Wikipedia:Reference desk/Archives/Science/2012 January 24

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January 24[edit]

T-shaped versus trigonal planar[edit]

Why do molecules such as ClF3 adopt a T-shaped geometry instead of a trigonal planar geometry (which would not only place the three substituent atoms as far away from each other as possible, but also place the two lone pairs as far as possible from each other)? Whoop whoop pull up Bitching Betty | Averted crashes 01:42, 24 January 2012 (UTC)

As always for "why", we can only hand-wave and rationalize, but here goes. The geometric terms you use are a simplistic VSPER type that ignore the lone-pairs entirely, but then you are specifically interested in their role. That's a problematic start. Trigonal–planar of the 3 F ligands is trigonal–bipyramidal overall, with the two lone-pairs at the polar positions. Those two pairs would indeed by further from each other but would become only 90° each from each of the three F ligands. T-shaped is still trigonal–bipyramidal overall, but with the two lone-pairs at equatorial positions. Not sure what the specific energy contributions of each type of interaction are, but "obviously (to nature) it's better that way". DMacks (talk) 01:53, 24 January 2012 (UTC)
Further hand-waving, but we can also say that the T-shaped geometry provides a lower-energy solution to the wave function than does the trigonal planar geometry, taking into account all electrons in the molecule. --Jayron32 15:23, 24 January 2012 (UTC)
We can also hand-wave at the atomic/molecular-orbital level. By geometry, the 5-coordinate center would appear to be approximately sp2 hybridized--three equatorial ligands in the traditional σ plane and two axial in the remaining unhybridized p atomic-orbital positions. In a weird sense, this structure (with T geometry, one F at each axial position) is like an SN2 transition state with one F displacing the other--the two axial Cl–F bonds are noticeably longer than the equatorial one.
But back to orbitals... let's try three Cl–F σ equatorial: each is formed using one of the sp2 atomic hybrids on Cl. That leaves p on Cl: only a single atomic orbital available for the two axial lone-pairs (all other atomics are orthogonal to those directions). With the T geometry of the three F, one Cl–F σ forms using one of the sp2 on Cl and the two lone-pairs are the other two of those hybrids. That leaves the p on Cl for bonding to the two axial F. No problem--a three-center four-electron bond is a standard bonding model that does not require exceeding fundamental orbital occupancy rules. Heck, it even means the chlorine does not exceed the octet rule, despite being drawn as "three σ and 2 lone-pair == 5 valence pairs". DMacks (talk) 15:49, 24 January 2012 (UTC)
Yes, but you can do the same for the trigonal planar geometry....! Chris (talk) 16:48, 24 January 2012 (UTC)
No, you can't, because for a trigonal planar geometry, with all of the same stipulations that DMacks explained regarding the p-orbital, would require both lone pairs to occupy the same p-orbital. Four electrons in one p-orbital is a no-no per the pauli exclusion principle. Indeed, IIRC, the 3-center 4-electron bond model is preferred because expanded octet hybridization schemes (like sp3d required for 5 bonds to the central atom) don't work well; the d-orbital is too far off in energy from the s and p orbitals to effectively hybridize with it. Multiple-center bonding schemes are perfectly allowable (if very cumbersome on paper) per molecular orbital theory, indeed the standard molecular orbital model of benzene utilizes a 6-center 6-electron pi-bonding scheme to explain the aromatic bonding; likewise diborane famously uses a pair of 3-center 2-electron bonds. --Jayron32 19:43, 24 January 2012 (UTC)
Yes, you can. Consider the Cl atom to be sp hybridised, with a lone pair in each of the hybrids in the axial positions. That leaves one Cl-F bond using a p orbital on the chlorine and a fluorine orbital, and the other p orbital on the chlorine to form the central portion of a 3c-4e bond with the remaining two fluorine atoms. Anyhow, this is irrelevant. Chris (talk) 23:00, 24 January 2012 (UTC)
Yeah, you could do it that way; excepting that that is apparently NOT the lowest energy solution to the system; the lowest energy solution is the T-shaped one. That's what the result comes down to: we could come up with a near infinite number of orbital organizations which could account for any number of possible shapes; however the molecule is still T-shaped, regardless. Any explanation which says it shouldn't be is, by necessity, a faulted explanation. Comeing up with an explanation that says the sun should rise in the west tomorrow wouldn't make it so. --Jayron32 02:47, 25 January 2012 (UTC)
Though I think I recall some differences of opinion on this, the use of d orbitals to make ClF3 work still seems to be respectable.[1][2] (While three-center bonds are possible, throwing the d orbitals into the mix certainly explain how you manage to fit five electron pairs around an atom, and also why this happens with chlorine and not fluorine) Somehow the d orbitals explain the structure whereas only s and p do not, according to that latter source. The d orbital geometry is more complex, and I haven't worked it out in my mind. Wnt (talk) 03:41, 25 January 2012 (UTC)
I think this is one of those clear cases of having two workable models, and trying to decide which bit to "fudge" to make one model better than the other. Hybridizing d orbitals into s and p doesn't work well according to the basic principles of orbital hybridization, which says that orbitals need to be close in energy to hybridize, but if you ignore that then a 5-orbital hybridization system gives the Trigonal bipyramidal molecular geometry predicted by VSEPR as the simplest geometry needed to fit 5 domains around the central atom. So that model is useful insofar as it works to give you the correct geometry. However, if trying to hybridize something like sp3d does bother one (and it does bother some chemists), then you can get the "correct" result using the molecular orbital method using the 3-center 4-electron bond method described above. Ultimately, it's two different models that give the same end result, so its probably moot which model one uses. --Jayron32 03:54, 25 January 2012 (UTC)
Maybe, yet... the quantum numbers really should be different for the two models. I'd think there must be some experiment that could distinguish which is right. Wnt (talk) 04:40, 26 January 2012 (UTC)
The quantum numbers would be different for the T-shaped vs. the trigonal planar shape. Quantum numbers tell you the geometry of the electron cloud, and for a molecule like this the relevent quantum numbers are those given by molecular orbital theory. The whole point I have been trying to make all along is that the T-shaped result is the one that the quantum numbers give you if you do the rigorous work. Any conclusion which states "But it should be <some other shape I like better for aesthetic reasons>" is faulty, and it isn't worth exploring any explanation that arrives at those conclusions. --Jayron32 14:06, 26 January 2012 (UTC)
Well, what I meant was that I would think, regardless of the observed geometry, an electron in a d orbital should be able to have an angular momentum number of 2. So if there are d orbitals hybridized in, shouldn't there be a way to detect that? Wnt (talk) 15:12, 26 January 2012 (UTC)
I'm not sure what you mean by "detect"? Bonding angles and distances are generally experimentally confirmed via things like X-ray crystallography, but you can't image an individual molecule and detect what the electron cloud "looks like" via any known direct means. You can plot the wave function, but that isn't an experiment, per se, its just a really detailed model, and isn't any more "experimental" than a dot diagram is, just more detailed. --Jayron32 15:17, 26 January 2012 (UTC)
I'm not well versed with this, but I'd think that there would be a way to use nuclear magnetic resonance or electron spin resonance to get at the data. I mean, the electrons really do carry angular momentum, which ought to be possible to detect somehow, if not in the compound itself then in some ion or derivative of it. I think.... Wnt (talk) 15:40, 26 January 2012 (UTC)

