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November 13

my camera for my experiment recorded video with a 0.80x resampling factor. When calculating centripetal acceleration and curvature, how do I scale?

I run the video through a program and it plots the trajectories of tracked objects over time (based on timestamps). The only issue, my video is 25% faster than reality. Video is taken for 5 min and 35 seconds, but only 4 min 30 seconds of video is recorded. The video is being time-scaled to match, i.e. it is not truncation. I have since corrected this problem, but I have two months of data that were taken in this "time-compressed" fashion.

Am I right to assume that curvature will remain unchanged, that is, the scale corrections applied to velocity and x' * y - y' * x will cancel out?

However for centripetal acceleration, since it is v^2 / r or v^2 * curvature, and curvature will not change with time scale, my values that I am calculating for centripetal acceleration will be (1.25)^2 faster, that is I have to apply a correction factor of 1/1.56 to my values. 137.54.63.225 (talk) 00:12, 13 November 2012 (UTC)[reply]

(EC) Why would you assume that that the time scaling would cancel out? Have you tried the calculus yourself? I think to say anything definitively, you'd have to know exactly how the camera samples the frames in time. Can you document that it's uniform? If not, then we cannot say for sure without knowing how frames are sampled over time. Also, is this the fruit fly project that IP questioners have discussed here before? SemanticMantis (talk) 02:13, 13 November 2012 (UTC)[reply]

I know it's uniform because VirtualDub explicitly says the resampling rate is 0.80x. When I disabled a time compensation feature (it was because the camera takes 15 FPS in the dark, rather than 30 FPS), it goes back to 1x. Yeah it's the project, but I just want to confirm my intuition. Using dimensional analysis, s^-1 implies one-fold framerate dependence, and s^-2 implies two. Curvature is specified as a inverse length scale, and has no time component, so it won't be affected by a sped-up framerate. However centripetal acceleration has a two-fold framerate dependence, as seen by c = v^2 * k and also looking at the dimensions of centripetal acceleration. Can somebody confirm my intuitions? 71.207.151.227 (talk) 08:35, 13 November 2012 (UTC)[reply]

Chemistry problem

I was wondering how to do part B of the following problem.

1) A saturated solution of sulfur dioxide was prepared at 27 °C and 760 torr (millimeters of mercury). A 5.00-milliliter sample of this solution was diluted and an excess of potassium iodate solution was then added. The iodine liberated was equivalent to 32.80 milliliters of 0.100-molar sodium thiosulfate solution. The reactions that occur are given in the equations below:
5 SO2 + 2IO3¯ + 4 H2O --> 5 HSO4¯ + 3H+ + I2
2S2O32¯ + I2 --> 2I¯ + S4O62¯
(a) Calculate the moles of I2 liberated.
(b) Calculate the volume of sulfur dioxide needed to prepare 1.00 liter of the saturated sulfur dioxide solution.

I got .00164 moles of I2 for part a (I am almost positive this is correct, if someone would like to double check feel free to do so and let me know).

NOTE:

The answer for part B is 40.3 L, but I am not sure how to do this.
Part B is not a limiting reaction, it is a gas law problem, and may or may not involve stoichiometry.
You don't need to write a detailed explanation for your steps, if you just write the equations as you progressed and the answers for them, I should be able to figure it out on my own. If I need more help I'll contact you. Or you can just write the steps you'd take.

Alternatively, if someone can't solve it but can just explain what Part B is asking, that'd be appreciated. SO2 and sulfur dioxide are the same, so isn't 1.00 Liter of satured sulfur dioxide the same as the volueme of SO2? I don't understand what they're looking for.

Please, if anyone can solve Part B and explain their steps here, it would be much appreciated! Or offer even some steps or help. I've tried this multiple times, but have never been able to get 40.3 (and I know this is the answer). --Jethro B 02:45, 13 November 2012 (UTC)[reply]

Part b is ambiguous. Does it assume that all the gas you have is dissolved?--Jasper Deng (talk) 03:46, 13 November 2012 (UTC)[reply]

Part b isn't ambiguous. I'm a little sleepy right now, and just got done a bunch of chemistry tutoring, so I'm a little fatigued, but you can calculate the moles of SO2 in solution using simple stoichiometry, and you can use Henry's law to calculate the partial pressure of SO2 needed to form a solution of SO2 at that concentration. I don't feel much like doing the math, but Henry's law is the key to solving this. You would need to know the Henry's law constant for SO2 dissolving in water, but there are tables of values for that, availible on the internet via google; type "Henry's law constant" into google to find a bunch. --Jayron32 03:58, 13 November 2012 (UTC)[reply]

Unfortunately, I want to do this without Henry's Law, as it was done in 1969 (when the problem was published) without Henry's Law. Henry's Law could work, but it's possible without. Some equations that may be of use here that I can think of are PV=nRT (or v=nRT/PV), and MM=gRT/PV where MM=Molar Mass. --Jethro B 04:22, 13 November 2012 (UTC)[reply]

I'm really not sure how to do it without Henry's law, it is the only way I know to calculate the relationship between the concentration of a gas in solution based on the pressure of the gas over the solution. Also, I'm not sure what 1969 has to do with it, Henry's law had been established for 166 years at that point. It was hardly unknown. Most general first-year chemistry classes teach it today, and though I was not yet alive in 1969, I can't imagine it wouldn't have been covered if the above problem had been given in class. --Jayron32 04:29, 13 November 2012 (UTC)[reply]

Addendum: I thought perhaps that Raoult's law would be applicable here, but that only works at ranges where the solute is below its boiling point. In this case, it isn't, as SO2 is a gas at room temperature. --Jayron32 04:34, 13 November 2012 (UTC)[reply]

Addendum #2: The key here is saturated is the word in the problem that leads me to Henry's law. But OK. Let me play around with the numbers to see what I can get. --Jayron32 04:37, 13 November 2012 (UTC)[reply]

.00164 x 5 = 0.0082 moles SO2. At the above conditions, PV=nRT gives (1 atm)(V) = (0.0082)(0.08205)(300) so V = 0.201843 liters for a 5 ml sample. To prepare a 1-liter sample of the same concentration, you'd need .201843 * 1/.005 = .201843 * 200 = 40.36 L. QED. So you were correct. Henry's law wasn't needed. Does this all make sense? --Jayron32 04:55, 13 November 2012 (UTC)[reply]
- Thank you, it looks good. I'm just going to write out what you wrote here on paper and make sure I see each step, and then I'll let you know. But thank you so much for the help. --Jethro B 04:59, 13 November 2012 (UTC)[reply]
- Yes, I got it. Thank you so much for your help! --Jethro B 05:02, 13 November 2012 (UTC)[reply]

Momentum

Why is p the symbol for momentum? --168.7.228.22 (talk) 06:18, 13 November 2012 (UTC)[reply]

Nobody knows. Seriously. —Tamfang (talk) 06:29, 13 November 2012 (UTC)[reply]

What he said. The other letters were already taken up, and p was still left over. It was basically like being the last guy picked for the kickball team. If it were "m" or "M" it would have been confused with mass, which would have been bad in equations. Since we couldn't have 2 Ms, it got hung on p. Otherwise, there is no other reason. --Jayron32 06:48, 13 November 2012 (UTC)[reply]

If you have as much trouble remembering silly things like that as me, think of it as "pomentum". StuRat (talk) 07:14, 13 November 2012 (UTC)[reply]

If you have less time, think at it as p for 'pull'. OsmanRF34 (talk) 17:38, 13 November 2012 (UTC)[reply]

We are a reference desk, not a plead our own ignorance desk. So here are several references [1][2][3] each claiming that "p" was inspired by the Latin "petere", which is the root of the English word "impulse" (Latin: "impetere"). It is claimed that the concept we refer to as "momentum" was called "impulse" at the time of Newton, and that this led to the choice of "p" to represent this concept (noting that "i" would have been a poor choice since it was already in use for other purposes). Now, I don't know for sure that the "petere" story is accurate, but it seems plausible enough and clearly has some support on the interwebs. Dragons flight (talk) 12:01, 13 November 2012 (UTC)[reply]

"Do You Feel Like An Insider Or Outsider?"

When asked the above question, is this a psychologically sound question? Is it a diagnostic question, or criterion for a DSM illness?Curb Chain (talk) 06:27, 13 November 2012 (UTC)[reply]

Can you define "Insider" and "Outsider"? You may be interested in our articles on solitude and emotional isolation. Someguy1221 (talk) 07:24, 13 November 2012 (UTC)[reply]

Innie and outie? 124.150.45.187 (talk) 09:00, 13 November 2012 (UTC)[reply]

The Outsider (Colin Wilson). μηδείς (talk) 17:37, 13 November 2012 (UTC)[reply]

electrons and time-varying electric fields

There is a free electron placed in an electric field that varies as E = r-> * cos(w*t) where r-> is some constant vector. How do I come up with a complex function for its velocity? If the time-varying electric field is external, won't that set up a time-varying magnetic field that will radiate EM waves away from the source? But now the electron will be moving too, and will be radiating away its own energy. 71.207.151.227 (talk) 09:40, 13 November 2012 (UTC)[reply]

You seem to be overthinking this. Work out the force f=Ee (vectors here). Then work out acceleration of the electron based on force and its mass. Velocity is time integral of acceleration. Assuming this is homework, you can calculate the resultant formula. You may have to consider initial velocity as well. Graeme Bartlett (talk) 11:31, 13 November 2012 (UTC)[reply]

But if the electron accelerates, won't it produce its own changing electric field, that it will respond to? 71.207.151.227 (talk) 12:07, 13 November 2012 (UTC)[reply]

In an introductory physics class, which I assume this is, second-order concerns such as how the electron's field will influence the experiment are almost always assumed to be negligible. Dragons flight (talk) 12:42, 13 November 2012 (UTC)[reply]

I'm studying graduate-level optics, where EM waves are very important. I last took E&M two years ago. I need to know these secondary effects. Eventually, I have to estimate very small effects like light pressure. However, I missed two lectures and now the homework is ver confusing for me. 137.54.43.148 (talk) 19:27, 13 November 2012 (UTC)[reply]

For a lot of electrons as in a plasma you can treat it as a differing dielectric constant k=1-(Ne²)/(ω²mε₀) m-mass of electron ε₀ is permittivity of free space, ω angular frequency N number of electrons per unit volume, e is electron charge. The dielectric constant is real unless you start to worry about collisions between the particles. This is from my textbook on radiowave propagation, where electromagnetic waves meet ionised gas. It says the positive ions can be ignored because they are so heavy. The book does not mention magnetic effects. These will become comparable if the electron is moving at close to the speed of light, which it might do if the wave has extreme intensity. But will not be big normally. To workout magnetic effect you can consider the magnetic field from that electromagnetic field and velocity of electron. However since the electric field is location independent, it is not travelling anywhere, so is not actually an electromagnetic wave. Graeme Bartlett (talk) 11:23, 14 November 2012 (UTC)[reply]

My professor asked me to "employ complex form" to compute the velocity of the electron. But the velocity of the electron doesn't have two components. How is this possible? 76.123.35.31 (talk) 23:04, 14 November 2012 (UTC)[reply]

You can do a Fourier transform of the electric field and the electron location/velocity. The complex result will reflect the amplitude and phase of the electron response. And you can express the electric field with e^jωt and if you want the real part is the real part of that magnitude or cos (ωt). The Fourier transform of this would be δ(ω). And one more clue I will give you is that the velocity of electron is 90° or π/2 out of phase with the field. So that would give you a factor of i, making the velocity purely imaginary. This situation happens in the night time ionosphere with radiowaves being reflected back to earth. When you transform the derivatives or integrals with time factors of ω will pop out, but since I can't think of what they are without looking them up, I won't write them here. Graeme Bartlett (talk) 11:06, 15 November 2012 (UTC)[reply]

0.45 vs 0.2 micron filters (for life sciences)

I'm looking for information regarding the use of 0.45 vs 0.2 micron filters? I guess the 0.45s are cheaper? Can bacteriological sterility be assured with a 0.45 micron filter? — Preceding unsigned comment added by 129.215.47.59 (talk) 10:04, 13 November 2012 (UTC)[reply]

0.2 micron filters are usually considered adequate to ensure bacterial sterility, while 0.45 micron filters would generally not be adequate. There are occasional reports of ultramicrobacteria that may have slipped through 0.2 micron filters as well, but this would be pretty unusual. A 0.45 micron filter will still catch many bacteria, and are sometimes used for that purpose, but they are unlikely to stop all of the bacteria present under typical conditions, and hence the filtered fluid is unlikely to be sterile. Dragons flight (talk) 11:01, 13 November 2012 (UTC)[reply]

Thanks. Why then do we have 0.45 um filters? 129.215.47.59 (talk) 12:49, 13 November 2012 (UTC)[reply]

Bigger holes generally filter faster (everything else being equal), which is an advantage. For some experiments the filter residue (i.e. the material remaining on the paper) is of greater interest than the fluid that passes through, and in those cases the difference between catching all cells and catching most cells may not matter much. For example, if you care about larger cells, such as eukaryotes, then you wouldn't necessarily care what happens to the bacteria. Dragons flight (talk) 13:02, 13 November 2012 (UTC)[reply]

It's also worth pointing out that light scattering samples need to be filtered to remove dust (> 1um) and other static scatterers (really big things) from solution. If what you're looking at is bigger than 200nm than you wouldn't really want to use a 0.2 um filter because you'll be filtering out the thing you're trying to analyze. I guess the take-home message is that such filters aren't used solely for sterilization purposes. (+)H₃N-Protein\Chemist-CO₂(-) 15:06, 13 November 2012 (UTC)[reply]

One manufacturer says that their 0.45 um filter reduces Serratia marcescens titer by a million-fold. [4] One good reason to use them is for "pre-filtration applications", to reduce "bioburden" as they say (or, I suppose, to get rid of precipitates/insoluble matter etc.) In other words, as a first rough filter to keep all these unwanted goodies from clogging up the one you use for your actual sterilization. Wnt (talk) 05:05, 14 November 2012 (UTC)[reply]

Dyson sphere - "Gravity"

Hi,

I have a question regarding the "gravity" caused by a dyson sphere (or ring if you prefer) at the equator due to the centrpetal force of the spinning sphere/ring. I realise that such a structure in unfeasable, but view this just as a thought experiment. Lets assume that the spere is 1 AU in diameter and is spinning at such a rate that the centripetal force is equal to that of gravity, and there are Earth like features (cliffs/mountains etc) on the inside of the sphere. My question is such: What happens if you were to run off the edge of one of these cliffs, would you "fall"? Is this indestinguisable (physically) from the notion of "falling" on Earth (assuming effects of curvature inward/outward are negilgable on both at such a small scale)? Eg. distance travelled, speed attained etc. Also, what happens if you adjust the spin speed of the sphere/ the radius of the sphere with this problem (aside from the obvious heat/cold problems). I am just having trouble imagining this, thanks for any help! Matt 80.254.147.164 (talk) 13:28, 13 November 2012 (UTC)[reply]

Gravitational attraction of any object appears to act from a point source located at the centre of gravity. In the case of a thin narrow dyson ring this means that the centre of gravity is at the centre of the sphere. So if it is rotating just fast enough for centrepetal force (acting outwards) to be equal to the gravitational attraction (acting inwards) at the ring, folk located at the ring will appear to be weightless. If they climb an inward facing mountain, the gravitation distance is less but the rotational speed is also less. So gravity is stronger so they will fall towards the centre. Clearly, if the mountains face outwards, centrepetal force will be the stronger of the two. Gravity is weakening in proportion to the inverse square of the distance to centre and centrepetal force is increasing in proportion to distance to centre, that is, faster than gravitational attraction. So folk on the outside of the ring fall away.

