Science desk
< March 22	<< Feb \| March \| Apr >>	March 24 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 23

"Forces of Potential"?

Is there a special name for forces that arise from potential energy, such as spring forces, electric forces, or gravitational forces? My coined phrase is "forces of potential," but is there a real scientific term for these types of forces? Yakeyglee (talk) 00:19, 23 March 2010 (UTC)[reply]

All forces, in fact, arise from some form of potential but it's not always possible to describe the force in terms of a potential from a macroscopic point of view. For instance, there is no macroscopic description of the friction force in terms of a potential but from the microscopic point of view the friction force arises due to electromagnetic interactions which arise due to a potential. Said that, the word you are looking for is probabily "conservative force". Dauto (talk) 00:29, 23 March 2010 (UTC)[reply]

It is generally regarded as impossible to construct a potential field for the strong nuclear force or the weak nuclear force. Depending on your school of thought (and your quantum physics professor), a macroscopic approximation can be used, e.g. Yukawa potential or some variant thereof. In quantum theory and in particular in the Standard Model, the idea of a potential field is replaced by the more general exchange of virtual particles; it is preferable to think in terms of "interactions", rather than "potential fields". Nimur (talk) 01:26, 23 March 2010 (UTC)[reply]

Nimur, I have a PhD in particle physics. you don't have to tell me about weak interaction. But your description, while actually quite accurate, is not really helping the OP. Dauto (talk) 01:48, 23 March 2010 (UTC)[reply]

The OP may be several years out from their first QM class, but it never hurts to give them a heads-up to the different ways that physicists think about energy and forces. I'm glad that my description is "quite accurate" as interpreted by a particle physics Ph.D.! Nimur (talk) 14:10, 23 March 2010 (UTC) [reply]

Coriolis Force

Okay, if we write the acceleration of a point in planar coordinates, we get a = (r' - rθ')r + (rθ' ' + 2r'θ')θ, where r and θ are unit vectors, and ' denotes a derivative. the 2r'θ'θ is supposed to represent the Coriolis term. But I thought the Coriolis force is always perpendicular to the velocity, not just to the radius. Help? 173.179.59.66 (talk) 02:03, 23 March 2010 (UTC)[reply]

The Coriolis acceleration (and hence the Coriolis force) only exists when using polar co-ordinates. It is one of the two components in the θ direction. If the motion of the point is analysed using co-ordinates based on the direction normal to the velocity vector and tangential to the velocity vector there is no 2r'θ' term and so no need to resort to a Coriolis acceleration.

If an artillery shell is fired over a great distance, an observer in a fixed position and orientation on the Earth's surface sees the impact of the shell occur to the right of the aim point in the northern hemisphere, or the left in the southern hemisphere. He attributes the deviation of the shell's trajectory to a Coriolis force (or the absence of the Coriolis force necessary to achieve a direct hit.) However, if the observer is orbiting the Earth in a satellite and is able to observe the rotation of the Earth, he will see the shell achieve a direct hit at the point where the aim point was at the instant the shell was fired; he will also observe the Earth rotating and will understand why observers on the Earth's surface imagine the shell deviated from the desired path.

The observer in a fixed position and orientation on the Earth's surface is, without realizing it, seeing things using rθ co-ordinates, and the observer in the satellite is able to see things using nt co-ordinates. Dolphin51 (talk) 03:02, 23 March 2010 (UTC)[reply]

The formula you are using assumes that the motion is radial from the point of view of the rotating reference frame. BTW The correct expression is a = (r' ' - rθ' ²)r + (rθ' ' + 2r'θ')θ. Dauto (talk) 03:18, 23 March 2010 (UTC)[reply]

I'm sorry, I don't see how.

{\boldsymbol {r}}=r{\boldsymbol {\hat {r}}}

can represent any arbitrary vector, and so

{\frac {d}{dt}}{\boldsymbol {r}}={\boldsymbol {v}}={\frac {d}{dt}}(r){\boldsymbol {\hat {r}}}+r\omega {\boldsymbol {\hat {\theta }}}=v_{r}{\boldsymbol {\hat {r}}}+v_{\theta }{\boldsymbol {\hat {\theta }}}

should represent any arbitrary velocity, as the equation seems to imply, right? I trust that I'm wrong, but I don't see where. 173.179.59.66 (talk) 04:31, 23 March 2010 (UTC)[reply]

Yes, the motion is arbitrary, but you need two separate independent arbitrary motions. One for the motion of the particle and one for the rotation of coordinate system. As it is the formula is using the same

\omega \,

for both the motion of the particle and the rotation of the system which means that from the point of view of the rotating system the motion of the particle is radial and the coriolis force will be perpendicular to the radius. Dauto (talk) 12:56, 23 March 2010 (UTC)[reply]

Gotcha, thanks. 173.179.59.66 (talk) 14:30, 23 March 2010 (UTC)[reply]

Air pollution UK

I heard on BBC Radio Four yesterday morning that air pollution in some parts of London, for example, is well above EU limits and takes eight or more years off people's lives. Yet it is difficult to get maps of how average air pollution varies within the UK.

Are any more detailed versions of the maps in section 5 "How air pollution varies across the UK" of this pdf http://www.airquality.co.uk/reports/cat05/0408161000_Defra_AQ_Brochure_2004_s.pdf available anywhere please?