Does balsamic vinegar go bad ?[edit]

If so, what happens to it when it gets old ? StuRat (talk) 05:03, 24 January 2012 (UTC)

Unopened, it'll stay good just about forever. Opened, the presence of oxygen can reactivate the bacteria in unpasteurized vinegar, and slowly change the acetic acid to carbon dioxide and water. However, that takes quite some time (a year or more), and when it happens, the vinegar becomes cloudy or gets a gelatinous lump forming in it. Nothing to get bent out of shape about. Just filter or take out the lump, and the vinegar is probably still ok to use unless the smell or taste are noticibly off. Also, aromatic components will eventually evaporate and the vinegar will lose some of its flavor, but again, it's still useable. Dominus Vobisdu (talk) 05:29, 24 January 2012 (UTC)
And that gelatinous thing is called a mother (astoundingly, we have no article for a vinegar mother, apart from bacterial culture), which can be used to make more vinegar. Acroterion (talk) 14:41, 24 January 2012 (UTC)
We have Mother dough as a redirect too. But mother (fermentation) seems ripe for an article. DMacks (talk) 14:54, 24 January 2012 (UTC)
Balsamic vinegar is the "bourbon" or "scotch" of the vinegar world; it is aged in wood casks and carries a wide variety of complex (and sensitive) flavor compounds. As noted, an unopened bottle will keep pretty much indefinitely, but an open bottle will tend to degrade. Which is not to say that it degrades into something toxic, but it can degrade into something less desirable than the original stuff; I wouldn't be wary of eating such degraded vinegar, but I wouldn't expect it to be as tasty. --Jayron32 15:19, 24 January 2012 (UTC)

