Jump to content

Wikipedia:Reference desk/Archives/Science/2011 December 17

From Wikipedia, the free encyclopedia
Science desk
< December 16 << Nov | December | Jan >> December 18 >
Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


December 17

[edit]

Transatlantic travel

[edit]

By what means (what type of ship) - and for how long - would a Briton (specifically, a Scot or an Englisman) of high social standing have travelled from Great Britain to America in a.) 1767 and b.) 1909-1913? My query pertains to A.) John Witherspoon, signer of the American Declaration of Independence and B.) P.G. Wodehouse, author. How long did the trip take, and what was it like? 82.31.133.165 (talk) 00:41, 17 December 2011 (UTC)[reply]

For the latter, one would travel on a liner such as RMS Lusitania; that article details the accommodations on board (which varied in quality greatly by the class of ticket bought). In ideal conditions (such as an attempt to break her own Blue Riband record) Lusitania could cross the Atlantic in less than 5 days. -- Finlay McWalterTalk 01:00, 17 December 2011 (UTC)[reply]
During the age of sail, common Atlantic-going trading vessels were ships like Galleons and Fluyts and Carracks. I'm not sure there were ever dedicated passenger ships; they wouldn't have been very economical; instead people merely booked on any merchant ship that was headed in the right direction. If there were a large group traveling together, they would essentially book the whole ship (see Mayflower for an earlier example of such a usage). The Mayflower took 2 months to cross the Atlantic; though it was beset by some bad traveling conditions; under ideal conditions about 4 weeks seems to be more standard, especially once Transatlantic travel became more common. --Jayron32 06:16, 17 December 2011 (UTC)[reply]

Deploying lifeboats

[edit]

I think this is the right place to ask this question, kinda fits in with engineering. If not that's fine, I'll move it to miscellaneous or something. But at any rate, on a cruise ship, where do ship employees (if that's the right word) go to actually deploy lifeboats? I emailed a Queen Mary 2 website asking that question but no one had replied. Would it be deployed at the site of the lifeboats (possibly with a control panel near by or something) or would it be near the engine room? Any help would be amazing. 64.229.180.189 (talk) 01:08, 17 December 2011 (UTC)[reply]

I can't imagine it being controlled anywhere but where the lifeboat is. You want to be able to see what's going on, when it's level with the deck so people can get aboard, when to stop it if there's some problem, etc. This youtube video (part 1 of 3) seems to go through it step by step (for a US Navy oiler, not for a cruise ship, but the procedure should be about the same I think). Clarityfiend (talk) 01:46, 17 December 2011 (UTC)[reply]
From someone who has clocked in a good few nautical miles, and is the offspring of a seaman: I've never seen a lifeboat being controlled remotely from any location from where you couldn't see it. Be it small or large lifeboats, they are deployed directly from their physical location, where the control panel, winch, launchpad or whatever, sits. --Ouro (blah blah) 08:39, 17 December 2011 (UTC)[reply]
Another good reason for close proximity is it greatly lessens the risk of wires being cut in whatever calamity the ship is facing. Clarityfiend (talk) 01:39, 18 December 2011 (UTC)[reply]

How far can you pull quarks apart

[edit]

According to Color confinement, you can't pull them apart, but they will stretch and finally form quark antiquark pairs. How far can they be pulled till this happens? ScienceApe (talk) 04:53, 17 December 2011 (UTC)[reply]

I'm not sure you can measure such "distances" functionally, nor can you "pull" quarks apart, that would imply you had some means to grab them individually. See Jet (particle physics), which is the only way that the existance of quarks can be inferred, no quark has ever been directly observed. --Jayron32 05:46, 17 December 2011 (UTC)[reply]
No isolated quark have ever been observed, but deep inelastic scattering experiments on protons show, for example, that protons must be composite particles made from three subunits. This is similar to the way that the existence of an atomic nucleus was originally inferred. There is a lot of evidence for quarks beyond simply the existence of jets. Some of this evidence is described in quark. Dragons flight (talk) 14:52, 17 December 2011 (UTC)[reply]
If you could pull them apart, theory suggests you could go a distance on the order of a few femtometers (10-15 m). Dragons flight (talk) 14:52, 17 December 2011 (UTC)[reply]

Why is traffic on long open roads "lumpy"?

