Wikipedia:Reference desk/Archives/Science/2009 October 1

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October 1

Pauli Exclusion Principle, Electron Spin, and Magnetism

I am teaching a lot of atomic theory (including quite a bit of quantum mechanics) to my 5th grade gifted and talented students. Even though my BA is in English, I take science very seriously (in a fun way) and really strive to get them excited about science. People seem to assume that kids are largely incapable of learning the really cool, deep science.

That being said. Is magnetism caused by the, "uniform motion of electrons," or is it a result of a large number of electrons spinning in the same direction along axis that are parallel? Is either statement close to the truth? Am I right to assume that Pauli's Exclusion Principle is not violated as long as none of the electrons are occupying the same space, even if they are all spiining the same way?

I apologize for the ignorant oversimplifications, but I'm a self-taught, math-challenged, lover of physics. —Preceding unsigned comment added by 98.169.171.112 (talk) 00:54, 1 October 2009 (UTC)

Both can cause magnetism. a large number of electrons spinning in the same direction will give you a permanent magnet. ANd a massed flow of electrons will give you an electromagnet. Even in a magnet many of the electrons are spinning in opposite directions, the but there is a surplus in one direction. For most substances the electrons will pari up, with one spinning one way and one spinning another way. It is only in special elements which have unpaired electrons not involved in a bond that you will have a spare electron to have one spin without being counter balanced. These are likely to be the transition elements or rare earth elements. They have a d-shell or f-shell that is "inside" the atom. Someone else can probably explain this better. Graeme Bartlett (talk) 04:46, 1 October 2009 (UTC)

(ec)Our magnetism article has a section explaining the sources of magnetism in different types of materials (and there are actually several different types of magnetism, each with different electronic origins; even protons and atomic nuclei have magnetic properties, which your students may have experienced in an MRI scan). Lots of complicated stuff in there! But many of the ideas should make some sense (and of course ask specific questions about it if not!).

The Pauli Exclusion Principle says two electrons cannot be the same in all respects. So if you have two different places you are already "different", and the spin can be "same" and not be a problem. This is the origin of the idea of "2 electrons in each atomic orbital": the pair electrons are the same except for spin; two different spin possibilities, so 2 electrons per orbital. DMacks (talk) 04:51, 1 October 2009 (UTC)

We need to distinguish three different types of magnetism: ferromagnetism, diamagnetism, and paramagnetism. All of these are essentially modeled by a quantum interaction between an electron's intrinsic magnetic moment and an externally applied magnetic field; in the case of ferromagnetism, the field is due to local interactions with other electrons in a magnetic domain (resulitng in a strong feedback and mutual alignment). In general, it is not a clean description to say that the magnetic moment is caused by "uniform motion" of electrons - this implies bulk motion (net measurable current) which contradicts experimental observation. In general though, there is a quantum-mechanical interaction between an electron's orbit description and its magnetic moment, if a magnetic field is present. If the magnetic field is uniform, the effect on all electrons' bulk motions in the material will be a similar alignment - but this is not the cause of the magnetic moment. Even if it were possible to freeze an electron motionless - (which is not actually possible) - that electron would still have an intrinsic magnetic moment. This is a very non-intuitive, non-classical effect - but it has been repeatedly observed by numerous experimental methods. A "stationary" electron is always "spinning" - it can never be stopped from spinning - it has this property built into the definition of an electron and it cannot be removed. Nimur (talk) 06:11, 1 October 2009 (UTC)

'Spin' is not actually a rotation of the particles. Electron's don't spin about their axis. (They don't even have an axis, what with lacking any kind of volume or internal structure as far as we know.) Two electrons with the same spin can occupy the same space. Two electrons with the same spin cannot occupy the same orbital - which is not a point in space but an entire pattern of motion (although 'motion' is a somewhat ill-defined concept at that level as well), different orbitals do overlap spatially. --Pykk (talk) 05:58, 2 October 2009 (UTC)

It may be helpful to think of electrons in their orbitals like a three-dimensional standing wave, though the actual term is more properly called a wave function. Bonding theories like Molecular orbital theory basically work on treating bonding like constructive and destructive interference of these waves. I commend you for taking this on with 5th graders; most adults cannot wrap their heads around quantum theory. You need to completely abandon your intuitive understanding of how physics works to really "get" it... --Jayron32 05:26, 3 October 2009 (UTC)

Garbage

the most hygienic way of handling and disposing garbage —Preceding unsigned comment added by TAMIKA LINDSAY (talk • contribs) 01:16, 1 October 2009 (UTC)

It depends on the nature of the garbage. See our articles Waste disposal . Intelligentsium 01:20, 1 October 2009 (UTC)

Annihilation with antimatter. Sagittarian Milky Way (talk) 02:29, 1 October 2009 (UTC)

I wouldn't advise it - the explosion would scatter the bits that didn't get annihilated all over the place. Not very hygienic. --Tango (talk) 03:22, 1 October 2009 (UTC)

