Wikipedia:Reference desk/Archives/Science/2008 May 5

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1 May 5

[edit] May 5

[edit] Aresol Can

Why in Aresol Cans and spray paint cans is there ball inside that makes a noise when you shake it? x

Please reply the fate of the world rests upon it... —Preceding unsigned comment added by 80.176.73.246 (talk) 00:51, 5 May 2008 (UTC)

The ball is there to mechanically agitate the contents of the aerosol can when you shake it, that is, it helps you mix the contents. For instance, if you try spraying a can of red paint without first shaking it up, you get lots of propellant and very little paint. The world can rest safe... Franamax (talk) 01:12, 5 May 2008 (UTC)

It is very difficult and quite unsafe to insert a wooden stick into the aerosal can to stir up the paint medium and pigment, so the little ball is necessary. Edison (talk) 03:32, 5 May 2008 (UTC)

[edit] Other "dry ices"?

Moved from Talk:Dry ice: Is there any other substance that forms a "dry ice"? Samw (talk) 03:09, 5 May 2008 (UTC)

Most substances have some range of pressure and temperature at which sublimation happens. Read the article for other examples of normal temperature and pressure sublimation. SpinningSpark 03:21, 5 May 2008 (UTC)

Thanks! So would you call naphthalene a "dry ice"? Or does the concept of "dry ice" even make sense scientifically and it's strictly a marketing term (which it was originally)? Samw (talk)

I wouldn't call it "a dry ice". Actually I've never thought of "dry ice" as a generic description for anything, rather just a name for a specific thing (solid CO₂). Naphthalene is not "ice" in the colloquial sense (it's not cold), so I would just call it a "solid". Maybe a "volatile solid". DMacks (talk) 05:19, 5 May 2008 (UTC)

Usually "ice" refers to water ice, and the special exception of dry ice is the name for solid CO₂. Not all cold solid crystals are ice; otherwise, steel and copper and quartz could be "ice". Nimur (talk) 16:43, 5 May 2008 (UTC)

In astronomy, they call any solid an ice that is lighter than rock. Ammonia ice and methane ice for instance as well as water and carbon dioxide.SpinningSpark 16:50, 5 May 2008 (UTC)

Getting slightly off topic, but I think astronomy is a bad guidance for such definitions. For astronomers, stars consist of hydrogen, helium and "metal"---where "metal" is everything heavier than helium... --Dapete ばか 19:11, 5 May 2008 (UTC)

To have something resembling "dry ice" you need to have a substance with a triple point pressure that is higher than the atmospheric pressure and a sublimation point below room temperature (this second requirement is so that the substance feels cold). From the common substances listed in this table [1], only acetylene (and CO2) meets the requirements, assuming you want your dry ice to be dry at sea level, which is 101 kPa. If you are willing to move up to the mountains, a couple other substances could work. For example, xenon and nitrous oxide could form dry ices in Mexico City, where the ambient pressure is around 78 kPa. And if you use a vacuum pump with low enough pressure, many other substances can be made to behave like dry ice. A couple of the substances in the table, graphite and uranium hexafluoride, have high enough triple point pressures so that they will never be a stable liquid under normal pressure, but their sublimation point is higher (for graphite much higher) than room temperature, so I am reluctant to call them "ices". --Itub (talk) 12:52, 6 May 2008 (UTC)

[edit] Power to weight ratio

When examining the power to weight ratio of an automobile, comparing it to another, horsepower per lb. or kg. is considered. I'm wondering if a torque to weight ratio should be given consideration. If so, should it be given equal consideration as horsepower, a certain amount of consideration, or none at all? Note: peak RPM, aerodynamics, and other values are not important to me at this time, just power to weight ratio. Thanks. MoeJade (talk) 04:41, 5 May 2008 (UTC)

That's a good question/comment. I venture to say yes. One of the most important factors when considering the acceleration of an automobile is the power to weight ratio, however HP is given as peak. Torque peaks at lower RPMs, so I would say that an important variable would be low end torque to weight ratio. That would be an excellent criteria for determine the cars ability to accelerate quickly. Wisdom89 _{(_T / ^C)} 04:45, 5 May 2008 (UTC)

Depending on the circumstance, it should definitely be given consideration. Wisdom89 _{(_T / ^C)} 04:46, 5 May 2008 (UTC)