[edit]

Why do cars on long road sections occur in groups with relatively long gaps between them, instead of being more evenly spread? Roger (talk) 09:42, 17 December 2011 (UTC)[reply]

There a couple of reasons for this. First, cars will get on and off of the road (not sure if this counts on long open roads, but it does on highways), thus you get areas where new traffic "spawns"; since these new cars are added at discrete intervals, this leads to some lumping. On the same vein, faster cars will get stuck behind slower cars; new slower cars added, even if going faster than some slower car ahead, may never reach it, thus, giving you clumps. Also, wrecks, or other such, will cause people to cluster, most people will stick to the cluster, thus, causing it to take a while to disperse. Finally, most people seem to have an interval of speed in which they feel safe, so when coming up on a slower driver/cluster, that is not going too much slower than them, they would probably not use the passing lane. There are other reasons, I'm sure; I would imagine a lot of them center around the fact that if some group of traffic is forced to slow down and another not, then the faster group is going to become closer to the slowing down group; depending on the course and reasons, this should lead to various bunching ups of groups. Phoenixia1177 (talk) 11:22, 17 December 2011 (UTC)[reply]
A good summary there, and you might be interested in following the links at traffic wave. Yes, there's even an article on this one. And I did an assignment on it once, too, but I wouldn't call it revolutionary - it is definitely part of applied mathematics. IBE (talk) 14:51, 17 December 2011 (UTC)[reply]
Traffic engineers call such a lump a "platoon" (Wikipedia has a platoon (automobile) article, but that only discusses proposals to deliberately engineer the formation of platoons, and as ramp meter shows, they also use the term for platoons that aren't wanted). In addition to the scenarios Phoenixia1177 mentions, specific road designs can induce platoons (pedestrian crossings, level crossings, traffic lights) and some can break them up (ramp meters, roundabouts, 4-ways stops). Traffic engineers are clearly in two minds about the desirability of platoons. They're desirable in some cases - if you have a complex signalled junction like a SPUI the traffic lights form platoons and the junction is designed to efficiently push the platoons around the junction - platoons are an efficient use of very limited road space, and they're deterministic (so the traffic modelling software like TRANSYT can predict how a junction will behave). And on highways platoons can be desirable - consider overtaking a platoon of slow lorries in one go, rather than moving back and forth to overtake a succession of widely spaced lorries. But platoons can be undesirable too - if you're waiting at a junction for traffic to pass, you get stuck when a platoon comes along (whereas you might find a space if the traffic was more even), and joining a motorway (on short urban ramp) and encountering a platoon of trucks can be a scary thing. Traffic engineers use queueing theory to try and model dense traffic flows, and (as with other applications of queueing) there's often a balance to be reached between efficiency and fairness. -- Finlay McWalterTalk 15:06, 17 December 2011 (UTC)[reply]
Sometimes the traffic may just be randomly distributed, with only the illusion of grouping. Mitch Ames (talk) 03:48, 18 December 2011 (UTC)[reply]
I also think there's a psychological reason. That is, when you've been passed, you feel like you're going too slow, so speed up. When you pass others, you feel like you're going too fast, so slow down. When everyone does this, the result in clumping. StuRat (talk) 05:34, 18 December 2011 (UTC)[reply]
I think that there is strong statistical evidence that traffic isn't randomly distributed. In fact, given the known limitations of a road network, a truly random distribution would be impossible, as long as vehicles were sufficiently close together to interact. As to how much is 'psychological' (per StuRat), and how much is due to 'physical limitations' (maintaining braking distance, needing excess speed to overtake etc), I'd not hazard a guess, though I suspect that the former may account for some of the more serious multi-vehicle collisions that have occurred on UK motorways: there seems to be some evidence that drivers encountering fog will speed up to keep the car in front visible...
(And yes, I know, 'citation needed' - I'm sure I've seen this somewhere...) AndyTheGrump (talk) 05:50, 18 December 2011 (UTC)[reply]
...And here's an abstract at least that supports this: "ISOLATION DUE TO THE LOSS OF THE VISUAL STIMULII REQUIRED IN ORDER TO REMAIN ALERT. SUCH A CONDITION OFTEN LEADS TO A SUBCONSCIOUS INCREASE IN SPEED, MOTIVATED BY AN INSTRUCTIVE NEED FOR VISUAL STIMULII (SIGNS, JUNCTIONS, OTHER VEHICLES ETC.) [1]. I've not got access to the original document, but it seems plausible. AndyTheGrump (talk) 06:07, 18 December 2011 (UTC)[reply]
A different but related effect is Bus bunching. Vespine (talk) 00:27, 19 December 2011 (UTC)[reply]

