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September 8[edit]

Nuclear process in the human body?[edit]

I'm having a discussion with someone and they seem to think that "neutron emission" is a normal part of some biological process, I don't think that's correct, in fact I think it sounds ridiculous. I thought pretty much every biological process that happens in the human body is a chemical processes. Apart from the occasional carbon 14 decay, (and exceptions like people exposed to ionizing radiation) are there any actual biological processes, cellular or otherwise (like the Citric acid cycle for example), that do or could generate neutron emission, or any other nuclear reaction for that matter? Vespine (talk) 00:26, 8 September 2011 (UTC)[reply]

No. (Sorry for the short answer, but I don't know how to find a source for something that doesn't exist.) Looie496 (talk) 00:31, 8 September 2011 (UTC)[reply]

Thanks, that is what I suspected.. I just have to work out a way to communicate to the person that the idea is nonsense. If we had nuclear processes happening inside us, humans would be radioactive, right? Vespine (talk) 00:34, 8 September 2011 (UTC)[reply]

Human beings are radioactive, of course— we consume naturally-occurring radioisotopes all the time, and cosmic ray bombardment and a few other processes can convert stable nuclei in our bodies to radioisotopes. Sleeping next to someone for a night means you're getting an extra 0.05 microsieverts. Potassium-40 and carbon-14 are the two biggest sources, followed by rubidium-87 (no known biological function, but probably accumulates through the same processes that bind potassium) and trace amounts of a number of others: [1]. Still, the point that there's no biological process which generates radioisotopes (or emits ionizing radiation as part of its function) is correct. Our biochemistry involves elements which sometimes happen to be radioactive; our bodies don't specifically seek out or require radioisotopes to function. TenOfAllTrades(talk) 00:46, 8 September 2011 (UTC)[reply]

Yes, we are, mildly - but only due to the stuff we ingest. Bananas are apparently noted for this, having a high potassium content - including naturally-occurring potassium-40. And of course, for smokers, there is lead-210 and polonium-210 in amongst all the other toxins. But I doubt very much that these isotopes are doing anything other than damage. AndyTheGrump (talk) 00:48, 8 September 2011 (UTC)[reply]

Ok that's some great answers, thanks. So I'll ask specifically about neutron emission then, I've had a quick look at potassium 40 and carbon 14 and I can't see that neutrons are emitted by the decay of either of those, so can anyone think of any neutron emitting process that might commonly happen inside the human body? Vespine (talk) 00:53, 8 September 2011 (UTC)[reply]

Unless you have volunteered as a human guinea pig for nuclear weapons testing -- no such processes happen in your body. 67.169.177.176 (talk) 01:49, 8 September 2011 (UTC)[reply]

Based on composition of the human body there would seem to be about 100 micrograms of uranium in the typical adult. Spontaneous fission of uranium, though a rare reaction, is probably your best bet for something occurring that emits neutrons. Dragons flight (talk) 02:47, 8 September 2011 (UTC)[reply]

Molten steel at the wreckage of the WTC?[edit]

A lot of 9/11 truthers say that the presence of molten steel at the wreckage of the WTC proves that it was a controlled demolition... or something like that. Now I know that steel doesn't need to melt in order to fail in supporting weight (it just needs to be heated up), so that's not an issue. But I am aware that steel has a very high melting point. So how was there molten steel at the wreckage (assuming the claim is true). I'm not a truther, but I would like to know how to respond to those claims. ScienceApe (talk) 03:05, 8 September 2011 (UTC)[reply]

The question should not be "how was there molten steel...?", but "was there molten steel...?". Though another question might well be "so what if there was?". As Steel#History_of_steelmaking tells is, we've been melting the stuff for about 4000 years or so, so it obviously isn't that difficult to generate the necessary temperature. AndyTheGrump (talk) 03:20, 8 September 2011 (UTC)[reply]

Don't forget that you've also go ~40 tonnes of aircraft aluminum that melts at half the temperature of steel. If one is only looking at the aftermath, I suspect many observers wouldn't be able to tell the difference between melted aircraft and melted steel beams. Dragons flight (talk) 03:51, 8 September 2011 (UTC)[reply]