Thanks, all. I found some balsamic vinegar of indeterminate age, and now I will feel free to try it on my salad. 17:45, 24 January 2012 (UTC)


Could it be that pets are evolving to understand language?[edit]

At least, if we prefer pets at our side which understand us better and better, and treat these pets as well as possible, that would make this trait more robust. (talk) 15:38, 24 January 2012 (UTC)

It's not clear that any of our current pets have anywhere near the types of neurological faculties to really understand language. But certainly one of the traits that has made dogs so domesticatable, and is prominent in basically every variety of domesticated dog, is the ability to differentiate human speech patterns enough to understand tone and a limited vocabulary. Can you go from that to full blown language? I doubt it, even over millions of years. But there's probably less known than not known about the evolution of language in humans, and that's really the only sample size we have go to on. (And evolution is not about treating anything "as well as possible." It's about passing on genes. You could be a horrible owner but a good breeder. Evolution only really cares much about the latter.) --Mr.98 (talk) 15:50, 24 January 2012 (UTC)
You are right about the fact that treating your pet better, even if it improves its life quality, won't spread its genes, and breeding them like chicken in cages could be more beneficial to its genes.. But I know from reliable source that dogs which show better understanding of human language are strongly preferred and some breeders strive to deliver these 'quality' dogs. That doesn't affect the fact that dogs as a whole might be getting stupid in average (just look at a chihuahua, to see how stupid a dog might be), but there seems to be possible to have a group of dogs with better linguistic abilities. If they are descendants of dogs that could understand two dozen words, maybe they can understand now simple sentences like go get x (x is one of 20 things). Take also into account that artificial selection is faster than natural selection at producing change. (talk) 16:05, 24 January 2012 (UTC)
Dogs? Languages per se? Not likely - and no particular reason they'd be doing it now, rather than over the entirety of our relationship with them. Besides, they're already quite expert at understanding other aspects of communication; they understand gesture, they understand nuance, they understand body language. But they'll never understand "stop barking at those raccoons, I'm trying to sleep!" (expletives deleted.) --jpgordon::==( o ) 15:57, 24 January 2012 (UTC)
No, but if you yell "Stop !", they might understand that you want them to stop barking, although them understanding your motivation (that you can't sleep with them barking) is probably well beyond them. StuRat (talk) 17:40, 24 January 2012 (UTC)
Of course, there are other pets. People keep African Grey Parrots as pets and the jury is still out on how much actual language they understand, but it's probably more than a dog. Cats are not even pack animals so I wouldn't expect much out of them (as usual). Fish, nope. --Mr.98 (talk) 16:02, 24 January 2012 (UTC)
Delphines are also a good candidate for understanding human language, and apes too. — Preceding unsigned comment added by (talk) 16:09, 24 January 2012 (UTC)
Yes, though humans have done very little selective breeding of dolphins, apes, and parrots. I understood the OP's question as largely pertinent to pets which have been domesticated, not exotic pets, which are usually effectively of the wild type. SemanticMantis (talk) 16:47, 24 January 2012 (UTC)
(WP:EC) I think it's fair to say that our selective breeding of domestic pets has been influenced in part by a desire to select individuals that are better at communicating with people. Not language per se, but it does go both ways, meaning e.g. that we have tended to breed dogs that are more responsive to us, and also dogs that we are more easily able to understand (tail wagging, barking to go out, etc.). SemanticMantis (talk) 16:04, 24 January 2012 (UTC)
This is definitely true. One of the interesting differences between dogs and wolves is the dogs know to look at human faces for information; wolves generally do not. (Or so I read, somewhere.) --Mr.98 (talk) 20:57, 24 January 2012 (UTC)
The NOVA series had a great episode on this: Dogs Decoded. It's impressive to see the tests about how different animals react to humans pointing. Only the dogs consistantly understood the human was pointing where the treat was:
Also impressive was the multi-decade project breeding foxes selectively for either aggression or docile behavior. Well worth a watch. — The Hand That Feeds You:Bite 22:08, 24 January 2012 (UTC)
The Border collie Betsy is supposedly capable of understanding at least 340 words, and seems to have at least a crude grammar (distinguishing between objects and commands). The three dogs with the largest vocabularies are all collies, and this Nat Geo article suggests it may be due to the fact that collies seem to have such good language skills comes from the fact that they were bred/evolved in the especially demanding field of herding. I believe the book The Intelligence of Dogs suggests a similar explanation. Smurrayinchester 22:55, 24 January 2012 (UTC)
I think you're onto it. Shepherd dogs have been selectively bred to understand commands. The great apes have not. So it shouldn't be surprising that dogs understand language better than apes do. ←Baseball Bugs What's up, Doc? carrots→ 23:02, 24 January 2012 (UTC)
In inter-species communication (as in many other intellectual endeavors), patience and compliance are almost as important as intelligence. Collies happen to be intelligent and compliant. German shepherds are just as intelligent, but they tend to be more free-spirited and independent, it takes more effort to teach them. Cats are independent to the point of being antisocial, and that makes it very difficult to train cats the way we train dogs, regardless of their level of intelligence.--Itinerant1 (talk) 02:12, 25 January 2012 (UTC)
This dog certainly seems to have sussed human language out. Especially when the subject is food... Enjoy: [3]. You need to have sound switched on so that you can hear the dog's heart felt comments.--Aspro (talk) 23:13, 24 January 2012 (UTC)
Your dog can talk? Bah, this dog can sing. He is called "Menchi" ("edible"), his owner keeps him around in case she runs out of provisions. --Enric Naval (talk) 00:01, 25 January 2012 (UTC)