-Thanks for the fast response, Wisdom89! So would you give torque to weight ratio equal consideration, averaging power to weight and torque to weight together? Or does power to weight take precedense (sp?), with torque considered, but not given as much weight? MoeJade (talk) 04:54, 5 May 2008 (UTC)

That's tough. I think if you were going to calculate both ratios, they should be kept separate. You'll probably find a positive correlation between the two anyway. In other words, high power to rate ratios correspond to high torque to weight rations. With such a relation, the power to weight would probably take precedence. Wisdom89 _{(_T / ^C)} 05:35, 5 May 2008 (UTC)

Another thing that we have to bear in mind is that at any given torque, the engine is producing a certain amount of (horse) power, so that reinforces our positive correlation. This is probably why you never see the torque to rate ratio advertised or touted about. Wisdom89 _{(_T / ^C)} 05:46, 5 May 2008 (UTC)

Technically that is true only if the vehicle is moving. If it isn't currently moving, mechanical work is not being done (and thus output power is zero). But that may not be a very important measure for most people anyway. Most of the time I think you only have to worry about overcoming inertia (and some small amount of static friction), so overall torque or force doesn't matter all that much. --Prestidigitator (talk) 08:29, 5 May 2008 (UTC)

Even if an engine is idling and the crankshaft is being spun, then power and torque are still being generated. Wisdom89 _{(_T / ^C)} 12:44, 5 May 2008 (UTC)

I was trying to point out a case where the two would not be correlated. That wouldn't be at idle but instead when the engine is engaged and trying to move the vehicle when there is a significant force resisting initial acceleration from a stand-still. That is a case where force (torque if we want to ignore gearing and wheel characteristics for the vehicle) is going to be significant, but output power is zero until it "breaks free". While weight might increase the amount of static friction that has to be overcome a bit, I think significant resistive force would really come if the vehicle is stuck or starting on an up-hill gradient.

Besides, aside from that needed to overcome a small amount of friction, no torque/power is required to keep the crankshaft moving at idle. Your efficiency for any engine during that condition is going to be very close to zero, and attempting to measure the amount of torque/power is likely to significantly change the value anyway unless you are very careful about how you measure/deduce it. --Prestidigitator (talk) 18:55, 5 May 2008 (UTC)

O.k. Good input. Basically, a friend and I were comparing two vehicles for potential acceleration. Car A has 221 HP, 236 f/lb. of torque, and weighs 1430 kg. Its power to weight ratio is 0.154, or 0.154 HP per kg. Car B has 216 HP, 152 f/lb. of torque, and weighs 1180 kg. Its power to weight ratio is 0.183, which is notably better than car A. However, car A has 236 f/lb. of torque, or a torque to weight ratio of 0.165, which is substantially better than car B's torque to weight ratio of 0.128. Observing power to weight ratio only, car B wins easily. But, with car A's significant amount of torque,and its significant torque to weight ratio, it's a tougher choice than it looks, to me, anyway. Any additional comments on this are welcome. MoeJade (talk) 13:52, 5 May 2008 (UTC)

I think Car A would have better low-end acceleration, but Car B would eventually go faster. However, gearing ratios would also come into play. If you can find the cars listed in a Road & Track or Car & Driver magazine test, they often show detailed full-on acceleration numbers. Franamax (talk) 16:29, 5 May 2008 (UTC)

[edit] Does an Electron Occupy Space? Why?

There is a paragraph in article Electron I don't get it well:

"The electron is currently described as a fundamental or elementary particle. It has no known substructure. Hence, for convenience, it is usually defined or assumed to be a point-like mathematical point charge, with no spatial extension. However, when a test particle is forced to approach an electron, we measure changes in its properties (charge and mass). This effect is common to all elementary particles. Current theory suggests that this effect is due to the influence of vacuum fluctuations in its local space, so that the properties measured from a significant distance are considered to be the sum of the bare properties and the vacuum effects (see renormalization)."

As my understanding, it tells us that an electron DOSE occupy space (its radius is 2.8179 × 10⁻¹⁵ m according to the article) and also it somewhat explains why an electron occupies space. But I'm afraid it doesn't help for understanding the reason why electron occupies space, especially:

"when a test particle is forced to approach an electron, we measure changes in its properties (charge and mass)"
"the properties measured from a significant distance are considered to be the sum of the bare properties and the vacuum effects"

the sentences above are confusing to me...dose anyone know what do they mean?

