Wikipedia:Reference desk/Archives/Science/2011 February 24

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February 24

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What is the effect of alcohol and drugs on non-human animals?

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If you were to give a dog or a grizzly bear vodka, or LSD, what would happen?--X sprainpraxisL (talk) 01:01, 24 February 2011 (UTC)[reply]

We have an article on the Effect of psychoactive drugs on animals, though it doesn't look like it has much on the specific animal/drug combinations you mention.
Or, here's an article from 2000 in Alcohol Research & Health: Animal Models in Alcohol Research, if that's more the sort of thing you're interested in. WikiDao 02:08, 24 February 2011 (UTC)[reply]
There was a great part in The gods must be crazy I remember from my childhood where some animals got drunk on the fermenting fruit of some African tree. You can find the segment on youtube. Vespine (talk) 07:32, 24 February 2011 (UTC)[reply]
I think that was in Animals Are Beautiful People (by the same writer-director). -- BenRG (talk) 09:23, 24 February 2011 (UTC)[reply]
From personal experience, if you give a dog a couple of beers, he pretty much gets sleepy. Not especially entertaining. Googlemeister (talk) 14:18, 24 February 2011 (UTC)\[reply]
This article [1] is about a serious case of drunk monkey business. 10draftsdeep (talk) 17:49, 24 February 2011 (UTC)[reply]
Bees can sometimes get intoxicated when they dine on naturally occuring fermented nectar. The guard bees at the hive act like bouncers, an intoxicated bee is turned away and refused entry into the hive. If that bee is persistent, then it will be treated as hostile, and will be quickly dispatched by the guards. Plasmic Physics (talk) 22:25, 24 February 2011 (UTC)[reply]
A cocketiel can get intoxicated from wine, they have a preference of red over white wine. Our cockatiel may lose it's balance and coorination, develop an attention deficit, and puffs up it's feathers, and just becomes generally entertaining. It gets annoyed when we remove access to wine if it hasn't already forgotten about it. Plasmic Physics (talk) 22:25, 24 February 2011 (UTC)[reply]
There are videos on YouTube of spiders and cats on LSD, caffeine, cocaine, alcohol, and other drugs. Be aware though, that dosage, set and setting vary the effect. Also see Rat Park. 75.138.198.62 (talk) 20:44, 26 February 2011 (UTC)[reply]
Mice exposed to coccaine will prefer the drug over food. Dimethyltryptamine is secreted natually in small amounts by humans and other mammals. ~AH1(TCU) 20:53, 26 February 2011 (UTC)[reply]

Question on collision between hydrogen atoms (momentum / energy question)

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Hi all,

I was doing my physics mid term exam today and I came across a question that I couldn't solve. Turns out it's from the textbook. Here it is:

When two hydrogen atoms of mass m combine to form a diatomic hydrogen molecule (H2), the potential energy of the system after they combine is -Δ, where Δ is a positive quantity called the binding energy of the molecule.

(a) Show that in a collision that involves only two hydrogen atoms, it is impossible to form an H2 molecule because momentum and energy cannot be simultaneously be conserved. (Hint: If you can show this to be true in one frame of reference, then it is true in all frames of reference.)

(b) An H2 molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed 1.00 * 10^3 m/s, and they are approaching at 120 degree angles so that at any instant, the atoms lie at the corners of an equilateral triangle. Find the speeds of the H2 molecule and of the single hydrogen atom that remains after the collision. The binding energy of H2 is 7.23*10^-19 J and the mass of the hydrogen atom is 1.67*10^-27 kg

Any help would be appreciated. —Preceding unsigned comment added by 164.67.66.113 (talk) 01:16, 24 February 2011 (UTC)[reply]

Consider the simplest possible case for (a), where the final product has no net velocity. To conserve momentum, each atom must approach with the same speed. Now calculate the kinetic energy before and after; and account for the binding energy, and determine if the reaction is possible.
For the second case (b), I think the problem is actually more straightforward, because they give you all initial conditions, and most of the final conditions; (so you don't need to "intuit" anything - just calculate). Nimur (talk) 01:31, 24 February 2011 (UTC)[reply]
Is there an assumption that hydrogen can not store excess energy as anything except kinetic energy? Can't it store the binding or collision energy as excited electrons (assuming the energy levels precisely match)? Can't diatomic hydrogen "bounce", i.e. stay bound but oscillate? Or slightly miss each other and rotate at high speed? Ariel. (talk) 05:08, 24 February 2011 (UTC)[reply]
That was the implicit assumption; energy is either kinetic or the fixed-value of the "binding energy." The validity of this assumption is at least partially predicated on the "perfect" alignment of inbound particles. (They can't start rotating if they had no rotational momentum in the initial condition). And electron energy might be negligible; you can actually calculate whether the kinetic energy is orders of magnitude larger than the hydrogen ionization potential, 13.6 eV; while the supplied binding energy is about 4 eV; so neglecting energy in the electron-orbitals might be a bit of a tenuous assumption. You can similarly compare order-of-magnitude to determine if diatomic oscillatory motion is "negligible." This is, after all, a hypothetical, simplified treatment. In practice, we should apply a nuclear scattering treatment and solve quantum-mechanical wavefunctions, and represent the collision as a quantum mechanical operation; but from the wording, I suspect that material is outside the scope of the physics class. Nimur (talk) 05:21, 24 February 2011 (UTC)[reply]
I suppose I feel the same as Ariel. The fact is, it is possible for H2 to form in a collision of two hydrogen atoms, with the excess energy being shed somehow (probably as photons, but there are other ways). I guess it's unlikely, but it doesn't violate any conservation laws. Many physicists would describe this as "a collision that involves only two hydrogen atoms". I think there's no worse question you can pose to a student than "show X" where X is not really true. Aside from being misleading, it penalizes the students who understand the material best, since they are more likely to notice the hole in the obvious (intended) proof and waste time searching for a better one. -- BenRG (talk) 10:08, 24 February 2011 (UTC)[reply]
True. I agree that this is a drastic oversimplification of the problem. But I am curious if this simplification is a valid "approximation" to the actual chemical kinetics that are experimentally observed. If so, I can understand why the simplified model might have been developed; in a sense, the "assumptions" would be the "reasonable conclusions" that must theoretically follow if the observed experiments found that only 3-atom collisions resulted in hydrogen; but it has a lot of serious drawbacks, and absolutely can't be considered a general case solution to reaction mechanics. This 1960 paper describes dissociation (the reciprocal reaction of formation): The kinetics of dissociation of a diatomic gas - and its opening line is: "One of the most unsatisfactory aspects of the theory of chemical kinetics has been the failure to describe reasonably the simplest of all reactions, the dissociation of a diatomic gas." The paper proceeds to develop a chemical kinetic description that extends the kinetic-energy/momentum conservation to treat the observed data more accurately. Nimur (talk) 20:57, 24 February 2011 (UTC)[reply]

