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July 6

I don't get gravity

It's supposed to apply to everything right? Even light. And yet according to the article on helium it was light enough to fly away from the cloud that the early Earth condensed from. Is helium somehow exempt? —Preceding unsigned comment added by 68.69.73.195 (talk) 01:12, 6 July 2010 (UTC)[reply]

Helium floats, so it'll always be on top of the atmosphere. Maybe you know that the velocity of a atom of gas depends on its temperature? (See Kinetic theory if not.) The hotter it is the faster it moves. The thing is, this speed is only an average - in any given sample of gas, some are moving faster than this, and some slower. It can happen that the faster ones are moving fast enough to escape the gravity (like a rocket). Ariel. (talk) 01:16, 6 July 2010 (UTC)[reply]

I don't think this is actually a question of heat, per se. gravity decreases the farther you get away from the core of the earth. Helium will tend to rise above other gases (it is lighter and less dense), and in the upper reaches atmosphere it will be exposed to the solar wind, which can strip it away from earth's weakened gravity. --Ludwigs2 01:38, 6 July 2010 (UTC)[reply]

I'm not sure it could be that, according to the article at weightlessness gravity is only 6% less at 200 km up. Unless the solar winds are really strong anyway. 68.69.73.195 (talk) 01:40, 6 July 2010 (UTC)[reply]

No, Helium is not exempt. It's just light and flies away due to a combination of two factors mentioned above. 1.) High thermal speed. 2.) Solar wind. Same thing happens to hydrogen, by the way. Dauto (talk) 02:17, 6 July 2010 (UTC)[reply]

According to our atmospheric escape article, atmospheric loss due to solar wimd is not significant for the Earth, on account of our magnetic field. So the primary mechanism for the loss of hydrogen and helium from our atmosphere is thermal loss, also known as Jeans escape. Gandalf61 (talk) 09:57, 6 July 2010 (UTC)[reply]

The simple answer is that gravity is weak, weak, weak, as far as the forces go, but acts over a loooonnng distance. Consider that two magnets from your refrigerator can hold themselves together, suspended in the air, and doing so can contradict the gravitational pull of the ENTIRE planet. But their force diminishes quickly with distance, whereas gravity can have effects for hundreds of millions of miles distant. Gravity is good at doing things like keeping the planets aligned, and for roughly keeping heavy things like you and me on the face of the Earth (and even then, with just a little effort we can liberate ourselves from it, at least temporarily). But when you talk about things that are really light—like gases—other forces can be a lot stronger on a local scale, even if gravity rules at planetary and galactic scales. --Mr.98 (talk) 12:59, 6 July 2010 (UTC)[reply]

Gravity certainly applies to everything with mass. But lighter things float on denser things just as wood floats on water. Hydrogen and helium (being the least dense gasses) naturally float to the top of the earth's atmosphere. Hydrogen usually doesn't make it that far without running into some oxygen atoms and turning into water - but helium is a noble gas and doesn't react with anything, so it can reach the upper atmosphere fairly easily. The solar wind is a fairly significant force at those altitudes and is quite capable of defeating gravity and gradually blowing away the upper layers of the atmosphere out into space...which makes helium a fairly rare commodity.

The only helium we have is that which is created by radioactive processes and becomes locked underground in natural gas deposits. Ironically, although helium is one of the most abundant substances in the entire universe, it's quite rare down here on earth - and we're actually in danger of running out of the stuff sometime in the next 200 years. SteveBaker (talk) 13:02, 6 July 2010 (UTC)[reply]

We can also produce Helium using the Fusion_power#D-T_fuel_cycle. 157.193.175.207 (talk) 13:28, 6 July 2010 (UTC)[reply]

So we only need to make our helium stocks hold out for another 50 years? -- 58.147.52.199 (talk) 13:56, 7 July 2010 (UTC)[reply]

There's probably still some primordial helium left on Earth, but not much, as evidenced by sources that are rich in helium-3: see this review at p. 735 (p. 53 of the PDF file). Physchim62 (talk) 18:05, 6 July 2010 (UTC)[reply]

Yes - but both natural sources and fusion-reactor by-product are truly microscopic quantities compared to industrial needs. A 100MW fusion power plant, operating continuously, would take something like 300 years to produce a kilogram of helium (I hope I got my math right! But that sounds about right to me.) - all of the future fusion power in all of the world wouldn't make enough helium to keep the Goodyear blimp flying...let alone provide inert gasses for welding, birthday balloons for your kids, etc!

Primordial helium is certainly going to be rare because it leaks out into space. Naturally occurring helium is a by-product of natural breakdown of radioactive substances in the ground and ocean - which is indeed being slowly replenished. But think about that for a moment: If it takes just a few hundred years from first human exploitation of helium to when we are likely to run out of the stuff - then it would take a billion years for natural processes to replenish what we use in just a hundred years...and that assumes that the rate of radioactive decay is constant over all time - which it certainly isn't. It's safe to say that when we've used it all up...we're screwed. Other resources (such as copper) are becoming alarmingly scarce too - but at least pretty much every scrap of copper we've ever used is sitting in land-fill someplace waiting to become valuable enough to dig up and recycle. But when you 'consume' helium, it's gone. VERY gone! The nearest place to get more is the Sun...and there are considerable engineering difficulties associated with that! SteveBaker (talk) 19:22, 6 July 2010 (UTC)[reply]

The sun might be closer, but I bet it would be easier to harvest the helium from Jupiter, which also has plenty. Could we use tritium to produce light helium (He-3)? I mean tritium is expensive, but is it more expensive then strolling over to Jupiter to pick up some Helium? Googlemeister (talk) 19:43, 6 July 2010 (UTC)[reply]

We're all missing the Archimedes principle - Helium is less dense than air, therefore as bundled in a balloon it rises as long as the weight of that balloon doesn't exceed the weight of air that the helium pushes out of the way? It can also be imagined in terms of higher air pressure below the balloon and lower pressure above it, or multiple other concepts, but they are all fundamentally the same physical process. The stuff about molecular collisions has to do with gases that are intermixing, where buoyancy, as the balloon effect is described as, has little effect on a microscopic scale. Of course, the atmosphere gets less dense as you go higher, and the temperature and pressure changes at different layers, so you have to recalculate the effect of buoyancy continuously. That said, simple helium weather balloons can fly to over 10km in height. SamuelRiv (talk) 09:31, 9 July 2010 (UTC)[reply]

Why does an acidic oxide reacts with water to form an acid?

