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August 30

Star positions in the past

Is there a website or a software that's capable of telling me where were certain stairs on the sky in a given year (e.g. in 1400)? 193.224.66.230 (talk) 13:11, 30 August 2012 (UTC)

This page will do it, but it's a bit tricky to navigate. Page down to the "virtual telescope" and select the "controls" link to set the date and time you wish to look at the sky for. The Earth's view of the stars has changed over time, but I think that a difference of 600 years isn't going to have changed it much; excepting the locations of the planets and comets and things like that. The star map would only look noticibly different on scales of 1000s of years or 10,000s of years. --Jayron32 13:18, 30 August 2012 (UTC)

Celestia. If you haven't used it before, and you're at all interested in space, you're missing out. I agree that 600 years isn't enough to see the stars move noticeably. To get a sense of how slowly they move, Barnard's star, the star with the highest proper motion, moves 3 degrees in 1000 years. Almost every other star moves many times slower than that. You can use Ptolemy's star charts today and they wouldn't seem wrong. --99.227.95.108 (talk) 15:40, 30 August 2012 (UTC)

As far as I can tell, Celestia doesn't model proper motion. TenOfAllTrades(talk) 21:23, 30 August 2012 (UTC)

I'd go with Stellarium as it's probably the easiest program to get the hang of. (There are download links on the article page.) But as said above, there wouldn't be a noticeable difference in the positions of stars over a 600 year period. FlowerpotmaN·(t) 19:04, 30 August 2012 (UTC)

Attractiveness

Is attractiveness subjective? From my understanding, attractiveness is based on certain dimensions an proportions of the face and body of a person. However is this subjective? And if so, are there certain faces, which people universally find attractive or unattractive? Clover345 (talk) 15:16, 30 August 2012 (UTC)

Physical attractiveness has both objective and subjective components. That is, there are some trends and nearly universal principles, but there are also many factors that are determined by individual taste, and will vary culturally or even person to person. --Jayron32 15:54, 30 August 2012 (UTC)

You may be interested in research on facial attractiveness [1]. You can also participate in real research at that site, making your preferences part of the process. Recently, many different studies have demonstrated that averaged faces are rated as more attractive than the average of many facial attractiveness ratings. This starts to get at "universally attractive" faces, but as Jayron points out, there will always be a large degree of subjectivity in attractiveness. Google /average face attractive/ for many news and pop-sci accounts. SemanticMantis (talk) 18:09, 30 August 2012 (UTC)

Physical attraction is an objective quantity.

F = G m₁ m₂ / r².

However the Greater Internet Uncertainty Principle states that you can know either m₁ or m₂ but not both. Good luck, as escape velocity is a known issue with internet dating. ;) 217.251.154.242 (talk) 07:02, 31 August 2012 (UTC)

Forget it, I will NOT add a link to "Spherical Cow Approximation."

You shouldn't ever call a girl that, you know... --Jayron32 14:15, 31 August 2012 (UTC)

A friend of mine once described fellatio from a spherical cow as like sodomizing a bowl of room temperature oatmeal. That seemed rather objective to me. μηδείς (talk) 21:59, 31 August 2012 (UTC)

Star Gazing Hobby

I want to start this hobby but with the meager savings I have. I will like to purchase a used telescope on eBay however I don't know where to begin to look or what to look for. Please give me any suggestions on how to start my hobby the right way Reticuli88 (talk) 16:25, 30 August 2012 (UTC)

One option is to see if a local astronomy club exists. Near where I live, the Morehead Planetarium and astronomy department at the University of North Carolina sponsors a monthly stargazing night at Jordan Lake, far enough away from light sources but close enough to see some cool stuff. They bring some really powerful telescopes and set them up on the beach, and it doesn't cost anything to come by and use their telescopes. This google search turns up a wealth of other similar arangements. --Jayron32 16:49, 30 August 2012 (UTC)

I strongly second Jayron's suggestion. Speaking as someone who read Astronomy at University and helped to run a local Astronomy Club myself, I'd say it would be near-impossible for someone new to the activity to choose with confidence an appropriate instrument from eBay or similar sites. You may be able to track down a local club from the back pages of Sky and Telescope or a similar magazine. Alternatively, if you can get to a specialist Astronomical Telescope vendor, they're likely to give you reliable advice because they'll be hoping to get repeat business from you when, as is likely, you trade up to better/more expensive equipment. {The poster formerly known as 87.81.230.195} 84.21.143.150 (talk) 17:16, 30 August 2012 (UTC)

The beauty of astronomy is that it can be done purely with the eyes alone. Learn the patterns of the constellations: learn the planets, what they look like in the skies: try and see how many of the Seven Sisters you can see with the naked eye (northern hemisphere only). If you get a cheap pair of binoculars from Ebay second hand, you can see details in some of the globular clusters. Be warned, however: you will never see the range or depth of colours you see in the professionally produced deep sky photos. I have known some enthusiastic amateurs give up when they realise that. --TammyMoet (talk) 18:38, 30 August 2012 (UTC)

Other people have written much more about this exact subject than we possibly can. See, for example, this Sky&Telescope article. --99.227.95.108 (talk) 03:48, 31 August 2012 (UTC)

I can tell you from childhood experience that a small refracting telescope will probably be worth far less than you would pay for it. You might as well get a good pair of binoculars. A friend's father had a six inch reflecting telescope. Due to its mass it was much more stable, and you could see the disks of planets which were mere dots with my "telescope". μηδείς (talk) 05:28, 31 August 2012 (UTC)

Check out Ice In Space, it's a huge website full of amateur astronomers and astro photographers. They have lots of "beginners" articles and a massive forum for support ranging from the very basics all the way to people who have had pictures published on NASA's APOD. I now own a 12" dob, but I started with a pair of 8x56 binoculars which I still use every time I go observing. That's the thing with a pair of binoculars, getting a telescope later won't make them obsolete, you'll still use them regularly to spot things or get a wider field of view, so they really are a good way to start rather then a cheap telescope which at best will become obsolete if you decide to buy a better one and at worst will be crap and will put you off entirely. Vespine (talk) 23:17, 2 September 2012 (UTC)

Raptor factoid

Concerning fish-eating raptors, such as Ospreys and Sea eagles that hunt by swooping down and grabbing fish from the water with their talons... any truth to the oft-quoted fact(oid?) that because the bird is allegedly physically unable to unclench its feet whilst they're embedded in the prey's flesh and let go, unless it returns to dry land, that if the bird happens to 'attach' itself to a fish that's too large to carry, that it will probably either drown on the surface due to becoming waterlogged - still attached to a dead fish, or be dragged under by the fish as it tries to escape - and drown that way?

I'm aware that these birds can only swim a little and do sometimes drown after miscalculating/missing their prey, getting too wet to fly and being trapped on the water's surface but that would seem to be a different scenario. --Kurt Shaped Box (talk) 20:06, 30 August 2012 (UTC)

I call shenanigans. What's so different about being on land? Does the bird have to eat the fish away from its claws before it can use them again? I think not. I think the claim is based upon a nugget of truth: the way raptor talons work. But the claim takes it to a silly extreme, and is also influenced by your observation that yes, these birds do sometimes drown.

The key principle is that raptor talons are similar to ice hooks, like this one [2], in the sense that they convert downward force (supplied by gravity and the ice/fish) into a compressive force of the hooks on the load. This means they can carry heavy things without using any more energy to grip than they would for lighter things (of course it still takes more energy to lift). This does mean that it's difficult to release under load, but it doesn't mean it's impossible. And if the fish is still in the water, it's not putting much load on the talons yet anyway. You may also be interested in this nice (freely accessible) paper "Predatory Functional Morphology in Raptors: Interdigital Variation in Talon Size Is Related to Prey Restraint and Immobilisation Technique" here [3]. SemanticMantis (talk) 17:24, 31 August 2012 (UTC)

Why did people in the late 1800s think an aether was necessary for the propagation of light, when Maxwell's equations show that this is unnecessary?

Maxwell described light as an electromagnetic disturbance of self-propagating E and H fields. Surely people in the late 1800s knew that magnetic and electric fields could exist in vacuum. So why did they think an aether was necessary to "carry" light waves? 71.207.151.227 (talk) 20:11, 30 August 2012 (UTC)

It is not self-evident that electric and magnetic fields can exist in a vacuum, even if you have Maxwell's equations in front of you. This requires careful, meticulous understanding of the equations, and of course the equations require a lot of experimental effort to validate. Among the relevant work that helped establish this fact: the Michelson-Morley experiment, which you've undoubtedly read about; and the experiments that established Vacuum permittivity and Vacuum permeability constants. Maxwell's original work, A Dynamical Theory of the Electromagnetic Field, which you can read online, was a study of the relationships between electric fields and magnetic fields and charges. It is not immediately obvious, just from that work, that his work implies propagation of energy in a vacuum. The science developed; the theories became more perfect, and the experiments continued to validate assertions; and eventually the majority of scientifically-minded people accepted as fact that there need not be any transmission medium to hold the fields. We can say that Faraday and Coulomb discovered these facts; but that's almost irrelevant: at what point did the consequences of these facts become well-known? There's hardly a single instant when we can say "this was the moment that people understood that electromagnetic waves propagate in vacuum." Nimur (talk) 20:34, 30 August 2012 (UTC)

The fact that waves propagate in a vacuum feels intuitively wrong when you consider what a wave is: it is a rhythmic motion that propagates through something. When light displayed wave-like properties, people began to ask "what is it propagating in". The Maxwell equations don't require a medium, but people's expectations did. --Jayron32 21:32, 30 August 2012 (UTC)

"It is not self-evident that electric and magnetic fields can exist in a vacuum.." What decade of the 19th century were the experiments done, with vacuum pumps and bell jars, to prove that magnets worked just fine in vacuum, and that electric fields in a vacuum affected charged objects? I suppose one could cavil that the bell jar vacuum is imperfect compared to the vacuum of outer space. Faraday found magnetic effects in a Torricellian vacuum (like the inside of a mercury barometer above the mercury) (p 194) in his research published as a collection in 1855, but much of the research was done and published earlier. Didn't anyone experiment in the mid or early 1800's with electrified objects in a high vacuum jar to [rove the functioning of electric fields? The writings of Brewster suggest so, though he wrote mostly about discharge and glow effects in vacua. rather than basic attraction and repulsion experiments as done in air. They knew by the 1830's that a charged object would discharge and that charge could be transferred in a Torricelian vacuum. Anyone calling himself a physicist or "natural philosopher" by 1860 would likely have been familiar with this. Edison (talk) 00:40, 31 August 2012 (UTC)

Lots of people experimented with discharges and vacuums, but it was harder than you let on to get a good vacuum in the 19th century. If your vacuum was poor, you got very messy results — the gas inside the tube would ionize, for example, insulating the discharge stream, and give you completely the opposite conclusion. The fundamental work done that established that indeed, cathode rays were affected magnetically was done only in 1896, by J.J. Thomson, and was part of the discovery of the electron. The real breakthrough experimentally was creating an unambiguously good vacuum, something that evaded even the great German experimentalists like Hertz and Lenard. All of which to say is — the point at which it became unambiguous was really quite late into the game, just a few years prior to Einstein and all that. The limitations were for the large part technical. --Mr.98 (talk) 00:56, 31 August 2012 (UTC)

I think two different meanings of vacuum are used here; "It is not self-evident that electric and magnetic fields can exist in a vacuum.." refers to an "absolute" vacuum (without aether). But in those days it was assumed that aether was still present in a vacuum, only air (gasses) was removed. As Newton wrote: "a much subtiler Medium than Air, which after the Air was drawn out remained in the Vacuum" Ssscienccce (talk) 15:12, 31 August 2012 (UTC)

Because it was known that according to Maxwell's equations, EM waves travel at a specific, finite speed c. The question immediately asks itself: relative to what? Usually, if you have a speed, you must have some definition of a stationary point with v=0. Defining stationarity at every point in space gives you a sort of aether. Of course, we know now that the speed of light is invariant and comes out the same for all observers. However, this requires us to give up any notion of absolute time, stationarity, and a lot of other notions in order to formulate special relativity. It seemed far more parsimonious to keep Newtonian physics and to suppose that light travels through an aether, than to eliminate the aether and also throw out all the known laws of mechanics. Given what was known at the time, I think this was a reasonable application of Occam's razor. --Amble (talk) 00:47, 31 August 2012 (UTC)

Just a point: the constancy of c was not in fact known until the late 19th, early 20th century. This was part of the point of Michelson-Morley. --Mr.98 (talk) 00:56, 31 August 2012 (UTC)

The constancy of c (or $1/{\sqrt {\epsilon _{0}\mu _{0}}}$ or however you want to write it) is a property of Maxwell's equations. To someone ignorant of later developments (such as Maxwell himself) the theory looked mathematically like a description of transverse sound waves in a solid, and it was natural to wonder about the properties of that solid, including its state of motion. -- BenRG (talk) 02:54, 31 August 2012 (UTC)

As I said, the invariance of c was not known until later. The fact that it's a constant comes out of the equations, and it was natural to take it as a constant velocity relative to some medium, as with other waves. Hence, the aether. --Amble (talk) 07:04, 31 August 2012 (UTC)

Sorry, what I meant was, this wasn't experimentally confirmed until much later than Maxwell. There were plenty of folks at the time who still thought this might be an open question until the measurements had come in. It was a hard thing to measure, obviously. --Mr.98 (talk) 14:46, 31 August 2012 (UTC)

That's why I said "according to Maxwell's equations." --Amble (talk) 15:26, 31 August 2012 (UTC)

Maxwell himself used the idea of the aether as a way to think through how the world worked. It was a visual metaphor to use as a means of interrogating nature. (Maxwell's own visualizations of the aether are amazing — pulleys and gears and whatnot. See figure 2 here. Very industrial age U.K.) Maxwell derived all of his original equations from these sorts of models, and had tremendous success with them. (The version of Maxwell's equations you are now familiar with are not Maxwell's own; they were re-written by Heaviside and others some time later.) When you get success out of a model, you don't turn around and throw out the major components of it immediately afterwards.

