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March 15

feeding plants carbon

What are some ways to feed plants carbon? Could I administer malic acid to CAM plants through the stomata? There's a product out there on the market that supposedly is an artificial carbon source for plants-- what are some possible mechanisms to "help out" plants with carbon fixation? (This is for ornamental plants, where the large-scale boost of organic material is desired.) I can't use ammonium (bi)carbonates because of the ammonium ion's toxicity to fish. Could I administer an organic acid + bicarbonate, or maybe boric acid? 74.65.209.218 (talk) 06:01, 15 March 2012 (UTC)[reply]

Sugar works pretty well. Plasmic Physics (talk) 07:09, 15 March 2012 (UTC)[reply]

Googling the topic of feeding plants sugar, I doubt this is a beneficial solution. It seems to encourage bacterial, rather than plant, growth. 74.65.209.218 (talk) 08:14, 15 March 2012 (UTC)[reply]

What makes you think your plant is deficient in carbon ? Doesn't it get enough from carbon dioxide in the air and/or organic molecules in the soil ? StuRat (talk) 08:17, 15 March 2012 (UTC)[reply]

You could always put an animal in with the plants as they produce carbon dioxide, maybe more fish? Or a hamster. SkyMachine ⁽⁺⁺⁾ 08:29, 15 March 2012 (UTC)[reply]

I'm trying to speed up carbon fixation. Furthermore, these are aquatic plants in a fish tank, which seem to grow slowly. 74.65.209.218 (talk) 09:10, 15 March 2012 (UTC)[reply]

First, are they growing more slowly than is typical for their species ? Second, what makes you think that carbon is the limiting factor ? Perhaps something else is deficient, such as light. If you don't accurately determine the problem, any "solution" is likely to cause more harm than good. StuRat (talk) 09:59, 15 March 2012 (UTC)[reply]

Did your search just focus on sucrose or did you include a variety of sugars. Plasmic Physics (talk) 08:32, 15 March 2012 (UTC)[reply]

My search included mentions of glucose. My search shows that glucose is actually an herbicide. 74.65.209.218 (talk) 09:11, 15 March 2012 (UTC)[reply]

I suppose if the sugar was colonized by yeast you would produce carbon dioxide. SkyMachine ⁽⁺⁺⁾ 08:43, 15 March 2012 (UTC)[reply]

This doesn't help. I can't have too many microbes proliferating in the tank water. Furthermore, I am not sure if roots are meant to absorb carbon dioxide or even sugar. I am looking at more sophisticated ways of adding carbon. Does administering malic acid to CAM plant stomata speed up carbon fixation? 74.65.209.218 (talk) 09:10, 15 March 2012 (UTC)[reply]

You could always use compressed CO2 like you can get at home brew stores or soda stream canisters. Modify a switch to slowly relase CO2 to bubble through the tank. SkyMachine ⁽⁺⁺⁾ 09:29, 15 March 2012 (UTC)[reply]

That's going to produce some carbonic acid in the water and make it more acidic, which might not be good for the fish. StuRat (talk) 09:56, 15 March 2012 (UTC)[reply]

Seems to be a mix of factors; light, CO2, fertiliser.

plants need light to grow, but the more light they get, the more CO2 and trace elements they will need.
CO2 diffusion in water is much slower than in air. There can be a CO2 depleted layer of water around the plants. CO2 injection is one of the techniques used, with a CO2 tank, valves, regulators and controller, measuring the pH to adjust the CO2injection. Seems to be a bit expensive.
Trace elements, especially iron it seems, may be lacking. Add some trace element mix for water plants.

air bubblers, biofilters, plants will remove CO2 from the water. Fish add CO2. Yeast generators are low cost way of adding CO2.

Adding CO2, when the lighting is adequate, will increase the oxygen in the water due to more photosynthesis from the plants.