I've read the air pollution article. Some of the links are broken. I did find some links to what were called "maps" but were just CSV files of raw data, not graphical. Thanks 92.24.91.12 (talk) 02:23, 23 March 2010 (UTC)[reply]

http://www.airquality.co.uk/ itself has more detail. Example: click on London, select the monitoring sites, check out the Last Hour's Data and the Weekly Graphs. You simply do not get much more granular than 16 or so monitoring stations in London. --Tagishsimon (talk) 02:29, 23 March 2010 (UTC)[reply]

Those links do not provide any maps, just graphs over time unfortunately. Am I being paranoid or is it a sinister cover-up that this death information is not being made available, except in obscure ways that you need a PhD to appreciate? 78.149.198.14 (talk) 00:44, 27 March 2010 (UTC)[reply]

Question about blackbodies and emissivity.

I recently asked a question about light particles and how they react with ionization. I have a new question, however, apropos emissivity and prisms.

Imagine a cubic box —approximately 15 cm X 15 cm X 15 cm— closed on 5 sides and open on the remaining side. The inside surface of the "bottom" of the box (the surface opposite the open side) is covered with a large number of microscopic prisms. Each of said prisms is oriented so that the "flat" end faces the open side of the box, and the other two ends lie at a 45-degree angle to the "bottom" of the box.

Also, each of those ends (but NOT the flat end of each prism facing the open end of the box) is coated with "Super Black" or some other ultra-low-emissivity substance.

My question is, would somebody looking inside the box see nothing but total darkness? To wit, would virtually all light entering the box fail to reflect back outside the box? Would the light refracted by the prisms be absorbed or "bounced around" to the point where it would become nothing but low-grade heat?

Also, could such a setup be engineered —with extant equipment and scientific methods— which could minimize both reflection of the right from the "flat ends" of the prisms and ensure that enough photons are either absorbed or "exhausted" (for lack of a more scientific term) that no visible light would exist in the box —regardless of how powerful the external light source?

Thank you for reading this! Pine (talk) 05:52, 23 March 2010 (UTC)[reply]

what the main focus of this artical or the purpose of the work?? —Preceding unsigned comment added by 58.165.81.107 (talk) 06:01, 23 March 2010 (UTC)[reply]

Sort of. At first you can absorb light, however over time the box will get hotter, and will start to emit blackbody radiation. You can make a box to absorb light, but not as you described. If you coat the outside of a prism, the inside can still bounce light. So your prism will actually bounce the light right back at you, and will ignore the coating on the outside (but see frustrated total internal reflection). Also the flat face of the prism will bounce light. It's not possible to make a surface that will "redirect" the light inward forever, at best you can absorb the light with carbon black and the like. See Anechoic chamber for the basic design, but be aware that it works only because the light source is 'inside' the room. You can't make an open room like that. The pyramids cause the light/sound to be bounced many many times, each time absorbing more of the energy - but it's not forever, eventually it bounces to the bottom of the pyramid and then starts bouncing back out, and then it goes to the other wall and starts again. You can make it bounce so many times that it absorbs a lot, but not everything. Ariel. (talk) 07:33, 23 March 2010 (UTC)[reply]

Yep - I agree. From a 'thought experiment' point of view, your box sits in a perfect insulating vacuum and no light or heat can escape. Light is energy - so if you shine light into the box and doesn't come out again, then you're adding energy into the box. Eventually, it's going to heat up - and then in the real world would likely start emitting infrared radiation until the IR energy emitted by the box equalled the energy of the light going in. If you imagined your "super-black" material to be SO black that it didn't emit anything then the box would continue to heat up until it started getting red-hot and glowing that way. If you somehow prevent that (or re-absorb the red light) then it'll get hotter and hotter until it starts emitting in some yet higher energy waveband - and if you prevent that, it'll melt! The point is that the energy has to go SOMEWHERE. The laws of thermodynamics don't allow energy to be destroyed. Maybe you could make your box turn energy into matter so that it could just get slowly heavier! SteveBaker (talk) 17:02, 23 March 2010 (UTC)[reply]

It's not clear to me what is the purpose of the prisms in this setup. Dauto (talk) 20:19, 23 March 2010 (UTC)[reply]

Wow, thanks for the prompt replies!

Let me first say that I'm finding this conversation to be extremely eye-opening! I'd just like to clarify a couple of points, if I may:

"[I]n the real world would likely start emitting infrared radiation until the IR energy emitted by the box equalled the energy of the light going in."

Obviously, the energy (as per the Law of Conservation) must go SOMEWHERE. At first, I figured it would all —or nearly all— be lost as low-grade heat. I realize now that I was very mistaken about that.

As for the "output", however, could such be engineered so that all (or virtually all) light emitted back out of the "box" would be in the IR spectrum —or some other wavelength invisible to the naked eye? If so, then (1) how would somebody engineer such a setup? And (2) if practically all of the light were emitted outside of the ROYGBIV frequencies, would an external observer see total darkness or some other visual effect?

"It's not clear to me what is the purpose of the prisms in this setup."

That likely was a stupid idea on my part. Perhaps I ought to more closely model this after the Anechoic chamber concept.