For the physicists of the day, the aether was actually considered a brilliant theoretical innovation. It was more than just the medium through which light moved — it was the interface between the world of electromagnetism and the world of matter. The theorists of the day came up with beautiful, elegant, and totally physically sensible models about how all matter arose out of whorls and knots in the aether; how this sublime otherworldly realm of pure energy lent itself towards the manifestation of all that we are and can observe. They saw it not as an appendage onto Maxwell's equations, or as an extremity on to the physical universe, but as a supremely elegant concept that also made good intuitive sense (light is a wave in something, right? Who's heard of a wave in nothing?). Beautiful physics came out of it. The problem was, well, it didn't exist, it wasn't necessary, and in fact they were going about it all the wrong way. We know that now, and it's hard to put yourself in the head of someone then, but look at the aether as the original theory of everything, the one-stop theoretical shop for All of the Rest of Physics. --Mr.98 (talk) 01:08, 31 August 2012 (UTC)

Honestly, not much has changed. The modern quantum vacuum has a lot of structure, and any particle physicist will tell you that particles are oscillations of the vacuum, not objects in the vacuum. The vacuum is similar enough to a solid that many phenomena of fundamental quantum field theory appear also in solid-state physics. The vacuum undergoes phase transitions, like the Higgs field condensation that breaks electroweak symmetry. Mechanical analogies haven't died. Quantum Field Theory in a Nutshell by Anthony Zee introduces the field as a "mattress" of weights connected by springs. It seems unlikely that the world really is a mattress simply because Lorentz invariance unifies space with time, and it would be strange if at a deep level space and time are aspects of the same thing but at an even deeper level they're not. There are hints (from gravitational holography and AdS/CFT) of a radically different underlying structure for the world. But the best modern theories of fundamental physics (general relativity and the standard model) are still mattress-based. In fact the next thing Zee does after introducing the mattress analogy is complain about the lack of anything better. -- BenRG (talk) 02:54, 31 August 2012 (UTC)

would maxwells aether be able to penetrate a black hole or its event horizon?GeeBIGS (talk) 04:04, 31 August 2012 (UTC)

Read Luminiferous aether and it should clear up any questions. The problem with the idea was that it had to be infinitely dense and infinitely malleable, which kinda doesn't work. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:27, 31 August 2012 (UTC)

Again, anyone who replies, "obviously it doesn't work" doesn't really understand it. Henri Poincaré and Hendrik Lorentz thought it worked and made sense; they were much smarter, and much deeper, than you or I. It is unilluminating to dismiss in a flip way ideas that were highly persuasive to luminaries in the not-even-that-far-back past. (Additionally, that article does not clear up the question about black holes. I have no idea how you'd make an aether world view compatible with a black hole. The former worldview was abandoned before General Relativity and my guess is that it is probably not easily compatible with it.) --Mr.98 (talk) 14:51, 31 August 2012 (UTC)

I do. The black hole is so dense that it forces the aether out of that space entirely.165.212.189.187 (talk) 15:04, 31 August 2012 (UTC)

I'm pretty ignorant about this, so take with a grain of salt, but my impression/recollection is that...

• versions of the aether that "work" with changing perspective are based on Fresnel's 1818 Aether drag hypothesis, which postulates that there is more "aether" inside glass than inside air.
• Some people, somewhere, I know not who, have tried to describe electromagnetic forces as curvature of space in some other dimension, just as gravity is described, so that objects subject to such forces actually follow straight paths (geodesics) just like objects subject to gravity.
• I would assume ...... that just as gravitational waves are transmitted as waves within the underlying spacetime (note that that is definitely a speed-of-light, inverse square transmission which by all contemporary formulations does occur in a sort of medium, i.e. spacetime!), that so electromagnetic forces are transmitted as fluctuations in this postulated electromagnetic dimension...
• And as such, the notion of the aether remains current.

Feel free to poke holes. :) Wnt (talk) 16:07, 2 September 2012 (UTC)

What are the differences between male and female human faces? In words...

This may be easy for lots of editors but not for me. Forgetting facial hair, which not all men have, how can we describe the general differences between men's and women's faces? Tom Haythornthwaite 22:41, 30 August 2012 (UTC) — Preceding unsigned comment added by Hayttom (talk • contribs)

Look at Eric Roberts and Julia Roberts and tell where you see differences - if any. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:49, 30 August 2012 (UTC)

Thank you, I think, but either you didn't understand my question or I didn't understand your answer or we have different senses of humour. It is easy to describe the general differences between male and female bodies because men are generally taller and have penises and women generally have breasts, and so on. But I cannot think of words to generally differentiate between their faces. hayttom 23:11, 30 August 2012 (UTC)

What I'm saying is that you can't, necessarily. To my mind, forgetting any facial hair, Eric and Julia look a lot alike. If you were to somehow remove all the hair anywhere on their heads and put their faces side by side, it would be hard to tell the difference. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:57, 30 August 2012 (UTC)

See this article on the role of testosterone in the development of male facial features. Alansplodge (talk) 23:34, 30 August 2012 (UTC)

My impression is that women are more likely to have the so-called "heart-shaped face", where the sides of the face slope inward as you go from the temples towards the mouth, whereas men are more likely to have straight-sided faces. Not sure I'm right; would be interested to hear about any real studies. --Trovatore (talk) 23:37, 30 August 2012 (UTC)

Leaving out individual variations, the basic difference is that female faces are more round, and male faces are more rectangular. Looie496 (talk) 00:07, 31 August 2012 (UTC)

That would be the reason rectangular, straight-side faced women, Ann Coulter for example, are sometimes referred to as "horsey" looking. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:11, 31 August 2012 (UTC)

Males tend to have more prominent brow ridges - the easiest way to tell from just a skull, as I learned in the palaeoanthropology lab at uni - though even that can be misleading sometimes. As for the heart-shaped vs straight-sided face, that is mostly due to males having more developed jaw musculature, I'd assume - and that will be rather variable. Combined, these two traits tend to make a male face look 'squarer', though I think that if you remove the cultural clues, it is often difficult to be sure just from the face. With children, it is even harder, and I suspect pre-puberty next to impossible from a casual inspection. AndyTheGrump (talk) 00:12, 31 August 2012 (UTC)

How would you characterize the brow ridge of Roy Blunt? 20.137.18.53 (talk) 13:58, 31 August 2012 (UTC)

Just to add in, those are the two things they told us in my forensic anthropology as well. The square jaw is not from musculature; it is skeletal. Women on the whole have large jaw angles than men do. Again, as you note, none of these skeletal features — even the non-facial ones, like hip bone shape — are a sure-thing, because there's a lot of variation in the species. --Mr.98 (talk) 00:47, 31 August 2012 (UTC)

How would you characterize the jaw structure of this Savannah they have on the Today show now? There's something odd-looking about her - like a squirrel with its cheeks stuffed, or something. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:30, 31 August 2012 (UTC)

The Adam's apple is one differenece. Not 100% of the time, but on the balance, men tend to have a more prominent Adam's apple than females. --Jayron32 04:11, 31 August 2012 (UTC)

Yes, always check the neck before necking. And, if going further, check for other lumps, too. :-) StuRat (talk) 09:02, 31 August 2012 (UTC)

August 31

Is there a way to help someone run faster?

Suppose a man and woman are both running away from a scary monster, and the man wants to help the woman run faster. This is a common scene in movies, and in said movies, the man usually holds the woman's arm to drag her forward. I've tried this before, thankfully without the monster, and it's an extremely uncomfortable way for both people to run. Neither person can swing their arms in synchrony with their legs, thus slowing both people down. Not only that, if the faster person pulls too hard, the slower person tends to fall forward instead of running faster. So, is there actually a way of helping someone run faster than she normally could? --99.227.95.108 (talk) 05:51, 31 August 2012 (UTC)

Being chased by a monster should do it. Emotional states and high adrenaline can cause people to outperform their own personal standards, but I don't know that there's a way, short of attaching rocket skates to her feet, for another person to aid her in running faster. --Jayron32 05:53, 31 August 2012 (UTC)

I agree that holding her arm won't work. But, if the man runs behind the woman, and with both hands on her sides, pushing her, she'll go faster. I've tried that and it works, balance is good, but there is a danger of feet contact, so the man has to run with his feet further to the sides. I should point out that my lady has longer legs than I do, so if there really was a scary monster, she most likely could outrun me. In any case, if the man is making the woman go faster by either towing or pushing, then he must be going slower than he could on his own. So, only do this if either a) you actually do love her, or b) you know for a fact that the scary monster only likes girls. I also point out that any test of strength, running speed, etc with girls is misleading. They do what they think will attract you, i.e., play the poor defenless female. Every girlfriend I've ever had has asked me to open jars etc. But they seemed to consume the contents just as much when I wasn't around. Wickwack121.221.84.138 (talk) 06:35, 31 August 2012 (UTC)

At the risk of generalizing, this reminds me of the saying/song 'the female of the species is more deadly than the male' (especially if her offspring is in danger). I think they pretend to be helpless and defenceless but when the chips are down, they are just as (if not more) capable than men to help themselves out of socially perceived 'male-only' situations. I did a test a few months back... there was a faulty handle in the bathroom and I left a screwdriver and pliers in the bathroom, telling my wife how to use it if she couldn't get out. The first time she screamed hell and highwater and I had to let her out (save her from death by extended bathroom time). The second time I was not around and she screamed/cried a bit, then actually used the tools to make her way to freedom. I wasn't popular that week but I had proved a point to myself :) Sandman30s (talk) 08:15, 31 August 2012 (UTC)

@Wickwack and Sandman: That's because you actually tolerate women who pretend to be helpless and defenceless, like by refusing to open jars. Some people view that as disgusting, not attractive. --99.227.95.108 (talk) 04:29, 2 September 2012 (UTC)

When in Rome, do as the Romans do. This aspect is culturally dependent. In Australia, if a girl asks you to do something, and you want the girl, do what she asks. I am aware that in Iran, for instance, no girl would dream of asking such things. Its a bit like opening doors for a lady. In Australia, a man opening a door for a lady is considered good manners. In some countries it may be an insult. Wickwack121.215.159.205 (talk) 10:11, 3 September 2012 (UTC)

In the case of running, women really do tend to be slower. It's not just due to shorter legs. Wider hips also make running less efficient, and a higher fat-to-muscle ratio means less sprinting power (although it might help in a marathon). StuRat (talk) 08:57, 31 August 2012 (UTC)

Skinny is good when it comes to marathon running. Sprinters should be 2.5% lighter than the average person while long-distance runners should be 15% lighter according to this article. Just look at these ladies. Alansplodge (talk) 17:56, 31 August 2012 (UTC)