It's all a balancing act it seems, check out some forums like forum.aquatic-gardeners.org for more info. 84.197.178.75 (talk) 11:25, 15 March 2012 (UTC)[reply]

What if you combine malic acid with a buffering agent? Plasmic Physics (talk) 11:28, 15 March 2012 (UTC)[reply]

Note about carbon fixation: plants take up CO2 via photosynthesis. That's the ONLY way they take up carbon in significant amount. Forget about trying to feed them carbon any other way. Do you want to boost carbon fixation because you want the plants to grow or because you want to reduce the amount of CO2 in the water?? For the first you want to add CO2 and light and trace elements if needed. For reducing CO2 you would add light and more plants and again trace elements if needed. But from what I understand, faster growing plants by CO2 injection will result in more O2 in the water for the fish, and under 30 ppm, the CO2 does not hurt them. 84.197.178.75 (talk) 12:19, 15 March 2012 (UTC)[reply]

If this is for aquatic plants, they make commercial CO₂ injectors specifically intended to introduce extra carbon dioxide into planted tanks. It's a whole category of products on the specialist sites (e.g. here). These get carbon dioxide from pressurized tanks, available either from a welding supply company or from paintball supply companies. You can also put together a DIY system with a homebrew reactor based on sugar and yeast (you don't put the sugar and yeast in the tank, you put it in a separate tank, and pipe the gas that comes off into the tank. (Search /diy co2 aquarium/ or /diy co2 planted tank/ on Google, and you'll get plenty of results, including many step-by-step instructions. Try also /co2 system for aquarium/). While adding the CO₂ will depress the pH a little due to the carbonic acid formed, when the plants take in the carbon dioxide, they'll reverse that process, neutralizing the acidity. And the pH drop can be mitigated by making sure your tank has enough buffering capacity (usually referred to as "KH" in the test kits). If you want your plants to really take off once you start adding CO₂, you may want to add some additional aquatic plant fertilizer. Try to avoid using regular plant fertilizer, as depending on formulation, it may produce algae blooms. You'll probably also want to invest in a better water chemistry test kit, as keeping acidity/buffering/nitrogen/phosphate/iron/etc. in balance in a planted tank maintained as such, especially with CO₂ injection, is more important than for a tank maintained just for the fish. -- 71.217.13.130 (talk) 16:44, 15 March 2012 (UTC)[reply]

I'm trying to find methods cheaper than CO2 injection. Carbon supplements already exist on the market. Specifically, my relative wants me to find an alternative to this product for him, one that he could mass produce. Would feeding plants 3-phosphoglycerate work, or would it just trigger algal blooms?

Also I know from reading the literature (I'm a chemistry student) that plants appear fix the bicarbonate they absorb; presumably if they can absorb fulvic acids in organic fertiliser (in mulch for example) then they could absorb sugars or organic acids. Do the organic acids in fertilisers boost the growth of plants directly or do they simply improve the growth of symbiotic fungi and so forth? 74.65.209.218 (talk) 19:38, 16 March 2012 (UTC)[reply]

A quick web search (/Flourish Excel ingredients/) turns up posts claiming that the active ingredient of Flourish Excel is glutaraldehyde (or rather a polymerized version of flutaraldehyde -which is good, as glutaraldehyde by itself is volitile and highly toxic). An MSDS on the Seachem website [1] says it's specifically polycycloglutaracetal, which matches what's written at Glutaraldehyde#Algaecidal_activity. I've seen some forum posts discussing people using aqueous solutions of glutaraldehyde as a DIY replacement for Flourish Excel, but caveat emptor and all that. -- 71.217.13.130 (talk) 03:56, 17 March 2012 (UTC)[reply]

Bullet through the brain

I'm under the impression that shooting a bullet through the brain almost always causes instant death. Is this true, and if so, why? Phineas Gage had a huge tamping iron driven through his skull, yet he remained mostly unaffected. Lobotomies remove the entire prefrontal cortex, yet leaves the patient mostly functional. In literature, I routinely read about studies of what happens when this or that area of the brain is lesioned/damaged. Why would a bullet, which is physically small and unlikely to take out a major portion of any brain structure, be so likely to cause death after penetrating the brain? --140.180.5.239 (talk) 06:49, 15 March 2012 (UTC)[reply]