Exactly why can't momentum and energy be simultaneously conserved in a collision? Energy is always conserved. Kinetic Energy isn't conserved. And why can't you create H2? If you send them towards each other at a high enough speed, assuming nuclear fusion does not occur, then they should collide inelastically, and you can calculate the required relative velocity with:   (where μ is the reduced mass of the system). ManishEarthTalkStalk 12:30, 24 February 2011 (UTC)[reply]

Ah, you've hit the nail on the head! An inelastic collision is defined by the fact that energy and momentum are not both conserved! Only in elastic collisions are both conserved. Of course, energy is always conserved if you do your bookkeeping right, and in the inelastic case the excess energy is "released" as "heat".
To explain that hand-wavy definition, in a two-particle inelastic system, that "heat" is a spring-like binding energy, so with no place else to go, the inelastic collision tries to store this heat in the two hydrogen atoms themselves. But since this energy was precisely sufficient to bind them in the first place, it is sufficient for them to rip apart too. When hydrogen gas burns, an initial spark makes H2→H+H, but then a lot of firey heat is released by the bond-breaking itself. We can't reverse that process and form bonds unless something can take away the excess heat, which is why you need at least one extra "helper" particle (or catalyst to the reaction) to carry it in kinetic energy. I hope this helps you understand how chemistry and physics describe the exact same thing in different terms. SamuelRiv (talk) 21:34, 24 February 2011 (UTC)[reply]
Careful: the breaking of bonds absorbs energy. If the freed atoms then form different bonds, they might release even more energy than the breaking absorbed. --Tardis (talk) 22:48, 24 February 2011 (UTC)[reply]
That equation seems right but isn't relevant here.   is the energy released by forming the bond. There's no minimum approach velocity because that energy isn't being taken from the kinetic energy. Rather, the molecule can only be formed permanently if whatever initial kinetic energy and some of   can be dispersed during the collision (via photons), so that the resulting system then lacks the energy to separate. In some cases with an off-axis collision, it may be possible to produce a stable, spinning molecule without shedding all of the initial kinetic energy, but I don't know if they occur in practice. --Tardis (talk) 22:48, 24 February 2011 (UTC)[reply]

What is this compound?

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  Resolved

I'm actually doing a course in organic chemistry right now and yet I still can't figure out the name of this. Any help? -- 04:23, 24 February 2011 (UTC)[reply]

Looks like 5-hydroxytryptamine, or serotonin. Here's the standard numbering system for tryptamines [2], which is a little confusing to the uninitiated... and needs periodic rechecking by the initiate as well, I'm afraid. Wnt (talk) 04:53, 24 February 2011 (UTC)[reply]
Thanks! Thinking of it, that should have been one of my first guesses for "organic compound someone would tattoo on their body". -- 07:09, 24 February 2011 (UTC)[reply]
I don't see the appeal. Sounds painful... Wnt (talk) 16:18, 24 February 2011 (UTC)[reply]
It's a nice idea, perhaps a reminder of a terrible depression in her life that was overcome with the help of an SSRI. To face one's darkness is to transcend it, hopefully. Of course, it could also mean that she's just really happy and wanted a happy tattoo instead of, say, the oft-used caffeine or ATP or capsacin. SamuelRiv (talk) 21:44, 24 February 2011 (UTC)[reply]

Consequences of (admittedly magically) preventing every cell in your body from ever dividing again?

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I understand that many parts of the body rely upon rapid cell division to maintain state (at least what we perceive at the macro scale, it's a Ship of Theseus thing). What would happen if you could snap your fingers and prevent every cell in your body from ever dividing again - in effect "freezing" the current state? Cells that die would not be replaced, clearly. I'm curious what would start to breakdown first from a medical perspective. I imagine death would not be long in coming, but I'm unsure as to how one would die? The Masked Booby (talk) 05:13, 24 February 2011 (UTC)[reply]

This should be pretty much like radiation poisoning - hematopoiesis (production of red and white blood cells) fails, the intestinal lining no longer regenerates, skin and hair can suffer, etc. I'm not sure offhand why headaches are a symptom of radiation poisoning and I'm not sure if the magical mechanism would replicate them. Wnt (talk) 05:29, 24 February 2011 (UTC)[reply]
See also Labile cell and Hayflick limit. Ariel. (talk) 05:35, 24 February 2011 (UTC)[reply]
I think your stomach juice would make short work of you. Vespine (talk) 07:26, 24 February 2011 (UTC)[reply]
Well, red blood cells have a lifetime of about 100 days, so in a couple of months you would die of anemia if nothing else killed you sooner. Looie496 (talk) 07:31, 24 February 2011 (UTC)[reply]
Yes, it's more the immune effects of lethal irradiation that tend to be a problem - though completely disabling gastrointestinal regeneration would probably work the quickest end here, barring extraordinary interventions. I suspect that lethal irradiation is relatively less damaging to the slowest-growing stem cells of the intestine than a magic stop to all cell division would be. Though there's also a certain uncertain ability of hematopoietic stem cells to seed the intestine when need be.[3] Wnt (talk) 16:17, 24 February 2011 (UTC)[reply]
A human being's body experiences about 10,000 trillion cell divisions in a lifetime, see Cell division. The OP's magic snap, if its effect could be focussed, would incapacitate Cancer cells. Cuddlyable3 (talk) 10:35, 24 February 2011 (UTC)[reply]
Your digestive system would start digesting itself. I'd give you 2 or 3 hours (at most 5 or 6) to get your affairs in order and say your goodbyes before you bleed out. Roger (talk) 18:48, 24 February 2011 (UTC)[reply]