Implications of the Higgs boson?

[edit]

Are there any practical applications or implications of the discovery of the Higgs boson or its mass? If so, what are they? 67.6.163.68 (talk) 10:16, 17 December 2011 (UTC)[reply]

There are none, yet. --Mr.98 (talk) 13:10, 17 December 2011 (UTC)[reply]
And there will probably be none for a long time to come. Dauto (talk) 16:06, 17 December 2011 (UTC)[reply]
There may not be any direct applications any time soon. The point of looking for it is to figure out how particles in general work and, by extension, how everything works. That could be quite useful. It is difficult to say what the answer will be useful for because we don't yet know what the answer is, but particle physics has had lots of useful results so far and I'm sure it will continue to do so. Take Positron emission tomography, for example. Doctors find it extremely useful and they wouldn't be able to use it if particle physicists hadn't discovered the positron. --Tango (talk) 00:11, 18 December 2011 (UTC)[reply]
The Higgs is thought to carry all mass. See also Higgsless model. ~AH1 (discuss!) 03:21, 18 December 2011 (UTC)[reply]

How does the GRW theory in quantum mechanics handle multiple hittings?

[edit]

In the Ghirardi–Rimini–Weber theory of quantum mechanics, spontaneous localisation of quantum particles occurs whenever there is a "hitting", which is a random event happening once in a hundred million years or so for a single particle. It is proposed that this explains how stable macroscopic states emerge from quantum measurements, because any large body (including a measuring apparatus) will have a huge number of particles, and hence many hittings. What happens if, in a rigid body, two hittings occur simultaneously, each one tending in a different direction? I suggest that this might happen for a measuring apparatus of relatively few particles, which would get into an entangled state when measuring a quantum particle. It would enter a superposition for a split second, then have two spatially separated particles undergoing hittings. The spontaneous localisation thus induced could easily be different for the two particles. Which one wins? IBE (talk) 13:56, 17 December 2011 (UTC)[reply]

Thankfully the Relativity of simultaneity keeps different events from sharing the same absolute time reference.
BTW: There is no Measurement problem. The so-called Wave function collapse is simply a process of averaging quantum "weirdness" over enough particles, that the effect becomes minimal. Classical mechanics is a delusion. Hcobb (talk) 14:39, 17 December 2011 (UTC)[reply]
Re: the measurement problem, I think you'll find it has occupied a good deal of physicists' time over the last 80 years or so, hence I think it is still debatable at least. As far as relativity is concerned, a good thing to note, but in this case they are not moving with respect to each other, so they can agree 1. on simultaneity, and 2. that neither had time to confer when the hitting happened. It does present on the surface as a problem, but it would have been thrashed out quite a bit, so I'm wondering what they've come up with. IBE (talk) 14:46, 17 December 2011 (UTC)[reply]
Wavefunction collapse can propagate faster than the speed of light Dauto (talk) 16:03, 17 December 2011 (UTC)[reply]
The Invisible Pink Unicorn also travels faster than light. BTW, what's with this mythical "rigid body"? If it isn't held together by electromagnetic forces then what's keeping it rigid? Hcobb (talk) 16:07, 17 December 2011 (UTC)[reply]
I confess I'm not an expert on unicorns, but the rigid body is something that is held together so that if one part of it goes one way, so must the rest of it, or there will be a strong force somewhere. It is an idealisation, not a precise description. Dauto - thanks for the explanation, but does that mean it must propagate instantaneously? Otherwise, there would seem to be a chance that at least some pairs of hittings would happen at almost exactly the same time, and thus give rise to the problem I mentioned. IBE (talk) 17:39, 17 December 2011 (UTC)[reply]

