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October 12

Missing reference/article

Where/How do I find "the more complicated exact equation" is "derived without using any approximations" enticingly alluded to in the last paragraph of

http://en.wikipedia.org/wiki/Doppler_effect#Analysis

At LEAST a computer accessible reference NEEDS to be provided !!

What does the "gadget at the end of the URL mean ? I would like to contact the author of this URL to be sure to get the complete derivation. RARE —Preceding unsigned comment added by 83.226.97.246 (talk) 03:10, 12 October 2010 (UTC)[reply]

I don't usually do this, but I've fixed the question. The link was mistyped, and I figured I might as well clean up the other mess while I was at it. Looie496 (talk) 03:14, 12 October 2010 (UTC)[reply]

Do you mean the #Analysis part of the link? If so, it's an URL anchor. This makes your browser scroll to the 'Analysis' section of the web-page when you follow the link. CS Miller (talk) 10:16, 12 October 2010 (UTC)[reply]

The more complicated exact equation appears to be the second last equation in that section. It is then simplified to the last equation presented. The derivation shown appears to be uncomplicated math, so doesn't necessarily need a source. The Lord Raleigh book is mentioned and likely should be properly referred in the "Further reading" section. It is computer accessible - use your computer to look up the address of your local library. If by "gadget" you mean the little box-on-box thingy right after the link, that is automatically put in by the software for any link that starts with http:, as these links usually lead off the Wikipedia site. Franamax (talk) 15:58, 12 October 2010 (UTC)[reply]

Nuclear isomers for chemical transmutation of gold i.e. alchemy?

I think that for purposes of sci-fi, there are very few things so splendidly versatile as nuclear isomers. Even so, this application is likely a bridge too far. Yet...

Many nuclear isomers are known, which have slightly different masses than other isotopes with the exact same number of protons and neutrons.

Nuclear isomers can have very different stability - for example, 106 days for 177mLu versus 6 days for 177Lu by beta decay.

Nuclear isomers may be increased or decreased in energy by absorption of photons of the right frequency, or internal transitions with electron potential differences of the right frequency.

Nuclear isomers can have a half-life more than comparable to the age of the universe, e.g. 10¹⁵ years for ^180mTa.

Now what all this means is that, while it is by no means certain, it is possible that exposing a compound of lead to just the right excitation in a flame, for example, might cause the nucleus to jump to a slightly higher metastable state. Or perhaps lead as we know it is in a metastable state and it can be decreased in energy to some as yet unrecognized "stable" state? And if the effect is to increase the chance of alpha-decay, maybe it wanders over to be mercury... and maybe (perhaps with more excitations) it can emit a positron and then an alpha and end up as gold?

It's all very unlikely sounding, and yet, based on what I've read of these nuclear isomers, it seems like modern science can't preclude the conceivability of chemical alchemy, i.e. the possibility of rare but meaningful effects of chemical ionizations and other electron transitions on the breakdown of the nucleus. Can you prove me wrong? Wnt (talk) 06:22, 12 October 2010 (UTC)[reply]

Science cannot prove you wrong, that's not what scientific evidence does. It can show zero support for your ideas, but it cannot prove you wrong. Proof is an illusion, all we have is the existance or nonexistance of evidence. --Jayron32 06:25, 12 October 2010 (UTC)[reply]

Granted... yet before I thought of this just recently I'd have said that "alchemy is impossible" like anyone else. Now I look at a figure like the one above and I wonder, can you go below "7/2+"? If you can, why don't we see Lu or Hf with the lower energy that should correspond to that - does it spontaneously decay into something else? (I forgot to mention above that usually the higher energy isomer is more stable than the lower, though I don't know how consistent this is) And if such a transition should be triggered, it looks like it should take well under 100 keV (though admittedly, this is actually well over the ionization energy - I don't know why internal transition is so important for nuclear isomer decays...) So I feel like I've gone from thinking "how" to "how not?" quite abruptly. Wnt (talk) 06:44, 12 October 2010 (UTC)[reply]

I have to say I must completely disagree with Jayron. Science is better at showing something must be false than it is as showing that something must be true. I don't know enough about isomers to say anything useful, but if this were a question about nuclear reactions or about conservation of energy or whatever, I am very confident that people would be happy to pipe up to say "that is wrong and here is why." Science can definitely say that some ideas are wrong according to all current theories, no problem. One just has to know a bit about the subject at hand. On the subject of isomers, I've got no idea, personally, but surely others do. --Mr.98 (talk) 18:20, 12 October 2010 (UTC)[reply]