See Biological_timeline_of_radiation_poisoning for the (grisly) details. The "walking ghost" phase in particular gives me chills. —Preceding unsigned comment added by 12.186.80.1 (talk) 21:26, 25 February 2011 (UTC)[reply]

See Laron syndrome, telomere and telomerase. ~AH1(TCU) 20:39, 26 February 2011 (UTC)[reply]

Battery rack earthing

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Whether battery rack has to connected to the safety earth pit or not? If it has to be connected what is the reason behind this? If not then why it should not? — Preceding unsigned comment added by Magesh1581 (talkcontribs) 12:01, 24 February 2011 (UTC)[reply]

You might want to tell us which sort of battery rack you are talking about, but in short, the answer is probably yes. The reason is that earthing the rack prevents harm should an inadvertent connection be made between the rack and the positive terminals of the batteries. Charge will be routed to earth, rather than to the next poor soul who touches the rack. --Tagishsimon (talk) 12:31, 24 February 2011 (UTC)[reply]
I think it is much more complicated, it depend on application and regulation. Depending on situation the answer can be yes or no. --Gr8xoz (talk) 23:59, 24 February 2011 (UTC)[reply]

The application which we connected the battery is for the UPS system. Batteries are 2volt 1100 ah vrla batteries. My question is if we connect the safety earth to the battery rack when there was any connectivity or short circuit between the positive terminal to the body or the negative terminal to the body or in between any terminal connected to the body, will the circuit close though the body to safety earth? will there be the current flow though the body to earth from positive or negative or inbetween terminal which get connects to the body? If there wasn't any current flow then whether it will affect the human who touches the body of the rack at this condition? — Preceding unsigned comment added by Magesh1581 (talkcontribs) 14:16, 25 February 2011 (UTC)[reply]

We cannot really tell you how to connect your battery, since that would be like telling you how to treat medical symptoms or giving legal advice. But we can provide some information and insights. Contact a licensed professional engineer with design questions. Connections must comply with the applicable electrical safety code, and it may well dictate what has to be grounded and how. You ask about current flow in several situations. Try drawing the electrical diagrams and analyzing them, based on the various possible intentional and accidental grounding of positive, negative, and rack. You have not said if your DC system is grounded, and you have not stated the total voltage. I have seen utility 100 volt or higher batteries which are operated in a floating (unearthed) system, with a battery ground detector relay. If somewhere a single positive or negative lead becomes frayed and touches ground, nothing happens except the battery ground alarm sounds, and someone finds and corrects the ground. If the negative lead were intentionally grounded, then a ground anywhere in the system could cause equipment to operate unintentionally, or to fail to operate. As for the battery rack, it would take some work to prevent it from having at least a high resistance earth ground. If a metal rack sits on a concrete floor, or is bolted to a brick or concrete wall to prevent it falling over in an earthquake, then it is grounded to some extent, though perhaps with many ohms or hundreds of ohms of resistance. For it to be ungrounded, insulators would be needed below the feet and it would have to be isolated from the wall by insulating supports. The isolation would need testing on installation and periodically in service, since insulators can crack or grounds can otherwise occur. Then leakage current from the batteries due to acid spills, dirt on the cases, or stray wires would likely cause the rack to be at some undetermined voltage from ground. Depending on the battery voltage, this could present an electrocution hazard or at least a painful shock for anyone standing on the floor who touched the battery rack. Intentional grounding would require a substantial cable or cables permanently connected to the rack in a way that leaking acid will not corrode and make ineffective the ground connection, then connected to an effective ground, such as a system of ground rods driven deep enough to have a low resistance earth connection. The ground cable must be large enough to carry any fault current, as if a battery lead breaks and the battery gets connected to the grounded rack while there is an accidental or intentional ground somewhere. The available fault current from the battery you describe will be very high. Edison (talk) 19:02, 25 February 2011 (UTC)[reply]

Electric Field strength at a distance x from an infinite charged sheet

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I was solving a problem based on Electric Field, and I noticed, that, in my derived formula for electric field at a distance x from a charged disc of radius R and charge density σ,  
When R tends to infinity, the surface becomes an infinite charged plane, and E is:  

Note that it's no longer dependent on x, i.e. the magnitude of the field is the same at all points in space (Except those on the plate)
I rederived the same formula for the infinite case with Gauss' theorem (the surface is a cylinder with its height perpendicular to the plane), and by summing up infinite strings.
My question is, why is E no longer dependant on x? Thanks, ManishEarthTalkStalk 13:49, 24 February 2011 (UTC)[reply]

I fixed your formula for you. Your answer is correct. The field doesn't depend on x. Just use Gauss' law in a cylinder of length 2x with the charged plane placed symmetrically at the center of the cylinder to get the same result. Dauto (talk) 15:47, 24 February 2011 (UTC)[reply]
The reason that it no longer depends on x (distance) is that the farther away you are from the infinite plane the more charge you "see", that extra charge exactly balances the loss of charge force due to distance. Ariel. (talk) 19:20, 24 February 2011 (UTC)[reply]
Another way to look at it is that for an infinite sheet the field lines are all parallel to each other and perpendicular to the sheet. The flux density is therefore constant at all points. SpinningSpark 20:26, 24 February 2011 (UTC)[reply]
Thanks! I understand it better now! ManishEarthTalkStalk 09:27, 25 February 2011 (UTC)[reply]

Semen omelet

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Unpalatable question, but curiosity is king: Could you make an omelet or a quiche or something out of semen? For the sake of decency, let's say it comes from a free-range bull fed on alfalfa and organic blueberries. Perhaps I'm wrong to suppose that the substance would have the culinary properties of egg. If not, what would it be like when cooked? (Bonus question: Have the Japanese already done this?) LANTZYTALK 14:05, 24 February 2011 (UTC)[reply]

Not a direct answer, but are you aware of Rocky_Mountain_oysters? SemanticMantis (talk) 15:30, 24 February 2011 (UTC)[reply]

