Wikipedia:Reference desk/Archives/Science/2012 November 15

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November 15

the material rising in the thermoscope \ thermometer

I would like to know if when the material rises in the the thermoscope meter (or in the thermometer), it indicates that the air in the thermoscope is shrunk or spreading. According to what I understand, the material (in example water ) spreading becaouse the warm and that is the couse of the material rising of the material inthe meter. What doyou think about? 46.210.166.245 (talk) 01:40, 15 November 2012 (UTC)

If the thermoscope is sealed, the liquid expanding will simply compress the air. If the temperature increases the air would also be warmed up at the same time as the liquid and if it was free to do so it too would expand, but since the liquid is far more dense then the air, the resistance the air applies to the fluid would be negiligable and the pressure inside the container would increase. I don't know for sure but I wouldn't be surprised if thermometers were created in such a way as to minimize the amount of air in the space before they are sealed. Vespine (talk) 02:36, 15 November 2012 (UTC)

I think it's more about the compressibility of a liquid (not very compressible) vs gas (easily compressible) rather than density itself as the underlying reason that the liquid expands to compress the gas even when ∆T says they should both be trying to expand. Our Alcohol thermometer article makes mention of the headspace above the liquid level. DMacks (talk) 07:10, 15 November 2012 (UTC)

Suppose, we have a thermometer of range 0 to 100 degree celsius. The thermometer is taken to a region of 100 degree celsius, I don't think that it will show 100 degree celsius if the glass cap of thermometer is attached at the same height as of 100 degree celsius mark. Air in the thermometer will compress to a certain amount, but will not get dissipate. If the cap is attached above the highest temperature scale (about 1 inch), the material inside is able to manage air compression and the thermometer will show correct temperature.

According to Vespine, pressure exerted by air on mercury will be negligible but what will happen if the mercury has expanded to the highest temperature. Will the air exert the same pressure at that temperature ? Air will also expand as the mercury (due to the heat of surrounding medium) and hence the kinetic energy of air molecules also increase and this time air will exert more pressure on mercury. Think about mercury barometer, it has vacuum in its mercury column but in case of mercury thermometer it has air. I do not know why ?Sunny Singh (DAV) (talk) 10:39, 15 November 2012 (UTC)

The air is confined to the volume left by the rising mercury. So the air pressure will increase due to 2 reasons 1) the pressure increases due to the reduced volume available, and 2) it increases due to the increased temperature. However, the increase in air pressure has negligible effect on calibration, as the mercury, being a liquid, is vitually incompressible. Floda 121.221.27.52 (talk) 11:14, 15 November 2012 (UTC)

I'm sorry, but I don't understand what does the meter shows when the material rises in the meter. the question is: Is the air in the meter is shrunk or spreading? (the thermoscope is open to get influences of atmosphere) 176.13.83.244 (talk) 18:17, 15 November 2012 (UTC)

The volume of the space at the top of the thermometer is smaller. That is the definition of "shrunk". --Jayron32 18:31, 15 November 2012 (UTC)

(edit conflict) As explained above, it doesn't make much difference whether the meter is sealed or open because it is the expansion of the material (usually coloured alcohol or mercury) that is being observed as the temperature rises. The only occasion when the compression of air is important is when a thermometer is heated well above its maximum range, and the air gets compressed almost to zero volume. The high pressure usually causes the bulb to explode. There may well be a partial vacuum above the material in a sealed thermometer, but this is not essential in a thermometer (though it is important in a barometer). Dbfirs 18:34, 15 November 2012 (UTC)

If you have a sealed container half full of fluid, when you heat it, the fluid will expand. The container with its contents will still have the same weight. But the fluid will be less dense, so another container within that fluid might sink. You should be able to make a Galileo thermometer which consists of concentric spheres, each containing a slightly different amount of fluid, so that each sphere rises or sinks within its surrounding sphere independently of the effect on all the other spheres. Though it would be terribly slow to respond unless made out of a very good heat conductor. I don't know if anyone has made such a thing but I would expect so. Wnt (talk) 02:52, 16 November 2012 (UTC)

Ever been to a gift shop Wnt? ;)

They are a bit on the expensive side but quite easy to find. Each glass orb has a punched sheet metal label, both to show the temperature and to calibrate the orbs. Clipping/filing the label will remove some weight and make the orb float more easily. - ¡Ouch! (^{hurt me} / _{more pain}) 09:27, 16 November 2012 (UTC)

