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March 26

recycled paper

I went to an office supply store tonight to get paper for my printer. In their store brand, they had regular paper, 30% recycled, and 100% recycled. Otherwise the descriptions were pretty much the same. The 30% recycled cost somewhat more than the regular paper and the 100% recycled cost nearly twice as much! Does using recycled paper really cost them that much more? If so, is recycling paper economically feasible? We recycle paper, plastic, glass, CFBs, batteries, and old electronics - is any of it worthwhile economically? Bubba73 ^{You talkin' to me?} 00:41, 26 March 2013 (UTC)

You've ignored the other half of the pricing equation, which is demand: are people willing to pay more for recycled paper. If they are willing to pay more, they will be charged more, regardless of what it costs to make. That is, a company will not charge less for a cheaper product if people are willing to pay more for it. Companies aren't in the business of making less money. --Jayron32 00:45, 26 March 2013 (UTC)

Perhaps you are right, but if recycled paper cost them less, they could charge the same for it and make more money. Of course, probably not, if they can get away with it. Bubba73 ^{You talkin' to me?} 00:58, 26 March 2013 (UTC)

The idea that they could charge the same for it and make more money isn't borne out by the evidence. If they could make more money by charging less they already would be doing that. The Price point for any product is a heavily researched concept, and one that retailers are constantly studying and experimenting on and gathering data about. It isn't random in any way, this is pretty much the major thing that ALL retailers do CONSTANTLY (at least, if they want to stay in business). The evidence that they are maximizing their profit at a particular price point is that the price point has remained consistent across time and space: The fact that all retailers use such price points (recycled paper is more expensive in nearly all stores) and it has pretty much always been that way. If you want to understand how pricing works, look at the pricing of bottled soda. A 2 liter bottle of Coca-Cola in a grocery store runs between $0.99 and $1.50; a 20 ounce bottle of Coca Cola in the same store, but being sold in the cooler by the checkout line is between $1.59 and $1.79. Why? Because will pay that. Obviously, the smaller bottle of soda is NOT more expensive to produce. The price of an object is very tenuously tied to the cost of producing it. The production and distribution costs set the floor for the price, but there is no ceiling at all: the ceiling is the price that will maximize profits. --Jayron32 02:27, 26 March 2013 (UTC)

If using recycled paper cost them less than regular paper, then they could charge the same price and make more off the recycled paper than the regular paper. Bubba73 ^{You talkin' to me?} 02:34, 26 March 2013 (UTC)

Why? If they can charge more and make more money with lest customers (that is, if the increase in revenue from customers who will still buy the paper at the higher price offsets the lost revenue from the customers that won't) then why would they charge less? That would make them less money. Look, consider these example, just making up numbers. Lets say that regular paper costs $1.00 per ream to make, and they sell it for $5.00 per ream. That's $4.00 per ream in profit. Let's, for the sake of argument, say that recycled paper costs $0.90 cents per ream and they sell it for $5.50 per ream. That's $4.60 per ream in profit. Now, lets say that 1000 people buy the regular paper, and 800 people buy the recycled paper. That's $4000 in regular paper and $3680 for the recycled paper. Now, let's say they lower the price of the recycled paper to the same price point. Now we've lowered the profit to $4.10 per ream. Let's say that now 100 people switch from regular to recycled, so now there's 900 buying each. That's $3600 for regular and $3690 for recycled. Under the older pricing scheme, the company made $7,680 in paper sales. Under the new pricing scheme, the company made $7,290 in paper sales. So, it isn't advantageous to lower the price of recycled paper because you make less money. Real world examples are going to be more complex than this, but this at least demonstrates that lowering the price of a product to sell more of it does not always make you more money, even if that product is cheaper to make. --Jayron32 02:47, 26 March 2013 (UTC)

Maybe they can charge more for recycled paper, but if it cost them less, they don't have to. They can charge the same price as regular paper and make more per unit. And if regular paper and recycled paper are the same price on the shelf, I think more people would chose the recycled paper, which could drive down the cost more, but that is another consideration. The bottom line is if recycled paper cost them less, and they sell it for the same price, they will make more per unit on recycled paper. Bubba73 ^{You talkin' to me?} 04:17, 26 March 2013 (UTC)

... But keeping the Coke cool is more expensive. And the customer may want one that is already cold because they want to drink it soon. Bubba73 ^{You talkin' to me?} 02:38, 26 March 2013 (UTC)

The second reason, and not the first, is your answer. --Jayron32 02:47, 26 March 2013 (UTC)

So, not taking the profit motive into consideration, can a company use recycled paper cheaper than cutting down trees? And what about the other things that are recycled? Bubba73 ^{You talkin' to me?} 02:07, 26 March 2013 (UTC)

If we want to apply microeconomics to this, an assumption should be made that the higher price does in fact reflect higher costs. There isn't a lot of differentiation between various forms of the same level of recycled paper, so in the long run the price is equal to the Average Total Cost. The price of recycled paper is more expensive in the long run so the average total cost must be higher. Ryan Vesey 02:26, 26 March 2013 (UTC)

Not necessarily. See my example above about bottled soda. The price will, of course, be affected somewhat by costs, insofar as no company will sell an item at a loss. HOWEVER, if a company can charge more for a product and make more money, they will. 1) If the potential loss of some customers at the higher price point is offset by the increased revenue from the customers who are willing to pay more, companies will charge the higher price. This has nothing to do with costs. 2) Some products actually sell better (i.e. move more units) at higher price points. Premium pricing involves using artificially high prices to increase the "stature" of a product and thus sell more units to people who buy it only because it is expensive. That is, lowering the price of such a product would actually move less units because people will have the perception that lower price = lower quality. This ALSO has nothing to do with actual costs. --Jayron32 02:30, 26 March 2013 (UTC)

The bottle example above isn't analogous. When you're dealing with pop, you're dealing with monopolistic competition where firms are price setters rather than price takers. When you're dealing with paper, you're in a perfectly competitive market and firms are price takers. If they refuse to accept the price decided by the market, their product will not be purchased. In the perfect competition, the long run price is at Average Total Cost so the only deciding factor in the long run price is changing costs. That being said, this is a bunch of economic bunk and requires a way of thinking that doesn't necessarily mesh with reality. Ryan Vesey 02:38, 26 March 2013 (UTC)

Yes, most people want a particular brand of soft drink. I don't know of any brand loyalty when it comes to a ream of paper for the printer, but there probably is some. Bubba73 ^{You talkin' to me?} 02:40, 26 March 2013 (UTC)

Anybody who doesn't believe that prices will rise to what consumers are prepared to pay, and doesn't believe in premium pricing just needs to look at few good examples:-

• I worked for a while in a building next door to the Nevada Shirt Company (now driven out of the market by cheaper Asian product), and got to know some of the girls working there. They were into a form of badge egineering, that is, they made much the same shirts using the same material on the same machines, but with several different brands and logos sewed on. Obviously the cost per shirt to the factory was the same in all cases. Some some logos were cheap brands eg Glo-Weave and some were more premium brands like Piere Cardin. The price you paid for a shirt in retail shops reflected the brand image.
• Some years ago, I was shown around a razor blade factory (to learn about their quality control). They made two lines - a cheap line and a premium line. The only difference was the branding on the blades, and the wording and colours on the packaging.
• In the 1970's, Mercedes cars sold small but steady numbers of cars here in Australia, for about 3 times the cost of a Holden (local GM variant roughly equivent to Chevy). But it was a MUCH better car, lasting about twice as long, 4-wheel independent suspension, full function heating and aircon, very quiet and many other features the Holden didn't have. Worth it if you could afford it. By the 1980's though, GM/Holden had caught up, and today Holdens last just as long and have all the features of a Merc. So what did Mercedes Australia do? They substantially increased their price, thus reinforcing the concept of exclusivity and something you can have because you are better than the ordinary peasant. It worked - their sales increased. It's a bit sad that GM's Australian division have never made such a good product as they do now, yet sales are falling - due to brand perceptions.
• Up until the 1970's pretty much all clothing sold in Australia was made in Australia, protected by an import tarrif. In 1974 there was a change in Federal Government, and the newly elected Govt (Labor Party) ended the tarrif protection, on the theory that it was causing high prices by preventing overseas competition. Somehow they thought that local manufactuers would improve their act under competivive pressure. Didn't happen. What happened is retailers started importing direct from cheap Asian suppliers (who sewed on the same brands and logos), rapidly forcing all the local manufacturers out of business. Did the retailers lower prices because they paid less money to factories? Of course not! People were accustomed to paying so much for a shirt, pants, or whatever, and the retailers kept on charging that much.

Wickwack 58.164.230.22 (talk) 03:11, 26 March 2013 (UTC)

All very good examples. Thanks for providing those. --Jayron32 03:18, 26 March 2013 (UTC)

This is elementary economics (not "a bunch of economic bunk"), and I'm surprised that nobody has linked to supply and demand. In particular, look at the first graph on the article, which illustrates exactly the scenario we're talking about. At any given price, consumers will want to buy a certain quantity of paper, given by the demand curve D1. At any given price, suppliers will be able to produce a certain quantity, given by the supply curve S. The intersection of D1 and S gives the price and quantity of paper exchanged in the economy.

Now, consider what happens if suppliers decide to sell recycled paper instead. If it costs the same as normal paper to produce, S wouldn't change. But consumers are more willing to buy recycled paper because it helps the environment, so at any given price, quantity demanded goes up. The demand curve shifts right, from D1 to D2, and intersects S at a higher price. Therefore the price of recycled paper is higher, even if the cost of production is equal. --140.180.254.209 (talk) 03:46, 26 March 2013 (UTC)

You're dealing with the short run. If firms are making profits, more firms are going to enter the market increasing supply. The price will be reduced. Again, long term price in a perfectly competitive market is dependent on ATC. Ryan Vesey 03:49, 26 March 2013 (UTC)

He's also dealing with a simple two-class market, comprising of only sellers and buyers. Real markets aren't that simple - theres's usually at least one layer of middle men, and often more. Take my example of Australian clothing prices above. Asian factories entered the market, which might be expected to drive prices down. They did, but not at the retail level - the shops just simply took a greater profit, knowing what the consumer was accustomed to pay. Economic/pricing theory is a bit like psychology - there's a bit of sound theory, but you are dealing with human buyers, and humans often don't do rational things. Like those who buy a $120,0000 Mercedes when a $35,000 GM/Holden is just a good, technically - but the GM doesn't say to the neighbours "look at me - I draw such a good salary I can waste it on image." Mercedes increased their sales by substantially increasing their prices. If GM did that sales would most likely nose-dive. Wickwack 58.164.230.22 (talk) 04:09, 26 March 2013 (UTC)

All of this discussion of pricing and economics is beside the point of my question. The question is about whether or not producing paper from recycled paper is more or less expensive than producing it from regular paper. And the question applies to other recycling as well. Bubba73 ^{You talkin' to me?} 04:21, 26 March 2013 (UTC)

This isn't the most scholarly source I've ever found, but it states that the extra steps required in the production of recycled paper increases the costs. Ryan Vesey 04:29, 26 March 2013 (UTC)

(edit conflict) [1]. --Jayron32 04:30, 26 March 2013 (UTC)

Bahaha, I love it. Ryan Vesey 04:39, 26 March 2013 (UTC)

Oh, haha. Looks like Google worked well for both of us. LOL. --Jayron32 04:40, 26 March 2013 (UTC)

(e.c. I haven't read the above yet.) Here's another case. We saved up aluminum cans for a couple of years and then I took them to be recycled and got a little money. We did this twice. Both times the gas for me to take the cans to the recycle center was about half of what I got for the cans. And that isn't counting other car expense, the environmental impact, and my time and trouble. Is is worth it - economically and envionmentally?

Similarly, when we lived in the Atlanta area, the county charged people to pick up their recycling. Is it worth it if they don't get enough out of the recycled materials to pay for the cost?

Now we put things in a recycle bin which is picked up every two weeks. Can the county send one of those big trucks down my street to pick up what is probably a few cents worth of recycling at each house? And what about the environmental impact of the truck, etc? Bubba73 ^{You talkin' to me?} 04:42, 26 March 2013 (UTC)

Thanks - that link is to the point of my question regarding paper. Bubba73 ^{You talkin' to me?} 04:46, 26 March 2013 (UTC)

Another consideration with products made from recycled material is that some organizations have policies that require the use of recycled products. I used to work for a government agency which was under the directive of a law that required the use of recycled products wherever one was available. The administrative staff had to order recycled paper for the office regardless of its higher cost. — Preceding unsigned comment added by 148.177.1.210 (talk) 14:38, 26 March 2013 (UTC)

Maybe the idea is that we have to do more recycling to get the cost down to where it does pay off. Bubba73 ^{You talkin' to me?} 16:45, 27 March 2013 (UTC)

Chemical reaction and water vapor related questions

Some questions related to chemical reaction and water vapor. Feel free to answer any question.
1. Is there any method (or trick) by which one can predict what will form after a reaction? Consider the following two reactions-

          Cu + H2SO4 → 

          C2H6 + O2 → 

How can one know what product will form after these reactions?
2. Which is the most reactive metal and which is the most reactive non-metal?
3. Clouds hover in the sky (carrying huge amount of water) even when water has higher density than the air. What is the reason behind this?
4. Why does water vapor go up in the sky even when water has higher density than the air? They should remain on the earth. Yellow Hole (talk) 08:32, 26 March 2013 (UTC)

I can offer an answer on question 3. Cloud is a colloid (or mixture) of microscopic droplets of liquid water dispersed among the millions of molecules of nitrogen, oxygen and other gases comprising the air. Even though these microscopic droplets of liquid water are much denser than the surrounding air, the force exerted on each one by Brownian motion (violent impacts from colliding gas molecules) is much greater than the weight of the droplet, so the weight becomes insignificant. Any droplets that do wander out the bottom of the cloud (into air of less than 100% relative humidity) promptly evaporate so this process is invisible. If and when the interior of the cloud cools sufficiently, droplets increase in size and coallesce with others. When this process passes a critical point, large water droplets accumulate and begin falling because their weight becomes much more significant than the forces due to Brownian motion. When these large water droplets emerge from the bottom of the cloud we say it is raining. Dolphin (t) 08:55, 26 March 2013 (UTC)

.

In regard to questions such as what happens when you allow Cu + H₂SO₄ (raeacting a metal and an acid) to react, chemists learn that this is a ceratin class of reaction, and rules apply that lead you to Cu + 2H₂SO₄ -> SO₂ + 2H₂O + Cu⁺⁺ + SO4^-- at ordinary temperatures. Ratbone 120.145.20.58 (talk) 10:21, 26 March 2013 (UTC)

In regards to what happens when you react C₂H₆ with O₂, i.e., combusting a hydrocarbon fuel (Ethane in this case) the problem is a lot more complex. The rate of reaction and what results you get is very highly temperature dependent, especially at low temperatures for rate, and high temperatures for products. For complete combustion, you'd need not C₂H₆ + O₂ as you've written, but C₂H₆ + 3.5O₂ -> 2CO₂ + 3H₂O, but this will only happen at low temperatures, and does not occur in a single step, but in a multi-step chain reaction. In general, for complete combustion of a fuel CnHm into steam and carbon dioxide, you need obviously need n + 0.5m of O₂. At very high temperatures, what you will get is a misture of mostly CO, O₂, O, H₂, and H.

The combustion of hydrocarbons is not fully understood at present, though good progress is being made. An approximate answer can be made by identifying a writing the individual steps in the chain reaction, and the modified arrhenius equation coeficients (if known) for each step, and all the possible products. Then solve the set of reactions over time. As this involves a large number of non-linear simultaneous equations that don't usually converge very fast, considerable computer time is required. And I do mean considerable. Fortunately, published data for arrhenius coeficients exist for most reaction steps, more gets published as time goes on, and we are at the stage where non-critical reaction steps can be guessed and overall accuracy is still tolerable. The rate of any single step gas phase chemical reaction may be predicted by the modified arrhenious equation:-

R = a.T^b.e^[-c/RoT] [A]ⁿ[B]^m

where R is the rate, a, b, and c are constants (generally determined by measurement), T is temperature, and [A] and [B] are the concentrations (gas partial presssure) of the two reactants A and B. Ro is the universal gas constant, 8.3143 kJ/kmol.K. Constant C is often termed the Activation Energy. n and m for this application are integers corresponding to the species formula prescripts. Considerable enginuity is required to measure the constants and, just to make it interesting, they generally vary with temperature.

To give you a feel of the problem, I list below the most important chain steps for a much simpler combustion reaction, xH₂ + yO₂. For maximum accuracy 136 reaction steps need to be considered.

The products of reacting hydrogen H₂ and oxygen O₂ are-

H, H₂, O, O₂, O₃, OH, HO₂, H₂O, and H₂O₂.

In theory, there are an infinite series, but products not in the set of 9 above only occur in less than parts per million and can be ignored.

The most important reactions (out of a total of 136) between these nine products are:-

H+O2>OH+O

H+OH>H2+O

H+H2O>H2+OH

H2+O>H+OH

H2+O2>HO2+H

H2+OH>H2O+H

Using only these 6 reactions in a calculation will result in considerable concentration errors. Calculations will be within 10% over a wide temperature range if the most critical 23 equations are used. Included in the 23 equations are reactions of the form X + Y + M > Z + M where Z is a combination of X and Y and M is each of the 9 species, with different arrhenius coefficients for each species. Al these reactions proceed simultaneously at different and changing rates until equilibrium is reached.

In one sense, the case of reacting H₂ and O₂ is simple, as H2, O2 and the other 7 species are all transparent. So there is no black body absorption of light, adn no emision of light. As soon as you add carbon atoms, and carbon is an ideal black body, you get both the emission of light, and light accelerating the reactions. This can be handled by tweaking the arrhenious constants, but the problem is knowing how much tweaking is needed.

In short, to find the products of reacting a hydrocarbon fuel with oxygen, we can say what it is for complete combustion is (its steam and carbon dioxide), but if anyone says to you he can work out a practical case with a page or two of figuring, he's having you on.

Ratbone 120.145.20.58 (talk) 09:25, 26 March 2013 (UTC)

If only the end-point concentrations of the products is required, then there is another way, at first glance much simpler: You can use the technique of dissociation mathematics. This relies on the fact that any chemical system will head for relative concentrations that have the lowest energy at the given temperature. In this technique, one writes out the princple reactions between the expected products, and from them derive equilibrium reactions. Knowing the thermodymanic properties of each product (you can find them in published tables for most products, and calculate for the ones not published), you can then balance the equations. This is then an application of solving a set of non-linear simultaneous equations. The trouble comes because typically the solution converges extremely slowly. Common numerical methods don't work, giving false convergence dure to rounding error. Again you need a lot of computer time. Personally, I find that it is all too easy to make mistakes writing the equilibrium reactions, causing much waste of time solving simultaneous equations that are not valid. By contrast, the modified arrhenious method given above, while tedious, is easily checked at each step and thus proceeds error free. Ratbone 124.178.132.17 (talk) 02:03, 27 March 2013 (UTC)

.