Having an evolutionary pressure is not enough. You still need to wait for random mutations to appear. Dogs are very far from understanding the nuances of language, so you will to wait until a lot of cummulative mutations appear, one after the other. And you will have to inter-breed those exemplars that have the "correct" mutations.

And then inter-breed their descendants until you make the mutations estable in a big population of dogs. Enjoy all the recessive genes popping up because of the inter-breeding between near relatives. --Enric Naval (talk) 00:01, 25 January 2012 (UTC)

Interspecies communication is certainly possible on many fronts. Some humans at least claim to be able to communicate to animals, though the other way around is not as likely. ~AH1 (discuss!) 00:18, 25 January 2012 (UTC)
My pets have certainly been able to communicate information to me. Usually, that information is "I'm hungry"... --Tango (talk) 12:59, 25 January 2012 (UTC)
There was a Far Side a decade or two ago in which some scientist had supposedly invented a machine to translate the barking of dogs into English. It turned out that they were yelling "Hey!" "Hey!" "Hey!" to each other. Jump ahead to the near-recent, and there was a Foster's Beer ad with Aussies yelling "Oi!" "Oi!" "Oi!" to each other... "Australian for 'Networking'." ←Baseball Bugs What's up, Doc? carrots→ 13:45, 25 January 2012 (UTC)
Ever hear of the BowLingual? Apparently, it's not really that good though... Anyway - from what I've noticed, a large proportion of gull squawkings and wailings do appear to be variations of (or literally) "Mine! Mine! Mine!". --Kurt Shaped Box (talk) 14:08, 25 January 2012 (UTC)
I've seen numerous plowmen whose horses responded reliably to "Giddap! Whoa! Gee!(for "right") Haw!(for "left") Back" and other standard commands, while in front of the teamster, thus unable to see any body language. They still might just be sensing tone, I suppose. I could not find an article on this practice of controlling horses by voice, because I wondered how many centuries back they had been used, and what corresponding commands were used in other countries. Edison (talk) 14:52, 25 January 2012 (UTC)

Airplane pressure[edit]

I'm sure that this question actually indicates that I lack a certain understanding of "how flying and pressure work" but hey, that's why I'm asking!