I've seen them with multiple orbs in a single tube, which the OP wonders whether it is open or closed (which won't matter to the degree that the liquid is incompressible relative to the gas in the outer chamber anyway). But I've never seen them with one orb inside the next inside the next etc.; the tags would hinder the aesthetics of this and their weight would need to be well calibrated without them. Wnt (talk) 17:14, 16 November 2012 (UTC)

I thought that the OP was asking about a thermoscope (or ordinary thermometer), not a Galilean thermometer. Dbfirs 00:37, 17 November 2012 (UTC)

You're right - we had a different question about that one a little while ago... Wnt (talk) 05:43, 17 November 2012 (UTC)

Omg - I didn't get that you were talking about all orbs being concentric, not all merely within one outer containment. - ¡Ouch! (^{hurt me} / _{more pain}) 18:11, 17 November 2012 (UTC)

Electromagnetic radiation

At the beginning of the statement, it says: "electromagnetic radiation (EM radiation or EMR) is a form of energy emitted and absorbed by charged particles". So are charged particles mean it is either positive or negative? How about neutral particle with no charge? Can neutral particle emit and absorb electromagnetic radiation?174.20.41.202 (talk) 03:17, 15 November 2012 (UTC)

Consider blackbody radiation. It can be emitted by matter that has, in bulk, no net charge. But, if you zoom in all the way to the fundamental interactions that are responsible for the thermal emission of radiant energy, you will discover that in the very microscopic sense, the radiation is emitted by a process that could be classically described as a movement of charged particles. We might say that warm objects have electrons in excited atomic orbitals, and that the photons are produced when the electrons decay. That explanation treats the photon as an emitted particle, and doesn't concern itself with the electrodynamics of the photon-as-a-wave. The exact details are actually somewhat subtle; for most purposes, we gloss over this detail; and if we want to study it closely, we have to use some very heavy-duty mathematical physics. Often, when we consider blackbody radiation, we are not interested in electrodynamic interactions; we only care about the energy that is being conveyed. But, any emitted electromagnetic radiation - whatever its source - is attributable to microscopic movement of charge.

When we discuss subatomic particles, we have to generalize the way we think about emitted radiation. Energy can be emitted in other forms besides electromagnetic waves, especially if it's radiated in a way that isn't related to wiggling electric charges. Sometimes, certain "exotic" energetic interactions between particles result in an emitted particle - which carries energy - that is not an electromagnetic wave. A neutrino can be emitted and can convey energy, without any interaction with electric charge. A phonon is energy that is emitted that takes the form of propagating wave in bulk condensed matter. There are many such examples where the radiated energy is not electromagnetic in nature, because the source of energy did not participate in any electromagnetic interaction. Nimur (talk) 06:19, 15 November 2012 (UTC)

I think Nimur is wrong. Consider a quantity of non-reactive gas at some finite temperature. The gas molecules, which are not charged particles are, according to the kinetic theory, continually flying about in the x, y, z tanslational axes and also rotating about these axes. The gas will be emitting, in all directions, heat in the form of infrared radiation, which is electromagnetic radiation. If there is incoming infrared radiation of total wattage greater than that being emitted, the temperature will increase - that is the gas will absorb energy by increasing the translational velocity of the molecules. If the molecules are not symmetrical in all x,y,z dimensions some of the heat energy will be converted into increased rotational movement. Some will be absorbed by lengthening the atomic bonds, thus increasing the "flywheel effect" Charged particles have no part in this beyond forming the atomic bonds, which are not changed by the increse in temperature. It is possible to have a "gas" in this thermodynamic sense consiting not of complete atoms or molecules, but instead consisting entirely of one sort of particle (electron, nuetron, etc), which may or may not be a charged particle. Floda 121.221.27.52 (talk) 11:08, 15 November 2012 (UTC)

Um, "Charged particles have no part..." Is completely wrong. If, as you say, there is "lengthening of the atomic bonds" then, by definition, that is a change in shape of molecular orbitals which involves a change in energy of electrons, which requires the electrons to change energy levels in some way, which requires the emission or absorption of photons, QED. It is inescapable: if photons are produced, something about the motion/location/energy of electrons changed (which "property" of electrons changed depends on your perspective, but the initial statement that all EM radiation originates in changes to charged particles is fundamentally true.) --Jayron32 13:50, 15 November 2012 (UTC)