For a dyson sphere, the rotation velocity falls off toward zero asr the rotational poles are approached. So folk on the sphere at the poles can only fall inwards, regardles. Everywhere else, they must be between being balanced as on the equator, or at the poles where they always fall inwards. So, everwhere else, they must fall inwards readless of whether the moutains poke inwards or outwards, unless the mountains approach a height just touching an imaginary cylinder tangent to the equator and of radius equal to the sphere radius. On any moutain exceeding that cylinder, they must fall outwards. Floda 120.145.145.21 (talk) 14:45, 13 November 2012 (UTC)[reply]

For the full sphere, the shell theorem applies, so the gravity of the sphere cancels out. Only the gravity of the sun (and potential planets) is relevant. That does not change the core statement - rotating a Dyson sphere would be a bad way of generating (pseudo-)gravity, since would change from the equator to the axis of rotation. On a Ringworld, the effect of the centrifugal force is very nearly the same as that of gravity on Earth. The major difference is that the Coriolis force affects things slightly differently. But this effect is rarely noticed directly in daily life. It might cause weather to do interesting things in a ringworld. --Stephan Schulz (talk) 15:33, 13 November 2012 (UTC)[reply]

OK, that makes sense, but what if we assume that the mass of the ring is zero and the ring is spinning fast enough so standing on the inside of the ring the centripetal force is "pulling" you towards the ring (ie the star would appear to be directly above you as if you were on earth) with the same force as gravity on earth. Also the height of the "mountains"/"valleys" is negligable compared to the distance to the central star. I am mostly interested with what happens when you jump off a "cliff" on the ring, really. I would think that since you are on in contact with the ring, the only forces acting is the force of the star, so you would orbit it or fall towards it. 80.254.147.164 (talk) 15:27, 13 November 2012 (UTC)[reply]

No, you would still "fall" towards the ring, because even though you're no longer touching it, you are still moving at the same speed as it, due to inertia, so centrifugal force will still act on you. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:29, 13 November 2012 (UTC)[reply]

Okay, call me Mr Pedantic, but someone has to say it. There is no centrifugal force. When you jump off the cliff you move in a straight line at constant velocity (ignoring air resistance and the effect of the star's gravity, which will be much smaller than 1g) until that straight line intersects the ring. You appear to be falling towards the ring but in reality the ring is rising to meet you. Gandalf61 (talk) 17:26, 13 November 2012 (UTC)[reply]

No, Mister Bond, I expect you to die! (as linked above ;-) --Stephan Schulz (talk) 17:42, 13 November 2012 (UTC)[reply]

Given the equivalence principle, the centrifugal acceleration is just as real as the gravitational acceleration that it's intended to replace. -- BenRG (talk) 20:42, 13 November 2012 (UTC)[reply]

That makes much more sense, the idea of the ring exerting force on a body it wasn't in contact with was confusing me. If someone wants to be extra nice , they could calculate the RPM/ velocity of the disc to require an earth like "gravity" - although when my brain is working i'll get round to it 46.64.43.100 (talk) 19:29, 13 November 2012 (UTC)[reply]

Or you could just read Ringworld, which is worth doing anyway. IIRC Niven got 770 miles/second. Of course his world wasn't exactly 1 AU in radius and the surface gravity wasn't exactly 1 g, but there's an order of magnitude for you. --Trovatore (talk) 19:57, 13 November 2012 (UTC)[reply]

To specifically answer what would happen when you fall off a cliff, you would fall as normal. Everything would seem equivalent. Until you fell far enough to notice the curvature of the ring, nothing would seem strange. Someguy1221 (talk) 22:59, 13 November 2012 (UTC)[reply]

This question is suspiciously coincidental with the release of Halo 4, which features both a set of concentric dyson spheres, and the usual ringworld the series is known for. That game has some crazy stuff in it, like solids that aren't; fake stars, objects that float in place. Apparently, it can all be explained by ideas like spacetime manipulation, quantum phenomena, exotic particles, etc. Plasmic Physics (talk) 09:08, 14 November 2012 (UTC)[reply]

No doubt, a lot of it is pseudoscience and quantum fruitloopery, but it is interesting to think about. Plasmic Physics (talk) 09:10, 14 November 2012 (UTC)[reply]

Certain electric weapons, use micro wormholes to teleport electrical current instead of using wiring, resulting in strange concepts such as the weapon fireing a few atoseconds before you actually pull what counts as a trigger. Plasmic Physics (talk) 09:53, 14 November 2012 (UTC)[reply]

Not sure what the grandfather scenario says about that one. If I pull the trigger the same instant as he does, will we both get shot before we pull the trigger? And more importantly, does that count towards a kill streak??? :D - ¡Ouch! (^{hurt me} / _{more pain}) 07:49, 16 November 2012 (UTC)[reply]

Atmosphere and oceans on Dyson ring

In the scenario above, with the ring rotating quickly enough to produce 1 g for anyone standing on the interior edge, I wonder about oceans and atmosphere on such a ring. Clearly, you would need edges on the ring to prevent them from flowing over the edge. But, would you need an inner ring or would centrifugal/centripetal force be sufficient ? StuRat (talk) 18:31, 13 November 2012 (UTC)[reply]

It depends on the time scale. The classical Ringworld had no cover, being open to the sun (with a secondary ring of shades providing day and night). If you have not read the book, do it now! Niven's ring had rims 1000 miles high - more than two times higher than the limits of the exosphere, and about four times the altitude of the ISS. Of course some atmosphere spills over, but little enough to be only noticeable over geologic time periods. And then you probably have atmosphere plants to fix it... --Stephan Schulz (talk) 18:58, 13 November 2012 (UTC)[reply]

Sufficiently high walls will do the trick (I think Niven's Ringworld called for 1000 km walls). You'll still get some atmospheric loss over the edges (disproportionately weighted towards non-critical stuff like hydrogen and helium), but if you can postulate "create a Dyson ring", "replace lost air" is comparatively trivial. — Lomn 19:00, 13 November 2012 (UTC)[reply]

Earth's atmosphere is gradually losing air too - as long as your ringworld doesn't lose air faster than Earth, I don't think anyone can really complain. --Tango (talk) 23:14, 13 November 2012 (UTC)[reply]

Well, at least Niven's Ringworld does not have tectonic processes that replenish the atmosphere with new volatiles. But on the other hand, if it's only good for 10 million years or so, that should be good for most civilisations ;-). --Stephan Schulz (talk) 23:21, 13 November 2012 (UTC)[reply]

You could catch comets ... —Tamfang (talk) 08:15, 17 November 2012 (UTC)[reply]

Do they even let people graduate high school without having read Ringworld? μηδείς (talk) 03:12, 14 November 2012 (UTC)[reply]

Yes. Smaller rings are all the rage nowadays. Someguy1221 (talk) 03:24, 14 November 2012 (UTC)[reply]

OP here, I ironically graduated with a maths degree as Niven did, and haven't read ringworld Oops! 80.254.147.164 (talk) 09:27, 14 November 2012 (UTC)[reply]

Fatigue=drunkenness?

Is it true that fatigue produces most of the same effects as drunkenness? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:22, 13 November 2012 (UTC)[reply]

No. OsmanRF34 (talk) 16:15, 13 November 2012 (UTC)[reply]

Fatigue, however, can be one of the sequelae of drunkenness. μηδείς (talk) 17:35, 13 November 2012 (UTC)[reply]

... and some of the symptoms of ordinary fatigue could be mistaken for those of drunkenness (e.g. impaired driving ability). Dbfirs 18:00, 13 November 2012 (UTC)[reply]

No, not fatigue. But Oscar Wilde made an important discovery: that alcohol taken in sufficient quantities produces all the effects of intoxication. :) -- Jack of Oz ^[Talk] 21:51, 13 November 2012 (UTC)[reply]

Also, lack of sleep due to to staying awake for a long time (say 40 hours). Count Iblis (talk) 03:39, 14 November 2012 (UTC)[reply]

Ah, yes, that reminds me of summer camp when I was twelve. Several of us stayed up all night twice in a row playing cards and were giddy on the third day so that it would indeed have looked like drunkenness. μηδείς (talk) 17:10, 14 November 2012 (UTC)[reply]

Two of the assistant MythBusters (Kari and Tory) drank almost enough beer to get busted for DUI, tested their driving skilz, then stayed awake for thirty hours and tried again. Sleepy was much worse than buzzed. (MythBusters (2010_season)#Episode 152 - Arrow Machine Gun) —Tamfang (talk) 08:14, 17 November 2012 (UTC)[reply]

Why are flat screens not flexible?

Are they too thick? Conductive materials can be made flexible, so why aren't they flexible? OsmanRF34 (talk) 18:49, 13 November 2012 (UTC)[reply]

"Can be" and "must be" are not synonyms. Most flat screens are constructed with one or more layers of glass which needs to be thick enough to maintain structural integrity and thus inflexible. Roger (talk) 18:55, 13 November 2012 (UTC)[reply]

Well, I see a couple of advantages for a product with a flexible screen, or a foldable screen. If some screens are an array of LEDs, why didn't we get yet the iPhone with a huge foldable screen, but in pocket size? OsmanRF34 (talk) 19:03, 13 November 2012 (UTC)[reply]

LCD screens are made on a silicon substrate, which is not flexible. In the not-too-distant future there will probably be OLED displays, which can be built on a flexible substrate -- see also our rollable display article. Looie496 (talk) 18:57, 13 November 2012 (UTC)[reply]

Nice link. OsmanRF34 (talk) 19:03, 13 November 2012 (UTC)[reply]

The problem with such displays will be durability. After you fold a map a few times, it tends to want to tear at the folds. The same would be true of a screen. So, folding is out of the question. On the other hand, rolling up a screen inside a tube could protect it, much as it does with engineering drawings. However, if you want a screen 2 meters by 3 meters in size, you still end up with a roll over 2 meters long, so it's not all that portable. Another problem is that after having been rolled up for a while, it won't want to lie flat, but a fold-out frame, possible stored inside the tube, could hold it flat. Another option is sufficient weight so that gravity will hold it flat, but, in the above example, that would mean a 3 meter long roll, if you want a "landscape" display. Also, lugging around that weight might be unpleasant. StuRat (talk) 21:06, 13 November 2012 (UTC)[reply]

The United States Army funds the Flexible Display Center, where the goal is to advance the technology for reliable, full-color video displays. Like most Army research, this technology would be incredibly useful if it worked. There are plenty of neat videos on their website, and you can even visit in person. If you're flush with cash, you can even buy in to the technology. Nimur (talk) 22:11, 13 November 2012 (UTC)[reply]

And note that the calculus is a bit different for the military, where they are willing to pay quite a bit more for portability than the average consumer. For example, a front-line infantry unit could use one to view the video feed from a surveillance drone. StuRat (talk) 22:21, 13 November 2012 (UTC)[reply]

November 14

Observation in quantum mechanics

Since wave function collapse requires someone to observe the system, does this mean that before humans (or other life), the Universe didn't exist? --168.7.231.3 (talk) 02:05, 14 November 2012 (UTC)[reply]

"Observation" is just the word we use for the event that occurs when a wavefunction collapses. It simply means that a certain kind of physical interaction has occurred. It does not require that any living thing be present. The confusion is understandable, but wavefunction collapse can happen whether or not humans are looking at the results. In fact, even physicists have a hard time defining what interactions will or will not constitute measurement. See measurement in quantum mechanics, wavefunction collapse and quantum decoherence. Dragons flight (talk) 03:26, 14 November 2012 (UTC)[reply]

Right. Just to be redundant, tyhis relies on the principle that there is no action at a distance, and that observation means somebody's bounced a photon off it or something similar. Which naturally means that the wavefunction is no longer what it was before you observed it. Gzuckier (talk) 01:44, 16 November 2012 (UTC)[reply]

Perfect number

I think I've found an error that make the information in the article contradicting. In the "even perfect numbers" section, it says: "As of June 2010, 47 Mersenne primes and therefore 47 even perfect numbers are known." Then later in the section it says: "It has not yet been proved that there are (or are not) others after the 41st." So I think it should have said "...other after the 47th."?174.20.101.190 (talk) 06:17, 14 November 2012 (UTC)[reply]

This is science reference desk, you should have asked the same on mathematics reference desk. Sunny Singh (DAV) (talk) 07:48, 14 November 2012 (UTC)[reply]

The explanation is quite simple. There are 47 known Mersenne primes, but only the first 41 are known to be consecutive - not all possibilities after the 41st Mersenne prime have been checked, so there may be as yet undiscovered Mersenne primes in the gaps between the 41st and the 47th on the known list. There is a 1-1 correspondence between Mersenne primes and even perfect numbers, so there are 47 known even perfect numbers but only the first 41 are known to be consecutive. This is what the perfect number article is trying to say - but I agree it is not very clear, and the wording could be improved. Gandalf61 (talk) 10:41, 14 November 2012 (UTC)[reply]

Thanks, Gandalf61. I've clarified it there. Duoduoduo (talk) 14:53, 14 November 2012 (UTC)[reply]

Heat energy vs thermal energy

Is correct to say that heat energy and thermal energy are different ? Sunny Singh (DAV) (talk) 07:36, 14 November 2012 (UTC)[reply]

There is a difference in a subtle way. Note that the term heat energy is bad english - heat is energy, so the term is like saying energy energy. The terms heat and thermal energy are often used interchangeably. However, more correctly in a scientific or engineering paper, heat is energy being interchanged from one media to another (e.g, the heat rejected by a gas to its surroundings when it is compressed), whereas thermal energy is the thermodynamic energy in a system. You may like to carefully read the WP articles Heat and Thermal Energy. Wickwack 58.169.249.183 (talk) 09:00, 14 November 2012 (UTC)[reply]

I was taught that heat was another term for waste energy. Plasmic Physics (talk) 11:46, 14 November 2012 (UTC)[reply]

In most engineering applications, heat is waste energy, i.e., unwanted or unusable, but in some cases it is the heat that is wanted. For example, the heat rejected in an internal combustion engine is waste, unless it is utilised in co-generation. The heat rejected by the burning fuel in a power station is not waste heat, but what's left over after boiling the feedwater and has to be lost in the cooling towers is waste. You may have mis-remembered or misunderstood your teacher. Wickwack 120.145.170.99 (talk) 15:00, 14 November 2012 (UTC)[reply]

No, I didn't mis-remember or misunderstand my lecturer, he used the second law of thermodynamics to explain it. Plasmic Physics (talk) 22:30, 14 November 2012 (UTC)[reply]

The 2nd law is essentially that in the conversion of heat into mechanical work, not all the heat can be so converted. How does that mean that that heat is waste energy? Not only is not all the heat "wasted", heat may be the desirable output as I said. Wickwack 120.145.143.165 (talk) 00:48, 15 November 2012 (UTC)[reply]

He didn't use "heat" in that sense. Plasmic Physics (talk) 05:04, 15 November 2012 (UTC)[reply]