Certainly they can't be obese, as that would interfere with efficient running (fat thighs would have a similar effect to wide hips, resulting in more of a waddling motion, and loose, bouncing fat is bad, too). However, exactly how thin they should be would be affected by the precise conditions of the marathon. Specifically, cold weather runs (especially in strong rain and wind) without the ability to consume energy drinks along the way are conditions where having a bit more fat to burn might be helpful. StuRat (talk) 20:40, 31 August 2012 (UTC)

Everyone burns fat during a marathon, whatever the weather. Top athletes now reduce carbohydrate intake to maximise fat burn Fuel On Fat For The Long Run. But the amount of body fat needed for this is really small and I've never seen an elite marathon runner who wasn't as thin as a stick. See Wilson Kipsang, Paula Radcliffe or Deena Kastor. But we digress. Alansplodge (talk) 01:12, 1 September 2012 (UTC)

You can help someone run faster without slowing yourself down by letting them run in your slipstream. It's not a particularly large effect for humans running (it's more important for lobsters walking along the seabed, and cyclists), but every little helps. --Tango (talk) 11:37, 31 August 2012 (UTC)

As for saving your g/f from a monster, how about getting the monster to chase you, say by throwing things at it ? Also, many shoes women wear are less than useless when it comes to running, so get her to kick off her stiletto heels before running. And, as a bonus, the monster may then decide she's no longer pretty enough to chase. :-) StuRat (talk) 08:57, 31 August 2012 (UTC)

Throwing things at it assumes that the monster is too dumb to go for the easier target and then use her as a shield. Wickwack120.145.10.162 (talk) 10:42, 31 August 2012 (UTC)

In the initial sprint of a few hundred meters, pulling or pushing the slower person might not help. but urging them to kick off any inappropriate shoes and run like hell might speed things along. In movies. the woman usually falls down, to add suspense and an occasion for the film to show a view of the monster gaining on them. Similarly, in movies, the runners always stop occasionally and look back, for cinematic effect. In military and some sports training intended to build a team, the more fit runners actually do physically help the exhausted runner along once he/she is winded and unable to continue the pace. At that point, if they were ascending a grade, for instance, putting an arm around the tired runner could materially speed them along and keep them moving. So in the initial sprint its "Feets, do your stuff!" with some benefit from vocal encouragement, but when exhaustion sets in an arm around the person's back can help them move along. Edison (talk) 15:46, 31 August 2012 (UTC)

In this YouTube video, exhausted soldiers are pushed, pulled and shouted at with varying degrees of success (from 5:40). Having an impressive moustache seems to help too. Alansplodge (talk) 18:05, 31 August 2012 (UTC)

Good-looking

Duplicate Ankh.Morpork 14:19, 31 August 2012 (UTC)
The following discussion has been closed. Please do not modify it.
What makes people good-looking; is there any objective criteria such as symmetrical features which influence people's perception? Why are we predisposed towards selecting better looking people as partners? Isn't there an increased potential of infidelity? Has people's perception of 'good-looking' changed over time? Ankh.Morpork 14:11, 31 August 2012 (UTC) Didn't we just answer this question above or on another desk? See physical attractiveness and the prior discussion. --Jayron32 14:13, 31 August 2012 (UTC) Indeed. Count up 6 threads. It was asked yesterday. --Jayron32 14:14, 31 August 2012 (UTC)

Corneal vs Cataract Transplants

Hello. How are opthalmologists able to determine whether a patient has had a cornea or cataract transplanted just by looking at the eye? Thanks in advance. --Mayfare (talk) 14:13, 31 August 2012 (UTC)

As a non-expert with a dangerously small amount of knowledge I would suggest that following a cornea transplant there would be a detectable (perhaps under magnification) circular scar on the cornea. When a lens with a cataract is removed it does not require the removal of the whole cornea, just a small incision, usually near the outer edge of the iris, big enough to remove the damaged lens and (nowadays) insert a replacement. Richard Avery (talk) 21:03, 31 August 2012 (UTC)

SR & QM unification

How plausible is it that the two theories have not been unified because of some minute error way back in the 1800 -1900's that has since been so baked into the equations and laws that it is virtually undetectible now? — Preceding unsigned comment added by 165.212.189.187 (talk) 14:19, 31 August 2012 (UTC)

Special relativity and quantum mechanics have been unified for ages; there is no difficulty there. The problem is with general relativity. Looie496 (talk) 16:56, 31 August 2012 (UTC)

The article Quantum gravity describes quite well some of the issues surrounding the marriage of general relativity with quantum mechanics. As noted, special relativity has no such problems, because the sticking point is gravity, which is the domain of GR. Theory of everything and Physics beyond the Standard Model dontains some of the theories and development at the vandguard of this area. --Jayron32 17:36, 31 August 2012 (UTC)

OK General Relativity then, same question.165.212.189.187 (talk) 17:57, 31 August 2012 (UTC)

OK, same answer then. --Jayron32 17:58, 31 August 2012 (UTC)

I missed the part where you say how plausible my question is?165.212.189.187 (talk) 18:14, 31 August 2012 (UTC)

Plausibility is not something we have the ability to provide references for. You can make up your own opinions about that, for anyone here to do so would be pure speculation. I have done what the mission of this page is for, which is to provide references to information. After reading about the difficulties that exist between marrying the two theories into a single cohesive theory, you can decide for yourself how plausible that marriage is. If you have any questions about the text of those articles, something you don't understand or need explained, we would be happy to help clarify or provide more references. But questions about how "plausible" something is are not answerable in the way this forum works. It is what it is, and the two theories have not been maried as yet. The articles already cited explain why. --Jayron32 18:24, 31 August 2012 (UTC)

Relativity wrong in back in time, and QM wrong in randomly , it should have united, thanks — Preceding unsigned comment added by 81.218.91.170 (talk) 18:19, 31 August 2012 (UTC)

hello water nosfim -- 203.82.95.168 (talk) 08:44, 2 September 2012 (UTC)

Our lack of knowledge of statistics comes from the movement during the time dimension, and it affects slightly on the wording theories, BELL talked little about causality and localty and it should incorporate . yes ? water nosfim , thanks — Preceding unsigned comment added by 81.218.91.170 (talk) 09:41, 2 September 2012 (UTC)

And the twin paradox we Supposed be solved through parallel universes and not relativistic time. — Preceding unsigned comment added by 81.218.91.170 (talk) 10:20, 2 September 2012 (UTC) , thanks water nosfim .--81.218.91.170 (talk) 10:22, 2 September 2012 (UTC)

Special relativity is basically just Lorentz symmetry. There's no minute error you can make in defining Lorentz symmetry; if it doesn't hold, its replacement will have to be something a lot more complicated. Quantum mechanics also has a very special form and it's either right or it isn't. There's more freedom in defining a relativistic theory of gravity. General relativity is a remarkably simple theory and it's still in perfect agreement with experiment, but it has always had rivals. Still, if there were any simple alternative to GR that led naturally to quantum gravity, I'm pretty sure quantum gravity researchers would have found it by now. So I would rate this as highly implausible. -- BenRG (talk) 20:19, 31 August 2012 (UTC)

There actually isn't a real problem here, the issue is simply that the field theory you get is nonrenormalizable. Count Iblis (talk) 02:32, 1 September 2012 (UTC)

Ebola Carrier Found

I recall reading a story that bats were determined as the carrier of Ebola Virus. Apes, as well as indigenous natives apparently eat dead bats and get the virus in that manner. Why hasn't Wikipedia updated this latest scientific discovery into the Ebola web page? — Preceding unsigned comment added by 66.8.226.55 (talk) 19:33, 31 August 2012 (UTC)

I see it in Ebola#Risk factors. Well I see about eating partially-eaten fruit, rather than eating the bats themselves. It's from references going back to at least 2007. If there is newer information available, someone would have to post the actual reference/citation to support it. DMacks (talk) 19:38, 31 August 2012 (UTC)

I'm currently on holiday on an island where even the restaurants have bat on the menu. Greglocock (talk) 00:41, 1 September 2012 (UTC)

GOD, I hope for your sake your room doesn't smell of sulfur. See (or, better yet, don't see) Popo Bawa. μηδείς (talk) 01:40, 1 September 2012 (UTC)

I would ask for it to be "well done". Alansplodge (talk) 01:20, 1 September 2012 (UTC)

The explanation for why Wikipedia is not expounding on "the great new scientific finding you just read about" is always either: A) No one has volunteered to do it yet; or B) the media totally overstated the significance of a recent finding. I'm sure if you post a link to the story you read, someone here will be happy to update the article, if appropriate. Someguy1221 (talk) 01:25, 1 September 2012 (UTC)

September 1

Alkane isomers

What's the general formula for the number of isomers of the alkane with n carbons? --168.7.232.202 (talk) 01:04, 1 September 2012 (UTC)

There is no known mathematical relationship between the number of carbons in an alkane and the number of possible isomers of that alkane, but it rises as a rate that resembles an exponential function (see here for a list of alkanes 4-15,20]. The problem is further complicated by the fact that some isomers that can be represented by a line diagram are not actually stable molecules (you start running into these at C16). People have written algorithms to calculate the number of isomers, and you can find one here that can be told to remove unstable isomers. Someguy1221 (talk) 01:12, 1 September 2012 (UTC)

See our article entitled "Alkane", which has a section "Isomerism" that discusses the theoretical calculation issue. Might be a good place to add a mention that the calculated result is a theoretical maximum, whereas stable reality is more limited. DMacks (talk) 04:41, 1 September 2012 (UTC)

Delta formation

I am not able to understand how is delta formed. Why is delta formed near the mouth of the river, not in its middle course? I want to know the reason for this according to geographical as well as scientific view. Sunny Singh (DAV) (talk) 02:16, 1 September 2012 (UTC)

Who says that Deltas don't form midcourse? Have you read the article River delta? It covers inland deltas and how they form. --Jayron32 02:18, 1 September 2012 (UTC)

I know but most of the delta is formed near mouth.Sunny Singh (DAV) (talk) 03:15, 1 September 2012 (UTC)

Delta formation requires the flow speed of the river to slow enough that it can no longer keep sediments suspended. The land formation that permits such a thing to happen is simply unlikely to exist near a river. In my own original researchy hypothesis, I would assume this is because rivers tend to erode themselves a riverbed through which to travel, preventing a sufficient loss of speedy. Someguy1221 (talk) 02:40, 1 September 2012 (UTC)

I read the following statement in a Chemistry book -

"Delta is formed when river water comes in contact with sea water for a long period.
River water is mostly muddy. These mud particles are charged colloidal particles. When river comes in contact with sea, the dissolved salts present in sea water provide ions with charge opposite to the charge on mud particles. These particles get uncharged and combine with each other to form bigger particles. They settle as solid mass. Over the years, deltas appear at these places."

This is a short explanation of delta formation. I want to know more about this. Sunny Singh (DAV) (talk) 03:15, 1 September 2012 (UTC)

This is the first time I've heard that theory. The far simpler explanation above seems more likely, that river water slows when it hits the ocean, and all the sediments it carried then settle to the bottom. StuRat (talk) 03:35, 1 September 2012 (UTC)

The theory sounds silly on the surface, anyway. If the river is filled with charged colloidal particles, it must also be filled with oppositely charged particles, colloidal or otherwise. The same goes for the ocean. Either Sunny misparaphrased, or whoever came up with the theory doesn't understand even the most fundamental of chemistry or physics. Someguy1221 (talk) 03:53, 1 September 2012 (UTC)

I have written the same thing written in the book. According to you, book's statement is wrong. This means the answer you gave earlier is the real cause of delta formation. Sunny Singh (DAV) (talk) 04:16, 1 September 2012 (UTC)

For the muddy particles in suspension in river water, flocculation does occur when it encounters seawater, and is one of the reasons that mud is deposited in deltas [4]. Mikenorton (talk) 08:07, 1 September 2012 (UTC)

For coarser particles (not muds), then it is the reduction in flow velocity that causes the sediment to be deposited, as already stated. Mikenorton (talk) 09:47, 1 September 2012 (UTC)

You might be interested in this paper, "Modeling river delta formation". Also, the River delta article isn't bad. Sean.hoyland - talk 10:28, 1 September 2012 (UTC)

What Sunny Singh said is exactly right (Someguy1221 jumped the gun). The colloidal are water soluble in fresh water but become insoluble when the charges present at their surfaces is cancelled out by charges present in the salty water. Dauto (talk) 23:50, 1 September 2012 (UTC)

I still hate that explanation. It's not the presence of any oppositely charged particles that renders the suspension insoluble, but specifically ones that can stably bind to the suspended particles. Someguy1221 (talk) 00:56, 2 September 2012 (UTC)