Doesn't omelet/quiche rely on being made of Egg and don't the female of the species produce the eggs (Ovulation) or am I missing something here? ny156uk (talk) 18:21, 24 February 2011 (UTC)[reply]

Well, it's a bizarre question, but you do seem to be missing the point that the OP already knows what omelettes are traditionally made from. The idea of a semen omelette may sounds pretty unpalatable (literally), but I don't see how it's any more horrid than eating a tarantula egg omelette (which I've seen done). I guess the thought (and I don't mean to speak for Lantzy, but this is what I assume to be the case) is that semen, like eggs, contains a lot of protein and therefore might cook in a similar manner. My gut tells me (well, it's telling me not to take part in this thread, but I'm ignoring that) is that it wouldn't coagulate in the same way as even an all-whites omelette. An egg is a discrete object with a structure that gets remade/denatured during cooking while semen is a fluid lacking that kind of cohesiveness; I'm thinking it would turn out grainy or crumbly, but I'll definitely leave it to the OP to do whatever experiments they think would help. Matt Deres (talk) 18:57, 24 February 2011 (UTC)[reply]
We have an article on this: Milt. And we also have an article on Roe—hence Row, Row, Row Your Boat. Bus stop (talk) 19:03, 24 February 2011 (UTC)[reply]

Being the empirical sort I decided to try it. Fried it was tough, very much like goose eggwhite but with a stringier texture. Considering the time and effort required to obtain the materials, I will not be doing it again.

Focusing camera mounted on biological gel imager

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On the top of our gel imaging equipment is a camera, allowing us to capture digital images of fluorescent bands of (usually DNA or RNA). The camera has a ring that we can adjust for zoom. It also has another ring which appears to alter exposure, which I surmise is the aperture. If I adjust the zoom, must I also adjust the aperture for a properly focussed image? Or will the image be equally focussed at all zoom level?

Changing the zoom on a lens is pretty fundamental, and you'll probably have to change the focus by a lot. Changing the exposure (time) or aperture (diaphragm) depends more on how desperate you are to get that faint little band you see to show up on the picture. As a rule, running the gel is enough of an expenditure of time and effort, even money, that you shouldn't hesitate to fiddle with every control on the camera trying to make it better (especially when it's digital!) ... just remember what it was before you started and put it back again so that people don't start murmuring. Wnt (talk) 16:05, 24 February 2011 (UTC)[reply]
Knowing what camera would help to give a better answer. Aperture affects the depth of field and the best (and same) f stop should satisfy all images in this application. This leaves only the speed to consider. So in other words: use the recommended f stop and don't fiddle with it!--Aspro (talk) 16:34, 24 February 2011 (UTC)[reply]
There are conceivable reasons to fiddle with an f stop - for example, depending on the thickness of your gel and thus the thickness of your bands, or if the surface of the gel is showing up too much for some reason and you want to try to blur that out a bit - but admittedly most of the time this isn't that important to fool with. Wnt (talk) 17:17, 24 February 2011 (UTC)[reply]
Note that if you zoom in more, the image will get darker. Some fixes for this are:
1) A brighter light. But too much might cook your sample.
2) A wider lens or lens opening (aperture/f-stop).
3) A longer exposure. Could cause motion blur if the subject moves.
4) Post-processing to brighten the picture. There's a limit to how much can be done this way, though. StuRat (talk) 17:11, 24 February 2011 (UTC)[reply]
1) When is flash going to cook a southern blot
2) What's wrong with keeping Aperture#Optimal_aperture
3) Just how fast do you think a southern blot needs to gallop along to cause blur?
--Aspro (talk) 17:28, 24 February 2011 (UTC)
[reply]
Who says the OP is imaging a southern blot? Might be a regular agarose gel and the DNA might be going to further downstream applications, eg sequencing. In that case I at least wouldn't lengthen the UV-exposure. Albval (talk) 19:35, 24 February 2011 (UTC)[reply]
Nobody. Yet... Unless s/he is in the habit imagining places where the sun don't shine, the lens's best performances will be in a very small depth of field - if its a off the self purchase. Of course, the camera/lens system may be specially chosen, but we have not been informed of this yet. So southern blot blah, blah, blah, will it cover most lab camera applications until --or unless-- the OP provides more detail.--Aspro (talk) 22:57, 24 February 2011 (UTC)[reply]

Where is KT?

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This question is about the iridium rich K-T boundary. Since the chalk of the Chiltern Hills is Cretaceous and the London Clay is Tertiary I am guessing that the K-T boundary lies at the junction of these two beds. Are there any outcrops around North London where the K-T boundary layer is visible? SpinningSpark 18:14, 24 February 2011 (UTC)[reply]

The Chalk Group extends up into the Paleocene, so the boundary lies within the upper chalk. According to this the Sarsen stones are derived from Danian age limestone (although our article calls them sandstone - something to check out), so they come from above the boundary, but I can't find anything about an exposure of this boundary in the Chilterns. Mikenorton (talk) 18:42, 24 February 2011 (UTC)[reply]
They may be formed in the Danian, but the Sarsens are apparently an example of silcrete, a superficial deposit resulting from weathering [4], so even if you could find the base of this, it wouldn't be a stratigraphic contact. Mikenorton (talk) 18:54, 24 February 2011 (UTC)[reply]
Finally, and sadly for your purposes, it seems that all of Britain was above sea level in the Paleocene (the Danian chalks I referred to are found in the North Sea) and there is definitive statement from here that "The earliest Tertiary deposit in Britain is the marine Thanet Sand Formation (Thanetian), a series of slightly glauconitic sands that rest directly on the eroded transgressive surface on Chalk". Sorry, Mikenorton (talk) 19:06, 24 February 2011 (UTC)[reply]
On a brighter note, if you fancy a trip to Denmark, you can see the boundary at Stevns Klint, probably the closest to the UK [5], see File:KPg stevns.png. Mikenorton (talk) 19:34, 24 February 2011 (UTC)[reply]
Thanks for the information. I was, of course, hoping the answer was going to be "it's at the bottom of your street, just past the pub." Thanks again. SpinningSpark 20:52, 24 February 2011 (UTC)[reply]