It seems odd to read that heat is waste. Although most of the heat produced throughout the world become waste, yet it is a useful form of energy. It is true that we don't notice most of the heat produced, but it is not like that heat is waste. I don't think so because heat is used in steam engines to evaporate water, in ovens, in heating effect of many devices and many more. Talking about the second law of thermodynamics, it says all the heat cannot be converted into mechanical energy but some can. Sunny Singh (DAV) (talk) 09:16, 15 November 2012 (UTC)[reply]

It isn't that "heat is waste", which is a rather trite saying and like all trite saying has a tiny bit of truth packed inside too much approximation to have any real meaning. We should avoid triteness when describing things scientifically. Heat is waste in the sense that when two substances are in contact they exchange heat: the warm one heats up the cold one, until such time as the temperature between the two is equal. Here's the thing: The energy transferred from the hot body to the cold body cannot be utilized again to do work. That's because commensurate with the energy transfer is an increase in entropy, and you cannot recover that energy without "stealing" entropy from somewhere else in the universe, which would require you to heat up that part of the universe by more than you recover. That's the inescapable part of the second law of thermodynamics. Now, you can use heat to do work, as it is in the process of moving from the hot place to the cold place, but once the two places are in thermal equilibrium, you can't use that energy to do work again in an absolute sense: that energy is lost to the universe, i.e. there is a loss of free energy. So, in one sense, the trite statement "heat is waste" is correct, however it is not wholly correct because you can do work with heat; you just can't recover the energy after the work is done. --Jayron32 14:02, 15 November 2012 (UTC)[reply]

First, that is off the mark. Take a red hot body and a cold body, say a large fired up rod and a cup of cold water. Let's put these together. Energy flows from one to the other, and yes, that energy can still do work, since it will create steam that can be used in a heat engine. As for other misconceptions regarding the second law, I intend to knock it out of its ivory tower too. --Modocc (talk) 14:22, 15 November 2012 (UTC)[reply]

Um, why do you start a response with, "that is off the mark" and then say something that agrees with everything I say 100%. I am confused as to what parts of my explanation are incorrect. --Jayron32 14:33, 15 November 2012 (UTC)[reply]

To clarify, you said this, "The energy transferred from the hot body to the cold body cannot be utilized again to do work." Since you said "cannot be utilized again" I assumed you understood that the hot body already did work on the cold body, by heating it up, with the kinetic energy of the molecules being transferred. Its incorrect to conclude that this energy that was transferred cannot do any work again as you said though. In any case, entropy itself is elusive in the sense that it is inherently system dependent. In the 70's, entropy and Gibbs free energy were used to explain why diamonds, at the time, could be created only under extreme pressure and temperature, but in the 80's new catalysts circumvented the presumed entropy. -Modocc (talk) 15:01, 15 November 2012 (UTC)[reply]

Really? Once the rod and the water are the same temperature, how do you presume to use the thermal energy that moved from the rod into the water to do work again? Energy which is transfered from one form to another, or from one location to another, can be used to do work indefinitely except energy which is transferred as heat. You can only use that energy once; as it makes its trip from the hot stuff to the cold stuff you use that transfer to do work; but only that one time. Other energy transfers can be used indefinitely to do work along the way. That's the difference between the first law and second law of thermodynamics. --Jayron32 18:12, 15 November 2012 (UTC)[reply]

Let the temperature of the rod be 200celius. Let the rod cool to 199degrees and the water heat up to 199degrees. Assuming my cup has a lid on it, the water has turned to steam and is in equilibrium with the rod. Now, some of the heat of that rod has done work and you have said its not able to do work again, but it can. Open the valve on the lid, so the steam enters an engine. The heat energy of this steam gets transferred to the engine such that it is again does work. -Modocc (talk) 18:45, 15 November 2012 (UTC)[reply]

When you open the cup, in order for the steam to do work, it needs to be opened into a region of a different temperature/pressure than the inside of the cup. When you do that, you introduce a new heat transfer (from the hot inside to the cold outside). So you're dealing with different heat. You no longer have a closed system of just the rod and water, and if you take your system to be cup+rod+steam engine, the heat transfers have not been complete until all three are in thermal equilibrium. Once all three are, your system loses its ability to do work. My point still stands as 100% correct. --Jayron32 05:41, 16 November 2012 (UTC)[reply]

Once the steam leaves the cup, and I shut the valve, the rod should not rapidly lose much more heat if the rod and cup are insulated well. But I'm not sure what you mean by "dealing with different heat." I do mean a new heat transfer, but this basically just means that this heat gets utilized again. Also, your choice of closed/open systems are arbitrary. Its my understanding that work performed on any object is equivalent to its change in kinetic energy. Let the water initially begin with an ambient temperature and the water is heated by the rod, the kinetic energy of the water molecules are increased (and energy is absorbed too in order to break bonds with its heat of vaporization). Then, when this steam is used in the engine, the water will lose this extra energy, with some of it doing work on the engine's pistons, but the rest will not, with these amounts depending on the engine's efficiency. But all of that extra energy over and above the water's ambient heat content which the engine receives, but which it does not fully utilize, was provided initially by the rod's heat doing work on the water to raise its temperature above the ambient temperature. Thus, the water's initial ambient heat content plus the transferred heat energy that does work on the water prior to opening the valve equals the sum of the engine's waste heat energy and the energy of the work done on its pistons. A portion of this rod's heat energy does work on the water, and then as a part of the water's heat energy, it either gets wasted or performs work again on the piston. Your statement that "The energy transferred from the hot body to the cold body cannot be utilized again to do work." is wrong. -Modocc (talk) 06:42, 16 November 2012 (UTC)[reply]

Once you've redefined the system, it can. The difference is that you're working from a different defined system each time. You can keep passing heat energy from one place to another until the heat death of the universe and keep doing work at each transfer. My statement never claimed you couldn't do that. What you cannot do is use heat energy twice within a closed system. --Jayron32 18:53, 16 November 2012 (UTC)[reply]

Err, this assertion regarding what can or cannot happen within closed systems is wrong too, but I'm done with this topic for now. -Modocc (talk) 01:50, 17 November 2012 (UTC)[reply]

Seems to me this discussion has lost the plot with all this talk of triteness, reusability, and diamonds. Seems to me Jayron has merely established that heat can be waste (and in any thermodynamic process, some will be waste in a component process of a larger process), but heat isn't necessarily waste. Maybe that is a trite (= not novel) thing to say, but it is clearly different to saying heat IS waste (always), which is what PlasmicPhysics said. Plasmic said something wrong. Wickwack 120.145.0.141 (talk) 16:06, 15 November 2012 (UTC)[reply]

Incidentally, in Engineering Thermodynamics, D B Splading and E H Cole, Edward Arnold publ, a standard undergraduate text for many years, in Chaper 5 Heat, it gives a formal definition of heat: "the interaction between systems which occurs by virtue of their temperature difference when they communicate" (page 86 in 3rd Edition). It then goes on with 7 more pages about what heat is and what it is not. There is not a word about waste or any similar word like it. I have several other textbooks on thermodynamics and they all give the same definition, albiet in their own words, without any statement about "waste" until they come to actual examples, where for example, in a Carnot process some of the heat must be unusable (wasted). Wickwack 120.145.0.141 (talk) 16:06, 15 November 2012 (UTC)[reply]

All too true, and the above usage of "waste" is simply being used synonymously for the concept of entropy. So to define entropy is to define waste. In my neighborhood, we do a great deal of recycling though, so as to reduce the amount of waste that gets wasted. We've been binging on fossil fuels for some time now, but its time to sober up and do a far better job of recycling energy than we have. -Modocc (talk) 16:31, 15 November 2012 (UTC)[reply]

I also said the same thing heat is not waste. According to me, the overall discussion goes in favor of Modocc and Wickwack. Sunny Singh (DAV) (talk) 12:08, 16 November 2012 (UTC)[reply]

Calling heat "waste" is kind of silly when you think that the Sun, the furnace in your house etc. are all providing "waste". Best to call heat heat and leave the rest to semanticists. As I understand it, given that it appears the universe never ceases to expand and thus to cool, the heat should never become unrecoverable to work at any time in the future; there is no heat death of the universe, though the article seems a bit unsatisfying on the point. Wnt (talk) 20:59, 17 November 2012 (UTC)[reply]

The fundamental definition of heat as given in modern textbooks is energy transfer from one body to another that is not due to macroscopic work. This definition does not involve thermodynamic concepts, because you need to have this definiton to build up the thermodynamic concepts. Macroscopic work is defined as energy transfer due to a change in the external parameters of a system. What one chooses as the external parameters (e.g. the volume) is entirely arbitrary in priciple, so this makes the separation of energy transfer into a work part and a heat part entirely subjective. If you choose to describe the system exactly in terms of all its fundamental degrees of freedom, then all of the energy transfers will be work, and the entropy will be the so-called fine grained entropy which is always equal to zero (which expresses the fact that there is no loss of information at the fundamental level). Count Iblis (talk) 19:17, 16 November 2012 (UTC)[reply]

New Sub-Question:-
Leave this long discussion as it doesn't answers my question. Talk about the question. I read the article heat and thermal energy for another time as suggested above, but I have some confusions. I'll thankful if someone happily resolves my confusions.

When two bodies having different temperature are brought in contact with each other, both bodies attain thermal equilibrium after sometime. Here during energy transfer, heat is transferred between bodies or thermal energy i.e., which one of the latter two is being transfered. Sunny Singh (DAV) (talk) 05:45, 18 November 2012 (UTC)[reply]
Section 'Distinction of thermal energy and heat' of the article thermal energy mentions "Statistically, thermal energy is always exchanged between systems, even when the temperatures on both sides is the same, i.e. the systems are in thermal equilibrium. However, at equilibrium, the net exchange of thermal energy is zero, and therefore there is no heat". In the first sentence it says thermal energy can be exchanged when the systems are at thermal equilibrium and in the second sentence it says thermal energy cannot be exchanged. How is it possible ? Sunny Singh (DAV) (talk) 06:21, 18 November 2012 (UTC)[reply]

Answer to #1. Your english is a bit cryptic, so I hope I'm answering what I think you must be trying to ask. The correct term for what gets transfered is Heat. The reason why heat is the correct scientific term is bound up in the theory of thermodynamics. However, as I said in my first post, you will often find the terms heat and thermal energy used interchangeably in the literature. Thermal energy can mean other things, but whether an author means heat or something else can be determined by the context. The heat (which is a form of energy) gets transfered by one or more of the following ways, depending on just wht the bodies in contact consist of: Conduction (thru inter-molecular forces and collisions, and Radiation (electromagnetic radiation eg infra-red). If one of the bodies is a gas, then heat can be transfered within it by convection as well, sometimes referred to as mass transfer.

Answer to #2. What the author is trying to say is this: There is a two-way flow of heat between bodies when both are at a finite temperature - for bodies A and B there is a flow of heat from A to B (which would decrease the temperature of A ind increase it for B), and also a flow from body B to A (thus increasing A temperature and decreasing B). If the two bodies are at the same temperature, the two heat flows are equal, and since they are in opposite directions their effects in altering the temperatures cancel. You are perhaps surprised that there are two flows, and not just a flow from the hotter to the colder. To see why there are two flows, consider two bodies separated by a gap. Because each body is at some temperature (ie not at absolute zero), it must be emitting radiation, which is a function only of its temperature and its own properties (area, emissivity), which propagates across the gap as electromagnetic energy. After a propagation delay, the radiation hits body B, which must absorb some of it, as a function of its transparency. Meanwhile, Body B must be also radiating, since radiation is a function only of its own temperature and properties. If the two bodies are touching, then heat can flow by conduction (and perhaps convection) - that really doesn't change the fact that as both bodies are at some non-zero temperature, both must be emitters.

I am sorry that the discussion went away from just answering your original question, and I'm partly responsible of causing that when I corrected the erroneous assignment of heat as "waste" by PlasmicPhysics. Why he brought in the term "waste", which is not generally used in thermodynamic textbooks, is something only Plasmic can know.

Wickwack 121.215.50.212 (talk) 10:15, 18 November 2012 (UTC)[reply]

Two Last Sub-Questions:-

Suppose I have two iron block of 1kg. I, somehow, increased the volume (mass remaining the same) of first block and increased the mass (volume remaining the same) of the second block. Assuming all blocks have the same temperature, which one has more thermal energy ?
Suppose I have a solid block of iron of 1kg and gaseous nitrogen in a large container, nitrogen has also the same mass of 1kg. Which one - iron block or gaseous nitrogen- has more thermal energy ? Sunny Singh (DAV) (talk) 03:59, 19 November 2012 (UTC)[reply]

It would be better to ask new questions as new questions, not as sub-questions to an old question. This is because (a) you are more likely to get answers from various contributors, and (b) given the length of discusion on this one, probably nobody wants to contribute any more, even if they are still looking.

Answer #1: This question cannot sensibly be answered, as you cannot in any real way take a block of iron and alter its mass or its volume without changing both in proportion. By increasing the temperature, you can increase the volume a tiny bit without changing the mass, but the tiny change possible is not relavent to thermodynamics, and you excluded a change in temperature anyway.

Answer #2: Each element and each chemical compound has a specific heat that can in theory be deduced from its molecular and electron orbit structure. Specific heat is the amount of heat that a unit mass of a substance absorbs when its' temperature is increased by 1 unit (1 kelvin, 1 degree farenheit, etc). Particularly for gasses, specific heat varies with temperature. From standard tables, iron at 25 C (ie solid iron) has a specific heat of 450 J / kg.K; nitrogen has a specific heat at constant volume of 742 J / kg.K. Nitrogen, weight for weight, holds the greater amount of energy.

When talking about heat contained within a mass, it is necessary to know what this heat could be transfered to. The above answer is relavent if any heat flow in or out of the 1 kg iron or nitrogen is with respect to something else at close to 25 C. If the nitrogen was coolled so that it liquifies, for example, it would give up additional heat according to its latent heat of vaporisation.