The main problem here may be one of definitions. You say you want "chemical" transmutation, but when exactly does something stop being "chemistry" and start being "nuclear technology"? I cannot imagine how you would excite a nucleus to the tune of even a single keV (~ about 10 million kelvin!), by methods that anyone would reognize as "chemical". –Henning Makholm (talk) 19:18, 12 October 2010 (UTC)[reply]

Agreed. This is a matter of definition. "Chemistry" deals with that domain of atomic interactions related to the electron cloud; while nuclear physics deals with that set of interactions related to the atomic nucleus. A series of famous papers was published during the 1930s and especially during the buildup to Manhattan Project, as the science of nuclear transmutation shifted from alchemical nonsense to scientific reality. Among these were infamous nuclear transmutations of lead to gold; but these "nuclear chemistry" tricks turned out to be among the least useful applications of nuclear science. In any case, the historical papers do make for interesting reading; and the processes to convert heavy elements into noble metals are well-known (just not very practical). You won't be inducing nuclear reactions with an ordinary candle-flame. Nimur (talk) 19:43, 12 October 2010 (UTC)[reply]

In chemistry there are all sorts of catalysts that will overcome absurdly high kinetic barriers. Isn't it possible perhaps to have some sort of "strange matter" catalyst that will use a bunch of stabilising interactions to lower the kinetic barrier, in a sort of nuclear version of an enzyme or Zeolite? John Riemann Soong (talk) 19:50, 12 October 2010 (UTC)[reply]

That really has been the domain of cold fusion research, and it has largely been unsuccessful at convincing the scientific community that any such "catalyst" does exist (or even that it theoretically could exist). Modern "cold fusion" research is often published with the key-phrase "low energy nuclear reaction" - you can evaluate the status for yourself; so far, no "useful" reactions with low activation-energy have been found under any conditions, including the presence of catalysts. Conventional theories of nuclear interaction (and by "conventional" I also include "relativistic quantum mechanical theories") require huge activation energies to overcome the electrostatic repulsion of the nucleus; or uncharged, fast neutron triggers in the form of a nuclear chain reaction. Nimur (talk) 20:27, 12 October 2010 (UTC)[reply]

If a transmutation is inherently exergonic, wouldn't you be able to supply an investment of activation energy that would be recouped later, in a controlled fashion? What about very strong electric or magnetic fields? John Riemann Soong (talk) 22:09, 12 October 2010 (UTC)[reply]

Yes, but it gets hot, because it's a lot of energy: nuclear fission. The process yields a net release of energy, in great quantities, even though the "activation energy" is high. Nimur (talk) 23:30, 12 October 2010 (UTC)[reply]

Ha, why gold? Gold is easy to procure. I need a process that will give me rhodium or osmium. I'm really glad that no chemical process can cause transmutation though -- if that were so, biological life would be very much under threat. John Riemann Soong (talk) 19:37, 12 October 2010 (UTC)[reply]

Part of my confusion is that there are low energy nuclear isomers like ^229mTh with a transition of 4.5-2.5 eV with a "nuclear gamma emission in the optical range".[1][2] (Does anyone have access to this one [3]? They didn't trigger it with a laser, did they??) I didn't realize till I'd gone on a bit that this other example I gave involved many KeVs, but I'm still not sure it's irrelevant, because I don't know if by causing an optical-range shift in nuclear energy, you might make an isotope go on to emit a multi KeV gamma ray. It is true that for such an improbable process as lead to gold you'd at least hope to be allowed to juxtapose various isotopes and transmit these gammas back and forth directly rather than relying on purely chemical reactions - I suppose I was counting it as "chemical alchemy" if it merely looks like you're doing a chemical reaction, from some 18th-century point of view, without an obvious nuclear reactor or particle accelerator being used. Wnt (talk) 20:24, 12 October 2010 (UTC)[reply]

Oh, here's where I got into hundreds of KeV by "chemistry" - [4]. Another paper I can't access, but it describes getting up to 150 KeV by creating "autoionization states" in which two or more electrons are promoted from holes in deep inner electron shells. From this I suppose that ionizing an ion takes a whole lot of energy. Admittedly though, reversing this process is not exactly what I'd call "chemistry" in the normal sense of the word... isn't quite exactly nuclear physics either though. Wnt (talk) 20:36, 12 October 2010 (UTC)[reply]

Your IOP article says

`Nuclear light', or the gamma radiation emitted by an atomic nucleus in the optical range, will probably be discovered experimentally in one or two years.