Jayron, you did not mention clouds of sub-atomic particles. So, if you (& Nimur) are correct in saying charged particles are necessarily involved in the emission and absorption of EM radiation, then a cloud of nuetrons either cannot exist, or cannot have a temperature? I always thought that the Kinetic Theory of Gasses applies to sub-atomic particles as well as atoms and molecules when they are not forming a solid or liquid. (A cloud of particles should obey the KTofG if the following are true: Sum of particle volume << cloud volume; number of particles is very large; except in collisions, interactions are negligible; Number of collisions between particles is large compared with collisions of particles with container). Floda 60.230.192.18 (talk) 15:34, 15 November 2012 (UTC)

I don't follow the connection that is being made here between kinetic theory and thermal radiation. A "gas" of sub-atomic particles will always have a temperature and other thermodynamic properties, but it will only emit or absorb thermal radiation if the particles carry an electric charge or have components that carry an electric charge. Atoms and molecules have no overall charge but they still emit and absorb thermal radiation because they have charged components and hence they have an electric dipole moment. A "gas" of neutral fundamental particles would have a thermodynamic temperature but would not emit or absorb thermal radiation. Gandalf61 (talk) 15:57, 15 November 2012 (UTC)

Such neutral fundamental particles will have to be completely decoupled from electromagnetism, otherwise there will be higher order effects leading to the emission of photons. This means that the particles must be hidden sector particles that are completely decoupled from the Standard Model. And even then there will be extremely small effects mediated by gravity. Count Iblis (talk) 17:37, 15 November 2012 (UTC)

To expand on Iblis's point: neutrinos don't have any electric charge, and are not mediated by photons, so interactions involving neutrinos only involve exchange of particles like the W and Z bosons. If something neutrino does causes the release of a photon, it is only of the "higher order effects" that Iblis mentions: the weak bosons that neutrinos release interacting in some way with a charged particle, and THAT charged particle releasing a photon. It should be noted (not mentioned yet as a point of confusion, but it could be given the way the discussion is going) that even neutrons are not fundamentally neutral; they are composed of charged quarks, and thus are subject to EM interactions just as other charged particles are. --Jayron32 20:18, 15 November 2012 (UTC)

God, I'm so lost... Didn't expect the answers to be this lengthy. What you guys saying are beyond what I can understand, I'm just a curious ordinary person who doesn't know physic very well. Can you someone simply answer my question in simply way? Can neutral particle emit and absorb electromagnetic radiation? If it can then how the article only say "emit and absorb by charged particles"?174.20.41.202 (talk) 21:43, 15 November 2012 (UTC)

The answer depends entirely on what you mean by "neutral particle". If you mean "any particle with no net charge" then the answer is "It can still emit and absorb EM radiation if the particle is made up of smaller particles that themselves have electric charges". IF you mean "a fundamental particle which is not composed of smaller particles, and which is fundamentally neutral at all levels of organization" then the answer is "No, it cannot emit and absorb EM radiation" Thus, both the hydrogen atom and the neutron, which are each neutral but which are also each composed of smaller bits that themselves are charged (a proton and electron in the former, up and down quarks in the latter) CAN interact via electromagnetic radiation. However, a particle which is both indivisible and neutral, like a neutrino cannot. Does that make any more sense? --Jayron32 22:15, 15 November 2012 (UTC)

[Edit conflict]Electrically neutral particles that are composed of equal amounts of positive and negative charge emit and absorb radiation, so yes, neutral particles do, but only because of the equal presence of both kinds of charge which can emit or absorb the radiation. Does that help? -Modocc (talk) 22:23, 15 November 2012 (UTC)

Jayron and Modocc's explanation make perfect sense to me. So basically if an elementary particle that has no charge then they won't emit or absorb radiation.174.20.41.202 (talk) 03:22, 16 November 2012 (UTC)

This isn't exactly true, only a good approximation. Neutral particles can have magnetic dipole moments, even if they are elementary. E.g. the neutrino has a magnetic dipole moment, despite it being a neutral elementary particle, see e.g. here, and they can then emit electromagnetic radiation. This happens due to quantum effects. In quantum mechanics, if A can interact with C via B, then A will also be able to interact with C if B isn't present due to virtual B popping out of the vacuum and disapearing again. This is how the neutrino gains a magnetic moment despite not consisting of charged particles. Count Iblis (talk) 03:40, 16 November 2012 (UTC)

A related question: why don't neutral objects create EM fields all the time as they move about? (Like say, a human running and sprinting and then slowing down to catch his breath and then running again?) Individually, the objects are made of charged particles, so shouldn't they generate their own fields, that then would mostly cancel out? Photons would be emitted though. 71.207.151.227 (talk) 23:42, 15 November 2012 (UTC)

You're blasting out electromagnetic radiation right now, without doing anything. See Night vision device and Thermography for some applications of this. --Jayron32 23:55, 15 November 2012 (UTC)

They wouldn't cancel out because the positive and negative charges are not located at the same place. See electric dipole moment for a simple example of an overall-neutral system of charges that has a very real electric field about itself. Someguy1221 (talk) 23:50, 15 November 2012 (UTC)

Jayron is completely right.