--Again, thank you! Pine (talk) 21:09, 23 March 2010 (UTC)[reply]

Here is an anecdote from the early days of television when the Test card was an actual printed card at which a camera was pointed. It was difficult to print a true black swatch at the dark end of the grey scale. The solution was to cut through the card a hole that lead to a tiny chamber built behind the card and lined with black velvet. Cuddlyable3 (talk) 22:43, 24 March 2010 (UTC)[reply]

aircrafts

is it possible to make a propeller of airplanes to fly using superconductors on the basis of magnetic levitation principle...... —Preceding unsigned comment added by Ravi3s31 (talk • contribs) 06:11, 23 March 2010 (UTC)[reply]

The propeller? Do you mean a Magnetic bearing maybe? Or do you mean the airplane itself? If you mean the airplane, it's probably far too high for a magnet to be possible. (It would have said practical, but there is a limit to how strong a magnet you can make using superconductors, so above a certain height it's impossible.) Ariel. (talk) 07:16, 23 March 2010 (UTC)[reply]

You might want to read our article on Maglev. These (not hugely successful) craft fly at a height of about 15 millimeters.--Shantavira|^{feed me} 08:53, 23 March 2010 (UTC)[reply]

Perhaps like the old propeller-driven monorail proposals: [1][2]. You would just replace the maglev's linear induction motor with a propeller but would still need an magnetic track so I don't know that you would gain any advantage. 75.41.110.200 (talk) 14:18, 23 March 2010 (UTC)[reply]

I seem to interpret this Q differently than the rest of you. I think they are asking if the friction in the bearings for a plane propeller in a prop plane can be reduced or eliminated using superconductivity. I would guess that it's possible, but that the additional weight needed to provide coolant to maintain the low temps needed would make the cost far more than any benefit. So, possible ? Yes. Practical ? No. StuRat (talk) 16:18, 23 March 2010 (UTC)[reply]

You can do Magnetic levitation without superconductors though - it's just that you need to control the magnetic fields more carefully. Check out Magnetic bearing for example. This requires some power - but nothing like what you'd need to maintain a superconducting setup. SteveBaker (talk) 16:52, 23 March 2010 (UTC)[reply]

Pressure in a liquid

Part One

So suppose we have a liquid, say water. Now, at a certain depth, the upwards force acting on a small point (say a molecule of water) will be equal to the weight of the water above it. That I understand: the water molecule must be at equilibrium, and the force above it will be the weight of the water compressing it. Now, when an object (for simplicity, say a cube) is placed submerged in the water, what is the force acting on the bottom side of the cube? I would have thought, by the same logic as above, that the water below the cube would be pushing up with a force equal to the weight of the water above it plus the weight of the cube, but apparently that's not correct? 173.179.59.66 (talk) 02:18, 23 March 2010 (UTC)[reply]

The pressure acting on the bottom side of the cube is the pressure acting everywhere at the depth of the bottom side of the cube. Pascal's law is relevant. The weight of the cube is irrelevant for the following reason. If the density of the cube is greater than the density of water there must be a force supporting part of the weight of the cube - for example, that force might be the tension in a string supporting the cube. Alternatively, if the density of the cube is less than the density of water there must be a force holding the cube under the water - for example, that force might be the compression in a rod holding the cube under the water.Dolphin51 (talk) 02:31, 23 March 2010 (UTC)[reply]

So, to finish that thought, if the cube is in equilibrium (neither rising nor sinking) the pressure on the bottom of the cube equals the pressure due to the weight of the water above the cube plus the weight of the cube, just as you said. This also equals the pressure due to the weight of water above that surface, if the cube wasn't there. In other words, since the density of the cube is the same as water, it acts just like there was water in it's place. StuRat (talk) 02:44, 23 March 2010 (UTC)[reply]

Oh, I didn't say the cube was in equilibrium. It could be less or more dense than water. I know that the cube will rise or fall, due to Archimedes' principle. If my question is unclear, let me know. 173.179.59.66 (talk) 04:57, 23 March 2010 (UTC)[reply]

Your question is perfectly clear. Is our answer clear? Dolphin51 (talk) 05:24, 23 March 2010 (UTC)[reply]

Apparently not. You said that "if the density of the cube is greater than the density of water there must be a force supporting part of the weight of the cube". Why? 173.179.59.66 (talk) 05:33, 23 March 2010 (UTC)[reply]

Pressure depends on depth. For any real object the pressure pushing down on the top is slightly less than the pressure pushing up on the bottom. Since we know that a volume of water having the same shape as the object would be stationary, we must conclude that net force pushing on a submerged object is equal the weight of a packet of water occupying the same volume (this in the principle behind buoyancy). Hence the net force of the water is always upward and equal to the weight of the water displaced. Whether this is enough to counteract the pull of gravity will depend on the density of the object in question. Dragons flight (talk) 07:27, 23 March 2010 (UTC)[reply]

Well, see, that logic is circular. This assumes that the net pressure on any given region doesn't depend on what's inside of it (ie it's the same if it was water or lead), which is what I'm trying to understand. Normally, I would have thought that pressure exerted by a point would equal the total weight above it, and so would depend on what's above, but that appears not to be the case. 173.179.59.66 (talk) 14:36, 23 March 2010 (UTC)[reply]