The claim is rather simple: An office fire cannot get hot enough to deform and possibly melt steel. Therefore, the government must have used something to melt the steel and used the planes as a cover-up. Well, the first statement is completely false. An office fire can (and often will) get hot enough to deform and, sometimes, melt steel. See Charleston Sofa Super Store fire as a textbook case of creating such a fire. Office furniture catches fire rather easily and, once burning, increases heat a lot. Because this was a warehouse full of office furniture, the heat was so hot in that particular fire that furniture nowhere near the source of the flames began to burn. I took many photos after the fire and I was allowed to post ones taken from outside the building in that article. The steel frame of the building was heavily warped - as can be seen from the outside. I suppose the "truthers" would claim that the government purposely burned the building. -- kainaw ™ 03:25, 8 September 2011 (UTC)[reply]

I suppose, that a fire fuelled by highly flammable jet fuel would be even hotter than the warehouse fire. Plasmic Physics (talk) 03:35, 8 September 2011 (UTC)[reply]

(edit conflictI believe there should have been found two depositss of molten metal on the site. The first deposit consists mostly of aluminium that would melt faster and flow out the building at a lower viscosity. The second deposit consists mostly of more viscous steel that rained down with the collapse of the tower. The steel layer should be overlying the older aluminium layer. Plasmic Physics (talk) 04:18, 8 September 2011 (UTC)[reply]

Usually those folk don't seem to understand that steel weakens long before it reaches its melting point. They think that if the beams failed or warped, then they must have melted. I don't know why they think that, we've all seen blacksmiths making things out of soft, but not molten, metal.

Perhaps their only experience with "melting" is water/ice, which behaves differently because it's a crystal. APL (talk) 03:55, 8 September 2011 (UTC)[reply]

Very true - I've managed to generate enough heat burning scrap wood in a bonfire to make steel rebar nicely malleable - I could probably have tied it in knots if I'd had the tools to handle it properly. I think a lot of the misunderstanding is down to a lack of practical experience with materials, and with the heat that even a small fire can generate. AndyTheGrump (talk) 04:06, 8 September 2011 (UTC)[reply]

Get it red hot and it should be fairly easy to bend. During the US Civil War, troops doing sabotage in enemy territory would pile up wood to make a bonfire, remove railway rails and toast them over the bonfire, then bend them around trees to prevent their easy re-installation. An open fire softened the steel enough to make it bendable. Red hot steel is far from molten steel. Edison (talk) 04:12, 8 September 2011 (UTC)[reply]

Of course, WP:WHAAOE, in this case Sherman's neckties. --Stephan Schulz (talk) 16:19, 8 September 2011 (UTC)[reply]

What I find harder to believe is that anyone would think it more likely that there was a complex, secret, government led conspiracy, inevitably needing to involved hundreds of highly skilled people, using complex equipment and materials, with no leaks, rather than a couple of dozen young, naive, ideologically driven terrorists with ultra cheap box cutters. HiLo48 (talk) 04:45, 8 September 2011 (UTC)[reply]

Hanlon's razor seems very applicable. Also, if I remember rightly, they never even tested the molten metal found at the base of the towers, it was just assumed to be steel. Silliness. Pascal (talk) 05:56, 8 September 2011 (UTC)[reply]

Given how much politicians love to talk, I fail to see how any kind of government conspiracy could exist in secret for more then a week. Googlemeister (talk) 12:57, 8 September 2011 (UTC)[reply]

It wasn't tested because each time an "eye-witness" was pressed to divulge details of the molten steel, the person would suddenly admit that he or she didn't actually see the molten steel. Instead, it was just mentioned that there was some down there. -- kainaw ™ 13:04, 8 September 2011 (UTC)[reply]