On top of that, the human body is both a conductor and a capacitor. It might be a coil, if you look at the blood vessels which contain in a coarse approximation a conductive NaOH solution. A few feet of copper wire and some simple electronic components are sufficient to recognize other bodies with some accuracy. Nothing with a snowball's chance to stand in court, but... I tuned one of these to a jogger I encountered regularly, during years when I had time to kill and easy access to the components needed.

When he passed me, the circuit made a nice "blip" sound. There were some false positives, but less than one in 100. I had to retune the circuit while I was losing weight, tho. My own signature was screwing with its detection ability. Boy was she scared...

But that's about it. You can't do anything really evil, like electrocute her, or plant a neat "Kill your boyfriend" thought. But hey, there are other methods to get that done nowadays. - ¡Ouch! (^{hurt me} / _{more pain}) 07:49, 16 November 2012 (UTC)

NaCl, dammit. Self-whack. <°|>>><< - ¡Ouch! (^{hurt me} / _{more pain}) 08:32, 16 November 2012 (UTC)

Recharging smartphone

Why does it take up to an hour to go from 99% charged to 100% on my HTC One S? 67.243.3.6 (talk) 15:48, 15 November 2012 (UTC)

See Lithium-ion battery#Battery charging procedure. The charging is not linear, but consists of three stages. Dbfirs 17:45, 15 November 2012 (UTC)

Doppler effect: two moving objects at an angle

What's the formula for the Doppler effect if two moving objects are moving at an angle subtending the velocities between the two? For example, an airplane overtakes a motor vehicle on the ground-- how would you calculate the frequency shift of the engine roar emitted or received at any specific moment? 128.143.100.179 (talk) 18:41, 15 November 2012 (UTC)

First, you would subract the velocity vectors (assuming you have an external frame of reference) to get the relative velocity between the two points. Then you use the Dot product to project the velocity vector onto the vector between the two points. This will tell you how quickly the points are moving towards or aways from each other, which can then be fed into the Doppler formula. 209.131.76.183 (talk) 18:55, 15 November 2012 (UTC)

That's fine for light in vacuum, but the nonrelativistic Doppler formula takes the velocities of emitter and receiver relative to the medium. Their relative velocity isn't enough. -- BenRG (talk) 05:23, 16 November 2012 (UTC)

You could also use a full-wave-equation modeling technique, such as the Finite-difference time-domain method, to estimate the wavefield at every point. Then, you could sample the simulator to approximate the signal that would be observed for any trajectory at any velocity. This method is a lot more difficult to implement in practice, but if performed correctly, can account for many non-ideal effects, like the interaction between the sounds of the two vehicles; or imperfections in the propagation material. (Our article on FDTD has a strong bias toward the application for solving electromagnetic wave equations, but you can equally-well apply the numerical technique to solve the acoustic wave equation, and many other multidimensional PDEs). Nimur (talk) 20:05, 15 November 2012 (UTC)

Okay, I would have to do this in an exam. (It's a problem that I anticipate, as my professor likes to give us "climax of all that we know" questiosns. Also, according to my lecture notes, converting the two velocities into a relative velocity doesn't really work as v becomes on the same scale as c (e.g. 0.1c) . 71.207.151.227 (talk) 23:38, 15 November 2012 (UTC)

Use the lorentz transform to calculate the relative velocities when you suspect that the effects will differ from the classical case. Then, use that intermediate result and apply your favorite variation on the relativistic Doppler effect. It sounds like this work is related to a class; so check your notes or textbook for your professor's formula; or you can use the formula provided in our article. Nimur (talk) 00:23, 16 November 2012 (UTC)

I meant c as in the speed of sound, not the speed of light. 71.207.151.227 (talk) 22:28, 19 November 2012 (UTC)

A general formula for the nonrelativistic Doppler shift is

${\frac {f_{r}}{f_{s}}}={\frac {c+\mathbf {v} _{r}\cdot \mathbf {x} }{c+\mathbf {v} _{s}\cdot \mathbf {x} }}$

where

$f_{s}$ and $f_{r}$ are the emitted and received frequencies,

$c$ is the speed of sound,

$\mathbf {v} _{s}$ is the velocity of the source at the time it emits the sound,

$\mathbf {v} _{r}$ is the velocity of the receiver at the time it receives the sound,

$\mathbf {x}$ is a unit vector pointing from the location of the receiver at the time of reception towards the location of the source at the time of emission,

and all the velocities and positions are computed in the rest frame of the medium.