so they didn't do it (but that was in 2002). In order to use an optical-to-keV "amplifier", you'd have to have an excited isomer already (as your second SpringerLink article says), in which case you're not using true "base metal"; moreover, the lower energy states should be less likely to then conveniently emit alphas. As for ionization, it's just that the core electrons have much higher ionization energies than the valence electrons, as well as that each successive ionization energy is (typically, at least) higher. But you'll have to use un-alchemical processes like lasers or ion beams to preferentially excite/remove the inner electrons; normal chemistry will always go after the outermost ones. --Tardis (talk) 14:25, 13 October 2010 (UTC)[reply]

You're probably right. Even so, I wonder if some of the "ground states" we know and love are actually higher-energy nuclear isomers with half-life >> 10¹⁵ years. And if they can be triggered to drop to a lower energy ground state that is not so stable.

Also, I wonder if you can mix just the right combination of nuclear isomers so that the emissions from one are tuned perfectly to trigger a transition in the next. For example, to break up nuclear gammas down into bits of usable chemical energy, or in this case, to drive an alchemical transition. Wnt (talk) 20:04, 13 October 2010 (UTC)[reply]

I suggest taking a look at island of stability, Bismuth, neutrino theory of light, deuterium, double beta decay, proton decay, delta decay and Radium. ~A H 1^(TCU) 02:19, 15 October 2010 (UTC)[reply]

Calculate the heat supplied

This example problem in my class XI chemistry book proceeds as follows : A swimmer coming out from a pool is covered with a film of water weighing about 18g. how much heat must be supplied to evaporate this water, if it is at 298K? given, enthalpy of vapourisation of water at 373K = 40.66kJ/mol Solution : 18g is equivalent to 1 mole. heat supplied is equal to 40.66kJ/mol * 1mol = 40.66kJ

I feel that this solution to the problem is wrong, as the water on his body is not at 373K, but only at 298K. So, to raise the temperature of the water from 298K to 373K, 18*(373-298)*4.19 = 3771J = 3.771kJ of heat must also be supplied. So the final answer would be 40.66 + 3.771 = 44.431kJ. Am I right?? This is not a homework problem, I'm just asking out of doubt.. Can the enthalpy of vapourisation of a liquid be used at any temperature, or only at its boiling point? How can we assume that the enthalpy of vapourisation of water is same at both 373K and 298K?? Thank You. harish (talk) 10:57, 12 October 2010 (UTC)[reply]

Are you confusing evaporation with boiling? I evaporate solutions at 288K, so it doesn't have to be boiling. Also, boiling water would make the swimmer very uncomfortable. Finally, how does the water get hot enough to boil? Evaporation is AFAIK only a random escaping of high-energy liquid molecules at the surface of a liquid. --Chemicalinterest (talk) 11:13, 12 October 2010 (UTC)[reply]

Does the book problem really say about 18g ? Cuddlyable3 (talk) 11:35, 12 October 2010 (UTC)[reply]

What you need is the heat of evaporation at 298 K. If the only figure you have is for 373 K, you can adjust for the temperature difference using conservation of energy: It should take the same energy to evaporate the water at 298 K and then heat the vapor to 373 K as it should to heat the water to 373 K and evaporate it there. At the precision you're working at here, you can probably get away with assuming that the specific heat of liquid water (resp. water vapor) is not temperature dependent. –Henning Makholm (talk) 12:32, 12 October 2010 (UTC)[reply]

Boiling is always a kinetic phenomenon, because that's when vapor pressure of the liquid equals vapor pressure of the atmosphere, allowing bubbles to form within the water -- below this temperature, nucleation is simply unfavourable because bubbles cause surface energy. Evaporation is a thermodynamic phenomenon. The heat of vaporisation should be nearly constant (it probably changes dramatically when you have really large differences, but within the range of 0-100C I suspect it's rather the same). Notice that you cool down when your sweat evaporates! Your body has to supply heat for it to evaporate -- evaporation and boiling are inherently, endothermic processes. John Riemann Soong (talk) 15:53, 12 October 2010 (UTC)[reply]

The first figure at Enthalpy of vaporization shows a thermal dependency that appears to give about a 10% difference over that interval. I wouldn't call that "rather the same". –Henning Makholm (talk) 16:25, 12 October 2010 (UTC)[reply]

is it a contradiction to know something about something we can't know anything about?