So the cabin of an airplane is pressurized, yes? That way we don't experience the lower psi of the higher atmosphere and feel...whatever bad effects those would be.

What I'm wondering is: if the cabin is "set" at a certain pressure for the comfort of the humans on the inside, why do our ears pop during ascent and descent? (talk) 16:32, 24 January 2012 (UTC)

Because the cabin pressure isn't kept constant - it is decreased during ascent and increased again during descent. Our article on cabin pressurization explains as follows:

The pressure inside the cabin is technically referred to as the equivalent effective cabin altitude or more commonly as the cabin altitude. The cabin altitude is the equivalent altitude having the same atmospheric pressure, so that if the cabin altitude were set to zero then the pressure inside would be the pressure found at sea level. In practice, it is almost never kept at zero, in order to keep within the design limits of the fuselage and to manage landing at airfields higher than sea level. The cabin altitude of an aircraft planning to cruise at 40,000 ft (12,000 m) is programmed to rise gradually from the altitude of the airport of origin to around a maximum of 8,000 ft (2,400 m) and to then reduce gently during descent until it matches the ambient air pressure of the destination.

Gandalf61 (talk) 16:39, 24 January 2012 (UTC)
OP here, so if I understand correctly---If I were theoretically taking off and landing in the same place, we could keep the pressure at a constant and do away with that pesky popping? The reason they raise and lower the pressure is to compensate for the difference in elevation of origin/destination? (talk) 16:43, 24 January 2012 (UTC)
You could in theory, but most planes are not designed for that. The fuselage is only designed to maintain a certain amount of pressure difference between the outside and the inside of the plane. If you tried to maintain sea level pressures at 30000 feet, then there is good chance that your typical commercial aircraft would explode. Even if you wanted to try, the air compressors probably wouldn't be capable of maintaining sea level pressures at high altitude. By lowering the cabin pressure at altitude, manufacturers don't have to build as much pressure resistance into their aircraft, which allows them to be cheaper and lighter, and hence more economical. Dragons flight (talk) 16:55, 24 January 2012 (UTC)
Of course it would be even cheaper and lighter to have no pressurization in planes at all. However, that would make flying at high altitude potentially fatal to the occupants, which is of course unacceptable. So the minimum cabin pressure is a sort of compromise between what is needed in order to be reasonably comfortable for air travelers and what is not too costly and heavy for airline manufacturers. Dragons flight (talk) 17:05, 24 January 2012 (UTC)
Lowering the pressure inside the airplane should also make it lighter, although I don't know if this is a significant factor in fuel economy. StuRat (talk) 17:30, 24 January 2012 (UTC)
Not pressurizing the aircraft might save on some weight but also may result in an increase in some weight due to the need to carry more oxygen. Private pilots fly unpressurized at high altitudes all the time, but they do need an oxygen bottle when doing so.
A point of interest -- I recall reading that the Boeing 787 maintains a 6000-foot cabin pressure rather than 8000-foot. ~Amatulić (talk) 21:18, 24 January 2012 (UTC)
The cabin pressurization article says "the Airbus A380 features 5,000 ft (1,500 m) when cruising at 43,000 feet (13,000 m)", for what it's worth. Pfly (talk) 01:07, 25 January 2012 (UTC)
In case you are interested, I believe that the FAA says that if pilots are flying a non-pressurized aircraft 12,500 feet for more than ½ an hour at a time, they have to use oxygen. If they are flying a non-pressurized aircraft above 14,000 feet, they must use oxygen at all times and oxygen must be available to the passengers (though interestingly the passengers don't seem to be obligated to use the oxygen). That is quite a bit higher than the 8,000 foot pressure altitude cited above, showing that there seems to be a large margin between what is deemed safe and what is put into practice. I have not had the training to fly a pressurized aircraft (yes, you need separate training, at least if you are going above 25,000ft), so I'm not too familiar with it. Falconusp t c 10:04, 25 January 2012 (UTC)