As long as the brain stem is intact, there is a possiblity of survival. If you want to not be brain dead, then you'll have to miss a few more sections. In addition, a bullet doesn't always make a clean wound, sometimes (depending on specs) the bullet liquifies tissue around it. There is a youtube video somewhere of what can happen, although demonstrated on an apple. Plasmic Physics (talk) 07:07, 15 March 2012 (UTC)[reply]

Curiously, all of the good reviews on gunshot wounds to the brain happen to be in journals my library doesn't have a subscription to. But no, this is certainly not true. Without those reviews, I couldn't come up with many numbers. What I was able to glean from abstracts is that over 2000 American soldiers in Vietnam managed to make it to a hospital alive despite taking a bullet through the brain. As for why a bullet causes so much damage, it's fast and spinning. It doesn't simply poke a hole through the tissue in front of it; rather, a bullet effectively pulls and drags the tissue around it, potentially causing catastrophic trauma. See Zapruder film for a famous example of what that means. If the bullet stops in the brain, which is more likely if it's a hollow point bullet designed to slow down after hitting its target, the brain has to absorb all of that kinetic energy very quickly. Finally, getting shot in the head can cause severe bleeding and can easily send a person into respiratory arrest, none of which will typically happen in the controlled setting of a surgical room. Someguy1221 (talk) 07:09, 15 March 2012 (UTC)[reply]

I would guess that most deep penetrating brain injuries result in death... but there are the rare exceptions to that and bullets are no exception. The shooting of Gabrielle Giffords is a salient recent example. In few of these cases, whether that shooting or the case of Gage, or in lobotomies, is there no damage. In fact, the damage is often quite profound. What's remarkable is that the victim doesn't die immediately.

What makes a bullet different from many of the other kinds of head injuries, Gage's probably included, is the sheer velocity of a bullet. A low velocity bullet, say a .45 ACP caliber, moves at almost 1,000 feet per second (680 miles per hour/1000 km/h). A bullet from a modern rifle (military or hunting) is about 3x that speed.

Take a look at hydrostatic shock and stopping power. Hydrostatic shock describes why "remote", i.e. not directly to the brain, bullet impacts can incapacitate almost instantly. You don't have to be a scientist to extrapolate those findings to what direct brain injuries do. Also look at terminal ballistics (not a great article, a lot of it looks like one person's production, but should give you some context). The short answer is that while a bullet is small, the shockwave it creates as it enters an object, particularly an object with features like tissue, create temporary disruptions much larger than the projectile itself. I actually doubt there's too much tumbling in brain tissue, although I could be very wrong about that. But there are a lot of very morbid journal articles (the above articles reference some of them) that talk about how occasionally supposedly more "humane" bullets have counter intuitive effects. (sidenote: the ~~Geneva Convention~~Hague Convention requires militaries use full metal jacketed bullets, however there's some debate over the differences between hollow point and full metal jacketed rounds). The brain is particularly sensitive in this respect, which is why these injuries are usually fatal. Shadowjams (talk) 07:26, 15 March 2012 (UTC)[reply]

A few points:

1) A modern rifled weapon, unlike an ancient one, has a spiral groove inside it, designed to spin the bullet to keep it from tumbling. This reduces air resistance and makes it go faster, farther, and straighter. If it continues like this through the brain, it may cause less damage than a tumbling bullet.

2) A slower bullet may actually cause more damage, by ricocheting around in the brain, rather than just going in and out.

3) As mentioned above, there are hollow-point bullets and other types, designed to rip apart on impact, causing much more damage. Such bullets are often illegal.