But these so called "collapses" WILL be observed to occur in different orderings by different observers. Action_at_a_distance_(physics)#Quantum_mechanics, etc. All this FTL garbage means is that quantum physics does not every violate the Conservation laws. If you measure a certain angular momentum then I will measure exactly the opposite change in angular momentum on my member of our shared particle pair, in order to keep from violating the law of Conservation of angular momentum. It doesn't matter if I measure my value first or you measure your value first. In fact if neither of our measurements are in the past Light cone of the other one, then different observers could see mine first, yours first, or both at the same instant. Hcobb (talk) 22:09, 17 December 2011 (UTC)[reply]

My interpretation of the wavefunction collapse is that multiple realities are relatively-simultaneously superimposed, and the observer causes the wavefunction to "collapse", thereby making one of the possible realities "real". Is this correct? ~AH1 (discuss!) 03:19, 18 December 2011 (UTC)[reply]

Summing up some threads here, I think, Hcobb, you are exactly right about the reference frame, as per your first sentence. From my understanding of the rest of your post, it develops this point further. So, my understanding of GRW is that it is at odds with special relativity, for the reason cited, and the solution is that it needs a preferred frame, which no one likes, but they are also trying to solve the quantum measurement problem. As far as AstroHurricane's comment goes, I'm not sure here, but it sounds like you are advancing a particular interpretation of QM, which would not be agreed upon by all physicists. You are saying that observation causes collapse, but the question is how and why. I think many physicists would agree roughly with your statement, without saying it solves the QMP, and some would disagree, such as Many Worlds theorists. Perhaps Dauto can help more? I'm still curious about my further question, whether collapse must propagate instantaneously. IBE (talk) 05:15, 18 December 2011 (UTC)[reply]

Angle of the shadow across the moon

[edit]

Hello. I've looked across our articles on the moon, and done some internet searching, but can't find an answer for this one. The line between the light and dark sides of the moon during it's various phases seems very horizontal recently, compared to the more vertical division I see in most diagrams on here. Is this because at the moment we are near mid-winter/summer and the earth's tilt is at it's maximum (layman's wording) so the view of the moon is different? Can you therefore tell the season from the angle of the moon's light/dark divide? Or am I being foolish? Ta, S.G.(GH) ping! 15:01, 17 December 2011 (UTC)[reply]