It should be easy to see how this reduces to the one-dimensional case given in the article.

For the relativistic Doppler shift of light in a vacuum, there's a simpler formula: ${\frac {f_{r}}{f_{s}}}={\frac {\mathbf {v} _{r}\cdot \mathbf {x} }{\mathbf {v} _{s}\cdot \mathbf {x} }}$, where $\mathbf {v} _{s}$ and $\mathbf {v} _{r}$ are now four-velocities and $\mathbf {x}$ is a four-vector pointing from the spacetime location of reception to the spacetime location of emission. You're no longer limited to a particular reference frame (of course) and $\mathbf {x}$ doesn't have to be a unit vector (in fact, it can't be, because it's lightlike).

Should I add these formulas to the articles?

One could also derive a relativistic formula for sound or light in a medium, but I'm not sure how useful it would be. -- BenRG (talk) 05:23, 16 November 2012 (UTC)

Origin of the Nebraska sand hills

Hi. Currently, I'm researching the Sand Hills (Nebraska) for a world geologic location and hazard research assignment, and I'm exploring the geologic origins and proneness to mass wasting of these ancient sand dunes. What I've found so far is that the sand hills region is composed of Quaternary-age "Eutric" regosols, bounded by much sediment and/or uplift of Eocene to Mesozoic (think Niobrara Sea) age, located at the edge of the Denver Basin, with loess deposits extending to its east and southeast (yes, I have citations for all this info), reaching all the way to the Mississippi River and beyond, while the loess sediment also reaches down to the mouth of the Mississippi River - perhaps this makes sense, since the Platte River runs through the southern sandhills. I'm asking anyone for more relevant information about this area that they can find, including the depth of the sand (soil and rock core profiles would really help), the presence of any halite or gypsum salt deposits (I'm noticing that Nebraska is downwind of Lake Bonneville of similar age), and the effects of past explosive volcanic eruptions including at Yellowstone caldera on the region, and any evidence of past vegetation or absence thereof - for instance, the area's dunes were active during the Medieval Warm Period. Might the dunes creep east of Omaha and Lincoln, NE in the not-too-distant future? Also, what is the source of the sand dunes scattered throughout the Lake Michigan shoreline - are they direct deposits from the Laurentide ice sheet, as suggested by the USGS map on the "loess" article, or might the Nebraskan sandhills contributed to their origin sometime during the Medieval Warm Period? Any information on the influence on the stability of the dunes from the underlying artesian aquifer - the Ogallala Aquifer, which has its thickest portion underneath the sandhills and which is used for irrigation in mile-wide circular drip irrigation systems, which has dropped several metres in the past decades. Also, what about tornado alley - can you get sand tornadoes over devegetated portions of the dunes? What type of dunes are they - Barchan, Transverse, Longitudinal, Concave, Parabolic or Star? Is there any evidence of erosion by water on the dunes, and what are their permeability, porosity and density like? What are the differences and similarities between Nebraska's sand hills and other sand hills regions across North America and worldwide?

I'm also willing to contact certain universities and research/environmental organizations who might have more relevant information or current research on the topic, either by email or by phone - can you recommend any groups I might contact?

Thanks. ~AH1 ^(discuss!) 21:57, 15 November 2012 (UTC)

I found my answer to the Michigan sand dunes at Lake Chicago - by the way, remember that any extra research done here could also go into improving the article(s) discussed. ~AH1 ^(discuss!) 22:10, 15 November 2012 (UTC)

Answering as to the tornado part, tornadoes in the Nebraska sand dunes will act just as tornadoes anywhere else. As rapidly rotating columns of air, tornadoes are not necessarily visible; the only reason they are visible are if a condensation funnel forms or if they pick up debris from the ground. Most often the debris picked up that makes a tornado visible is dust/dirt (thus how you can have red tornadoes in Oklahoma). Because of this I would expect at least some dust/sand to be in any tornado over the sand hills, but this doesn't make them any different from other tornadoes. Ks0stm ^{(T•C•G•E)} 01:22, 16 November 2012 (UTC)