In answer to (4), water vapor mixes with and rises up through the air because it is not liquid water but a gas, that is individual molecules darting about. Liquid density has no relavence for a gas. Any substance in liquid form (eg water pooled on the ground) has at any given temperature a vapor pressure. Molecules leave the liquid in order to establish the vapor pressure. In a sealed rigid container containing nothing but H₂O, at temperatures high enough to prevent freezing (~0 C) there will be a portion of water in liquid form and the remainder in gasseous form. Just enough will be a gas in order to establish the vapor pressure for that temperature. There cannot be a full vacuum. Vapor pressure rises sharply with temperature, so at high temperaures there will be more gasseous H₂O and less liquid in the container. In the atmosphere the H₂O molecules mix thoroughly with the air accodring to the laws of diffusion, driven by brownian motion, until low temperatures at altitude cause the water to precipitate out to form clouds. Ratbone 120.145.32.100 (talk) 11:07, 26 March 2013 (UTC)

As to number 2, it would depend on how you define "reactive", and depends on ambient conditions and what the substance is reacting with. Cesium has the lowest first ionization energy (table here), which makes it a good candidate for the most reactive metal. Below 28C, Cesium is solid and is less reactive than when it is in liquid form. NaK is also very reactive in terms of ionization potential and remains liquid down to -11C. (Both will react violently with water.) Fluorine is perhaps the most reactive non-metal, both because it has the highest electronegativity (except for neon, I suppose, but neon is inert) and because it spontaneously reacts with nearly every other element.--Wikimedes (talk) 11:39, 27 March 2013 (UTC)

Why does light follow straight line?

We have a number of proofs regarding the speed and path followed by light. My question is why does light always follow straight line and its speed in vacuum is constant? 106.215.97.55 (talk) 08:50, 26 March 2013 (UTC)

Science makes observation of the universe, and the natural world. Science rarely addresses the question why? For example, science observes that light mostly travels in straight lines; and the speed of light is constant in a vacuum, regardless of latitude, altitude etc. Theologians might be interested in speculating why it is so, but scientists are not. However, scientists are interested in making observations to find conditions under which light does not travel in straight lines; and in trying to find conditions of vacuum in which there is a discernible difference in the speed of light. Dolphin (t) 09:07, 26 March 2013 (UTC)

Actually, science continualy asks, as a famous scoffer of chocolate (http://en.wikipedia.org/wiki/Julius_Sumner_Miller once said multiple times at each appearance, "Why is it so?".

It can be proven mathematically that the speed of light in a vacuum can be calculated from the permitivity and permeability of free space, which are constants, and by other mathematical means.

In saying light travels in straight lines, what we really mean is that the wavefornt is not tilted. Tilting can only occur where there is a transition from one permitivity or permeability value to another - this can only happen if what the light is passing through is not a vacuum. Light does travel in curved lines when passing through material that has a gradient of permitivity or permeability. Ratbone 120.145.32.100 (talk) 11:48, 26 March 2013 (UTC)

I agree that the speed of light in a vacuum can be determined by taking account of the permitivity and permeability of free space, and nothing else, but that doesn't answer the question "why is it so?" Scientists have little interest in why vacuum displays this particular property. Theologians might say that vacuum is a concept created by God for the benefit of mankind, and that God dictated all the properties that vacuum will display. However, scientists are not much interested in such an explanation. Dolphin (t) 12:28, 26 March 2013 (UTC)

That's because it offers nothing, it offers no understanding. It's just saying it is because it is. What's the good of that? You can say scientists are not interested in why as much as you like - that doesn't make it right. Ratbone 120.145.32.100 (talk) 13:11, 26 March 2013 (UTC)

Saying "God made it that way" is the same thing as saying "the forces of nature made it that way." Hence, as you say, it adds no new facts. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:56, 26 March 2013 (UTC)

The problem is that people have two different things they mean when they use a word like "why". One of them is "by what mechanism did this come to be", as in "Why are the mountains on Earth located where they are?". Science can totally answer those questions. The other use of "why" is "For what purpose is it this way", for example "Why am I here?" or "Why are the properties of the universe the way they are, rather than some other way?". Those are fundamentally unanswerable by science because they don't present falsifiable concepts. Science can't give you purpose or meaning. It can tell you how things work, just not for what purpose (even if there is no purpose). Such questions must be answered by other methods.--Jayron32 16:01, 26 March 2013 (UTC)

There are no "other methods" that can discover what purpose a mountain has if it has absolutely no purpose. If something does have a purpose--for example, if an aspect of the universe was designed by an intelligent entity--that has observational consequences, and would most definitely be a scientific question. There's simply no sense in which science cannot answer "why" questions. The most we can say is that we don't have enough knowledge to answer some "why" questions, but saying they can never be answered is arrogant and an insult to future generations. --140.180.254.209 (talk) 17:35, 26 March 2013 (UTC)

There are many non-falsifiable propositions, and questions like "Why am I here?" does not have falsifiable answers. That is, even if your answer is "You have no purpose", that answer itself is nonfalsifiable, and as such, not testable. There are many nonfalsifiable things we deal with as humans all the time. Questions of asthetics or faith, for example, are not really falsifiable. That one person prefers to listen to death metal and another likes Mozart is not determined by any laws of science, and not subject to falsifiable experiments. --Jayron32 20:25, 26 March 2013 (UTC)

"You have no purpose" is most definitely falsifiable. Namely, anyone can falsify it by discovering the purpose that you were made for. The fact that we haven't falsified it as of now does not mean that it's unfalsifiable.

I don't know of any respectable biologist who would seriously argue that aesthetic preferences are not subject to the laws of science. It's not as if your brain decides to obey Schrodinger's equation when regulating breathing rate, but not when enjoying or being annoyed by music. I agree with you on faith, but most people use faith as an excuse to believe in one fairy tale with absolute conviction while rejecting slightly different fairy tales with equal conviction. Of course, all these beliefs are not just wrong; they're not even wrong. --140.180.254.209 (talk) 20:41, 26 March 2013 (UTC)

Silly me. Of course you are 100% correct. --Jayron32 21:38, 26 March 2013 (UTC)

For the first part of your question - "why does light follow straight lines" - see Fermat's principle (and note that in general relativity you must replace "straight line" with "geodesic"). Gandalf61 (talk) 13:22, 26 March 2013 (UTC)

For the first part, an alternative explanation is Newton's first law--everything moves in a straight line at constant speed in the absence of external forces. If a photon doesn't encounter anything, it moves in a straight line, just like everything else.

For the second part, see electromagnetic wave equation. From Maxwell's equations, which encompass all of classical electromagnetism, you can derive an equation that represents a wave propagating through space. The speed of the wave can be directly read off from the equation. --140.180.254.209 (talk) 17:35, 26 March 2013 (UTC)

Science could try and answer why there is something instead of nothing and why does matter and energy has the properties that it has. All the rest would follow from that. As to why, in the sense of "what for", I agree with Jayron32 above, that it's not a question science could dwell into. OsmanRF34 (talk) 18:17, 26 March 2013 (UTC)

It doesn't; it is affected by the curvature of space-time - see gravitational lensing. That said, in a localised area any curvature can hard (or impossible) to detect, and bear in mind that the thing you use to measure the straightness of a beam of light might also be curved by space-time as well. Astronaut (talk) 20:12, 26 March 2013 (UTC)

To me, the more interesting question is, "What is it about the nature of light that compels its speed to be what it is?" ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:53, 26 March 2013 (UTC)

I agree with Jayron32. His appeal to the concept of falsifiability is very persuasive. Dolphin (t) 22:01, 26 March 2013 (UTC)

That is a very valid scientific question, possibly with falsifiable consequences. For one possible answer, see anthropic principle. Basically, we don't really know, which does not translate to "we will never know", or to "it's impossible to know", or to "God did it". --140.180.254.209 (talk) 22:38, 26 March 2013 (UTC)

Some days I wonder how a particle can "know" that it is traveling in a straight line. That implies some form of interaction with the surrounding empty space that affixes a particular direction and kinetic energy. I haven't seen a satisfactory explanation of how this can happen at the quantum scale, for example. Praemonitus (talk) 02:09, 27 March 2013 (UTC)

Why not turn it around? There's nothing to tell it /interact with it to cause it to diverge. Wickwack 124.178.132.17 (talk) 03:25, 27 March 2013 (UTC)

Conservation of momentum, which follows from the fact that the laws of Nature are invariant under transations. Of course, the momentum of a system is not conserved if the system is interacting with another system, so a photon can change directions in a medium or in a gravitational field. Under a translation of only the system you are looking at with the other things present, you don't have translational invariance. But if you are locked up in a box that is completely isolated from the environment, then the result of experiments done within the box won't depend on the position of the box in space. Under this assumption it can be proved that the total momentum of a system is conserved. Count Iblis (talk) 00:04, 28 March 2013 (UTC)

Relative speed thought experiment

Hi all, Imagine 2 spaceships of about the same size of the space shuttles (to avoid quantum effects mostly). The ships both set off with two very accurate clocks which both start off at the same time t=0. The only constraint on their movement is that they both have to observe the same relative speed of each other (ie if ship A records a velocity of v then so must B at all times). The observations are also constrained by modern interpretations of physics (such as they can't record the other ships velocity instantaneously).

Under classical mechanics their clocks will both record the same times at all points. Is there a way to use relatavistic effects to make this not the case? Like is there a way to use a large mass to distort spacetime to make the clocks record different times, whilst the ships both observe the same velocities? Or a large moving third mass? Or close to light speed travel?

Thanks! 80.254.147.164 (talk) 11:14, 26 March 2013 (UTC)

One thing is noticeable that mass as well as velocity, both distort space-time. Technologous (talk) 12:10, 26 March 2013 (UTC)

Just for phrasing clarity, I'd suggest that you go with the two ships needing to remain at rest relative to each other. That's what "observe the same velocity" amounts to, and helps illustrate why you then don't have to further care about speed. But as above, mass will impact the clocks, per gravitational time dilation. The clock in the deeper gravity well will run more slowly. If your clocks and measurements are sufficiently precise, this doesn't even have to be a "large" mass. — Lomn 13:50, 26 March 2013 (UTC)

Ah that makes sense, I had forgot that relative rest is the same thing, I think I had just overcomplicated things. 80.254.147.164 (talk) 13:59, 26 March 2013 (UTC)

Product of two scalars

The product of two vectors may be scalar by dot product or vector by cross product. But when we multiply two scalars what would be the result - a scalar or a vector ? Technologous (talk) 12:05, 26 March 2013 (UTC)

A scalar multiplied by a scalar is also a scalar. Dolphin (t) 12:15, 26 March 2013 (UTC)

I was also thinking the same, but here is a contradiction. We know, Pressure(P) = Force(F)/Area(A), hence it can be re-written as F = P*A. Here, both 'P' and 'A' are scalars, but their product 'F' is a vector. So, what is your opinion about this. Technologous (talk) 13:09, 26 March 2013 (UTC)

I would dispute that force, in F=P*A, is a vector. Given a pressure P and an area A, what direction is the force acting in? Based on that information, all you derive is the magnitude of the force, i.e. a scalar. — Lomn 13:53, 26 March 2013 (UTC)

Force can be a vector if you use an area vector for the area. No contradiction here. Gandalf61 (talk) 13:59, 26 March 2013 (UTC)

This means that force in few cases could also be direction-less. Thanks for correcting me. Technologous (talk) 15:06, 26 March 2013 (UTC)

I disagree that force can be direction-less. I believe the error in the above posts is the suggestion that pressure and area in the above equation are both scalar. In the equation F=PA it is the area that is a vector. The magnitude of the vector is simply the area of the shape on which the pressure is acting, and its direction is orthogonal to the shape, and it acts at the centroid of the shape. Wherever water pressure acts on a surface, or a small part of a surface, the resulting force always acts orthogonally to the surface. So F=PA is a scalar (pressure) multiplied by a vector (area) and the result (force) is also a vector. Dolphin (t) 21:37, 26 March 2013 (UTC)

I would also disagree, but isn't it the pressure, not the area, which is a vector? Imagine a book dropped on top of a rheopectic liquid? It exerts pressure on the surface, but only in one direction.--Gilderien Chat|List of good deeds 22:54, 26 March 2013 (UTC)

No, Pascal's law says fluid pressure acts equally in all directions. Resnick and Halliday say "Pressure is a scalar quantity." I am looking at Physics by Robert Resnick and David Halliday, section 17-2 Pressure and Density. Elsewhere in section 17-2 they say "An element of the surface can be represented by a vector delta S whose magnitude gives the area of the element and whose direction is taken to be the outward normal to the surface of the element." The force delta F on an element of fluid is equal to p times delta S. Dolphin (t) 05:11, 27 March 2013 (UTC)

You need to describe this using tensors. The stresses in a material are described by the stress tensor, the pressure contributes to this via the term $p\delta _{i,j}$. The i, j component of the stress tensor is the jth component of the force per unit area exerted accross a plane that has its normal in the ith direction (so that plane divides the medium into two parts and you can consider the force that one part is exerting on the other). Count Iblis (talk) 23:45, 26 March 2013 (UTC)

men body hair

why have some men too many hair in body? what is the efficient and cheaper way to remove them?? is this way is suitable or it has some side effects?? — Preceding unsigned comment added by 14.102.26.19 (talk) 16:25, 26 March 2013 (UTC)

Bird never make nest in bare tree. ;-) Anyway our article on this is Hair removal Dmcq (talk) 16:41, 26 March 2013 (UTC)

Here's the article on Body hair. Some men have more (or more visible) than others, but I don't think there's any medical definition of "too much". thx1138 (talk) 17:29, 26 March 2013 (UTC)

There is such a thing as too much hair, Hypertrichosis, but I doubt that's what the OP is talking about. Anyway, having a lot of body hair should help you get in touch with your more distant ancestors :P 109.99.71.97 (talk) 18:01, 26 March 2013 (UTC)

The colour of your body hair can make a big difference to the perception of how hairy you are. Dark hair against lighter skin (and to a lesser extent light hair against a darker skin) can make you look more hairy. Other factors include how long the body hair is, and your age (older men just seem to more hairy). Hair can be removed by shaving and costs only the price of a suitable blade and perhaps a lubricant. Other methods cost more. Permanent removal can be expensive. Astronaut (talk) 18:55, 26 March 2013 (UTC)

Let's be honest here. The concept of "too much hair" is largely based on fashion. When I was younger, hairy chests were all the go, at least for men. Men only removed hair from their faces, and not all did that. I haven't done it since 1970. There was a brief period in the '70s when it was OK for (some) women to not shave anywhere. (Maybe it paralleled bra-burning). Current fashion for both sexes (guaranteed to change) seems to be the pre-pubescent, totally hairless look. All at a time when we are more concerned than ever about paedophilia. Go figure. HiLo48 (talk) 04:54, 27 March 2013 (UTC)

A prowl of 4chan might suggest it is not so out of fashion for men after all, though it seems files like Image:Nude_doorwaygirl.jpg are not so common. Wnt (talk) 14:22, 27 March 2013 (UTC)

whales

can a whale swollow a person?? — Preceding unsigned comment added by 99.104.97.215 (talk) 18:46, 26 March 2013 (UTC)

No, although their throats measure some inches. They are not much bigger than ours, and not in proportion to their mouths . OsmanRF34 (talk) 19:04, 26 March 2013 (UTC)

Even a sperm whale? thx1138 (talk) 19:11, 26 March 2013 (UTC)

The OP didn't say anything about eating whole, though their question might have been inspired by the story of Jonah. A killer whale is an apex predator, quite capable of killing a person and presumably eating them. That said, it is rare that a killer whale will attack a person. Astronaut (talk) 20:01, 26 March 2013 (UTC)

The story of Jonah actually says he was swallowed by a "great fish", not by a whale as such. The assumption of it being a whale is a folk extension, like the "apple" eaten by Adam, and the "three" wise men. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:46, 26 March 2013 (UTC)

Of course, at the time that was written, the distinctions between a whale (mammal) and fish were unknown. StuRat (talk) 23:42, 26 March 2013 (UTC)

I think they would know a whale when they saw one. That presupposes that the story is literally true, which I wouldn't bet the family jewels on. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:24, 27 March 2013 (UTC)

They wouldn't know a whale isn't a giant fish. That requires science they didn't have yet. To them, if it lived in the sea and swam like a fish and looked like a fish, then it was a fish. StuRat (talk) 00:52, 27 March 2013 (UTC)

Well, the point has been made that it's not possible for Jonah to have been swallowed by a whale. But a "great fish", sent specifically by God to punish Jonah, could do the trick (within the confines of the story). ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:51, 27 March 2013 (UTC)

But, if we're going to allow the impossible by saying "God can do anything", then why not just let God enlarge the throat of an actual whale ? StuRat (talk) 04:32, 27 March 2013 (UTC)

• The English term whale and the Latin for shark, squalus are cognates and the PIE root (s)k^walos is cognate with various other Eurasiatic languages, from Proto-Uralic through Eskimo-Aleut, all meaning "big fish". I remember reading that sea predators find humans generally unpalatable because of our boniness compared to fish and pinnipeds. μηδείς (talk) 01:05, 27 March 2013 (UTC)

Are you saying that it was a shark that swallowed Jonah and then spat him out? 24.23.196.85 (talk) 03:33, 27 March 2013 (UTC)

What she's saying is that there are different possible translations from the original Hebrew into English as to the exact English word or words used to describe the thing that swallowed Jonah, and either "Whale" or "Large Fish" would, from a linguistic point of view, be entirely consistent. That is, there is no distinction in the original Hebrew which would lead one to pick the word "whale" or "large fish", so whichever the translator chooses would not be constrained by linguistic concerns. --Jayron32 03:38, 27 March 2013 (UTC)

And, as I pointed out above, it's not just a case of having a common word for two different concepts. To them, a whale was a large fish. StuRat (talk) 03:45, 27 March 2013 (UTC)

Which is why they didn't have a separate word. Medeis's point is that in early Indo-European languages, there was ALSO not a separate word, which is why both Whale and Shark are cognate words: not to point out that the bible is saying that it was a shark that swallowed Jonah, but merely to point out that linguistically, different words have evolved from the same earlier word that means "large fish". --Jayron32 03:48, 27 March 2013 (UTC)

I have absolutely no knowledge of Hebrew. My point is that big fish/vs whale is not a conceptual distinction shown to be made before the modern era. Semitic is not very closely related to Eurasiatic, so the comment on (s)k^walos (*kala in PU, *iqaluq in Eskimo (if I remember correctly)) is more cautionary than informative. μηδείς (talk) 03:52, 27 March 2013 (UTC)

Our article Shark#Etymology says nothing about it being cognate with whale or anything else. It says its origin is uncertain. If you know different, you'd better update the article pronto, with references. -- Jack of Oz ^[Talk] 05:29, 27 March 2013 (UTC)

You didn't read me carefully; I said whale and squalus were cognates. I said nothing about what the word shark is cognate to, and was aware its etymology is obscure. μηδείς (talk) 16:40, 27 March 2013 (UTC)

Wow, my first mistake this century. It had to happen eventually. Thanks for setting me right. -- Jack of Oz ^[Talk] 21:10, 27 March 2013 (UTC)

I'll keep a list of every error I come acrost so you needn't trouble yourself. μηδείς (talk) 01:06, 28 March 2013 (UTC)

Whale shark problem solved. Article says they are known to "cough" things that they unexpectedly ingest. any more brain busters, kidding165.212.189.187 (talk) 13:46, 27 March 2013 (UTC)

Great info, Medeis! I should note generally though that the use of the word doesn't really mean that there was no distinction drawn. For example, we still use the term "slime mold" even though it may refer to as wide a range of organisms as are to be found in nature. Many of our common usages are polyphyletic like that, based on general qualities rather than true genetic affinity. I wonder if there is something political in the careful distinction of fish from cetacean - the notion being that those of the Moby-Dick generation might have thought of a "fish" as anything you could gaff or harpoon in the water, while many in modern times might prefer a monophyletic term to distinguish mammal from an unprotected species.