4) If you survive the initial trauma of the bullet, then infection becomes a major concern. StuRat (talk) 07:37, 15 March 2012 (UTC)[reply]

Well you're wrong on pretty much every point StuRat. All modern handguns are by definition rifled (this is a pretty elementary point to anyone with a cursory familiarity with firearms)... smoothbore guns are generally considered shotguns or muskets (if black powder)... ever wonder why a handgun that shoots .410 shotgun shells is legals? answer is... because it has a rifled barrel. Btw, none of that has anythign to do with my point... if you could get a musketball to do 1000 fps it'd do substantially more damage too. Again, "straight through" if it was slow that might be true, but the high velocity of a bullet has affects on tissue that are disproportionate. As for point 2, there's a huge debate over "energy delivered" and "stopping power" and "hydrostatic shock" and other similar concepts. Many modern militaries have shifted to high velocity, smaller rounds. I doubt there's much chance for "richochet" inside the skull with most modern rounds. I've heard that mafia tale that a .22 was used for assassinations for this reason, but I get a strong suspicion that's urban legend. Your point 3 is again subject to the intense debate about the effectiveness of particular round types Point 4, i doubt that's true with modern medicine. I think swelling is probably the greater risk. Shadowjams (talk) 09:07, 15 March 2012 (UTC)[reply]

I've modified my point 1 accordingly, but don't think you've made your case that my points 2-4 are wrong. On point 4, swelling may also be a major concern, but that doesn't mean that infection isn't. StuRat (talk) 09:51, 15 March 2012 (UTC)[reply]

A treatment of last resort for severe cerebral oedema (swelling of the brain) is a decompressive craniectomy, which is basically cutting a hole in the skull. Since the bullet has already done that, swelling may well not be that serious an issue. --Tango (talk) 21:01, 16 March 2012 (UTC)[reply]

I've noticed that the old method of immediately sealing any wound has been replaced by a newer method of leaving it open to allow it to drain, so this would help prevent swelling, but does pose a challenge as far as preventing infection. StuRat (talk) 23:25, 16 March 2012 (UTC)[reply]

This looks like another great opportunity to mention Mike the Headless Chicken on the science desk.--Shantavira|^{feed me} 08:55, 15 March 2012 (UTC)[reply]

Or even Roland the Headless Thompson Gunner. --Jayron32 14:06, 15 March 2012 (UTC)[reply]

Or Carlos Rodriguez. --Itinerant1 (talk) 18:41, 15 March 2012 (UTC)[reply]

You may also be interested in this man, who lost 43% of his brain in the Falklands War and is still with us: Robert Lawrence (British Army officer). --TammyMoet (talk) 09:35, 15 March 2012 (UTC)[reply]

The amazing part is that he's not only with us, but he managed to lead an active life and even to get married after the injury. I'd have expected him to be a vegetable (or at least severely mentally disabled.) --Itinerant1 (talk) 02:30, 16 March 2012 (UTC)[reply]

"Am I serious about her ? I have half a mind to marry that girl !". :-) StuRat (talk) 23:27, 16 March 2012 (UTC) [reply]

Car radio

I asked this question a few years ago on another forum, but didn't get any replies I felt answered it. I used to own a car where the car radio would sometimes go quiet. A quick push on the front panel of the radio would restore the sound level. So far so straightforward. However, I noticed that driving under high-voltage power lines would also sometimes restore the sound level. Any ideas why this would happen? 86.134.43.228 (talk) 20:00, 15 March 2012 (UTC)[reply]

Metal contacts can oxidize, and such an oxide layer can be an insulator, or have semiconductor properties. Pushing the radio may shift the contacts a bit, breaking through the oxide layer. A high voltage can also break through a thin semiconductor layer, and once it does it causes avalanche breakdown: the electrons are accelerated, collide with atoms which get ionized by this creating a chain-reaction. That's how zener diodes above 5.5 volt work, see avalanche diode and avalanche breakdown. Usually there's an hysteresis effect, meaning that the voltage at which the conduction stops will be lower than the one where it started. That could be an explanation, with the power lines inducing a higher b-voltage over the contacts, enough to break through. Also the micro-weld phenomenon seen with coherers could be involved. In general, it would be some thin insulating layer that can withstand the 12 volts over the contacts but breaks down at a higher potential. That's my best guess. 84.197.178.75 (talk) 21:25, 15 March 2012 (UTC)[reply]