And why does an acidic oxide reacts with a base to form a salt?

What is the principle behind them? Thank you. 114.247.10.148 (talk) 04:30, 6 July 2010 (UTC)[reply]

I assume you mean metal oxides and non-metal oxides. A non-metal oxide consists of a non-metal bonded via covalent bonds to oxygen. The addition of water generally created an oxyacid, which consists basically of hydroxide groups now being attached to the nonmetal. Since the covalent bond to the oxygen withdraws some of the electron density away from the H-O bond, it makes the H very easy to leave. The definition of an acid is something that loses H⁺ ions to solution. Thus, non-metal oxides produce acids. A metal oxide consists of metal ions and oxide (O^2-) ions held together by ionic bonds. When the O^2- ions come into contact with water, it tends to remove one of the hydrogen atoms from the water, producing two hydroxide (OH^1-) ions. The definition of a base is something that produces hydroxide ions in solution. Thus metal oxides produce bases Its basically the difference in reactivity between a covalently bonded oxygen atom, and the oxide ion. --Jayron32 05:07, 6 July 2010 (UTC)[reply]

The farther to the left on the periodic table you go, the more basic the oxides are. The farther to the right, the more acidic. Also, higher valence oxides such as chromium(VI) oxide tend to form acidic solution. Lower valence chromium(III) oxide is only very weakly acidic. --Chemicalinterest (talk) 10:51, 6 July 2010 (UTC)[reply]

Thank you. I see the reason why non-metal oxides produce acids, not something else. But I start to wonder why non-metal oxides react with water, rather than being nonreactive? Thanks. 114.247.10.133 (talk) 12:44, 6 July 2010 (UTC)[reply]

Not all non-metal oxides react with water: take carbon monoxide as an example of a neutral oxide. But that's really something which is in the "character" of each oxide. Physchim62 (talk) 12:55, 6 July 2010 (UTC)[reply]

Hmm... Maybe on some of them the bonds are strong between the nonmetal and oxygen so they wouldn't accept a hydrogen and as a result, wouldn't dissolve in water to form an acidic solution. Very few nonmetal oxides are unreactive toward water, though. --Chemicalinterest (talk) 13:10, 6 July 2010 (UTC)[reply]

It depends on what you mean by "react". Aldehydes such as formaldehyde for example, are hydrated by water (they form hemiacetals and acetals). It all comes down to whether your compound/element is an electrophile or a nucleophile. Lewis acids tend to be electron-withdrawing / electrophilic; Lewis bases, electron donating / nucleophilic. John Riemann Soong (talk) 16:04, 6 July 2010 (UTC)[reply]

You with your organic chemistry again... over my head --Chemicalinterest (talk) 16:31, 6 July 2010 (UTC)[reply]

Rifled slug?

Copying the attached pic from a few threads above this one. Why would a shotgun slug be rifled? I didn't think shotguns were rifled. Comet Tuttle (talk) 05:56, 6 July 2010 (UTC)[reply]

There are rifled shotguns designed specifically for slugs. 68.69.73.195 (talk) 06:11, 6 July 2010 (UTC)[reply]

http://www.rogueturtle.com/articles/shotguns.php

Normal bullets are not rifled. The "rifled" slugs do not, to my understanding, cause significant spin. As mentioned above, modern slug guns often have rifled barrels, but are more often used with sabot slugs. The part of the article at Foster_slug#Foster_Slugs says that these fins on the slug were more about reducing friction than imparting spin. Friday (talk) 15:05, 6 July 2010 (UTC)[reply]

I'm no expert but might it be that the 'fins' cause spin due to aerodynamic forces AFTER the slug has left the barrel? That would make sense because you don't need fins to make a bullet spin when the bore is rifled and since so many shotguns are smoothbore, it would make perfect sense to put fins on a shotgun shell for the odd occasions where you want a more accurate shot and less scatter. SteveBaker (talk) 18:48, 6 July 2010 (UTC)[reply]

There is no scatter at all from a slug - it is a solid chunk of lead. Rmhermen (talk) 20:40, 6 July 2010 (UTC)[reply]

Statistical scatter perhaps (multiple shots?)

Actually the links above note that the centre of gravity of such a slug is close to the tip (like a shuttlecock) so there's (probably) not need for a spin to be imparted to prevent tumbling, since the drag does that. It also claims the fins are there to reduce friction in the barrel, and that there is no spin on the slug in flight. That's what it says, I'm just repeating it.77.86.6.186 (talk) 21:00, 6 July 2010 (UTC)[reply]

If the fins are just to reduce friction - surely you'd want them to be parallel to the direction of flight? SteveBaker (talk) 00:12, 7 July 2010 (UTC)[reply]

That would increase the edge length for potential abrasion though? from just the leading part of the fin (and to a lesser extent the shallowly angled sides) to multiple circular edges. (Assuming that edge abrasion is a larger contributor to gun fouling than side abrasion - since the sides are more solidly supported - 2pi steradians of support vs pi steradians of support). Spiraling the fins ensures even distribution of support around the circulat cross section of the barrel compared to hypothetical straight fins.

According to book mentioned here the fins impart a slight spin of 1 turn in 24feet - I'm fairly certain this is not effective cf spin distance mentioned in Rifling (though much mass is on the outside so the rotational inertia is greater than a similar solid bullet). There's also an video here [1] (might be a solid slug, not clear).. but it may not convince either way.