I should further add that the mouth of a blue whale is indeed enormous [2] I see no reason why someone couldn't be "swallowed" in such a cavity without being literally in the stomach. The story of Jonah itself, of course, is not consistent with any mundane interpretation - three days and nights! - but a skeptic might plausibly argue this could be an exaggeration that began with some such brief swallowing episode and subsequent rejection by the baleen plates. Wnt (talk) 14:14, 27 March 2013 (UTC)

Yes, it is only negative evidence that there doesn't seem to be any distinction inherited within the Indo-European languages that shows a common early distinction between whales as a group and fish as a group. For example, we can be sure the PIE's had separate concepts for the head and the body even if the Greek, Russian, and Italian kephali, golova, and testa and soma, telo, and corpo are not cognates. But we can be sure on positive evidence that they did distinguish consistently between dogs and wolves, given the separate PIE roots *wlk^wos and *kunos and their descendants, such as wolf and hound and lupus and canis and λύκος and κύων. (Incidentally there does seem to be a distinction between two possible Eurasiatic roots, approximately *qal- and *tik- (think squalus and ichthys) meaning "big fish" and "little fish".) μηδείς (talk) 16:40, 27 March 2013 (UTC)

OK, maybe more to the point: Is there any known sea creature that is capable of swallowing an adult human whole? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:24, 27 March 2013 (UTC)

It's known that great whites don't swallow prey that large whole, they take chunks, so fish are out. Killer whales don't appear to do so either, they toss large prey about to soften and tear it up, and I am unaware of any case of a human having been eaten by one. Sperm whales eat giant squid, and it is unlikely they chew them. This guy from NOAA says a Sperm Whale could do so, and that sounds possible to me. But about as likely as a human trying to swallow an egg whole. The biblical story is so silly as not to merit consideration, and the fact that it is often told with the sea monster leviathan as the antagonist speaks toward the total unreality of it. μηδείς (talk) 22:51, 27 March 2013 (UTC)

It's still a reasonable question to ask. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:07, 28 March 2013 (UTC)

You mean whether a whale can swallow a person? I don't expect any reasonable person would call that a ridiculous question. μηδείς (talk) 02:42, 28 March 2013 (UTC)

The article on Jonah[3] gets into that question a bit, and pretty much answers the OP's question. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:49, 28 March 2013 (UTC)

In so far as misleading and relating bullshit to someone counts as answering him to some extent. Interesting to learn that dag is Hebrew for fish. μηδείς (talk) 03:53, 28 March 2013 (UTC)

The OP asked if a whale can swallow a human being whole. The first answer said "No" but didn't provide a source. The Jonah article has a sourced answer to the OP's question, which, while not specifically asking about Jonah, is a question often asked in reference to the Jonah story, which is why the Jonah article talks about it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:00, 28 March 2013 (UTC)

It's a reasonable link, just not at all a good article. μηδείς (talk) 16:06, 28 March 2013 (UTC)

It looks pretty good to me. It's written in good English, and it gets into various possible origins and references to the story. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:26, 28 March 2013 (UTC)

It has a lot of "some people think" type statements and speculation, and poorly supported claims such as great whites eating people whole, which is obvious nonsense, unless you count people who win the Nathan's Hot Dog Eating Contest as eating hotdogs "whole". The simple fact is the story is an obvious myth on the level of Leda and the Swan, that no whale or fish could possibly swallow a man whole, let alone keep him alive for three days, rather than biting, crushing, drowning, and killing him within a few minutes if not much less. The question of what real beasts may have served as partial inspiration for the mythical beast the unnumbered authors and retellers of this story may have had in mind is an open one. But our article doesn't provide more than armchair analysis toward that end. μηδείς (talk) 19:14, 28 March 2013 (UTC)

That being said, we do have documentary evidence of fish swallowing prey whole.

March 27

Are there any solvent of cyanoacrylate that does not dissolve polystyrene?

I had just found a bottle of dried cyanoacrylate. Is it likely to be still useful? I also wondered if I can use another solvent instead, because propanone dissolves polystyrene that I would like to glue.--Inspector (talk) 02:50, 27 March 2013 (UTC)

No, it isn't merely dried out, it has probably also polymerized. That is, cyanoacrylate glues don't work by merely drying out; there's a chemical reaction that occurs when they set, and the chemical reaction isn't really reversible. The idea behind cyanoacrylate (Super Glue, among other names) is that it spontaneously polymerizes when it is exposed to the humidity in the air. The monomers are dissolved in a highly volatile solvent (like acetone/propanone) so that they remain as monomer units; but as the acetone evaporates, the remaining monomer containing solution reaches the critical concentration and bingo: polymerization. This is a one-way trip: you don't get a solution of monomer units back when you dissolve it in acetone. Now, the polymer will soften some in acetone, often enough to break it up and remove it from, say, your skin. But it won't magically become monomer units again if redissolved in acetone. So, if you have a bottle of spoiled cyanoacrylate, the only way to get it back is to buy a new bottle. That one is toast. Regarding glueing polystyrene: you're also screwed there with any cyanoacrylate glue. There isn't, as far as I can think of, any solvent which will dissolve cyanoacrylate and not polystyrene. If you're trying to glue polystyrene to anything (either another piece of polystyrene, or another substance) you need a special polystyrene glue. Regular "Elmers" school glue should work too, but it usually doesn't hold very strong. Here is a good set of instructions on how to do it. --Jayron32 03:16, 27 March 2013 (UTC)

Thanks. I just got idea when I remembered some websites vaguely say about special types of CA that might be used on polystyrene.--Inspector (talk) 08:51, 27 March 2013 (UTC)

Flatulence

1) What is the average speed of a fart, in litres per second?

2) An unrelated question: why is a bad idea synonomous with the phrase "brain fart"? Plasmic Physics (talk) 06:05, 27 March 2013 (UTC)

2) A "brain fart" is an occasional undesirable result of thinking, just as a fart is an occasional undesirable result of digestion. StuRat (talk) 06:28, 27 March 2013 (UTC)

Our article flatulence is surprisingly good, referencing 5-375 ml for the typical volume - and, well, you know how long it goes on for. Wnt (talk) 14:32, 27 March 2013 (UTC)

Is it possible to control the pitch of a fart? What's the usual pitch range? Double sharp (talk) 15:51, 29 March 2013 (UTC)

It's hard to guess, since it contains many subharmonics, but the primary could cover he lower two thirds of the audible range. Plasmic Physics (talk) 03:59, 30 March 2013 (UTC)

Isn't the pitch proportional to sexual experience? — O'Dea (talk) 20:59, 31 March 2013 (UTC)

Heat released or absorbed in a reaction

I read these statements in an encyclopedia. Energy is conserved in chemical reactions. If stronger bonds form in the products than are broken in the reactants, heat is released to the surroundings, and the reaction is termed exothermic. If stronger bonds break than are formed, heat must be absorbed from the surroundings, and the reaction is endothermic. Earlier I was thinking just opposite. I think that the term 'exothermic' should be replaced by 'endothermic' and the term 'endothermic' should be replaced by 'exothermic' in the above italic statements. So, someone, please, explain the above statement. Concepts of Physics (talk) 06:05, 27 March 2013 (UTC)

I'm not sure why you think that's wrong, but our articles endothermic and exothermic explain it all. Rojomoke (talk) 06:13, 27 March 2013 (UTC)

Exothermic means means rejecting (pushing out) heat, in this case from hidden away in chemical bonds to becaome apparent in raising the temperature of the reactant/product mix - heat that will flow out into the surroundings. Endothermic means the temperature of the system is lowered in order to provide bond energy. Stronger bonds are so because they need less energy to make. Hence the statement you read is correct. It can be confusing to some because you often need to apply heat to break the bonds of the reactants in order to start the reaction off. But the heat you get back in an exothermic reaction is greater than the heat you put in. Wickwack 121.215.47.30 (talk) 06:23, 27 March 2013 (UTC)

The above explanation is very good, but try this one on for size as well: Bond strength is measured by how much energy you need to put into it to break it. Breaking bonds is always endothermic. Thus, by symmetry, forming bonds is always exothermic. If the bonds you break (the reactant side) are stronger than the bonds you make (product side), that means you need to put more energy into the reaction (because breaking is more energetic than making), so the whole process is endothermic. If the bonds you break (the reactant side) are weaker than the bonds you make (product side), then you get extra energy out, because the making step releases more energy than the breaking step needed to break its bonds, and the whole process is exothermic. Thus, if stronger bonds get made (products) the process is exothermic, while if the stronger bonds need to be broken (reactants) the process is endothermic. That's why the statement as written "If stronger bonds form in the products than are broken in the reactants, heat is released to the surroundings, and the reaction is termed exothermic. If stronger bonds break than are formed, heat must be absorbed from the surroundings, and the reaction is endothermic" is perfectly correct. --Jayron32 13:06, 27 March 2013 (UTC)

Others have discussed the science at play, but if you are hung up on the words, note that there is a bit of arbitrariness involved. See e.g. Exothermic#Contrast_between_thermodynamic_and_biological_terminology, which explains how the biological usage of these terms is basically the opposite of the thermodynamic usage. SemanticMantis (talk) 14:00, 27 March 2013 (UTC)

Car battery chargers with different amperages, what's the point?

I got a car/motorcycle battery charger for charging either 6V or 12V batteries with the choice between 2A, 4A, and 6A, and on the back there's a chart saying how many hours it will take (assuming the battery's 50% charged at the start). Obviously, when you select 2A or 4A on the device, it takes longer. Why would anyone select the lower options and voluntarily take longer? 20.137.2.50 (talk) 14:41, 27 March 2013 (UTC)

The chargers with smaller output cost less. Pay only for what you need. Battery chargers have different purposes.

(response to above unsigned line) This one charger has settings to choose between 6A (fastest) 4A (medium) and 2A (takes longest). 20.137.2.50 (talk) 14:56, 27 March 2013 (UTC)

If you take a car out of service for an extended period (e.g., a 6 or 12 month working holiday in another country), the battery may be flat and somewhat ruined when you get back. To prevent this, put the battery on continuous trickle charge. You need only a 2 Amp charger to do this, and most of the time it won't actually be delivering more than a fraction of that, unless the car security system draws a fair bit, which some do.

However, a 4 to 6 amp charger gives you some flexibility. If, say, you go to start your car and find the battery flat becasue you left the headlights on, connect a 4 to 6 Amp charger, wait 30 to 45 minutes, and your car will now start. Vehicles with larger batteries and larger starter motor draw (eg diesels) will need more amps or more charge time.

4 or 6 amp charging can be excessive for the small batteries used in light motorcycles that do not have electric start (50 to 125 CC class).

Wickwack 121.215.47.30 (talk) 14:52, 27 March 2013 (UTC)

OK, if the benefit of voluntarily choosing lower amperage and longer time is that it is less damaging to deeply drained batteries, then that is the answer to my question of why anyone would choose 2A. I wonder what situation makes 4A preferable to the still faster by a few hours according to my chart 6A. 20.137.2.50 (talk) 15:02, 27 March 2013 (UTC)

It is not a question of how deeply the battery is drained. You have misread or misunderstood if you think that. Ideally charging should be tapered off as a battery gets near full charge. Use the 2 Amp seeting if you are trickle charging a car battery - that is you are preventing the (fully charged) battery from going flat over time, as distinct from recharging a battery that has been flattended by some accident, such as leaving the lights on, or a defect in the electrical system. Use a lower charging rate on small motorbike batteries because any more will stress them - because they are samll, not because they are deeply discharged. Wickwack 121.215.47.30 (talk) 15:09, 27 March 2013 (UTC)

Thanks for the correction. The ad copy on the back of my thing says "fully automatic microprocessor-controlled battery charger 6 amp fast charge rate automatically adjusts charging rate" I thought that would be a one-size-fits-all that senses when not to give so much amperage, and was planning on setting it to 6 when I get home today, as the car was fine for the past 2 months since just replacing the battery and alternator until this morning when the low battery light came on and I took the other car to work but want to charge this battery up so I can get the 40 miles to the mechanic tomorrow as I suspect I got a short-lived alternator 2 months ago. 20.137.2.50 (talk) 15:24, 27 March 2013 (UTC)

Here are some references about battery charging current:

http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

http://www.evdl.org/pages/hartcharge.html

--Guy Macon (talk) 15:14, 27 March 2013 (UTC)

W boson in a hypothetical particle accelerator

I know that with a half-life of 10^-25 seconds, it is difficult to study a W in detail, but picturing one in some hypothetical super-strong particle accelerator seems theoretically valid. (Besides, who knows what devices people living shortly after the end of the electroweak era, composed of God only knows what kind of exotic matter, might have created?)

I have a feeling this house of cards will fall at an early step, but I'll ask all these at once for your convenience, and perhaps your amusement.

• Suppose a W+, whose temperature is just a little below the point of electroweak symmetry breaking, is accelerated in an evacuated chamber.

• Is it correct that the "temperature" of the isolated W+ is based simply on its velocity, per the Boltzmann relation?

• If it transitions to become massless at higher temperature, it should start moving at the speed of light, and be viewed as moving at the speed of light from any reference frame. Does that mean that there are two energy states, which I will ignorantly call W_+ and W*+, one of which is massed and one massless, and doing work on the particle involves the chance of promoting it to the more energetic state?

• Suppose we have a massless hot "W*+" moving toward a huge concentration of positive charge. It seems difficult to picture how relativistic electromagnetism applies to a positively charged particle moving at lightspeed... I get this notion that due to Lorentz contraction all the positive and negative charge in the cosmos (in the line of travel) is pancaked onto the W+, which makes it hard to figure, but there must be a mathematical treatment that avoids singularities?

• Anyway, let's suppose (dubiously) there's some way to do work against the "W*+" based on its charge. Presumably that should reduce its energy because it can't reduce its speed... since the charge doesn't go away, does that mean you could reduce its energy somewhere past the critical point, conceivably even reduce it to zero, before the "W*+" would decay back to a "W_+"? With a further push, could you abruptly stop it from moving forward at the speed of light and send it backward instead?

Wnt (talk) 15:03, 27 March 2013 (UTC)

I don't know much about the details, but I can make some general comments. First, Lorentz symmetry (the principle of relativity) implies that a rapidly moving particle has the same behavior as a particle at rest. For the same reason, a single fundamental particle can't have a temperature. You need a soup of particles with comparable kinetic energies and randomly directed momenta (relative to the center-of-mass frame) to have a temperature. Second, I don't think there's any phase of the standard model where all particles propagate freely at the speed c. At temperatures above the electroweak unification temperature the Higgs field no longer breaks the electroweak symmetry, but it doesn't disappear; there's a ton of energy in it (and every other field) at that temperature and all kinds of interaction going on, and no particle will get very far in any one direction at the speed c. If the zero state of the Higgs field is a pencil balanced on its point, and the low-temperature symmetry-breaking state is a pencil that fell on its side in a random direction, the high-temperature state is a pencil bouncing wildly off the walls. Third, electric charge doesn't make sense before electroweak symmetry breaking, so it's not a good choice for this example. Regarding charged massless particles in general, the gluons are a low-energy example in the standard model, and in gravity any massless particle (including a graviton) qualifies, but I have no good intuition for it either. Lorentz symmetry implies that you can't associate a time scale with a massless plane wave, and I think such waves don't self-interact. But if you have waves propagating in more than one direction there is a center-of-momentum frame and a nonzero total energy in that frame that gives you a time scale for interaction, even if they're individually massless. -- BenRG (talk) 17:31, 27 March 2013 (UTC)

Are you implying that if you take a W+ and heat it to the critical temperature, the positive charge is actually destroyed? Or does it become somehow "inactivated" but still conserved? Because the third option would seem to require it still be there on the W+, provided that particle hasn't decayed somehow to emit it somewhere else ... right?

I suppose one thing to check is... does the Higgs field as described have a definable velocity? Can a particle be moving "relative" to it? If not, what defines how it "bounces around" in relation to a single W suspended within it? Wnt (talk) 02:52, 28 March 2013 (UTC)

Electric charge is a certain combination of the U(1) fundamental gauge field and part of the SU(2) fundamental gauge field, which are still there at high energy, but the SU(2) is probably very different from its low-energy version (confining, like the strong force? This is the part I don't know much about). At any rate, that combination of fields still exists, but no longer behaves like electric charge, and there are a lot of other analogous combinations you could write down that happen not to be the electric charge at low energy, but seem to make as much sense at high energy. It's not an important point and I probably shouldn't have said anything.