Is it a contradiction if you prove, physically/mathematically, that we cannot know ANYTHING about a given "other Universe", and go on to prove, physically/mathematically, something that must be logically true in any "other Universe" -- I'm talking pure logic there, so that if a mathematician in that other Universe were to explore the property of the natural numbers in that Universe... -- , implying that you DO know something about it?? Thank you. 84.153.253.103 (talk) 13:18, 12 October 2010 (UTC)[reply]

We don't know anything about the potential other universe, we know something about maths and logic (namely, that they are universal in a sense that our "universe" possibly is not). From an information theory point of view, there is not information gain (if you hear that a guaranteed event has happened, you gain no new knowledge). If we prove or disprove a mathematical result, we do not gain any new information - we just make previously available information explicit. --Stephan Schulz (talk) 13:50, 12 October 2010 (UTC)[reply]

It is tempting to say that the truths of mathematics are ~~synthetic truths~~ analytic truths i.e. true by virtue of the definitions of the terms involved. Thus the truth of 1+1=2 is implicit in the definitions of "1", "+", "=" and "2" - it can only be false in a context in which at least one of these definitions is changed. And therefore an analytic truth, such as 1+1=2, must be true in all possible universes. However, there is a philosophical objection to this stance. To determine whether a proposition such as 1+1=2 is ~~a synthetic~~ an analytic truth, you have to use the laws of logic. But then you need to establish that the laws of logic are themselves ~~synthetic~~ analytic truths ... and you run into an infinite regress. Whether or not there are any analytic truths, true in all possible universes, is vigorously debated among philosophers. Gandalf61 (talk) 14:49, 12 October 2010 (UTC)[reply]

I think you've got your terminology a little mixed up. "True by virtue of its meaning" is "analytic", not "synthetic". Synthetic truths are true because they correspond to the state of things. "True in all possible worlds", on the other hand, sounds more like "a priori". --Trovatore (talk) 18:39, 12 October 2010 (UTC)[reply]

Yikes, I got that post really scrambled. Fixed now, I think. Gandalf61 (talk) 09:49, 13 October 2010 (UTC)[reply]

Um, thank you for your feedback guys but neither of you is answering my question, which includes the hypothetical that we do prove something necessarily true (and not by definition) that must be true in another Universe. This part of my question is not really what I wanted feedback on. I want you to assume that we can know, for sure, that if there is a being in another Universe who explores his universe somehow, then he will find something true (and not because he must define it the same way we do). If we do make that assumption: is this state now inconsistent with the fact that, in point of fact, we know nothing about the other Universe? That is, is there a contradiction or paradox here? (I am not asking if my assumptions are correct in your opinion: in both of your opinions Gandalf and Stephan, they are not; rather, I am asking if my assumptions lead to a contradiction/paradox in any sense). 84.153.253.103 (talk) 16:44, 12 October 2010 (UTC)[reply]

The answer is obviously yes, it's a contradiction to prove both that we can't know anything and that we know something. Why do you even need to ask? But the contradiction is meaningless, because the statement that we can't know anything at all about a different universe will never be true. Looie496 (talk) 17:11, 12 October 2010 (UTC)[reply]

And for me (OP again), it is just as obvious that 'we can't know anything at all' is in some sense true: we have no channel of information. The other Universe could be any way at all (it has perfect informational entropy in Shannon's sense -- we have perfect uncertainty about it), and we have no informational channel; yet, even though it could be "any way at all", and without our having any channel to learn anything about it, what would happen if we presumed to know something about that Universe anyway? You say this could be a paradox: can you make the paradox more explicit for me, like mathematical? Thank you. 85.181.51.248 (talk) 18:51, 12 October 2010 (UTC)[reply]

If you know that you "can't know anything about a given other Universe" then you know that much about it already. Can you imagine a universe with slightly different values of universal constants? With that assumption, you can derive likely properties of that other Universe. But that also assumes that there would be any physical laws at all in other universes, whatever they might be. And it would seem likely that every universe must have laws, but that is again an assumption being made just by employing the concept of "universe" itself: something as distinct from nothing, or from "pure chaos" which would amount to the same. I think the answer is yes, the paradox arises from your assumptions. WikiDao ☯ (talk) 12:15, 13 October 2010 (UTC)[reply]

I belive your apparent paradox arises because you postulate a separate universe which we can't know anything about, but by then you have already stated one property of it, that it exists. (As opposed to bigfoot and the loch ness monster.) If you drop this assumption, saying that the universe may or may not exist, it follows that you can't make any a-priori true statement. EverGreg (talk) 08:25, 15 October 2010 (UTC)[reply]

Yikes! Spooky overlapping of Non-overlapping magisteria: this is somewhat close to the ontological argument for the existence of god! --ColinFine (talk) 09:34, 16 October 2010 (UTC)[reply]