The following explanations are much much more likely: The strength of radio waves often changes dramatically near and under high voltage power lines. Usually the signal strength falls under power lines but it also can increase. AM Radios incorpoarte and automatic volume control system (AVC)(the more correct term is automatic gain control) so you don't notice the change as tune form one station to another, move around, go under power lines etc etc. My guess is that there is a bad solder joint in an area affecting the AVC. When the radio passes under the power lines perhaps the change in signal causes a sufficient change in voltage in the AVC circuit to overcome the oxide layer in the faulty solder joint. FM radios often incorporate a Mute circuit, as without it you get a full volume blast of noise when tuning between stations. Maybe the mute circuit is affected by a crook solder joint, making it mute at too high a signal strength, and the change in signal when passing under power lines is overcoming it. Keit124.178.61.156 (talk) 00:40, 16 March 2012 (UTC)[reply]

The question didn't indicate that it specifically affected weaker stations, and it's an unlikely defect for an AVC circuit imo, since these reduce the gain of strong signals, not boost a weak one. A faulty solder joint affecting the mute/squelch circuit is a possibility,

fixing a bad connection on a PCB with a quick push would only work reliably if that put significant force on the PCB. Pushing the old fashion volume or tuner knob would be most effective (good way to damage it too). So yes, it can be caused by a bad solder joint on the board. But I'm thinking that in that case, it would often be triggered by touching the controls, changing volume or station. The way it's described, it didn't sound like something that would happen several times a day. And it's easily fixed. Makes me think, it's likely a spring leaf type connector rather than a "force-less" contact. If there's any movement between the contacts due to vibrations, they will cause rapid fretting corrosion of the contacts. You get a buildup of oxide material because the oxide layer gets scratched, exposing fresh metal, increasing the layer thickness and accumulating metal and oxide particles. A "normal oxide layer on metal conductors will withstand about 0.2 volt. Typical open-circuit voltage somewhere in a radio would be from >1 to several volt I think, so you need a big oxide layer. A loose connection with that much build-up would behave more like a typical coherer, with the high resistance state easily triggered by vibrations, more so than when the contact surface is under pressure. Vibration won't separate the contacts, it takes time to build up the oxide layer and a bit of random chance to get to the high resistance state. And it won't be very stable. Movement or RF noise could return it to the conducting state.

Power lines develop faults with age, The insulators crack or get covered with dirt causing leak currents that emit high amplitude RF noise. And power companies only fix those faults if they have to; to quote "An Important Rule" for technicians resolving power-line RF noise given in an industry publication:

"Perhaps the most difficult hurdle to overcome in this process is to ignore those noises not affecting the customer's equipment. An important rule for efficient and economic RFI troubleshooting is to locate and repair only the source causing the complaint." (Transmission and Distribution World, sept 2004)

But I'm just speculating; anything is possible ;-) 84.197.178.75 (talk) 18:49, 16 March 2012 (UTC)[reply]

You are right - almost anything is possible. It is possible with some types of AVC circuits to go faulty so as to cut off an IF stage rather than leave it full on. Some signal still gets thru due to stray capacitance. Not enough to keep the owner happy, but maybe enough under high signal conditions to joggle the fault. I used to do car radio, stereo, and TV repair. Two things I learnt very solidly:- (1) Intermitant faults are the sneakiest and trickiest things. Prodding in one place can make it come and go, but the dry joint is somewhere else. (2) if a customer says their set is faulty, you can (usually) assume it IS faulty, but don't rely on what they say about it. Non-technical people have funny theories and often leave out or confuse vital information. Like saying their TV is crook only on Channel 2, when in fact it is faulty on all channels but they only watch Channel 2. Keit120.145.40.231 (talk) 03:15, 17 March 2012 (UTC)[reply]

Excellent answers, thanks for the replies everyone 86.134.43.228 (talk) 20:10, 18 March 2012 (UTC)[reply]

Hydrogen scattering length density?