In your second paragraph, are you talking about my wildly bouncing pencil? At high energy you're going to have a soup of particles/waves including Higgs bosons, and particles/waves have a state of motion. However the low-energy Higgs field is like the cosmological constant: it has a constant value everywhere which is locally Lorentz-invariant (so has no state of motion). -- BenRG (talk) 05:11, 28 March 2013 (UTC)

I suppose this is like an electron which modifies a field nearby it that normally is zero in every frame of reference... still, there are some things that seem hard to understand. The Higgs field now has a non-zero expectation value everywhere, but it is also described as being like a "Mexican hat". Why aren't some regions of space on the "north" side of the hat and others "south", "east", and "west"? And if the Higgs particle changes the field....... how? It has no spin, it is its own antiparticle, so how does the field know which way to change when it is nearby? Wnt (talk) 13:15, 28 March 2013 (UTC)

I happen to have a PhD in Grand Unified Theories, so a probably can answer some of those question better than Ben RG (Not a common situation). Let me start by answering BenRG's question. Both etectroweak and strong interactions are asymptotically free at high energy. That means the strength of the interactions decreases at higher energies/temperatures. That is the opposite of confinement, which requires an increasing strength of interaction (which happens at low energies/temperature). Something just came up here. I'll post another answer later. Dauto (talk) 17:56, 28 March 2013 (UTC)

The relation between atoms' kinetic energy and the energy levels of their electrons upon excitation

In other words, what really happens when an atom is excited, either by radiation or by collisions, or otherwise ? What are the mechanisms under which the transferred energy goes to exciting the electronic levels, the atom's kinetic energy as a whole, or both ? in the latter case, which is supposedly the most common case, what's the ratio between the two (elec energy levels & atomic kinetic energy) ? Any elaboration on these aspects will be great. BentzyCo (talk) 18:08, 27 March 2013 (UTC)

The kinetic energy (for a clound of atoms kinetic energy is translational kinetic energy and is its themperature) of an atom can be any value on a continuum from zero to whatever. However electron orbitals exist in discrete levels - they can only "jump" from one level to another.

When an atom encounters radiation, the radiation is best modelled as quanta - discrete packets of energy. If an atom recieves a quanta of energy matching the difference between one electron orbital and a higher one (not necessarily the next orbital in the series) then it absorbs the incoming quanta of energy by jumping to the higher orbital, without changing the atom kinetic energy. Conversely, if there is an orbital decay, a quanta/packet of radiation will be emitted. Decays occur at random intervals.

Atoms can also drop orbitals back towards the ground state by transferring the energy to their kinetic energy at random times. In this way kinetic energy increases in discrete steps. Once the electrons are in the ground state configuration, no more transfer to kinetic energy can take place. Conversely, an atom with sufficient kinetic energy (~thermodynamic temperature) to make a jump from one orbital to the next will do so at a random time, reducing its kinetic energy in a discrete step. If kinetic energy is insufficient (ie thermodynamic temperature is too low) to match an orbital change above the ground state, no transfer can take place.

Atoms can change/interchange their kinetic energy by collision with other atoms by whatever value satisfies the kinetic energy available already in the colliding atoms.

A atom at a given temperature ie kinetic energy below that required to change orbitals remains at that temperature indefinitely and does not require heat input. An atom at a temperature sufficient to jump orbitals will keep spitting out quanta of radiation at random intervals and thus looses energy. It requires never ending input of energy (either in variable amounts from collisions or in discrete jumps from incomming radiation of sufficent quanta size) in order to maintain temperature.

In a cloud of atoms, there is a random distribution of kinetic energy amongst the atoms. Some atoms will be more energetic than others. So in practice you don't see things suddenly change as you bring temperature up to a critical value - what you see is a gradual increase in radiation as you bring the temperature up. Similarly, with incomming radiation, you see a gradual apparently continously increasing temperature, as some atoms will be ready to jump an orbital level (and then to decay back to increased kinetic energy) and some won't.

I have not dicussed what happens with molecules or atoms combining at collision to form molecules - this is a further complex topic, where temperature is not solely determined by translational kinetic energy.

Wickwack 121.221.78.59 (talk) 00:30, 28 March 2013 (UTC)

Spacetime

Is it possible that time and space are separate but we cant physically do it? Or has that been unequivocally proven165.212.189.187 (talk) 18:36, 27 March 2013 (UTC)

It is possible that time is a purely human construct and not a physical dimension. On that basis, the answer to your question is yes it is possible. --TammyMoet (talk) 18:56, 27 March 2013 (UTC)

No, it is not possible, at least not on the basis of that article. Physics Essays is a journal that doesn't hesitate to publish crackpot ideas. I've managed to track down the original paper, and there's so many things wrong with it that I don't know where to start, but here's a small sample of the nonsense:

1. Most of it has no discernible meaning

2. The paper makes no predictions, whether testable or otherwise

3. The authors have absolutely no understanding of relativity: "the idea of time being the fourth dimension of space did not bring much progress in physics and is in contradiction with the formalism of special relativity". Time is not the fourth dimension of space; it is not a space dimension at all. Special relativistic formalism led to the idea that time is a dimension in the first place, because it transforms both space and time in a similar fashion.

4. The authors completely misunderstand Zeno's paradox, specifically Achilles and the tortoise. The basic paradox was solved by Aristotle 2400 years ago, and calculus put the solution onto a rigorous mathematical footing. Today, any 6th grader has the mathematical tools to plot distance vs. time for both Achilles and the tortoise, and see exactly why Achilles surpasses the tortoise. --140.180.254.209 (talk) 20:11, 27 March 2013 (UTC)

I don't understand the question. Time and space definitely are separate--you measure one with a clock, the other with a ruler. What you might be referring to is that according to Einstein's theories of relativity, time and distance are both relative. That means that if I'm on Earth and you're on a fast spaceship, we won't agree on how long the same event takes, or how big the same object is [insert bad joke here]. That doesn't negate the fact that both of us can take out a watch, take out a ruler, and measure both distance and length without any ambiguity.

You might also be referring to the fact that time is often considered a "dimension". First, even from a purely mathematical standpoint, time is different from the other dimensions of spacetime. The metric signature is either + − − − or - + + +, with the "special" sign always corresponding to the time dimension and the other 3 corresponding to the 3 space dimensions. Second, all the dimensions are "separate", in the sense that you can refer to them individually in any reference frame you want. --140.180.254.209 (talk) 20:11, 27 March 2013 (UTC)

I don't understand the question either. I wonder if the OP has read or misunderstood the article on spacetime. There's possibly some explanation of the nature of time at quantum level (Brian Cox's winding clocks), but I can't explain it because I don't really understand it. (Tammy might like to read The Unreality of Time for an older "theory", but it's philosophy, not science. ) Dbfirs 21:37, 27 March 2013 (UTC)

Matter and energy are in essence the same thing but you can't measure them with the same instruments. Electricity and magnetism were also thought to be different things until it was demonstrated by Maxwell that they were the same. Vespine (talk) 23:18, 27 March 2013 (UTC)

Electricity (more correctly electric field) and magnetism (magnetic field) are NOT the same. Maxwell showed how they interact to do various things. By suitable devices, energy can be transfered from one to the other. But that does not make them the same. It is often said that both make up the Lorentz force (the force on a point due to both) but this does not make them the same either. Neither is matter (more correctly mass) and energy the same. One can be converted into the other by certain means but that does not make them the same. I can convert food into poo, and other organisms can convert poo back into food, but please don't mix food and poo up.

Boffins talk quite correctly about mass-energy equivalence, but it is sloppy reading or sloppy use of the English language if you don't see the use of the word "equivalance" here as the mathematical sense of the word. I can say that there is an equivalence between heat energy and mass of a gasoline fuel, as from knowing one you can calculate the other with a standard formula, but that does not mean gasoline and heat are the same thing. This sort of sloppy language use seems to be very common in Ref Desk posts, but that doesn't make it right either.

Wickwack 121.221.78.59 (talk) 01:13, 28 March 2013 (UTC)

Electromagnetism:Originally electricity and magnetism were thought of as two separate forces. This view changed, however.... As for mass energy, well, again, you can't have mass without energy and you can't have energy without mass, wrt to the OP, he is asking could time and space be separate? One reply suggested that they could be separate since you can measure them with different instruments. Mass is measured with a different instrument to energy yet neither can exist without the other. I think the same can be said for electricity and magnetism. Apologies if I was sloppy with my language. But I still think my point in context is valid. Vespine (talk) 02:59, 28 March 2013 (UTC)

I have difficulty in determining context here as like others I don't understand what the OP was trying to ask.

You have directly quoted from the Wiki Electromagnetism article (History Section). But is the article right? You certainly CAN have an electric field without a magnetic field, and vice versa. Happens all the time, when, whichever one it is, is static (isn't moving). I have sitting in front of me right now a ring magnet that conveniently holds paperclips. There's no electric field around it. I've also got a 1.5 volt cell with nothing connected to it. There's an electric field around it. But absolutely no magnetic field. As to whether you can have mass without energy or vice versa - that's a bit harder to say. We'll have to wait until the boffins figure out matter at, or if required beyond, the sub-sub particle level. At any practical (measureable) level today, you CAN have one without the other, just as you can't realise the heat energy in gasoline unless you burn it in a heat engine. Wickwack 60.230.200.148 (talk) 03:29, 28 March 2013 (UTC)

By the time of Maxwell, people already knew that Electricity and magnetism were related phenomenons. That was obvious from Ampere and Oersted's work, and I don't think Maxwell added anything to that. What Maxwell really showed was that the potential-model of electromagnetism and the force-field-model of electromagnetism were equivalent. Someguy1221 (talk) 03:04, 28 March 2013 (UTC)

(ec) No, Tammy should not read The Unreality of Time unless he (she? it?) has a lot of time to waste. I've studied philosophy before, and most of it is intriguing, provocative, and mind-opening. The Unreality of Time, on the other hand, is pure bullshit dressed up as philosophy. The upshot is this: whenever anyone makes a claim about reality, the first question should be "does it fit with what we already know?" and the second should be "does it explain more than what we already know?" If the answer to either question is no, the theory is worthless. You can call it philosophy or religion all you want, but that doesn't save it from being wrong.

To add to what Wickwack said, you can unambiguously distinguish between matter and energy. You can also unambiguously distinguish between electric and magnetic fields. It makes perfect sense to say "the electric field here is 2 V/m, and the magnetic field is 0", or "an apple is matter, but a light beam is energy". --140.180.254.209 (talk) 03:18, 28 March 2013 (UTC)

Fair comment. I wasn't really recommending the article, and I accept your judgement that it's not even good philosophy. Dbfirs 07:49, 28 March 2013 (UTC)

UGH.. Mass–energy equivalence:In this concept, mass is a property of all energy; energy is a property of all mass. Saying gasolene has energy that is only released when you burn it is completely missing the point. Gasaolene has CHEMICAL energy which is released when it is burned, but ALL matter has/is energy as per E=mc^2. Electric field An electric field that changes with time, such as due to the motion of charged particles in the field, influences the local magnetic field. That is, the electric and magnetic fields are not completely separate phenomena; what one observer perceives as an electric field, another observer in a different frame of reference perceives as a mixture of electric and magnetic fields. I don't dispute the statements in these articles, if you do, feel free to find better references and change them. 05:18, 28 March 2013 (UTC)

All mass can be CONVERTED into energy per E=mc². The energy is not there anyway, just as the there is no heat coming out of a can of gasoline. In each case you have to DO SOMETHING to get the energy, and then you no longer have what you started with. Wikipedia is a valuable resource, but if I was to go through and fix all the wonky bits, I would need a team of hundreds if not thousands working full time to do it. Re your quote on Electric fields: The first sentence says the same as what I did, but the rest of it is a bit mucked up. An electric field cannot be percieved in another frame of reference as a magnetic field. If you are in a static electric field, it matters not whether you are still or moving, you still have only an electric field (unless you introduce a conductor). The nature of electric and magnetic fields as different things is clearly demonstrated by the fact that they decay off dramatically with distance, but electromagnetic waves propagate on a much more gradual inverse square law basis. Wickwack 60.230.200.148 (talk) 06:11, 28 March 2013 (UTC)

Look, we've gotten to the point where we're debating semantics instead of physics. Here's the physics. If you have some matter with mass m, you can turn it into mc^2 of any form of energy: photons, thermal energy, kinetic energy, whatever. If you have some energy, you can turn it into matter. If you have some energy, say potential, trapped inside an object, the object has greater inertia. That is, it's harder to move, and so more massive. General relativity treats rest mass energy and other forms of energy the same way; they all go into the stress-energy-momentum tensor. Does that satisfy your definition of "sameness"? If it does, that's fine; if not, also fine. (From 140.180.254.209)

You said: If you have some energy, say potential, trapped inside an object, the object has greater inertia...and is more massive. That's not right. A 1 kg object still has a mass of 1 kg whether it is moving or not, and whther it is raised to a height or not. I make no appology for delving into a discussion of language, because unless and until you understand the terminology, you cannot properly understand the subject, and anything either of us says will just get sillier and sillier - as shown by your statement I repeated in italics. You have confused potential and kinetic energy, and mixed up mass and kinetic energy. Wickwack 124.182.18.126 (talk) 07:29, 28 March 2013 (UTC)

Take two systems. System A is just like Earth, with box sitting on it. System B is just like Earth, but that same box is sitting at a slightly higher altitude. Everything else is the same. Well, System B is slightly more massive. That mass is not "in the box", but it has to be in the system somewhere. Someguy1221 (talk) 08:07, 28 March 2013 (UTC)

Nope. Both systems have the same mass - the sum of box and earth. Nothing is hidden. You can't use energy to increase mass with going through a conversion process. If the box System B is allowed to come to the Earth surface somehow (say because it was nudged over the edge of a platform, then the potential energy will be converted to kinetic on the way down. (Let's assume there is no atmosphere to much things up with drag) Upon impact, the kinetic energy will be converted into heat (and a bit of shock wave propagating in the eath until it too is converted to heat). So System B will see a temperature increase. But unchanged mass, unless the height, size of the earth, and consequent impact caused ionisation or atom smashing or some such, but that is just one of the ways to CONVERT energy into mass, and is in this example a special case. And a most unreliable one, because you'll more probably get a flash of EM radiation and a reduction in system mass. If you don't think so, explain your thinking with a bit more that just E = mc². Wickwack 121.215.45.143 (talk) 08:44, 28 March 2013 (UTC)

You seem to deeply underappreciate the mass-energy equivalence. I can use my boxes to turn a generator as they fall, and have that generator fire a laser that shoots photons into space. Those photons carry mass away from the Earth. Potential energy --> kinetic energy --> Light = energy == mass. It would be wrong to say the extra mass was in the box - it is in the Earth-box system. But it is there and you can't pretend it's not. You could also just look at Mass-energy_equivalence#Practical_examples to see where you're just flat-out wrong. Any change in the total energy of a system is associated with a change in the mass of the system. Someguy1221 (talk) 09:03, 28 March 2013 (UTC)

Well, I could say your good self deeply underappreciates the meaning of the word equivalence in this context. But calling each other dopey (albiet in a slightly more polite way) doesn't help much does it? Unfortunately the operation of a laser involves ionisation - as I said this is a method of CONVERTING mass to energy. In any case it is not the case that additional mass due to the box's potential energy was used to create the photon energy (and presumably you think that that could leave System B mass the same as System A, it is the case that some of the mass in the laser (part of the earth) got converted into photons. So System B then end ups with slightly LESS mass than System A (unless more energy from outside the system was added to the laser to allow the laser working gas to come back to ground state). The practical examples in the Wiki article do I think include one with a common misconception. Note that no references are cited for the examples given. Did the author make them up? I have come across the Grand Coolee Dam example before, in popular press, but not proper physics or engineering textbooks. As Wikipedia says, don't take Wikipedia articles as verified fact, use it to get new ideas and review the references. Have a carefull read of this: http://www.weburbia.com/physics/mass.html. Wickwack 121.215.45.143 (talk) 09:36, 28 March 2013 (UTC)

I honestly don't know if you're simply wrong, or arguing a semantic point. But let me try to clarify: are seriously telling me, that you believe, that if I show you two otherwise physically identical systems, and I heat one of them up by pumping energy in, that no mass was added to the system? That the increased thermal energy bears no associated increase in mass? That the box's inertia doesn't change? That the box's gravitational footprint doesn't change? I think you are misunderstanding your own link, Wickwack! It does not say that the mass does not increase. "In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself. In the final analysis the issue is a debate over whether or not relativistic mass should be used is a matter of semantics and teaching methods." And many more quotes like it. Sure, the mass of the box does not change, but that is not the entire system. As I said, the energy of the system, the mass of the system, changes. You can go down this semantic road where energy does not have mass, but adding energy to a system alters its gravity and inertia. I call that mass, whatever word you want to use. Someguy1221 (talk) 09:59, 28 March 2013 (UTC)

In this case, semantics are important - it is at the root of the trouble. Yep - I'm saying that if you add mechanical forms of energy, or electrical energy, to a system, there is no inherent increase in mass. Neither does removing energy inherently reduce mass. An increase or decrease in mass does occur if there is a conversion. Conversion occurs in nuclear processes, emission and absorption of photons, ionisation, and the like. If the energy is applied in such as way as to add velocity to a mass, the time compression occurs to the mass as observed externally. The time compression produces an apparent (to the external observer) increase in inertia. Energy can be added without necessarily increasing velocity. The increased aparent inertia is seen in devices such as television picture tubes. The electrons are accellerated to something like 1/10th the speed of light over a distance of 300 - 400 mm. Tube design engineers take into account the apparent increase in inertia in calculating the beam deflection force required (delivered magnetically), slightly larger than would be expected from electron mass alone. There is a notional equivalent mass to any quantity of energy. Nowhere did I focus on the mass of the box alone - I referred to the system mass. Have a look at this one: http://henry.pha.jhu.edu/mass.pdf from John Hopkins Uni. Wickwack 58.170.139.239 (talk) 10:07, 28 March 2013 (UTC)

Someguy1221 is correct. There is no consistent way to maintain a separation of mass from energy gated by "conversion" operations. There is a real distinction between four-momentum and rest mass (the former is a vector, the latter is its length), but there is no sense in adding the rest masses of components of a system; it's the same as adding the lengths of a bunch of vectors that point in different directions. Someguy's System A and System B have different masses. The Wikipedia example of the hydroelectric power plant is accurate. -- BenRG (talk) 17:16, 28 March 2013 (UTC)

Merely repeating statements, as you've done, doesn't help much. The artcle has no references cited for the examples given. Do you have an authoritive reference for the power generation example? One that supports the tranport of mass over the grip, and not just one that calculates m = E.c^-2 where E is the output of the dam? Wickwack 124.178.52.113 (talk) 00:38, 29 March 2013 (UTC)

Wickwack, adding and subtracting energy from the total energy of an object always increases/decreases its mass. For instance this. -Modocc (talk) 01:13, 29 March 2013 (UTC)

Emission and absortion of light involves a conversion. Wickwack 124.178.52.113 (talk) 01:48, 29 March 2013 (UTC)

You wrote: "I'm saying that if you add mechanical forms of energy, or electrical energy, to a system, there is no inherent increase in mass". No. These bulk forms of energy transfer are primarily a consequence of the interactions of fundamental charges and photons. -Modocc (talk) 02:11, 29 March 2013 (UTC)

For incoming energy to atoms in the ground state (eg a noble gas) below the amount necessary to jump electron orbitals, what interaction occurs? Wickwack 120.145.140.148 (talk) 04:39, 29 March 2013 (UTC)

Electromagnetic induction which is mediated by photons. -Modocc (talk) 05:06, 29 March 2013 (UTC)

Which is the displacement of free electrons in metals and the like, which are not a noble gas, and need not be an absortion of energy anyway. Wicwack 58.167.232.210 (talk) 05:43, 29 March 2013 (UTC)

I suppose I could have been clearer when I said electromagnetic induction, since I mean the Maxwell–Faraday equation, because it describes the propagation of the light which gives rise to Van der Waals forces, covalent bonds, electrostatic attractions and even phonons. If you are asking what the possible interactions can occur within a medium or a reflector which does not absorb the radiation, my answer is the same, because its my understanding that this equation applies to this light too as it also interacts with the noble gas. Bound charges tend to be shielded though and have greater effective mass, nevertheless the atoms' charges do contribute to the radiation's propagation. Its very late here and I'm going to be busy tomorrow and this is off-topic to the OP's question, thus perhaps we can conclude and should hat/hab some of this at some point? -Modocc (talk) 07:14, 29 March 2013 (UTC)

Fair enough. I might re-introduce this as a question of my own in a few weeks - hope that is ok. It won't be good etiquette to pose a question and debate the answers though. Wickwack 124.178.141.34 (talk) 08:19, 29 March 2013 (UTC)

Look, Wickwack, I honestly don't know why you're disputing basic physics, but I'll show the evidence you're looking for. Look at Einstein's field equations, which I'll paste here for convenience:

$R_{\mu \nu }-{1 \over 2}g_{\mu \nu }\,R+g_{\mu \nu }\Lambda ={8\pi G \over c^{4}}T_{\mu \nu }$

The left hand side describes the curvature of spacetime. The thing on the right, T_uv, is the stress-energy-momentum tensor. It contains exactly what its name implies: energy density, momentum density, and stress. Do you see any distinction whatsoever between mass and energy? In fact, do you see mass anywhere in the tensor? You might be worried that Einstein's field equation describes the "very local" environment because it has derivatives, which is illogical if we're discussing the mass of a system. But consider this: in the low-energy limit, general relativity must reduce to the Newtonian limit, correct? In such a limit, there's still going to be no distinction between mass and energy, right? So in terms of gravity and inertia, general relativity does not discriminate.