Electrode potentials at red heat

Why do electrode potentials change at high temperatures? For example, charcoal can reduce sodium carbonate to sodium at a high temperature. Oxide can reduce protons to hydrogen at high temperatures, decomposing water. Iron(III) chloride releases chlorine when heated. Why do oxidizing agents and reducing agents get so much stronger at high temperatures? --Chemicalinterest (talk) 14:10, 12 October 2010 (UTC)[reply]

Electrode potentials measure the equilibrium point of the chemical reaction that takes place at the electrode. Such an equilibrium changes with temperature because it is in principle just an expression of the Boltzmann distribution as applied to different binding states. –Henning Makholm (talk) 15:04, 12 October 2010 (UTC)[reply]

Thanks. A more specific question: Why is carbon such a strong reducing agent at high temperatures? --Chemicalinterest (talk) 15:42, 12 October 2010 (UTC)[reply]

See Gibbs free energy. Unfavourable reactions can sometimes become more favourable at high temperatures, because of entropy. Notice in all these reactions you release a gas that also escapes at that high temperature! John Riemann Soong (talk) 15:47, 12 October 2010 (UTC)[reply]

Also, it isn't just the "T" in the G=H-TS equation that makes the Gibbs Free Energy temperature dependant. In general chemistry, they teach you to "assume" that H and S are temperature independant. This is not strictly true; at the small ranges of temperature near room temperature, say from 1-100 degrees celcius, H and S show only small variances with temperature, enough to be ignored. At very high temperatures, however, H and S show measurable temperature variation. --Jayron32 20:09, 13 October 2010 (UTC)[reply]

Chile mine rescue

How much did the rescue operation for the 33 Chile miners cost to date? I realize it is ongoing, but is there an estimate for a final cost? Googlemeister (talk) 15:30, 12 October 2010 (UTC)[reply]

Codelco is a nationalized corporation, so it will be hard to differentiate between public- and private-sector expenses. Any labor or equipment that was needed for the rescue could be diverted from another mine at "no market cost." (Read Chilean nationalization of copper for historical background on revolutionary Chileanized Copper politics). As another example, the paramedic to be sent down is a member of the Chilean Navy - are his expenses to be counted? He'd have to receive his training and salary whether he were on a rescue mission or not. Probably the most clear-cut expenses are the purchases (or leases) of three large, advanced drills which otherwise would not have been needed (a 27-inch diameter bore drill (Schramm TX130) is pretty expensive - but it is both a rescue expense and a capital investment for the mining company - and our article notes that the American skilled-labor to operate it was "all volunteer"). Keep in mind of course, you can't walk into any old store in Chile and buy a TX130 - only one authorized dealer exists, and it's priced on a case-by-case basis... (like most large, special-purpose industrial equipment - and in this case, there is exactly one buyer, exactly one seller - "market economic price" does not apply). Finally, don't forget to factor in monetary losses - the opportunity cost of suspended mining operations while the rescue effort diverted labor-force (and presumably, the actual cave-in interfered with normal operations as well, both to re-evaluate safety and simply because the cave-in disrupted normal operation of the mine). In summary, this rescue operation is large and it is nigh-impossible to clearly account for every attributable expense that it incurred, especially the intangibles like labor costs and loss-of-productivity. I would recommend an approach similar to estimating the cost of a space-shuttle launch - where you evaluate the total project-size, number of man-hours consumed, and the fixed- and variable costs (see this economic evaluation of Space Shuttle). This way you can end up with two numbers, loosely translated as: "amount spent" and "amount we would have spent anyway." Similarly, consider reading Codelco's financial reports for 2010 and comparing originally projected- (pre-accident) versus actual- (post-accident) earnings for 2010. The difference, after you account for other market factors, should give you a good approximation of losses due to the mining accident. Ironically, the reduced production of copper will drive the price of copper up - economics is a complicated topic. Nimur (talk) 16:57, 12 October 2010 (UTC)[reply]

Don't forget that there are also gains from the rescue operation that go beyond the lives of the 33 miners. In the short term, all the rescue workers, families and journalists have to be fed and lodged, which must be a boom for the general economy of the area (although local miners who weren't traped have lost wages). In the long term, the whole operation is great publicity for Chile as a whole: it will be hard to quantify the effect of that publicity, but if you imagine that winning the World Cup in soccer is said to be worth a gain in 0.5% of the winning country's GDP, you can imagine that the economic benefits of the mine rescue will be very substantial indeed. Physchim62 (talk) 17:52, 12 October 2010 (UTC)[reply]