It also depends on your definition of "hygienic." You can incinerate all traces of garbage "out of existence" if you're willing to release the vapors it produces into the atmosphere. List of solid waste treatment technologies is relevant. Sanitary landfill and incineration are both listed there. See also hazardous waste - final disposal. Nimur (talk) 03:30, 1 October 2009 (UTC)

The most hygienic manner in which to handle garbage would undoubtedly be with out direct body contact -- you can use impermeable gloves or chopsticks. The most hygienic fashion in which to dispose of garbage would follow suit, making sure whatever you encase the garbage in does no disintegrate until after the garbage disintegrates -- or you could decontaminate the garbage in an autoclave. DRosenbach ^{(Talk | Contribs)} 03:42, 1 October 2009 (UTC)

The most hygienic fashion in which to dispose of garbage (including nuclear waste) would be to load it into a reliable rocket and Set the controls for the heart of the sun. Dbfirs 09:05, 1 October 2009 (UTC)

While not a bad idea on the face, what happens if you have your rocket malfunction and crash, or worse, explode in midair? Then instead of having your nuclear waste contained in a nice small barrel, you scattered it across 500 square miles. Googlemeister (talk) 13:09, 1 October 2009 (UTC)

I did say a reliable rocket. Also, nuclear waste is regularly enclosed in strong containers that will withstand explosions and falls from great heights, but I agree that there is a small risk. Perhaps it would be safer to transport it in very small quantities up to an orbiting space station just outside the atmosphere, then send the rocket from there. Of course, once we get nuclear fusion going properly, there will be no nuclear waste, and we can use the fusion reactor to get rid on non-nuclear toxic waste. Dbfirs 16:25, 1 October 2009 (UTC)

The Earth is moving sideways at 30 km/s relative to the Sun — it'd need to drop like a stone to fall into the Sun. If ever feasible you'd probably want to do gravity assist(s) to help. Considering that we have what, 70,000 tons of nuclear waste and a thousand tons of the highest grade waste alone, at the current prices (hundreds of millions of dollars per ton) this would be a very expensive method of disposal. Sagittarian Milky Way (talk) 18:59, 1 October 2009 (UTC)

Solar sails can do the job, I believe (in a manner equivalent to a sailing boat tacking). --Tango (talk) 22:17, 1 October 2009 (UTC)

Getting it into space is the expensive part, and all we have for that is rockets. — The Hand That Feeds You:^Bite 17:05, 6 October 2009 (UTC)

Quantum paradox

Let's say you're doing a single particle double-slit experiment with detectors. You shoot the electron at the screen and measure which slit it goes through. Your detector has a unique design. It stores the information somewhere and prevents it from decohering. If you ask it for which slit the particle went through, it tells you. However, you can also tell it that you don't want to know. Then, it makes a conjugate measurement of it's information storage, erasing the original observation. Obviously, whether you see interference patterns depends on what question you ask your detector. The paradox is - what if you ask your detector after you have run all the trials? What if you pick your question to be the one that would cause the set of probabilities that is not the one you observed (ie. see interference + ask which slit or don't see interference + and ask to forget)? 76.67.76.19 (talk) 04:23, 1 October 2009 (UTC)

Read Wheeler's delayed choice experiment and Delayed choice quantum eraser Graeme Bartlett (talk) 04:39, 1 October 2009 (UTC)

The short answer is that the pattern on the screen will be the same (no interference) regardless of what you do to your stored information and when. Your choice of measurement basis affects what you learn from the measurement, but the pattern on the screen is only affected by the initial act of copying. -- BenRG (talk) 12:58, 1 October 2009 (UTC)

But how is this a measurement? The detector prevents the information from decohering, so it isn't really a classical measurement. 76.67.73.124 (talk) 04:47, 3 October 2009 (UTC)

It's not a measurement, but the interference pattern disappears if any information about the path remains anywhere, in quantum or classical form. The path from source to slit A to position X on the screen interferes with the path from source to B to X, but the path from source to A to |A>|a> to |X>|a> doesn't interfere with the path from source to B to |B>|b> to |X>|b>, because the final states are different. (|a> and |b> here being orthogonal states of your stored qubit.) -- BenRG (talk) 23:41, 5 October 2009 (UTC)

Thanks. 76.67.74.18 (talk) 22:31, 7 October 2009 (UTC)

temp. & pressure depdendence of heat of fusion/vaporisation

The heat of vaporization article tells me that enthalpy and entropy of a phase change does not change significantly with T, but the graph on that page shows different! Which is it? John Riemann Soong (talk) 04:54, 1 October 2009 (UTC)

Clearly, it depends on the material. The graph illustrates several volatile fluids which have a dramatic enthalpy-of-vaporization dependence on temperature. Most materials do not have such a large variation. Nimur (talk) 05:56, 1 October 2009 (UTC)

So for most metals (for phase changes like changing from alpha-manganese to beta-manganese, or from solid to liquid lead) I can assume that the enthalpy of phase change is constant? John Riemann Soong (talk) 07:58, 1 October 2009 (UTC)

Yes, as an approximation assuming the volume change on phase change is relatively insignificant.83.100.251.196 (talk) 12:18, 1 October 2009 (UTC)

Homework check

Per the "no homework" rule, I'm posting on here to check that my thinking is correct, not just fishing for answers. Anyway, among others, I was given these two problems for a Physics class:

"You are sitting in a boat in the middle of a lake. Also in the boat with you is an anvil. If you throw the anvil into the lake, what will happen to the level of the water in the lake?"