I presume that it makes a difference if the cube is falling, and the force adds to the pressure immediately below the cube, like you said. I'm not an expert though, I'm just butting in because you didn't get any more replies. (Seems pretty clear that if the cube is held in place by strings then its weight is irrelevant, but that's not what you're considering, is it. Oh, and I suppose the falling cube would also lower the pressure on its upper side?) 81.131.36.119 (talk) 20:09, 23 March 2010 (UTC)[reply]

Your initial question was based on a cube is placed submerged in the water. This creates an image in the mind of a cube, stationary at the nominated depth in still water. All the answers that have been provided have been based on fluid statics, and they are correct for zero relative velocity between the cube and the water. If we now consider the cube falling through the water so that there is a non-zero relative velocity, we need to consider fluid dynamics and the situation becomes more complex. For example, the pressure on the underside of the cube would be the stagnation pressure and that depends on the relative velocity - how fast is the cube falling? Dolphin51 (talk) 05:27, 24 March 2010 (UTC)[reply]

Part two

I've been looking up on pressure in a liquid, and I've noticed a recurring trend: equations and rules are generally stated without proof or justification, with the exception of heuristic arguments that convince you that the results are plausible. For instance, the hydrostatic pressure of a liquid is given as P = ρgh, although I don't see why that would be true in general, or why it would take the same value regardless of direction. Same thing for Pascal's principle. [I did, however, find a delightfully clever proof for Archimedes' principle.] Any insight would be helpful. Thanks! 173.179.59.66 (talk) 06:48, 23 March 2010 (UTC)[reply]

In case what I said was confusing, allow me to clarify my question. 1) If you would consider a column of water, the pressure at the bottom of the column (or at any height) would be P = F/A = mg/A = ρAhg/A = pgh. But why is this equation valid in general. And say we were considering a point (in the same column) at some height h, and that this point is touching the container. Then apparently P = ρgh is still true, but I can't see why. Finally, why is Pascal's principle true? I saw a little clip that explained it for a U-bend with different areas (courtesy of http://www.youtube.com/watch?v=lWDtFHDVqqk&feature=SeriesPlayList&p=AD5B880806EBE0A4), but it's validity in general remains a mystery to me. 173.179.59.66 (talk) 07:05, 23 March 2010 (UTC)[reply]

Consider an infinitesimal cube of fluid with dimensions dx, dy, dz, and density ρ. Assuming a static equilibrium, that packet of fluid must be at rest. Which means that the net force on the packet must be zero. Consider the force on the cube. The available forces are the pressures pushing on all six sides, and the downward force of gravity. Specifically:

(P(x,y,z)dydz){\hat {x}}+(P(x+dx,y,z)dydz)(-{\hat {x}})+

(P(x,y,z)dxdz){\hat {y}}+(P(x,y+dy,z)dxdz)(-{\hat {y}})+

(P(x,y,z)dxdy){\hat {z}}+(P(x,y,z+dz)dxdy)(-{\hat {z}})-(g\rho dxdydz){\hat {z}}=0

It follows immediately that

P(x,y,z)=P(x+dx,y,z)=P(x,y+dy,z)=P(z)

, i.e. that the pressure doesn't change as one moves horizontally. While the last term implies:

P(z)-P(z+dz)=\rho gdz

\Rightarrow {\partial P(z) \over \partial z}=\rho g

\Rightarrow P(z)=\rho gz+C

Dragons flight (talk) 07:50, 23 March 2010 (UTC)[reply]

Dragons flight explanation is correct but is unlikely to dispel the OP's confusion since the OP seems to be asking why the pressure used for the vertical force is also the same pressure used for the horizontal force. I think he wants an explanation for hydrostatic pressure. Dauto (talk) 13:13, 23 March 2010 (UTC)[reply]

A couple of comments which, I hope, complement the answers above:

You asked "For instance, the hydrostatic pressure of a liquid is given as P = ρgh, although I don't see why that would be true in general,". It isn't true in general, only in the specific case of an incompressible fluid.

Then you said, "or why it would take the same value regardless of direction. Same thing for Pascal's principle" . I think both of these state the same question. But it seems to me that neither are things to be proved or derived. The first is an observation (which you could test by putting a manometer under water, and putting the nozzle at any orientation you like, though I never have.) The second, Pascal's law, seems to be a postulate of mathematical physics, that can only be verified because it is able to predict experimental results, a little like Newton's law of universal gravitation. I may be wrong, but Pascal's law looks like a definition of an ideal static fluid. However, unlike gravitation, we know that it is caused by fluid molecules, so I guess that somewhere in the literature, there is an explanation or derivation hydrostatic pressure starting from jiggling marbles or statistical mechanics, perhaps. If so, it would be interesting to see it, and perhaps to add it to the article.

These are actually laws for ideal situations, where fluids are uniform and highly symmetrical. In real life, some fluids won't obey this law: in particular some fluids that are in between a liquid and a solid. As a fluid starts to crystallise, it may transmit stress preferentially in some directions over others. I suspect there will also be problems with moving fluids, such as in viscous flow, or in boundary layers, but I don't recall any examples, though we know, for example, that Bernoulli's equation doesn't work for some non-ideal fluids.