Uncertainty Principle - Doubt

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I have a doubt about Uncertainty Principle. What does it really means ? It means that We cannot measure / know position / speed of anything with acuraccy better than h/4pi due to influence by at least photons in measurements or it means that electron (for example) doesn´t have position and speed precise and in consequence we cannot measure it ? — Preceding unsigned comment added by Futurengineer (talkcontribs) 20:04, 24 February 2011 (UTC)[reply]

We have a fairly detailed article with the obvious title "Uncertainty principle". There is some dense math and physics in there, but the very first paragraph or two directly answer your questions in pretty simple language. Idea of position of an electron is confusing because it's not purely a "particle", so you have to account for that as well when measuring it. The third and fourth paragraphs of the article address this idea. DMacks (talk) 20:13, 24 February 2011 (UTC)[reply]
Unfortunately, though, there is no simple and agreed answer as to what it really "means". It's connected with the general quantum weirdness, which poses problems for our intuitive realist approach to the physical world. If you read the answers provided in the usual references and feel that they don't really answer the question you were getting at — you're probably right. But I think we have an article on interpretations of quantum mechanics that at least provides a starting point for various ways in which physicists and philosophers approach the problem. --Trovatore (talk) 20:28, 24 February 2011 (UTC)[reply]
To put it very simply: it is most like the second proposition of yours than the first. It is not just about the inability to measure it. The measurement issue is a secondary one to the true quantum uncertainty. --Mr.98 (talk) 20:54, 24 February 2011 (UTC)[reply]
I believe it started with realizing that both could not be measured (your first interpretation), but over time people came to realize it's not just a measurement issue the values/concept does not actually exist for a particle (your second interpretation). And because of that particles can do some really strange things. Ariel. (talk) 22:00, 24 February 2011 (UTC)[reply]
Trovatore and Ariel, that's silly, and you shouldn't equate fairly simple quantum mechanical concepts with philosophy, as QM is no more an epistemological problem than anything in the classical-mechanical realm.
Futurengineer - when things get very small, we can't "see" them like we see everyday objects, simply because we really can only send one photon or electron at a time to look at an object that's of similar size (as opposed to, say, a blade of grass which we observe with literally billions of photons at once. So basically, we only get a "flash" of what the tiny object looks like at a particular time.
Now take an ordinary camera, which also can only take a "flash". If we want to photograph a moving car, we can make that flash exposure extremely quick, giving us a clear imagine of the car, except in the photo the car seems extremely still and lifeless. You could also let the exposure last a little longer, giving the car a "motion blur", but the image is much less sharp (see examples). So in the sharp image, we see very clearly where the object is, but it's still and lifeless, while in the blurry image, we can actually get an impression of the movement of the object, but it's position is blurry and less defined. The mathematics of the Uncertainty Principle article is a very strict, a priori definition of this phenomenon which occurs whenever we can only take "flashes" of an object we want to measure: we either get good position or good velocity, but we can never get both exactly. That's really all there is to it, quantified for the scale of our observation by Planck's constant, which is the only defining QM effect. You don't have to "believe" the principle - you can see it for yourself. SamuelRiv (talk) 22:10, 24 February 2011 (UTC)[reply]
I didn't say "epistemological"; it's more ontological. But yes, QM is absolutely more of a philosophical problem than classical mechanics. --Trovatore (talk) 22:38, 24 February 2011 (UTC)[reply]
(ec)That doesn't really explain the other uncertainty relationships - like energy and time. It also doesn't explain other effects like how if you constrain a particle in a box (i.e. the location is known) it's momentum becomes fuzzy. It's not just a measurement issue it's fundamental to the nature of reality. Ariel. (talk) 22:40, 24 February 2011 (UTC)[reply]
That's an incorrect explanation of the uncertainty principle. The HUP is a consequence of the fact that the momentum representation for a wavefunction is the fourier transform of its position representation (or the same could be said for any pair of conjugate variables). Since it's impossible to arbitrarily localize both a function and its frequency spectrum, you cannot accurately determine the value of two conjugate variables simultaneously. Also, there are quite a few philosophical difficulties with quantum mechanics in general. Truthforitsownsake (talk) 00:18, 25 February 2011 (UTC)[reply]
There is no such a thing as philosophical difficulties with quantum mechanics. Some people have philosophical issues with quantum mechanics, true. But that is problem for those people, not for the theory. The theory is completely self consistent and has been tested ad nauseam. That's all that is required from a physical theory. Dauto (talk) 02:42, 25 February 2011 (UTC)[reply]
Well, it's not just a problem for those people, as though they're making stuff up. There are genuine difficulties in turning QM's predictions into explanations. Explanation most certainly is part of the function of a theory (see e.g. Nagel on this).
Now, that's completely different from saying that these difficulties refute QM, which is a possible (though not necessary) reading of Truthforitsownsake's contribution. I personally do not make any such claim. I do however think that QM is sufficiently hard to reconcile with our intuitive metaphysics that people have a right to be confused. I think that a lot of times people who ask such questions get answers to different questions, and the person answering should at least acknowledge that it's not precisely an answer to the question that was asked. --Trovatore (talk) 05:19, 25 February 2011 (UTC)[reply]
Some philosophical problems are listed here: measurement and quantum theory[6], different interpretations of quantum theory[7][8][9]. Philosophical issues include: what constitutes an observer or a measurement, which entities in quantum mechanics actually exist and which are merely convenient fictions, what is the nature of quantum collapse and how it can exist alongside a smooth wave function, how many universes there are, whether objects exist independent of an observer, and how come there are imaginary numbers in the formulas. The question of whether this matters is itself a philosophical problem: are scientific theories meant to explain the world (in which case the philosophical issues are very relevant), or simply to produce useful outcomes (in which case they are less relevant) - as an analogy, if you had a box that seemed to predict the exact lottery numbers every week, would you want to know how it works, or would you simply accept that it works? --Colapeninsula (talk) 14:35, 25 February 2011 (UTC)[reply]
It means that there a movment back and forth to the futher and it afect now /thanks water nosfim . —Preceding unsigned comment added by 84.228.225.214 (talk) 05:42, 25 February 2011 (UTC)[reply]
See also quantum zeno effect. ~AH1(TCU) 20:25, 26 February 2011 (UTC)[reply]
Colapeninsula: I will address your philosophical problems in turn: what constitutes an observer is anything that interacts with a particle, be it a photon, another atom, a strong-acting force, etc - the questions of "conscious observer" really aren't valid anymore. Functions vs existence have as much to do with classical mechanics as QM. Quantum collapse and QM in general have multiple equivalent interpretations, as does classical mechanics, but the physics is the same - what constitutes "reality" is a question for all of physics and math and epistemology - QM is nothing special in that sense. How many universes has nothing to do with basic QM (like the uncertainty principle) - it's associated with various cosmologies and particle theories. Whether objects exist independent of a conscious observer again has nothing to do with QM - it becomes a "tree falling in the woods" question, unless you're completely fringe. Imaginary numbers exist in QM theory because they exist in all wave theories as a convenience (though QM has a special complex term in the classical Lagrangian, but this again is a formulation that is nothing special about imaginary numbers). QM doesn't predict lottery numbers - it predicts the probability that a given lottery number will occur. I can do that right now - a pick-three lottery has a 1/1000 chance of giving 1-1-1. The point is that the uncertainty principle is not some kind of mystical magical entity - the ∆x∆y≥hbar form can be derived completely with classical wave theory (though the ∆E∆t form is a bit of trickery) - it exists for any wave that is measured by a forceful interaction. SamuelRiv (talk) 21:26, 26 February 2011 (UTC)[reply]
Samuel, you're glossing over some very difficult issues and not even acknowledging that they are issues. Whether a tree falling in the woods with no one to hear it makes a noise, may not be an issue for a logical positivist, but a realist (that is, most of us) considers it a real question that must have an answer. Similarly for the issue of which, if either, slit in the two-slit experiment the electron has really gone through. (The realist would accept the answer, "it's not a particle at all, but a wave, and therefore goes through both slits", but that then appears to be inconsistent with the granular nature of the response of the film.)
The different interpretations try to address these questions in various ways, which have various problems in being reconciled with our intuitive metaphysics sufficiently well to be called "explanations". If you just want to predict pointer coincidences (let's see if we have an article on that!) then QM does fine. But that's not enough for most of us. --Trovatore (talk) 21:55, 26 February 2011 (UTC)[reply]