Ignoring all the above the reason to have non-circular fins is swaging - see ~~Swage#Firearms_and_ammunition~~ (not that link - doesn't mention gun chokes see [2] instead) - It would be great way to acchieve consistent jamming if the slug wasn't perfectly lined up. So longitudinal fins are needed, and non-twisted longitudinal fins are not the best solution for reasons explained above.

This gives rise to a bullet that looks like it's designed to rifle.. However if you think the description on the page Foster slug is wrong I'd take it up there. It's not difficult to find sources that say they are rifled to make them more stable if you look though; I'm suggesting it's just consequence of other design parameters.

It would be interesting if anyone can be bothered to calculate the stabilisation acchieved from the slow rate of spin that is imparted, and say whether or not it is actually significant.77.86.6.186 (talk) 00:39, 7 July 2010 (UTC)[reply]

I'm fairly certain the number one factor behind the design here is making a shell that goes through a Choke (firearms) with no problems.77.86.6.186 (talk) 01:55, 7 July 2010 (UTC)[reply]

It's possible to assemble a thousand opinions on this subject. Here's one

"Heavy external "rifling" was cast into these Foster type slugs, allegedly to allow the air they flew through to impart a slow spin that would help stabilize the slug. Like most something for nothing schemes, the rifling proved ineffective, but it did provide some space for some compression if the slug had to squeeze through a tight choke.."[3]

77.86.6.186 (talk) 02:01, 7 July 2010 (UTC)[reply]

Duh - just realised that's a Brenneke slug and not a Foster slug - the shuttlecock effect doesn't apply to the brenneke slug.77.86.6.186 (talk) 02:08, 7 July 2010 (UTC)[reply]

Seeing as it's a Brenneke slug I can link this:

"The Brenneke design with angular ribs is not to create spin, but to ensure problem-free choke passage" www.brenneke-munition.de

That's the company website, hopefully it can be taken as fact.77.86.6.186 (talk) 02:15, 7 July 2010 (UTC)[reply]

Hyrachyus

Hi, I wish to know where lived the Hyrachyus? Because from this two articles I can't find out, where lived this animal.

http://en.wikipedia.org/wiki/Hyrachyus

and

http://en.wikipedia.org/wiki/Rhinocerotidae#Evolution

One say Europe and the other North America. Please answer. —Preceding unsigned comment added by 89.123.165.191 (talk) 13:24, 6 July 2010 (UTC)[reply]

Here's a cite for Hyrachyus in North America: [4] (it's in there somewhere). There might also have been Hyrachyus in Europe, not sure. That 1883 report mentions Lophiodon, a European cousin. There's a document here: [5] which I think says "Lopliiodon is limited to Europe; Hyrachyus, Hyracotherium and Pliolophus to Europe and North America" ... but I'm not allowed to read it unless I want to pay thirty quid. I had to extract those words from google in two searches. I think it might be volume III of the same report, which is also available here [6], but I can't find those words in it. However, it does say "Hyrachyus is the American Lophiodon, the difference between them being but slight; both are found in France ; the former in the Lower Parisian, the latter in the Phosphorites." So that's pretty definite: the answer is both. (Not sure how that jives with Laurasia being already separated a million years previously?) 213.122.47.102 (talk) 14:28, 6 July 2010 (UTC)[reply]

Thanks anyway! —Preceding unsigned comment added by 92.86.247.63 (talk) 10:30, 7 July 2010 (UTC)[reply]

Hawaii

Would Hawaii's climate be considered tropical, or sub-tropical? Googlemeister (talk) 16:45, 6 July 2010 (UTC)[reply]

Probably tropical according to File:ClimateMap_World.PNG. --Chemicalinterest (talk) 17:51, 6 July 2010 (UTC)[reply]

The state as a whole is tropical, but the actual climate varies quite a bit. Indeterminate (talk) 20:25, 6 July 2010 (UTC)[reply]

Cooking Pizza

Not sure if this belongs here or the Misc. desk, but I think this is the best place to start. Buying a pizza from a supermarket and cooking it at home is drastically cheaper than buying one ready-cooked from a pizza takeaway, I don't know why, but it is. However, a pizza cooked at home is never as good as that from a shop - but why? I am sure that the method of cooking is the problem here. When I cook a pizza, the centre is lukewarm but the edges look they've been incinerated but pizzas cooked in a shop are crisp and even.

I know the pizza is slightly thinner towards the edges but we're talking less than a centimetre so I don't see why the difference should be that profound. I also follow the cooking instructions to the letter - put on the top shelf of an oven and cook for 12-14 minutes at 200C. After 12-14 minutes, the outside is just about right but the middle needs at least another 5 minutes (minimum) otherwise it is barely cooked at all. By this time, the edges are hard, dry and burnt. I don't think it's a problem with the oven as I have experienced this with other ovens. I know pizza shops have special ovens but that can't be that much of a factor.

Deep-pan pizzas are just about edible after this cooking ordeal (if you leave the crusts) but thin crust ones? Forget it, the crust is so hard it's virtually inedible.

Is there any way I can cook a pizza more evenly? I would assume I am doing something wrong, due to a. the fact that pizzas for home cooking are available and b. the picture on the box always looks perfectly cooked. Would cooking it more slowly on a lower heat help? Covering up the edges in foil? Anything else? —Preceding unsigned comment added by 94.196.158.83 (talk) 18:01, 6 July 2010 (UTC)[reply]

First, frozen pizzas have a number of constraints that make them less than edible (they are usually pre-baked and frozen, and they need to be very cheap. They are also much thicker than real pizza). Fresh pizza is made from unfrozen dough. Secondly, real pizza ovens are a lot hotter than 200°C, and the pizza is placed on a hot stone, cooking from the top and the bottom very quickly. Oh, and thirdly: Do you thaw the pizza before cooking? --Stephan Schulz (talk) 18:19, 6 July 2010 (UTC)[reply]