If you're not convinced by this argument, see this paper. You claimed that a hotter object is not more massive, meaning harder to move. This paper claims the opposite: "many balk at the application to kinetic energy. Can it really be true that a hot brick weighs more than a cold brick?", and "We can thus tell our students with confidence that kinetic energy has weight,not just as a theoretical expectation, but as an experimental fact" --140.180.254.209 (talk) 07:44, 29 March 2013 (UTC)

I have a lot of trouble with the concept of a hot brick weighing (overlooking for moment that's not the right word - the paper actually has a distinction between inertial mass and gravitation mass. Weight by definition is determined by gravitation ) more than a cold brick, but am happy with a fast travelling brick being harder to deflect than a slowly travelling brick. That's like the electrons in cathode ray tube (oscilloscope tubes, TV tubes) example I mentioned elsewhere, which is well known to tube engineers. The electrons don't change their mass (which would affect what happens when they hit the screen) but to calculate the deflection force required for scanning, engineers need to, and do, take into account an apparent increase in electron inertia, expected as the flight time as seen by the deflection electromagnets is out compared to that "seen" by each electron, due to the relatavistic consequence of these electron bean going typically at 1/10th the speed of light or more. One of the few, and perhaps, only, situation outside nuclear reactors where relatavistic effects affect pratical Engineer's usual 3- or 4- place accuracy calculations. Wickwack 124.178.141.34 (talk) 08:36, 29 March 2013 (UTC)

I think I know what the Electric fields article was trying to get at. If you have a charge at rest, there's only an electric field. If you use a moving reference frame, there's an electric field and a magnetic field, because the charge is now moving. This is discussed extensively in classical electromagnetism and special relativity. Actually, electric and magnetic fields transform in a similar way as spacetime in special relativity, which shouldn't be surprising because Maxwell's equations were Einstein's inspiration for SR. --140.180.254.209 (talk) 07:17, 28 March 2013 (UTC)

You may or may not have noticed that I've been trying steer the discusion into talking about electric fields rather than point charges. Fields are more fundamental. You may get greater clarity if you think about electric fields (and magnetic fields). Consider a pair of parallel plates, large in area, spaced 1 m apart. One plate is held by some means a steady +1 MV with respect to the other. There is therefore an electric field strength at any arbitary position between the plates of 1 kV/mm. And I can measure that with a suitable instrument. There is clearly no magnetic field. If I move the instrument about within the field, say at 100mm/mSec in any x, y, or z direction, by some means that is non-conducting and no magnetic, I'll still always get 1 kV/mm. Now, tell me, if this instrument is also a magnetometer, what magnetic field will I get? Let's say I reapeat the experiment, this time moving the plates steadily apart while at the same time increasing the voltage on the plates at the same rate, maintaining the same 1 kV/mm electric field strength. What readings will I get off the dual electric field meter and magnetometer now? What if I hold the instrument at a fixed spot and move the two plates around, keeping them at the same voltage and distance apart? Think before writing please. Wickwack 124.182.18.126 (talk) 07:55, 28 March 2013 (UTC)

As 140.* said, Lorentz transformations mix electric and magnetic field components, and that's why people say you can't have an electric field without a magnetic field or vice versa in special relativity. Of course, in another sense of those words, you can.

If you move your electric field detecting instrument parallel to the electric field lines it will measure an unchanged electric field and no magnetic field, but if you move it in any other direction it will measure a slightly increased electric field and a small magnetic field. -- BenRG (talk) 17:16, 28 March 2013 (UTC)

(ec) Finally, some questions I can answer with mathematical precision! Suppose you have an instrument that measures both electric and magnetic fields, and you move it perpendicular to the electric field lines at 100 mm/ms, or 100 m/s. Then the new E' and B' you will measure are:

${\begin{aligned}&\mathbf {E} '=\gamma \mathbf {E} \\&\mathbf {B} '=\gamma \left(-{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {E} \right)\end{aligned}}$

where

$\gamma \ {\overset {\underset {\mathrm {def} }{}}{=}}\ {\frac {1}{\sqrt {1-v^{2}/{c}^{2}}}}$

Plugging in the numbers, I get E' = 1.0000000000000556 kV/mm and B' = 1.11e-9 Tesla, pointing perpendicular to the electric field. This calculation comes directly from the first equations in the article I linked to, classical electromagnetism and special relativity. If you hold the instrument still and move the plates, you get the exact same answers, because only relative speed matters. --140.180.254.209 (talk) 17:26, 28 March 2013 (UTC)

Yes, that follows from equations 2 and 3 in the Wiki article, simplified for zero initial magnetic field. I asked for three cases: a) fixed plates and moving probe, which you've answered, c) holding teh probe fixed, and moving the plates about, which you have reasonably inferred is really the same as a), and b) moving the plates steadily apart while changing the voltage on them so as to maintain the same field strength, which you did not answer. Can you please explain what measured values would occur in this case. Wickwack 124.178.52.113 (talk) 01:42, 29 March 2013 (UTC)

Since (by assumption) the field strength doesn't change as you separate the plates, the measured value is also unchanged. Did you have some reason for asking these questions? -- BenRG (talk) 04:03, 29 March 2013 (UTC)

Yes, I have two specific reasons. I would like 140.180.254.209 or someone else to come back on this before revealing both reasons - no offence to you Ben. However, nothing in the World is perfect, certainly not me, so I might therefore have made a mistake, though if so I don't know what it is (if I did, I would admit it and then shut up). So, if challenged, vigorously but without new logic (almost all this debate has been mere re-statement) I ask questions which hopefully might lead to a new insight, for me and you. I hope that makes sense. Wickwack 120.145.140.148 (talk) 04:34, 29 March 2013 (UTC)

Ben's answer is correct, with one caveat. If you move the plates apart, the plates carry charge, and moving charges generate magnetic fields due to the Biot-Savart law. Exactly in between the two plates, it turns out that the magnetic field is 0 due to the symmetry of the problem. (It might be zero elsewhere too; I'm too lazy to work out the exact answer, but it's just an area integral.) --140.180.254.209 (talk) 07:44, 29 March 2013 (UTC)

To clarify my question: I first thought about how to measure or explain time without using some aspect of space, and how to measure or explain space without using some aspect of time. If the state of t=0 is identical to t=1 then how can you be sure time passed? How do you differentiate 1km vs 1000km without using time to measure each?165.212.189.187 (talk) 16:01, 28 March 2013 (UTC)

There is an unavoidable circularity in the definition of time and distance and everything else, since physical concepts can only be defined in terms of other physical concepts. Physical theories can nevertheless make objective statements about the world by imposing more constraints than there are concepts. Then you can take n of those constraints as definitions of the n concepts, and the rest as testable experimental predictions. However, there's nothing forcing you to choose any particular set of n constraints as the definitions. For example, is F=dp/dt subject to experimental test, or is it merely the definition of force, or maybe of momentum, or even of time? Depending on the interpretation you pick, you may end up with different ideas of what a given experiment is really measuring, even though the experiment itself and the predicted outcome are both objectively unchanged.

That said, I think physical theories constrain the "reasonable" choices of definitions. Given the importance of Lorentz symmetry in physics, for example, it seems unreasonable to use a certain definition of time without using a symmetrical definition of distance. Professional physicists break this particular "rule" all the time, though, especially in relativistic quantum mechanics. -- BenRG (talk) 19:23, 28 March 2013 (UTC)

Detonators

Were radio detonators available in 1944? How much did they typically weigh? For comparison, how much did a typical clockwork fuze from the same time period typically weigh? Thanks in advance! 24.23.196.85 (talk) 22:47, 27 March 2013 (UTC)

Take a look at Pencil detonator for a simple timed detonator. These would have weighed only a couple of ounces and are literally the size of a pencil. Here's a patent for a radio controlled mine from 1942. Nikola Tesla had a radio control boat by 1898 and detonating an electrical detonator would be trivial compared to controlling a boat.Tobyc75 (talk) 01:18, 28 March 2013 (UTC)

See our Wikipedia article on the proximity fuze for good detail on what was developed during and before World War 2. Wickwack 60.230.200.148 (talk) 03:55, 28 March 2013 (UTC)

Well, I was asking specifically for the weight of a clockwork fuze -- not a time pencil, which is a type of chemical detonator. Also, it would be nice to know the approximate weight of a radio detonator -- I want to make sure that Alfred, my demolition man, doesn't get a loadout that's too heavy (he already has to carry a heavier loadout than anyone else in the team, because of the 21 lbs. of Composition C that he has to lug on his back). But thanks for the info about the radio-controlled mine -- that's just what I needed to know. 24.23.196.85 (talk) 04:51, 28 March 2013 (UTC)

Well, perhaps you should be more specific about what you want, so we don't have to guess. What do you mean by "radio detonator"? A proximity fuse? A radio controlled bob or mine? If the latter, a factory produced militarty specified device or an improvised device? People (terroists, resistance fighters, etc), have always been able to improvise things like model aircraft RC control devices to set explosives off remotely, and even cruder things requiring minimal technical skill, such as contacts glued to the speaker cone of an ordinary radio tuned to a quiet frequency. Wickwack 60.230.200.148 (talk) 05:47, 28 March 2013 (UTC)

By "radio detonator", I mean a device that sets off a demolition charge remotely by means of radio waves -- certainly not a proximity fuze. And I do mean a factory-produced device (such as those used by the OSS or the SOE), not an improvised device -- I don't think it's even possible to generalize about the weight of an improvised device because it varies so widely. 24.23.196.85 (talk) 06:13, 28 March 2013 (UTC)

The smallest WWII commercially available device that could work as a radio detonator would probably be the RBZ Receiver. See

http://www.cryptomuseum.com/spy/rbz/index.htm and

http://www.virhistory.com/navy/xmtr-ww2-port.htm (about 2/3 down the page).

The bottom part of Artillery fuze#Time fuzes has some info on WWII clockwork fuses. --Guy Macon (talk) 09:47, 28 March 2013 (UTC)

The RBZ was of course a militarised standard general coverage reciever for portable communications. Rather a waste to use it to explode bombs, which was definitely not its intended application. If standard commercial product was to be used as a radio fuse, it would make more sense, for cost, factory prioritisation, and operational reliability, for a modified model aircraft/boat radio control set to be used. Particularly in Britain, authorities made every effort to find a war production use for every factory, regardles of its peacetime role. For example, my mother as a young lady worked in a factory making toy electric trains, which involves making tiny electric motors. Mum wound the coils in the motors using some sort of hand operated winding machine. When the War started, toy train production stopped. The factory was converted to making an only slightly better quality tiny "motor" that was used with a tiny propellor as a generator to arm air-dropped bombs. Factories that made furniture, therefore having expertise in metal and plywood got to make parts for aircraft. None of this means of course that model RC derived equipment was used in WW2 as radio fuses, but it might be useful to research to find out or build into a plot. Ratbone 58.170.139.239 (talk) 12:01, 28 March 2013 (UTC)

First, find a radio receiver -- any radio receiver -- that [A] was available in WWII, [B] ran without a power cord, and [C] was smaller than the the RBZ Receiver. The various radios made for spies are a good place to look. If you can't find one smaller, then you have your minimum size and weight. --Guy Macon (talk) 12:39, 28 March 2013 (UTC)

A reasonable suggestion, I think. But choose one designed for morse code operation - they could easily set up to offer better immunity to false triggering from noise, and more easily adapted to close a contact to detonate the bomb. Small radios intended for spies in WW2 may not be a good idea. They were often regenerative or superegenerative. This means that the recievers put out a continuous radio signal of their own as a byproduct of their circuit technique. Regeneration and superregeneration gives very high sensitiviy with very simple compact circuits. The self radiated signal was judged to be acceptable where the opeartor was expected to only switch on at scheduled "listening out" times and switch off imediately the message was recieved. But the self radiated signal meant 1) the enemy could detect it and direction find onto it, and 2) the radiated signal could itself detonate some types of electric detonators. No good going bang just as you switched on! Also, these types were not very selective, meaning that false trigerring could occur from lightning, man-made noise, and transmitters on other channels. Last, but not least, they could require continual readjustment of the reaction control and tuning as the battery ran down, otherwise the sensitivity would drop off rather dramatically. Not a problem in recieving short messages, but may be a problem is left on for several hours. Best go for a radio described as "TRF" or better, "Superheterodyne". The RBZ is a sensitive selective superheterodyne state of the art for battery radios in WW2. Ratbone 124.182.145.28 (talk) 13:15, 28 March 2013 (UTC)

Would a "Sweetheart" (RCD Type 31/1) work? It says in the War Office catalogue that it's a TRF set, and the weight is given as 3 3/4 lbs (less than the RBZ set, but still heavier than the explosive charge itself). Oops, the 3 3/4 lbs. figure is for the total packed weight -- the weight of the actual receiver with power unit is 2 lbs. 2 oz., which is actually less than that of the explosive charge itself. 24.23.196.85 (talk) 00:59, 29 March 2013 (UTC)

Not a good choice. Despite the claim of it being TRF, the circuit given at http://www.cryptomuseum.com/spy/sweetheart/ shows it to be a regnerative 3-tube set of inferior quality, with a regenerative stage followed by 2 stages of low power audio amplification. It will have all the disadvantages of regenerative sets that I described above. It was designed to recieve broadcasts from the BBC etc in the then standard HF broadcast band. It is design to drive a piezo earphone, which makes conversion to closing fuse contacts not simple. A fully trained radio technician or engineer could certainly do it though. It appears that battery life and performance were considerably sacrificed to reduce battery size. In a radio fuse application that may require the receiver to be left on continuously for many hours, I would suggest using 3 telephone No. 6 batteries (which are physically much larger), in series for the filament supply, or several of the normal 4.5V batteries in parallel. Anyone using a raio set in enemy territory would have to buy batteies at frequent intervals. Radio tubes then drew a fair bit of current, while batteries back then were much inferior to what you buy now. It's difficult to tell from the manual, but it appears that the designers intended it to be run on the then standard bicycle headlamp battery. This would be readily available in enemy territory, and purchase would not attract attention like purchasing a specific radio battery would. Note that this radio is designed for use with an elevated 30 foot wire antenna. Performance would drop off rapidly if the wire is shortened for concealment purposes. However a simple modification would permit operation with a short wire, say a few metres long. NOTE: Since the method of tuning is somewhat uncertain, it would be necessary for the demolition team to set it all up at the site, with the actual detonator disabled, and have the transmitter send a test signal, with which the demolition team could tune the radio and verify that the fuse contacts close. Pre-tuning the reciever would not work. In contrast, pre-tuning an RBZ (a vastly better set technically in all respects) and locking the tuning in place with some sort of glue or wax before going anywhere near the site would be completely satisfactory. How far away does your plot require the transmitter to be? What are its concealment requirements? There is one advantage of the sweetheart set, and regenerative sets like it. Your team could take two. Use one as the transmitter - all that needs to be done make it transmitt is turn the regeneration control (the right hand knob accoding to the manual available at the crypto museum website) all the way up. Range would be rather poor and difficult to predict, though it should be reliable over a kilometre or so. Ratbone 121.215.43.129 (talk) 02:29, 29 March 2013 (UTC)

A more reasonable choice might be the so-called biscuit tin radio, MCR-1. See http://www.cryptomuseum.com/spy/mcr1/index.htm. This is a properly designed superheterodyne, far above the Sweetheart rubbish in ease of use and performance, but not up to the standard, cost, and bulk of an RBZ. It would have been well known to any SOE person. It also would require modification by a tech to fire a detonator, but the modification would be quite simple. Don't be misled by the size of the biscuit tin - your operator would not need all the various attachments eg for pwer main operation and would not need all the coil packs for different bands. It can work effectively with a metre or so of wire as an antenna. It can be trusted if pre-tuned and tested and the tuning locked with glue back at base before going anywhere near the demolition site. It should hold its' tuning and sensitivity as the battery runs down to a certain extent. The team can still use a Sweetheart as a transmitter to trigger it. As the Sweetheart's tuning is uncertain, the transmitter operator would just rock the tuning to and fro around the selected frequency until the bomb goes bang. Ratbone 120.145.25.156 (talk) 05:20, 29 March 2013 (UTC)

Thanks! The requirement here is to simultaneously blow up 3 parked Tiger tanks with 3-pound shaped charges, and the team doesn't have to move all that far -- just a hundred yards or so down the railroad tracks. Also, they won't be able to go back to the tank cazerne to retrieve the radio, even if it's left intact -- this is intended to be a feint to draw the Germans out, so once the tanks blow up, the whole place will be full of field-grays (exactly as intended). Would it be possible to use a single MCR to set off all 3 demo charges? 24.23.196.85 (talk) 05:54, 29 March 2013 (UTC)

I don't see why not. I, having an electronincs engineer background, know zilch about explosives and detonators, but the modification required for MCR-1 radios, which were designed to feed low impedance magnetic headphones, is to alter the output stage to operate a small relay. The contacts on this relay would be arranged to close a circuit, using the radio HT battery as a energisation source, to fire electric detonators and so set off the charges. I would think that all three detomators could be wired in parallel and fed from the same relay contact. As the required range is ony 100 m, things should be quite trustworthy, even if a Sweetheart is used as the triggering transmitter. Turn the Sweetheart reaction control all the way up, a quickish turn of the tuning knob past the selected frequency that the MCR is pre-tuned to, and up she goes! I presume the reason why you don't want to describe the team running a wire back 100 m is so that the Germans run around in indecision, and not follow the wire back, which would give an officer time to think maybe this is a feint - it being the natural tendency of a person in command to bark out some quick orders to keep the subordinates occupied and off his back while he has a think. Maybe the team should ensure the MCR radio is obliterted in the bang in order to increase the uncertainty and stress in those Germans. Ratbone 124.178.141.34 (talk) 07:50, 29 March 2013 (UTC)

Yeah, that's what I'm thinking -- rig up the MCR on the middle tank, and run wires (or maybe even explosive cord) to the other two. This will keep the Germans guessing, while the Maquis team is well on the way to its second diversionary objective (namely, blowing up a munitions train). 24.23.196.85 (talk) 04:52, 30 March 2013 (UTC)

March 28

Silver Sulfide Decomposes At Boiling Point?