Additional question: What if we force two electrons collide? Will they just overlap with each other and then pass through without collision?

(solving the questions above might involve some QFT and/or QED but they are too tough to me) - Justin545 (talk) 06:06, 5 May 2008 (UTC)

While that paragraph does leave much to question, I think the answer is definitely yes. The electron does occupy space. It is just really small compared with the nucleus of the atom. Thus, when a test particle is forced to approach the electron, it causes changes that can be measured. The problem with small things is that trying to measure and define their properties actually changes their behavior. So, we can only guess about their real behavior and measure from a distance so as to not affect the results.

In answer to the second question, since they occupy space, they can't just go through each other. They will collide and bounce apart. Leeboyge (talk) 07:22, 5 May 2008 (UTC)

I'm not sure. But I think things collide that's because of the fundamental interactions (strong interaction, weak interaction, electromagnetic force and gravitation) between them. If we are able to overcome the repulsion and all other forces between the two electrons, I think they will overlap and go through each other when they approach to each other. Even though they occupy space! (please correct me if I'm wrong) - Justin545 (talk) 08:14, 5 May 2008 (UTC)

Sticking my nose temerariously into a question well outside my expertise: I think the answer is that, as of 2008, no one really knows whether the electron (in the sense of the bare charge, not the cloud of virtual electrons and positrons that surround it) occupies space. My understanding is that QED takes it to be a point charge, occupying no space, and having therefore infinite density (and by the way also infinite charge -- the cloud of virtual particles surrounding it "shield" most of that charge). QED works extraordinarily well, but the extent to which it faithfully represents the underlying reality, as opposed to simply being a collection of hacks that get the right answer -- again, nobody knows. --Trovatore (talk) 08:28, 5 May 2008 (UTC)

Telling is the sentence after the statement of the electron's "radius": "This is the radius that is inferred from the electron's electric charge, by using the classical theory of electrodynamics alone, ignoring quantum mechanics." I take that to mean it helps in some physical problems (collision cross sections maybe?), but shouldn't necessarily be interpreted as a literal "size" as we think of it in intuitive macroscopic terms. --Prestidigitator (talk) 08:40, 5 May 2008 (UTC)

I'm afraid it depends on what you mean by "occupy space". If you want the everyday-scale notion of occupying space to extend down to the quantum level then I think you're pretty much forced to say that electrons do occupy space, since most of the space supposedly "occupied" by solid objects only has electron orbitals in it. The space occupied by an electron in this sense has nothing to do with any intrinsic size. A hydrogen atom and a He+ ion both have a single electron orbiting a nucleus, but the former is larger than the latter because of the different nuclear charge, even though it's an identical electron that occupies most of the space.

Protons and neutrons do have an intrinsic size: they're bound states of more fundamental particles (much like an atom is), and the bound state has a characteristic radius (much like an atom does). People have proposed preon theories where the electron is a composite particle, but none of them have had much success. If an electron were composed of preons then I suppose it would have a size in this sense, but I don't know much about this.

Strings (from string theory) have a characteristic size but don't really occupy space, being one-dimensional subsets of a three-or-more-dimensional space.

It seems fairly likely that future physical theories won't have a concept of "space" any more (except as a large-scale approximation), which will make this question even harder to answer.

2.8179 × 10⁻¹⁵ m is the classical electron radius. Ignore it, it's meaningless. -- BenRG (talk) 10:28, 5 May 2008 (UTC)

I could be wrong, but I think what happens if you collide electrons is Møller scattering. --98.217.8.46 (talk) 15:17, 5 May 2008 (UTC)

To make the conclusion, I presume:

(classical and modern view) An electron occupies NO space. It is just a macroscopic illusion to say that an electron occupies space as the article Electron did.
(classical view) An electron is just a tiny, movable electric field with definite position. Because the movable electric field has mass $m_e=9.1\times 10^{-31}\mbox{ kg}$ and inertia, when it feels a net force $\bold F$ it will accelerate with $\bold a=\frac{\bold F}{m_e}$ .
(classical view) Because the electron is nothing more than an electric field with mass $m e$ , when we force two electrons move toward to each other they will finally pass through each other WITHOUT collision. Moreover, they will overlap (perfectly) halfway while they move toward each other. (however, Pauli exclusion principle tells us "no two identical fermions may occupy the same quantum state simultaneously" so I'm not sure if it holds in modern physics)
Electrons are not solid/stiff objects/entities so they don't collide. The collision is just an illusion due to the repulsion Coulomb forces between them. So saying an electron occupies space is meaningless.
The sentences "Hence, for convenience, it is usually defined or assumed to be a point-like mathematical point charge, with no spatial extension. However, when a test particle is forced to approach an electron, we measure changes in its properties (charge and mass)." in Electron should be removed or rewritten as it misleads the reader to think an electron occupies space.

the conclusion above may sound ridiculous. Please point out my errors if any. - Justin545 (talk) 01:53, 6 May 2008 (UTC)

Two electrons can't be forced to move towards eachother until the overlap, since that would require infinite energy (the force follows an inverse square law, so approaches infinity as they electrons approach each other). --Tango (talk) 16:56, 6 May 2008 (UTC)

This is false. Because electrons are ultimately probability clouds, the amount of charge at any point in space is infinitesimal. Infinitesimal charges can be seperated by neglible distance without requiring infinite energy, and as a consequence electron clouds can overlap (and do routinely in all atoms with more than 1 electron). Dragons flight (talk) 18:51, 6 May 2008 (UTC)

True, but Justin listed that item as being the classical view, and it's incorrect from the classical viewpoint. --Tango (talk) 12:33, 7 May 2008 (UTC)

From the classical viewpoint, two electrons could not overlap because $V = {1 \over 4\pi\varepsilon_0}\frac{q_1q_2}{r}$ . If

r = 0

there will be infinite potential.

From the quantum viewpoint, two electrons could overlap. For example, suppose

Ψ = ψ(x 1, x 2)

is a wavefunction of two electrons with position

x 1

x 2

respectively. Then $\int_c^d\int_a^b|\psi(x_1,x_2)|^2\,dx_1\,dx_2$ will be the probability of finding the first electron between interval

(a, b)

and the second electron between interval

(c, d)

. If

ψ(x 1, x 2)

is well-designed such that:

$\int_c^d\int_a^b|\psi(x_1,x_2)|^2\,dx_1\,dx_2\ne 0$
$(a, b)$ and $(c, d)$ overlap. (which implies $a = c$ and $b = d$ )
$a$ is very close to $b$ so they are almost at the same point

then we have a chance (with non-zero probability) of finding the two electrons arbitrarily close to each other! However, two electrons overlap may sound like to violate Pauli exclusion principle. But as long as the two electrons with different momentum they are not in the same quantum state. (again I could be wrong. please point out my errors) - Justin545 (talk) 02:11, 7 May 2008 (UTC)

I added scare quotes to "classical electron radius" in the electron article because it's not really an electron radius, but that doesn't mean electrons don't occupy space! There's no Platonic notion of occupying space that we can appeal to here. It's a matter of how you want to define the English phrase. If you define it in such a way that electrons don't occupy space, then neither does anything else, which makes the phrase useless. I think that's reason enough to use some other definition.

Also, it's pretty hard to make the case that electrons are at all point-like to begin with. In the original quantum mechanics (now taught to undergrads) you start with a quasiclassical theory where electrons are point particles and then "quantize" that, which introduces wave behavior. (The scare quotes here are because quantization (physics) isn't actually quantization. Horrible terminology.) But in quantum field theory you start with a quasiclassical field theory of electrons, so even at the classical level the electrons are spread out and occupy space to the same extent that electromagnetic fields do. When you "quantize" this theory you get field phonons which are referred to as particles, not because they are but because the term was already grandfathered in by the time QFT was developed. The only sense in which electron field quanta are particle-like is that they tend to spatially localize when they interact with a thermodynamically irreversible system (like a cloud chamber). I still don't grok why this happens, but it has something to do with quantum decoherence.