--Hroðulf (or Hrothulf) (Talk) 14:19, 23 March 2010 (UTC)[reply]

The OP might also like to read our article on continuum mechanics. The various examples presented are generalizations of the Newtonian mechanics formulations to fluids, taking into account that mass, position, acceleration, and so forth, all must be rewritten in terms of infinitesimal fluid elements. Rigorous derivation of the laws of fluid mechanics is sometimes ugly and requires a lot of calculus. As for "proof", Newtonian mechanics has been demonstrated countless times; and so physical principles that reduce to the Newtonian case by mathematical manipulation are generally accepted as valid. Nimur (talk) 14:26, 23 March 2010 (UTC)[reply]

OP here. First off, thanks, you've made this stuff much clearer. However, a few more questions: I understand ${\partial P(z) \over \partial z}=\rho g$ , but to go from that to : $P(z)=\rho gz+C$ , doesn't that assume that ρg is constant? If there were something other than water above the infinitessimal cube (say a rock), would the result change? 173.179.59.66 (talk) 14:52, 23 March 2010 (UTC)[reply]

If there was a dense rock above the cube, it would sink towards our pressure sensor, so we don't have the equilibrium we wanted.

But, we could fix the rock in place, for example we could glue it to the wall of our container. A naive integration over

z

with

x

and

y

constant wouldn't work. When we reach the rock we get to a discontinuity where

p(z)

doesn't apply. However, Pascal's principle allows us to move horizontally at will, and experience the same pressure. With a horizontal movement or two, we can find a way to the surface, so we might as well integrate over

z

after all. (The container is not symmetrical, but all that fluid mechanics requires is local symmetry: each infinitesimal element of fluid is symmetrical.)

~~What does the $\P$ symbol mean here?~~

By the way, I suggested looking into statistical mechanics, and I just noticed that in hydrostatic pressure, Wikipedia says that

\ p(h)=p(0)e^{-Mgh/kT}

can be derived from the assumptions of statistical mechanics. I guess if we consult any statistical mechanics textbook, we can add the derivation to the article, which would help answer your question (for the gas case, not a liquid). For small

h

this is equivalent to

P=\rho gh

.

--Hroðulf (or Hrothulf) (Talk) 15:32, 23 March 2010 (UTC)[reply]

It occurred to me that it might help you get to grips with this if you turn the thought experiment about the dense rock on its head. The rock is fixed to the sides of the container, so you can use P = ρgh above and below the rock to predict the forces on the rock. In other words, you can calculate how much upward force the container has to exert on the rock to support it, which will be its weight less its buoyancy. The rock doesn't interfere with the calculation of pressure, and if you use a balance or a strain gauge to measure the rock's buoyancy, you can use it to demonstrate that P = ρgh holds true. --Hroðulf (or Hrothulf) (Talk) 16:39, 23 March 2010 (UTC)[reply]

Everything makes sense now, much thanks. 173.179.59.66 (talk) 22:50, 23 March 2010 (UTC)[reply]

Damage due to loss of circulation in extremities

This question is inspired by the question further up the page asking about organ failure, but I'm curious about the other extreme. If one loses circulation in a non-vital region, say a finger or toe, or perhaps an arm or a leg, then how long does it take for permanent damage to occur? Similarly what tends to be first permanent consequences of such loss of circulation (e.g. what is first to be damaged)? I would assume, though I don't really know, that the extremities can be sustained longer without circulation than the internal organs generally can. Dragons flight (talk) 08:10, 23 March 2010 (UTC)[reply]

They can go quite a long time. See Reperfusion injury for a non obvious injury. Ariel. (talk) 08:19, 23 March 2010 (UTC)[reply]

I assume that you (the OP) are talking about an acute injury such as an arterial blockage due to an embolism (cholesterol embolism, fat embolism, etc.). The bottom line is that the timing of damage is going to depend on the metabolic rate of the organ involved, the degree of collateral circulation, and the amount of tissue that experiences ischemia. A finger/toe, which is mostly skin, connective tissue, tendons, nerve endings, etc., may tolerate temporary cessation of blood flow better than an arm/leg, which contains significant amounts of muscle, not to mention a great deal more tissue. The tourniquet article discusses some consequences of sudden loss of circulation to distal tissues but doesn't give any specific duration that I can see. Acute mesenteric ischemia can cause bowel infarction, and the article says that "prognosis depends on prompt diagnosis (less than 12-24 hours and before gangrene)" suggesting that for the gut, at least, permanent damage happens after about 12-24 hours, probably varying a great deal depending on the size of the infarcted area. You might also want to consider chronic peripheral circulatory problems such as those seen in diabetes or peripheral vascular disease, which can lead to necrosis over long time periods. Take a look at the gangrene article for some nice photos of the process (unless you happen to be queasy about that sort of thing). --- Medical geneticist (talk) 18:48, 23 March 2010 (UTC)[reply]

I'm not a medic, but my info is that a limb can survive for up to 2 hours with blood flow completely cut off before gangrene sets in. FWiW 24.23.197.43 (talk) 23:23, 23 March 2010 (UTC)[reply]

You may also be interested in frostbite and reflex cold urticaria. ~A H 1^(TCU) 01:59, 24 March 2010 (UTC)[reply]

What is the quantum model for refraction of light?

Could I ask you to recommend a website or text book that would introduce to me a quantum mechanical model for the transmission of light (photons) through a dielectric like glass or water?

I wish to explore the verbal description in this article more deeply: http://www.physlink.com/education/askexperts/ae217.cfm (Why does light slow down when entering a prism but speed up when it exits?)