My thought here is that the level will go down. When the anvil is inside the boat, it's displacing water equal to its weight. When it's thrown out, it's displacing water equal to its volume. Since the anvil is presumably less dense than the lake (barring a wooden anvil on a mercury lake), the water level will go down.

"You are attempting to lift a bar with a man grasping either end. Once off the ground, the men each perform a chin-up. Does this increase the amount of work needed to lift the bar? What if the men were jumping up and down on the bar, instead?"

Newton's third law would mean that as the men perform their chin-ups, the force they're exerting downwards on the bar would equal the force upwards on themselves. As I'm lifting the whole lot, the work required to lift the system wouldn't change. If they're jumping up and down on the bar, the initial jump upwards wouldn't change the work needed for the same reasons, but when they land, they're adding an additional downward force due to gravity. So the work in that case would increase, provided they complete at least one full jump and landing.

Is my thinking correct? 84.66.41.175 (talk) 07:01, 1 October 2009 (UTC)

I think you meant "more dense" instead of less dense in your answer for the first question. For your second question do they start performing their exercises during or after you have lifted the bar? --antilived^{T | C | G} 07:18, 1 October 2009 (UTC)

Your first answer looks good, although you made an error here: "the anvil is presumably less dense than the lake". In your second answer, for the bar to remain stationary, you must oppose the downward force on the bar with an equal, opposite force. As the man begins to pull up, he exerts another force on the bar due to his upward acceleration, in addition to his body weight. As he decelerates at the top of his pull up, he exerts less force on the bar than his body weight, so it becomes slightly easier for you to keep the bar still. Axl ¤ [Talk] 07:30, 1 October 2009 (UTC)

Yes, it's all a matter of timing. In the first example, as you throw the anvil slightly upwards to clear the side of the boat the average water level may rise marginally, then it will fall as the anvil is in the air, then rise again as the anvil enters the water, but the overall effect is a slight fall (too small to measure in a normal-size lake with a liftable anvil). In your second example, it is theoretically possible to reduce the amount of work done in lifting the bar if the chin-up men complete their upward acceleration before you start lifting, and decelerate (slightly pushing the bar upwards) as you lift. They would have to perform very athletic chin-ups to achieve this. With the opposite timing, the work needed would increase. Exactly the same applies to those jumping on the bar except that no upwards force is possible (unless they hook a toe under the bar), so work will nearly always increase. Remember that there is an increased force as they begin their jump, as well as on landing, so the work needed would always increase (over the empty bar) unless you lifted in mid-jump. Such split-second timings are unlikely to be achieved in practice. If the number of chin-ups or jumps is an integer during your lift, then the total work done is the same as it would be if the men were just hanging from or standing on the bar. Remember that if you lift whilst the jumpers are in the air, you are lifting only the bar, but if you lift whilst they are starting the jump, you are lifting the bar, plus their weight, plus an additional force needed to accelerate them upwards. Very hard work! You must be super-human! Dbfirs 08:46, 1 October 2009 (UTC)

The work needed to lift the bar and the gymnasts is equal to their combined weight times the height through which you raise their combined centre of mass. As long as the gymnasts end up in the same position relative to the bar at the end of each lift, the height through which the centre of mass is raised is the same, so the work done is the same. It does not matter whether they do chin-ups, jump up and down, or do a whole Olympic high bar routine while the bar is being lifted - as long as they always end up in the same position relative to the bar at the end of the lift, the work done on the bar and gymnasts is always the same. Gandalf61 (talk) 10:12, 1 October 2009 (UTC)

True, but only if they end up at rest in that position. Extra work needs to be done if they are still moving at the end of the lift. Also, the two jumpers or chin-uppers might have done some of the work. Dbfirs 16:18, 1 October 2009 (UTC)

If the gymnasts are still moving at the end of the lift, extra or less lifting work will be needed depending on the direction of their combined kinetic energies. Slightly extra work is needed in any case due to the energy they dissipate in causing air turbulence. Cuddlyable3 (talk) 23:38, 1 October 2009 (UTC)

Good point - I'd forgotten about air turbulance, though the amount of extra work is minimal at the speed most people do chin-ups! Kinetic energy is a scalar, so how can less work be needed if kinetic energy is given to the gymnasts? I'm still thinking about this ... Dbfirs 11:49, 2 October 2009 (UTC)

Followup (Sam)