Here's an account of an individual who tried to find a first-hand source for the 'pools of molten steel' allegation by tracking back news coverage of the claim. In most cases, it turns out that people are just repeating claims that they heard from someone else, who heard it from someone else, etc. The few cases where it seems likely that someone might have actually observed hot, molten material come from witnesses who weren't qualified to positively identify the material that they saw as steel. This site talks about the story further, suggesting that other molten materials could just as easily account for the observations by non-metallurgists. Among other possibilities, aluminum (extensively used in aircraft construction and office furniture) melts at just 660°C—easily within reach of a small campfire, let alone a massive jet-fuel-accelerated building inferno. TenOfAllTrades(talk) 15:15, 8 September 2011 (UTC)[reply]

How would molten steel (or anything else) argue for "controlled demolition"? When a bomb blows something up, the fragments are barely hot, certainly not melted. Wnt (talk) 18:17, 8 September 2011 (UTC)[reply]

I think that refers to the thermite part of the conspiracy theory. Mikenorton (talk) 21:06, 8 September 2011 (UTC)[reply]

Why use thermite, there are far more efficient methods to initiate a critical failure of a structure? Plasmic Physics (talk) 21:31, 8 September 2011 (UTC)[reply]

Actually, I truly believe that the "9/11 truth movement" is a paid disinformation campaign by some government agency - other examples being the second-gunman theory of the JFK assassination and alien origin of UFOs. In those cases attention was effectively misdirected from whether Oswald was the lone gunman but trained/sent/killed by some conspiracy (Communist or Republican...) and whether the UFOs were, say, weather balloons or other craft conducting surveillance missions over other countries. In the 9/11 case the issue buried was whether the Bush Administration had previous contacts with bin Laden from the muaheddin freedom fighter days, or allowed/arranged the attack to justify a power grab, etc. Of course, the other ideas suppressed don't actually need to be true to justify a disinformation campaign to suppress them, but then again, they don't have to be false either. Wnt (talk) 03:15, 9 September 2011 (UTC)[reply]

Note: Wnt is almost certainly a paid Illuminati agent, tasked with spreading misinformation on Wikipedia ;-) AndyTheGrump (talk) 03:22, 9 September 2011 (UTC)[reply]

@Wnt: Yes, you are correct that the 9/11 conspiracy movement is a paid disinformation campaign by "some government agency" -- that government agency being (get this) Pakistan's ISI intelligence service! 67.169.177.176 (talk) 00:21, 10 September 2011 (UTC)[reply]

(unindent) A lot of people seem to be unaware that part of the Twin Towers' structure was also made of aluminum, so this makes for a lot more than 40 tons of the stuff (which melts at a temperature only 2/5 of the melting point of steel). Of course, aluminum is also a component of thermite, which only adds to the conspiracists' insane claims of "controlled demolition". In fact, what happened was, when the aluminum and steel was melting from the fire, they reacted chemically to form a residue that resembles thermite because of similar reaction conditions. In other words: it looked like thermite, but it wasn't thermite. 67.169.177.176 (talk) 00:35, 10 September 2011 (UTC)[reply]

Chemical nomenclature[edit]

If an alpha amino acid is an amino acid with the amino group on the alpha carbon, what is the name of an amino acid where the amino group is attached to the final carbon?

What is the name of a compound that is both an aldehyde and an ether?

I there a name for a compound where there is a hydroxy group on both the first and the last carbon? Plasmic Physics (talk) 03:44, 8 September 2011 (UTC)[reply]

To the first question: It depends on how many carbons away it is. The general practice is to label the carbons by greek letters in order of the distance between the target functional group and the second one; thus an alpha amino acid is one with the amine group right next to the acid group while a beta amino acid would have the amine group two away. To the second question: If the aldehyde and ether are on the same carbon, then what you have is an ester. If the aldehyde and ether are seperated by some distance, than you just have an aldehyde, and an ether. To the third question: two hydroxy groups are called diols. The only two distinctively named diols are "geminal" diols (same carbon) and "vicinal" diols (neighboring carbons). There is no special name for other diols. --Jayron32 04:32, 8 September 2011 (UTC)[reply]

On the first question: is there not an umbrella term for amino acids like, carbamic acid, α-glycine, β-alanine, etc.? A functional group embues different reactivity when it is attached to a terminal carbon, take ketones and aldehydes as an example.