Publicity cuts both ways. One of the "messages" attached to this story is that Chile's industry operate using unsafe and archaic conditions, which could discourage investment in Chile. Obviously that is a gross generalization that may or may not be connected with the larger reality beyond this one company. However, in general I would expect that industrial disasters are not good for a country's image. Rescuing the men does help, but it isn't obvious to me that this story is a net positive. Dragons flight (talk) 22:54, 12 October 2010 (UTC)[reply]

I personally wonder what the miner's next paycheques are going to look like, considering they haven't yet punched the timeclock on the way out the gate after their shift ends. There's going to be an awful big number in the "overtime hours" column. ;) Franamax (talk) 20:00, 12 October 2010 (UTC)[reply]

Well supposedly their company is bankrupt anyways so they might not get a paycheck. Of course, they are stuck in a gold mine, so maybe there are some nuggets lying around for payment. Googlemeister (talk) 20:20, 12 October 2010 (UTC)[reply]

What's more, they can't have been awake 24/7 while trapped; surely they've been sleeping on the job for several hours every day. Nyttend (talk) 01:30, 13 October 2010 (UTC)[reply]

When the two miners were rescued from Australia's Beaconsfield Mine collapse after two weeks trapped underground four years ago, they made a deliberate and conspicuous act in front of the TV cameras of placing their time cards back into their relevant places. Don't know if they got paid any overtime for it. Both did subsequently do OK from media appearances. HiLo48 (talk) 06:13, 13 October 2010 (UTC)[reply]

Did the miners have a union contract? If their contract was anything like a union contract at a US company where I once had to work "around the clock" for days, you get straight time for the first 8 hours, then time and a half for the following 8 hours, then double time for the next 8 hours, which brings you to the start of the regular work day, and then the overtime pay stops because you are in an 8 hour "rest period" at regular straight time, even if still at the jobsite. After the 8 hour "rest period," (at straight time) you are back on a cycle of 8 hours time and a half, followed by 8 hours of double time, and then the cycle repeats. By that plan, it would be 60 days times the equivalent of 36 hours pay per day, or 2160 hours pay. A year's pay for two months in hell. Their contract might have additional benefits for missed days off, or having to work Sundays and holidays. Edison (talk) 16:26, 13 October 2010 (UTC)[reply]

Or, given the unique conditions and hazards of their work, they may have specific contractual line-items that specify payment-terms for emergency events such as these. While this particular cave-in was extraordinary, accidents of lesser severity happen often enough that a "shelter" was built inside the mine; it seems probable that the labor contract would include provisions for dealing with such conditions. Nimur (talk) 17:43, 13 October 2010 (UTC)[reply]

Back to the original question, the Santiago Times has estimated the cost of the rescue at US$20 million, mostly in the hire of equipment. Physchim62 (talk) 07:55, 14 October 2010 (UTC)[reply]

Unitary operator

Much like |±z> and |±x>, imagine we have two orthonormal set of eigenvectors given by |αi> and |βi> with i = 1, ..., n. They could be thought of as eigenvectors of two operators A and B respectively. Using these two set of eigenvectors, assume that I construct an operator U such that it has the following matrix elements: (U)_ij = <αi|βi>. Show that U is unitary.

This was my homework question, but I've done a pretty good job of convincing myself that U is not unitary. I figure that (U^†U)_ij = Σ<βi|αk><αk|βj>, which, if U is unitary, is supposed to equal δ_ij. But I don't see why it should. Did I do something wrong? 74.15.136.172 (talk) 18:48, 12 October 2010 (UTC)[reply]

Use the completeness relation : Σ|αk><αk|=1 and then use the orthogonality relation : <βi|βj> = δ_ij. 169.139.219.254 (talk) 19:25, 12 October 2010 (UTC)[reply]

But, if we said |α1> = |+x>, |α2> = |-x>, |β1> = |+z>, and |β2> = |-z>, then we get... oh wait, I forgot about the sum, nevermind! 74.15.136.172 (talk) 23:49, 12 October 2010 (UTC)[reply]

Emulsion paint - why not water-soluble when dry?