Many factors are at play here. The angle of the moon's orbit relative to Earth's horizon is the obliquity of Moon's orbit, determined by both Earth's axial tilt, and Moon's orbital inclination. The illumination is determined by Moon's obliquity relative ro the ecliptic, in other words by compositing the moon's orbit around Earth with Earth's orbit around Sun. We have articles on axial tilt, orbital inclination, and other orbit parameters.
These parameters are discussed in great detail in our article, orbit of the Moon. There is some relation to Earth's season, insofar as the angle of illumination relative to the horizon is affected by our axial tilt, and latitude, of course. But, there are many other lunar orbit periodicities that do not correspond to any Earth calendar phenomenon - because no Earth culture historically designed such a calendar. This is why lunar eclipses occur at "irregular" intervals during Earth years/seasons. Despite all the tidal locking, many of our Moon's orbit parameters have no integer relationship to Earth month, year, nor to lunar revolution period. So, you could not tell the angle of illumination simply by knowing the Earth season - but you could use a lunar almanac, chart, or astronomy software to calculate this. Don't misunderstand - the angle is not "random," it's just dependent on periodicities that few outside of lunar-watchers bother to track. Nimur (talk) 15:16, 17 December 2011 (UTC)[reply]
Wouldn't it change throughout the night, ie. throughout the moon's course across the sky?. Or am I totally muddled here? IBE (talk) 15:21, 17 December 2011 (UTC)[reply]
The most important factor out of all the ones mentioned above is of course the latitude of the location of the observer which has nothing to do with the moon's orbital parameters. Dauto (talk) 16:47, 17 December 2011 (UTC)[reply]
Does that mean I am wrong about it changing throughout the night? I would have thought it could be horizontal at the horizon, and near vertical when at its zenith. Of course, if it is directly overhead, then there can be no such thing as horizontal - just turn around and it will go through every orientation (unless my brain is really out to lunch). IBE (talk) 17:46, 17 December 2011 (UTC)[reply]
You're probably confusing yourself because the angle of the moon's solar terminator with respect to the Earth's horizon is poorly defined. The horizon is a plane; the terminator is, technically, also defined by a plane; technically, the angle between two planes can be defined as the dot product of each normal vector. But that's probably not the geometry you're thinking about: you're probably imagining the projection of the terminator on to the visible disc of the moon as observed from your location; and at moonrise, or moonset, trying to figure the angle that makes to the apparent horizon (again, a projection of the horizon plane onto the small region of the sky that you're looking at). You might start at Horizontal coordinate system to see some of the geometric transforms you need to work in a more general coordinate system. Compound this with the fact that the moon's solar terminator is changing, slowly, and you actually have a quite complicated time-varying function to define this "illumination angle."
Assuming spherical earth and moon, the solar terminator is defined strictly by the vector between Sun and Moon. Your local horizon plane is determined strictly by your latitude and longitude. The angle between these two vectors varies based on the moon's current position in space - determined by its orbital parameters and the current time. Moon's position in space varies slowly enough that it is reasonable to consider its position "fixed" over a short period of observation. The other parameter is the Earth's rotation relative to the sun - this rotates at 15 degrees per hour - so it's very unsuitable to assume it is "fixed" for even the most rudimentary astronomical observation). Nimur (talk) 18:57, 17 December 2011 (UTC)[reply]

Galactic collison

[edit]

What will happen when Andromeda collides with the Milky Way? --108.225.115.211 (talk) 16:19, 17 December 2011 (UTC)[reply]

See galactic collision. Bo Jacoby (talk) 16:35, 17 December 2011 (UTC).[reply]
There's an article on that: Andromeda–Milky_Way_collision. 79.148.65.247 (talk) 16:34, 17 December 2011 (UTC)[reply]

DIY antenna, starting from a USB adapter

[edit]

Starting with a USB Wi-Fi adapter (which has an SMA connector), how can you make an antenna for it? BTW, commercial antennas would cost something like $30, so it must be something cheaper. Which is a good source of DIY projects for such things? Would connect a long cable on the adapter work as an antenna? 79.148.65.247 (talk) 16:33, 17 December 2011 (UTC)[reply]

There are people making all kinds of wi-fi antennas. Google for something like make wifi antenna for lots of instructions. Although I don't know if you'll be saving a lot of money; people tend to make antennas to get directionality or other efficiency, rather than to save money. $30 might not be the best deal you can make; shop around, search the web. 88.112.59.31 (talk) 17:07, 17 December 2011 (UTC)[reply]
I've found this site to be helpful when I've been fiddling with Wi-Fi http://www.usbwifi.orconhosting.net.nz/ --TrogWoolley (talk) 18:17, 17 December 2011 (UTC)[reply]

Physiology of getting warm

[edit]