Somewhat related, perhaps someone could comment on the physics of the event portrayed at about 4:30 of this video: [1] →Baseball Bugs ^{What's up, Doc?} carrots 10:30, 1 October 2009 (UTC)

And to whoever it was that tried to delete the above, he needs to back off. I would, in fact, like to hear the explanation (if any) as to why the item at 4:30 is realistic or fanciful, as I've always wondered about it, and it's within the general realm of the original question here. More importantly, why it's realistic or fanciful. →Baseball Bugs ^{What's up, Doc?} carrots 11:04, 1 October 2009 (UTC)

Well, it pains me to break the news to you - but cartoons aren't real. I understand how this might come as a shock...would this be a good time to break the news about Santa Clause and the Tooth Fairy? SteveBaker (talk) 11:14, 1 October 2009 (UTC)

I didn't say it was "real", I asked if it was "realistic". You've got a heavy object (again, an anvil) falling from a significant height, and Sam catches it. Assuming all that's possible, does the boat drop in the water somewhat? It sounds to me like a cousin to OP's question. The boat sinking is obviously a gag. But would the boat's waterline drop at all? →Baseball Bugs ^{What's up, Doc?} carrots 11:23, 1 October 2009 (UTC)

Yes - of course it drops...the ship has to displace water equal to it's weight...including that of the anvil. In cartoon-physics, anvils have near-infinite mass - so the ship sinks. SteveBaker (talk) 11:35, 1 October 2009 (UTC)

If the anvil were heavier than the water displaced by the boat itself, presumably it could sink the ship, right? So if it were dropped from an airplane and had sufficient mass, combined with a sturdy deck and a Sam of super-human strength, it might sink the ship, right? But instead it's dropped from the crows nest - making the situation fairly similar to the OP's question. When the anvil is dropped from the crows nest, would the ship rise slightly? And would it then fall back to its original waterline once Sam catches it, but also drop a little farther due to the added force of the anvil's fall? →Baseball Bugs ^{What's up, Doc?} carrots 11:42, 1 October 2009 (UTC)

But the weight of the anvil is not invariant under situational transformation in comic physics. In this case, this is even true in real physics - the anvil up on the mast is further from the middle of the Earth than down on the deck. I'd guess that the mast is roughly 1.5 times the radius of the Earth, so on the deck it will have near infinite mass_{In comic physics, mass and weight are interchangable*blush*} (per Steven), while in the crows nest it will only have near (infinite/5.25) mass, obviously much less. --Stephan Schulz (talk) 11:51, 1 October 2009 (UTC)

The mast doesn't look that long, but it could be trick photography. The point being, if the ship were made of balsa, i.e. lighter than the anvil, then the anvil dropping probably wouldn't sink the ship by weighing it down, it would probably sink it by breaking all the way through it. But assuming the deck is strong enough to withstand the anvil hitting it (or to withstand it hitting Sam), would the anvil cause the waterline to at least momentarily drop somewhat lower than it was when the anvil was at rest in the crows nest? I say momentarily because I'm guessing the boat would have to eventually rise back up to its original waterline, with the anvil again being at rest. Right? →Baseball Bugs ^{What's up, Doc?} carrots 12:01, 1 October 2009 (UTC)

You can only gauge the relative mast height from long distance shots showing the curvature of the Earth. When the anvil is released, the ship will start an upward movement, and perform a dampened oscillation around the new waterline. When the anvil hits Sam, a downwards impulse is transmitted to the ship, which leads to a second oscillation. I think those two will simply add up until the ship settles again. --Stephan Schulz (talk) 13:44, 1 October 2009 (UTC)

That makes sense. And the more I think about it, the more similar this question is to the original, especially as Sam throws the anvil off the ship once it's underwater. It's really not too different from OP's hypothesis, except it's taken to a comical extreme. →Baseball Bugs ^{What's up, Doc?} carrots 13:52, 1 October 2009 (UTC)

The ship would rise when the anvil was released from the top of the crow's nest. Time would elapse before the anvil reconnected with the ship. Upon reconnecting with the ship there would result a lowering of the ship to the level it had been at prior the the anvil being released from the point of the crow's nest. There would be bobbing motion through all these steps, as the ship is free to move to beyond its point of equilibrium, in the process of finding its point of equilibrium. The only factor that I am aware that I am leaving out of this description involves the distance from the center of the Earth as that may be negligible in this instance. I think if the distance from the center of the Earth were factored in the result would be a lower level of the ship in the water with the anvil on the deck than with the anvil in the crow's nest. Bus stop (talk) 14:05, 1 October 2009 (UTC)

(I'm getting rid of my comment, (And it's followups) and making a similar comment on the talk page. ) ) APL (talk) 14:44, 1 October 2009 (UTC)

WHAAOE - see cartoon physics. hydnjo (talk) 15:35, 1 October 2009 (UTC)

Yeh, except I'm not talking cartoon physics, that's just an illustration. →Baseball Bugs ^{What's up, Doc?} carrots 16:57, 1 October 2009 (UTC)

Yeh, I know - I was just sayin'... hydnjo (talk) 20:58, 1 October 2009 (UTC)

calculating the entropy of a substance

So my problem set isn't making any sense for me. Say that below some phase change temperature T_c (this particular case is 993 K), an alpha phase is favoured over the beta phase. But apparently if I calculate the free energy of each phase (from enthalpy and entropy of each phase), I predict a much lower phase transition temperature. On the other hand, I'm given the enthalpy of the phase transition and the temperature it occurs at (so I could well, predict the entropy change of the phase transition, and I have this value).