Voice perceiving

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Is there any differences between how I hear my own voice and how it's heard by others? Specifically, does my voice sound to others exactly the way it sounds to me? I became a bit suspicious when my mobile phone had echoed my voice differently from what I perceive it. —Preceding unsigned comment added by 89.76.224.253 (talk) 20:31, 24 February 2011 (UTC)[reply]

In short, yes, there definitely is a difference. Your own voice vibrates your skull and chest cavity, and you perceive these vibrations as well as the sound waves in the air. When other people hear you speak, they only hear the vibrations you've induced in the air. (WP:OR follows) So most people express some confusion the first time they hear themselves recorded, saying "Do I really sound like that ?!". Also, cell phones are notorious for the artifacts they introduce because of the high levels of compression they use. SemanticMantis (talk) 20:43, 24 February 2011 (UTC)[reply]
Definitely. My voice, when recorded, sounds quite a bit higher than it sounds to me. Of course the problem is not the recording, it is the difference of how our own voice sounds. — Preceding unsigned comment added by GaryReggae (talkcontribs) 20:49, 24 February 2011 (UTC)[reply]
So my actual voice is that on voice-recording devices? Weird :)) —Preceding unsigned comment added by 89.76.224.253 (talk) 20:50, 24 February 2011 (UTC)[reply]
Indeed. To get a fair idea of what you really sound like, it's worth it to get a decent recording in a quiet room with a good microphone (i.e. not cell phone, voice mail, laptop built-ins, etc.). SemanticMantis (talk) 21:09, 24 February 2011 (UTC)[reply]
Is this at least a partial explanation as to why some people think that they are great singers when really they are awful? Googlemeister (talk) 19:21, 25 February 2011 (UTC)[reply]
Maybe, but when singing alone, people tend also to imagine instruments playing, which makes them sound in reality much amusing. Quest09 (talk) 13:58, 26 February 2011 (UTC)[reply]
A classic trick by radio announcers, to hear how their voice sounded to others, was to use a finger to close one ear. The change in resonance approximated the way the voice sounded over the microphone. Gary Owens on Rowan & Martin's Laugh-In did this announcer trick. Edison (talk) 03:29, 26 February 2011 (UTC)[reply]

Electric shower temperature fluctuations

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I have an electric shower that is frustrating me. I don't know whether the shower unit itself is knackered or if it is caused by outside factors.

I don't know exactly what the setup of the shower is but I am pretty sure it is one of those very common (in the UK at least) wall-mounted devices that go over a bath and heat water as it goes through. I can't see the inlet pipe(s) as it is concealed in the wall but I think these things only have a cold inlet. It works fine when first switched on but after a couple of minutes or so, starts to run cold. The shower has an LED marked with 'Reduce temp' on it and this illuminates when it starts running cold. There is also a very noticable change in the sound it makes - you can hear something (probably a thermostat) go click and then you can hear something turn off (a pump?). After about 30 seconds or so, the light goes out, the motor kicks back in and it gets back up to temperature before starting this whole cycle again.

What is causing it to do this? I have searched the web and looked at the WP page on showers but not been able to shed much light on the subject. Perhaps this type of shower is only used in the UK?

I can understand why there is a function to cut out as this is probably to prevent overheating. However, it should be regulating the temperature so that it doesn't get hot enough to invoke the safety limit. I am not sure that there is anything wrong with the actual shower as I have had this issue with other showers of the same type. But surely in the 21st century it should be possible to produce hot water at a constant temperature.

There are several other things that potentially complicate things. The first is limescale but it is not caused by a colleged shower head as I have unscrewed that and tried it with the hose pointed in the bath. It could potentially be furred up inside but I don't know. The second thing that could be an issue is whether there is a pump in there. I guess if the incoming water rate is inconsistent then it could cause the temperature to reach the limit. That said, I don't think there is a problem with the mains water pressure...it seems pretty good to me. The final point of note is that adjusting the temperature knob on the shower makes no difference to whether this problem occurs. The temperature when it runs is different but it still cuts out on a frequent basis.