What does it mean that silver sulfide decomposes when it boils?Curb Chain (talk) 00:57, 28 March 2013 (UTC)

I assume it means that the vapour phase is unstable compared to liquid phase. Plasmic Physics (talk) 02:56, 28 March 2013 (UTC)

See thermal decomposition and chemical decomposition for our articles on the subject. Tevildo (talk) 10:21, 28 March 2013 (UTC)

Water and pressure related questions

1. Does evaporation of water take place at all temperatures? Does it take place at -200 and 0 degree Celsius?
2. Is it possible to convert water which is polar into non-polar by any means?
3. Suppose there is an empty bottle on a desk. Atmosphere is applying such a huge pressure on it, according to this situation it should be crushed into a small volume, but it doesn't happens. Why? Yellow Hole (talk) 03:43, 28 March 2013 (UTC)

(1) Evaporation occurs at all temperaures and pressures that allow water to exist in liquid or solid states. Evaporation is driven by vapour pressure. Vapour pressure exists because some water molecules continually leave the water (or ice) surface (and some on contact rejoin). If you enclose a quantity of water in a sealed rigid chamber, which contains nothing except the water, some of it will occupy the space left over by evaporation to a gas, to the extent that the gas pressure equals the vapour pressure.

The molecules that leave the water surface do so because heat energy is randomly distributed between molecules - thus some are more energetic than others, ans some of those have enough kinetic energy to leave and dart about on their own. Vapor pressure, and evaporation, sharply increase with increase in temperature. Vapour pressure and evaporation are very weak at 0 C but it does continue very slightly down to absolute zero.

(2) No. The way that hydrogen atoms each bind to the oxygen means that a water molecule can only be polar.

(3) Truely empty containers do get crushed, if their strength does not withstand atmosphereic pressure. To get a truely empty container, you have to pump out the air or whatever that is in it, and create a vaccuum. Otherwise whatever is in it will have the same pressure as the air outside. There was a televison personality, Julius Sumner Miller, who effected the manner, in his voice and mannerisms, of a mad scientist and who made a name for himself crushing all manner of containers from small tin cans to huge oil drums by filling them with steam, sealing them, and then pouring cold water over them to condense the steam and create a vacuum ( usually swearing under his breath if the drum didn't collapse on cue ). He's dead now, unfortunately. He used to drive high school science teachers nuts by inspiring school kids to ask questions that teachers cannot answer but know that they should. Teachers hated him.

Wickwack 60.230.200.148 (talk) 04:06, 28 March 2013 (UTC)

If there is a complete vacuum inside the bottle, will the atmospheric pressure be able to crush the bottle? If the bottle contains air, the air will prevent the bottle from being crushed. But what prevents a page from being crushed, when one holds it vertically or horizontally. If the bottle containing air is taken on the moon (where there is no atmosphere), will it expand and ultimately burst? Yellow Hole (talk) 05:46, 28 March 2013 (UTC)

What do you think? You have all the information you need to figure it out. If you are too dumb to figure it out from my first sentence in (3) above, get someone to show you a tube from an old TV set. See that big glass thing that you look at to see the picture? It's the tube - really a large bottle, and it has a very near perfect vacuum inside, so that the electrons can get from one end to the other without being impeded by anything else. Look how thin the glass is at the back end (about 1 mm or so), but notice it's round, which is the strongest shape. Wickwack 60.230.200.148 (talk) 05:55, 28 March 2013 (UTC)

Enough with the nastiness! Nothing is obvious if you look at it hard enough to really understand it in the first place. This question reminds me of [4] ... if you don't think something through in the simple case you'll only get tripped up in the more complex one. It is hard to grasp intuitively at first that air is under such dramatic pressure while it yet has so little weight. The difference between the top of a page held in the air and the bottom is small, the difference in pressure is small, and the air on either side increases in pressure equally. Wnt (talk) 13:30, 28 March 2013 (UTC)

Sometimes it's better to challenge people to think, rather than just tell them the answer. If you think, you learn. If you fed an answer, you soon turn to something else and forget about it. This one was in fact really simple. He wasn't asking about paper, he asked would a bottle get crushed if it contains a vacuum. This OP has asked about very simple things (like under what cicumstance will air pressure crush things) obvious to anyone with ordinary schooling, mixed with significantly more advanced concepts (eg water being polar) which made me a tad suspicious, actually. Wickwack 124.182.145.28 (talk) 13:58, 28 March 2013 (UTC)

Challenging someone to think ==/== being rude and insulting. Those are not coincident activities. You can challenge someone without calling them dumb. --Jayron32 03:26, 29 March 2013 (UTC)

collapse bickering
The following discussion has been closed. Please do not modify it.
1) Note that evaporation directly from a solid is called sublimation. StuRat (talk) 16:05, 28 March 2013 (UTC) Not right. Sublimation is the conversion of a solid into a gas by adding heat to raise it above the sublimation temperature. (Many substances, including H2O, go directly from solid to gas without passing through a liquid stage, if the pressure is right). In pure substances (ie not solutions or alloys), sublimation occurs sharply at a specific pressure dependent temperature and only that temperature, just as melting and boiling does. Sublimation is not evaporation any more than boiling is evaporation. Evaporation, regardless of whether it is off a solid or a liquid, occurs at any temperature below the sublimation point, melting point, or boiling point, whichever is applicable at the operating pressure. Stick to what you know or have checked, please Stu. Ratbone 121.221.4.38 (talk) 03:05, 29 March 2013 (UTC) As usual, you're talking out of your ass, Ratbone. See Sublimation_(phase_transition)#Water for examples of water ice subliming. StuRat (talk) 03:14, 29 March 2013 (UTC) And as usual, StuRat, you've come back with a ridiculous attempt to justify your initial error. There's nothing in the article you cited that supports your claim. Look in any standard physic textbook. I have several, all saying essentially the same thing. One, a standard univerity text, says "sublimation in the process by which a solid passes from solid phase to vapour phase without going through a liquid phase, and occurs at low pressures. Sublimation occurs at the sublimation line in the pressure-temperature phase diagram, between the origin and the triple point." This means, of course, that sublimation occurs at a specific temperature for the given pressure. See Physics for Scientist and Engineers, Douglas C Giancoli. There's several editions of this. In mine it is on page 474. However, very old texts and standard dictionaries often do not make the distinction plain. Now be good for once, and learn from your error. Sublimation, an abrupt process, can only occur below the triple point pressure. Evaporation off a solid can occur at any pressure below the vapor pressure, though for many materials the vapor pressure off solid will be below the triple point value at all temperatures and evaporation may not occur. That is why glass does not evaporate on the inside of vaccuum tubes, for example. For water, vapor pressure off solid runs above the triple point pressure and evaporation does occur. You cannot make a vacuum flask out of ice. I hope you can see for all this, at least that there are two distinct processes going on, just as there are two distinct processes involved in going from liquid to gas/vapour (boiling being the abrupt one, evaporation being the other). Ratbone 124.182.24.177 (talk) 04:08, 29 March 2013 (UTC) Nothing you said in any way contradicts what I said. You're just trying to pick a fight, as always. Might I suggest you go to a pub and insult a man with a shaved head and multiple tattoos ? That's the most effective way to provoke a fight. StuRat (talk) 05:00, 29 March 2013 (UTC) As usual, you are the one who started with the insult. But what I said most certainly is in contradiction to yourself. You claimed sublimation is evaporation of a solid. I have said sublimation and evaporation are two diffrent processes occuring under difernet conditions. Ratbone 120.145.25.156 (talk) 05:35, 29 March 2013 (UTC) The question was "Does evaporation of water take place at all temperatures ?". I corrected their terminology, by pointing out that water going from the solid phase to gas directly is called sublimation. You come into every thread here looking for any possible way to misinterpret a response so that you can say it's wrong. Now go pick a fight elsewhere. StuRat (talk) 05:48, 29 March 2013 (UTC) If I may step in: it appears to the layman outsider that what's really at work is that one of you is using the colloquial form of "evaporation" while the other is using the formal definition. Both are acceptable within the right context; certainly neither of you is unconditionally "wrong". Had either of you bothered to consider or acknowledge such, this whole mess could have been useful to other readers. — Lomn 14:38, 29 March 2013 (UTC)

Why do we use the word 'infinity' in the definition of potential energy?

In almost every definition of potential energy, whether electric or gravitational, I see the word infinity. While defining electric potential, we write "The electric potential at a point in an electric field is defined as the amount of work done in bringing a unit positive charge from infinity to that point." Same definition applies to gravitational potential energy. What is the significance of the word 'infinity', here? Thanks in advance! 106.216.102.21 (talk) 06:33, 28 March 2013 (UTC)

It's a hypothetical point infinitely far away. To define potential energy, you need some point of reference. The natural choices seem to be the point at the center of the field, a point on the surface of the body creating the field, and infinity. The first one results in a mathematical problem, since it represents a singularity. The second is only well-defined for spherical bodies, and it results in different potential energies for different sized bodies with the same mass or charge. The third is not quite as obvious, but it works consistently in all cases, it has nice mathematical properties, and its the one we use in physics. In reality, you are often interested in the potential energy difference between two points - that is independent of your choice of reference point. --Stephan Schulz (talk) 06:45, 28 March 2013 (UTC)

"Infinity" means "beyond the range of effect of the charges we're talking about". Given how willing physicists usually are to make approximations, I'm surprised they don't just say "far away". Wnt (talk) 17:13, 28 March 2013 (UTC)

• The short answer is laziness. A proper definition makes it clear that potential energy is always defined with respect to an arbitrary reference point, but unfortunately people have difficulty understanding concepts that are relative in this way, and feel an urge to reword the definition in a way that makes it look less relative. I believe that this is a blunder, because a person who does not understand that potential energy is relative really doesn't understand it at all. The same issue applies to voltage (electrical potential), and I have repeatedly run into trouble in my efforts to maintain our membrane potential article from editors who try to reword the definition to remove mention of an arbitrary reference point. Looie496 (talk) 17:23, 28 March 2013 (UTC)

• The actual answer is less laziness and more about reasonable precision. As you bring electric charges farther and farther apart from each other, the effect they have on each other decreases with increasing distance; that is the effect of moving a distance of 1 inch to 2 inches is much greater than from moving 2 inches to 3 inches, and MUCH MUCH greater than moving 15 inches to 16 inches, and so on and so on. Eventually you reach a distance where incremental changes to the distance betweens the charges stops making a measurable difference in their effect, so, for example, you can't actually measure any difference in effect no matter how far you move it. Mathematically, you can always calculate such an effect, but when you reach the point where such an effect is dozens or hundreds of orders of magnitude smaller than the finest measurement than any device could ever make, it's not necessary to make a distinction between those distances and "infinity". That is, bringing a charge "from infinity" to a given point just means "bringing a charge from a distance far enough away that the incremental changes in distance are insignificant" to that given point. Since functionally, any distance that large would work, infinity is as good as any word, and useful in the sense that it takes less syllables to say than "a distance far enough away that the incremental changes in distance are insignificant" --Jayron32 20:28, 28 March 2013 (UTC)

• I must respectfully disagree. You can go clear to the other side of the universe, and still if you move the object next to a black hole, you'll get an enormous potential energy. That's a nitpick in a sense, but it touches the crucial point. I've taught this stuff a number of times, and I've come to believe that to teach it properly you must get the student to understand the concept of a reference point, and that the idea of "moving to infinity" only serves to obscure that concept. Looie496 (talk) 20:44, 28 March 2013 (UTC)

• Who said anything about introducing black holes into the problem? If a problem doesn't mention them, what's the point of bringing them into the discussion? --Jayron32 03:32, 29 March 2013 (UTC)

Well, it seems we have two teachers disagreeing on how to teach. I agree with Jayron though. He's in good company, I have several engineering and physics texbooks that set out his very reason. However, Ralph Morrison's textbook Grounding and Shielding Techniques (don't fret over the name, it is a book about mostly about electrostatics and electromagnetics fundamentals) carefully avoids using points at infinite distance - it develops everything from two arbitarily charged points a distance x apart. Morrison's book in its various editions was a standard undergrad text for electrical & electronic engineers for 40 years. So Looie is in good company too, though his reference to black holes is a red herring. It is undertood that a pint at infinite distance is a point in the middle of nowhere. Wickwack 120.145.31.249 (talk) 03:26, 29 March 2013 (UTC)

• More clearly, when we say "infinity", it implies that we are really taking a limit, as there is no such point.--Jasper Deng (talk) 03:51, 29 March 2013 (UTC)

If the OP is really asking about why physicists set the zero point at infinity, as opposed to why students are taught that, there's a number of compelling philosophical reasons. First, setting the zero point at infinity implies that a completely empty universe has 0 energy. There's an undeniable elegance about that, and it's certainly better than claiming that an empty universe has 5 joules--why 5, and not 5.01, or -151? Second, where else would you set the reference point? You can't set it at 0 distance from the point charge/mass, because the potential energy would be infinite there. If you set it at 4.3 meters, why are you choosing 4.3, instead of 4.8? Infinity is the only elegant solution.

Third, many equations become much simpler and more intuitive with the reference point set to infinity. For example, a Sun + comet system has positive energy if the comet can escape the Sun's gravity, and negative energy if it can't. The virial theorem says that in any bound system with no mean velocity, gravitational potential energy is -2 times the kinetic energy. Physicists are not in the business of making equations more complicated than they need to be--besides wasting time, that tends to hamper intuition more than it helps, and for no benefit. --140.180.254.209 (talk) 06:52, 29 March 2013 (UTC)

Note that in some cases you can neither use zero nor infinity, e.g. in two dimensions the Coulomb potential is proportional to Log(r). But even in these and other cases where you can't escape singularities anymore, you still want to use such points. A good example is quantum field theory, where it turns out to be much more convenient to define the theory at arbitrary small lenght scales and then use clever tricks involving counterterms that become infinite in the limit, rather than doing all the calculations such that everything stays finite. Count Iblis (talk) 15:24, 29 March 2013 (UTC)

How close must two particles meet for annihilation to take place?

--Inspector (talk) 08:26, 28 March 2013 (UTC)

See Cross section (physics) (not a very good article at the moment, to be honest). Calculating the cross section for a particular interaction is fairly complicated - see Bhabha scattering for an example, or this rather more accessible paper. Both require a reasonable university-level knowledge of particle physics to understand completely, I'm afraid, but science is sometimes like that. Tevildo (talk) 10:38, 28 March 2013 (UTC)

Flowing electrons

Electricity, which has invaded our lives from all directions, is nothing but the flow of electrons. One should think how these charged particles run a number of electrical equipment? What makes electrons so special that they produce electric current and light the bulbs, run the fans and motors, etc? Britannica User (talk) 13:36, 28 March 2013 (UTC)

Electric current is not solely comprised of moving electrons. It can be other sorts of charge carriers, such as ions. What makes all these things comprise electric current, when they collectively move, is that each one carries an electric charge. What is an electric charge? Something that provides an electric field - that is it can exert a force wrt another charged thing. Wickwack 124.182.145.28 (talk) 13:51, 28 March 2013 (UTC)

Electrons are quite light yet carry a large electric charge, and are located on the outside of each atom. The electrons in the valance shell (outside edge) are particularly likely to "break off" and are therefore the most mobile. StuRat (talk) 16:11, 28 March 2013 (UTC)

It is kind of a boring answer but the fans and motors in question are designed to run via electric current (which is translated, usually through magnets and so forth, into kinetic motion). You could make fans and motors that ran off of diesel fuel or even just falling water. There is nothing "special" about electrons in this way; they can be conveniently used as a means of transmitting energy relatively efficiently, and that is why we've built a world around them. They are not so much "special" as "convenient"; we could have a world built around falling rocks as well, but that would be much more costly. Electric currents are very easy to produce from mechanical force (e.g. using a dynamo), can be transmitted over long distances, and can be tuned pretty efficiently for the task at hand — this is what makes them so common in our current world. --Mr.98 (talk) 16:23, 28 March 2013 (UTC)

I think much of it has to do with the conduction band of metals and other conductors being such a remarkably easy place to move around in. Wickwack speaks of other ion carriers, but you couldn't pump sodium ions hundreds of miles through a cable. Maybe you could build a machine run by virtue of a proton beam in an evacuated space like in a particle accelerator? But I don't know if it has ever been done. It really isn't that obvious to me why electrons are able to move through a vast matrix of positive and negative charges with so little overall resistance - I suppose delocalization is somehow at the root of it, maybe chemical resonance in a way, all ultimately due to the low mass of the electron and its long Compton wavelength? Wnt (talk) 17:08, 28 March 2013 (UTC)

Google

In responce to what is Google I was surprised that the word only had the mathematical definition of 10^100. For us early 1960's engineering and science students the word was then used (may have had slight difference in spelling like googel) to represent the number of atoms in the universe which at that time was 10^85. It has since been lowered to around 10^83. How come in the discussion of Google or the answer to the number of atoms in the universe there is no reference to this association. The math definition (Wikipedia response to Google)refers to the limit of calculators being 10^99 which occurred after the 1960 time period.