Wickwack 58.170.164.182 (talk) 11:09, 19 November 2012 (UTC)[reply]

I was not talking about heat, I was talking about "thermal energy"; this is too late, so, close this discussion, I'll ask the same later. Thanks to those who contributed to this question and special thanks to Wickwack. Sunny Singh (DAV) (talk) 13:51, 19 November 2012 (UTC)[reply]

Vitamin D

Vitamin D is already present in cow's milk or it is added to milk by humans. Sunny Singh (DAV) (talk) 07:52, 14 November 2012 (UTC)[reply]

Most of the vitamin D in milk is artificially added. I can't find a source online that says exactly how much vitamin D is naturally present. Someguy1221 (talk) 08:20, 14 November 2012 (UTC)[reply]

Yes, I read the same thing in the last line of second paragraph of the article vitamin D. I am also confused how much vitamin D is naturally present in cow's milk. What about mother's milk (referring human), it contains vitamin D or not. Sunny Singh (DAV) (talk) 08:40, 14 November 2012 (UTC)[reply]

I found the same thing as with cows milk. It contains some vitamin D, but not enough to meet even a baby's daily requirements, but I didn't find a source that gave the exact amount. Someguy1221 (talk) 09:42, 14 November 2012 (UTC)[reply]

This site says that milk in the UK is not routinely fortified with Vitamin D. I also found this: "Whole cow’s milk was found to contain 38 i.u. vitamin D/I. Whole human milk contained 15 i.u. " from "The total content of vitamin D in cow's milk and human milk", Leerbeck (1980) - Journal of Human Nutrition. (There is a link on Google, but I've had a lot of trouble trying to get it to download and I'm not sure about putting the reference in here.) --TammyMoet (talk) 10:37, 14 November 2012 (UTC)[reply]

According to recent research, mother's milk contains enough vitamin D for the baby (more than 400 IU per liter) provided the mother's calcidiol level is above 120 nmol/l, which requires a daily intake for the mother of at least 4000 IU/day [5]. This is one of the arguments for higher calcidiol levels (120 nmol/l or higher instead of 50 nmol/l) despite there not being strong evidence for better health outcomes for adults, or that evidence being disputed. Obviously if babies would routinely need vitamin D pills because they would otherwise not even get the 400 IU/day for which we know their health is adversely affected, then that also implies that adults need more than 4000 IU/day. If we are right about this not being necessary for cancer or heart disease, then for other reasons we don't know anything about yet.Count Iblis (talk) 16:56, 14 November 2012 (UTC)[reply]

The use of ultraviolet light to enrich milk with vitamin D was discovered by Harry Steenbock. It's not so much that the vitamin D is added (there's is not some separate container of vitamin D that is mixed into the milk), rather the milk is vitamin D enriched/fortified, as there are pre-vitamin D compounds in milk that are transformed by the ultraviolet light into active vitamin D, in the same fashion sunshine (which contains ultraviolet light) can convert pre-vitamin D compounds to vitamin D in the skin. -- 205.175.124.30 (talk) 22:46, 14 November 2012 (UTC)[reply]

USDA's National Agricultural Library has all sorts of interesting information. This and this compare the nutrient content of 3.25% whole milk with and without added Vitamin D. Zoonoses (talk) 07:07, 15 November 2012 (UTC)[reply]

Forest fire vs candle flame

Candle flame goes off by the effect of wind but forest fire increases with increasing velocity of wind. Why ? Sunny Singh (DAV) (talk) 09:03, 14 November 2012 (UTC)[reply]

If you had a wind in proportion to the size of the flame, you could blow it out. For a forest fire, you'd need a hurricane, at least. StuRat (talk) 09:33, 14 November 2012 (UTC)[reply]

Also a forest fire is much, much hotter than a candle flame. Even if you could "blow out" the actual flames of a forest fire, the timber would still be burning. You can't really blow out any fire other than a tiny one such as a candle flame.--Shantavira|^{feed me} 09:45, 14 November 2012 (UTC)[reply]

I disagree. You'd just need a long and powerful enough wind to allow for the heat to dissipate. This might be several hours at several thousands of miles per hour, though. StuRat (talk) 09:49, 14 November 2012 (UTC)[reply]

I think Shantavira was talking about the real world. Caesar's Daddy (talk) 15:52, 14 November 2012 (UTC)[reply]

Is it correct to say that- wind in case of forest fire bring flame from burning tree to neighboring tree and in this way fire increases rapidly, but this doesn't happen in the case of candle flame. Will same thing happen if we stick some candles in a row and blow wind ? Sunny Singh (DAV) (talk) 11:09, 14 November 2012 (UTC)[reply]

Yes, wind helps a forest fire to spread, but as Shantavira says, pre-existing heat is a major factor. It's worth considering a blacksmith's forge, where moving air from a bellows or similar is used to increase the intensity and heat of a fire.

Oil well fires are typically put out using the "Wind" from a large high-explosive blast: the shockwave from a high-explosive charge "blows out" the flame in exactly the same manner as you blow out the candle: Bigger flames require bigger winds to blow them out. --Jayron32 15:40, 14 November 2012 (UTC)[reply]

See also the fire whirl article. Count Iblis (talk) 17:27, 14 November 2012 (UTC)[reply]

The analogy is flawed because one single candle has nothing near the geometry of a forest. Pack a shoebox with upright birthday candles, light one, wait a few seconds for it to warm up, and then blow on it with a speed scaled to a gale force as if the candles were 60 foot tall trees. Get back to us after you extinguish the charred box full of burnt wicks and molten wax. μηδείς (talk) 05:36, 16 November 2012 (UTC)[reply]

... and even a single candle in a light draught burns much more rapidly, often spilling wax in the process. Dbfirs 07:21, 16 November 2012 (UTC)[reply]

Parrot Drone or Quadrotor

Is there any project that try to develop Parrot Drone or Quadrotor that can lift a person or two persons? What's the obstacles to built it? Is it motor problem or battery problem? Is Parrot drone can fly very stable? What if another two motor added horizontally, can it fly faster? Will these two motors cause disruption in flight stability? Thanks... roscoe_x (talk) 09:10, 14 November 2012 (UTC)[reply]

What's a parrot drone ? I don't think a quadrotor is particularly safe. If any one of the four rotors fail, I assume it then crashes. StuRat (talk) 09:36, 14 November 2012 (UTC)[reply]

Parrot AR.Drone apparently. No flying machine weighing more than a few ounces can get off the ground it runs from a battery.--Shantavira|^{feed me} 09:52, 14 November 2012 (UTC)[reply]

Not so. The Sikorsky Firefly is a manned electric helicopter. For that matter, a bit more searching reveals the E-volo VC1 (see picture in the article), a manned electric quadcopter-style rotorcraft. That said, there are major design tradeoffs that make battery-powered helicopters completely impractical given current or near-term future tech, just as the quadrotor concept isn't really solving a problem in the field. — Lomn 15:45, 14 November 2012 (UTC)[reply]

I suspect that mostly it's a function of "no need" once you scale up to manned aircraft. Helicopters and gyrocopters are well-developed technologies that don't need the additional complexity that a four-rotor design introduces. For your final question, there's the compound helicopter, which adds horizontal thrust for additional speed -- the Sikorsky X2 is a recent example that reached 250 kts in level flight. There's also the V-22 Osprey, a tiltrotor. Both types of aircraft have considerably more engineering difficulties than traditional helicopters. — Lomn 14:43, 14 November 2012 (UTC)[reply]

I found an idea for a two-person quadcopter named The Skyflyer and a suicidal Chinese man. I think less is more when it comes to personal helicopters. Trio The Punch (talk) 19:36, 14 November 2012 (UTC)[reply]

It seems odd that those two quadrotor designs put the pilot above the rotors. This design is less stable and obscures the view of the ground, which is critical for landing. StuRat (talk) 19:55, 14 November 2012 (UTC)[reply]

Its probably much safer to sit on top of them in case of a crash landing and if the rotors are above you you'll need extra big and strong (=heavy) landinggear. Trio The Punch (talk) 22:20, 14 November 2012 (UTC)[reply]

Interesting answers, thank you. Especially the news about the farmer who built his own flying machine. roscoe_x (talk) 00:58, 15 November 2012 (UTC)[reply]

Some of those farmers are pretty innovative - see Richard Pearse. Zoonoses (talk) 07:10, 15 November 2012 (UTC)[reply]

This thing is pretty cool, but unmanned. Trio The Punch (talk) 15:08, 15 November 2012 (UTC)[reply]

chemical / molecular cross-section

What is the formula for the chemical reactivity cross-section of a molecule? The article cross-section only gives info about light absorption cross-sections and neutron cross sections. 128.143.1.142 (talk) 12:12, 14 November 2012 (UTC)[reply]

Reaction rates depend on the details of the chemicals involved as well as their physical states and abundance. In general, there isn't a single cross-section that would be useful for all molecules or all possible reactions, though you can use tables of things like standard electrode potential to determine what chemical reactions are likely to proceed. If you want more detail, you probably need to ask about specific molecules and reactions. Dragons flight (talk) 18:24, 14 November 2012 (UTC)[reply]

this is what is the compuound of Splenda, sugar and cloro form this, can we understand if it is a natural usefull compound for the human body?

1,6-dicloro-1,6-dideossi-β-D-frutto-furanosil 4-cloro-4-deossi-α-D-galattopiranoside o C12H19Cl3O8. — Preceding unsigned comment added by 195.110.143.127 (talk) 16:26, 14 November 2012 (UTC)[reply]

It looks like you've given the IUPAC name for sucralose. Splenda is the brand name of a sucralose-based sweetener that also contains dextrose and maltodextrin. Do the sucralose and Splenda articles answer your question? They both have referenced sections on health and safety. 209.131.76.183 (talk) 18:49, 14 November 2012 (UTC)[reply]

Artificial 2-toned lobsters

Is it currently possible to manipulate a lobster so that it looks like this one? Are we able to control the pigmentation selection process at the first cell division of the embryo (of certain species)? Is this technically possible in all multicellular beings (if you invest lots of time and money for research)? Are 2-toned humans theoretically possible? Trio The Punch (talk) 16:27, 14 November 2012 (UTC)[reply]

It's called heterochromia which in this case is caused by mosaicism, an early mutation in one cell after fertilization; but can also be caused by Chimerism, the merger of two embryos. There are people with such a condition, but due to differences in the splitting of cells in mammals and arthropods a symmetrical half-and-half mixture would be unlikely. See these images. Yes, a mad scientist could make one, but almost certainly not as striking as the lobster. μηδείς (talk) 17:04, 14 November 2012 (UTC)[reply]

As a rule, the kind of straight-down-the-middle color boundary so beautifully illustrated here is more common in arthropods. See gynandromorph for other such examples. (A Google Image search will be rewarding also) The split doesn't have to honor this midline boundary, however, even in arthropods. In mammals, the contribution of different cells in chimeras is pretty close to completely unpredictable. Wnt (talk) 18:29, 14 November 2012 (UTC)[reply]

Thanks a lot guys, very interesting links! Trio The Punch (talk) 19:14, 14 November 2012 (UTC)[reply]

Note again that the lobster in this case is a mosaic, not a chimera. Humans can be both mosaics due to a mutation in an early embryonic cell or chimeras of fused non-identical twins. Interspecies chimeras have been created, such as this horrific house mouse/deer mouse chimera. Note the monstrous lack of symmetry in the eyes due to the developmental gene expression. Do that with a Human and a Chimp and you'll get something out of H. P. Lovecraft. μηδείς (talk) 20:33, 14 November 2012 (UTC)[reply]

I hope our mad scientist creates a Cthulhu-like creature.

Trio The Punch (talk) 22:12, 14 November 2012 (UTC)[reply]

Yah, I've been daydreaming the human-chimp thing since the eighties... we could use an interpreter or two for interspecies communications. :) Wnt (talk) 22:44, 14 November 2012 (UTC)[reply]

Nyeah, bad idea. Reminds me of Asimov's The Ugly Little Boy. -- OBSIDIAN†SOUL 10:13, 15 November 2012 (UTC)[reply]

I don't know why would someone ever want to have 2 skin colored. It looked pretty abnormal and doesn't look good at all to me. Plus it is not ethical to do that anyway. You can't just test that on people.174.20.41.202 (talk) 21:58, 15 November 2012 (UTC)[reply]

Agreed, see wrongful birth and wrongful life. The picture of the chimerical mouse I linked to should scare the pants off any ethical experimenter. A human-chimp hybrid would be symmetrical and even-colored--but it's still not worth the risk. μηδείς (talk) 22:23, 15 November 2012 (UTC)[reply]

Named particles?

The Oh-My-God particle was a specific particle which was named for its surprisingly high speed. This is the first example I'd ever heard of a specific particle having a name. (Not a class of particles, but this particular proton.) Are there other examples? Staecker (talk) 16:35, 14 November 2012 (UTC)[reply]

Not a particle, but of the same idea: See Wow! signal. --Jayron32 17:09, 14 November 2012 (UTC)[reply]

This is strictly a case of notability. Anyone can arbitrarily name anything they want. For example, I name all of my protons, and some of my better mesons, but I'm not famous enough for any of my pets to have their own encyclopedia articles. Nimur (talk) 17:11, 14 November 2012 (UTC)[reply]

Yes I know. I'm asking if any other individual particles have achieved this level of notability. Staecker (talk) 00:15, 15 November 2012 (UTC)[reply]

gravity = side effect of vacuum?

Could it be that the displacement of vacuum by massive objects causes the density of the vacuum closest to the object to be denser than the vacuum any distace farther away from the object, thus causing the denser vacuum to "suck" smaller objects to that area, which just happens to be the surface of the object?165.212.189.187 (talk) 20:45, 14 November 2012 (UTC)[reply]

Gravity is the effect of mass causing curved space-time. (relax.... its non intuitive and complicated) Spacetime — Preceding unsigned comment added by Ap-uk (talk • contribs) 21:25, 14 November 2012 (UTC)[reply]

Thanks, (and no offense) but that sound like regurgitated double-talk165.212.189.187 (talk) 21:31, 14 November 2012 (UTC)[reply]

It's much easier to regurgitate sound-bytes describing our most non-intuitive physical explanations than to understand them! In any case, whether we understand it or not, gravity is a phenomenon that we observe in the universe. It is one of the most fundamental interactions we know, and it appears to contribute to the dynamics of almost everything we know about (with only a very few exceptions). We can describe the effects of gravity in many different ways. For many purposes, it is useful to describe gravity as a force, whose strength depends on a quantity we call mass. This is the "classical" gravity that physicists attribute to Isaac Newton. Einstein and more recent physicists used a different mathematical formulation to avoid the trouble with inexplicable action-at-a-distance, and to resolve some of the more subtle effects of gravity, like the way it interacts with electromagnetic waves. But, whichever formulation of gravity you use to model your observations, very few physicists ever talk about the "density" of a vacuum. And I have never heard any reputable physicist suggest that vacuums "cause" gravity as a side-effect; I can't really even figure out what that would mean. "Vacuum" is just a term we use to describe a region of empty space. It sounds as if you're trying to stretch an analogy of gravity to some type of buoyant force, and you're hypothesizing some type of gravitational phlogiston fluid - but if you carefully study all the implications of that hypothesis, I think you'll find that theory does not correspond to what we observe in nature. You are not alone; many great scientists throughout history have tried to use a hydraulic analogy to describe as-yet unknown effects of heat, electricity, and chemistry; but in each case, these hypotheses have been superseded by more parsimonious explanations. Nimur (talk) 22:26, 14 November 2012 (UTC)[reply]

Dammit, Jim, I'm a lawyer not a physicist, but 165.212.189.187 it seems to me that you've got the cause and effect wrong. Vacuum density is a measure of how much matter — gas and other particles — are within a space. Those things are denser near an object because of gravity, or to say it backwards, if it were not for gravity the density would not be higher near the object. Someone who's a real physicist will probably tell me I'm wrong, but that's my first impression. Regards, TransporterMan (TALK) 21:30, 14 November 2012 (UTC)[reply]

No offence taken , it's to do with perspectives. Mass, space and time are interlinked but our general default brain setting is one of 17th century mechanics. The proof of Einsteins curvature of space time explained 1) Mercury's weird orbit 2) Displacement of starlight around the sun. Both are shown here : Tests_of_general_relativity ... To show that time is not absolute a 1971 experiment is good fun :check out the "Hafele–Keating experiment" Ap-uk (talk) 23:00, 14 November 2012 (UTC)[reply]

One theory of gravity is that the vacuum exerts a positive pressure, and that the proximity of two bodies blocking that pressure from each other causes them to move together. I am not sure what that theory is called, so I can't get you references, but perhaps someone else here does. μηδείς (talk) 23:59, 14 November 2012 (UTC)[reply]
- You may be referring to Le Sage's theory of gravitation. Richard Feynman says it doesn't work, and that's good enough for me. Someguy1221 (talk) 04:37, 15 November 2012 (UTC)[reply]
Three things. First, gravity exists and can be measured in the absence of vacuum or pressure forces, or even in the presence of opposing forces - a heavy weight dropped in the ocean will sink, even though the water pressure surrounding it in every direction is roughly equal, and that the water pressure increases with depth rather than decreasing. If low pressure (or vacuum) caused gravity's attraction, the weight would more likely float up into the sky. Second, attraction due to vacuum cannot explain how a small, very dense object like a black hole or neutron star can exert a much stronger gravitational effect on larger but less dense objects. This observation also affects the positive pressure theory Medeis mentioned. Third, because pressure in a given area tends to like staying equal, vacuum 'sucks' on an object in space roughly equally in all directions. With equal force pulling the object in every direction, the net force effect is zero, not 'sucked toward the nearest other object' as the question suggested. So in short, no, vacuum cannot explain the observable properties of gravity. – NULL ‹talk›
‹edits› 04:32, 15 November 2012 (UTC)[reply]

Arent atoms' volume 99% empty space?165.212.189.187 (talk) 16:35, 15 November 2012 (UTC)[reply]

@ Someguy and Null. Le Sage's theory of little particles is not the same one I was thinking of, but rather a Casimir effect caused by vacuum energy. The geometry is the same, but in the second theory it is a force, not particles. This is mention in this section of the Le Sage article, and no criticism or refutation is given. Second, the vacuum energy is someowhat of a misnomer. It's supposed to exist every, including in relative vacuums, not just in or because of vacuums. Perhaps the OP means vacuum in a different way from vacuum energy, he'll have to speak for himself. μηδείς (talk) 17:54, 15 November 2012 (UTC)[reply]

Just to be sure ... how does Le Sage's notion of ultra-mundane particles differ from the notion of virtual particles filling all space, e.g. a Dirac sea? If gravity is an exchange of virtual gravitons, is this the same as the blocking of exchange of certain virtual antigravitons which would be Le Sage's particles? Wnt (talk) 17:11, 16 November 2012 (UTC)[reply]

A question on economics.