Emulsion paint goes hard after drying but does not soften when it gets wet. Why? What would make it soften - what solvents? Emulsion paint is used in the UK for painting interior walls and ceilings and may have a different customary name in other countries. Thanks 92.15.31.184 (talk) 20:59, 12 October 2010 (UTC)[reply]

Emulsion polymerization is a good place to start. Acroterion _(talk) 21:31, 12 October 2010 (UTC)[reply]

I'm very doubtful that applies to humble emulsion paint, despite having one word in common, as emulsion polymerisation appears to involve a lot of heat. Whereas emulsion paint just dries off at room temperature. 92.15.31.184 (talk) 21:57, 12 October 2010 (UTC)[reply]

From the lead section: "In other cases the dispersion itself is the end product. A dispersion resulting from emulsion polymerization is often called a latex (especially if derived from a synthetic rubber) or an emulsion (even though "emulsion" strictly speaking refers to a dispersion of an immiscible liquid in water). These emulsions find applications in adhesives, paints, paper coating and textile coatings." How the dispersion then acts is unclear to me. Acroterion _(talk) 01:04, 13 October 2010 (UTC)[reply]

You may find a useful start in acrylic paint, which is the term I'm familiar with. That article doesn't have much on the process, but following the links in that article, particularly to acrylic resin and emulsion provides a little more information. Basically it seems the acrylic resin suspended in the water based solution undergoes polymerisation as the water evaporates, creating an insoluble product. The polymerisation probably is mildly exothermic, but becuse its such a thin layer it may be hard to perceive (just me speculating). The article doesn't say what solvents would dissolve it, but acrylic resin goes on to talk about Poly(methyl_methacrylate) as another acrylic resin and I know that can be dissolved with xylene (and probably other similar organic solvents). Mattopaedia ^{Say G'Day!} 05:03, 13 October 2010 (UTC)[reply]

Yes, polymerisation can be somewhat favourable, in terms of energy. however an excess of water makes polymerisation unfavourable. Drive off water -- drive polymerisation forward. John Riemann Soong (talk) 08:13, 14 October 2010 (UTC)[reply]

... and, from my experience of cheap emulsion paint in the UK, it does remain partially soluble, especially if it is applied in an area where it fails to dry out fully. I haven't found a solvent for fully dried paint. Dbfirs 07:43, 13 October 2010 (UTC)[reply]

White spirit or turpentine are used for cleaning paint brushes, so perhaps they would be a solvent. 92.15.20.132 (talk) 15:46, 17 October 2010 (UTC)[reply]

No, these are solvents only for oil-based paint. They cannot be used for emulsion paint (or at least they are less effective than water). Dbfirs 16:57, 17 October 2010 (UTC)[reply]

why wrinkles around mouth

why do we get wrinkles around the mouth (like a cartoon beard)? I'm in my late twenties and it is one of the first places I'm getting wrinkles. I can't think of particular facial gestures / expressions I make that stress that part of the skin, unless it is from brushing my teeth... Is there something I can avoid doing so as not to exacerbate that... Thank you. 85.181.51.248 (talk) 22:14, 12 October 2010 (UTC)[reply]

There is no way to avoid wrinkles, but if you smile more often than you frown then you will have pretier wrinkles. 169.139.219.254 (talk) 22:25, 12 October 2010 (UTC)[reply]

Injections of Botox are used for cosmetic reduction of wrinkles, typically by celebrities. See Botulinum toxin#Cosmetic. Cuddlyable3 (talk) 08:49, 13 October 2010 (UTC)[reply]

See transverse muscle of the chin. ~A H 1^(TCU) 02:10, 15 October 2010 (UTC)[reply]

natural sciences v. finance

How feasible is it to transition from the natural sciences to finance? A lot of the skills seem similar. Because I haven't taken an econometrics class, I am curious when people complain about all-nighters in finance or accounting (this is at the McIntire School of Commerce) whether their complaints are really legit, compared to all-nighters in the study of something also intensively quantitative like ecology or physical chemistry for example.

I guess I want to ask is -- what do undergrad students of finance and accounting study? Is it really hardcore? Looking at some appropriate articles, I see things like stochastic models and fractals -- so I wonder if I'm missing out. John Riemann Soong (talk) 22:30, 12 October 2010 (UTC)[reply]

Around the science buildings at Stanford are many job-postings for banks, insurance companies, and brokerage agencies seeking to recruit "quants" - the industry buzz-word for "quantitative thinker." There are more postings for financial industry careers than for jobs in engineering and science. In the computer science building, the posters advertise with the catch-phrase "Hack Wall Street." The operational wisdom in the finance industry seems to be that anybody who can get (part-way) through a science or engineering curriculum has already well-satisfied the mathematical and cognitive requirements for a career in the financial industry. I imagine that if you sought a degree transfer you would have no trouble with the material. ~~Undergraduates~~ Graduate students in finance at Stanford study the Financial Math curriculum, (which is more rigorous than a Finance curriculum); their mathematical training includes elementary statistics and probability, some calculus, a differential equations class, and at least one economics course. Comparatively, undergraduates in the physical sciences are required to fulfill more rigorous mathematical training: advanced calculus and differential equations, advanced statistics, probability, and at least one economics class. Nimur (talk) 22:37, 12 October 2010 (UTC)[reply]