Probably the most misleading part of QFT is Feynman diagrams. They're not pictures of pointlike particles propagating and interacting in spacetime, they're graphs (in the computer-science sense) and their origin is combinatorial. Any integral of the form $\int\cdots\int e^{P(x,y,\ldots)} \, dx \, dy \cdots$ , where P is a polynomial, can be evaluated with Feynman diagram techniques (for some value of "any"). The idea is to write P = Q + R where Q contains only terms of degree 2 or less, and then Taylor expand R to get $e^P = e^Q (1 + R + R^2/2! + \cdots)$ , which gives you a series of relatively easy Gaussian integrals. The

R n

term of this series can be represented by a collection of Feynman diagrams with

n

interaction vertices. The diagrams always look like QFT diagrams with lines and interaction vertices, whether or not there's anything resembling space or time in the integral. There's a nice discussion of this in chapter I.7 of Quantum Field Theory in a Nutshell by A. Zee, which I just found out is available online. This expansion is extremely useful in QED, where the series converges rapidly, but not so useful in QCD, where it doesn't. There are other approaches to evaluating the integral which give you a totally different "picture of what's going on". I love the book QED, but Feynman went way overboard with his claims about the particle nature of light and electrons. In fact chapter 2, where he talks about photons only, is completely classical in almost every respect. The idea that pulses of light take every possible path from the source to the target, and every path contributes equally to the final amplitude, is just the path-integral form of Maxwell's equations. His discussions of reflection, refraction and diffraction are classical. It's crazy to claim that these arguments show the particle nature of light unless you're willing to argue that Maxwell's theory is also a particle theory. What does show the particle nature of light is the fact that the detector (photomultiplier) registers individual clicks of equal amplitude in low light. That's the only genuine quantum mechanics in the whole chapter. -- BenRG (talk) 16:33, 7 May 2008 (UTC)

BenRG: do you really think those scare quotes are necessary? I went back and saw your edit. I think "So called" is a little harsh as well. Granted, contemporary theory no longer describes the electron this way, but the paragraph qualifies the term as such. Scare quotes would imply that there is something intrinsically wrong with the phrase, which there is not. Qualifying the radius described as Classical gives it a timeframe during which the description was perceived as accurate theory. It would be like putting scare quotes around phlogiston or n-ray. Thoughts? --Shaggorama (talk) 19:57, 8 May 2008 (UTC)

[edit] Question regarding electricity

This isn't homework, it's part of my revision material and I need some help with understanding it. I'm not asking for an answer.

A filament bulb is connected to the 230V mains power supply and a meter indicates that 31 coulombs of charge flows in a period of 60 seconds. Calculate: a) the total electrical energy transferred to another form by the lamp b) the power dissipated by the lamp

Am I right in assuming that 31 coulombs of charge per 60 seconds is roughly 0.52 Amps since an amp is the flow of charge per second? My problem is i've forgotten whichever equation to use to calculate the electrical energy transferred.
As for part B, i'm right in assuming I need to use P = I^2 x V for this?

Regards, CycloneNimrod^Talk? 10:47, 5 May 2008 (UTC)

Your 230V mains is AC Alternating current.71.236.23.111 (talk) 11:38, 5 May 2008 (UTC)

Your revision question is badly worded, but your calculation is right. You need Joule's Law to work out the energy and power. --Heron (talk) 12:22, 5 May 2008 (UTC)

You are right on the first statement $\frac{Q}{t}=I$ but the second formula you give is wrong, it is $P= I \times V$ . Ignore the statement above about alternating current, that has nothing to do with it. SpinningSpark 12:28, 5 May 2008 (UTC)

Thanks for your replies :) Regards, CycloneNimrod^Talk? 17:18, 5 May 2008 (UTC)

A curiosity of the English usage difference between the UK and the U.S. is the British use of "revise" when we would say "review" or "study." On this side of the pond, to "revise" would only mean "to edit and make changes." Edison (talk) 18:44, 5 May 2008 (UTC)

You're editing and making changes to your knowledge of the subject, I suppose. --Tango (talk) 19:18, 5 May 2008 (UTC)

For once I'd agree that the U.S. grammar in this sense is correct, I think it makes more sense to 'review' than to 'revise', but alas, it's what i've been taught to say ;) Regards, CycloneNimrod^Talk? 19:38, 5 May 2008 (UTC)

Actually the word is French in origin (ultimately Latin) and literally means to "to look at again" so it is entirely cognate with review. I suppose that the modern sense of "altering" arose because it is natural to find fault and change something after reviewing it. Anyway, when I was at school I felt it was not insignificant that the word was next to "revile" in the dictionary which is exactly how I felt about the process. SpinningSpark 20:06, 5 May 2008 (UTC)

[edit] GMT and UTC

How different are these two times? Is the difference always the same or does it vary?