Here in the UK a lot of economics statistics and also general work statistics (ONS) are used by the media who report it as fact.

Many years ago I got a degree in science and struggle to see how accurate these statistics actually are especially with economics. The general trend is to make it all look too difficult to the average person so they don't question things and also there is a lack of a glossary and explanation as to what has been left out or included.

1) Is economics a pseudoscience?

2) Is there some kind of bias in the statistics to make things look slightly less intimidating to the public? I notice for example that with regard to wealth stats the 69% of people who die in debt are left off the table, the wealth stats are then tabulated to take the super rich out (in fact you are in the top percentile with a net worth of only £423,000).

If economics is a pseudoscience why are no top scientists complaining or is it safer (as regards future financial science funding) to just pick easy targets like religious people ;)

--Ap-uk (talk) 21:19, 14 November 2012 (UTC)[reply]

1) Economics is a social science. 2) It would really depend on what you are looking at, who compiled the wealth stats you are looking at? They might have reasons to present data in a particular way other then "making it less intimidating".. And of course, nothing stops an economists biasing data if they have an agenda to push. Vespine (talk) 21:38, 14 November 2012 (UTC)[reply]

Your answer 1) is open for discussion. I wouldn't call it a science at all, social or not. Unless you are testing and improving your theories, do not call it a science. Science operate on the principle of falsifiability. The most amazing thing about economists is that they don't even care if their theory is wrong or not. It's an ideology pretending to be a science.~On the other hand, they indeed do use, although poorly, scientific tools like statistics, but that's not enough. OsmanRF34 (talk) 23:42, 14 November 2012 (UTC)[reply]

This is one of the sillier generalizations I have seen on here in awhile. There are indeed many branches of economics that involves compiling huge datasets and testing theories against them. There are branches of economics that intersect heavily with behavioral psychology and involve running clinical studies. It is a very broad field in terms of methodology. I have never met an economist who didn't care of their theories weren't correct. Saying "economics is an ideology pretending to be a science" is just reflective of your ignorance or prejudice and nothing more. --Mr.98 (talk) 15:44, 17 November 2012 (UTC)[reply]

I got the wealth stats from the ONS (Office of National Statistics) however it was not straight forward to find out the answers that I wanted and I had to glean it from several documents even then it was not 100% clear. The ONS did become separate from the government recently however I still see traces of "non-controversial" bias. — Preceding unsigned comment added by Ap-uk (talk • contribs) 21:55, 14 November 2012 (UTC)[reply]

This reminds me of the control systems lecturer we had when I was an electrical engineering student. He loved to poke fun at fields that were not like engineering, which is 100% based on fundamentals. He used to say that when an Engineering lecturer writes this year's version of the exam, he changes the questions. When an economics lecturer rewrites the exam for the year, he keeps the questions and changes the answers!

More seriously, economics when compared to things like engineering is a bit like psychology. Both economics and psychology DO have fundamentals, but the subjects are not fully understood and day-to-day issues have to be solved by postulation. Examples of fundamentals are: In psychology - all learning is either clasical conditioning (http://en.wikipedia.org/wiki/Classical_conditioning) or operant conditioning (http://en.wikipedia.org/wiki/Operant_conditioning); in economics, in a closed economy, the total value of money is equal to the total value of goods, services, and property. But can a psychologist predict what sort of responses will be triggered by my post, or can an economist predict how the price of widgets will react to Hurricane Sandy? Well, yes they can, but not by a process solely involving calculation and logical deduction starting with fundamentals. Engineering is quite different. In engineering, all solutions are derived from calculation and logical deduction starting with fundamentals like ohm's law. Then you get climate change science, which is not even as solid as psychology or economics. Wickwack 120.145.143.165 (talk) 01:13, 15 November 2012 (UTC)[reply]

The answer is sort of complicated. When economics is quantitative, it is a science. But because economics is so closely tied to politics, there are aspects of it that are not scientific, as for example the work of Karl Marx and the Austrian School. Looie496 (talk) 03:31, 15 November 2012 (UTC)[reply]

The category of pseudoscience may not be as useful as you have been led to think. Classifying something as pseudoscience begs the question whether there is a generally applicable criterion by which to do so, the famous demarcation problem. Certainly no-one would argue that the economy is something beyond scientific study, and certainly most economists are quite earnestly trying to understand, quantify, and describe it. On the other hand, there are many scientists, including economists, who have made scathing critiques of the state of mainstream economics, for example that in economics, the theoretical side and the empirical side exist in fatal separation from one another (in a very interesting paper by an economist, but naturally I forget her name – I think she was from the University of Chicago). This, if true, would make economics bad science but not pseudoscience.

As to official government figures relating to the economy, you do well to take them with a good pinch of salt, since massaging the figures is usually the easiest way to make a problem go away. Inflation can be conveniently suppressed by adjusting the consumer price index, unemployment be kept down by defining people out of the ranks of the unemployed, etc. (I have no idea if and how much this is done in the UK, but would be surprised if it isn’t.) The “official” net worth of an individual person may not reflect their actual wealth with any degree of accuracy, since much of what the very rich own will often be nominally owned by some corporation or otherwise be distanced from them. But if you know how to read the figures correctly, they do provide useful information about real life. If the media report them uncritically, it’s the media’s fault, not the statisticians’.--Rallette (talk) 07:23, 15 November 2012 (UTC)[reply]

There are websites who specialise in checking out statistics, especially those quoted by politicians to back up their claims. check out Full Fact and Fact Check. (UK urls given as OP is UK) --TammyMoet (talk) 09:54, 15 November 2012 (UTC)[reply]

If there isn't enough information for you on how the official stats are collected, crunched and presented, there are always contact details of ONS staff who can answer your questions Some reasons why the methods might not be what you would regard as optimal include the need to ensure comparability with other countries, comparability over time, cost of collecting data and data protection. Journalists will never take any notice of the caveats that come with the data. Academic researchers should, but they also slip into professional conventions. The national productivity statistics are one of the most striking examples of the phenomenon you identify - they measure something, and they show that one country consistently "does worse" than another, but what they measure doesn't appear to be productivity. Itsmejudith (talk) 02:40, 17 November 2012 (UTC)[reply]

November 15

the material rising in the thermoscope \ thermometer

I would like to know if when the material rises in the the thermoscope meter (or in the thermometer), it indicates that the air in the thermoscope is shrunk or spreading. According to what I understand, the material (in example water ) spreading becaouse the warm and that is the couse of the material rising of the material inthe meter. What doyou think about? 46.210.166.245 (talk) 01:40, 15 November 2012 (UTC)[reply]

If the thermoscope is sealed, the liquid expanding will simply compress the air. If the temperature increases the air would also be warmed up at the same time as the liquid and if it was free to do so it too would expand, but since the liquid is far more dense then the air, the resistance the air applies to the fluid would be negiligable and the pressure inside the container would increase. I don't know for sure but I wouldn't be surprised if thermometers were created in such a way as to minimize the amount of air in the space before they are sealed. Vespine (talk) 02:36, 15 November 2012 (UTC)[reply]

I think it's more about the compressibility of a liquid (not very compressible) vs gas (easily compressible) rather than density itself as the underlying reason that the liquid expands to compress the gas even when ∆T says they should both be trying to expand. Our Alcohol thermometer article makes mention of the headspace above the liquid level. DMacks (talk) 07:10, 15 November 2012 (UTC)[reply]

Suppose, we have a thermometer of range 0 to 100 degree celsius. The thermometer is taken to a region of 100 degree celsius, I don't think that it will show 100 degree celsius if the glass cap of thermometer is attached at the same height as of 100 degree celsius mark. Air in the thermometer will compress to a certain amount, but will not get dissipate. If the cap is attached above the highest temperature scale (about 1 inch), the material inside is able to manage air compression and the thermometer will show correct temperature.

According to Vespine, pressure exerted by air on mercury will be negligible but what will happen if the mercury has expanded to the highest temperature. Will the air exert the same pressure at that temperature ? Air will also expand as the mercury (due to the heat of surrounding medium) and hence the kinetic energy of air molecules also increase and this time air will exert more pressure on mercury. Think about mercury barometer, it has vacuum in its mercury column but in case of mercury thermometer it has air. I do not know why ?Sunny Singh (DAV) (talk) 10:39, 15 November 2012 (UTC)[reply]

The air is confined to the volume left by the rising mercury. So the air pressure will increase due to 2 reasons 1) the pressure increases due to the reduced volume available, and 2) it increases due to the increased temperature. However, the increase in air pressure has negligible effect on calibration, as the mercury, being a liquid, is vitually incompressible. Floda 121.221.27.52 (talk) 11:14, 15 November 2012 (UTC)[reply]

I'm sorry, but I don't understand what does the meter shows when the material rises in the meter. the question is: Is the air in the meter is shrunk or spreading? (the thermoscope is open to get influences of atmosphere) 176.13.83.244 (talk) 18:17, 15 November 2012 (UTC)[reply]

The volume of the space at the top of the thermometer is smaller. That is the definition of "shrunk". --Jayron32 18:31, 15 November 2012 (UTC)[reply]

(edit conflict) As explained above, it doesn't make much difference whether the meter is sealed or open because it is the expansion of the material (usually coloured alcohol or mercury) that is being observed as the temperature rises. The only occasion when the compression of air is important is when a thermometer is heated well above its maximum range, and the air gets compressed almost to zero volume. The high pressure usually causes the bulb to explode. There may well be a partial vacuum above the material in a sealed thermometer, but this is not essential in a thermometer (though it is important in a barometer). Dbfirs 18:34, 15 November 2012 (UTC)[reply]

If you have a sealed container half full of fluid, when you heat it, the fluid will expand. The container with its contents will still have the same weight. But the fluid will be less dense, so another container within that fluid might sink. You should be able to make a Galileo thermometer which consists of concentric spheres, each containing a slightly different amount of fluid, so that each sphere rises or sinks within its surrounding sphere independently of the effect on all the other spheres. Though it would be terribly slow to respond unless made out of a very good heat conductor. I don't know if anyone has made such a thing but I would expect so. Wnt (talk) 02:52, 16 November 2012 (UTC)[reply]

Ever been to a gift shop Wnt? ;)

They are a bit on the expensive side but quite easy to find. Each glass orb has a punched sheet metal label, both to show the temperature and to calibrate the orbs. Clipping/filing the label will remove some weight and make the orb float more easily. - ¡Ouch! (^{hurt me} / _{more pain}) 09:27, 16 November 2012 (UTC)[reply]

I've seen them with multiple orbs in a single tube, which the OP wonders whether it is open or closed (which won't matter to the degree that the liquid is incompressible relative to the gas in the outer chamber anyway). But I've never seen them with one orb inside the next inside the next etc.; the tags would hinder the aesthetics of this and their weight would need to be well calibrated without them. Wnt (talk) 17:14, 16 November 2012 (UTC)[reply]

I thought that the OP was asking about a thermoscope (or ordinary thermometer), not a Galilean thermometer. Dbfirs 00:37, 17 November 2012 (UTC)[reply]

You're right - we had a different question about that one a little while ago... Wnt (talk) 05:43, 17 November 2012 (UTC)[reply]

Omg - I didn't get that you were talking about all orbs being concentric, not all merely within one outer containment. - ¡Ouch! (^{hurt me} / _{more pain}) 18:11, 17 November 2012 (UTC)[reply]

Electromagnetic radiation

At the beginning of the statement, it says: "electromagnetic radiation (EM radiation or EMR) is a form of energy emitted and absorbed by charged particles". So are charged particles mean it is either positive or negative? How about neutral particle with no charge? Can neutral particle emit and absorb electromagnetic radiation?174.20.41.202 (talk) 03:17, 15 November 2012 (UTC)[reply]

Consider blackbody radiation. It can be emitted by matter that has, in bulk, no net charge. But, if you zoom in all the way to the fundamental interactions that are responsible for the thermal emission of radiant energy, you will discover that in the very microscopic sense, the radiation is emitted by a process that could be classically described as a movement of charged particles. We might say that warm objects have electrons in excited atomic orbitals, and that the photons are produced when the electrons decay. That explanation treats the photon as an emitted particle, and doesn't concern itself with the electrodynamics of the photon-as-a-wave. The exact details are actually somewhat subtle; for most purposes, we gloss over this detail; and if we want to study it closely, we have to use some very heavy-duty mathematical physics. Often, when we consider blackbody radiation, we are not interested in electrodynamic interactions; we only care about the energy that is being conveyed. But, any emitted electromagnetic radiation - whatever its source - is attributable to microscopic movement of charge.