So: a clear winter's day, spend time outside, get cold, come in to a normal heated building. Have a nice cup of tea, and feel the warmth immediately, as far away from the esophagus as the legs. (The same effect can be noticed with alcohol; I am not sure if that has a separate physiological explanation.) What's going on? Is it a kind of learned reaction? The body saying "I know I was cold, and am cold, but from past experience I know I am going to be warm soon?" Also, what is the muscular reaction to warmth called -- the small involuntary contractions more discrete than continuous shivering? "Shudder" redirects there, though I think of them as distinct. BrainyBabe (talk) 17:11, 17 December 2011 (UTC)[reply]

Vasodilation, the expansion of blood vessels to change circulation, has both a voluntary and involuntary component. Nimur (talk) 19:05, 17 December 2011 (UTC)[reply]
For the opposite, see vasoconstriction. ~AH1 (discuss!) 03:14, 18 December 2011 (UTC)[reply]
Yes, I do think your body anticipates when it's warming or cooling at a rate that will make it too hot or too cold soon, and reacts accordingly. StuRat (talk) 05:27, 18 December 2011 (UTC)[reply]

Can an EMP shut down a diesel engine?

[edit]

I understand that an EMP can shut down a gasoline powered engine, but I'm guessing it's because it uses spark plugs to ignite the fuel air mixture. A diesel engine does not have this, so will a diesel engine be unaffected by an EMP? ScienceApe (talk) 18:08, 17 December 2011 (UTC)[reply]

The electromagnetic pulse generated by, say, a nuclear weapon, can supposedly shut down a modern gasoline engine because it fries the microelectronics in the engine control unit; with the ECU dead there's nothing to fire the spark plugs, so the car stops. I'd imagine an old car with a distributor (and no delicate ECU) should survive okay. Diesel cars don't need the ECU for spark timing, but modern diesels still have ECUs which are fitted to all kinds of sensor and actuators. I don't know how modern diesels are designed to run when their ECU is totally dead; they may have a "run rough" failsafe, but I wouldn't be surprised if "don't run at all" is the fail safe mode they've chosen. -- Finlay McWalterTalk 20:03, 17 December 2011 (UTC)[reply]
There is a government report here on the potential effects of EMP. The section on automobiles and trucks is on page 115 (PDF page 131) with some prior lead-in discussion. I found another article here that includes some commentary based on the first reference. In summary, both automobiles and trucks would be affected to some degree due to increased use of electronic controllers, but the report seems to indicate that the effect would be much less than some would expect. Traditional coil and distributor (non-electronic) ignition systems would be less vulnerable. I suspect my '47 Chevy pickup truck would survive, but the headlights might take a hit. -- 24.254.222.77 (talk) 20:48, 17 December 2011 (UTC)[reply]
Thanks, that's a really excellent source. So, not as bad as everyone thought; come friendly bombs and fall on Slough, then. -- Finlay McWalterTalk 02:11, 18 December 2011 (UTC)[reply]
If anyone doesn't get the joke - see Slough (poem). Alansplodge (talk) 19:14, 21 December 2011 (UTC)[reply]

Angular momentum of the observed universe

[edit]

Has anybody tried to actually calculate the Angular momentum of the Observable universe? If so, was there any net bias in the spin axis of distant galaxies or perhaps the polarization of the Cosmic microwave background radiation? If there was, what might this say about the the total size of the universe to get this number down to zero? Hcobb (talk) 22:20, 17 December 2011 (UTC)[reply]