Thank you. Wanderer57 (talk) 13:34, 5 May 2008 (UTC)

Try reading GMT and UTC, I expect you'll find your answer in them somewhere. --Tango (talk) 13:47, 5 May 2008 (UTC)

Here's a short answer. The term "UT" was basically invented as a replacement for "GMT" that, for political reasons, did not mention Greenwich. As time measurement improved, it was realized that GMT (UT) as it had been interpreted did not provide a suitable basis for timekeeping, and several kinds of UT were developed: UT0, UT1, UT2, and UTC. You can think of them as kinds of GMT if you like, or you can think of GMT as meaning a specific one of them. They all vary with respect to each other, but UT1 and UTC are never more than 9/10 of a second apart. Now see the articles for more detail. --Anonymous, 18:44 UTC, May 5, 2008.

[edit] White teeth

I've got a couple of questions I hope somebody can answer. Thank you.

if you drink black tea, and then brush your teeth immediately, how great is the risk that your teeth may become somewhat tainted anyway?
can your teeth become tainted to some degree even if you use a straw to drink black tea?
what kind of foods should one avoid in order to get whiter teeth? Mind you, I don't mean foods which damage teeth in general, just those which stain the teeth.

Thanks again. 83.37.4.200 (talk) 14:23, 5 May 2008 (UTC)

Look at Tannin to get a start. Eliminating all these foods from your diet doesn't seem a good idea. Antioxidants help your body stay healthy. Lisa4edit (talk) 14:54, 5 May 2008 (UTC)

No, it wouldn't be a good idea, it seems :) And as for my first question, has anyone got an answer please? 83.37.4.200 (talk) 14:58, 5 May 2008 (UTC)

I'm not sure avoiding foods will ever get you whiter teeth—what you mean is what kinds of foods can you avoid to avoid staining your teeth. That's not the same thing (your teeth will never "get whiter" on account of avoiding foods. You can, though, make them "get whiter" by various bleaches and etc., which are not at all expensive these days). Note that over-brushing is not a great idea in and of itself — you can damage the enamel, especially if you brush immediately after eating. Chewing sugarless gum after a meal is a better idea—the saliva cleans the teeth, restores the pH, and is non-abrasive. (I am not a dentist! But I read it in the New York Times not long back.) I have no clue though whether anything you do would have any effect on staining of that sort. --98.217.8.46 (talk) 15:09, 5 May 2008 (UTC)

Talking of chewing gum the Xylitol page might interest you. But that won't whiten your teeth either.Lisa4edit (talk) 15:58, 5 May 2008 (UTC)

I understand that peoples teeth naturally become less white as they grow older as the enamel thins and the inner part of the tooth shows through. 80.0.98.228 (talk) 10:01, 12 May 2008 (UTC)

[edit] Henry IV of France and gonorrhoea

I'm working on an account of Henry IV's love life. In his biography of Henry, David Buisseret writes: "Towards the end of the month he had another attack of gonorrhoea, which gave rise to a temporary but alarming heart condition". Could anyone tell me if temporary heart conditions can indeed be brought on by gonorrhoea? (Historians sometimes repeat this sort of thing without question, so I need to check this out.) The incident is important because Henry nearly died. Many thanks. qp10qp (talk) 15:06, 5 May 2008 (UTC)

If the bacteria spread through the body as "Disseminated Gonococcal Infection" endocarditis can occur. See http://www.ncbi.nlm.nih.gov/pubmed/1728091 William Avery (talk) 15:35, 5 May 2008 (UTC)

Many thanks. I will trust it. qp10qp (talk) 21:47, 6 May 2008 (UTC)

[edit] tuberculosis (moved from WP:HD)

what is the method of minimising resistance to treatment in tuberculosis?172.159.163.56 (talk) 16:27, 5 May 2008 (UTC)

Antibiotic resistance may be of use. Resistance to tuberculosis is pretty much the same as in other bactera. The best ways to reduce resistance are to be picky about who you give antibiotics to and to make sure that they finish their course of antibiotics. If bacteria aren't killed off completely, they'll likely come back with a resistance and hey presto, you have yourself a new strain of resistant bacteria. Of course it's not quite that simple, but it's along those lines ;) Regards, CycloneNimrod^Talk? 17:16, 5 May 2008 (UTC)