When we discuss subatomic particles, we have to generalize the way we think about emitted radiation. Energy can be emitted in other forms besides electromagnetic waves, especially if it's radiated in a way that isn't related to wiggling electric charges. Sometimes, certain "exotic" energetic interactions between particles result in an emitted particle - which carries energy - that is not an electromagnetic wave. A neutrino can be emitted and can convey energy, without any interaction with electric charge. A phonon is energy that is emitted that takes the form of propagating wave in bulk condensed matter. There are many such examples where the radiated energy is not electromagnetic in nature, because the source of energy did not participate in any electromagnetic interaction. Nimur (talk) 06:19, 15 November 2012 (UTC)[reply]

I think Nimur is wrong. Consider a quantity of non-reactive gas at some finite temperature. The gas molecules, which are not charged particles are, according to the kinetic theory, continually flying about in the x, y, z tanslational axes and also rotating about these axes. The gas will be emitting, in all directions, heat in the form of infrared radiation, which is electromagnetic radiation. If there is incoming infrared radiation of total wattage greater than that being emitted, the temperature will increase - that is the gas will absorb energy by increasing the translational velocity of the molecules. If the molecules are not symmetrical in all x,y,z dimensions some of the heat energy will be converted into increased rotational movement. Some will be absorbed by lengthening the atomic bonds, thus increasing the "flywheel effect" Charged particles have no part in this beyond forming the atomic bonds, which are not changed by the increse in temperature. It is possible to have a "gas" in this thermodynamic sense consiting not of complete atoms or molecules, but instead consisting entirely of one sort of particle (electron, nuetron, etc), which may or may not be a charged particle. Floda 121.221.27.52 (talk) 11:08, 15 November 2012 (UTC)[reply]

Um, "Charged particles have no part..." Is completely wrong. If, as you say, there is "lengthening of the atomic bonds" then, by definition, that is a change in shape of molecular orbitals which involves a change in energy of electrons, which requires the electrons to change energy levels in some way, which requires the emission or absorption of photons, QED. It is inescapable: if photons are produced, something about the motion/location/energy of electrons changed (which "property" of electrons changed depends on your perspective, but the initial statement that all EM radiation originates in changes to charged particles is fundamentally true.) --Jayron32 13:50, 15 November 2012 (UTC)[reply]

Jayron, you did not mention clouds of sub-atomic particles. So, if you (& Nimur) are correct in saying charged particles are necessarily involved in the emission and absorption of EM radiation, then a cloud of nuetrons either cannot exist, or cannot have a temperature? I always thought that the Kinetic Theory of Gasses applies to sub-atomic particles as well as atoms and molecules when they are not forming a solid or liquid. (A cloud of particles should obey the KTofG if the following are true: Sum of particle volume << cloud volume; number of particles is very large; except in collisions, interactions are negligible; Number of collisions between particles is large compared with collisions of particles with container). Floda 60.230.192.18 (talk) 15:34, 15 November 2012 (UTC)[reply]

I don't follow the connection that is being made here between kinetic theory and thermal radiation. A "gas" of sub-atomic particles will always have a temperature and other thermodynamic properties, but it will only emit or absorb thermal radiation if the particles carry an electric charge or have components that carry an electric charge. Atoms and molecules have no overall charge but they still emit and absorb thermal radiation because they have charged components and hence they have an electric dipole moment. A "gas" of neutral fundamental particles would have a thermodynamic temperature but would not emit or absorb thermal radiation. Gandalf61 (talk) 15:57, 15 November 2012 (UTC)[reply]

Such neutral fundamental particles will have to be completely decoupled from electromagnetism, otherwise there will be higher order effects leading to the emission of photons. This means that the particles must be hidden sector particles that are completely decoupled from the Standard Model. And even then there will be extremely small effects mediated by gravity. Count Iblis (talk) 17:37, 15 November 2012 (UTC)[reply]

To expand on Iblis's point: neutrinos don't have any electric charge, and are not mediated by photons, so interactions involving neutrinos only involve exchange of particles like the W and Z bosons. If something neutrino does causes the release of a photon, it is only of the "higher order effects" that Iblis mentions: the weak bosons that neutrinos release interacting in some way with a charged particle, and THAT charged particle releasing a photon. It should be noted (not mentioned yet as a point of confusion, but it could be given the way the discussion is going) that even neutrons are not fundamentally neutral; they are composed of charged quarks, and thus are subject to EM interactions just as other charged particles are. --Jayron32 20:18, 15 November 2012 (UTC)[reply]

God, I'm so lost... Didn't expect the answers to be this lengthy. What you guys saying are beyond what I can understand, I'm just a curious ordinary person who doesn't know physic very well. Can you someone simply answer my question in simply way? Can neutral particle emit and absorb electromagnetic radiation? If it can then how the article only say "emit and absorb by charged particles"?174.20.41.202 (talk) 21:43, 15 November 2012 (UTC)[reply]

The answer depends entirely on what you mean by "neutral particle". If you mean "any particle with no net charge" then the answer is "It can still emit and absorb EM radiation if the particle is made up of smaller particles that themselves have electric charges". IF you mean "a fundamental particle which is not composed of smaller particles, and which is fundamentally neutral at all levels of organization" then the answer is "No, it cannot emit and absorb EM radiation" Thus, both the hydrogen atom and the neutron, which are each neutral but which are also each composed of smaller bits that themselves are charged (a proton and electron in the former, up and down quarks in the latter) CAN interact via electromagnetic radiation. However, a particle which is both indivisible and neutral, like a neutrino cannot. Does that make any more sense? --Jayron32 22:15, 15 November 2012 (UTC)[reply]

[Edit conflict]Electrically neutral particles that are composed of equal amounts of positive and negative charge emit and absorb radiation, so yes, neutral particles do, but only because of the equal presence of both kinds of charge which can emit or absorb the radiation. Does that help? -Modocc (talk) 22:23, 15 November 2012 (UTC)[reply]

Jayron and Modocc's explanation make perfect sense to me. So basically if an elementary particle that has no charge then they won't emit or absorb radiation.174.20.41.202 (talk) 03:22, 16 November 2012 (UTC)[reply]

This isn't exactly true, only a good approximation. Neutral particles can have magnetic dipole moments, even if they are elementary. E.g. the neutrino has a magnetic dipole moment, despite it being a neutral elementary particle, see e.g. here, and they can then emit electromagnetic radiation. This happens due to quantum effects. In quantum mechanics, if A can interact with C via B, then A will also be able to interact with C if B isn't present due to virtual B popping out of the vacuum and disapearing again. This is how the neutrino gains a magnetic moment despite not consisting of charged particles. Count Iblis (talk) 03:40, 16 November 2012 (UTC)[reply]

A related question: why don't neutral objects create EM fields all the time as they move about? (Like say, a human running and sprinting and then slowing down to catch his breath and then running again?) Individually, the objects are made of charged particles, so shouldn't they generate their own fields, that then would mostly cancel out? Photons would be emitted though. 71.207.151.227 (talk) 23:42, 15 November 2012 (UTC)[reply]

You're blasting out electromagnetic radiation right now, without doing anything. See Night vision device and Thermography for some applications of this. --Jayron32 23:55, 15 November 2012 (UTC)[reply]

They wouldn't cancel out because the positive and negative charges are not located at the same place. See electric dipole moment for a simple example of an overall-neutral system of charges that has a very real electric field about itself. Someguy1221 (talk) 23:50, 15 November 2012 (UTC)[reply]

Jayron is completely right.

On top of that, the human body is both a conductor and a capacitor. It might be a coil, if you look at the blood vessels which contain in a coarse approximation a conductive NaOH solution. A few feet of copper wire and some simple electronic components are sufficient to recognize other bodies with some accuracy. Nothing with a snowball's chance to stand in court, but... I tuned one of these to a jogger I encountered regularly, during years when I had time to kill and easy access to the components needed.

When he passed me, the circuit made a nice "blip" sound. There were some false positives, but less than one in 100. I had to retune the circuit while I was losing weight, tho. My own signature was screwing with its detection ability. Boy was she scared...

But that's about it. You can't do anything really evil, like electrocute her, or plant a neat "Kill your boyfriend" thought. But hey, there are other methods to get that done nowadays. - ¡Ouch! (^{hurt me} / _{more pain}) 07:49, 16 November 2012 (UTC)[reply]

NaCl, dammit. Self-whack. <°|>>><< - ¡Ouch! (^{hurt me} / _{more pain}) 08:32, 16 November 2012 (UTC)[reply]

Recharging smartphone

Why does it take up to an hour to go from 99% charged to 100% on my HTC One S? 67.243.3.6 (talk) 15:48, 15 November 2012 (UTC)[reply]

See Lithium-ion battery#Battery charging procedure. The charging is not linear, but consists of three stages. Dbfirs 17:45, 15 November 2012 (UTC)[reply]

Doppler effect: two moving objects at an angle

What's the formula for the Doppler effect if two moving objects are moving at an angle subtending the velocities between the two? For example, an airplane overtakes a motor vehicle on the ground-- how would you calculate the frequency shift of the engine roar emitted or received at any specific moment? 128.143.100.179 (talk) 18:41, 15 November 2012 (UTC)[reply]

First, you would subract the velocity vectors (assuming you have an external frame of reference) to get the relative velocity between the two points. Then you use the Dot product to project the velocity vector onto the vector between the two points. This will tell you how quickly the points are moving towards or aways from each other, which can then be fed into the Doppler formula. 209.131.76.183 (talk) 18:55, 15 November 2012 (UTC)[reply]

That's fine for light in vacuum, but the nonrelativistic Doppler formula takes the velocities of emitter and receiver relative to the medium. Their relative velocity isn't enough. -- BenRG (talk) 05:23, 16 November 2012 (UTC)[reply]

You could also use a full-wave-equation modeling technique, such as the Finite-difference time-domain method, to estimate the wavefield at every point. Then, you could sample the simulator to approximate the signal that would be observed for any trajectory at any velocity. This method is a lot more difficult to implement in practice, but if performed correctly, can account for many non-ideal effects, like the interaction between the sounds of the two vehicles; or imperfections in the propagation material. (Our article on FDTD has a strong bias toward the application for solving electromagnetic wave equations, but you can equally-well apply the numerical technique to solve the acoustic wave equation, and many other multidimensional PDEs). Nimur (talk) 20:05, 15 November 2012 (UTC)[reply]

Okay, I would have to do this in an exam. (It's a problem that I anticipate, as my professor likes to give us "climax of all that we know" questiosns. Also, according to my lecture notes, converting the two velocities into a relative velocity doesn't really work as v becomes on the same scale as c (e.g. 0.1c) . 71.207.151.227 (talk) 23:38, 15 November 2012 (UTC)[reply]

Use the lorentz transform to calculate the relative velocities when you suspect that the effects will differ from the classical case. Then, use that intermediate result and apply your favorite variation on the relativistic Doppler effect. It sounds like this work is related to a class; so check your notes or textbook for your professor's formula; or you can use the formula provided in our article. Nimur (talk) 00:23, 16 November 2012 (UTC)[reply]

A general formula for the nonrelativistic Doppler shift is

{\frac {f_{r}}{f_{s}}}={\frac {c+\mathbf {v} _{r}\cdot \mathbf {x} }{c+\mathbf {v} _{s}\cdot \mathbf {x} }}

where

f_{s}

and

f_{r}

are the emitted and received frequencies,

c

is the speed of sound,

\mathbf {v} _{s}

is the velocity of the source at the time it emits the sound,

\mathbf {v} _{r}

is the velocity of the receiver at the time it receives the sound,

\mathbf {x}

is a unit vector pointing from the location of the receiver at the time of reception towards the location of the source at the time of emission,

and all the velocities and positions are computed in the rest frame of the medium.

It should be easy to see how this reduces to the one-dimensional case given in the article.

For the relativistic Doppler shift of light in a vacuum, there's a simpler formula:

{\frac {f_{r}}{f_{s}}}={\frac {\mathbf {v} _{r}\cdot \mathbf {x} }{\mathbf {v} _{s}\cdot \mathbf {x} }}

, where

\mathbf {v} _{s}

and

\mathbf {v} _{r}

are now four-velocities and

\mathbf {x}

is a four-vector pointing from the spacetime location of reception to the spacetime location of emission. You're no longer limited to a particular reference frame (of course) and

\mathbf {x}

doesn't have to be a unit vector (in fact, it can't be, because it's lightlike).

Should I add these formulas to the articles?

One could also derive a relativistic formula for sound or light in a medium, but I'm not sure how useful it would be. -- BenRG (talk) 05:23, 16 November 2012 (UTC)[reply]

Origin of the Nebraska sand hills

Hi. Currently, I'm researching the Sand Hills (Nebraska) for a world geologic location and hazard research assignment, and I'm exploring the geologic origins and proneness to mass wasting of these ancient sand dunes. What I've found so far is that the sand hills region is composed of Quaternary-age "Eutric" regosols, bounded by much sediment and/or uplift of Eocene to Mesozoic (think Niobrara Sea) age, located at the edge of the Denver Basin, with loess deposits extending to its east and southeast (yes, I have citations for all this info), reaching all the way to the Mississippi River and beyond, while the loess sediment also reaches down to the mouth of the Mississippi River - perhaps this makes sense, since the Platte River runs through the southern sandhills. I'm asking anyone for more relevant information about this area that they can find, including the depth of the sand (soil and rock core profiles would really help), the presence of any halite or gypsum salt deposits (I'm noticing that Nebraska is downwind of Lake Bonneville of similar age), and the effects of past explosive volcanic eruptions including at Yellowstone caldera on the region, and any evidence of past vegetation or absence thereof - for instance, the area's dunes were active during the Medieval Warm Period. Might the dunes creep east of Omaha and Lincoln, NE in the not-too-distant future? Also, what is the source of the sand dunes scattered throughout the Lake Michigan shoreline - are they direct deposits from the Laurentide ice sheet, as suggested by the USGS map on the "loess" article, or might the Nebraskan sandhills contributed to their origin sometime during the Medieval Warm Period? Any information on the influence on the stability of the dunes from the underlying artesian aquifer - the Ogallala Aquifer, which has its thickest portion underneath the sandhills and which is used for irrigation in mile-wide circular drip irrigation systems, which has dropped several metres in the past decades. Also, what about tornado alley - can you get sand tornadoes over devegetated portions of the dunes? What type of dunes are they - Barchan, Transverse, Longitudinal, Concave, Parabolic or Star? Is there any evidence of erosion by water on the dunes, and what are their permeability, porosity and density like? What are the differences and similarities between Nebraska's sand hills and other sand hills regions across North America and worldwide?

I'm also willing to contact certain universities and research/environmental organizations who might have more relevant information or current research on the topic, either by email or by phone - can you recommend any groups I might contact?