On the second question: what I am looking for, is a name like aldeho-ether, oxa-aldehyde, alkoxy-aldehydes, 1-oxidanylidene-oxa-alkanes, or some kind of umbrella term. Plasmic Physics (talk) 04:47, 8 September 2011 (UTC)[reply]

Again on the first: The only reason why aldehydes and ketones are treated differently is that hydrogen atom attached to the carbonyl group, however that hydrogen is pretty unreactive; and ultimately the chemistry of aldehydes isn't really all that different from the chemistry of ketones. Other than the odd case of carbonyl compounds, other functional groups don't make a distinction, nomenclature-wise, between primary (terminal), secondary, or tertiary subtituency. On the second question: I'm not sure (aside from the special cases of formic acid esters) that there is any umbrella term for compounds that randomly contain both aldehydes and ethers. --Jayron32 05:05, 8 September 2011 (UTC)[reply]

Omega amino acid could be used for opposite end amino group. You may also want to consider esters especially a formyl ester which has "ether group" next to an "aldehyde". The diol at terminal ends used to be called a x-methylene glycol with x being something like tri- or penta- or other numeric prefix for the number of carbons. Graeme Bartlett (talk) 13:27, 8 September 2011 (UTC)[reply]

Small tangent, but people have done some interesting stuff with Beta-Amino Acids. Being able to make Beta-peptides with secondary structure is pretty neat, I'm not well versed in the applications though. (+)H₃N-Protein\Chemist-CO₂(-) 13:52, 8 September 2011 (UTC)[reply]

Krill peeling[edit]

How do they peel krill? I couldn't find anything either at krill or krill fishery, except for the bare fact that krill must be peeled on account of the high level of fluorine in their exoskeletons. How on earth do they do it? Some chemical process? A clever machine? A not-so-clever machine? When krill is sold for human consumption, what does it look like? LANTZY ^TALK 11:05, 8 September 2011 (UTC)[reply]

I imagine that it's done mechanically as I once stayed at a small settlement in west Greenland that had a prawn packing factory (we were stuck there waiting for the weather to improve so that a helicopter could move us to our next field camp). We were shown around the plant and saw the bit of the machine where the peeled prawns appeared on small conveyor belts - the only apparent human intervention was from the lines of summer students (it only worked during the summer when the water supply was unfrozen) stood there removing the bits of the exoskeleton that had not been removed by the machine. As I recall someone referred to sets of small rotating wheels inside the machine that rubbed the shells off. Mikenorton (talk) 12:18, 8 September 2011 (UTC)[reply]

Here is a a patent for a krill peeling machine from the 1980s. According to this book, however, krill peeling machines were too slow in the 1980s for commercial purposes so modern ones may be substantially different. Spinning Spark 12:27, 8 September 2011 (UTC)[reply]

That looks similar to the workings of the machine that I saw (that was in 1976 I should have pointed out). Mikenorton (talk) 12:36, 8 September 2011 (UTC)[reply]

Evidently I underestimated the size of krill. I was thinking that they were very tiny indeed, and would require some sort of brute-force mechanism, perhaps involving crushing and straining. But evidently not. LANTZY ^TALK 13:14, 8 September 2011 (UTC)[reply]

Academic studies of cubicle impact on worker productivity / quality of life[edit]

We've had cubicles for nearly 60 years now, yet I'm not finding anything on Google Scholar regarding their impact (pos/neg) on worker productivity. The wikipedia article is also silent on this topic. Can someone help me out? The Masked Booby (talk) 12:25, 8 September 2011 (UTC)[reply]

I checked Google Scholar with this query and I see a few articles that discuss productivity of cubicle workers. -- kainaw ™ 13:06, 8 September 2011 (UTC)[reply]

See HowStuffWorks - Death by Cubicle (Warning: browser may crash) and search Google Scolar for [Subliminal Distraction Exposure]. ~AH1 ^(discuss!) 15:21, 11 September 2011 (UTC)[reply]

melting gold over a campfire[edit]

In the book A Game of Thrones as well as in the TV series, a character melts a bunch of gold in a soup pot over a campfire and dumps it on another character's head.