(EC) I must admit, I do not usually have these problems when cooking ready made pizzas except the crusts on the cheap frozen thin crust pizzas does become quite hard, but out of curiosity, are we talking about the frozen ones, or ones that you can find that are refrigerated, which I find are usually higher quality and more expensive? One other possibility, are you at a high elevation (1500+ m)? Googlemeister (talk) 18:22, 6 July 2010 (UTC)[reply]

Thanks for the reply. This seems to happen with the shrink-wrapped non-frozen pizzas from the supermarket (which they apparently make in-store from 'fresh' ingredients and yes, they are more expensive) as well as the boxed and shrink-wrapped frozen ones, although the frozen ones seem to cook slightly more evenly than the non-frozen ones for some reason (I usually cook these from frozen as that is usually what the instructions say). I am not at a high elevation, barely above sea level. I thought the frozen ones cooking better might have something to do with humidity (the burnt edges are very dry). 94.196.158.83 (talk) 18:27, 6 July 2010 (UTC)[reply]

Pizza stone says you can buy a small one, or use a bunch of unglazed tiles. 213.122.64.116 (talk) 18:33, 6 July 2010 (UTC)[reply]

I believe it has more to do with the mode of cooking than with the source of the pizza dough or whether it's frozen, thawed or fresh. My wife (who is an excellent cook) has made pizza completely from scratch and we've had the same disappointing problem of incinerated edges and half-cooked middles. Making 'thin-crust' style pizza seems a little easier than regular or 'deep-dish' styles - but still, it's tough to get it right 100% of the time even with thin-crust.

What seems to help is a proper pizza cooking tray. Ours is a round metal plate with a slightly concave surface and a few hundred half centimeter holes punched in the center third or so. The idea is (presumably) to let the radiant heat from the oven pass through the holes and thereby cook the center of the pizza a little faster than the edges. This seems to be partially succesful...but still, I have to say that it's hard to beat delivery pizza on crispness and crust texture (although it's entirely tivial to do vastly better on taste with fresh ingredients, proper spices and not spending an hour being kept lukewarm in a delivery pouch).

It's frustrating because a combination of the delivery crust style with homemade toppings would be a truly awesome thing to behold.

...and please stop asking these questions on days when I've just missed lunch!

SteveBaker (talk) 18:43, 6 July 2010 (UTC)[reply]

Surely the little holes are to allow steam to escape? --Sean 21:19, 6 July 2010 (UTC)[reply]

(1) It's a frozen pizza. It is contrary to all the Laws of Science that it will ever (a) taste as good as an average restaurant-made pizza, or (b) look as good as the picture on the box (that would also violate the Laws of Advertising); or (c) cook as good as the instructions lead you to expect -- IIRC, this is the Law of Frozen Pizza Imperfectability (where are those Wikipedia articles when you actually need them?); (2) The old "Red Baron" frozen pizzas had a metalic-like inside cover that customers were told to fold up and under the pizza before putting it in the oven (or maybe the microwave -- maybe it wasn't quite metalic, but it looked like it was -- it's been a long time since I reformed from my evil ways; as I recall, it cooked little mini-pizzas well enough); (3) the hallowed Frank Pepe's Pizzeria, when a new generation of the family ownership decided to do the unthinkable and expand to different locations than the one on Woooster Street in New Haven, Connecticut, arranged to have each new oven built to match the original, ancient oven at Pepe's on the unscientific understanding that you don't mess with success, so that's anecdotal, unscientific evidence that the oven must have something to do with it although this is pretty mystical for a science reference desk, I admit (see Frank Pepe Pizzeria Napoletana#The oven) (3) You do realize, I hope, that there's a special circle in hell reserved for frozen-pizza buyers and sellers? It's in Dante's Inferno somewhere, and I'm told modern translations just don't do it justice. Something about getting shoved inside a brick pizza oven. Or that might have been in The Godfather; (4) try a cookie sheet under the pizza. I encase it with aluminum foil first, shiny side up (at least it's easier to clean). Still doesn't work quite right. Usually, we buy it from a pizzeria; (5) Please realize that scientists all over the world have been working on this problem for decades, and while they haven't solved it yet, side benefits have included the microwave oven, the Tang drink and those diapers that look like blue jeans. Rest assured that while science sometimes progresses slowly (cure for the common cold, anyone?) it does progress. -- JohnWBarber (talk) 18:51, 6 July 2010 (UTC)[reply]

By the way, pizza ovens are sold to homeowners. -- JohnWBarber (talk) 18:57, 6 July 2010 (UTC)[reply]

The difference is basically in the oven: you can't cook proper pizzas in a domestic oven, and the one's you buy in shops are a compromise that lets you get something edible (and lets remember that pizza, like sex, is often disappointing but only very rarely is it truly bad) The official pizza napoletana is cooked at 485 °C for 60–90 seconds... Physchim62 (talk) 19:04, 6 July 2010 (UTC)[reply]

I've never tried his pizza, but Alton Brown has never let me down on anything else. The transcript and recipe from the Good Eats pizza show is here. He uses an unglazed quarry tile to bake on. There are a number of (to me) surprising bits (such as allowing a very long cooled rise for the dough), but he bases his recipes and instructions on food science and tests them empirically before airing them, so I would tend to believe him. As far as fresh/frozen, the problem is likely with the dough being 1) of poor quality and 2) blind baked for too long, making it difficult for the end user to balance the sogginess with the, um, burniness. Matt Deres (talk) 19:38, 6 July 2010 (UTC)[reply]

A very long cooled rise is generally the default for people who want to maximise the tastiness of their yeast product. In my experience, you can make good pizza with a short rising too: the only real difference is in the 'grain' taste of the crust, but I'm guessing you eat bread that hasn't been slow-risen, so I don't think you'll notice anything bad. A cool, slow rise also means you can start the dough and leave it overnight or while you go to work: if you don't want to do that, you can rise it in a cool (< 50 C, in a bowl covered with a towel, oven turned off) oven in less than an hour. 86.164.57.20 (talk) 21:00, 6 July 2010 (UTC)[reply]

A perpetual motion machine

Gravity is a force, and as such it does work. If I drop a stone from my window, the Earth's gravity will exert a force of roughly 10 newtons per second on it to bring it down.