So I read Refractive index#Speed of light.

However I am confused by the claim that though the phase velocity (wavelength divided by ~~frequency~~ period) decreases, the speed of the individual photons is said to be unchanged. If I understand the implications of the classical model correctly, the vibration of the electrons predicted by Maxwell's equations shortens the wavelength, similar to the way a sound wave has its wavelength shortened and refracted when it moves from a stiff solid to a less stiff one. In my mind, the classical model predicts that the actual velocity of the light will change, which seems to correspond to Fizeau's experimental results on the speed of light through a tube of water. Therefore, to help fight my way through the forest, I wanted to see a mathematical description of quantum light transmission through matter, if it can be boiled down to a small set of consistent equations.

Is there a single accepted quantum model for this effect?

Is it covered in Feynman's QED: The Strange Theory of Light and Matter? (If so I will seek out a copy or watch the lectures online)?

--Hroðulf (or Hrothulf) (Talk) 12:16, 23 March 2010 (UTC)[reply]

You are correct. The speed of individual photons decreases due to interactions with the dielectric. Dauto (talk) 12:33, 23 March 2010 (UTC)[reply]

Thanks, Dauto! I suspect that the two descriptions (1) of a photon slowing, and (2) a photon being absorbed, and then an identical one emitted at speed

c

with a time delay and phase shift, though they appear contradictory, are actually two different, correct, viewpoints of exactly the same phenomenon.

I will be interested to discover if, when a photon wavefunction overlaps the wavefunction of an electron, the mathematics produces the same outcome as when a light wave oscillates a classical point charge. I ordered Feynman's QED from the library.

--Hroðulf (or Hrothulf) (Talk) 14:57, 24 March 2010 (UTC)[reply]

You are in for a treat. Feynman's QED is a delight to read. I don't remember seen the answer for that specific question in there but it's been quite some time since I last read it. Either way, it's a book worth reading. Dauto (talk) 20:24, 24 March 2010 (UTC)[reply]

National Medal of Science

Did Bush forget to award 2008 and 2009 National Medals of Science? Why does our article only go to 2007? Why does http://www.nationalmedals.org/medals/laureates.php only go to 2006? Does the sad shape of these medals' records have anything to do with the censorship at the White House office of Science Policy? 99.56.137.254 (talk) 12:23, 23 March 2010 (UTC)[reply]

According to the NFS website, there were none in 2008 and nine in 2009. Someone should update our article with last year's winners. Speculation about political intrigue is out of scope for this ref-desk. DMacks (talk) 13:41, 23 March 2010 (UTC)[reply]

It seems unlikely that the lack of any winners would have to do with top-level politics. There is no shortage of scientists who do good work on topics with no political relevance, or relevance primarily on things that even Republicans like to take credit for (George Bush was as happy as any to talk about how great renewable energy research was). --Mr.98 (talk) 14:44, 23 March 2010 (UTC)[reply]

Well, there could be a scenario where the obvious winner is someone they don't want to award, like a global warming scientist. So, then the choices would be to hold the award ceremony, and snub that scientist, getting all sorts of negative press, or just cancel it altogether for some believable reason, hopefully garnering less negative press. I have no idea if this actually happened, just offering one possible scenario. And, looking at the list, it seems that there were many years when no medal was awarded, so I'd tend not to look for conspiracy theories. StuRat (talk) 16:09, 23 March 2010 (UTC)[reply]

I'm not sure if there is any difference between a medal recipient and a "notable laureate". I left a question at Talk:National_Medal_of_Science#2009_recipients.3F. --Enric Naval (talk) 16:59, 23 March 2010 (UTC)[reply]

I think the database is wrong, there are no 2009 recepients yet and won't be until later this year. The ones listed under 2009 are actually the ones for 2008, whoever manages the database probably screwed up, I've emailed them. Check out the article talk page for more details. Incidentally as noted in the article the 2005 & 2006 medals were awarded together, in 2007, I don't know why. Perhaps there was too much campaigning going on in 2006. Maybe they wanted to save money so decided to only have one ceremony for two years. You may be able to find out if you search around. Nil Einne (talk) 23:08, 23 March 2010 (UTC)[reply]

Thiomersal/Thimerosal

Why the two names? Is it a brand name thing, or just English variants (à l'aluminium/aliminum)? Our article doesn't seem to say. AlmostReadytoFly (talk) 14:47, 23 March 2010 (UTC)[reply]

I believe it's just a spelling variant. This MSDS lists a couple of other names for the same, including some trade names. SDY (talk) 14:53, 23 March 2010 (UTC)[reply]

Thiomersal is the International Nonproprietary Name designated by the World Health Organization. Thimerosal appears primarily in North American usage; I don't have access to a suitable database at the moment, but I suspect that is probably a United States Adopted Name. (Both are considered 'generic' names in their respective jurisdictions; manufacturers will have their own unique trade names for the product.) As to the 'why' — it comes down to different standards in jurisdictions with different governing bodies. Why is parecetamol sold as acetaminophen in the United States? ("Why can't you carry acetaminophen in the jungle?" "Because paracetamol!") TenOfAllTrades(talk) 15:11, 23 March 2010 (UTC)[reply]