The Incas used a similar of execution against the conquistadors, when they would kill the Spaniards by pouring molten silver down their throats.

But silver has a much lower melting point than gold. I was looking on Wikipedia and on Google at the temperature of wood fires and it seems that the maximum temperature is lower than that of gold's melting point.

I realize that it's a "fantasy" series, but still I would like to know if it's posisble to melt, say, 10-15 ounces of gold in a cast iron pot over a wood fire or wood charcoals and if so how long it would take.

--24.228.90.14 (talk) 12:30, 8 September 2011 (UTC)[reply]

The melting point of gold (~1000 C) is significantly lower than the potential temperature of a charcoal fire (~2700 C). Undoubtedly that 2700 is under ideal conditions (forge with controlled airflow, etc), but I suspect that trivial campfire construction techniques could create airflow sufficient to reach 1000 C in the field -- so yes, it's certainly possible with charcoal. As for how long it would take to melt -- depends on how big the fire is. A rough stab at fire size, plus the heating and melting characteristics available in the gold article, would let you put together a pretty good estimate. — Lomn 13:32, 8 September 2011 (UTC)[reply]

According to our articles the melting point of silver is only about 100 Celsius lower than the melting point of gold. That's not a huge difference. Dauto (talk) 13:48, 8 September 2011 (UTC)[reply]

The peak temperature reached is going to depend fairly strongly on the geometry of the fire, and whether or not it is insulated (to some extent) from its surroundings. This page makes the important point that the temperature in the fire and in the flames will depend quite heavily on how rapidly the fire can radiate away the heat produced by combustion. (If you surround a fire with a building's walls, or firebricks, or rocks, or sheets of steel, the fire will gradually heat those surfaces; it will be trying to come to thermal equilibrium with a container that's at several hundred degrees, rather than open air at 20°C. In principle, I suppose one could get a similar effect given a large enough bonfire; the outer layers could insulate the interior to some extent.) In simulations of building fires, the temperature inside the room can reach as high as 1200°C. This paper puts the measured temperature over a wooden 'crib' fire (essentially a matrix of wooden sticks with air gaps, used for fire testing) as rising steadily to over 1000°C near the back wall of the test compartment. (The temperature peaked after about 45 minutes as the fuel supply was exhausted; presumably it could have gone higher had additional wood been available.)

Based on that information, I would say that reaching the melting temperature of gold (1064°C) on an outdoor fire would be challenging but possible, provided one had an ample supply of wood, and probably requiring some sort of confinement (an ad hoc arrangement of stones or steel shields might do the job) to partially insulate the fire. TenOfAllTrades(talk) 14:56, 8 September 2011 (UTC)[reply]

Never tried gold, but you can easily melt aluminium (melting point 660 C[2]) on a small camp fire; more than half-way there! Alansplodge (talk) 02:27, 9 September 2011 (UTC)[reply]

That's one reason I ditched my old aluminium camping cookware... --Stephan Schulz (talk) 13:20, 9 September 2011 (UTC)[reply]

Hopefully, my gold camping cookware is still safe. Googlemeister (talk) 14:13, 9 September 2011 (UTC) [reply]

I'm just waiting for that gold beer can... Alansplodge (talk) 14:25, 9 September 2011 (UTC)[reply]

Replacing washed-out grease in hair[edit]

I find that my hair is a lot easier to manage and style when I haven't washed it for a few days, which I figure is due to grease. When I wash it, it goes all fluffy and I'd be better off getting it cut short than sporting that. I'd like to wash it more often but then is there something I can add afterwards that will mimic the grease in keeping it down and more or less together? Obviously I've got hair products but none of them mimic whatever oils it is my body is adding to my hair. — Preceding unsigned comment added by 129.215.5.45 (talk) 14:32, 8 September 2011 (UTC)[reply]