But then, is gravity really ethernal, or does it eventually run out?

If gravity lasts forever, that means any piece of rock can exert an infinite amount of force, and thus an infinite amount of energy. Wouldn't it be a perpetual motion machine then?
If gravity does not last forever, when is it calculated to "run out"?

I'm just having a hard time reconciling the idea of an ethernal gravity with the second law of thermodynamics, and I'd like someone to explain this to me. 88.1.139.43 (talk) 19:18, 6 July 2010 (UTC)[reply]

It is not perpetual motion though. Eventually, the rock will stop when it hits something and you would need to exert energy to lift the rock back up to drop it again. Basically, the rock has already stored energy and it is using it up while falling, but it will not fall forever. Googlemeister (talk) 19:31, 6 July 2010 (UTC)[reply]

Phrased alternately, this is the distinction in physics between force and energy (or work). You say "...an infinite amount of force, and thus an infinite amount of energy", but this incorrectly equates "force" and "energy". Rather, energy is (roughly) force multiplied by distance. If you want to turn that 10N of force into 10J of energy, you have to move the rock one meter (as the joule is a unit of Newton-meters).

Gravity is a force, but a force in and of itself doesn't do anything. A PM machine, on the other hand, has to be able to do infinite work, that is, the transfer of energy. As Googlemeister has already noted, gravity can cause a rock to fall -- once. Then it reaches a point of equilibrium (the surface of the Earth, the center of the black hole, the center of mass of the combined rock-attractor system, whatever) and it ceases to do any more work until you spend external energy to move it. Gravitational force is still there once the rock stops doing work; it doesn't need to "run out" for useful work to end. — Lomn 19:49, 6 July 2010 (UTC)[reply]

Basically agreeing with the other responses but saying it a bit differently: gravity only does work on things that move downward. It can't do infinite work because there isn't an infinite amount of down. If you move the object back up to where it started, gravity does negative work while you're lifting it, so there is no accumulation. Looie496 (talk) 20:15, 6 July 2010 (UTC)[reply]

A stone lying on a window ledge has potential energy. While falling the potential energy of the stone is converted to kinetic energy. When the stone has fallen as far as it can go, it hits something and its kinetic energy suddenly converts into sound and heat. Gravity is still there but a stone with no place to fall has no energy at all. I think you mean "eternal". Cuddlyable3 (talk) 21:16, 6 July 2010 (UTC)[reply]

Actually, your stone can "fall" forever, achieving a stable orbit around a planet for example (we omit interplanetary dust and solar wind for simplicity). However, this will still not make it a perpetual motion machine: if you try to extract energy out of it, you will slow it down / change its orbit to a lower one.--131.188.3.21 (talk) 23:41, 6 July 2010 (UTC)[reply]

Moreover, no orbit is truly "stable", not in terms of thermodynamics. — Lomn 02:24, 7 July 2010 (UTC)[reply]

The OP may check the height of their window against the article about Low Earth orbit. Cuddlyable3 (talk) 13:30, 8 July 2010 (UTC)[reply]

I'm sorry, we don't stock actual perpetual motion machines, but we do have a fine selection of these other models that might serve your purpose: Gravity can give you practically a perpetual motion machine in the form of a hydroelectric (or mill) dam, as long as the water doesn't run out (Mother Nature brings it uphill for you). Harnessing tidal forces might be more "perpetual". A geothermal pump is practically a perpetual heating/cooling machine (although you have to actually supply some motion for the water or liquid to circulate -- hook it up to your dam), and a solar panel (or other device dependent on energy from the sun) is practically a perpetual electricity-making or heating machine -- in space, anyay -- until you get an eclipse or the sun dies. (I think some utilities use solar and wind power to pump water back up into the reservoir of a hydroelectric dam and there's an idea out there about pumping air into a cave or mine and releasing it later, saving energy until the sun doesn't shine and the wind dies down, a more tenuated "perpetual motion" machine.) Of course, you're simply taking energy from some massive forces of nature (we always are) and these will all die out eventually, but eventually we'll all be dead. But you didn't actually want a machine? You just wanted an explanation of a concept? Sorry, we don't stock those either, but this week our models are all on sale ... -- JohnWBarber (talk) 23:42, 6 July 2010 (UTC)[reply]

All of your suggestions only work for very small values of "perpetual", of course! Unless you manage to answer The Last Question. --Stephan Schulz (talk) 08:31, 7 July 2010 (UTC)[reply]

The problem here is a confusion between a 'force' and 'energy'. This confusion is so common - and the question comes up here so frequently, that I've written an answer on my personal Wiki HERE so I don't have to keep repeating my answer every couple of weeks! I recommend you go and read that. SteveBaker (talk) 04:58, 7 July 2010 (UTC)[reply]

Without having read steve's answer yet where I am quite sure it would be mentioned, magnets is the other thing which is frequently subject to exactly the same confusion. Vespine (talk) 02:07, 8 July 2010 (UTC)[reply]

Stellar parallax and the movement of the sun

Stellar parallax is a way to measure distances to stars, comparing their position in the sky when earth is at opposite points in its orbit. This gives a distance of 2 AU. But the solar system is moving at 220 km/s around the galaxy, which is about 46 AU per year.[7] How does this square with the stellar parallax method? Are all nearby stars approximately at rest relative to our sun? EverGreg (talk) 19:38, 6 July 2010 (UTC)[reply]

For the most part yes. The fastest star near us, Kapteyn's Star, moves on the order of about 0.001 c relative to the sun and is 13 ly distant. Googlemeister (talk) 19:49, 6 July 2010 (UTC)[reply]