Scrambled eggs

When scrambling eggs, if you give them a vigorous stir after they begin to set, they will return to a liquid state. What chemical & physical processes are going on? DuncanHill (talk) 14:54, 23 March 2010 (UTC)[reply]

Denaturation (biochemistry) has a nice photograph at top of the article. It is probably a special type of Maillard reaction (which is normally discussed in the context of cooking meat). We also have a great article on Molecular gastronomy - again, there's a really nice egg photo at the top of that article too. Nimur (talk) 15:42, 23 March 2010 (UTC)[reply]

Towards the perfect soft boiled egg discusses several quantitative models for egg cooking chemistry. And Cooking for Eggheads says: "when an egg cooks, its proteins first unwind and then link to form a rigidifying mesh. But not all its proteins solidify at the same temperature. Ovotransferrin, the first of the egg-white proteins to uncoil, begins to set at around 61 degrees Celsius, or 142°F. Ovalbumin, the most abundant egg-white protein, coagulates at 184°F. Yolk proteins generally fall in between, with most starting to solidify when they approach 158°F. Thus, cooking an egg at 158°F or so should achieve both a firmed-up yolk and still-tender whites, since at that low temperature only some of the egg-white proteins will have coagulated." When you stir scrambled eggs, you're probably rehomogenizing the mixture, which has started to fractionally solidify. Nimur (talk) 15:48, 23 March 2010 (UTC)[reply]

Thermal Radiation

What's the difference between the average surface temperature and effective temperature when treating Earth as a Black body? I can find the new article has the following formula:

T_{E}=T_{S}{\sqrt {\frac {R_{S}{\sqrt {\frac {1-\alpha }{\overline {\epsilon }}}}}{2D}}}

But I remember I've read this article before (an old revision of the Black body article), with the following equation:

T_{E}=T_{S}{\sqrt {\frac {R_{S}}{2D}}}

Applying first equation gives about -18.8 °C and is called effective temperature, whereas the 2nd one gives about +14.5 °C.

Was that a correction or what?. Thanks in advance.--Email4mobile (talk) 16:33, 23 March 2010 (UTC)[reply]

It's an improvement that takes into account that the earth doesn't emit or absorb radiation as effectivelly as ablack body but you would know that had you acctually read the article. Dauto (talk) 18:41, 23 March 2010 (UTC)[reply]

Thank you very much, Dauto. I've another question indeed regarding the maximum theoretical Temperature T_E. If we assume a perfect black body but with a shape different from a sphere, let's say a sheet or disk in the space with one side exposed to the radiation and the other not. Can we estimate the maximum possible temperature in this case and can we say that it won't exceed $T_{S}{\sqrt {\frac {R_{S}}{D}}}$ ?--Email4mobile (talk) 21:10, 24 March 2010 (UTC)[reply]

Trailing off...

Is there a name for a speech impediment where the person starts out talking strongly, but then the volume drops off until eventually they aren't speaking at all ? For example:

"I NEED TO GO to the store to get ...".

Do we have an article on this ? StuRat (talk) 16:42, 23 March 2010 (UTC)[reply]

Blue for murmur and Prosody_(linguistics). Red for muttering and mumbling. The most common term for that I've heard is "trailing off". A reverse dictionary didn't find anything. There is a stub for enunciation. I'm almost sure this isn't what you are looking for but Sotto voce. If you want to be poetic about it you could use a musical equivalent and say someone is speaking "al niente", "calando", "decrescendo", "diminuendo", "perdendo", "perdendosi", "morendo", or "smorzando". I'm no music buff so I couldn't tell you which one of those would best fit the bill. 152.16.15.144 (talk) 17:37, 23 March 2010 (UTC)[reply]

Edit. Probably not aposiopesis. List of voice disorders? 152.16.15.144 (talk) 17:39, 23 March 2010 (UTC)[reply]

Doesn't so much sound like a speech impediment as it does a memory problem. I'd expect this sort of behavior in someone with either an physical injury to the brain, on some sort of drug or with some sort of memory based disease (either degenerative or attention based). Sorry I can't link you to anything in particular. Regards, --—Cyclonenim | Chat 21:46, 23 March 2010 (UTC)[reply]

Maybe the loss of one's train of thought, or a tip of the tongue phenomenon? ~A H 1^(TCU) 01:57, 24 March 2010 (UTC)[reply]

My first thought was ADHD. Second thought was Absent-mindedness. Third thought.. was.. ...was... Pfly (talk) 09:52, 24 March 2010 (UTC)[reply]

Is this a known symptom of ADHD ? StuRat (talk) 15:17, 24 March 2010 (UTC)[reply]

Well I meant to link to ADD, but it redirects to ADHD predominantly inattentive, which is apparently what ADD is called now. I piped it to say ADHD, for reasons that made sense in the moment but don't now. Anywa, the page actually linked to says "it is characterized primarily by inattention, easy distractibility, disorganization, procrastination, forgetfulness.." So, well, ..um, oh look, cows. Pfly (talk) 09:23, 25 March 2010 (UTC)[reply]

I don't think ADHD presents like this. I know I said an attention based disorder above, but I wasn't suggesting ADHD. In fact, I'd expect a person with ADHD to blurt stuff out rather than trail off. My best guess for this case would be traumatic brain injury which has affected either the person's Broca's region (responsible for producing speech, though I'm sure not sure why they'd trail off, usually one would just stop speaking completely) or perhaps one of the areas responsible for memory such as a part of the hippocampus. Regards, --—Cyclonenim | Chat 18:59, 25 March 2010 (UTC)[reply]