I was discussing SANS this morning with another grad student, and realized I don't actually know the answer to this question myself: Does anyone have a simple answer for why hydrogen has a negative scattering length density? The article says "neutrons deflected from hydrogen are 180° out of phase relative to those deflected by the other elements", but that's purely phenomenological. I know it's quantum mechanical in origin, but beyond that I don't have a really good grasp of why this the case. It's slightly counterintuitive to me that something akin to a scattering cross-section would be negative. I've also looked over the article on neutron cross sections, but it in turn just references back the the scattering length density article. Any thoughts? (+)H₃N-Protein\Chemist-CO₂(-) 21:55, 15 March 2012 (UTC)[reply]

*bump* (+)H₃N-Protein\Chemist-CO₂(-) 14:22, 16 March 2012 (UTC)[reply]

??? (+)H₃N-Protein\Chemist-CO₂(-) 22:42, 18 March 2012 (UTC)[reply]

To what extent is a fermion's position part of its quantum state for the purposes of Pauli exclusion?

Recently this controversy regarding a Brian Cox (physicist) lecture was brought to my attention. Although it is only touched on briefly by the many people objecting to May's interpretation of the Pauli exclusion principle, it is generally agreed that position is part of an electron's quantum state. But to what extent is that so? For example, two electrons orbiting the same helium nucleus are forced into different spins because they are close enough together, and similar things cause Pauli exclusion in much larger molecular orbitals. But how far apart do two electrons need to be before they can otherwise both exist in the same quantum state? Npmay (talk) 22:07, 15 March 2012 (UTC)[reply]

To do this correctly, you need to solve the wave function for interacting electrons, which is very hard. (Why is it hard? Because the potential energy is not constant - much like any non-quantum n-body problem - only, also add the complexity of quantized states.) If you take your ordinary quantum mechanics textbook, they'll walk through the solutions for a single electron around a highly-ionized atomic nucleus; and usually, they'll assume the potential energy function for a stationary, electrostatic potential well. But if you have multiple moving charged particles, you can't do this; the math becomes quite difficult. If you'd actually like to work it out, I can recommend several good texts to walk you through the math - but let's be honest: physics students (who are very smart people) usually spend something like a full year working the basic mathematics that describes the quantum-mechanically correct electron orbit, during the course of a two or three semester advanced physics class, and still do not even solve for two electrons. So, the probability that we can summarize this quickly or easily is very low.

If you're looking for a one-line answer, though, let's phrase it this way: "The farther apart the electrons, the greater the probability that they are non-interacting." Quantized states notwithstanding, electron-electron interactions are modeled by a Coulomb potential, whose strength falls off as the inverse of distance. Nimur (talk) 23:00, 15 March 2012 (UTC)[reply]

So, the extent to which the electrons interact, which is proportional to the strength of the electromagnetic force in accordance with the inverse square law, determines whether they are in the same position for the purposes of being in the same quantum state? That would make some sense. It would also resolve the controversy in that distant electrons only have a tiny but nonzero probability of being subject to Pauli exclusion. Is that good enough to avoid the math details? Npmay (talk) 23:29, 15 March 2012 (UTC)[reply]

Electromagnetic interaction doesn't really have anything to do with it—in everything that I wrote below, it's irrelevant whether the fermions are electrons or neutrinos or (hypothetical) particles that don't interact at all. -- BenRG (talk) 23:59, 15 March 2012 (UTC)[reply]

The key point is that there's one wave function for the whole system, not one per particle. In a system of two spinless identical fermions confined to a line segment, you can think of the wave function as defined on a square whose corners are "both fermions at the far left", "fermion A at the far left and fermion B at the far right", "both fermions at the far right", and "fermion A at the far right and fermion B at the far left". (In fact it's not fair to give the particles labels since they're indistinguishable, but I can ignore that here, so I will.) The exclusion principle says that the wave function is zero at all points that correspond to the fermions being in the same place, which in this case is the diagonal line from "both fermions at the left" to "both fermions at the right". Since the wave function is continuous, it also approaches zero as you approach that diagonal, but there's no particular bound on how large it can be except exactly on the diagonal. The exclusion principle doesn't make any difference when the wave function is zero (or nearly zero) near the diagonal—in other words, when there's no (significant) chance that the fermions are near each other.