So, are there any suggestions for sorting out this annoyance? Should I just get a new shower or is it pointless? Any ideas would be much appreciated! GaryReggae (talk) 20:47, 24 February 2011 (UTC)[reply]

These showers have a cut-out thermostat that prevents overheating. It is possible that the thermostat is working correctly but that limescale is causing it to register a higher temperature than was intended by the designers. If the thermostat is faulty then it can be replaced (many years ago I replaced one to give the shower unit years of further use), but this is probably a job for an expert who is both electrician and plumber because there are safety issues involved. If the cause is reduced water pressure as other cold-water taps are turned on then you will be able to see the change in water flow from the shower-head. Dbfirs 21:45, 24 February 2011 (UTC)[reply]
limescale would cause a lower reading, not a higher, or more accurately a slower reading, so instead of reacting quickly to temperature changes and cycling at a constant rate it's reacting slowly. Just a guess. Ariel. (talk) 22:03, 24 February 2011 (UTC)[reply]
Yes, that's true if the limescale is around the sensor and the heater is on the other side of the limescale. I was thinking of the scale impeding the flow, or the heater being on the same side of the limescale as the sensor. Dbfirs 07:42, 25 February 2011 (UTC)[reply]
Try replacing the shower head to see if that has any effect, as they raipdly get blocked. Although you tried the shower with the shower head removed, I think showers are often sensitive to pressure and so the open pipe may have made a difference to its behaviour. 92.28.246.36 (talk) 00:27, 25 February 2011 (UTC)[reply]

Pulsar Dyson ring

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So my EE friend and I were looking at the pulsar page and when we saw the image of a pulsar's magnetic field his first instinct was to put a wire around it and use it to draw off power. That got me to thinking. Pulsars are relatively small and could fit inside a spaceship or a Dyson sphere or ring relatively easily to be harnessed as a source of massive amounts of energy. Has science fiction explored this idea? -Craig Pemberton 21:29, 24 February 2011 (UTC)[reply]

They might be small, but they are astonishingly heavy. If you have the technology to move something that heavy you probably don't need their energy. Plus their gravity is immense because they are so small and yet heavy - it's about as close as you can get to a black hole without being a black hole. But it's a cool idea. Ariel. (talk) 22:22, 24 February 2011 (UTC)[reply]

100 Oct LL Aviation fuel in car engines

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My question is this: if I were to run 100 Octane LL aviation fuel in an average modern car, which is not supposed to take leaded fuel, would it be bad for the car? I ask this not because I want to run 100LL in my car, but every now and again I find myself having to dispose of aviation fuel, and am looking for environmentally friendly ways to do that (when testing fuel, we aren't supposed to put it back in the airplane). I figure that burning it in an engine is about the best thing to do with it, so why not in the car? Just as a note, 100LL is essentially normal automotive gasoline, except with a higher octane, lead, blue dye, and more stringent quality checks. Falconusp t c 21:53, 24 February 2011 (UTC)[reply]

Lead will destroy (foul up) your catalytic converter. The converter will tolerate a small amount, but eventually will stop working, and then you'll fail a tailpipe test. It may also physically clog, but I think it's more an issue of the active components getting coated. You'll probably cause more environmental damage from your bad converter than you will from just burning the extra fuel. Just make sure to have a lot of air when you burn it (i.e. no smoke, small fire) and that's the best thing to do with it. You can also pour the fuel into a plastic soda bottle and just put it in the garbage. It will either be buried, or incinerated and both are fine. As long as it doesn't leak it will do no harm. Fill the bottle only halfway and squeeze out the air, to give it a cushion against being compressed. Write "aviation gasoline 100LL" on the bottle in case some future garbage-miner finds it. Ariel. (talk) 22:09, 24 February 2011 (UTC)[reply]
Also the blue dye is designed to stain your car's fuel system. If the police/tax men find a dyed engine, you can get prosecuted for fuel-tax evasion. What the smallest amount of dye that is detectable, and how long it lasts is a different matter. CS Miller (talk) 22:19, 24 February 2011 (UTC)[reply]
[EC] Do you have a source for that? I most certainly pay fuel taxes for 100LL, and I would be surprised if they would try to dye an aircraft engine. That's just one more variable. I have always been told that the blue dye is solely to distinguish it from MoGas, JetFuel, other Octane AvGases (which are died Green, Red, etc), etc. Falconusp t c 22:28, 24 February 2011 (UTC)[reply]
Ah, I might have been mistaken. From our fuel dye article, some of the dyed fuels are dyed because they are low-rated duty, but for AvGas, as you said, it is to clearly indicate which grade of AvGas the fuel is. CS Miller (talk) 22:56, 24 February 2011 (UTC)[reply]
I'm not so sure about trashing fuel. I mean, if you ordered something like that to be shipped to you in the U.S., you know how it would come, in some fancy packaging with placards and whatnot. I doubt it would be ORM-D. I mean, if you trash it, some hydraulic ram or bulldozer is probably going to break open the container and spray fuel all over, and then, at best, you're likely to have some kind of trouble trying to find its way back to you. If you're going to do something faintly illegal with it, at least you could pour it all down a mine shaft and blow it up for fun. Wnt (talk) 22:53, 24 February 2011 (UTC)[reply]
How about using the fuel in a model aircraft? Nimur (talk) 22:25, 24 February 2011 (UTC)[reply]
I think it should be left at a collection site for harmful/dangerous waste. I do not know how it is handled in your state. The next best thing to do is probably to burn it in free air or in an engine without a catalytic converter. I would certainly not throw it in the thrash, especially not if the waste goes to a landfill, a plastic bottle will break sooner or later. --Gr8xoz (talk) 23:27, 24 February 2011 (UTC)[reply]
Actually soda bottles are quite strong - they are not easy to break, even on purpose. An ordinary garbage compressor will do nothing to them, especially if you only fill them halfway and squeeze out all the air. I think I will do an experiment involving a 5lb sledge hammer and an empty soda bottle. Ariel. (talk) 07:55, 25 February 2011 (UTC)[reply]
Most of them will probably not break immediately but I do not think a bottle will withstand the pressure if pressed between something hard and sharp and these [10] wheels on a 50 ton machine. I think most plastic bottles will begin to leak after some decades of exposure to gasoline. It will certainly leak some time in the future, in thousands or million years. It is highly irresponsible to recommend someone to throw toxic waste in the household garbage. There are also a very big difference between an empty bottle and an closed half filed bottle with uncompressable fluid.--Gr8xoz (talk) 13:31, 25 February 2011 (UTC)[reply]
While you may own a car with a catalytic converter and use an electric or push-powered lawnmower, perhaps you have a friend who owns a vintage automobile or uses gas powered lawnmower. -- 119.31.126.66 (talk) 11:38, 25 February 2011 (UTC)[reply]
Many modern gas powered lawnmowers have a catalytic converter.--Gr8xoz (talk) 13:31, 25 February 2011 (UTC)[reply]
Your local fire department will be happy to advise you on how to store and dispose of flammable liquids. Also, google for your local Hazardous Waste Disposal Centre and your local Recycling Centre might take it as well. In most civilized countries today it is illegal to dump it along with other trash. --Aspro (talk) 13:51, 25 February 2011 (UTC)[reply]
Wouldn't AvGas deplastisize a plastic soda bottle anyway? Googlemeister (talk) 14:17, 25 February 2011 (UTC)[reply]