I don't know if you're already putting conditioner in your hair after washing it. If not, you can get some which you leave in the hair and don't wash out, or you can get ones which you do wash out. Maybe try an anti-frizz conditioner? --TammyMoet (talk) 15:09, 8 September 2011 (UTC)[reply]

Many people in your position simply abandon the shampoo that strips the oils in the first place. Instead, you can wash your hair with baking soda. Unless your hair gets mud or paint in it, you don't really need all the fancy chemicals in shampoos. See e.g. here [3] SemanticMantis (talk) 15:52, 8 September 2011 (UTC)[reply]

Everyone washes his/her hair far too often. You dont need to wash your hair at all. after a few weeks the sebum production will level out and you'll have nice looking hiar that doesnt smell. — Preceding unsigned comment added by 89.243.143.217 (talk) 16:01, 8 September 2011 (UTC)[reply]

It certainly reaches a steady-state after about 10 days, but as I recall it was more that you got used to the smell (this was doing fieldwork in Greenland where the washing facilities - streams fed from snow-melt and man-eating mosquitoes - were not conducive to regular hair-washing or indeed thorough washing of any kind). That's odd I've ended up mentioning my fieldwork in West Greenland twice here in a couple of hours. Mikenorton (talk) 16:10, 8 September 2011 (UTC)[reply]

Correct. I had a coworker that repeatedly went on tirades about the evil soap companies and bragged that he never washed his hair with soap and he didn't smell at all. Well, he did. He smelled terrible. Nobody wanted to be anywhere around him. Everyone was just too polite to tell him how bad he stank. -- kainaw ™ 17:35, 8 September 2011 (UTC)[reply]

Sorry it's not so much that I want to wash it more often but I've been prescribed a medicated shampoo and need to. I don't have much choice re: the chemicals in it. --129.215.47.59 (talk) 17:12, 8 September 2011 (UTC)[reply]

In that case you will need to get a conditioner, as I described above. --TammyMoet (talk) 17:36, 8 September 2011 (UTC)[reply]

I agree, you need a good conditioner. You can get very cheap conditioners, but I have found this is one area where it makes a difference when you buy a decent (not bank-breaking) product. As in, I'll happily use own-brand shampoos, but I always end up going back to Aussie or TRESemmé conditioner. You only need to use a little bit each time, and avoid getting in on the roots because it can make your hair limp and greasy. 86.163.0.200 (talk) 17:32, 9 September 2011 (UTC)[reply]

Do the hair products you've tried include this venerable product? I wonder also if Macassar oil is still on the market? {The poster formerly known as 87.81.230.195} 90.197.66.202 (talk) 23:44, 8 September 2011 (UTC)[reply]

Hair wax? --Colapeninsula (talk) 10:36, 9 September 2011 (UTC)[reply]

How can i distribute elecrtrons of F-Block elements using Auf-Bau??[edit]

How can i distribute elecrtrons of F-Block elements using Auf-Bau?? because i want to solve some Problems in my HW which asked me to get Period , group of some elements in F-block but when i start to do it by simple way to distribute electrons i found i have made many mistakes so i think i have a problem in correct distributing of electrons in that block — Preceding unsigned comment added by Mido22 (talk • contribs) 14:43, 8 September 2011 (UTC)[reply]

F-block elements have frequent execeptions to the standard rules, and they aren't always readily predictable. Generally, if you follow the strict rules, you'd start with 4f¹ at Lanthanum and 5f¹ at Actinium, and count up from there. However, because the 4f, 5d, and 6s orbitals are so close in energy to each other, there are frequently times when the actual, experimentally tested configuration does not match the predicted aufbau configuration. As an aside, the term is the Aufbau principle, no hyphen. --Jayron32 16:55, 8 September 2011 (UTC)[reply]

In real chemistry the irregularities do not matter. Those ground-state configurations are for gaseous atoms sitting alone by themselves, hardly in chemical environments. You will not predict anything very wrong if you think of La as "ideally" 4f¹, and in fact it can have 4f occupancy in compounds. There are lots of low-lying excited states in the f and d metals that may become the ground states in chemical environments! Double sharp (talk) 04:38, 3 June 2020 (UTC)[reply]