Thanks. ~AH1 ^(discuss!) 21:57, 15 November 2012 (UTC)[reply]

I found my answer to the Michigan sand dunes at Lake Chicago - by the way, remember that any extra research done here could also go into improving the article(s) discussed. ~AH1 ^(discuss!) 22:10, 15 November 2012 (UTC)[reply]

Answering as to the tornado part, tornadoes in the Nebraska sand dunes will act just as tornadoes anywhere else. As rapidly rotating columns of air, tornadoes are not necessarily visible; the only reason they are visible are if a condensation funnel forms or if they pick up debris from the ground. Most often the debris picked up that makes a tornado visible is dust/dirt (thus how you can have red tornadoes in Oklahoma). Because of this I would expect at least some dust/sand to be in any tornado over the sand hills, but this doesn't make them any different from other tornadoes. Ks0stm ^{(T•C•G•E)} 01:22, 16 November 2012 (UTC)[reply]

I found out from this 32-page piece of research that the Sand Hills are composed of mostly complex transverse dunes, and that the aquifer does indeed indirectly stabilize the dunes: it feeds shallow wetland systems, that in droughts can release evaporation that sustains neighbouring prairie grasses. ~AH1 ^(discuss!) 04:45, 16 November 2012 (UTC)[reply]

Your apparent thesis that aquifer depletion could release these dunes and cause the desertification of the area is most intriguing, but as you've probably gathered by now, the Refdesk is having a lot of trouble rising to your level on this one (myself included). My gut feeling is that if you talk to someone in the right department at one of the colleges closest to this feature (no matter how small), you'll get far more information that we can provide. Though actually one source I found [6] which I think is saying the aquifer is not presently in trouble, as long as precipitation doesn't greatly decrease, was from a Chinese group. Wnt (talk) 18:24, 17 November 2012 (UTC)[reply]

Bok choy is the same species as the common turnip, really? It resembles nothing like a turnip

I know that artificial selection can really bring out the variety in a species, but this amazes me. How is it possible? It tastes nothing like turnip leaves. 71.207.151.227 (talk) 23:32, 15 November 2012 (UTC)[reply]

And yet a Chihuahua, a Newfoundland, a Boston Terrier and a Gray wolf are also all the same species of animal. It doesn't require your belief to be true, you know. The Brassica genus contains hundreds of different vegetables in about half a dozen species, and Bok Choy and Turnips are indeed, to the best of our ability to classify such things, the same species. --Jayron32 23:51, 15 November 2012 (UTC)[reply]

Sub question: Whats the importance or relevance of knowing that bok choy and common turnip are of the same species? (Of course im not asking specifically abouit bok choy and turnip alone) 203.112.82.129 (talk) 00:19, 16 November 2012 (UTC)[reply]

What's the point of knowing anything not immediately useful in daily life, really? But seriously, it could help with people trying to breed new strains to know that they can cross-breed related plants. It helps to keep from wasting time trying to narrow it down from all the other plants in the world. That's the first answer off the top of my head. I'm sure there are plenty of other reasons, too. Mingmingla (talk) 01:50, 16 November 2012 (UTC)[reply]

Trivial knowledge about various beet species provides for good conversational filler between more important, but perhaps less stimulating, quotidien tasks, such as programming iPads. On the statistical average, it helps us focus when we can rest our brains for a little while by analyzing simpler problems like taxonomic minutiae and behaviors of different types of muons. In other words, it's in the national interest to support beet research and discussion, even when the immediate benefits are not apparent. You wouldn't want to remove such stimulating discourse from our society, and through the consequent demotivation, preclude our most creative and productive minds from cranking out iPads at peak efficiency, would you? Nimur (talk) 02:35, 16 November 2012 (UTC)[reply]

Even now, sinister forces are hard at work trying to create an invincible army of bok choynips. Clarityfiend (talk) 01:54, 16 November 2012 (UTC)[reply]

And you'd be right. I used to work at an institute for fruit and vegetable research, and crosses like this are commonplace. Mainly to restore traits lost in the one with traits from the other, like disease and pest resistance. Dominus Vobisdu (talk) 01:59, 16 November 2012 (UTC)[reply]

My mistress' eyes are nothing like a turnip. --Trovatore (talk) 02:03, 16 November 2012 (UTC) [reply]

The knowledge might also be relevant since if you are allergic to one variety there's a good chance you'll be allergic to all. μηδείς (talk) 02:22, 16 November 2012 (UTC)[reply]

According to [7], Brassica rapa is a 6n (triploid) species. "The recurring genome duplications and triplication events have created massive genetic redundancy that quickly opens the possibility of sub-functionalization and neo-functionalization for duplicated or triplicated homeologs (Force et al., 1999; Shruti and David, 2005). It is likely that the extreme morphological diversity seen within the various Brassica species is due, at least in part, to the genetic redundancy and functional diversification permitted by these genomic events." Wnt (talk) 02:47, 16 November 2012 (UTC)[reply]

What Wnt's interesting statement means in less technical terms is that at some point the normal paired set of chromosomes found in the parent of the species got tripled (as if a human somehow got six sets of 23 chromosomes, in stead of just two sets), meaning that some of what were then extra genes got freed up to mutate and develop other functions that otherwise wouldn't have been possible if all the genes were busy serving their original functions. μηδείς (talk) 05:09, 16 November 2012 (UTC)[reply]

And, of course, we have an article on this!--Rallette (talk) 06:54, 16 November 2012 (UTC)[reply]

November 16

Graviton

At the beginning of the article, it says: "the gravitational force appears to have unlimited range"? What is that mean? Let say the Sun's gravitational force and if I'm like a million or a billion light years ago from the Sun, can the Sun's gravitational force affect me? If not then the statement in the article is a false statement.174.20.41.202 (talk) 03:30, 16 November 2012 (UTC)[reply]

Yes, gravitation has infinite range, as far as anyone can tell. The laws of gravitation, either using Newton's law of gravitation or the models of general relativity predict that there is no limit to how far away two objects can be while still influencing one another. It is always the case, however, that beyond a certain distance you will no longer be able to detect the influence of some given object, as the attraction gets weaker and weaker. Someguy1221 (talk) 03:35, 16 November 2012 (UTC)[reply]

Newton's law of universal gravitation tells us the gravitational force between 2 objects is inversely proportional, not to the distance between them, but the square of the distance between them. That is, double the distance and the force diminishes by a factor of 4. Triple the distance, and the force is now only one-ninth what it was, and so on. So, if 2 objects are a billion light years apart, there's still a gravitational force between them, but it's infinitesimally small. Way too small to be measurable. But not technically non-existent. -- Jack of Oz ^[Talk] 06:22, 16 November 2012 (UTC)[reply]

LOL at some distance, it is pretty much like nothing... The concept in unlimited range is pretty mind blowing. 174.20.41.202 (talk) 07:09, 16 November 2012 (UTC)[reply]

That's what I meant by "way too small to be measurable". The difference is that, while we could not measure it empirically at all, we could still work out on paper what the infinitesimally small force would be if we could measure it, assuming the paper was large enough to hold all the zeros in the calculation. -- Jack of Oz ^[Talk] 20:57, 17 November 2012 (UTC)[reply]

You might be a little careful, given that you don't know whom exactly you're talking to, with the word infinitesimal. I take it that you're using it in the informal sense of "very very small indeed", and not in the precise sense of "smaller than any positive real number". --Trovatore (talk) 21:01, 17 November 2012 (UTC)[reply]

There is a limit ultimately, because according to general relativity, fluctuations in gravity actually travel at the speed of light, instead of infinitely fast as Newtonian gravity incorrectly assumes. So gravitating bodies outside of the observable universe have precisely zero effect on us, because the metric expansion of space keeps the gravitational effects of those bodies from ever reaching us. But the distance scales under consideration implicitly in the graviton article when it uses the word "unlimited" are vastly smaller than the diameter of the observable universe. Red Act (talk) 07:29, 16 November 2012 (UTC)[reply]

You are applying gravitational force on a number of sun(s) in other galaxies as well as on these galaxies, but they are not being affected by your gravitational force. It doesn't matter whether they are being attracted or not, but you are attracting them. Sunny Singh (DAV) (talk) 11:26, 16 November 2012 (UTC)[reply]

lepton

There are 6 kinds of lepton, one of them is electron. I know electron flying around the nucleus, which made up of neutron and proton or more precisely quark, in an atom. So where can you find the other 5 kinds of leptons?174.20.41.202 (talk) 07:09, 16 November 2012 (UTC)[reply]

The charged leptons in the second and third generations, i.e. the muon and tauon, decay rapidly, so they don't occur in normal matter, and are only seen in extremely high-energy environments such as cosmic rays or particle accelerators. The electron neutrino would be found in the highest concentration around radioactive matter undergoing beta decay. Neutrinos of all three generations stream throughout the universe, but rarely interact with normal matter. Red Act (talk) 07:47, 16 November 2012 (UTC)[reply]

If you could go with a neutrino, you could travel the universe, too, once you lepton, of course. StuRat (talk) 18:21, 16 November 2012 (UTC) [reply]

W and Z bosons

W and Z bosons are part of the gauge boson. While gluon and photon are massless then how can W and Z bosons have mass? Where are the W and Z boson come from? What emit and absorb them? 174.20.41.202 (talk) 07:23, 16 November 2012 (UTC)[reply]

Read W_boson#Weak_nuclear_force for your second question. Sunny Singh (DAV) (talk) 10:50, 16 November 2012 (UTC)[reply]

The W and Z gain mass from the electroweak symmetry breaking. The W⁺ and W^- are the boson that is involved in beta decay. But all fermions interact with it. A Z⁰ can be produced in any interection that can produce a photon (as long as there is enough energy to form it's mass). Also just to clarify the W and Z are gauge bosons as are photons and gluons. Dja1979 (talk) 17:42, 16 November 2012 (UTC)[reply]

How's this bird called?

What's this bird? Picture: http://i.imgur.com/sDG5w.jpg 109.173.37.164 (talk) 07:31, 16 November 2012 (UTC)[reply]

European Robin (Erathacus rubecula) Richard Avery (talk) 07:58, 16 November 2012 (UTC)[reply]

Thanks. 109.173.37.164 (talk) 12:33, 16 November 2012 (UTC)[reply]

In the UK, it's a symbol of Christmas [8]. Alansplodge (talk) 10:56, 17 November 2012 (UTC)[reply]

Interesting. The Northern Cardinal is the Christmas Bird here in the US. We could export you some to go with the grey squirrels. μηδείς (talk) 19:59, 17 November 2012 (UTC)[reply]

Curiosity image

I think this image of Curiosity (found on found on this BBC page) seems to be a pretty sophisticated simulation - because I can't see anything like a boom extending from the rover out of the image that could be holding a camera.

Right?

Your Username 08:35, 16 November 2012 (UTC)  — Preceding unsigned comment added by Hayttom (talk • contribs) [reply]

It is a montage of pictures, see [9]], got from [10] Dmcq (talk) 10:05, 16 November 2012 (UTC)[reply]

Thanks very much. (There's a really interesting debate there.)

Resolved

Your Username 09:02, 17 November 2012 (UTC)[reply]

Is natural folate metabolized to 5-MeTHF?

"Folic acid" and "folate" are used interchangeably but are not exactly the same chemical compound and have some different properties.

Folic acid is metabolized to 5-MeTHF (aka 5-MTHF, 5-methyltetrahydrofolate, Levomefolic acid) by the human body.

Is natural folate also metabolized to 5-MeTHF?

Thanks. — Preceding unsigned comment added by 134.153.91.186 (talk) 15:31, 16 November 2012 (UTC)[reply]

The distinction between the two is only in dry form or on paper or in solutions when you can set the pH to what you want. Let folic acid, or folate, loose in an aqueous solution, and it can freely exchange away a proton (H+) to any passing water molecule, and pick one back up later on, at a very fast time scale. The counterion determines which form you see on the shelf in a lab, but as an ionic compound the folate and counterion are free to wander in aqueous solution, and in a biological solution there are so many other ions floating around for each to hang out with that they will act independently, never reuniting. Wnt (talk) 17:22, 16 November 2012 (UTC)[reply]

Thanks, Wnt, but I may have given the impression that I know more chemistry / biochemistry than I actually do. Does your answer mean, "yes, natural folate is metabolized to 5-MeTHF"? If I eat broccoli and bread in the same meal, will the folate from the broccoli and folic acid from the (fortified) bread undergo the same processes in my gut and/or elsewhere? — Preceding unsigned comment added by 134.153.91.186 (talk) 22:00, 16 November 2012 (UTC)[reply]

Yes. Wnt (talk) 02:53, 17 November 2012 (UTC)[reply]

What happens to the brains of the enlightened?

Is there some serious study about the effect on the brains of the enlightened (in a spiritual way)? If they feel different something must be different somewhere, I suppose. OsmanRF34 (talk) 17:02, 16 November 2012 (UTC)[reply]

[11][12][13] ... (the secret to getting these results is only to know the keywords "religiosity" and "fMRI" for a Google search, which spit them out as the first three of a long list of similarly relevant results!) Wnt (talk) 17:25, 16 November 2012 (UTC)[reply]

I wanted something more specific. Instead of 'religious' something that's the product of meditation. And instead of anatomy of the brain, something at the level of neurotransmitters and hormones, which is what makes us feel. Unhappily, a simple search for 'neurotransmitter' and 'meditation' pours out a wealth of dubious pages with dubious claims. OsmanRF34 (talk) 18:10, 16 November 2012 (UTC)[reply]

Oh Lord, first you're going to have to define enlightenment. I mean, you have to know who to put in your fMRI machine before you can look for differences in the pattern. Trying "Monk" and "fMRI" I get [14] which talks about soon to be published research by an adjunct professor ... surprisingly, here it is. I'm going to leave it at that for now - my wheels are spinning freely when I try to skim through this one. Wnt (talk) 19:25, 16 November 2012 (UTC)[reply]

Wnt: your "wheels are spinning freely" because you lost track in this and the question below. I never said religious, and I clearly say neurotransmitters and hormones. OsmanRF34 (talk) 21:23, 16 November 2012 (UTC)[reply]

Temporal lobe epilepsy is interesting. μηδείς (talk) 22:57, 16 November 2012 (UTC)[reply]

One of the most active scientists in this area has been Richard Davidson. Our article about him lists a number of his more important publications. Looie496 (talk) 00:08, 18 November 2012 (UTC)[reply]

Sleep more to grow more

I have read somewhere that our height increase only when we sleep. What do you think about this ? Is it true ? Sunny Singh (DAV) (talk) 17:10, 16 November 2012 (UTC)[reply]

Sitting, standing, etc. compresses the spine and reduces height. [15] So normally for most people height will decrease during the day and increase at night. This change in height will be much more than the change due to growth. What I don't know is whether more time in a prone position increases the rate of long-term growth - I can't think of a reliable way to do the experiment to look it up. Also I should add that the suggestion the paper I cited seems to be making about hyperextension as some sort of prophylaxis for occupational stress sounds nuts to me, and from a quick glance I don't think they prove it in any way; I suspect it should have some harmful effect on the discs if any at all. Wnt (talk) 17:34, 16 November 2012 (UTC)[reply]

I think the OP is asking about development from birth to adulthood, not decompression of the spine. The consensus is that sleep is when most growth occurs, and that disturbed sleep can cause stunting. Since sleep is a necessary but not causative factor getting enough sleep in for the days growth is important, but extra sleep won't lead to extra growth. See this article for sleep and puberty, and these for growth during sleep. If your own or a specific person's height concerns you, see a doctor. There are various causes and treatments for abnormal shortness, and abnormal tallness can be a sign of dangerous but treatable conditions. μηδείς (talk) 18:09, 16 November 2012 (UTC)[reply]

I think Wnt was saying it's difficult to study when growth occurs, since it's so overwhelmed by spinal compression. Also, even if you could establish that people who don't get much sleep are shorter, that doesn't necessarily mean the sleep is the causative factor. It might be that whatever stresses keep them from sleeping also stunt growth. StuRat (talk) 18:15, 16 November 2012 (UTC)[reply]

Actually, our article on sleep mentions a study of children finding no correlation between growth rate and the amount they slept. But as StuRat says, there are so many complicating factors... besides, just because the amount of sleep doesn't affect the amount of growth still doesn't prove the growth can't occur mostly when one is asleep. Wnt (talk) 19:13, 16 November 2012 (UTC)[reply]

Reference 10 of Wikipedia article Growth hormone says growth hormone is secreted during sleep. 'Biological regulation' section of the article growth hormone mentions deep sleep as stimulator of growth hormone. I think there is correlation between growth rate and sleep, but I am sure, so I have asked it here to get the correct answer. Thank you. Sunny Singh (DAV) (talk) 00:29, 18 November 2012 (UTC)[reply]

The question is not: if we sleep more, do we grow more. It's only whether we grow while we sleep, which seems to be answered by Medeis above. OsmanRF34 (talk) 21:27, 16 November 2012 (UTC)[reply]

Actually I don't think the question really is answered; it's complicated. Apparently in culture bone formation is enhanced by intermittent and even to some degree constant compressive force (such as one would expect while awake) - see PMID 3505768, PMID 22559784 - yet it also increases apoptosis (PMID 16368547). But the actual process of sleep, rather than inactivity, is a more complex phenomenon that can't be replicated in a dish, and I didn't find anything on the effect of it on extracellular matrix deposition or other effects in cartilage in a quick NCBI search. Wnt (talk) 18:04, 17 November 2012 (UTC)[reply]

I always see people on reference desk using OP while they answer, but I don't understand its meaning. Please, tell me its full form. Sunny Singh (DAV) (talk) 04:36, 17 November 2012 (UTC)[reply]

It is the Original Poster, the Other Person. It comes from internet talk, amounts to a polite genderless way to refer to someone already mentioned. If you see the IP it means the person editing with an IP adress as a user name. In the case above, since you and I haven't really interacted, and since Wnt should realize I was addressing him/her, and talking about you, using OP was not unusual. Had I said "I think Sunny..." it would have implied more certainty on my part in this case. But Sunny would have been fine. I like having more pronouns to use to avoid ambiguity and to make subtler distinctions. Our lack of a formal tu-vos distintion is annoying, but one can say yourself/yourselves and you all. Excluding you request for clarification and my answer, the term is shown used five times on the board as of this edit. In the future, google OP urbandictionary to find out the meaning of any new usage, cutural, internet, adult themed--it's all there. μηδείς (talk) 05:23, 17 November 2012 (UTC)[reply]

Acceleration and time dilation

Okay, about a week or two ago I asked about the maximum G-force a person could withstand and a very helpful user told me we can withstand an acceleration of 1g for months on end and could even theoretically speed up to the speed of light in this manor (neglecting the problems surrounding mass and energy obviously).

