really tough to disprove. SteveBaker 21:22, 12 November 2007 (UTC)


 * SteveBaker 21:22, 12 November 2007 (UTC)


 * Are you claiming that, in the deterministic version, the intersection of the topological closure of the red set, with the closure of the blue set, has positive Lebesgue measure? I can't refute that off the top of my head but I think it's most unlikely. Whereas if that intersection has measure zero, then the up-to-epsilon computability that I mentioned earlier would be true, and would probably match the "clockwork universe" idea well enough (though admittedly in a completely impractical way). --Trovatore 22:53, 12 November 2007 (UTC)
 * I don't know whether there is a positive Lebesgue measure - but that's not necessary. We don't know whether the Mandelbrot set has a positive Lebesgue measure either.  That doesn't prevent it from being infinitely crinkly which is all that is needed in order to guarantee that for some areas of the diagram, there are areas where we have infinite sensitivity to initial conditions. SteveBaker 00:15, 13 November 2007 (UTC)


 * Sure, but by itself that's not very interesting. Balance a pencil on its point -- will it fall to your left or to your right? Even in the deterministic version of the problem there's a boundary point, where you have your infinite sensitivity. But for almost all (in the sense of Lebesgue measure) possible positions of the pencil, in the deterministic version, you can predict where the pencil will wind up, given sufficiently accurate measurement. I don't think this would have upset Newton or Einstein too much. --Trovatore 00:28, 13 November 2007 (UTC)
 * That example isn't quite the same as this case. With the pencil, it will fall in the direction of the error in it's initial position - the consequences of even an infinitesimal error are extremely predictable.  With the magnets and pendulum case, you can't predict where it'll end up for an infinitesimal error. SteveBaker 15:25, 13 November 2007 (UTC)


 * Well, but the point is that you can predict it (in the deterministic version of the problem), except at a boundary point (that is, a point that's neither in the topological interior of the red set nor of the blue set). Of course how small the error has to be depends on how close you are to the boundary. (Note that if you're not on the boundary, then your distance to the boundary is nonzero -- that's because the boundary is closed.)
 * So the question is, how often are you at a boundary point as opposed to an interior point? My conjecture is that the boundary has measure zero and is nowhere dense. Since the boundary is necessarily closed, the latter condition is the same as saying the boundary has empty interior -- there isn't any region containing an open set such that the infinite sensitivity obtains over the whole region.
 * But the quantum case is different. If you take these extremely-finely-divided, but still cleanly separated, red and blue regions, and smear them by the tiny uncertainties imposed by quantum mechanics, you see that there are regions with nonempty interior where what you have is genuinely purple. --Trovatore 16:56, 13 November 2007 (UTC)


 * From the question: To the contrary, many quantum effects do not follow 'cause and effect' and are fundamentally unpredictable.  Maybe my understanding was wrong, but this isn't how I interpreted quantum mechanics.  Rather, as I learned it, the apparatus for measuring the effect operated on the cause and therefore it was impossible to distinguish the two cause and the measurement.  The proccess of observing affects what is being observed.  In quantum mechanics it really did matter if there was someone there to hear the tree fall in the forest.  Entanglement and the wave/particle duality experiments highlight this.   --DHeyward 07:17, 13 November 2007 (UTC)