Viscosity of an ideal gas only depends on the temperature[edit]

Hi guys, I have quite a basic question: I understand that the viscosity of a gas increases with temperature, but can anyone tell me a good answer for that? Most books I read attribute it to "increase in molecular momentum transfer", but I am having a hard time comprehending that; I'd be grateful if someone could explain a bit more clearly. Thanks! Lynch 7 15:28, 8 September 2011 (UTC)[reply]

Consider a gas with some flow velocity that depends on the position. Then imagine dividing the region occupied by the gas in two regions. Then the gas in region 2 exerts a force per unit area on the gas in region 1. The component of the force per unit area parallel to the plane that divides the two regions (the so-called shear stress), is the viscosity times -dU_x/dz, where U_x is the component of the velocity in the direction parallel to the plane and the z direction is perpendicular to the plane. We assume here that there is no component of the velocity perpendicular to the plane.

Now, the flow velocity U at some point is the average of the velocities of the molecules there. The molecules have some random velocity of order v = sqrt(3 k T/m) in addition to the macroscopic flow velocity. The flux of molecules that crosses from region 1 to 2 and vice versa (i.e. the number of molecules per unit time and per unit area) is of the order of n v, where n is the number density of the molecules (it's actually 1/4 n v, but let's ignore the numerical factors here). Now, molecules can collide with each other, and one can express this in terms of an effective cross section, sigma. If we denote by L the average distance a molecule travels before it collides with another molecule, then we have:

sigma L n = 1

sigma L is the effective volume the molecule sweeps out when traveling a distance L, if this multiplied by the number density is 1, then that means that after traveling a distance of L the molecule woill interact with one molecule.

So, we have a flux of n v molecules that travel an average distance of L = 1/(sigma n) across the boundary of the two regions. Since the gradient of the macroscopic velocity is dU_x/dz, the amount of momentum that is transfered per unit time and area is:

-L m dU_x/dz n v

On average the same number of molecules move from region 1 to 2 as in he pposite directin. On average a single molecule from region 1 will travel a distance L into region 2 where the flow velocity is larger by L dU_x/dz, so on average when two molecules are exchanged, an amount of momentum of -L m dU_x/dz is transfered from 1 to 2. And there are n v such molecules crossing per unit time and area. Now the momentum per unit time gives you the force, so -L m dU_x/dz n v is the force per unit area, the coefficient of - dU_x/dz is by definition the viscocity. This is thus given by m L n v. But L = 1/(sigma n), so it is of order m v/sigma = sigma sqrt(3 m k T).

So you see that the viscosity does not depend on the density, and it increases as a function of temperature. Count Iblis (talk) 17:36, 8 September 2011 (UTC)[reply]

~~Is there a question you need an answer to? Because I cannot find one here... --Jayron32 17:57, 8 September 2011 (UTC)~~[reply]

Count Iblis must have been quite irritated to see my question deleted after he'd finished typing out the long answer, sorry for that :) (I'll put back the question to avoid confusion). Lynch 7 18:03, 8 September 2011 (UTC)[reply]

Sorry, I didn't realize he was answering a question that got removed. My bad. Carry on. --Jayron32 18:10, 8 September 2011 (UTC)[reply]

First of all, thank you for the detailed answer! I really appreciate that. A couple of clarifications, if you don't mind?

What cross section exactly does sigma denote here? and how did you arrive at sigma L n =1? Thanks! Lynch 7 18:16, 8 September 2011 (UTC)[reply]

Particles, atoms or molecules can collide with each other, even though they are (or consist of) point particles. If you have a beam of particles and you shoot that at some target, then some fraction of the particles in the beam will interact with the particles in the target. Using quantum mechanics, you can compute this fraction or you can simply measure it. Suppose that the target is a single particle and we fire F particles per unit area per second on it. Then the number of collisions per second will be proportional to F. The larger you make the flux of particles in the beam, the more collisions there will be. You can e.g. keep the total number of particles in the beam per second the same and make the beam narrower.