I found out from this 32-page piece of research that the Sand Hills are composed of mostly complex transverse dunes, and that the aquifer does indeed indirectly stabilize the dunes: it feeds shallow wetland systems, that in droughts can release evaporation that sustains neighbouring prairie grasses. ~AH1 ^(discuss!) 04:45, 16 November 2012 (UTC)

Your apparent thesis that aquifer depletion could release these dunes and cause the desertification of the area is most intriguing, but as you've probably gathered by now, the Refdesk is having a lot of trouble rising to your level on this one (myself included). My gut feeling is that if you talk to someone in the right department at one of the colleges closest to this feature (no matter how small), you'll get far more information that we can provide. Though actually one source I found [1] which I think is saying the aquifer is not presently in trouble, as long as precipitation doesn't greatly decrease, was from a Chinese group. Wnt (talk) 18:24, 17 November 2012 (UTC)

Bok choy is the same species as the common turnip, really? It resembles nothing like a turnip

I know that artificial selection can really bring out the variety in a species, but this amazes me. How is it possible? It tastes nothing like turnip leaves. 71.207.151.227 (talk) 23:32, 15 November 2012 (UTC)

And yet a Chihuahua, a Newfoundland, a Boston Terrier and a Gray wolf are also all the same species of animal. It doesn't require your belief to be true, you know. The Brassica genus contains hundreds of different vegetables in about half a dozen species, and Bok Choy and Turnips are indeed, to the best of our ability to classify such things, the same species. --Jayron32 23:51, 15 November 2012 (UTC)

Sub question: Whats the importance or relevance of knowing that bok choy and common turnip are of the same species? (Of course im not asking specifically abouit bok choy and turnip alone) 203.112.82.129 (talk) 00:19, 16 November 2012 (UTC)

What's the point of knowing anything not immediately useful in daily life, really? But seriously, it could help with people trying to breed new strains to know that they can cross-breed related plants. It helps to keep from wasting time trying to narrow it down from all the other plants in the world. That's the first answer off the top of my head. I'm sure there are plenty of other reasons, too. Mingmingla (talk) 01:50, 16 November 2012 (UTC)

Trivial knowledge about various beet species provides for good conversational filler between more important, but perhaps less stimulating, quotidien tasks, such as programming iPads. On the statistical average, it helps us focus when we can rest our brains for a little while by analyzing simpler problems like taxonomic minutiae and behaviors of different types of muons. In other words, it's in the national interest to support beet research and discussion, even when the immediate benefits are not apparent. You wouldn't want to remove such stimulating discourse from our society, and through the consequent demotivation, preclude our most creative and productive minds from cranking out iPads at peak efficiency, would you? Nimur (talk) 02:35, 16 November 2012 (UTC)

Even now, sinister forces are hard at work trying to create an invincible army of bok choynips. Clarityfiend (talk) 01:54, 16 November 2012 (UTC)

And you'd be right. I used to work at an institute for fruit and vegetable research, and crosses like this are commonplace. Mainly to restore traits lost in the one with traits from the other, like disease and pest resistance. Dominus Vobisdu (talk) 01:59, 16 November 2012 (UTC)

My mistress' eyes are nothing like a turnip. --Trovatore (talk) 02:03, 16 November 2012 (UTC)

The knowledge might also be relevant since if you are allergic to one variety there's a good chance you'll be allergic to all. μηδείς (talk) 02:22, 16 November 2012 (UTC)

According to [2], Brassica rapa is a 6n (triploid) species. "The recurring genome duplications and triplication events have created massive genetic redundancy that quickly opens the possibility of sub-functionalization and neo-functionalization for duplicated or triplicated homeologs (Force et al., 1999; Shruti and David, 2005). It is likely that the extreme morphological diversity seen within the various Brassica species is due, at least in part, to the genetic redundancy and functional diversification permitted by these genomic events." Wnt (talk) 02:47, 16 November 2012 (UTC)

What Wnt's interesting statement means in less technical terms is that at some point the normal paired set of chromosomes found in the parent of the species got tripled (as if a human somehow got six sets of 23 chromosomes, in stead of just two sets), meaning that some of what were then extra genes got freed up to mutate and develop other functions that otherwise wouldn't have been possible if all the genes were busy serving their original functions. μηδείς (talk) 05:09, 16 November 2012 (UTC)

And, of course, we have an article on this!--Rallette (talk) 06:54, 16 November 2012 (UTC)