I'm curious as to how you would have enough valid data to even start with such a calculation? For example, looking at a star that's a thousand lights year away, you're seeing where it was a thousand years ago, as opposed to where it is "now". Likewise if you were near that star, looking at our sun. ←Baseball Bugs What's up, Doc? carrots23:07, 17 December 2011 (UTC)[reply]
This paper suggests that a universe with non-zero momentum is impossible (from a general cosmological viewpoint, this makes sense - the universe we believe to be isotropic, which would be impossible if there was any large scale rotation). That said, the fact are understanding of theory forbids it is no reason we shouldn't look. Galaxy Zoo, although a very simple crowdsourced experiment, looked at the spin directions of galaxies. They thought they found a small bias towards anti-clockwise galaxy rotations at first, but they now think that was just an interesting psychological error on the part of the volunteers and that there is actually no bias in galaxy spin. Smurrayinchester 00:56, 18 December 2011 (UTC)[reply]
The shape of the universe might determine the results. How would you calculate spin for an open non-spherical infinite universe of which only an infinitesimal portion is observable? ~AH1 (discuss!) 03:13, 18 December 2011 (UTC)[reply]
What if all we can see is an eddy? We'd see a net spin and that would give us some sense of how large the local flow would have to be, even if we couldn't see most of it. Hcobb (talk) 04:13, 18 December 2011 (UTC)[reply]
The OP specifically asked about the observable universe, so it's a perfectly well-defined question. I think the OP's idea was that we could draw conclusions about the size of the entire universe based on the theoretical requirement that the entire universe be isotropic (and therefore have zero angular momentum). If the observable universe had some spin by random chance then the size of that spin would allow us to estimate the size of the entire universe (the larger the universe, the less significant the bit we can see would be, so the larger its spin could be). I'm not sure if that idea would actually work, but it's moot anyway since the evidence we have is that the observable universe has no discernible angular momentum. --Tango (talk) 05:18, 18 December 2011 (UTC)[reply]
I'd disagree that the evidence that the universe has no discernible angular momentum. At this stage, it would seem perfectly reasonable to even assume the universe may have a minuscule net angular momentum. There are a lot of subtle processes that go on in the first microseconds of the Big Bang that have broken most symmetries that we might have otherwise assumed went unbroken. So for now I think it's a safe assumption that there's some net angular momentum left over. Is it measurable? As everyone has said so far, it's probably too subtle for current astronomy to confirm or deny. On the scale of fundamental particles, however, I have no idea, but I'm sure there's theories out there that break that symmetry. From what I understand, though, symmetries that we take for granted in the universe seem to be broken more often than not. SamuelRiv (talk) 07:45, 18 December 2011 (UTC)[reply]

From the point of view of classical mechanics the contents of the universe could rotate relatively to absolute space, but from the point of view of general relativity there is no absolute space. An inertial frame of reference is unrotating with respect to distant galaxies. So the univers has no angular momentum relative to an inertial frame of reference. Bo Jacoby (talk) 15:34, 18 December 2011 (UTC).[reply]

How much leaf-area does it take to support a human being?

[edit]

How much leaf-area does it take to support a human being? It takes a certain plant biomass -- some edible by humans, some edible by their prey, some not -- to give a human and his/her prey enough food to eat. On average, how much would the total area of the photosynthesizing organs -- the leaves -- of those plants have to be, at the leafiest time of year, to support one average human sustainably? Eldin raigmore (talk) 22:26, 17 December 2011 (UTC)[reply]

The problem with that is it would depend on if those leaves are facing the Sun or at an angle, and if they get direct sunlight or not. To get around this, a certain amount of land area is found in the calculations, rather than leaf area. Another problem is that this calculation is highly dependent on the diet of the human. A vegetarian diet takes many times less land area than an all meat diet. The location also matters, as a place on the equator with good weather can grow far more food. StuRat (talk) 05:18, 18 December 2011 (UTC)[reply]
Partial answer with rough data: A human requires very roughly the amount of energy intake that the sun shines on 1 m2 (about 150 W). Photosynthesis has an efficiency of roughly 1%. So that would mean 100 m2 of (human foodcrop) farmland, or rather more since, as you say, we don't eat all the parts of the plant, so maybe 200 m2. Plus some if we want to eat meat, as Stu pointed out. DirkvdM (talk) 09:37, 19 December 2011 (UTC)[reply]

Thank you, StuRat and DirkvdM, for your answers. StuRat got me to realize that related questions are "How much land does a person need?" and "How many people should the Earth support?". But I'd be interested in a science-fictional question; How much "leaf"-area would a photosynthesizing human being need to include in his/her body, if s/he were to survive without needing to eat for energy? (Obviously they'd need to eat for other things.) Eldin raigmore (talk) 16:11, 26 December 2011 (UTC)[reply]