I am however given their differential heat capacities ... which happen to be constant (not dependent on temperature). But don't the free energy lines (with dependence on temperature) on the phases cross ultimately because of temperature-differential changes in heat capacity? John Riemann Soong (talk) 11:46, 1 October 2009 (UTC)

"But apparently if I calculate the free energy of each phase (from enthalpy and entropy of each phase), I predict a much lower phase transition temperature." - as a guess - are you assuming that the entropy enthalpy is constant with T ?

Second part - the free energy lines cross because of the dependence of free energy with temperature ie don't you need to calculate

GT= Gstart/standard - ʃSTdT

Where S_T is the entropy as a function of temperature - the relationship includes the heat capacities..

Is this what you've been trying, orr something else (there's usually different approaches to the same answer...)83.100.251.196 (talk) 12:28, 1 October 2009 (UTC)

ok I'm a bit confused - have you tried calculating G using the entropy derived from the heat capacities? ie using dS = (C/T) dT as described at Specific heat capacity - does that give a different or closer answer? I think that's wrong.83.100.251.196 (talk) 18:48, 1 October 2009 (UTC)83.100.251.196 (talk) 13:43, 1 October 2009 (UTC)

How do electromagnetic waves propagate through a vacuum?

I'm looking for the simplest, most layman-friendly answer to be able to understand this. Thanks.20.137.18.50 (talk) 12:05, 1 October 2009 (UTC)

Electromagnetic waves are a form of energy, so unlike sound waves, they don't need a "medium" in order to propogate. All they need is space. That's what I recall from 9th-grade science, anyway. →Baseball Bugs ^{What's up, Doc?} carrots 12:36, 1 October 2009 (UTC)

You can visualize them as Photons, traveling through space at the speed of light. If you want to think of it as a wave, then consider it as electric and magnetic fields oscillating perpendicular to each other traveling at the speed of light. Look up Light. Rkr1991 ^{(Wanna chat?)} 12:44, 1 October 2009 (UTC)

And if you want to think of light as a wave - the varying magnetic field produc es and electric field (at right angles), and the varying electric field produces a magnetic field. These fields exist OK in a vacuum. Graeme Bartlett (talk) 14:03, 1 October 2009 (UTC)

What is a field in a vacuum if a vacuum is nothingness?20.137.18.50 (talk) 14:07, 1 October 2009 (UTC)

A vacuum isn't really nothingness. It's just the quantum state with the lowest possible energy. See vacuum state. Red Act (talk) 14:18, 1 October 2009 (UTC)

The electromagnetic field progates by medium of photons, the electromagnetic fields gauge boson. Photons are the field, and the field is photons, the two 'classical style' parts of the wave-prticle duality as described by QED. The field as such is not smooth as is taught in classical dynamics, but lumpy and made up of phtons as in QED. However for most purposes the smooth classical model suffices, and is useful for being drastically simpler in most cases. Elocute (talk) 17:37, 1 October 2009 (UTC)

Magnets work fine in a vacuum; they attract and repel just as in air. Electrostatic attraction and repulsion works fine in a vacuum. Thus there is no reason that electromagnetic waves would not propagate in a vacuum. Edison (talk) 04:47, 2 October 2009 (UTC)

Followup (Photon)

OK, here's a spinoff question, which I'm not so sure has an answer. Supposedly an object's mass increases exponentially as it accelerates toward the speed of light, right? So that, in theory, a massive object can never get very close to the speed of light, and certainly not equal to speed of light, because its mass would become infinite, assuming the math about light-speed is correct. Or that's what I recall from 9th grade science. However, an extremely small particle can be accelerated nearly to the speed of light. The built-in assumption is that photons are traveling at the speed of light, i.e. at this assumed upper-bound of speed. But are they really? Do photons have literally 0 mass, or do they merely have a mass that's too small to measure? In the latter case, are they really traveling at that upper bound of speed? Or are they just short of it? →Baseball Bugs ^{What's up, Doc?} carrots 12:52, 1 October 2009 (UTC)

Well, I'll give the annoying answer. Yes - the photons are moving at the speed of light, because their speed is what we know, call, and have measured to be the speed of light since, by definition, light moves at the speed of light (well, usually). ~ Amory (user • talk • contribs) 13:00, 1 October 2009 (UTC)