Is my recollection of old class notes incorrect?

Thank You Larry Oliva — Preceding unsigned comment added by 151.190.0.1 (talk) 14:06, 28 March 2013 (UTC)

Note that the number is spelled googol; "Google" is an alternate corporate spelling for (I assume) trademark purposes a corporate misspelling. A googol has always been defined as 10¹⁰⁰, without specific reference to any physical phenomena, but we note that it has frequently been used as a mental reference point for other very large quantities such as the number of subatomic particles in the universe. 10¹⁰⁰ isn't really a very good approximation for 10⁸⁵ (being off by a factor of, what, a quadrillion?), but it's a better approximation than any other number that you're likely to know a convenient name for (as illustrated by my uncertainty above regarding 10¹⁵, much less 10⁷⁰). — Lomn 14:14, 28 March 2013 (UTC)

For those who are interested: 10⁷⁰ is known as ten duovigintillion on the shotscale, and ten undecilliard on the long scale. Plasmic Physics (talk) 18:50, 28 March 2013 (UTC)

A googol is known as ten duotrigintillion and ten sexdecilliard respectively. Plasmic Physics (talk) 18:55, 28 March 2013 (UTC)

How do certain foods interfere with Levothyroxine?

Warnings on the bottle, our article, and everything I can find on the net indicates that levothyroxine should be taken on an empty stomach, half an hour before meals, because of interference with its absorption. I am wondering what the exact mechanism(s) of that interference is. Is, for example, the proper acidity of the stomach or intestine important? Does the drug have an affinity for certain foods they way fiber can absorb certain lipids?Is it only absorbed in the stomach or upper small intestine, with the presence of food preventing it from getting fully absorbed before it passes down the line? Does it react chemically with other substances, denaturing it? I can speculate, but am hoping for a referenced or authoritative answer, if anyone has one. Thanks. μηδείς (talk) 17:51, 28 March 2013 (UTC)

The fact that this drug binds to fibre etc., that can pass thought the human gut undigested is all perhaps the medic need to know (it is not explained in my pharmacology books either). Your only recourse (and perhaps a quicker one) might be to email or phone [5] and ask them: “what is it about this molecule that that inhibits it from being broken down once it has bonded to fibre”. I can guess, that enzymes can't get their hooks into the middle of it and release the iodine hormone due to to bonding strength, but as you say, you don't want a guess.Aspro (talk) 19:21, 28 March 2013 (UTC)

For some reason I couldn't open your url, but if you are telling me it binds to fiber that is fully comprehensible and the sort of answer I am looking for. I have also seen remarks about it interreacting with coffee and grapefruit juice. I am wondering if the drug itself is a peptide or peptide analog, which I assume would mean it's absorbed in the small intestine, or what other class of drug/nutrient it might be classified as or analogous with, such as an MAO inhibitor. μηδείς (talk) 19:28, 28 March 2013 (UTC)

Here is the link in long hand http://www.forestpharm.com/mic.aspx It states that For Product Inquiries Forest's Medical Information and Communication department can be reached at: 800-678-1605 ext 66297. According to this http://medlibrary.org/lib/rx/meds/levothroid-2/ Quote:”Synthetic T4 is identical to that produced in the human thyroid gland.” unquote. So, it looks like an exact peptide copy.--Aspro (talk) 20:02, 28 March 2013 (UTC)

Thanks, again, exactly what I was looking for. μηδείς (talk) 20:14, 28 March 2013 (UTC)

Resolved

What lifeform can regenerate the fastest?

^Topic. ScienceApe (talk) 19:29, 28 March 2013 (UTC)

The Doctor. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 05:56, 29 March 2013 (UTC)

James Bond 007. 24.23.196.85 (talk) 06:04, 29 March 2013 (UTC)

I think yeast would be hard to beat. Looie496 (talk) 19:47, 28 March 2013 (UTC)

My vote goes to some thermophilic anaerobics [6]. Even band Wikipedia editors, trolls and Madonna can't regenerate that fast. --Aspro (talk) 20:10, 28 March 2013 (UTC)

From [ http://www.britannica.com/EBchecked/topic/48203/bacteria/272364/Growth-of-bacterial-populations#ref955453 ]: ", Clostridium perfringens, one of the fastest-growing bacteria, has an optimum generation time of about 10 minutes; Escherichia coli can double every 20 minutes; and the slow-growing Mycobacterium tuberculosis has a generation time in the range of 12 to 16 hours." --21:01, 28 March 2013 (UTC)

What do you mean by "regenerate"? The previous (good) answers are focused on population growth/ generation time. But if you're interested in healing or regeneration of lost organs or tissue, things like planaria planarians can grow new heads and tails if they need to. SemanticMantis (talk) 23:08, 28 March 2013 (UTC)

Wouldn't viruses be among the leaders in rapid replication? ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:38, 30 March 2013 (UTC)

The OP said "lifeform". Whether viruses count as lifeforms is debatable. Brambleclawx 20:14, 30 March 2013 (UTC)

Do they replicate faster than what you would consider "true" lifeforms? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:20, 30 March 2013 (UTC)

March 29

Does Navier-Stokes work for solids?

I was looking at the article Derivation of the Navier–Stokes equations. Where in the derivation does it use the assumption that the substance is a fluid? Does Navier-Stokes work for solids?

150.203.115.98 (talk) 04:56, 29 March 2013 (UTC)

As far as I can tell it does. f = −∇·σ, where f is the body forces and σ is the stress tensor (including tension/pressure), is true for a solid in equilibrium. -- BenRG (talk) 05:27, 29 March 2013 (UTC)

Cracks could cause discontinuities in both density and velocity, making them undifferentiable.--Wikimedes (talk) 05:51, 29 March 2013 (UTC)

Ethical obligation of professional scientists toward open accountability

I'm looking for published pieces in authoritative sources on scientific ethics concerning the reasonable amount of time that a high-profile scientist in a moderately controversial field ought to spend answering questions about their work from (1) other scientists in their field, and (2) the general public, especially when a larger than ordinary number of such questions are forthcoming. Ideally, there's a Nature or Science essay out there suggesting a specific number of hours per week for both as being reasonable, but even an editorial in a third-tier journal or monograph on the topic would really help. Sadly personal opinion on this will not help me at all. Thanks in advance. 75.166.202.252 (talk) 06:05, 29 March 2013 (UTC)

Speaking as a scientist myself, I'm pretty certain there is no rule specifying an amount of time. The basic rule is that good work should be reproducible by other scientists, and a scientist is obligated to supply any information missing from a publication that is needed in order to reproduce the findings. However long that takes is however long you are obligated to spend. Looie496 (talk) 07:30, 29 March 2013 (UTC)

The questions I'm concerned about here are asking for explanations of model assumptions and interpretation of evidence, in a less experimental science, i.e. astrophysics. 75.166.202.252 (talk) 08:18, 29 March 2013 (UTC)

Most scientists are eager to answer questions if they seem at all reasonable. If you're asking questions and not getting responses, there may be something wrong either with the questions or the way they are being asked. However, there are also people who never look at their email, or whose spam filter malfunctions, so it's hard to say anything that applies universally. Looie496 (talk) 21:54, 29 March 2013 (UTC)

I don't think there are any such authoritative pronouncements on this. I know from my scientific colleagues in universities that many of their appointments include undefined requirements of "service," which is where one might think this sort of thing might be found, but usually those are explicitly with reference to service to the department (e.g. serving on hiring committees) and not towards the general public (in fact, spending too much time doing "public" activities — even when that means testifying before Congress or serving on important panels — is sometimes seen as a negative by university departments, as absurd as that may seem). I've had one friend even complain that being President of the major professional society for his discipline (and it's not a minor discipline!) was not considered "service" by his university, which is totally nuts, but there you have it. --Mr.98 (talk) 16:52, 29 March 2013 (UTC)

one issue is that scientists are reluctant to take data they have laboriously assembled over months or years of grant-sponsored work and the rewards of which they are now reaping in terms of publications containing analysis of said data, thereby enhancing their attractiveness for future grant money and employment, until they themselves have gnawed the last publishable flesh off the bones, before they hand it over to those who, being in the same field, are competitors for the aforementioned glittering prizes, in order that said competitors might crank out their own publications. The Ant and the Grasshopper and all that. On the other hand, data and analyses derived from publicly available datasets via various transformations etc. without the same sort of investment of time and labor should be fair game. If they can't or won't cough up the methods, intermediate calculations and results, etc. of how they got from alpha to omega, time to be skeptical. [7] Gzuckier (talk) 19:34, 29 March 2013 (UTC)

Since when is astrophysics a moderately controversial field?? Dauto (talk) 16:55, 29 March 2013 (UTC)

Open Yale Courses had a course about controversies in astrophysics. Add to it the Drake equation and there is the Solar neutrino problem too. There you have your moderately controversial field. Less controversial than climatology, but more than mechanics. OsmanRF34 (talk) 00:43, 30 March 2013 (UTC)

I do research in astrophysics, and I've yet to see a single scientific paper that takes the Drake equation seriously. The solar neutrino, as the article you linked to says, was resolved in 1968 with the discovery of neutrino oscillation. --140.180.254.209 (talk) 16:14, 30 March 2013 (UTC)

I think the approach varies from university and from field to field. At the university I graduated from, teaching staff and researchers used to be most helpful in answering queries from outside about their work. I have sometimes studied a journal paper and not been able to fully understand it or thought there was some key information missing. I'm no boffin, just a practicing Engineer. I've sent emails to the authors and been amazed at the thoughtful and most useful replies I've got back. But back to my own uni: Some time ago they formed a marketing division, with a mission to commercialise and extract money from as much of their activities as possible. Since then, it's just a total waste of time to ask them anything.

It used to be that if a paper was published that drew conclusions from a great mass of numerical data, that the raw data was not published in with the paper in the journal, as it took up too much space. But if you wrote and asked, they would supply. Some journals have the data and additional information to enable replication ready for download on their website. That's great!

Trouble is, these days, in many fields, the author's conclusions were determined by computer program, or the paper is about a computer program. I am interested in 4-cycle engine combustion. There is a certain group of researchers who have claimed in a journal paper that they can accurately simulate combustion with a computer program they devised, and have drawn certain conclusions about combustion from the program results. But they will not, release the source code or even a usuable description of the algorithms. So, nobody can replicate their work; nobody can benefit from what they learnt. I think that is disgusting. Ratbone 121.215.20.145 (talk) 02:39, 30 March 2013 (UTC)

I doubt that they are publishing in high-quality journals. Looie496 (talk) 15:06, 30 March 2013 (UTC)

Actually, they have published in a well known peer reviewed proffessional journal, one I would think would, otherwise at least, be regarded as high quality. However, this is not the place to set out just who they specifically are (and they just might read Ref Desk, and work out from emails and Ref Desk posts who I am). Given that I have set out the subject, and you would assume their papers are recent, if I identify the journal, you will know who they are. Actually, if you have a similar interest in combustion, you probably should have guessed who they might be anyway, but there's no need for me to make it specific here. Ratbone 120.145.158.214 (talk) 15:28, 30 March 2013 (UTC)

Freak event

What would happen if there was a freak event where all electrical generators in the world ceased to work for an extended period of time? All generation technology is included including internal combustion engines and those torches that you shake and then they produce light. 105.227.178.237 (talk) 11:37, 29 March 2013 (UTC)

According to Revolution (TV series), the collapse of government and public order, followed by the rise of militias and warlords where for whom ability to generate electricity becomes the most powerful weapon available. - Cucumber Mike (talk) 12:11, 29 March 2013 (UTC)

I think that also happened in the Tom Cruise version of War of the Worlds. Although that wasn't exactly a freak event, it was an invasion. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:55, 29 March 2013 (UTC)

For some strange reason, in Revolution, War of the Worlds and other such shows, bicycles no longer exist and everyone has to walk or rid a horse. --Guy Macon (talk) 16:42, 29 March 2013 (UTC)

In the near future when Revoution is set, every bicycle has a generator built in for headlights, and humanity has lost the ability to make a bicycle without one. Gzuckier (talk) 20:07, 29 March 2013 (UTC)

It depends on whether you're talking about science or magic. If it's science, then anything that shuts down all electromechanical generation kills everything, among other various apocalyptic consequences. If it's magic, then it can do whatever you want it to do. The latter is obviously more interesting from a storytelling perspective, but there is no referenceable "right" answer. — Lomn 14:27, 29 March 2013 (UTC)

Note that the best real-world analogy, electromagnetic pulses, have the ability to destroy a lot of equipment, but not the ability to inhibit the function of new equipment. It's the magical "nothing new can ever work, either" part that breaks things. — Lomn 14:29, 29 March 2013 (UTC)

The problem is that you're trying to eliminate all forms of electrical generation - and that is one hell of a stretch. Waving a magnet in front of a coil of wire produces enough electricity to light a flashlight bulb. You don't need a nice man-made magnet - a simple lodestone will do...and you don't even need a manufactured "wire" - a coil of anything that conducts electricity would do...a garden hose full of seawater, for example. If that doesn't work in your hypothetical world, then very fundamental laws of physics have to be broken...and the consequences of that are very likely to be a sudden, immediate end of all things! I doubt that human brains could function in that scenario.

You can't really couch the question with enough hypotheticals and caveats to limit the damage to "What would happen to human society without electricity?" without it boiling down to being "magic"...and once you have that, all bets are off when it comes to predicting the outcome. SteveBaker (talk) 14:38, 29 March 2013 (UTC)

Well said, Steve. And so that I'm not just replying to myself in this thread, I'll stick this point here instead: the OP may also be interested in S.M. Stirling's Emberverse series, a set of novels that more or less provide the inspiration for the new Revolution show. The effect itself is "magic", to the extent that any explanation at all is attempted, but the thrust of the books is "post-apocalyptic alt-history" rather than straight-up "fantasy". — Lomn 14:43, 29 March 2013 (UTC)

There are other such books which deliberately don't use technology, but generally there are social, rather than technological or physical, reasons for doing so. For example, in the Dune franchise, there is no use of computers or any other "thinking machines"; the novel explains their non-use as a result of the Butlerian Jihad rather than any specific physical reason why they couldn't work. Similarly, by the time of the "Galactic Empire" and "Foundation" period of Isaac Asimov's extended universe, there are no working Robots, despite the fact that many earlier works in the same universe featured them. Asimov explains this as being a ban on the use of such Robots, similar to the Dune universe. So, if you want to create a future world where electricity isn't used, it may be more realistic to invoke a strong social reason (such as a religious proscription or a governmental ban or something like that) to explain why it isn't used, rather than any physical reason which, as noted, is simply ridiculous to anyone with even a basic understanding of physics. --Jayron32 16:05, 29 March 2013 (UTC)

Continuing in the theme of 'social' pressures, science fiction writers have also been known to invoke a need for concealment – either from other humans, angry aliens, or both – as a compelling reason to curtail or completely suppress the use of electricity (and the associated nearly-unavoidable leakage of electromagnetic radiation). To pick just a very few more recent works, this concept comes up in David Brin's Brightness Reef, David Weber's Safehold series, and John Scalzi's The Last Colony. TenOfAllTrades(talk) 17:33, 29 March 2013 (UTC)

well, for one thing, the ability of the earth's iron core to generate a magnetic field will cease, and we will be fried to death the next time the Sun has indigestion and burps up a stream of high energy ions. Gzuckier (talk) 20:07, 29 March 2013 (UTC)

• There are also The Waveries and Blood Music if you want good sci-fi treatments of an end to electricity. μηδείς (talk) 22:35, 30 March 2013 (UTC)

March 30

Medical procedure and patients' gender

Is not knowing the gender of a patient in some contexts life-threatening? Traditionally, I suppose doctors would just know, but transsexual operation appear to be getting better and better, so could it happen that doctors don't know the biological gender of a patient? OsmanRF34 (talk) 00:31, 30 March 2013 (UTC)

That sort of thing comes up when a doctor gets a patient's medical history. It would be rare (though not impossible; for example if an unconscious patient needed an emergency procedure) for a patient to walk into a doctor's office and not have the opportunity to give a medical history. There are some issues; for example IIRC, pregnancy tests, if they give a positive result for men, can be an indication that the man may have cancer:[8] Now, it is not recommended that men take pregnancy tests for that purpose, as false positives can lead to unnecessary and possibly dangerous further procedures and tests. But there are biological differences between men and women which it would be important for a medical doctor to know, and post-operative transsexuals could generate complications for themselves if they did not reveal that fact through discussions with their doctors. Per the ideas of medical confidentiality, that sort of information is for the use of the doctor only, and only for treating the patient. --Jayron32 00:51, 30 March 2013 (UTC)

I distinctly remember that this same question has been posted on Ref Desk a few times before. However I was unable to come up with a way of searching the archives to find them. If someone knows, perhaps they could share it. Once imaging is used though, no doctor should be fooled. The difference between a female pelvis and a male pelvis is obvious visually and on X-ray, and any surgery to alter it would show up. Surgery can create a pseudo-vagina in a male, but he'll still have no ovaries and tubes etc to show up on ultrasound. The geometry of the chest, back, and neck skeleton are somewhat different too, though you do come across women with very short necks and men with very long necks, as with Canadian Jenna Talackova. Wickwack 120.145.158.214 (talk) 03:37, 30 March 2013 (UTC)

I must assert here that we have genetic sex, rearing sex, gonadal sex, and assignment sex. When we say Gender, which gender are we referring to? Even in a transgender person with no ovaries, the chromosomal (genetic) sex might still be female; similarly, in a feminised male, the genetic sex would be a male; some communities with high illiteracy and non-access to hospitals might wrongly assign the sex of a virilised female as male and rear her as such. These are valid problems in medicine. Genetic sex can be discovered by doing a FISH test on a scraping from the oral (buccal) mucosa. ----

Sex differences in medicine states "More recently, medical research has started to understand the importance of taking the sex in to count as the symptoms and responses to medical treatment may be very different between sexes." but doesn't go into any details. Women's health has a short section that says that proton pump inhibitors can interfere with calcium uptake, leading to bone fractures, presumably much more with women than men. There are some difficult to diagnose diseases such as lupus and ovarian cancer that are more prevalent in women than men, so mistakenly identifying someone as a man could conceivably result in a life-threatening delay in diagnosis and treatment.--Wikimedes (talk) 21:29, 30 March 2013 (UTC)