Violating the Heisenberg Uncertainty Principle

[edit]

Would it be possible to determine simultaneously both the position and velocity of an electron, thus violating the Heisenberg uncertainty principle, by bouncing both a photon and a negative photon, with equal but opposite energies, off the electron at the same time, so the electron is located without disturbing it? Whoop whoop pull up Bitching Betty | Averted crashes 23:11, 17 December 2011 (UTC)[reply]

Photons are their own anti-particles; there aren't distinct anti-photons, I don't think. And in any case, wouldn't this require knowing, ahead of time, the position of the electron to a level of precision beyond that allowable by the UP? That is, doesn't this beg the question of how you knew where the electron was to begin with? --Mr.98 (talk) 23:30, 17 December 2011 (UTC)[reply]
An anti-photon would merely have the opposite CHARGE to a photon. A negative photon would have negative everything - negative energy, negative spin, negative wavelength... see negative mass.
And as for the knowing-ahead-of-time-where-the-electron-is, just keep firing photons and negative photons through the vacuum until an electron just happens to pass exactly through the midpoint between the photon and negative photon generators... Whoop whoop pull up Bitching Betty | Averted crashes 23:34, 17 December 2011 (UTC)[reply]

The Uncertainty principle is NOT based on the fact that it is impossible to detect an electron without disturbing it. On the contrary, the fact that it is impossible to detect an electron is based on the uncertainty principle (That's why it is called a principle - it comes first). The uncertainty principle is based on the fact that an electron (or any other particle) really doesn't simultaneously have an entirely defined position and momentum. Dauto (talk) 23:44, 17 December 2011 (UTC)[reply]

(Edit conflict) It seems a bit meaningless to ask about the properties of a "negative photon" in this case. One might as well ask whether we could determine an electron's position and momentum by measuring changes in the luminiferous aether. Anyway, I think UP would pose other, slightly better hidden problems that would mess up your result anyway. For instance, you'd need to know the precise locations that each photon was emitted from (impossible), and they'd need to be lined up perfectly to know the position of the electron perfectly. Uncertainty forbids us from knowing the position of the emitters precisely enough. We also don't know anyway of producing a stream of photons with exactly the same wavelength - even the best lasers will have a tiny bit of variation. Nor can you perfectly know the timing of the release. Plus, though I must admit I'm hazy about how this machine is meant to detect an electron, I'm pretty sure that, if it works as I'm imagining it (electron absorbs light, re-releases it) when the electron interacts with the photon there will be an essentially unknowable time lag between absorption and emission. All of this contributes to the uncertainty of the measurement.
In short, you can never know a precise position, time or momentum of the photons you release, so you can't know the precise position of the electron either. This part of what makes the uncertainty principle so powerful - as Dauto says, because every particle is subject to the principle, any piece of fantastic machinery you invent to circumvent the limit must itself be limited by the principle. You can't know precisely where the sensor is, so you can't know precisely where the sensee is.
(Oh, and if your plan is for an electron to absorb the positive and negative electrons at the same time, and their energies and momenta cancel perfectly, I'm not sure I see any reason for the electron to ever release them. The state of an electron + photon + negative photon = the state of an electron, and we never observe electrons in the lab spitting out "negative photons" (... or do we?).) Smurrayinchester 00:30, 18 December 2011 (UTC)[reply]
There are no negative photons, or at least, the photon is the negative photon. Also, a photon is chargeless, unlike as is implied above; moreover, antiparticles do not have opposite mass, they have equal mass. As for the actual question, no, it would fail, for exactly the reasons already given. Phoenixia1177 (talk) 04:55, 18 December 2011 (UTC)[reply]
I don't quite understand the question either, except that it seems to involve knowing both the energy of the photons and the time to infinite accuracy. That's impossible: UP also states that energy and time cannot be simultaneously known, just as it says that position and momentum can't be simultaneously known. --140.180.15.97 (talk) 19:27, 18 December 2011 (UTC)[reply]