As far as we know, photons have a rest mass of zero (or, more preciely, the concept of rest mass does not apply to them since they can never be at rest). If they had a non-zero rest mass then the speed of light would depend on its energy, and hence on its frequency - but observations of gamma-ray bursts in distant galaxies show to a high degree of accuracy that the speed of light is independent of its frequency [2]. Usenet Physics FAQ lists some other theoretical and experimental evidence for zero rest mass. Gandalf61 (talk) 13:10, 1 October 2009 (UTC)

By definition, the speed of light is the speed that photons travel. That follows. The speed of light also varies depending on what it's traveling through. A partial or spinoff answer to the OP's question, ironically, is that the presence of a "medium", e.g. clear glass, actually slows down the speed of the photons traveling through it, if I recall correctly. When "the speed of light" is said by itself, it usually implies "in a vacuum". But I wonder if the speed of light itself, in a vaccuum, is not the actual upper bound of speed, or whether light-speed is just short of the actual upper bound - maybe so slightly short of it that it's unmeasurable. →Baseball Bugs ^{What's up, Doc?} carrots 13:18, 1 October 2009 (UTC)

Gandalf's above bit about photons having zero rest mass is how we assert that c is in fact the upper bound. We have no evidence that speeds above that exist. Of course, once you get to "so slightly short that it's unmeasurable"... well, that's why 0.999...=1 — Lomn 13:25, 1 October 2009 (UTC)

OK, I think that covers it. Danke. →Baseball Bugs ^{What's up, Doc?} carrots 13:33, 1 October 2009 (UTC)

Photons always (under every conceivable circumstance) travel at c, this is in all meaningful sense a tautology, as when we measure c, we measure the speed of photons. However the speed of light is a different issue for reasons relating to the difference between physical and effective (optical) path lengths. That is, a photon going through a medium doesnt go slower, it just travels further at the same speed. It is correctly noted that it cannot be taken as given that all photons (and thus all light) travels at the same speed, no purely logical reasoning will deduce this, but that as Gandalf said, the evidence to support it is strong. A side note that the mass increas is not exponential, it has a pole at c.Elocute (talk) 17:33, 1 October 2009 (UTC)

How fast is a photon going at the moment that it hits a mirror and is about to be reflected? →Baseball Bugs ^{What's up, Doc?} carrots 23:46, 1 October 2009 (UTC)

There's not really such a moment since there's an uncertainty relation between the energy of the photon and the point in time it hits the mirror. That aside, there's no reason the velocity of a photon has to be continuous in time (unlike for a massive particle). In other words, it can go from c to -c without ever being in between and in fact it's required to. You could call the value at that point of discontinuity zero, but what would that mean? Rckrone (talk) 00:07, 2 October 2009 (UTC)

OK, so it's always either headed toward the mirror or headed away from the mirror. The instant it touches the mirror is essentially 0 time, hence it doesn't count, i.e. it doesn't "compress" and bounce off like a spaldeen or something. So let's say the room is full of mirrors. The photons would bounce all over the place, forever, wouldn't they? Or if not, then where would they go? →Baseball Bugs ^{What's up, Doc?} carrots 02:16, 2 October 2009 (UTC)

Yeah, in the ideal case with perfect mirrors they would bounce around forever. What happens in practice is that mirrors absorb some of the photons that hit them and the photons' energy becomes heat (or gives some electrons enough energy to make a break for it). The other issue is that photons have momentum, so when they bounce of a stationary object (like an electron of some atom), they have to transfer some energy to the object for momentum to be conserved. The effect is pretty minimal for visible light but a comes into play with high energy photons like X-rays (see Compton scattering). Rckrone (talk) 02:57, 2 October 2009 (UTC)

OK, that was the missing piece. The photon travels at light speed while it's a photon, but it can be transformed into another state, yes? I think this exhausts my questions on this topic. I probably knew all this stuff in 9th grade, but that was some time back. I appreciate your help and your patience. →Baseball Bugs ^{What's up, Doc?} carrots 03:17, 2 October 2009 (UTC)

Here's a paper that goes through a lot of experiments that have been done to show that photons have zero rest mass [3] to high accuracy. As others have mentioned, this isn't something scientists can take for granted, despite the fact that the universal constant c that crops up in relativity is also called "the speed of light." They're only the same under the assumption that photons have zero rest mass, although that seems to be a pretty good assumption. A story that worked out differently was neutrinos which were first assumed to be massless but experiments showed this to be wrong. Rckrone (talk) 19:25, 1 October 2009 (UTC)

The em wave creates its own virtual particle pairs as it propagates? just an idea. —Preceding unsigned comment added by 79.75.110.116 (talk) 23:41, 1 October 2009 (UTC)

cooling question

If I take an airstream which is at 500K and vaporize 1 gallon of water to cool it to 400K, will evaporating the same amount of water in the same volume of airstream which is at 600K cool it to 500K, or would it be more like 480K? Googlemeister (talk) 15:14, 1 October 2009 (UTC)

It depends how exact you want to be. See specific heat capacity. Evaporating the same amount of water requires the same amount of energy - but that assumes that the only relevant energy transfer is into the latent heat of vaporization. It also assumes a constant heat-capacity of the air stream with respect to temperature - again, this is probably a fairly invalid assumption, because at higher temperatures, the density may be lower due to thermal expansion - so there may be less mass present. Heat capacity is usually quantified in terms of mass, so a moving airstream needs some way to parameterize its mass flux. The intricacies are going to depend on what approximations you want to make - but since you've only specified so few parameters, it might be safe to use a very simple model. Nimur (talk) 16:35, 1 October 2009 (UTC)

Name of integrated foreign DNA?