Put in perspective, if the solar system is moving with respect to a target star at 46 AU per year, that's less than .000001 c. However, I'm not sure that's really the point here. In the time it's taken the Earth to complete half an orbit (to allow the 2 AU parallax), the solar system has moved about 23 AU, which is to say, 0.00036 light years. On the other hand, we're using data from the Hipparcos satellite to measure stellar distances of up to 1600 light years, a factor difference of about five million. So the motion of the solar system is messing with parallax measurements by thousandths of a percent, which is probably well within the margin of error. Even Proxima Centauri, the closest star, would have its distance mis-measured by only one part in 12000 if we didn't account for the motion of the solar system. So it's not whether nearby stars are at rest relative to us as much as a few dozen AU being insignificant when you're talking light years. — Lomn 19:57, 6 July 2010 (UTC)[reply]

Note that it's pretty easy to check on this. All you have to do is repeat the measurement when the Earth comes back to its starting point. If the distant star is moving enough to cause a problem, you will get a substantially different result on the second measurement. Looie496 (talk) 21:07, 6 July 2010 (UTC)[reply]

Are all nearby stars approximately at rest relative to our sun? Yes, in fact this 220 km/s frame is called the local standard of rest. -- Coneslayer (talk) 11:25, 7 July 2010 (UTC)[reply]

Hmmm - interesting - so I have a followup question: At what time of year does the earth move in the same approximate direction as the sun's motion around the galactic center? SteveBaker (talk) 23:31, 6 July 2010 (UTC)[reply]

December. Physchim62 (talk) 23:51, 6 July 2010 (UTC)[reply]

So right now, we're moving slower than average relative to the galactic core? Cool! SteveBaker (talk) 04:40, 7 July 2010 (UTC)[reply]

Bass energy sucking

Is it possible for a bass drum to suck my bass energy from my bass cab and make it sound bad if they are too close? —Preceding unsigned comment added by 88.104.83.202 (talk) 20:29, 6 July 2010 (UTC)[reply]

It would definitely resonate or buzz when bass notes were played, and that sound would likely degrade the quality of sound you heard in the room. So it could easily make the bass cabinet sound different. I just knew that some good would come out of having a bass drum occupying space in the living room. I tried putting a large bass drum in front of a speaker and playing classical or rock music with the bass turned up. I could feel the drumhead vibrating, but couldn't detect much difference in the sound. "Bad" is in the ear of the listener, since musicians often want distortion in a guitar amplifier. Edison (talk) 20:41, 6 July 2010 (UTC)[reply]

Yeah, I'd expect that a bass drum could attenuate the sound energy, but not that it would significantly alter the characteristics of that energy. Beyond that, there's no "energy sucking". You might get lousy sound quality, though, if the bass guitar is playing a note that's near, but not at, the note played by the drum -- but that's just being out of tune, not anything specific to these two instruments. Similarly, the room might have lousy acoustics for that frequency range, but again, nothing particular to the instruments themselves. — Lomn 21:22, 6 July 2010 (UTC)[reply]

But doesn't a bass drum have a fairly low Q factor, so that it can absorb energy over a reasonably wide frequency range and thus distort the bass from my bass cab? —Preceding unsigned comment added by 88.104.83.202 (talk) 21:52, 6 July 2010 (UTC)[reply]

If the bass drum absorbed acoustic energy, should it convert the energy to heat, and either radiate heat or have its temperature rise? Edison (talk) 04:47, 7 July 2010 (UTC)[reply]

Ok then, say I had an unenergised bass cab next to my active bass cab, would my bass energy be sucked by the passive cab? —Preceding unsigned comment added by 88.104.91.80 (talk) 00:02, 8 July 2010 (UTC)[reply]

What you call energy sucking is just the absence of normal room acoustics. Ways to experience your bass cab (I assume this means "loudspeaker cabinet") without room reflections are to play it outdoors or in an Anechoic chamber. These have far more noticeable effect than a single item such as nearby drum or cabinet. If you noticed a bass loss you would presumably increase your volume to compensate, and would hardly call the effect "bad". However there is another possible effect from a resonator that posesses both high Q and non-linearity. This might buzz at certain frequencies. Cuddlyable3 (talk) 13:51, 8 July 2010 (UTC)[reply]

So what about these? http://www.customaudiodesigns.co.uk/basstraps.htm

Sprite freezing

Have a question guys. I just took a sprite out of my mini fridge (pretty much a freezer) and when I opened it the liquid froze from top to bottom.

What is this magic?

I thought it might have something to do with the pressure being released, but doesn't high pressure raised a liquid's freezing point. 74.15.137.192 (talk) 22:04, 6 July 2010 (UTC)[reply]

Pressure tends to lower the freezing point of a liquid. However, what I think is more likely is that the Sprite was supercooled. In fact, our article notes that Sprite was, at one time, specifically sold in the UK in a supercooled state such that it turned to slush when opened. Supercooled liquids tend to freeze rapidly when they're disturbed, which is why opening the can/bottle caused the freezing -- both the physical shock of opening the can and the pressure change associated with it. A similar, and significantly more dangerous, phenomenon is superheating, which is relatively easy to accomplish with water in a microwave and can lead to severe scalding. As such, it's recommended that you not try to boil water in a microwave unless you've added some rough surface (such as a wooden spoon) to the container so that superheating won't occur. — Lomn 22:18, 6 July 2010 (UTC)[reply]

Pressure tends to raise the melting points of liquids – except for the essential exception that is water. Otherwise, I agree completely, the phenomenon seems to be supercooling. As you release the pressure, you get tiny bubbles of carbon dioxide forming, and these act as nuclei for the surrounding liquid to freeze around. Physchim62 (talk) 22:47, 6 July 2010 (UTC)[reply]

You're absolutely correct; I should have said "water-based liquid" (as I expect Sprite to be sufficiently water-like for water's general pressure-temperature relationship to hold). — Lomn 02:21, 7 July 2010 (UTC)[reply]