For spin ½ particles (like electrons) you can use four copies of the square, one for "both particles spin-up" and so on. The diagonals in the two squares where the particles have the same spin are zero, but the diagonals in the two squares where they have different spins don't have to be zero.

Regarding Cox's lecture, see WP:Reference desk/Archives/Science/2011 December 18#Pauli exclusion principle and speed of light. His words can be interpreted in various ways, but basically he was just wrong. I mostly agree with Sean Carroll's blog post, but even he seems to believe that every quantum object is spread out over the entire universe, an idea which I mocked in my last post to that old Ref Desk thread. -- BenRG (talk) 23:59, 15 March 2012 (UTC)[reply]

How do you decide which particles to include in the "complete system" wave-function? I own some beachfront electrons in Tucson and I feel that their interactions should be included in the wave-function for your two-particle system. Clearly, there must be some sanity in deciding when a particle is "far enough away" that it no longer matters. If this criteria isn't based on the magnitude of potential-energy function of the interaction, (i.e., electrostatic potential, for an electron-electron interaction), then what else would it be? Nimur (talk) 00:11, 16 March 2012 (UTC)[reply]

True, you need some criterion to separate system from environment. But electromagnetism has nothing to do with the Pauli exclusion principle, so I don't think it's relevant here. I had two particles in my system because one wouldn't be enough and three would be an unnecessary complication. The two particles are isolated from all outside influence because it's my thought-experiment and I say they are. Electromagnetism is relevant if you're specifically talking about atomic orbitals, but that's complicated enough (as you said) that I couldn't have given anything like the answer I did. -- BenRG (talk) 00:54, 16 March 2012 (UTC)[reply]

That doesn't make sense to me. Two adjacent helium atoms have between them two pairs of electrons, each pair of which is in the same quantum state except for its position. If their electromagnetic interaction determines whether they are in the same quantum position as well when they are near, then what determines whether they are in the same quantum position as well when they are further away? Npmay (talk) 01:11, 16 March 2012 (UTC)[reply]

Now you're getting into some of the really messy Schrodinger's cat territory, and you've basically asked the vital question: The two electrons, two protons, and two neutrons around a single helium atom represent a single "system" which can be analyzed by a single "wave function" which describes, among other things, the relationship between the two electrons around that atom. This requires quantum mechanics to do accurately. Once you introduce the idea of "two adjacent helium atoms" now you really need to define "adjacent". If the two atoms interact meaningfully, then what you have is essentially a helium molecule of some sort, which is dealt with quantum-mechanically by molecular orbital theory, and the math is identical in spirit to the math to calculate the orbitals around a single helium atom, excepting that it is more complex, as you have 4 electrons and 2 nucleii to deal with now. If you're dealing with two atoms sitting in a box together, occasionally colliding; well now you're into that fuzzy "Schrodinger's cat" area; which is that QM has a real problem in describing with behaviors of particles interacting classically. Which is not to say that it cannot/is not done. Its just that, at some level, the quantum mechanical solution to a problem and the classical mechanics solution to a problem converge, so there's no need to go through the exhaustive QM mathematics, which is almost impossible, and instead you can just use the Newtonian math to solve it. Two helium atoms bouncing around a box can basically be described in Newtonian terms and get the same result as using QM terms, so there's no need to do the messy bit... --Jayron32 13:20, 16 March 2012 (UTC)[reply]

Thanks for taking the time for that clear explanation. If the Pauli exclusion principle is what makes two helium atoms bounce off each other instead of pass through unaffected, then perhaps the thermodynamic gas compression information inherent in Boyle's law explains how close is close enough to be more in the same system than not. Npmay (talk) 23:05, 16 March 2012 (UTC)[reply]

Real gas#Models and Equation of state are certainly complicated and indeterminate enough to fit the bill. Npmay (talk) 11:10, 17 March 2012 (UTC)[reply]