Water - oil mixture

Why does water not get uniformly mixed with oil? One reason I read is that water is heavier than oil. Then, why does water get mixed with alcohol and doesn't make separate layer as in case of oil. Sunny Singh (DAV) (talk) 02:24, 1 September 2012 (UTC)

We have an article on this phenomenon, at Hydrophobe. You may also be interested in Chemical polarity. The general way I have seen this presented to beginning chemistry students is that polar molecules, such as water, bind extremely well to themselves and other polar molecules, very much unlike non-polar molecules such as oils. The immiscibility of water and oil is a consequence of the water molecules basically binding to themselves to the exclusion of the oil molecules. Someguy1221 (talk) 02:36, 1 September 2012 (UTC)

And no, differences in density are not the reason, in the case of water and oil, although two fluids of different densities might take a long time to mix, just because one would be at the bottom and the other on top, with a limited area of interface, which would slow the mixing. This can happen, for example, with fresh water and salt water. Even though they would mix completely if stirred up, under the proper conditions they can stay in layers for a long time, due to density differences. StuRat (talk) 03:31, 1 September 2012 (UTC)

Given that they are not miscible, they form layers even if stirred up because of different density: they "separate" rather than dissolve because of immiscibility, and the separation is top-vs-bottom because denser stuff sinks and less dense stuff floats. Two separate (sorry:) issues that have various combined effects. If you pick two immiscible materials that happen to have the similar density and mix them well, they can form an even suspension or colloid that does not visibly re-separate. Milk (especially whole homogenized milk) and Mountain Dew (thanks to [[brominated vegetable oil) are two common examples. The closer the density and/or the better you mix them, the longer time or stronger gravity it will take for them to separate. DMacks (talk) 04:51, 1 September 2012 (UTC)

Even with mixtures where the densities are somewhat different, you can create metastable suspensions that last a very long time with the help of extreme agitation or emulsifiers. But regarding the water-oil problem, let me confirm StuRat has already said: the bonding energy you would gain back by forming water-oil bonds is not enough to overcome the energy needed to seperate water molecules from each other, or to get fancy, the energy of solvation is too low between oil and water. Part of the situation with the immiscibility of water and oil is that oil is composed of essentially nonpolar molecules, and the polarized water needs to have some sort of electric charge to "hold on to". There are no localized charges in oil molecules, so water is far happier holding tightly to other molecules. --Jayron32 05:00, 1 September 2012 (UTC)

It's a shame there's no such thing as butter or mayonnaise. μηδείς (talk) 23:30, 1 September 2012 (UTC)

Exactly. Butter is a metastable mixture of fat and water, and it forms via extreme agitation. Mayonaise contains eggs, which contain lecithin, which is an emulsifier. It seems odd that you would confirm exactly the point I made about metastable mixtures of fat and water, and then take a tone as though you would be disagreeing with me. That is quite weird. --Jayron32 01:38, 2 September 2012 (UTC)

It's a shame you're not American, or you would recognize kidding sarcasm when you come acrost it. But yer not. Some sorta weird island of literal zombies called Carolina, apparently. μηδείς (talk) 01:55, 2 September 2012 (UTC)

The individual molecules in a liquid are electrically polarised - they have an unequal distrabution of electronic charge across the molecule. This means that different portions of a molecule is more negatively charged than the rest. I assume that you know that oppositely charged objects attract; molecules attract each other due to their polarity.

Water molecules are more strongly polarised than oil molecules; the attraction force between charged objects increases as the charge on the object increases. Meaning, water tends to associate more strongly with itself than with oil. Relative to water, oil doesn't associate strongly with itself at all, there is just no other alternative.

The differen density between water and oil has nothing to do with it, density just determines the order of the fractions as it sepperates due to gravity or centrifugation. Alcohol mixes with water, as even though it has a weaker polarisation, it is comparable to that of water.

Note that nothing is truely insoluble, just diminishingly soluble. Plasmic Physics (talk) 01:33, 2 September 2012 (UTC)

A lot of beginner chemistry students make the mistake of thinking that "polar" and "non-polar" are opposite terms, more accurate terms should be "weakly polar" and "strongly polar". Plasmic Physics (talk) 01:45, 2 September 2012 (UTC)

Tumor Lysis Syndrome seen in antibody dependent cell mediated cytotoxicity?

Tumor Lysis Syndrome (TLS) is of course a major adverse event in anti-cancer treatments such as Rituxan. As the monoclonal antibody kills B-cells, their contents are spilled into the bloodstream.

What I'm trying to determine is whether TLS would be a possible consequence of anti-cancer treatment that is based in antibody-dependent cell-mediated cytotoxicity, such as the kind you might find when using a therapeutic cancer vaccine.

If not, what is the difference between what activated T-cells are doing to the targeted cancer cells and their contents vs what an immunotherapeutic like Rituxan is doing?

Or God forbid am I even making sense? I'm a health/medical writer but cred-wise I'm just a lay person.

--Wolfgangus (talk) 06:25, 1 September 2012 (UTC)

Yes, TLS is listed as a possible consequence of Rituxan, see http://en.wikipedia.org/wiki/Rituxan#Adverse_events. You should also read the manufactuer's Patient Information sheets, and note the advice given. Wickwack58.164.230.22 (talk) 06:46, 1 September 2012 (UTC)

A few mechanisms of action (including ADCC) of rituximab (anti-CD20) have been proposed but it's not clear which one dominates, as described in this recent review PMID 22621628. It's not clear that the mechanism of action will be the same for different antibodies and cell types. Much may be learned from situations in which these treatments stop working, e.g. PMID 22368276 -- Scray (talk) 20:16, 1 September 2012 (UTC)

I may not have been entirely clear in my question. TLS is generally a consequence of the manner in which an anti-cancer drug kills a cancer cell--the cell's contents are released into the bloodstream. My question is whether TLS might be expected when the treatment involves enlisting the body's own T-cells for an immune response, as is the case in the emerging area of therapeutic anti-cancer vaccines. If the body's T-cells are responsible for killing a cancer cell, what does it do with the cell? If chemotherapy induces cell death during the cell cycle and the contents spill into the blood, what does an activated T-cell do with a cancer cell that it has killed? Does it just carry it back to the lymph nodes like any other foreign pathogen? If so, TLS doesn't seem possible. Wolfgangus (talk) 22:18, 2 September 2012 (UTC)

See the WP article I linked in my 1st post. TLS is unlikely with the usual run of chemotherapy agents, which interfered with a cell's divison/reproduction process. In standard chemo, if a cell is not dividing, it does not die, it just carries on. If it would otherwise divide, chemo either prevents it from dividing, or makes the divided (daughter) cells unviable. All cells are affected, not just cancer cells - fortunately almost all cells in the body only divide during growth. Cells that do divide, such as hair cells, finger nail beds, etc tend stop dividing during chemo, and because the brain cannot make new neurons, learning a complex subject may be difficult. Rituxan/Rituximab is quite different. It is not a chemotherapy. It is an antibody targeted to a specific cell type, whether cancerous or not, and leads to all cells reached getting destroyed, subject to imune system capability. Thus TLS is quite unlikely with chemo, but can occur with Rituxan. Wickwack58.170.164.118 (talk) 23:01, 2 September 2012 (UTC)

Insulation value of aluminium foil

Hi, I'm currently building an ice-box in my boat, and I intend to insulate it on all sides with 100mm of rigid urethane foam. Friends have suggested I include a layer of reflective foil around the outside of the foam. If I did, the foil would be tightly sandwiched between urethane foam on the inside, and plywood on the outside, and I can't understand how this would reduce the heat flow into the ice-box. Can someone explain this to me please?124.191.178.49 (talk) 07:02, 1 September 2012 (UTC)

I think the idea is that the aluminum foil will reflect radiated heat. If so, it should be placed on the foam, with an air gap between it and the plywood (a vacuum would be better, but I assume that's not possible here) . However, I agree that jamming it in so it fills the air gap isn't good, as that will allow the aluminum foil to conduct heat directly from the plywood to the foam. So, if there isn't room for it, leave it out (or make the plywood box a bit bigger).

Another concern is that you could get condensation on the foil, which may then soak the wood, allowing it to rot. If the air inside was completely dry and sealed from the outside, this wouldn't happen, however, an airtight seal would be difficult, so an air gap and drainage holes at the bottom are a good alternative, to allow the water to drip down and out. Put a pan under the drainage holes to collect the water. StuRat (talk) 08:48, 1 September 2012 (UTC)

Agree with Stu, foil won't do much in this situation. Careful construction of the insulation with seams minimized (and staggered if using several layers) should give best results. Ssscienccce (talk) 14:19, 1 September 2012 (UTC)

Perhaps your friends were confusing aluminum foil with mylar foil, which IS used effevtively as a thermal insulator. Though mylar is often shiny and metallic looking, it isn't interchangable with the stuff you wrap your leftovers in. Mylar may be useful for this, but I can't imagine Reynolds Wrap as being a good idea. --Jayron32 14:50, 1 September 2012 (UTC)

Aluminium foil works just as well as insulator, it reflects about 98% of radiant heat and light (source). However, aluminized Mylar is a better choice because it does not tear as easily. - Lindert (talk) 15:30, 1 September 2012 (UTC)

Yes, but foil's value as a radiative reflector is more than compensated by its value as a contact conductor of heat. That is, while it may reflect radiated heat, it is also an excellent conductor, and as such, if it is in contact with what you are trying to keep cold/warm, its reflective ability doesn't really come into play. --Jayron32 15:49, 1 September 2012 (UTC)

Yes, if it not separated by air or empty space, it is basically worthless. In this case the space should not be between the cooled object and the foil, but between the object/foil and its environment. The reverse is true if keeping something warm. The same is true if you use mylar foil though (which btw is just plastic foil with an aluminium layer). - Lindert (talk) 15:52, 1 September 2012 (UTC)

No, no, no. The aluminum foil will be useful (By reflecting radiation) even if touches the plywood layer. It would conduct heat, off course but it would still reflect radiation back to the plywood. You don't need to separate the foil from the environment with vacuum or air. You can separate with urethane foam and plywood which are good insulators. Dauto (talk) 23:40, 1 September 2012 (UTC)

It will reflect radiation, but it will be of no significant use because the amount of heat transfer due to conduction far exceeds the amount due to radiation. - Lindert (talk) 12:32, 2 September 2012 (UTC)

Many answers on Ref Desk are very good answers. However, this question is a good example of well-meaning folk jumping in without thinking, and giving opinions on things they know little about. The reflective properties of aluminium foil are of no help here, as wood is not transparent to heat radiation. Rigid urethane foam also has negligible transparency to heat radition. Therefore, only heat conduction is important, and, of course aluminium is an excellent heat conductor. However, Dauto may have intened the aluminium foil to go on the inside of the foam, not between the foam and the plywood. In that case, the low emissivity of the aluminium may help, provided none of the contents of the ice box touch it, as Jayron said. However, the OP's friend may have been confused, not by aluminsised mylar, whose properties in this regard are not much different to alauminium foil, but by the use of aluminium foil as thermal insulation in houses and shops. Here the aluminium is istalled uder the rafters, so there is a 125 mm air space between the aluminium and the roof outside material, and below it is the air in the ceiling space. Here, the low emissivity of aluminium does work with the air above and below it to lower heat transmission. It is not anywehre as good in this respect as 100 mm of foam or fibreglass batts on top of the ceiling. I can't see condensation affecting the outer wood being an issue in the way StuRat said. Rather, the aluminium will act as a moisture barrier (which is a common use of it) and protect the wood. However the OP will need to prevent condensation from inside of teh ice box from getting into the fowm, ruining its insulative properties. Ratbone121.221.217.156 (talk) 03:33, 2 September 2012 (UTC)

Re: "negligible transparency to heat radiation", yes, this is true, but irrelevant. The heat passes through the plywood (slowly) by conduction, not radiation. Once on the inside edge of the plywood, if there is an air gap, it can not efficiently transfer to the foam by conduction (and a vacuum would stop convection/conduction completely). However, radiation of the heat from the plywood to the foam can still occur. This is where the foil comes in. Yes, a non-conductive material which is reflective would be better, but aluminum foil would still be better than nothing, provided there is an air gap between it and the plywood to prevent conduction. StuRat (talk) 03:57, 2 September 2012 (UTC)

It's very relavent - its the crux of what diffrent posters are on about. The OP specifically said "tightly sandwiched" between the foam and the wood. However, you are correct in that if an air gap was intrduced, the aluminum would help.