Is it possible that the unopened Sprite doesn't freeze because of the dissolved carbon dioxide, and the sudden loss of CO2 upon opening changes the freezing point? I would imagine that this has a small effect, given that there is a ton of sugar dissolved, and still considerable amounts of CO2 remaining in solution, but it may contribute. What if there was a tiny imperfection in one of these supercooled cans? If it was sitting in a store for weeks, wouldn't it be possible to turn to slush before opening?24.150.18.30 (talk) 00:57, 7 July 2010 (UTC)[reply]

I would expect that most stuff isn't stored supercooled, though -- generally any soft drinks I buy are at room temperature in the store, and I therefore see no reason that they'd be stored otherwise in the warehouse. In the case of the sold-supercooled link above, I expect that the Sprite was only supercooled once it was in the vending machine, protected from further disturbance. — Lomn 02:21, 7 July 2010 (UTC)[reply]

I've seen it happen with bottled water, so dissolved CO2 is not required. (It's a neat trick. Makes you feel like a super hero.) APL (talk) 03:22, 7 July 2010 (UTC)[reply]

Unless you didn't want it to go slushy, and just put it in the freezer to cool fast then forgot about it and thought you pulled it out in time but ooops maybe not Nil Einne (talk) 06:04, 7 July 2010 (UTC)[reply]

Thanks, everyone, for the quick replies. I've actually always been interested in questions like this. I don't know what the field this would be called (maybe physical chemistry?), but can anyone recommend some books on this subject? Or are there standard university courses that teach material like this? 74.15.137.192 (talk) 02:55, 7 July 2010 (UTC)[reply]

This is probably general chemistry; see if the chemistry article has anything good. Also take a look at Portal:Chemistry. Any introductory chemistry course should carry the concepts, including high school and college books. --Chemicalinterest (talk) 20:48, 7 July 2010 (UTC)[reply]

I don't understand quantum mechanics!

I was reading the article "uncertainty principle" in an attempt to find out why it is not possible to know a particle's position and its momentum at the same time. The probability that a particle is in a certain position is given by the amplitude of the wave packet at that point (correct?). It also says that to obtain an accurate reading of position, this wave packet must be 'compressed' as much as possible, meaning it must be made up of increasing numbers of sine waves added together (what does this mean?), and that the momentum is proportional to the wavelength of one of these waves, but it could be any of them. —Preceding unsigned comment added by 203.22.23.9 (talk) 22:38, 6 July 2010 (UTC)[reply]

First, don't be ashamed you don't understand it. It's genuinely hard and unintuitive when you start. Here is my attempt to break it into relatively easy concepts — it is extremely simplified but hopefully will get the gist across and is, I hope, not so wrong that it will be just laughed at by those who have deeper understandings of it. (Or, if it is as bad as that, hopefully they will write up similarly straightforward corrections!) I am not a physicist but I dabble in its history and concepts.

As for waves. If you combine two waves together, they can either destructively or constructively interfere. See the diagram here. That part is easy. If you add two waves together of different frequencies, you can get interesting results, like so. The bottom sine waves have been added together and constructively/destructively interfere to make the wave packet on top.

Now imagine that we wanted to make that wave packet one infinitely small wave—just a single high point with nothing around it. Well we could keep on adding more and more sines together and we'd get something that looked more and more like a single wave in the middle of calm. A nice illustration of that is figure 7.4 on this page.

Now in QM, the wave packet's amplitude is the probability of finding a particle in a particular region of space. So at the parts of the sine wave that are high, you have a high chance of finding the particular. Now the issue is that if you want that uncertainty to be zero—you want to know exactly where it is—that means you have to add up a very large number of waves, so that you have basically one giant spike in the middle of a bunch of nothing. The more you confine that wavepacket, trying to nail down the position perfectly, the more waves it has to be made up of. As you add more waves to it, you are adding more uncertainty about the momentum of the wavepacket as a whole. So you can get the position down perfectly—no problem. But what do you lose at the same time? You lose the ability to know the frequency of the wave, the more you nail down its amplitude into a tiny space. That's the momentum issue.

Does that help at all, at least with the understanding of what the waves are supposed to indicate? The take-away message of the wavepacket description of UP for me is that UP is not a problem of measurement in and of itself (it is not just that when you physically "touch" one property, it physically modifies the other); it's a fundamental property of the universe, if the various properties of the universe (like a particle's position and momentum) are correctly described as these sorts of wave forms. --Mr.98 (talk) 23:40, 6 July 2010 (UTC)[reply]

I struggle, too, but I like Mr 98's explanation. You might be interested in these Wikiquotes. Dbfirs 07:46, 7 July 2010 (UTC)[reply]

But how does one "add" waves, in practice, by taking measurements? —Preceding unsigned comment added by 203.22.23.9 (talk) 03:41, 7 July 2010 (UTC)[reply]

He refers to the Fourier decomposition of the wavefunction.

Anyway, say we have an electron. To measure its position we can bounce photons off it (i.e. shine a light). The accuracy of the measurement is roughly the photon's wavelength. Photons have energy and momentum, so the resulting collisions destroy information about the particle's momentum. As p = hc/λ for photons, we can only improve precision in position at the expense of momentum. MER-C 10:44, 7 July 2010 (UTC)[reply]

Right, but that's actually not a good explanation of UP, because it makes it look like it is just an artifact of the measurement system, whereas the wavepacket model makes it clear it is a fundamental property of the universe. It's not that we don't measure it well—it's that the information is simply not there. (At least in the Copenhagen interpretation.) UP as put forward by Heisenberg was originally a measurement issue; Bohr is the one who said, "ah ha! It's actually about the fundamental limits of physical knowledge." Which is a much more interesting and important interpretation, and the wave approach is the only way I really know how to illustrate that. (It also makes more clear why EPR and Bell's theorem are important, and why it is not just position/momentum that are at issue, but lots of other states as well.)--Mr.98 (talk) 13:17, 7 July 2010 (UTC)[reply]

One reason why quantum mechanics is difficult to understand is that several arguments and concepts, which should be kept apart, are put together.