Yes, sorry, I was just saying what my first thoughts were, and perhaps trying to say that the question was a bit vague. For something more related to neurology, perhaps some kind of expressive aphasia? That term is broad and includes all kinds of things, but it appears that there are a number of things that might cause a trailing off of speech type symptom. Pfly (talk) 20:46, 25 March 2010 (UTC)[reply]

Including stroke ? StuRat (talk) 23:59, 25 March 2010 (UTC)[reply]

Strokes tend to create incoherent speech rather than trail off well produced speech, but it could be a post-stroke symptom if damage was severe enough to memory regions. Regards, --—Cyclonenim | Chat 12:02, 26 March 2010 (UTC)[reply]

UH-60L data busses ....

I have a need to know what data busses are employed by the UH-60L copter. For instance, what bus does the Honeywell H764 navigator use? My guess: 1553. But I can't confirm that. Is there a site I can look at or publication that I can determine what data busses are used on the various subsystems?

Thanks, Whatsamatteru (talk) 17:47, 23 March 2010 (UTC)[reply]

This Honeywell Document says that the Honeywell H-764 features both MIL-STD-1553 and RS-422 busses. -- Finlay McWalter • Talk 18:21, 23 March 2010 (UTC)[reply]

And this and this both say that the M version of the Blackhawk use 1553-based glass cockpits. I can't immediately find comparable info for the L version. -- Finlay McWalter • Talk 18:26, 23 March 2010 (UTC)[reply]

Yeah. I'm trying to find a technical document that details the data busses like the doc I have that details the electrical busses. Difficult to find. But thanks for checking into it for me, Finlay.

Whatsamatteru (talk) 19:06, 23 March 2010 (UTC)[reply]

Such technical documents are probably ITAR-controlled. Nimur (talk) 09:16, 24 March 2010 (UTC)[reply]

Rh disease and James Harrison

Is this real? And if it is, how come I can't find a thing about it? Nor any details of what exactly they do with his blood? And what they plan to do after he dies? Except for a paragraph someone plopped at the bottom of Rh disease I couldn't find anything on wikipedia.

After some more searching I found this. So apparently this is an Australian thing only? i.e. it's only in Anti-D, but not RhoGAM? Kinda confused, maybe someone know more. Ariel. (talk) 21:47, 23 March 2010 (UTC)[reply]

I don't think it's Australia only. Our article says "Rho(D) Immune Globulin is a derivative of human plasma". According to the second article, there are about 200-300 donors in Australia producing the required antibodies which are sufficient for the Australian requirement, the idea that this single person is supplying antibodies for the whole of Australia is I presume wrong (hardly surprising, Daily Mail isn't particularly notable for the accuracy of their science articles and it is a feelgood story) although he was one of the earliest suppliers. I presume most other countries have similar setups. I also don't think these are the only special plasma components collected from a small subset of donors, e.g. this discusses the issue when it comes to people selling their blood plasma [3] and there are also rare blood types e.g. [4] which similarly come from a small number of donors Nil Einne (talk) 22:39, 23 March 2010 (UTC)[reply]

(post Edit conflict) Basically the treatment of Rhesus disease consists of injecting the Rh- mothers with an antibody Rho(D) immune globulin (why is it written immune globulin instead of immunoglobulin?) which binds to and will lead to the destruction of fetal Rh positive blood cells which have crossed the placenta. I think the idea behind it is to keep the mother's immune system from becoming sensitized to Rh antigens. The sensitization and subsequent reaction storm can lead to the mother's antibodies crossing the placenta and attacking the fetus's blood. If they are using this guy's plasma for its antibodies, then this guy probably has Rho(D) immunoglobulin and thus is Rh- and sensitized to the antigens. The blood types article noted that "It is common for D-negative individuals not to have any anti-D IgG or IgM antibodies, because anti-D antibodies are not usually produced by sensitization against environmental substances." That's one reason the guy's blood may be uncommon then. Rho(D) immune globulin is made from human plasma, so I'm guessing this guy's plasma is used to make this stuff. His blood might also have other useful attributes, like being AB type, which means the plasma lacks the IgG anti-A or IgG anti-B antibodies and can be donated to A, B, and O types safely. This is all a guess so hopefully this helps... 152.16.15.144 (talk) 22:44, 23 March 2010 (UTC)[reply]

Avalanches

Hello everyone, I think I might have asked a similar question some time earlier, but I'll ask it anyway: What is the latest in the spring that a major avalanche had been recorded in the Rockies of central Alberta (in the area around Jasper, or thereabouts)? By "major", I mean an avalanche big enough to destroy/badly damage one or more buildings or kill/bury multiple victims. If someone could point me to where I can find this information, I'll be very thankful. Thanks in advance! 24.23.197.43 (talk) 23:33, 23 March 2010 (UTC)[reply]

OK, looks like nobody here knows the answer. Looks like I got to look it up on my own. Oh, well... 24.23.197.43 (talk) 03:20, 27 March 2010 (UTC)[reply]

Wikipedia:Reference desk/Archives/Science/2010 March 23

Contents