Agreed, and I said so in my initial response. So I'm baffled at why you said this Q is an example of "well-meaning folk jumping in without thinking, and giving opinions on things they know little about". StuRat (talk) 04:52, 2 September 2012 (UTC)

You had it right in your 1st post, but others did not:-

Jayron32 said that mylar film would be good, implying it would be good using the way the OP said, sandwiched between the foam and the wood. It would not be good. While the thermal resistivity of mylar is very much greater than that of aluminium, in film form it is too thin to matter.

Lindert thought that the reflectance of the film was important. It is not, because the foam and the wood will block radiant heat. Only thermal conductance is important, unless there is an air gap, or vacuum gap. But after prompting from Jyaron, he got it right, only for....

Dauto said, no, film without an airgap will work. No, it won't. And he went on to imply that an airgap between the foam and the wood will improve it because they are insulators. Well, an airgap will improve things, but NOT because the foam and wood are insulators. An airgap will always improve things, regardless of materials, because it is itself a thermal insulator. An airgap will be less important if teh materials are insulators.

Ratbone58.170.164.118 (talk) 11:20, 2 September 2012 (UTC)

As for condensation, this happens when moist, hot air seeps into the air gap, hits the cool aluminum foil, and the water vapor condenses on the foil. If it's in contact with the plywood, the wood would then get wet. If not, then drops will form and fall to the bottom of the aluminum foil, and, hopefully, down to a drainage tray below. If the foam is thick enough to keep the foil from getting too cool, or the box is sealed well enough to keep out hot, moist air, then this won't be a problem. Also, if this was in a desert, then moist air would be rare, but, in a boat, it should be common. StuRat (talk) 04:01, 2 September 2012 (UTC)

Actually, although wood is not a good conductor of heat, a likely thickness of wood will conduct vastly better than the OP's 100mm of foam. Therefore nearly all the temperature drop will be across the foam, not much across the wood. Only if the foil temperature is below the dew point will condensation on the wood side occur. It should be easy to seal the wood with varnish of whatever, as you would anyway on a boat, to reduce the diffusion of moisture to a level that does not matter. Wood ALWAYS has a significant moisture content anyway, as any cabinet maker will tell you. Wickwack124.182.13.142 (talk) 04:45, 2 September 2012 (UTC)

Sealing it doesn't seem so practical, to me. First there's the door to consider, where you would need to seal between the foam and plywood. Then there's thermal expansion to consider, as well as the wood swelling, shrinking, and warping with humidity changes (say dry dock versus out on the water). Loading up the freezer will also tend to cause it to sag under it's own weight. All of these factors will make maintaining an airtight seal rather difficult. If an airtight seal is the goal, then plywood is the wrong material. StuRat (talk) 04:56, 2 September 2012 (UTC)

As I explained, you don't need to seal it to air tight / gas tight degree. You only need to seal it to bring down the moisture diffusion to a tolerable level. Plywood does not have significant swelling/shrinking in the legth and width, only in thickness. Plywood does not warp, as the plys balance each other out. Wickwack124.182.13.142 (talk) 06:09, 2 September 2012 (UTC)

I disagree that plywood does not warp: [5]. To balance out, all of the plies would need to be exposed to identical temperatures and humidity, which is definitely not the case here. StuRat (talk) 06:13, 2 September 2012 (UTC)

Thank you all very much for your responses, now I can clearly understand the differance between radiant heat transfer and conduction, so I won't be wasting effort on foil. (OP)124.191.178.49 (talk) 05:05, 2 September 2012 (UTC)

You're welcome, but don't overlook the suggestion to add an air gap. Any good thermos will have a gap, although they are able to maintain a vacuum inside it. StuRat (talk) 05:25, 2 September 2012 (UTC)

Weight of atom

What will be the weight of a atom if its size expanded to a cricket ball?--Sunny Singh (DAV) (talk) 10:46, 1 September 2012 (UTC)

It would depend on the atom, since they each have different masses and sizes. The mass varies by element and isotope, and to a much lesser extent by charge. The size of an atom is a bit harder to define, as the outside edge is an electron probability envelope.

But, if we assume the atom in question is a typical atom from which a cricket ball is made, then the mass would be about the same. This is because in a solid, the atoms are packed essentially right next to each other. There is still lots of empty space, between the nucleus of each atom an the electron cloud, but that exists in both the individual atom and the collection of atoms in the cricket ball, in the same proportion. I'd say the single atom would have a slightly greater mass, since packing is never perfect with a collection of atoms. The atomic packing factor runs around 68-74% in a crystal, and might be a bit worse with an amorphous solid, let's say 50%. In that case, the single atom would have twice the mass, and so would weigh twice as much, too. StuRat (talk) 11:17, 1 September 2012 (UTC)

This is a none question because expanding atoms is not possible. It can only be answered hypothetically and depends what one assumes for the properties of the hypothetical atom. In other words the answer can be anything you like. For instance, if one postulates an atom that composed of normal protons and electrons, but has somehow contrived to place the electron shell at the diameter of a cricket ball, then the mass would be that of a single "ordinary" atom - almost nothing. On the other hand, if one postulates an atom with a nucleus the size of a cricket ball but with the same density, which is about the density of a neutron star, the weight would be something like that of the Alps. SpinningSpark 20:01, 1 September 2012 (UTC)

You seem to be assuming an atomic nucleus the size of a cricket ball--which would weigh 100,000^3 the mass of a cricket ball--but that is not what the OP said. μηδείς (talk) 20:09, 1 September 2012 (UTC)

I think we can assume that the question was asked hypothetically. 86.146.107.246 (talk) 21:03, 1 September 2012 (UTC)

An atom the size of a cricket ball (let's assume 10 cm radius for convenience, and a Helium atom, since that is the example at our atom article--the proportions of carbon twelve are the same) would have a "solid" nucleus 1/1000th of a millimeter across, containing 99.94% of the atom's mass (pretty extremely dense, huh?) and the remainder would be made of a cloud of electrons. It would in no sense resemble a cricket ball with a uniform density and a solid surface. μηδείς (talk) 20:06, 1 September 2012 (UTC)

This is an easy one, just simplify it to finding the atomic density for a given atom using its atomic mass and Van der Waal volume, and multiplying it by the standard volume of a cricket ball. Plasmic Physics (talk) 22:11, 1 September 2012 (UTC)

The Van der Waal volume being based on the size of the electron cloud, yes, that's what I said. μηδείς (talk) 23:27, 1 September 2012 (UTC)

Just a small aside, his name is Johannes Diderik van der Waals. With the 's'. Otherwise, yeah, that's how i would solve it as well. --Jayron32 00:30, 2 September 2012 (UTC)

• The atomic radius is n atomic mass units and something on the order of 100 picometers. A men's cricket ball is about 160 grams and 226 mm circumference = 36 mm radius. Taking 100 pm as our benchmark, expanding it to 36 mm is a 0.36 x 10⁹-fold increase in diameter; if we imagine that mass increases per the increase in volume, that's 0.047 x 10²⁷-fold increase in mass multiplied by 1.66 x 10^-27 kg per amu. This gives us 78 grams per amu mass of the atom - to be corrected by dividing by the cube of the atomic radius over 100 for the element chosen. So phosphorus has 100 pm diameter, and would weigh 78g * 31 = 2.4 kg. Hydrogen has only 25 pm radius, so we take its 78 gram mass but multiply by a factor of 4³ (because we have to blow it up 4x larger in diameter) getting us 78g * 64 = 5.0 kg. Likewise radium is 78g * 216 / 2.15³ = 1.7 kg. Now what's odd about this is that they're all coming up at least 10 times heavier than an actual cricket ball, and StuRat's logic seems convincing - true, the article says cricket balls are made from cork, which is not a solid material, but even so, seems a little off... there may be some aspect of the definition that I'm not properly considering. Wnt (talk) 06:15, 2 September 2012 (UTC)
  • Here's how I tried to figure it out. Imagine if a cricket ball were made of solid lead. Using your numbers, I get the volume of a cricket ball to be about 40.6944 cubic centimeters. If it were made of lead, that would weigh about 461 grams. Lead is face centered cubic crustal structure, so that's a packing efficiency of 0.74. So, a single lead atom, scaled up to the size of a cricket ball sized chunk of lead, would weigh 461/0.74 = 662 grams. I did this because a single atom would have a packing efficiency of 1.00: there would be no gaps between atoms to consider if the cricket ball were a single atom, but the individual atoms would have the same mass and density as in the real example. If we choose a different atom, like carbon, then carbon has a hexagonal crystal system with a packing efficiency of 0.74 also. Using the density numbers for graphite, I get a ball of graphite to weigh 92 grams, which makes our mega-atom of carbon would weigh 92/0.74 = 124 grams. So it really depends on which atom you're using. This makes some sense, given that atomic weight and van der Waals radius doesn't scale, Antimony is about 10x the mass of carbon, but the van der Waals radius of Carbon atom is 170 pm , and that of an Antimony atom is 206 pm. So heavier atoms should be a LOT more dense than lighter atoms, as this calculation bears out. Someone should check my math, but this feels intuitively right. --Jayron32 06:38, 2 September 2012 (UTC)

Hmmm! the way I did it, the mass of lead is 207 amu and its radius is 180, so taking 78g*207/1.8³ gets me 2.77 kg, which is high by a factor of 4.18 (or a radius off by 1.6 fold). Trying to get at the discrepancy, I find [6] which I take to indicate that 495.08 pm * sqrt(2) / 4 is the closest distance between lead atoms in it (along a diagonal on a face of the cube). That's 175.0 pm .... which doesn't account for any difference! And as you say the fcc packing is efficient at 0.74-fold difference in mass. Dang it, I just went over the math a third time and I still haven't found the bug. Wnt (talk) 15:49, 2 September 2012 (UTC)

Balanced wheel

Could I check a couple of things?

Suppose a wheel is oriented horizontally (i.e. the rim of the wheel lies in a horizontal plane), and it balances perfectly under gravity when it is resting stationary on a point (say on the tip of a spike). Does that also mean it's perfectly balanced when it rotates on that axis at any speed in any orientation?

Is it always possible to balance any wheel by adding one point mass to the rim?

86.146.107.246 (talk) 19:27, 1 September 2012 (UTC)

A wheel that was balanced on a point when horizontal will only still be balanced when tilted if it has zero width. A real wheel with thickness will have its cg shifted slightly to the side which is tilted down causing it to tilt further and then fall. The same applies to a wheel rotating slowly. If the wheel is rotating fast enough, gyroscopic action can overcome the unbalance and hold it in place. SpinningSpark 19:49, 1 September 2012 (UTC)

The guy talked about rotation along the plane of the wheel (I think) - to be precise, along the horizontal plane that is perpendicular to down - not about tilting, which wouldn't make sense. So, he says: if you can get the wheel to balance when it's still, can you rotate it 'along the horizontal plane' and it it still balanced? For me, the answer is obviously "yeah, you can still rotate it" as this is no different from "rotating the point" that it is on, since a point can rotate and it is the same. (Alternatively, you can simply walk around the wheel, rotating your view, and the effect is the same as if you had rotated the wheel along the horizontal. So, you can always rotate it and it will remain balanced.)