But this means that the number of collisions per second, S, is of the form S = sigma F, and because F has dimensions of Time^(-1) Length^(-2), and S has dimensions of Time^(-1), sigma must have the dimensions of an area. This despite the fact that we're dealing with point articles. So, you can say that you can pretend that particles have a cross sectional area equal to the interaction cross section to compute the probability of collisions. Sigma is not constant, it depends on the energy of collisions. For hydrogen atoms at low energy it is of the order of the Bohr radius squared.

Then to find the average distance a molecule has to travel before it suffers a collision, all you need to do is to find the volume it needs to sweep out with its interaction cross section sigma such thatthis volume will contain one molecule. If the number density is n, then a volume of V contains n V molecules, so V need to be 1/n. But V is also sigma times L, so this is where L = 1/(n sigma) comes from. Count Iblis (talk) 21:04, 8 September 2011 (UTC)[reply]

Thank you :) It is cleared now :) Lynch 7 02:13, 9 September 2011 (UTC)[reply]

Did Earth have two moons?[edit]

I've looked in the obvious places on Wikipedia to see if this theory has been reported here. Here is a link to a newspaper article I saw about the theory.Vchimpanzee · talk · contributions · 20:06, 8 September 2011 (UTC)[reply]

That was only reported last month and is still pretty actively being discussed; I think it needs to be grokked a bit more deeply before we should be ready to cover it. Looie496 (talk) 20:13, 8 September 2011 (UTC)[reply]

Okay, thanks.Vchimpanzee · talk · contributions · 20:15, 8 September 2011 (UTC)[reply]

It's mentioned in the last two sentences at Giant impact hypothesis#Impact (added on August 4). Deor (talk) 20:17, 8 September 2011 (UTC)[reply]

non-sphere stars?[edit]

I've read about Vega and I wondered, whether there are much bizarre stars' shapes? If not, is it possible theoretically? When I say bizarre, I mean of course for a non-sphere shapes (cubic for example). Exx8 (talk) —Preceding undated comment added 20:29, 8 September 2011 (UTC).[reply]

Gravity makes all large objects roundish (Assuming that pressure and gravity are in equilibrium and rotation is negligible). Dauto (talk) 20:34, 8 September 2011 (UTC)[reply]

We've been here before a number of times - for instance [4]. Hydrostatic equilibrium#Planetary geology probably refers. --Tagishsimon (talk) 20:36, 8 September 2011 (UTC)[reply]

If rotation is important, than you get a oblate spheroid. Dauto (talk) 20:38, 8 September 2011 (UTC)[reply]

Donor stars in binary stars will be a vaguely flattened teardrop-shape due to the tidal pulling (or actual outflow) of matter to its Roche lobe (or beyond) by the primary star.-- Obsidi♠n Soul 22:54, 8 September 2011 (UTC)[reply]

Me am thinking cube too bizarro. Clarityfiend (talk) 00:59, 9 September 2011 (UTC)[reply]

Or borg. ;) -- Obsidi♠n Soul 04:38, 9 September 2011 (UTC)[reply]

Theoretically, a cubic star would require a nonic system where there is a central star surrounded by eight white dwarfs, giving eight Roche lobes for the eight corners of the cube, which is so improbable, you can call it impossible. What about a triple system, where a star is surrounded by two white dwarfs? Plasmic Physics (talk) 01:20, 9 September 2011 (UTC)[reply]

That would be unstable. Dauto (talk) 01:41, 9 September 2011 (UTC)[reply]

If you consider accreting new stars, then they may have a disk and jets. Expanding stars can form a planetary nebula which can have odd shapes such as squares. Graeme Bartlett (talk) 11:02, 9 September 2011 (UTC)[reply]

Many fast-rotating stars such as Achernar are very oblate. ~AH1 ^(discuss!) 15:12, 11 September 2011 (UTC)[reply]