Falconus is correct. The US Federal Aviation Administration (and equivalent authorities in other countries) mandate that supplemental oxygen must be used by the pilots above specified altitudes in non-pressurised aircraft. See §91.211. Above 14,000 feet, supplemental oxygen is required, regardless of how short the time at that altitude. For a long flight, supplemental oxygen is required above 12,500 feet. (Many countries specify 10,000 feet.) These are minimum requirements for pilots and they all have to be fit and in good health. The passengers in airline aircraft are not all in such good health. There are some medical conditions that would not tolerate extended exposure to 12,500 feet cabin altitude. Manufacturers of pressurised aircraft generally provide a cabin pressure altitude around 8,000 feet so that represents a better situation for all on board - pilots and passengers. The FAA doesn't mandate what amount of cabin pressurisation must be provided in pressurised aircraft. The decision as to whether to pressurise, and how much pressurisation to provide, is left entirely to the airplane designer. Dolphin (t) 12:18, 25 January 2012 (UTC)
Another option might have been for them to let the pressure drop a bit further and supplement the oxygen percentage, to maintain the same partial pressure of oxygen. Of course, oxygen tanks are dangerous, as they can easily start fires, but perhaps oxygen concentrators could be used to pull in atmospheric oxygen. They might also need to vent the excess oxygen as they descend to avoid getting too high of a percentage at normal pressure, unless people would use it up quickly enough to make this not matter. I wonder if anyone has attempted this. StuRat (talk) 21:20, 25 January 2012 (UTC)
I agree that the partial pressure of oxygen is what it is all about. Human beings are affected by the partial pressure of oxygen rather than the total pressure of their surroundings. Your suggestion is implemented in many combat aircaft which are designed with a partially pressurised cockpit and the crew must also use face masks and supplemental oxygen. I am not aware of any commercial aircraft that uses this approach.
The problem with oxygen bottles in commercial aircraft is not because they are dangerous but because they are heavy. Cabin pressurisation is achieved by bleeding air from the early compression stages of the engines, conditioning it to the desired temperature, and feeding it into the cabin. Bleeding air for this purpose reduces the efficiency of the engines a little. The most critical time of a flight is the takeoff and during this time cabin pressurisation is not required so there is no cost in terms of loss of engine efficiency but if heavy oxygen bottles are carried they must be carried throughout the flight, including takeoff. Dolphin (t) 22:38, 25 January 2012 (UTC)
At some point the absolute pressure becomes a problem, too. Those climbing Everest may have blood burst out of their alveoli and foam up in their lungs, for example, even while on supplemental oxygen. I believe astronauts also breath oxygen-enriched, pressure-reduced air directly from the cabin, not from a mask, as I envision. StuRat (talk) 22:54, 25 January 2012 (UTC)
Not all astronauts. The ISS is at full atmospheric pressure (see International Space Station#Life support), for example. --Tango (talk) 23:41, 26 January 2012 (UTC)
Or, see Life support system#Atmosphere (WP:WHAAOE). --Tango (talk) 23:50, 26 January 2012 (UTC)

Diminishing star brightness curve[edit]

I came across this on [4], it is a totally different structure to all of the other light curves, what is going on? Is this star dying?

It's just 30 days. Possibly it's a picture of a star being slowly occluded by an object. ~Amatulić (talk) 21:11, 24 January 2012 (UTC)
It could just be a long-period variable star. -- (talk) 22:16, 24 January 2012 (UTC)

Cold gas thruster and heat[edit]

Lets say you have an object in space like a small meteor and you rig up this thruster that uses cold gas to propel it through space. Is there any way to detect it using some kind of heat sensor? ScienceApe (talk) 20:44, 24 January 2012 (UTC)

Such thrusters should still be subject to adiabatic cooling as the escaping gas expands against (near) zero pressure; thus the gas jet should be colder than the gas in the tank. --Jayron32 20:54, 24 January 2012 (UTC)
However, if attempting to make a "stealth" propulsion system, you could calculate how much it will cool and heat it just enough, prior to release, so the temp will cool to match the ambient temperature in that area. StuRat (talk) 23:04, 24 January 2012 (UTC)
In addition, the gas in the vacuum of space could be subject to ionization if near enough to the sun. This is why a comet tail glows. Ionized gas would likely be visible to a heat sensor too, if the ionized gas emits photons in the infrared bands. ~Amatulić (talk) 21:08, 24 January 2012 (UTC)