Is there a technical name for foreign DNA (from some kind of pathogen, for example) that has been integrated into the DNA of the host? I'm not asking about integrons or retroviral proviruses, but rather some non-directed integration. Thanks. —Preceding unsigned comment added by 129.49.7.150 (talk) 15:31, 1 October 2009 (UTC)

Exogenous DNA. If it was artificially done, the organism is a transgenic organism and the "chimeric" DNA is called recombinant DNA. 152.16.15.144 (talk) 19:24, 1 October 2009 (UTC)

Moving with a cat

Can't you move to a new home with a cat, or is it just a legend? Quest09 (talk) 17:36, 1 October 2009 (UTC)

Sure you can. It might take them a little while to get adjusted to their new territory, but they will. It depends on their disposition. Some will hide for awhile, others will adjust quickly. Just be as good to them as you can, and over time they'll get used to their new surroundings. →Baseball Bugs ^{What's up, Doc?} carrots 17:40, 1 October 2009 (UTC)

(EC)Google it? This is the first result I clicked on. Stressful but not impossible. Vimescarrot (talk) 17:41, 1 October 2009 (UTC)

I've personally done it. Twice. With the same cat. I had no problems with it, and neither did the cat ;-) J.delanoy^gabs_adds 17:43, 1 October 2009 (UTC)

Just make sure you know if the property permits pets if you are renting. Googlemeister (talk) 18:11, 1 October 2009 (UTC)

That's a must. →Baseball Bugs ^{What's up, Doc?} carrots 21:05, 1 October 2009 (UTC)

I've moved house with a cat. It is important to keep them indoors for a week or two so they learn that this new place is home, otherwise they may try to go home to the old house. We shut the cat in the bathroom for the first day or two so as not to expose her to the chaos of moving. It all went very smoothly. --Tango (talk) 19:00, 1 October 2009 (UTC)

Indoors vs. indoors/outdoors vs. outdoors cat is a key issue. And "sealing them off" in a room with everything they need will likely ease their discomfort (along with lessening the likelihood of them running away). →Baseball Bugs ^{What's up, Doc?} carrots 21:05, 1 October 2009 (UTC)

We had a cat that used to jump up on the back shelf of the car to come with us when we went off on weekends. Didn't half surprise a person who was admiring our nice soft toy at the back when he stretched and yawned. Dmcq (talk) 22:49, 1 October 2009 (UTC)

Turns out that we had heavy rain the day after moving day and our outdoor cat sat like a soldier waiting to be let in, Our old house had "cat doors" but I hadn't yet installed them in our new house. Anecdotally, they seem to tolerate the move better than →hydnjo (talk) 22:59, 1 October 2009 (UTC)

If your cat doesn't seem to cooperate, there are other remedies -- penubag  (talk) 02:55, 2 October 2009 (UTC)

While I don't suggest that particular model of cat carrier, it is important to have one, as the average cat will go nuts inside a moving car. For long trips with some difficult cats, you may need to get tranquilizers from the vet. Once you arrive, I agree to lock them in a room with familiar items, including food, water, a blanket to sleep on, and a cat pan they've used before (it will have "their smell" on it). Some outdoor cats may need to give up that lifestyle, if the outdoors are not as safe in the new location. StuRat (talk) 01:23, 5 October 2009 (UTC)

I strongly advise you to keep the cat locked up at the new place for a week or so to make it clear that this is now "home". Otherwise, there's a reasonable chance it will attempt an "incredible journey" to its old place, as our cat did after a move. --Sean 14:54, 2 October 2009 (UTC)

I suggest you get rid of your cat, and get a dog instead. -- Coneslayer (talk) 17:26, 6 October 2009 (UTC)

im about to do the same thing actualy but with kittens, is there any diference? --Talk Shugoːː 16:02, 6 October 2009 (UTC)

Why are clouds so "lumpy"?