Now, my question is this. As you speed up, due to time dilation, time for you, relative to an observer travelling at a constant velocity lower than yours, slows down. Acceleration is a vector quantity and is calculated using time.

If you were in a space ship accelerating they could calculate your acceleration using basic suvat equations. But would they have to take time as the time they experience pass or the time you experience pass. And if it is based on their time can you withstand 1g acceleration by their time or is this measurement based on their time? — Preceding unsigned comment added by 109.153.175.182 (talk) 19:13, 16 November 2012 (UTC)[reply]

1 g refers to the acceleration in the spaceship's frame of reference. Ruslik_Zero 19:37, 16 November 2012 (UTC)[reply]

(EC) Humans are comfortable with a 1g acceleration in their own frame of reference, i.e. a 1g proper acceleration, which is observed as being an acceleration of less than 1g in any inertial frame of reference in which your direction of acceleration is in the same direction as your velocity. You don't use the SUVAT equations, because those are Newtonian. Instead, the relativistic equation of motion for an object with constant proper acceleration is

x^{2}-c^{2}t^{2}=c^{4}/\alpha ^{2}

, as per the article Hyperbolic motion (relativity). Red Act (talk) 19:41, 16 November 2012 (UTC)[reply]

Thank you! This is extremely useful, however I am unfamiliar with this equation. What does the alpha and the x represent?

α is the acceleration, x is the position of the rocket (measured in some inertial frame), and t is the coordinate time measured in the same frame. The elapsed time for the person on board the rocket (proper time) is

\tau =(c/\alpha )\sinh ^{-1}(\alpha t/c)

, where sinh is the hyperbolic sine.

Whoever said that people can withstand 1g for "months on end" was presumably joking. Anyone who lives at sea level withstands 1g acceleration for their whole life. But if, hypothetically, a person could only stand a particular acceleration for a few months, that would be a few months of proper time (tau). You could plug that value of tau into the reverse equation

t=(c/\alpha )\sinh(\alpha \tau /c)

to get t, then plug that into Red Act's equation to get x (which is the distance traveled, more or less—actually, it's the distance traveled plus c²/α). -- BenRG (talk) 01:20, 17 November 2012 (UTC)[reply]

Okay, so what is the absolute limit of acceleration a person could withstand? Also by x being the position of the rocket measured in an inertial frame, do you mean some measurement of distance? — Preceding unsigned comment added by 31.54.166.31 (talk) 23:34, 17 November 2012 (UTC)[reply]

Also which time is T and which is t? — Preceding unsigned comment added by 31.54.166.31 (talk) 23:38, 17 November 2012 (UTC)[reply]

Unfortunately the body's tolerance for acceleration does not seem to be the limiting factor. At 1g, we could get to foreign star systems in quite reasonable (ship's) times, assuming you're willing to spend years in a tin can. But first, no one knows how to achieve sustained acceleration of 1g. And if we did know a way, we still don't know any way to shield the occupants from the interstellar medium, which at relativistic speeds appears to you as hard radiation.

Larry Niven addressed these problems in some of his works (both in Known Space and otherwise) by some fairly optimistic hypotheses about the workability of the Bussard ramjet). Unfortunately, according to our current state of knowledge, those do not appear to be true. --Trovatore (talk) 02:22, 18 November 2012 (UTC)[reply]

Neurons in the cochlear nuclei

How many neurons are there in the cochlear nuclei? — Preceding unsigned comment added by 144.96.215.130 (talk) 21:52, 16 November 2012 (UTC)[reply]

The book Hearing in Children, by J. L. Northern and M. P. Down, gives a total number of 8800, but doesn't make it clear how that number was obtained. Looie496 (talk) 00:04, 18 November 2012 (UTC)[reply]

November 17

Two problems involving angular motion

I do not know how to approach these problems. The first is finding the initial angular velocity/speed of a wheel that rotates 5.3 radians to a complete stop in 3.4 seconds. The second is finding the angular acceleration of a 26 centimeter thick 2.9-kilogram cylinder that has a force of 16 Newtons being applied to it. --Melab±1 ☎ 05:02, 17 November 2012 (UTC)[reply]

The first problem isn't well-defined as stated; what's missing is presumably an assumption that the wheel is subjected to a constant torque, or equivalently, that it undergoes a constant angular acceleration. This problem might be easier for you to solve with a well-chosen change in variables. Think of a movie of the decelerating wheel that's played at the same speed as it was recorded, but backwards. The movie being played backwards will still be 3.4 seconds long, and will still show the wheel rotating 5.3 radians, still at the same constant angular acceleration except for a change in sign. What constant angular accelereration will produce a rotation of 5.3 radians in 3.4 seconds? And then given that value for the constant acceleration, what will be the speed of the wheel at the end of the reversed movie? Red Act (talk) 05:59, 17 November 2012 (UTC)[reply]

The basic procedure for the second problem is to look up the cylinder's moment of inertia at List of moments of inertia, determine the torque from the equation at Torque#Moment arm formula, and then find the angular acceleration from the equation at Angular acceleration#Constant acceleration. Red Act (talk) 06:19, 17 November 2012 (UTC)[reply]

However, we are missing some geometry info on the cylinder. Is that a hollow cylinder ? If so, and it's rotating about it's axis, we need an inside radius or diameter and an outside radius or diameter. Given that we have the thickness, we could find either one, if we had the other. We also need to know if it's made of a uniform density material. The length of the cylinder is not important, since we are given the total mass (it would be, however, if we were instead given the density). And where is this force applied ? Tangent at the outer diameter ? StuRat (talk) 22:32, 17 November 2012 (UTC)[reply]

civil engineering related

why moment of inertia of a hollow circular cross-section is more than a solid circular cross-section? — Preceding unsigned comment added by 49.137.37.116 (talk) 13:14, 17 November 2012 (UTC)[reply]

The weighted average distance from centre of all small mass elements making up a thin ring is equal to its radius, but the wesighted average distance from centre of all small mass elements in a solid cylinder must obviously be less that its' radius, as elements exist from the centre outwards. However, this does not mean that the moment of inertia will be less for a solid cylinder as its' total mass will be considerably greater. See the formulae in the List of Moments of Inertia article. Floda 121.221.79.138 (talk) 13:53, 17 November 2012 (UTC)[reply]

The moment of inertia for a single point is mr^2. Turning that point into a ring, or a cylinder, doesn't matter - you still have the same mass, rotating around the same axis at the same length. But making that ring solid means you take the mass that was at distance r, and distribute it in a column from the center point (moment of inertia zero) to the full distance r. However - the distribution is not even, because there's only one point at the very center. So a small segment of the ring, when distributed toward the middle, forms a narrow triangle, with the average point in the triangle at a distance of r over the square root of 2 away from the center. Twisted any which way, the result is half the moment of inertia. Wnt (talk) 20:49, 17 November 2012 (UTC)[reply]

if the eye changes focal length to focus on an object, why isn't our visual perception like a zoom lens?

I don't notice my visual field at infinity (f=17 mm) being that much wider than my visual field at f=22mm, at the minimum focusing distance, at least not on the order that I would expect a 30% change in focal length to achieve. 71.207.151.227 (talk) 17:45, 17 November 2012 (UTC)[reply]

That's because your eyes are not cameras and your brain is not a .jpg file. Visual perception doesn't really work like a camera. Of course, the laws of optics still apply to the lenses in your eyes, but beyond that how you perceive the electrical signals that get sent from your retina to your visual cortex is very different from what any camera does, and you can't really reliably draw analogies between the two. --Jayron32 19:46, 17 November 2012 (UTC)[reply]

In the eye, the image plane (retina) never moves in relation to the aperature (iris). Only the shape of the lens changes. Since the light moves in a relatively straight line from part of an external feature to a spot on the retina, the external feature remains about the same size. In a camera with a zoom lens, the distance from a lens to the film changes, but the lenses themselves are of solid and unalterable glass. Wnt (talk) 20:33, 17 November 2012 (UTC)[reply]
A zoom lens is not "solid and unalterable glass," but a collection of glass lenses which changes the spacing between elements to alter the focal length. The distance of some elements to the film may change both to alter the focal length and to adjust the focus. Edison (talk) 02:11, 18 November 2012 (UTC)[reply]
But if the focal length changes, the angle of view should change shouldn't it? A focal length is a measure of how strongly it bends rays overall, which in turn affects angle of view. It shouldn't matter by what method the focal length is changed. 71.207.151.227 (talk) 09:51, 18 November 2012 (UTC)[reply]
A more relevant comparison with a camera would be using manual focusing, which does not change the size of the image, but allows you to choose whether the foreground or the background should be in focus. --NorwegianBlue^talk 22:36, 17 November 2012 (UTC)[reply]

Where is the Heisenberg cut article?

I searched for Heisenberg cut but didn't find it. So does it exist as a subsection somewhere or is it just a dead cat? Hcobb (talk) 01:10, 18 November 2012 (UTC)[reply]

There is certainly enough information out there to create an article: see [16]. Wikipedia only exists because people no different than you created and expanded all of the articles. That is, if you find something is missing from Wikipedia, you are literally the best person in the world to add it; or you're no worse than anyone else. So, feel free to create that article! --01:44, 18 November 2012 (UTC)

Voltage across an inductor

I read about the self inductance phenomenon and about the equation V=L(di/dt). But I am confused about the direction of V. My confusion is that current flowing through an inductor causes a 'back-emf', and the back emf is equal to V(or so I think). If so, shouldn't the direction of V be in the opposite of that of I? Also, in the Voltage and Current graph sketch across the inductor, I saw that the phase difference is 90 degree, but the voltage and the current are infact, in the same direction. I just don't get the fact that how V is in the same direction of the current that causes it. According to Lenz's law, it should be in the opposite direction. — Preceding unsigned comment added by 210.4.65.52 (talk) 04:17, 18 November 2012 (UTC)[reply]

Voltage doesn't have a direction; I guess you're talking about the direction of the voltage drop, which is what's meant by V. I don't know if this helps, but in the hydraulic analogy, an ideal inductor is like a heavy but frictionless paddlewheel. Its inertia opposes any attempt to either increase or decrease the current, which leads to a downstream pressure drop (positive V) or gain (negative V) in the case of a forced current increase or decrease respectively. -- BenRG (talk) 07:23, 18 November 2012 (UTC)[reply]

URGENT INFO NEEDED

Dear all, I need to know how thymoquinone actually acts! I mean does it pass through the cell membrane? Does it have a receptor? How does it expert its effect? attaches to a protein to change its activity? ... Any info regarding that will be appreciated. Best kukubah 04:51, 18 November 2012 (UTC) — Preceding unsigned comment added by Kukubah (talk • contribs)

PubChem is a good place to start for stuff like this (see here). Also, note the warning at the top of this page about timeliness - this really isn't the best place to ask urgent questions. Zoonoses (talk) 05:46, 18 November 2012 (UTC)[reply]

Hmmm, oddly enough I'm finding a source that the antinociception and anticonvulsant effects work via the kappa opioid receptor. [17][18] Wnt (talk) 06:04, 18 November 2012 (UTC)[reply]

When did dark energy start to dominate?

There's a graph currently on the Hubble's law article that shows universal expansion. It's hard to read the inflection point, and I don't find it mentioned in the article or any related articles.

That is to say: after inflation ended, the universal expansion was decelerating. Now it is accelerating. When was the critical time when deceleration became acceleration?

(Let's say, treating the dark energy as a cosmological constant... do other proposals imply a different critical time?) — Preceding unsigned comment added by 174.118.1.24 (talk) 08:07, 18 November 2012 (UTC)[reply]

Ignoring radiation (which only matters at early times) and assuming the universe is flat (no global curvature), the evolution of the cosmological scale factor is:

\left({{\dot {a}} \over a}\right)^{2}=H_{0}^{2}\left(\Omega _{M}a^{-3}+\Omega _{\Lambda }\right)

Where

a

is the scale factor,

H_{0}

is the current Hubble constant,

\Omega _{M}

is the fraction of the closure density in mass (including dark mass) and

\Omega _{\Lambda }

is the fraction of the closure density in dark energy. (See also: Lambda-CDM model)

The inflection point occurs at

{\ddot {a}}=0

.

Rewriting the above we get:

{\dot {a}}^{2}=H_{0}^{2}\left(\Omega _{M}a^{-1}+\Omega _{\Lambda }a^{2}\right)

2{\dot {a}}{\ddot {a}}=H_{0}^{2}\left(-\Omega _{M}a^{-2}{\dot {a}}+2\Omega _{\Lambda }a{\dot {a}}\right)

0=\left({{\dot {a}} \over a}\right)\left(-\Omega _{M}a^{-1}+2\Omega _{\Lambda }a^{2}\right)

Given that

{\dot {a}}>0

for all times since the creation of the universe, it follows that inflection occurs at:

\Omega _{M}=2\Omega _{\Lambda }a^{3}

a=\left({\Omega _{M} \over 2\Omega _{\Lambda }}\right)^{1/3}

Using current values for

\Omega _{M}

and

\Omega _{\Lambda }

, gives

a\approx 0.57

. Which implies that the inflection occurred when the universe was about 57% of its current size. Getting the corresponding time will require integrating the equations above with respect to time, but since the expansion was roughly linear, the 57% of size is approximately 57% of time, implying that the inflection point occurred roughly 7.8 billion years after the Big Bang, or roughly 5.9 billion years ago. Dragons flight (talk) 09:36, 18 November 2012 (UTC)[reply]