The cornerstone of quantum mechanics is that energy $\scriptstyle E$ , measured in joule, and frequency $\scriptstyle \nu$ , measured in hertz, is the same quantity. The conversion factor is the planck constant $\scriptstyle h$ , having the dimension of joule per hertz. So $\scriptstyle E=h\nu$ .
Now what is frequency? The function $\scriptstyle \psi (t)={\overline {A}}e^{2\pi i\nu t}$ has a well defined frequency $\scriptstyle \nu$ . (See Circular motion#Description of circular motion using complex numbers). This function satisfies the differential equation: $\scriptstyle {\frac {d\psi }{dt}}=2\pi i\nu \psi$ or $\scriptstyle \nu \psi ={\frac {1}{2\pi i}}{\frac {d\psi }{dt}}$ . 'Multiplying by the frequency $\scriptstyle \nu$ ' is thus identified with 'applying the operator $\scriptstyle {\frac {1}{2\pi i}}{\frac {d}{dt}}$ '.
Fourier series like $\scriptstyle \psi (t)=\sum _{\nu =-\infty }^{\infty }{\overline {A_{\nu }}}e^{2\pi i\nu t}$ (where $\scriptstyle A_{\nu }=\int _{0}^{1}{\overline {\psi (t)}}e^{2\pi i\nu t}dt$ ), or Fourier integrals like $\scriptstyle \psi (t)=\int _{-\infty }^{\infty }{\overline {A(\nu )}}e^{2\pi i\nu t}d\nu$ (where $\scriptstyle A(\nu )=\int _{-\infty }^{\infty }{\overline {\psi (t)}}e^{2\pi i\nu t}dt$ ), do not have well defined frequencies. Still the operator $\scriptstyle {\frac {1}{2\pi i}}{\frac {d}{dt}}$ applies to such functions, thus generalizing frequency.
Combining these points gives that (multiplying by) the energy $\scriptstyle E$ is identified with (applying) $\scriptstyle {\frac {h}{2\pi i}}{\frac {d}{dt}}$ .
The differential operator $\scriptstyle {\frac {d}{dt}}$ , and the multiplication by $\scriptstyle t$ , do not commute: $\scriptstyle {\frac {d(t\cdot \psi )}{dt}}-t\cdot {\frac {d\psi }{dt}}=\psi$ , or, isolating the operator: $\scriptstyle ({\frac {d}{dt}}t-t{\frac {d}{dt}})\psi =\psi$ , or shortly: $\scriptstyle {\frac {d}{dt}}t-t{\frac {d}{dt}}=1$ .
So time and frequency does not commute either: $\scriptstyle \nu t-t\nu ={\frac {1}{2\pi i}}$ .
Nor does time and energy commute: $\scriptstyle Et-tE={\frac {h}{2\pi i}}$ .
Scaling time up means scaling frequency down: if $\scriptstyle \psi (t)=\int {\overline {A(\nu )}}e^{2\pi i\nu t}d\nu$ then $\scriptstyle \psi (kt)=\int {\overline {A(\nu )}}e^{2\pi i\nu kt}d\nu$ $\scriptstyle ={\frac {1}{k}}\int {\overline {A({\frac {\nu }{k}})}}e^{2\pi i\nu t}d\nu$ .
The argument that the commutation relation $\scriptstyle \nu t-t\nu ={\frac {1}{2\pi i}}$ results in the uncertainty principle $\scriptstyle \sigma _{\nu }\sigma _{t}\geq {\frac {1}{2}}$ is complicated, but scaling up time and scaling down frequency conserves the uncertainty product: $\scriptstyle ({\frac {1}{k}}\sigma _{\nu })(k\sigma _{t})=\sigma _{\nu }\sigma _{t}$ .
The uncertainty between time and frequency is a mathematical fact dealing with fourier transforms. Only after multiplication by planck's constant does it become the quantum mechanical uncertainty relation between energy and time: $\scriptstyle \sigma _{E}\sigma _{t}\geq {\frac {h}{2}}$ .
The case $\scriptstyle \sigma _{E}=0,\sigma _{t}=\infty$ describes a stationary state, having well defined energy but not defining a point of time. The case $\scriptstyle \sigma _{E}=\infty ,\sigma _{t}=0$ describes a quantum mechanical starting pistol defining a point of time but having no precise energy.
Substituting the differential operators for the physical quantities in the classical mechanical equation $\scriptstyle E=H(p,q,t)$ (where $\scriptstyle E$ is the energy, $\scriptstyle t$ is the time, $\scriptstyle p$ is the momentum, $\scriptstyle q$ is the space coordinate, and $\scriptstyle H$ is the Hamiltonian function), produces a quantum mechanical equation, $\scriptstyle {\frac {h}{2\pi i}}{\frac {d}{dt}}\psi =H({\frac {h}{2\pi i}}{\frac {d}{dq}},q,t)\psi$ , the Schrödinger equation.

Hoping this is helpful. Bo Jacoby (talk) 22:11, 7 July 2010 (UTC).[reply]

No offense meant, but how could the above be helpful to someone who doesn't understand how you add waves? I don't mean to be glib but unless you already have extensive understanding of the mathematical methods, formulations, and concepts of QM, the above might as well be in Chinese. It doesn't illuminate anything—unless you already understand everything—and it certainly won't help the original poster. --Mr.98 (talk) 22:24, 7 July 2010 (UTC)[reply]

Point taken. The OP had two questions:

The probability that a particle is in a certain position is given by the amplitude of the wave packet at that point (correct?). Well, the absolute square of the amplitude, actually.
it must be made up of increasing numbers of sine waves added together (what does this mean?). It means that the sound of a starting pistol does not have a well defined frequency.

Bo Jacoby (talk) 08:11, 8 July 2010 (UTC).[reply]