Sure. If a person breaks a law or embarrasses a powerful and violent antagonist because of deceit involving some gender classification, it might be life-threateing-- at least that happens regularly on TV. There are countries in which the penalty for cross-dressing might be death if homosexuality is inferred, so any unplanned medical encounter that revealed that the person was trying to deceive that country's social rules might put the patient at risk for serious harm or death. alteripse (talk) 01:09, 31 March 2013 (UTC)

Proton current

Normal electric current is the flow of electrons through a conductor. Is t possible to create a proton current, by using a custom designed zeolite fibre to conduct a flow of protons or protonated small cations? How would the properties of such a hypothetical circuit be? Plasmic Physics (talk) 02:54, 30 March 2013 (UTC)

Since the Wiki article on zeolites says they can absorb a variety of cations and exchange them with adjacent substances, then yes, they can carry an electric current. Since the fibre would be positively charged, this would have to be nuetralised. A very thin fibre with an insulated coaxial screen negatively changed (not to be confused with using a coaxial screen to carry a return current) should do that. I have no idea how you would estimate the electrical resistance of such a conductor, but think it would be inconveniently high, and so the current carrying capacity would be uselessly low. Transient phenomena such as inductance and capacitance could be easily calculated from the physical dimensions (principally the geometric mean radius) using standard formulae as for any conductor. Since conductionover a long thin fibre would be dependent of the charge on the coaxial screen, the assembly could function as a Field Effect Transistor. However, the two key physical dimensions crucial to operator as a FET, geometric mean radius, and screen to conduction path separation, would I think have to such that operation as a FET would be absolutely hopeless. As a teenager, I once decided to make a FET using a long length of thin polyurethane varnished copper wire ("magnet wire" the thickness of a hair) soaked in water, using the water as the control gate. I was able to show that it worked, but the gain was something like 1 microamp per kilovolt, the absolute limit of measurability with what I had, whereas a typical commercially made junction FET using doped silicon is about 4 milliamps per volt - about 4 million times better. I expect a zeolite FET would be even worse. Protons on their own will not work in zeolite. Wickwack 120.145.158.214 (talk) 03:26, 30 March 2013 (UTC)

OK. What if the zeolite itself is carries a negative counter charge, effectively creating an inverse electride: H⁺(zeolite subunit)^-? Plasmic Physics (talk) 03:37, 30 March 2013 (UTC)

Won't they react and destroy the conduction? Wickwack 120.145.158.214 (talk) 04:06, 30 March 2013 (UTC)

Won't what react? Plasmic Physics (talk) 04:11, 30 March 2013 (UTC)

Why not just a tube of acidified water. Wouldn't that be functionally a proton conductor? --Jayron32 04:38, 30 March 2013 (UTC)

That be true, however, then the conductor would be equivalent to a molten salt, and not to the required solid metal. Plasmic Physics (talk) 04:48, 30 March 2013 (UTC)

So you're looking not simply for a conductor of positive charge, but system wherein a positive charge moves through a stationary lattice? Would you say I was cheating if I suggested a hole conductor? Someguy1221 (talk) 04:55, 30 March 2013 (UTC)

That is indeed a correct interpretation, however, a constraint of the system is that the charge carrier must be a proton or proton containing cation. Plasmic Physics (talk) 05:57, 30 March 2013 (UTC)

Are you only interested in zeolites? We have a short article on proton conductors and the fast ion conductor article has a few examples of proton conductors as well.--Wikimedes (talk) 20:57, 30 March 2013 (UTC)

Wonderfull links, thank you. Plasmic Physics (talk) 21:35, 30 March 2013 (UTC)

These are used as membranes though, where an electric field acros the membrane is sufficient to drive the hydrogen ions through it. You asked for fibres, which might have indicated you wanted some sort of proton carrying wire. Wickwack 124.178.140.181 (talk) 00:47, 31 March 2013 (UTC)

Yes. Plasmic Physics (talk) 02:21, 31 March 2013 (UTC)

Law of accelerating returns

Question about Kurzweil's Law of accelerating returns. Has the rate of technological change this century (so the last 13 years) followed the pattern predicted by this? Or are there any examples of technologies that support this? Any examples that refute it? --105.227.178.237 (talk) 10:40, 30 March 2013 (UTC)

Just an hour ago there was a program on BBC radio thattook issue with the sceptical paragraph about Moore's law in that Wiki article. In the BBC program this was compared to the famous example of doubling the number of grains on each chessboard; during the first 32 squares, it actually is still managable, it only becomes mind boggling large on the last 32 squares. Similarly, we have now had a bit more than 32 Moore's law doublings and we are only now seeing mind boggling appplications fo technology that the experts even a few years ago didn't expect to see in their lifetimes. An example given in the program was the automated cars of Google. A few years ago, most experts would have thought that autonomous cars driving in real life chaotic traffic would not happen for many decades, this used to be the typical example cited by sceptics of AI to point out how the brain is superior to computers (you have a lot of information to prcess in real time, you need to decide what is relevant and what not, you can't capture all of that using simple rules). Count Iblis (talk) 13:05, 30 March 2013 (UTC)

Some technologies that might support it:

1) TVs/computer monitors. We had only CRTs for many decades, then we suddenly had an explosion of new flat-screen technologies.

2) Light bulbs. Both CFL and LED bulbs are recent additions to home lighting options.

Some technologies that might refute it:

A) Medications. We tend to not get new medications which actually cure diseases, but only those which treat them. In antibiotics, specifically, our medications are actually becoming less effective, as the microbes adapt quicker than we can come up with new meds.

B) Economics. We don't seem to be getting any better at managing the world economy. We may even be getting worse, at least in the West. StuRat (talk) 13:21, 30 March 2013 (UTC)

Developments in biotechnology have or will have the potential to defeat microbes. And libertarians will argue that "managing" economics is inherently bound to fail. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:34, 30 March 2013 (UTC)

By "managing" I'm including the option of letting free markets do their thing completely unregulated. This approach doesn't seem to work all that well, either. We end up with Bernie Madoff scams. StuRat (talk) 13:46, 30 March 2013 (UTC)

Prevention and punishment of fraud or other criminal activity doesn't really qualify as "managing the economy". Punishing successful individuals and companies by taking away their incentive to produce, definitely qualifies. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:10, 30 March 2013 (UTC)

Bizarre statement, Stu. I have to agree with Bugs that fraud is considered a crime, not free trade, even under laissez-faire capitalism. μηδείς (talk) 22:31, 30 March 2013 (UTC)

The problem is, in laissez-faire capitalism, requiring companies to report on their activities is considered unwanted interference. And, without this requirement, it's easy to get away with fraud. So, while it doesn't outright legalize fraud, it does make it impossible to prevent. StuRat (talk) 02:34, 31 March 2013 (UTC)

Some factors which prevent technology from increasing exponentially are the speed of information transfer and patents. For the information transfer, publishing an article on a new technology isn't necessarily much faster than it was a century ago. It still involves writing it, submitting it to legit publication, and having them approve it. Yes, we can also put new info out directly on the internet, but there's so much bad info there, it's difficult to discern the wheat from the chaff. As far as patents go, new technologies often remain prohibitively expensive until the patents expire, and the length of time patents last isn't decreasing (see the case of the CFL: [9]). StuRat (talk) 13:33, 30 March 2013 (UTC)

One reason new technologies are more expensive is that the developers of those technologies need to recoup their research and development costs. Shortening the length of time a patent is in effect could reduce the incentive to create those new technologies. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:37, 30 March 2013 (UTC)

Yea, patents are a problem no matter what you do. Perhaps government funded research programs, where the results are free for all to use immediately, are the answer. StuRat (talk) 13:44, 30 March 2013 (UTC)

That means an increase in taxes. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:06, 30 March 2013 (UTC)

True. StuRat (talk) 02:39, 31 March 2013 (UTC)

What you're saying could be true (as there in a 1975 patent), but the citation (which is also in our article) doesn't say anything about patents. Nil Einne (talk) 00:55, 31 March 2013 (UTC)

Yea, that link implies the patent expired, but doesn't outright say it. Perhaps it was changed after the citation was added to our article. StuRat (talk) 02:39, 31 March 2013 (UTC)

How can it imply something about patents when it doesn't mention them at all? Were it not for the fact that patents have been brought up here and that one seems to exist, the obvious conclusion from the link would actually likely be there were no patents involved but it was more one of cost and market potential, no one bothered until someone figured out it could be done sufficiently cheaply in China to make it worthwhile. Nil Einne (talk) 03:02, 31 March 2013 (UTC)

The slowness of the patent system does not have the braking effect on innovation that you might think it has, for two reasons: a) You cannot have a valid patent for some development that is the result of reasonably expected refinement by a practitioner of the art. For example, if a steel garden tool as typically made by several competitors sufers from rust, and one of them decides to use galvanising, that would not be patentable, becuase any reasonable product engineer should think of that, and its mearely a question of manufacturing costs. But if he has a brainwave and thinks up a radically new way to apply galvanising at 1/10th the normal cost, that could be patentable. It's true that a lot of patents are taken out and paid for that are just such reasonably expected refinements, but if/when it comes to court, such patents should, and often do, fail; b) a heck of a lot of innovation is not patented anyway - the inventors just go ahead a make and sell, because the patent system expense and complexity is just not justified in may cases. In many cases innovators can get protection against competitors by the Registered Design system, which is cheap and simple. Wickwack 60.230.245.220 (talk) 01:39, 31 March 2013 (UTC)

a) Even if they might eventually win, many companies would still shy away from using any technology that would require patent litigation. StuRat (talk) 02:42, 31 March 2013 (UTC)

The one place I can think of very plain accelerating returns is in genome sequencing costs. They have been decreasing at a jaw-dropping rate over the last 6 years or so (those graphs are each log charts — so a straight line would already be exponential decrease, and those lines ain't straight). But I would caution, counter Kurzweil, that in the real world, exponential processes always level off or die off eventually. Always. That doesn't mean the last generation or so before they end isn't spectacular, or that we aren't a long way away from it, or whatever, but it's worth keeping in mind. --Mr.98 (talk) 13:42, 30 March 2013 (UTC)

Largest science

Whats the largest scientific organization in the US? Whats the largest in Europe? Pass a Method talk 14:29, 30 March 2013 (UTC)

The American Chemical Society is the largest scientific society in the world. That would then also qualify it as the largest such group in the U.S. as well. --Jayron32 15:33, 30 March 2013 (UTC)

OP: Do you mean the largest professional association/society of members (like the ACS), or the largest company or enterprise run by a CEO or Chancellor that does scientific research as its main activity? Or the largest research organisation that is owned by an entity (the entity not necessarily focussed overall on research), such as Bell Laboratories? Ratbone 120.145.158.214 (talk) 15:41, 30 March 2013 (UTC)

It all depends on what you call organization. Are governmental agencies like the NASA included? And what means by large? Large number of people or turn-around? OsmanRF34 (talk) 16:12, 30 March 2013 (UTC)

In Europe its Helmholtz Association of German Research Centres i belive. --Kharon (talk) 16:29, 30 March 2013 (UTC)

I have in the past heard its the National Academy of Sciences, or maybe thats the most prominent? Pass a Method talk 16:42, 30 March 2013 (UTC)

That has 2,100 members and 380 foreign associates; compare this with the 163,000 (!) members of the ACS. -- Jack of Oz ^[Talk] 19:56, 30 March 2013 (UTC)

thermodynamics

while calculating total entropy change in an isothermal reversible process,change in entropy of surroundings is calculated by the following formula= heat absorbed or released by the surroundings/temp of surroundings.But in text books,the authors have equated temp of system with temp of surroundings.Can u please explain why? — Preceding unsigned comment added by 49.201.252.13 (talk) 14:35, 30 March 2013 (UTC)

Well, eventually the temperature of any system will equal that of it's surroundings, if we assume the surroundings have a constant temperature and the thermal capacity of the system is insignificant compared to that of the surroundings. StuRat (talk) 14:41, 30 March 2013 (UTC)

Good old StuRat to the rescue! An isothermal process, by definition, is a constant temperature process from start to finish, and heat flows in or out of it without a change in temperature anywhere. There's no "eventually" about it! Entropy is reversible heat divided by system temperature and for a reversible process, total entropy is constant. Ratbone 120.145.158.214 (talk) 15:36, 30 March 2013 (UTC)

Isothermal usually means that the system doesn't change temperature as it expands, contracts, or whatever. It doesn't necessarily mean that the system and environment are at the same temperature. --140.180.254.209 (talk) 16:32, 30 March 2013 (UTC)

If the temperature of surroundings were different, the process would be irreversible. Ruslik_Zero 18:11, 30 March 2013 (UTC)

Derp. Yes, you're absolutely right. If the system and environment had different temperatures, change in entropy would be positive, so the process would be irreversible. --140.180.254.209 (talk) 21:09, 30 March 2013 (UTC)

What speed is needed so I'm always at the same time on the earth?

More specifically, what speed is needed on the surface of the earth so the sun is always on the same spot, because I'm rotating the earth aside the sun, so there will be never night. 181.50.183.182 (talk) 17:11, 30 March 2013 (UTC)

The circumference of the earth is 24,901 miles, so a little over 1,000 mph at the equator. μηδείς (talk) 17:15, 30 March 2013 (UTC)

Zero if you're at the North Pole -- all you have to do is very slowly turn. More generally, the farther you are from the equator, the lower the speed. Looie496 (talk) 17:19, 30 March 2013 (UTC)

I should know this, but it's not coming to me: What does the graph look like for the speed at each degree of latitude? At the pole it's technically 0, and at the equator it's 1,000 MPH. But what is it at 45 degrees north or south? 500 MPH? Larger? Smaller? Another way to put it: What is the circumference of the circle at 45 degrees of latitude? ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:31, 30 March 2013 (UTC)

If the Earth were a sphere, you would just multiply the equatorial radius by the cosine of the angle of latitude. The actual reference ellipsoid length of the 45th parallel will be marginally longer than 24,901 times 0.7071067 miles. Dbfirs 17:39, 30 March 2013 (UTC)

For a perfectly spherical Earth, which the Earth very nearly is, a graph of surface speed versus latitude in degrees would be a half-sine shape. Wickwack 124.178.140.181 (talk) 00:52, 31 March 2013 (UTC)

Hmm. I thought we had an article (or part of one) concerning a performance artist that stood on the north pole for 24 hours, carefully shifting himself against the turn of the earth so that he "stood in place" rather than revolving around the earth as every other human has always done. I was going to link it in reference to Looie496's post above. Now I can find nothing about it (but a remarkable lot about pole dancers who are also performance artists). Hoax article? Ring a bell? Matt Deres (talk) 01:04, 31 March 2013 (UTC)

Um, aren't we missing something (or perhaps I am mistaken)? At the north pole, you cannot turn so that the sun doesn't move in the sky, because of axial tilt. See Axial_tilt#Obliquity_of_the_ecliptic_.28Earth.27s_axial_tilt.29. Even an equatorial path would not keep the sun stationary, for the same reason. There is probably some path one could take to keep the sun at the same place in the sky, but it does not have constant latitude. SemanticMantis (talk) 01:21, 31 March 2013 (UTC)

Yes you are mistaken.

Prominent science

Whats the most prominent and reputable scientific organization in the Americas? What about in Europe? Pass a Method talk 21:18, 30 March 2013 (UTC)

Internationally, the ICSU. The USA's representative is the National Academy of Sciences, although other countries of the Americas have their own representatives. Of the European representatives, the DFG (Germany) is the largest, and the Royal Society (UK) is the oldest - the Académie des Sciences (France) is (probably) equally prominent and reputable. Tevildo (talk) 21:58, 30 March 2013 (UTC)

This question attracted answers from Jayron32, Ratbone, OsmanRF34, and JackOfOz. There posts were deleted at 21:18 by the OP, PassAMethod. See History. Deletion or editing other people's posts is very poor ettiquete, and may result in questions going unanswered. If you don't like answers or think people went off track, just politely say so and clarify your question. Wickwack 60.230.245.220 (talk) 01:21, 31 March 2013 (UTC)

Um, I didn't answer this question, if I recall. I answered a different question asked above. And my answer is still there. Are you sure you aren't mistaken? --Jayron32 01:31, 31 March 2013 (UTC)

Ooops! You are right. He didn't delete anything. He asked much the same question twice with different headings and different wording, and I didn't spot that for some reason. My appologies, especially to the OP, Pass A Method. Wickwack 60.230.245.220 (talk) 01:43, 31 March 2013 (UTC)

Co-transport... again

I asked a while ago about co-transport and how it works. I got a whole ton of stuff from you guys, which was helpful, but I'm still confused somewhat. So I'm going to lay out my question step by step, and I would appreciate if I could get a step-by-step from-the-basics type of answer too.

Now. I have a cell membrane. I also have sucrose, inside and outside my cell. Sucrose cannot diffuse through the membrane. Sucrose moves randomly (Brownian motion). I have learned that some membrane proteins facilitate passive diffusion when a molecule binds to it, leading to a conformational change which moves the substrate into the cell. Assuming that the cell interior has a higher concentration than the exterior, why is it necessary to use energy to move sucrose in (through cotransport with a sodium ion), when theoretically, the external sucrose moves randomly (as opposed to being "repelled" by its concentration gradient, it only collides with the membrane and any proteins on its surface less regularly than the sucrose on the interior). Shouldn't a protein whose active site is on the exterior and changes shape when sucrose binds to it suffice? Or is my perception of membrane proteins wrong in my belief that they only work in one direction? I've been told it has something to do with free energy and entropy. Could someone explain how this means that energy input is required (without math, using a real explanation of why energy is required?) Brambleclawx 22:48, 30 March 2013 (UTC)

I can't make sense of this question. Did you mean to say that the exterior has higher concentration than the interior? (You said the reverse.) Looie496 (talk) 00:24, 31 March 2013 (UTC)

It looks to me like he is trying to make the cell membrane act as a variation of Maxwell's demon, which will fail for the same reasons that Maxwell's demon fails. --Guy Macon (talk) 00:37, 31 March 2013 (UTC)

Brambleclawx, if you have a protein that adopts a shape such that sucrose is forbidden from moving in the other direction, you will need energy to "reset" the protein, such that it may take in another molecule. This reminds me vaguely of Richard Feynman's brownian ratchet, a device powered by brownian motion that only allows a wheel to turn in one direction. And it doesn't work without an additional power source. Someguy1221 (talk) 01:40, 31 March 2013 (UTC)