I've been thinking a lot about clouds recently (as one does when trying to write software to make pretty pictures of them) - and it occurs to me that it's a bit odd that clouds form such dense, compact 'lumps'. The basic idea is that water evaporates off the ocean or something - the water vapor moves up to some altitude where it condenses into droplets...but why isn't there just a more or less uniform sheet of grey 'stuff'? Why does it form clumps that look like bunny rabbits? (Except that one over there that looks like a cartoon fish). Some of them are uniform-ish sheets of grey - but some of them are really dense - with relatively hard edges. Mutual gravitation of the water droplets? Nah...something like surface tension? Van-der-waal's forces? None of these seems plausible. SteveBaker (talk) 21:50, 1 October 2009 (UTC)

My daughter asked me this a little while ago. I think it is because (1) when warm and cold fronts of air mix you get edges between air of differing temp and humidity (2) when one side of the interface starts to form water droplets it rapidly changes its absorption and radiation of radiation which extenuates the difference between it and the other side of the boundary. I don't know though it is a guess, but some feedback seems needed to give the clean edge versus natural mixing, and the upper side of clouds are often sharper which fits with radiative heat loss once droplet form. I have also noticed that sometimes jet trails of planes disappear but sometimes they just spread out and out thicken. Again, my assumption is that in the right marginal conditions the heat loss from radiating droplets can be a positive feedback cooling air and increasing droplets but generally the mixing of the water and heat just reverts the trail to the surrounding conditions. --BozMo talk 22:07, 1 October 2009 (UTC)

Jet trails can actually generate clouds, if the conditions are right, by providing nucleation sites for water droplet formation. StuRat (talk) 01:11, 5 October 2009 (UTC)

"compact lump" sounds to me like some mechanism is minimising surface area, that would tend to result in a sphere (like surface tension, but it isn't a liquid so I guess it isn't actually surface tension, I'm not sure what it would be [certainly not gravity, it will be something based on electromagnetism]). I imagine it is flattened by atmospheric effects - the atmosphere is full of layers with different winds, temperatures, humidities, etc., the cloud is restricted to one layer (if it crossed the boundary between two layers with different winds it would get ripped apart). That flattening restricts your degrees of freedom, which increases the chance of spheres meeting and becoming bunny rabbits. I'm not sure what stops those bunny rabbits becoming a larger sphere... --Tango (talk) 22:10, 1 October 2009 (UTC)

While you are looking at the topic Steve the WP article on Double diffusive convection could do with some work. There was a whole graduate course on this when I was a lad, including atmospherics...I don't think there is any attraction involved, only feedback but I could be wrong. --BozMo talk 22:14, 1 October 2009 (UTC)

There are essentially two types of cloud (well there are in climate models, and probably in reality): convective and "large scale". Large scale clouds are when the entire atmospheric layer becomes saturated (typically because the whole lot cools, either in place or because of forced ascent) and clouds form - these are the grey slabs. Convective ones are the bunny rabbits. In this case, the entire layer isn't supersaturated, but the column is unstable with respect to overturning, once you take into account heat of condensation. But in that case the ascent is organised, and this is where the clouds form. So ascending air cools, hence vapour condenses, hence the air warms and ascends, and so on (i.e. since the air is being heated as it ascends the moist adiabat is different to the dry, so the plume stays buoyant). And the plume is what makes the clouds lumpy (it can be a tall plume if its a thundercloud or a rather squat one if only a little puffy cloud) William M. Connolley (talk) 22:46, 1 October 2009 (UTC)

Here's a not-too-incorrect explanation of why daytime clouds are lumpy.[4] The lumpiness is fundamentally caused by atmospheric turbulence. There's a lot going on in clouds -- rumor has it there are people who make their living trying to predict them. Short Brigade Harvester Boris (talk) 00:16, 2 October 2009 (UTC)

I think there's really another question here. Besides being "lumpy", some clouds, including the standard cumulus humilis cloud have an oddly discrete outline to them, which I think adds to the impression of lumpiness. "Non-lumpy" clouds like nimbostratus, or even fog obviously also have all kinds of structural variation due to slight differences in temperature and pressure, we just don't think of them as lumpy because they're indistinct. So maybe a better question would be to ask why some clouds, like cumulus, are so sharply delineated, thus letting us see their lumpiness. I have a couple of hypotheses of my own on that, but no evidence to back them up and so will keep them to myself in hopes somebody else can come forward with something more concrete. I would also like to record for posterity that this is easily the most times I've typed the word "lump" in a single paragraph! Matt Deres (talk) 03:39, 2 October 2009 (UTC)

The pedant in me wished to point out that you didn't use the word "lump" even once in the preceding paragraph. "lumpy"=3, "lumpiness"=2. It's OK though - I have my 'inner pedant' under control again now. SteveBaker (talk) 20:34, 2 October 2009 (UTC)

Cumuliform clouds are buoyant parcels that are distinct from the air around them. They are the same as "thermals" in the lower atmosphere that as birds and glider pilots use. If the thermal ascends so high that it cools enough for the water vapor in it to condense, it becomes visible as a cloud. (This height is called the "lifting condensation level" and corresponds to the flat bottoms of these clouds.) The distinct boundary of the cloud is essentially the interface between the rising parcel and the outside environment. Stratiform (layered) clouds have less distinct boundaries because they are large masses of air that get lifted rather than distinct blobs, as William explains above. The high-level wispy clouds are made of ice and have fuzzy boundaries because ice crystals doesn't evaporate as readily as water droplets. Short Brigade Harvester Boris (talk) 13:10, 2 October 2009 (UTC)