The principles specific to antituberculosis treatment: [1] never use only one antibiotic; combination therapy is mandated. It is important to assess the resistance pattern of the specific strain of tuberculosis infecting the patient, as using an ineffective antitubercular may leave one effectively treating with only one (or no!) effective antibiotics. This is an exception to the usual admonition to avoid polypharmacy. [2] directly observed therapy: patient compliance with prescribed medication is monitored and enforced by public health officials. - Nunh-huh 22:54, 5 May 2008 (UTC)

[edit] Gondwanan distribution and Laurelia

There are two species of Laurelia, one native to the Andes - Laurelia sempervirens (Chilean Laurel, Peruvian Laurel, Peruvian Nutmeg) and the other found in New Zealand - Laurelia novae-zelandiae (Pukatea) .

When I saw one of these trees, growing at Enys in Cornwall, UK, yesterday, the Head Gardener said they were examples of "Gondwanan distribution". I have looked at the Gondwana article, but this is very brief on the topic, which is a REDIRECT to Godwana.

I would like to know more about the differences between the two species and when, approximately, they diverged, due to the separation of New Zealand and South America from the Gondwanan continent. Vernon White ^{. . . Talk} 19:24, 5 May 2008 (UTC)

I shall leave the differences to the botanists. As for the divergence, as I understand it, dating the Gondwanan distribution is not easy. For example, in a discussion on ferns, Te Ara Encyclopediaof NZ suggests that some distribution was windborne. this Te Ara page also discusses the time span, ocean distribution and later isolation. Googling Pukatea + Gondwanan yields 39 results which might bear investigation. Gwinva (talk) 05:17, 6 May 2008 (UTC)

[edit] engineering question

Which field fo engineering is the most multidiciplinary i.e. which concerns the widest range of topics from maths, physics, chemistry, geography, biology etc. Clover345 (talk) 20:53, 5 May 2008 (UTC)

Possibly constructions with biopolymers while exploiting advantages geological features. You know these disciplines are just groups into which we categorise knowledge, they interconnect depending on how do you define there disciplinesBastard Soap (talk) 21:21, 5 May 2008 (UTC)

Environmental engineering would be my suggestion. I'm not certain I'd recommend it as an area with stellar career opportunities, though. Lisa4edit (talk) 21:33, 5 May 2008 (UTC)

I would propose Electrical Engineering. After discussions with many of my colleagues, I have reached this conclusion. In 2008, "electrical engineering" may refer to the base science and the applied technologies in fields as diverse as:

computer architecture and computer programming
chemical processing for silicon and semiconductors
theoretical mathematics for information processing and communication theory
physics and electromagnetics for radio and signal processing
bioengineering, including electronic circuits and electromechanical or electrofluidic machines
mechanical design for electronics, as well as packaging, stress and thermal relief
nanotechnology, for semiconductors and a variety of other things
circuit design, which is what I used to think exclusively comprised the entire discipline
project management including financial and business-analysis for electronics projects

I personally know "electrical engineers" who work in all of the above areas. The job market is huge - software companies, aerospace, communications, academia... Nimur (talk) 00:50, 6 May 2008 (UTC)

An interesting question. I would suggest that mechanical engineering would be as this covers a wide range of diciplines including electrical / electronics engineering, both on the micro scale, for example nanobots and macro scale, for example, lifts and ventilation in buildings. Civil engineering mainly covers all static objects while mechanical covers moving objects. Civil may also be multidiciplinary but usually only involve large scale building projects. Environmental engineering is a sub-dicipline of civil engineering. Tbo ¹⁵⁷(talk) 18:04, 6 May 2008 (UTC)

Where they put Environmental Engineering depends on your U. MIT and Stanford put it with the Civs but lots of the smaller ones either have a separate department or put it somewhere else. In many other countries it is a separate discipline, too. 71.236.23.111 (talk) 06:56, 7 May 2008 (UTC)