As to the other question I should think the answer is yes but I'm not sure. --80.99.254.208 (talk) 20:14, 1 September 2012 (UTC)

To clarify, by "when it rotates on that axis at any speed in any orientation" I am talking about putting the wheel on an axle in the same position and orientation as the spike on which it was resting. So, it's two different setups: the first resting horizontally on a spike which is where the axle will be positioned, the second with the wheel in a normal configuration on an axle, with the axle oriented in any direction. Sorry that was not very clear. 86.146.107.246 (talk) 21:01, 1 September 2012 (UTC)

A wheel running on a perfectly rigid axle fixed in space does not need any balance weights, as any twisting torque on the axle will be absorbed, by definistion. But a real wheel of finite width on a real axle can be a very different story. To see why, visualise two coupled wheels, near each other, on a common vertical axle. Let's say the lower wheel is perfectly balanced, but the upper wheel is not. As the wheels rotate, the lower one will have no effect on the axle, but the upper wheel will obviously try to displace the axle horizonatally, twisting it round. Therefore, a real wheel of a certain width may require balancing with more than one weight. For this reason, while most car tyres are balanced by adding a single weight to the inside of the rim, you sometimes see that the tyre fitter has used a larger weight on one side of the rim, and a smaller weight on the other side. A wheel that is unbalanced on the center plane (the plane orthogonal to the rotation axis) causes a paralllel displacement of the axle. It is called static unbalance because it can be detected in situ without continuous rotation. A wheel having an unbalance mass to one side of the center plane causes a twisting force on the axle - this is called dynamic unbalance. Wickwack58.167.241.235 (talk) 02:43, 2 September 2012 (UTC)

There's also an effect that the force of gravity is different at different altitudes. So, theoretically it should be impossible to balance a vertical wheel without moving weights around as it rotates, since the top will always weigh a different amount than the bottom. However, even in the case of the largest diameter wheels (a Ferris wheel ?) this effect is so negligible as to be completely overwhelmed by the gyroscopic effect. StuRat (talk) 04:10, 2 September 2012 (UTC)

Not true. Its' mass that has to be balanced, not weight. If mass is balanced, then gravity can not upset that, as no matter what the rotatory position of teh wheel is, the pull on each part of it will always summ the same. Wickwack124.182.13.142 (talk) 04:34, 2 September 2012 (UTC)

I disagree. Imagine a wheel 10,000 miles in diameter, with a lump of extra mass causing a slight imbalance. Horizontally, at a uniform height of 5000 miles, it would work fine, with the gyroscopic effect overcoming the imbalance. But when rotated vertically, so the top is 10,000 miles into space, the gravity acting on the bottom would be far more, causing a magnification of the imbalance as the lump went around the bottom, and a reduction when it went over the top. This variable imbalance would be worse than a constant imbalance. StuRat (talk) 05:20, 2 September 2012 (UTC)

The question is whether a object, perfectly balanced when horizontally rotating about, is then perfectly balanced rotating in any plane on that point, and whether in order to balance an object, only one added weight is required. And the answer is 1) yes it is, and 2) but it may need more than one balance weight. You have it wrong, Stu. Wickwack124.182.13.142 (talk) 06:04, 2 September 2012 (UTC)

I think we're just talking about different things. You're talking about perfect balance, while I'm talking about the real world, where that is impossible. StuRat (talk) 06:07, 2 September 2012 (UTC)

No is the answer (if I have understood the question). If the stationary wheel will balance (in a uniform gravitational field) on a point at the centre of its shaft, then the wheel's centre of gravity lies on the centre-line of its shaft. This means the wheel has static balance. If it doesn't have static balance, then this can be corrected by a single weight. However this does not mean that spinning the wheel about its shaft will not result in shaking or vibration. For smooth running of the wheel you need dynamic balance, which means that (1) the centre of gravity is centred on the shaft and (2) one of the wheel's principle axes of inertia is aligned (parallel) with the axis of rotation. You can't in general correct dynamic balance with a single weight - in particular you can't balance the wheel with a single weight if it already has static balance, since any additional weight would destroy the static balance. --catslash (talk) 14:46, 2 September 2012 (UTC)

Almost Correct. You are entirely correct in the context of the OP's question. However (as a fine point), in parctice many wheels on axles have their axles pratically constrained against turning at right angles to their axis, while not so constrained against lateral dispalcement. In such cases only static balance is necessary, particularly because in practice static unbalance is larger than dynamic anyway. Wickwack58.170.164.118 (talk) 23:04, 2 September 2012 (UTC)

Could this "dynamic imbalance" versus "static imbalance" distinction arise if the wheel has a negligible thickness (i.e. a thin disk, approximately like a CD for example), or is it only a function of the non-uniformity of the wheel over the "thickness" dimension? 86.177.105.185 (talk) 23:49, 2 September 2012 (UTC)

A wheel with negligible thickness that somehow still has significant mass cannot by definition have any dynamic unbalance. The vibration consequences of dynamic unbalance of wheel mass is proportional to the offset of the mass unbalance from the plane orthogonal to the raotaion axis and passing thru its centre of gravity - yes, it is a function of the uniformity over the "thickness dimension". Wickwack121.215.159.205 (talk) 01:58, 3 September 2012 (UTC)

Thanks, that makes sense. Thanks everyone for the replies. 86.177.105.185 (talk) 02:01, 3 September 2012 (UTC)

can I see the observer effect in action?

Is there an experiment like the double-slit, where there is an interference pattern unobserved, but it disappears observed, and if so I would like to watch a Youtube video of someone (i.e. a piece of machinery) alternately either not observing the single photons and observing htem.

I would like to see the observer effect "in action" like toggling the interference pattern by observing or not. To be clear, I don't want to myself observe: I want to observe the *experiment (+ observer) OR (+ 0) = outcome* where the middle term is switched on and off and I can see, by the changing outcome, the observer effect. --80.99.254.208 (talk) 20:11, 1 September 2012 (UTC)

This is an interesting question. Note that if you have a photon-detecting device at one slit with, say, its readout hidden behind a door so you can't see it, opening the door is not going to make the interference pattern disappear. All that's necessary to make the interference pattern disappear is recording the passage of the photon somewhere, even if only as electrical signals in the bowels of the machine. That aside, the idea makes sense in theory, but there are practical problems:

• A milliwatt laser at a visible wavelength emits upwards of 10¹⁵ (1,000,000,000,000,000) photons per second. That's pretty dim light. To make this experiment work you need photons to pass through the slits at a rate slow enough that you can detect each one individually. So you're not actually going to be able to see the light on the screen at the far side of the slits.
• Practically speaking I don't see how you could detect a photon's passage through the slit without absorbing it, but that makes the experiment pretty boring since you're simply blocking one of the slits. The detector would need to be connected to another laser at the slit which would emit a photon whenever one was detected. Quantum mechanically it doesn't matter, but you might find the experiment less convincing this way.
• Practically speaking I think the only way to prevent measurement of the photon's passage would be to physically remove the detector (and re-emission laser) from the slit. Again, this might make the experiment less convincing, especially since single-photon detectors are not very accurate in the first place and the mere presence of one in the slit is going to make a big difference, measurement effect or no.

-- BenRG (talk) 22:33, 1 September 2012 (UTC)

Some of the difficulties pointed out by BenRG would be less problematic if you use electron instead of photons in the double slit experiment. Dauto (talk) 23:22, 1 September 2012 (UTC)

I may not know the actual way of things, but as far as I was taught in university level Special Relativity, they observer theory concerning quantum superposition of states, appears to be incomplete or faulty - everying in the universe is an observer, everything contantly communicates information through the fundamental forces. Observation should more likely to be statistically determined than definitively determined, but how? Plasmic Physics (talk) 02:12, 2 September 2012 (UTC)

Note that quite ordinary photomultipliers (http://en.wikipedia.org/wiki/Photomultiplier) can detect single photons. Measuring the precise time of photon arival is another matter. Wickwack58.167.241.235 (talk) 02:50, 2 September 2012 (UTC)

There's a logic error in your idea, since you are an observer, too, so can't watch the pattern change based on who observes it. Anytime you are watching, it is being observed. StuRat (talk) 04:13, 2 September 2012 (UTC)

Stu, which poster are you refering to? The OP? Plasmic? Me? Wickwack124.182.13.142 (talk) 04:33, 2 September 2012 (UTC)

This Q actually has proper indentation, so, since I indented once from the OP, it should be apparent that my response is to them. StuRat (talk) 05:11, 2 September 2012 (UTC)

This whole business of wave function collapse resulting from mere observation is simply folly. Superstition in the guise of science, what it is. 66.87.127.241 (talk) 05:07, 2 September 2012 (UTC)

I do believe that it is a result of mere observation, however, I also believe that observation is currently ill defined - the collapse of wavefunctions would continue without the presence of life in the universe. I sounds like flat earth logic, full of holes. Plasmic Physics (talk) 05:27, 2 September 2012 (UTC)

Yes, I never understood why we should suppose that Schrödinger's cat is both alive and dead, as opposed to definitely having one state, but that state being unknown to us. StuRat (talk) 05:40, 2 September 2012 (UTC)

It should be noted that Schrödinger thought the exact same thing. He devised the cat problem to show how silly quantum mechanical principles would be if applied to something like a cat. It was supposed to be reducing the problem to absurdity. --Jayron32 06:03, 2 September 2012 (UTC)

(edit conflict)That is not what I mean, I definitely believe in superposition. According to my understanding, the box would be observing the state of the cat. If the box was a closed experiment, then the superposition should collapse on its own.

My question is: what amount of exchange of information qualifies as an "observation"? Clearly, in the double-slit experiment, a superposition persists, despite the environment of the experiment exchanging information. Plasmic Physics (talk) 06:08, 2 September 2012 (UTC)

See double-slit experiment. Note that observing which hole the photon goes through is different from observing the pattern which is why the experiment is meaningful, and the effect of this is indeed really observable; it's not just a thought experiment. Wnt (talk) 05:46, 2 September 2012 (UTC)

Yes, but only under certain interpretations, such as the Relational interpretation, does this mean the observation actually changed the system. StuRat (talk) 05:59, 2 September 2012 (UTC)

OP here - stating my assumptions

The above responses are on a bit of a tangent. Let me state my underlying assumption. 1. The double slit experiment shows a point source of photon emisions passing through two slits like this: if the equals sign is a photon gun it goes pew-pew-pew = · · · · x | where the vertical bar at the end is a screen, and the x is either a | with two small slits in it parallel to each other or one.

2. If we repeat the procedure on a large scale with marbles, the marbles will hit the screen in two different locations: one where the ray passing from the emission point source through one small slit hits the screen, and the other where the ray passing from the emission source through the other small slit hits the screen. (i.e. the finer the slit's width / closer the screen to the slits the more narrow the bands are.) This makes sense from a particle interpretaiton

3. If we repeat the produce on a large scale with the slits in water and the point source bein ga source of ripples, the water goes like this: = ) ) ) ) ) x 3 3 3 3 | where instead of a point · we get a ripple, and the ripple passes through the double-slitted x and turns into two ripples which interfere with themselves. At the point of | you get an interference pattern.

4. Returning to the photon scale, I am told that if you slow the photon gun to emit one photon at a time (and not a stream) it will still produce a wave pattern (as if each individual photon were interfering with itself) like this: x 3 3 3 3 | showing an intereference pattern on the screen. But if you measure which slit of x it goes through it turns into · · · · x · · · | simply landing on one of the two bands that you get from connecting the point source with the slit and continuing to draw the ray. In other words, no ripple effect.

These are my assumptions.

5. Therefore, if one were to alternately check which slit the photon goes through and not check, you would alternate from the effect x · · · | to x 3 3 3 3 |

Is this right? If so I would like to see number 5 in action. I don't care if it's time-lapsed over hours. I just want to see the observer effect. --80.99.254.208 (talk) 12:55, 2 September 2012 (UTC)

That's almost correct—the only mistake is that the particles diffract at the slits whether or not they're measured there, and therefore you get one wide blob on the screen when there's no interference, not two narrow blobs. What appears or disappears is the modulating pattern of interference fringes within the blob. You could get the same visual effect by using a bright monochromatic light and putting linear polarizing filters at each slit. If they're oriented the same way, there will be an interference pattern. If you rotate one of them by 90°, the interference pattern will disappear. That's exactly what you would see in a real double-slit experiment if it were possible to see it directly. It's a classical effect (predicted by Maxwell's equations), but you can explain it in the language of the quantum double-slit experiment: using different polarizations at the two slits leaves a record of which slit the photon chose, and that's enough to destroy the interference pattern.

Do you just want a teaching video that could be shown in a classroom (in which case a simulation would be fine), or are you not quite convinced that the effect is real and want to see more direct proof (which as I said above is hard and will depend on what counts as "direct")? -- BenRG (talk) 18:22, 2 September 2012 (UTC)

Yeah, it's stupid and I don't believe it and want to see the effect -- JUST the observer efect - in action. I don't believe you could make an interference appear or disappear just by looking and not looking (even in not quite so literal terms) at the thing that produced it. let me ask this way. If the slits are far enough from each other, we could put a repeater that is activated by a photon and release a single photon "along the same" trajectory. (I don't personally care if the repeated trajectory is a bit off). Alternatively, the repeater could be moved out of the way, so that the photon is not caught. If I understand the claim correctly, then if you measure the repeater's signal (and this may not even be necessary as long as the signal is there somewhere) then if we do a time-lapse video where a die is thrown, even means the repeaters are in the way, odd means they're in the way, and a 3 is thrown, we see the pattern from 6 hours of time-lapse (no interference pattern, 3 means they repeaters measure and are in the way), then a 5 is thrown, again no int. pattern, a 1 thrown, again no int. pattern, a 4 is thrown, they're put in the way and the effect suddenly disappears, a 3 is thrown, they're moved again, etc. This would be quite convincing. Not QUITE as convincing as if the electron had been measured in transit without a repeater, but if that's the best science can do it will do. Basically I want to see the observer effect "in action" like a 'minimum difference' type setup. --80.99.254.208 (talk) 19:38, 2 September 2012 (UTC)

It's much easier to "passively" detect an electron than a photon because it's electrically charged, so this experiment might actually be possible without the "repeater". Unfortunately all I can find online is an experiment by Akira Tonomura and collaborators (web site, Youtube), but it makes no attempt to detect electrons at the slits. They don't seem to have published, either.