Haha I don't want to transfer degrees. I just want to know that if grad school / med school / TFA doesn't pan out or whatever, how difficult would a transition be after graduation. (Also I happened to sign into an honors physics track despite being a biochemistry major -- for the sole reason that the honors track was more interesting. So in the face of uncertainty I wonder if I will be punished careerwise instead of rewarded for my risk-taking.) John Riemann Soong (talk) 22:45, 12 October 2010 (UTC)[reply]

As Richard Feynman pointed out on numerous occasions, as long as there are wars, there will be demand for physicists:

“

After the war, physicists were often asked to go to Washington and give advice to various sections of the government, especially the military. What happened, I suppose, is that since the scientists had made these bombs that were so important, the military felt we were useful for something.

”

— Richard Feynman

This is still true in the 21st century. Nimur (talk) 22:52, 12 October 2010 (UTC)[reply]

If you do any kind of math based natural science at a well-known university, you will be actively courted for consulting and financial jobs. --Mr.98 (talk) 13:04, 13 October 2010 (UTC)[reply]

What's the usual GPA cutoff for a tier 1 school? John Riemann Soong (talk) 21:15, 13 October 2010 (UTC)[reply]

expansion of the universe

Does the correlation of greater speed at which bodies travel away from us and the greater distance just mean that the greater the distance away from us not only the older the body is but the faster that time is ticking? --96.252.213.127 (talk) 22:53, 12 October 2010 (UTC) In other words does a second near a Quasars or other distant object that is moving away at faster speed than closer objects, ticking faster than a second here? --96.252.213.127 (talk) 04:19, 13 October 2010 (UTC)[reply]

You will perceive that any object with a GR redshift (from motion or gravity or the expansion of the universe) has a "slower clock" than yours, by definition. Let's take a very common clock, which is light itself. Red light "vibrates" slower than blue light does. So any light that has been red shifted will vibrate less quickly than its source would otherwise indicate. Hence the events encoded in that red shifted light will take the same number of "light vibrations" as usual, only those vibrations will be spread over a longer period of time from the observer's point of view. Hcobb (talk) 04:39, 13 October 2010 (UTC)[reply]

Okay then that seems to confirm that whatever is going on now where the object is, is occurring at a faster rate than we can observe. In other words explosions that occur at greater distances will appear to take longer to complete. Can we use percent redshift to determine how much slower our observation tiem frame is for these events versus the time they actually take to complete? --96.252.213.127 (talk) 05:37, 13 October 2010 (UTC)[reply]

The universe (as far as we can see) is roughly homogeneous; that means that nothing else is going faster or slower in any absolute sense, since that would make it different from us. That said, yes, we see cosmological events slowed down by the same ratio that their light frequencies are lowered. This is possible because the object emitting the light is moving away from us; see Doppler effect for some pictures. For example, type Ia supernovas all look similar to one another, and the redshifted ones also happen slower by just the right ratio. This is an important piece of evidence that the cosmological redshift really is a Doppler shift and not, say, a light scattering effect like the one that makes the sun appear red near the horizon. -- BenRG (talk) 07:15, 13 October 2010 (UTC)[reply]

... and don't forget (as Ben implied above) that any observer on the far side of the observable universe would see events here slowed down by the same amount. The whole effect is thought to be caused by the metric expansion of the space in between. There is no "real" difference in the speed of events (or the speed of objects) here and at the extreme, just a symmetric difference in our observations. Dbfirs 07:35, 13 October 2010 (UTC)[reply]

"...just a symmetric difference in our observations." okay that is the terminology and understanding I was looking for. In other words a second is a second here and a second is a second there but an even that takes a second to happen there that is observed from here only appears to take longer than a second when observed from here. --96.252.213.127 (talk) 13:11, 13 October 2010 (UTC)[reply]

I suppose we should exclude regions near to large black holes or similar super-massive objects, because a large gravitational field will upset the symmetry. Dbfirs 15:36, 13 October 2010 (UTC)[reply]

Would gravitational microlensing in this case bend not only space, as viewed from a faraway observer, but also distort the apparent speed of time relative to the observer? ~A H 1^(TCU) 02:06, 15 October 2010 (UTC)[reply]

I'm not sure about that (we await an expert in GR), but my intuition is that it wouldn't if the large mass was well away from the two observers. Dbfirs 13:05, 15 October 2010 (UTC)[reply]