Multidisciplinary with regard to engineering usually means it contains many aspects of engineering - e.g. electrical, civil, mechanical, electronic, aeronautical - rather than the multiple subjects you mention. However, from my experience, I would suggest that civil most closely fits your definition of multidisciplinary. There's the maths and physics, along with a lot of work on environmental/geographical elements, and fluid dynamics (wind flows). It lacks chemistry and biology though... LHMike (talk) 09:12, 8 May 2008 (UTC)

[edit] Atomic vibration

Can anyone please explain to me what makes the little buggers vibrate about?Bastard Soap (talk) 21:16, 5 May 2008 (UTC)

What kinds of atoms, in what situation? If you're talking about air at room temperature, then most of the molecules are not vibrating. They're moving and rotating, but not vibrating, because the first vibrational energy level above the non-vibrating ground state is too high compared to the average thermal energy. If you're talking about some solid material like a metal, then it is vibrating, because unlike vibrations of gas molecules, there is effectively no lower limit for the energy of a phonon.

So to sum up, the answer to your question is that when things are vibrating, they're usually vibrating because of heat. Heat is the microscopic motion of atoms. —Keenan Pepper 23:55, 5 May 2008 (UTC)

That seems a little circular to me. They're vibrating because of heat, which is vibration? --Tango (talk) 00:35, 6 May 2008 (UTC)

Heat is a form of energy. If you put energy into a system the atoms will move more, i.e. they will have more heat.--Shniken1 (talk) 00:44, 6 May 2008 (UTC)

In thermodynamics (as opposed to everyday usage) temperature and heat have very precise definitions and are not synonymous. Temperature is a measure of the thermal energy of an object, whereas heat is the energy exchanged between an object and its surroundings if they are at different temperatures. If an object is at a lower temperature than its surroundings then heat energy will flow into it, but this energy does not necessarily increase the object's temperature - the object might instead do work by expanding against an external pressure, or it might change state. Conversely, it is possible to increase an object's temperature above that of its surroundings by doing work on it - by compressing it, for example, or by boring a hole in it - or to reduce its temperature by making it do work - by allowing it to expand, for example. Gandalf61 (talk) 09:34, 6 May 2008 (UTC)

Heat is the macroscale result of atomic level vibration. They are the same. The faster the atomic vibration, the higher the bulk sample temperature. —Preceding unsigned comment added by 131.111.100.44 (talk) 16:30, 6 May 2008 (UTC)

I knew that heat makes them vibrate, my question is why is it so?Bastard Soap (talk) 18:32, 6 May 2008 (UTC)

Heat is energy. If you add heat to a molecule, either it becomes higher in kinetic and/or potential energy. Motion is kinetic energy... DMacks (talk) 21:48, 6 May 2008 (UTC)

Indeed. In thermodynamics, heat is just a type of energy flow. If you add energy to an object, some of it may go into doing work that we can measure at a macroscopic level - expansion, for example. Some of it may go into breaking inter-molecular bonds - a change of state. And some of it may go into increasing the vibrational and kinetic energy of each individual molecule. We cannot observe this molecular energy directly, but we can observe it indirectly, because the object then tends to transfer heat energy back to its surroundings in accordance with the second law of thermodynamics. What we call "temperature" is just a measure of the average molecular energy of an object. Gandalf61 (talk) 22:26, 6 May 2008 (UTC)

My question is still why and how do the molecules vibrate? I know that the energy of heat has to go somewhere, I just don't understand what system causes it to vibrate.Bastard Soap (talk) 17:26, 7 May 2008 (UTC)

This is one of those "just-because" answers. When a molecule increases in energy, that has to be reflected somehow. The somehow's include increases in translational, rotational, vibrational and excitational modes. Vibration is just one of the ways to store the extra energy. How? Energy is transferred to the molecule and taken up by the particular mode. Why? Because that is the most appropriate mode to accommodate the energy. There are no easy answers when it comes to really tiny things. Franamax (talk) 10:00, 8 May 2008 (UTC)

If you are asking about the mechanisms that transfer heat energy from the surface of an object or container into the molecules that may lie deep inside it, then the canonical answer involves the three mechanisms of conduction, convection and radiation. Radiation involves energy transfer via electromagnetic waves; conduction involves the transfer of energy via inter-molecular forces; convection involves the macroscopic movement and mixing of a fluid. There are other, less frequently quoted, mechanisms such as thermodynamic phonons in crystals. Gandalf61 (talk) 10:34, 8 May 2008 (UTC)