As far as the theory goes, it's a property of all waves (quantum or not) that you only get interference between paths that lead to the same final state. For light that means the same location and same polarization. In the example above with polarizing filters, when the polarization is different there's no interference (and there's never interference between light reaching one part of the screen and light reaching a different part). What's unique about quantum wave behavior is that the final state includes not just the position and polarization of the light but also everything else. If there's something in the world that ends up in one state when the light goes through one slit and another state when it goes through the other, there's no interference on the screen even when the final position and polarization are the same. On the other hand, if your repeater's final state is unaffected by whether it absorbed and re-emitted a photon/electron (which is theoretically possible), the interference pattern will remain. I know this is strange, but it's not that much stranger than classical waves... -- BenRG (talk) 00:00, 3 September 2012 (UTC)

September 2

Non-fatal viruses

Are there any viruses that are not potentially fatal to humans first-hand? One that I can think of is HIV (which can allow other diseases to cause death), but are there others? Thanks. 64.229.153.184 (talk) 00:17, 2 September 2012 (UTC)

There are millions of viruses which don't use humans as vectors at all, i.e. the Tobacco mosaic virus doesn't even infect animals. Bacteriophages are a class of viruses that infect bacteria. --Jayron32 00:28, 2 September 2012 (UTC)

In before this gets removed as a request for medical advice. What Jayron said, basically. I'm not even sure that a significant portion of the viruses in existence are potentially fatal to humans. That would seem to be rather oddly genus-specific, and I think most viruses tend to use a rather small number of animal vectors. As I understand it, the principle of natural selection would also make killing the majority of your hosts something of a genetic/evolutionary dead end, wouldn't it? Evanh2008 ^{(talk|contribs)} 00:34, 2 September 2012 (UTC)

It's not intended to be for medical advice, just out of curiosity. 64.229.153.184 (talk) 00:52, 2 September 2012 (UTC)

Yes, it is not evolutionarily stable for a virus to kill most of its hosts (or at least to kill them too quickly). Among viruses that infect humans (which I think is what the question is intended to focus on), I believe there are many that are rarely lethal (I do not have a handy reference, a little help here?). As for references, the OP may be interested in Virulence#Evolution, and optimal virulence. Both of these articles describe the factors that promote and inhibit lethality in pathogens. Lastly, this is a perfectly legitimate (and interesting) question. It is in no way a request for medical advice, see User:Kainaw/Kainaw's_criterion SemanticMantis (talk) 01:52, 2 September 2012 (UTC)

Are you asking for viruses that simply can't kill humans, viruses that infect humans but don't kill, or viruses that make humans more likely to die but can't kill on their own? Someguy1221 (talk) 01:46, 2 September 2012 (UTC)

Infect but don't kill, sorry for not being clearer. 64.229.153.184 (talk) 01:55, 2 September 2012 (UTC)

A couple of examples: GB virus C and Transfusion transmitted virus. -- Scray (talk) 02:08, 2 September 2012 (UTC)

We all get influenza from time to time. Only in rare cases do flue viruses kill. Various forms of the herpes virus either produces no symptoms, just a rash when the human is under stress, or just a mild rash. Wickwack124.182.34.199 (talk) 02:14, 2 September 2012 (UTC)

That does not answer the original question. -- Scray (talk) 03:23, 2 September 2012 (UTC)

It depends on what the OP meant by "first hand" I suppose. Then he clarified it by saying "infect but don't kill". One could ask "is there any food known that doesn't kill?" Strickly speaking, it is possible I could choke on a piece of apple, therefore apple can kill. I know a fried who is alergic to nuts. She inadvertently ate a nut in a cake at a party once, and had to have an emergency traecheotomy. So, do we say apples and nuts are lethal? That would not be of any value. My friend's root problem is not nuts, it's a genetic mishap. Similarly, flu viruses should not be considered lethal, though in rare cases they will if there is some other factor. I doubt that significant numbers of folk die from even the worst of the herpes types. This quite different to ebola virus, which will kill you regardless of other factors. So, yes, the OP's question has been answered. — Preceding unsigned comment added by 124.182.13.142 (talk) 04:30, 2 September 2012 (UTC)

Most viruses kill some, usually small portion of people who get them, such as chicken pox. Flu kills 5-50,000 a year in the US, not all that rare. http://www.cdc.gov/flu/about/disease/us_flu-related_deaths.htm Chicken pox around 100. Any infection can kill if you have a weakened immune system or a particularly virulent mutant. See Viral disease for a start at their variety. 02:37, 2 September 2012 (UTC)

No, that is incorrect. As I noted above, GB virus C and Transfusion transmitted virus are non-virulent. There are zero reports that they've killed. -- Scray (talk) 03:22, 2 September 2012 (UTC)

Are you telling me, Scray, that you are unfamiliar with the meaning of the word "most"? μηδείς (talk) 15:49, 2 September 2012 (UTC)

I should have been more specific in stating my disagreement: I was referring to your statement that "any infection can kill..." - because the viruses I cited are highly prevalent, many episodes of infection in immune-compromised hosts have been studied, and no deaths have been observed (citations in the respective articles, which I've already linked). I'll also point out that the original question was about viruses that are "not potentially fatal" - hence your answer overall was not responsive to the initial query (which implicitly acknowledged that most viruses can be lethal). -- Scray (talk) 21:51, 2 September 2012 (UTC)

Okay, that makes quite a bit more sense. It is true, however, that any infection that potentially causes an immune response can kill, although "infections" that fly under the radar, and neither evoke a response nor incapacitate their hosts obviously won't. But I am not sure that they would then be called infections. I was taking Tammy's response below as a great example. Warts don't usually kill. But papilloma virus infections of the lungs in AIDS patients do kill, and the virus does induce various cancers. (Oh, a point which I see on further reading you have explicitly mentioned, minus rectal cancer.) μηδείς (talk) 22:34, 2 September 2012 (UTC)

Warts are transmitted by a virus. --TammyMoet (talk) 08:33, 2 September 2012 (UTC)

True, but with respect to the original question, the viruses that cause warts can be lethal (e.g. PMID 10197157, PMID 21242344) to people with reduced immunity. More importantly, HPVs causes cancers of the cervix and the head and neck - also relevant: PMID 19646562. -- Scray (talk) 22:03, 2 September 2012 (UTC)

Adeno-associated virus "is not currently known to cause disease and consequently the virus causes a very mild immune response." Not sure our immune response really is a consequence of our knowledge, but that's what the article says. Consider also the article Viral vector, although this is a bit cheaty since those viruses have typically been engineered to be harmless.  Card Zero  (talk) 14:53, 2 September 2012 (UTC)

Some 95% of the adult population is infected with Epstein-Barr Virus yet it is rarely fatal unless it leads to for example Burkitt's lymphoma or HIV-associated lymphomas. Wolfgangus (talk) 22:22, 2 September 2012 (UTC)

what is a linear dielecric material

explain. --150.203.114.14 (talk) 04:23, 2 September 2012 (UTC)

Well, linear, in this sense, means an output variable varies proportionately with an input variable, therefore, if the input doubles, so does the output (although it could be slightly more complicated if the line on the graph is offset from the origin). However, I'm not sure what the input and output variables are, in the case of a dielectric. StuRat (talk) 05:32, 2 September 2012 (UTC)

A dielectric is a material that can maintain charge seperation; that is if I apply an external charge to it, the charges in the material shift around; applied negative charge will push negatives away, leaving positive in its place. What makes something a dielectric is, if the external charge is removed, the charge seperation stays for a time. Ever rub a balloon on your hair and stick it to the wall? That's the dielectric effect. I'm not exactly sure what property is linear, but my guess is that one could have a dielectric whose charge response is linear with respect to the input charge; double the input charge creates double the charge storage on the dielectric. --Jayron32 05:56, 2 September 2012 (UTC)

Jayron has given a physicist's answer. Here's a perhaps more useful and complete engineer's answer:-

What makes a dielectric a dielectric is that it displays a dielectric constant, normally termed permitivity. Permitivity ε in a linear dielectric is given in farads per meter by D/E where E is the applied electric potential across the dielectric, volts per meter, and D is the resulting electric field (flux), ie charge , Columbs per square meter. This is analogous to ferromagnetic materials, which have a permeability, in henries per meter, μ = B/H. Just as real magnetic materials are non-linear, that is permeability varies with the strength of the applied field, and also displays hysteresis, a real dielectric's permitivity varies with the applied volts per meter, and displays hysteresis - that is, you get a different permitivity at any given voltage when increasing it, to what you get when passing thru that same voltage while decreasing it.

A perfect vacuum has a permitivity of 10⁷/(4 π c²) F/m where c is the vacuum speed of light. All real substances have a permitivity above the vacuum value.

Just as a linear magnetic material allows the construction of an inductance having no electrical energy loss, a linear dielectric allows the construction of a capacitor with zero electric energy loss. It being a real (imperfect) world, you can't have either. However in practical electrical calculations, you can generally get a useful result with minimum effort by assuming they are linear, and then if necessary applying simple corrections to allow for the energy loss.

Keit60.230.231.168 (talk) 08:34, 2 September 2012 (UTC)

How to know that a drug is safe?

I know that clinical trials test new drugs in rats or other animals and then in people, who are paid for that. But, Independent of the model, human or animal, you are using, you'll be always extrapolating the results and taking risk, won't you? So, how could we know whether a drug is safe for pregnant women or children? OsmanRF34 (talk) 13:37, 2 September 2012 (UTC)

Actually, until you test it and wait for a couple of years or so, you can't know with total certainty if a new drug is safe for you. There can always be side effects that haven't been noticed during the trials. Here in Sweden, we have children suffering from narcolepsy after a vaccination against the swine flu that was supposed to be safe. Lova Falk talk 14:31, 2 September 2012 (UTC)

Furthermore, clinical trials seldom are longer than at the most half a year. Some meds are taken daily for years and years, and long-term effects are assessed while patients take the meds. For instance SSRI. They were tested for a relatively short time, put on the market, and now we can see if they are safe also for long term use. Lova Falk talk 14:35, 2 September 2012 (UTC)

• This is really just a specific case of the problem of induction, which applies to all scientific research. Looie496 (talk) 17:04, 2 September 2012 (UTC)

In many cases we don't know whether a drug is 'safe' (whatever that means) for pregnant women, many drugs ([7] has some numbers) are untested in pregnant woman and the information sheet that comes with the drug warns of that. As mentioned in the earlier source, the Food and Drug Administration Amendments Act of 2007 has a number of amendments that relate to drug testing in children contionuing and extending from earlier legislation, [8] mentions only about 20% of drugs approved by the FDA were labelled for pediatric use before they began a pediatric program (from that source I think that was in 1997). Nil Einne (talk) 17:45, 2 September 2012 (UTC)

I can add to that that a lot of research is done on men, because you know, women have their changing hormones and periods and that could confuse results, and then we just happily generalize the results to women, not bothering about possible differences. Also, a lot of the times, no trials are done with children. Yet they do get the meds. Lova Falk talk 18:31, 2 September 2012 (UTC)

As the sources I linked to and text of my reply hinted at, if no trials were done on children then usually the drugs won't be approved for use pediatric use and therefore any such usage is off label in the US and I think many other countries with decent regulatory regimes (see e.g. [9] for NZ and Australia). Nil Einne (talk) 19:31, 2 September 2012 (UTC)

Sunrise progression

I took a series of sunrise pictures from an east-facing beach in Florida last spring, including this closeup of the sun just as it rose above the horizon. My question is this: what causes the wedge-shaped base between the sun's disk and the horizon? It's definitely not a camera artifact. Juliancolton (talk) 15:02, 2 September 2012 (UTC)

It's a mirage, more specifically a mirage of an astronomical object caused by the bending of light. -- MacAddct1984 ^{(talk • contribs)} 16:55, 2 September 2012 (UTC)

Notice as well how the sun is not round. That bending of the light is caused by refraction due to the atmospheric inhomogeneity. Dauto (talk) 21:06, 2 September 2012 (UTC)

More specifically an inferior mirage, and quite pretty. μηδείς (talk) 21:07, 2 September 2012 (UTC)

"Pocket" flashlight that recharged from sun-light or room-light amout of photons?

automatically light-recharged flashlight, that lesser the natural need to replace a battery or that could even cancel it. just like solar calculators work. does anything like it or near it even exists? thanks. 109.64.151.131 (talk) 21:26, 2 September 2012 (UTC)

Sure. Google "Solar flashlight" You can buy one for under US$ 20.00 Dauto (talk) 22:36, 2 September 2012 (UTC)

Link? How long is the charge? μηδείς (talk) 23:48, 2 September 2012 (UTC)

Well, there are those night light things you put on the sides of your front walk, but they aren't very impressive. After absorbing sunlight all day on that tiny solar panel, they give a weak light for a few hours, at best. To get a bright light that lasts all night, you would need a huge solar panel. There are inefficiencies at every step that make this the case. When they are all considered together, you might be lucky to get 1% of the light that hits the solar panel back in light from the flashlight. StuRat (talk) 03:27, 3 September 2012 (UTC)

What sort of skin diseases are there that look like melanomas?