Thought experiment: the walking stick problem revisited (Pt. 2 - rotation)[edit]

Hi. Recently, someone asked a question about a rod that was light-years long, tugged at one end. We approximately concluded that the limit for the information travelling from one end to the other, would be at the speed of light for a "perfect" rigid material. Now, let's assume we have the same rod, but its inertia is not so great, meaning that any person or other "tugger" could move it, at least locally. What would become of the rod, assuming it could exist (no semantic or ontological refutations, please), if the end upon which we had control were swung, in an angular motion, 180 degrees from one side of the Earth to the other? Since the speed of information transfer is limited to the speed of light (possibly slower), there is no way for the observer, or swinger, to see any part of the rod bend, as any visible bend apparent to an omniscient observer would take at least twice the time for a light photon or wave to travel to the observer that is holding the rod. Thus the rod would appear to be perfectly rigid according to the person holding it, even when the rod is pointing the opposite way it was originally. Now, assuming there is enough time for the person swinging the rod to do so before the rest of the rod caught up, what would happen to the shape of the actual rod, as per an omniscient instantaneous observer? After discussing the problem with others, including teachers/professors, I and other humans have arrived at the following conclusions:

  • The stick would warp spacetime around itself, so that it can remain locally straight and not violate either relativity or its own notion of rigidness.
  • The stick would instantaneously "existify" through the core of the Earth, so the person holding it would now be holding a straight stick pointing down rather than up.
  • The person holding the stick would spontaneously cease to exist as the situation is impossible, except for frames of reference that assume quantum immortality.
  • A "shock wave" consisting of spacetime (or whatever medium light travels through) rather than air, as in the case of sound, would wrap around the moving parts of the stick.
  • For a rigidity limit of c, the stick would assume the shape of a treble clef.
  • As the information is travelling at light speed, energy is converted to mass and the stick becomes a longitudinal black hole or wormhole.
  • The stick would have so much mass, and therefore inertia, that the Earth would stop spinning.
  • The stick would not move, "knowing" it is to be spun rather than tugged.
  • The stick would now be under the influence of all sources of gravity in its vicinity, including the Earth, being warped at a limit of c·t.
  • A mathematical function for the stick would reach or approach infinity.
  • Spacetime would fold, rather than bend.
  • The Moon would collide with the stick, eliminating any notion of the continued existence of either entity.
  • God (arriving via an electromagnetic disturbance).

Obviously, the problem seems to have no solution, though I'm sure some dedicated and iconoclastic wikians will enlighten me.

Although space is empty, would the stick encounter any resistence, and if so, what would be the fluid (because usually it is a fluid) that causes this? There is yet another problem: if straight-line information travels at light speed, then when the other end of the stick finally does catch up, its speed will have to be π times slower (if my visualization is correct) than it otherwise would be, in order to travel below the universal speed limit. If this is true, however, then the stick would be bent in every frame of reference, unless spacetime is somehow curved over large distances. If it does travel along with the body of the stick and maintain rigidity, then spacetime will break the Theory of Relativity. Thus, near-infinite rigidity (as defined by a limit of c) and infinite mobility (to overcome inertia) seem to contradict each other, or the stick itself might break, despite its rigidity, due to tidal forces incurred by the speed limit itself. Perhaps there would be no issue because the stick would morph into space itself, which can travel faster than light. Unless, of course, the relativity of simultaneity somehow also figures into this, thus annulling the problem...

Let's continue this thought experiment even further. Say that the stick is brought not only 180 degrees, but a full 360° back to its original position (and yes, on the same plane). Relative to an omniscent observer, what will the stick look like then (would the Earth be destroyed by the stick's catch-up motion)? If all information being passed in the stick were significantly slower than the speed of light, then the original stick in its original position, or at least the far visible end of it, would still be visible to the swinger in its original orientation! Or maybe, there would be a second observer who has been observing the stick the whole time: would this person see something different from the swinger? If this is true, however, then maybe the visible end of the former position (or actually, the correct position when the light information was sent) would be significantly redshifted, disappearing into an edge of invisibility of some sort. Now, the stick is still being held, and because information is travelling at relativistic speeds, the rod or stick would still appear completely rigid to the Earth-bound observer (there would be, at maximum, some slight visible bending apparent). Thus, the stick might actually seem to meet its old position, being in two different places at once. After this, what would happen to the shape of the stick? Now for a different experiement, assume there is a being holding the stick at either end, on two different planets, bringing the stick 180, then 360 degrees. What would be the final result? Thanks. ~AH1 (discuss!) 23:46, 24 January 2012 (UTC)