I know there are cancers and other diseases of skin vascularization that are red-brown and have irregular outlines, but haven't been able to find anything on line by looking at our article for melanoma or for diseases that look like melanoma. I am aware of the various sarcomas. Are there any other suggestions? Thanks. μηδείς (talk) 23:47, 2 September 2012 (UTC)

September 3

For how long has Schroedinger's Cat been rotting?

So, I have set up the cat experiment with enough air, food and water to keep the cat alive for 10 days. Only then do I open the box. Given that he has neither been alive nor dead for 10 days, he certainly cannot have been rotting for 10 days. Indeed, he should still be warm, whether alive or dead. If he has rotted, who observed him ten days ago to kill him? I don't want arguments in response to this question, I want previously published considerations of it. Thanks for the help. μηδείς (talk) 00:10, 3 September 2012 (UTC)

The premise of the question is wrong. It's not that he's been neither alive nor dead, he's been both at once (supposedly). 86.177.105.185 (talk) 00:53, 3 September 2012 (UTC)

Yes, so he has been rotting or not for ten days? Or just since the box was opened? Maybe just rotting for five as a compromise? A published source which discusses this dilemma would be appreciated. μηδείς (talk) 00:57, 3 September 2012 (UTC)

The physical processes of rotting (if it occurs) will introduce all kinds of interference with the supposed superposition of dead/non-dead states, vaguely akin to opening the box. I appreciate that you want published sources, so please excuse my unsupported replies. 86.177.105.185 (talk) 01:07, 3 September 2012 (UTC)

Isn't the whole point of the cat thing that at the macro scale it IS a paradox? It doesn't make logical sense. So trying to infer other logical conclusions from it is kind of pointless. Vespine (talk) 01:39, 3 September 2012 (UTC)

It makes conceptual sense for the cat to be alive and dead at the same time if you believe that interpretation; however, in reality the cat is a big and vastly complicated thing, and once you start taking into account physical process and interactions such as rotting, breathing, eating, etc., the original concept of the thought experiment becomes overwhelmed with other factors not originally relevant. 86.177.105.185 (talk) 01:54, 3 September 2012 (UTC)

Inasmuch as the scenario makes sense at all, he has been rotting (or not) for ten days. More specifically, over a ten day period, |alive> + |dead> evolves to |alive for ten days> + |dead for ten days>. A quick search of the arXiv and Google Scholar failed to turn up any discussions of this, though. -- BenRG (talk) 01:42, 3 September 2012 (UTC)

I believe you're thinking about the cat too literally. It is a metaphor, and if you were to try it with an actual cat, you're not going to observe any particularly strange quantum effects. The problem with your question comes about with the addition of an additional variable (the rotting process) that really has no analogous feature in quantum physics, but the cat isn't a great analogy to begin with. (If the cat is alive, he damn well knows it, whether anyone on the outside of the box does or not.) If you were to forget about all the various problems inherent with the paradox and the question, though, I believe BenRG's answer is correct. Evanh2008 ^{(talk|contribs)} 02:12, 3 September 2012 (UTC)

My personal interpretation is the same as BenRG's, and is expanded as follows: The problem is that the thought experiment is indeed usually portrayed as literally true, yet a cat's being dead is not just a yes-or-no thing--death as variously defined is a moment that has a temporal location. If a cat is dead, it means it has been dead for a finite time. That raises the question of how long he's been dead before the door was opened to check. If he's rotten, it implies he did die, not when the door was opened, but when the atom did indeed decay ten days ago, before anyone measured it. If he has only just died when the door opens, then not observing an atom's decay would delay the cat's death by ten days. But we all know unobserved atoms decay regardless of whether they are observed. (And we can even add that the atom has to decay within a certain time frame as a condition.) One could, as BenRG seems to be doing, posit that opening the door now reaches back into time and kills the cat ten days ago. That's what I would guess is the orthodox position. But none of these answers quite seems to fit with what is supposed to be going on. I cannot imagine that this situation was never suggested before, so I am quite interested in knowing what the literature suggests. I cannot believe that physicists don't know the difference between a freshly dead and ten-day dead cat. 02:26, 3 September 2012 (UTC)

Before you open it, the contents of the (perfectly isolated) box is a mini-many deadly cat worlds. Count Iblis (talk) 02:59, 3 September 2012 (UTC)

Unless I have missed it, that article doesn't seem to answer the "when did the cat actually die?" question. Are we supposed to assume it didn't actually die at any specific time, and hence hasn't rotted? Or that it did die, if it died, ten days before anyone opened the box? This seems like such an obvious question to a biologist that I cannot believe ther is no mention of this question in the literature. μηδείς (talk) 03:13, 3 September 2012 (UTC)

The only way the "paradox" is in any way useful or logical is if you have a perfect box- that is, there is no way to tell if the cat is "rotten" since any information cannot escape the box. Asking about rotten cats implies you have not understood what the thought experiment is supposed to demonstrate. 70.162.10.166 (talk) 03:44, 3 September 2012 (UTC)

While I have reservations about the seriousness of this question, no one else does, so I might as well through in my 2 cents. The problem with your above remark is that the cat is not a single particle, it is a whole bunch of them, this muddies up the whole affair, so let's try to simplify. Replace the cat with what we will call chemical X and instead of poison we have chemical Y. X and Y combine molecule for molecule to make a molecule of Z, for all X and Y to disappear and us to be left with just Z, it will take 10 days. Then, we can measure the amount of Z in the box to determine the length of time since the radioactive substance went off. So, what is the state inside the box before we look? In this case, the state is a super position of each possible count of Z molecules, the coefficients of which evolve in time so that higher numbers of Z are more likely at later times. The reason this question seems difficult is because "how long was the cat dead" is an entirely different question than "did the cat die", it's an entirely different Hilbert Space, looking at one in terms of the other confuses the issue and stretches the analogy past what it is meant to do. Phoenixia1177 (talk) 04:06, 3 September 2012 (UTC)

It died, if it did die, ten days before you opened the box. The whole situation is exactly the same as if you had a classical probabilistic distribution. If there's a 50% chance that the cat just died, ten days from now there will be a 50% chance that it died ten days ago. There is no interaction between the possible living cat and the possible dead cat. They are distinct possibilities that evolve independently. That's why in quantum mechanics if you start with |X> + |Y> and evolve it forward ten days, you always get |X after ten days> + |Y after ten days>. It's a manifestation of the probabilistic nature of the theory. Quantum mechanics is only different from classical probability if you later observe interference effects by doing something like a double-slit experiment. But that's not something you can do with cats, or any macroscopic system at room temperature. It's strictly more difficult than bringing the cat back to life.

You probably won't find a discussion of this in the literature because the part about the time evolution of |alive> + |dead> is too obvious to bear mentioning, and the rest is just cocktail-party philosophy. -- BenRG (talk) 04:12, 3 September 2012 (UTC)

Is D necessarily zero?

Suppose you have a sphere. Suppose $\int \int {\vec {D}}\cdot d{\vec {A}}=0$. Is D necessarily zero? Or, is there some symmetry condition that forces D=0? --130.56.91.41 (talk) 02:47, 3 September 2012 (UTC)

D is a vector field and dA is the area element, right? Then if the vector field points along the surface of the sphere at every point, the integral will be zero even though D is not zero anywhere. In fact, Maxwell's equations of electromagnetism say that the surface integral of the magnetic field is ALWAYS zero, across any surface, even though magnetic fields themselves are clearly not zero everywhere. --128.112.70.89 (talk) 03:51, 3 September 2012 (UTC)

D is Electric displacement field--130.56.84.156 (talk) 04:39, 3 September 2012 (UTC)

Minimal food requirements

Suppose we want to send a small human, perhaps 10cm shorter and quite a bit lighter than the average woman, to Mars. It'll be a short 545-day trip, and the only requirement is that the astronaut is alive at the end. Being emaciated, insane, and suffering from osteoporosis are all OK, because no matter how cruel the conditions, there won't be a shortage of people (even small women) willing to go. What's the absolute minimum amount of food, in weight and volume, the astronaut would require? Does being in space change the food requirements from what they would be on Earth? --128.112.70.89 (talk) 03:48, 3 September 2012 (UTC)

Not sure about earth, but the minimum average calories a person needs to maintain their current health is about 1,800, per Food energy, but that is quite activity dependent. --Jayron32 03:55, 3 September 2012 (UTC)

But that's for an average person, not for a small woman or really small man. Also, an Earth-dweller needs to walk around, think, pump blood, drive, and eat, all under Earth gravity. I'm thinking that the astronaut could be confined to a small space or chained to prevent any physical activity. That would be a serious animal cruelty if done to any animal on Earth, but as I said, there's be no shortage of people willing to endure that to go to Mars. Finally, how much mass and volume does 1 calorie translate to, given the most compact possible storage? --128.112.70.89 (talk) 04:04, 3 September 2012 (UTC)

If you don't care about the health of the individual and just want to send them so you could say you did it, why not send an animal instead ? StuRat (talk) 04:26, 3 September 2012 (UTC)

I think the story you've made up as the premise is a bit too silly. Of course no sane person would want to endure 545 days of near starvation to get to Mars. What happens AFTER the 545 days? They stop eating all together? Obviously one of the 1st requirements for any planetary mission would be to provide enough food to keep a person healthy. I think saying it'll be someone who is "small" or "woman" won't play a significant enough part, I think far more important qualifications would be "smart" and "sane". Look at sailors in the days of discovery, they surely endured harsh conditions, probably surviving on minimal rations, but even wealth and fame weren't enough incentive to willingly starve yourself as part of the plan. CErtain people in the army also probably know quite a lot about "minimums" required to keep people alive and healthy. Besides, if you're taking thousands of kilos of fuel and cargo and spending billions on R+D, it's hard to not justufy a few extra kilos of food, especially if you dehydrate it and use recycled water, then maybe you could recycle food too by growing some of your own? It seems absurd that anyone going on that sort of mission would be expected to starve as part of the "price". As for calorie density, lard, butter and nuts are about as high density as you get. Not sure how long you could eat just those three before the mere sight or smell of them will make you vomit. Vespine (talk) 04:37, 3 September 2012 (UTC)

Bauxite

What is "soundiferancic" in http://en.wikipedia.org/wiki/Bauxite?

It's under "Processing"

"Bauxite rocks are typically classified according to their intended commercial application: metallurgical, abrasive, cement, chemical, and refractory, cosmetic, electric, soundiferancic."

Thanks — Preceding unsigned comment added by 66.177.68.235 (talk) 04:26, 3 September 2012 (UTC)

I think it is a longstanding typographical error or something like that. A quick google search: [10] shows that the word exists only in the Wikipedia article on Bauxite and websites that directly copied the Wikipedia article. I'm not sure what it is trying to say, but I think you'd be safe removing the word entirely. It looks like gibberish. --Jayron32 04:36, 3 September 2012 (UTC)

It was part of a particularly nasty little vandalism edit by an anonymous editor in June [11] that apparently nobody ever caught. It's a good thing that editor added the clearly nonsense word "soundiferancic," because the other things changed in that edit were not so obvious. I've reverted those changes. —Bkell (talk) 04:51, 3 September 2012 (UTC)