That's a lot of words, but what you're really asking is "why can the phase velocity be faster than the speed of light?" The answer is, "because the group velocity is the part that carries information, energy, and so forth; and it is always slower than the speed of light." Nimur (talk) 23:54, 24 January 2012 (UTC)
Is the phase-to-light velocity ratio equal to some rational fraction of pi? ~AH1 (discuss!) 00:20, 25 January 2012 (UTC)
No. In your hypothetical example, you have set up a constraint: one end of the rod is fixed (at a distant point, far from Earth); and the other end is at a specified point on Earth's surface. This is a boundary value problem. There are numerical solutions to the boundary value problem for a semi-rigid pole or rod. In their simplest forms, these solutions depend on the assumption that the rod may deform elastically, and this is usually predicated by stating that the deformation is small (so as to avoid snapping the rod). You never specified what the rod is made of - so we can't really judge whether such assumptions are valid. Nonetheless, one of two possible scenarios will play out, because you are applying a forcing function to one of the boundary conditions. So, either the rod will deform elastically, and continue to satisfy the constraints; or the rod will snap.
Can you follow the mathematics in the derivation of the 1-D wave equation? Do you see those ≈ symbols? Those mean, "this equation is approximate." By introducing ridiculous material properties, or ridiculous scale lengths, those ≈ will turn into ≠ signs. This means that your rod will snap, or deform. We don't really even need to use advanced modern physics or relativity. Nimur (talk) 00:27, 25 January 2012 (UTC)
This is not all that complicated. Many of these "conclusions" are either wrong or overcomplicate things by involving factors (e.g. black holes) which can be safely ignored here.
Just like your perfectly rigid stick can't maintain constant length because of the limited speed of sound, it also can't remain perfectly straight, for the same reason. So, if you spin your end of the stick so quickly that you complete the move before the signal gets to the other end, it will bend (not in the form of a treble clef, but in the form of a question mark, with you at the upper end, or an Egyptian crook-staff.) After you complete the motion but before the end of the stick can travel to the new location, you will see that the stick is bent in an arc around you.--Itinerant1 (talk) 00:34, 25 January 2012 (UTC)
You go wrong right at the start - why can't the observer see it bend? It's just a rod; I've seen rods bend loads of times. That fact that it is really long isn't going to change anything. Yes, you can't see it bend faster than it takes for light to get from you to the bit that's bending and back again, but that doesn't mean you can't see it. It just means it takes some time before you see it. --Tango (talk) 00:50, 25 January 2012 (UTC)
(ec) The main mistake you're making is thinking that the stick has a "notion of rigidness" and must appear to be perfectly rigid even if it really isn't. In fact it is flexible and appears flexible. It's not fundamentally different from an object with a very slow "update speed", like a slinky, except in the time scales involved. -- BenRG (talk) 00:52, 25 January 2012 (UTC)
<offtopic>Interesting slinky experiment - what happens to a tennis ball tied to the bottom of a slinky after the top of the slinky is let go?</offtopic> Von Restorff (talk) 04:24, 25 January 2012 (UTC)
I think it's on topic. There are several of these slinky videos on YouTube—here's a nice one taken with a high-speed camera. It would be cool to see a variation of this trick with the slinky being whirled around in a circle (horizontally) before release. The outside end of the slinky should continue to follow a circular arc until the inside end catches up with it. -- BenRG (talk) 19:51, 26 January 2012 (UTC)
The rod simply bends. You just can't physically make an infinitely stiff material; relativity makes that impossible. Speculating about the relativistic behaviour of an enormously long perfectly-rigid rod is meaningless; there can be no such rod. --Srleffler (talk) 06:24, 28 January 2012 (UTC)