Is there any value in rethinking the fuel testing protocols -- for example, having a stock of clean sterile and inexpensive containers for holding your fuel samples -- such that you could afterward put the fuel back into the airplane?? Solves the problem by making it not a problem in the first place... DaHorsesMouth (talk) 23:41, 25 February 2011 (UTC)[reply]

The thinking is to get all of the potential water and contaminants out that were in the bottom of the fuel tank, whether visible or not. It would defeat the purpose to return them to the fuel tank. Thanks for all the responses guys! Falconusp t c 02:41, 28 February 2011 (UTC)[reply]

The fate of Voyager 1

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According to its article, it will run out of power in 2025, but what happens after that when it's a powerless block of metal floating through space? Assuming it doesn't slam into a star, what will happen to it in the long run? Will it be gradually eroded by the interstellar medium, or burnt away by radiation, or will it slowly evaporate, or will it fall into the centre of the galaxy, or will something else happen to it? Or, indeed, will it stay around long enough to experience a cosmological fate like the big rip/big crunch? Smurrayinchester 22:42, 24 February 2011 (UTC)[reply]

You might find this discussion from last October interesting. Nobody really knows what happens to technologically-formed metallic objects over billion-year timescales. Micrometeorite impacts, exposure to extreme cold, vacuum, and high-energy cosmic rays and energetic particles will eventually degrade the lattice-structure of any metals. The objects are too small to self-gravitate into a hydrostatic equilibrium spherical ball (the material strength is orders of magnitude larger than the self-gravitation). In a sense, interstellar space is a pretty darned good clean-room - it's cold, empty, and very little is happening; but there are the occasional sand-granule, large asteroid, or mega-electron-volt energetic hydrogen nucleus slamming into the object. Accumulated across billions of years, those impacts with the spacecraft might stress-fracture or deform the material macroscopically. If the spacecraft is subject to tidal or gravitational forces, it may be sheared or drawn into a collision with a massive object. Nimur (talk) 23:15, 24 February 2011 (UTC)[reply]
There is no reason to assume it will be eroded by the interstellar medium, or burnt away by radiation, evaporate, or fall into the centre of the galaxy. Interstellar space will treat it exactly the same way that space treats meteorites. Ever since the planets and galaxies formed, meteorites that strayed too close to a planet or moon were drawn into it and destroyed on impact. Meteorites that were drawn too close to a sun or something else with an atmosphere were burnt up entering that atmosphere. The remainder of the meteorites are still out there, wandering around. It is reasonable to assume Voyager 1 will do the same. Dolphin (t) 02:41, 25 February 2011 (UTC)[reply]
On its current course, how long untill it enter's another solar system with sufficient solar radiation to recharge its solar panels? Should it still be able to relay interpretable imagery? Plasmic Physics (talk) 03:03, 25 February 2011 (UTC)[reply]
It doesn't have solar panels, it has an RTG. The article says it will be 40,000 years before it even comes close to another star (1.6 light years, which isn't very close). But even if it did start transmitting, it would no longer be pointed in the direction of earth. Ariel. (talk) 03:21, 25 February 2011 (UTC)[reply]
See the links which I provided at Wikipedia talk:WikiProject Environment#The (outer) Space Environment (permanent link here).
Wavelength (talk) 16:18, 25 February 2011 (UTC)[reply]
Sorry, I must be confused with another probe. Plasmic Physics (talk) 21:58, 25 February 2011 (UTC)[reply]
It could possibly enter into a parabolic orbit when it crosses the outer limit of another star's sphere of influence (astrodynamics). ~AH1(TCU) 20:22, 26 February 2011 (UTC)[reply]

Pioneer Anomaly

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Hi. I was browsing the usually laughable Conservapedia website when I came across this quote regarding the Pioneer spacecraft anomaly by a Dr D Russell Humphreys: "The only non-standard assumption I used was that the matter of the cosmos is limited in extent, with a fair amount of empty space beyond the matter—an assumption supported by the Bible. With those relatively modest beginnings, I was able to explain the Pioneer anomaly — it’s due to a change in the ‘fabric’ of space. In fact, this anomaly could be the first local manifestation we have observed of the expansion of the cosmos, and the first evidence that expansion is occurring in the present, not just the past. The assumption I used violently contradicts the foundational assumption of the big bang, which says the universe has no centre and no edge. In that model, the fabric of space would not change. Consequently, the big bang model has been unable to explain the anomalous Pioneer acceleration"

I don't really understand what this guy is trying to say, and less whether it has any connection with real science. Can anyone enlighten me? —Preceding unsigned comment added by 81.129.125.167 (talk) 22:48, 24 February 2011 (UTC)[reply]

Frankly I don't really see how this would explain the Pioneer anomaly (which may not have anything to do with the fundamental laws of physics — it's likely something quite banal). It's unclear to me why this "fabric" of space (what, exactly, does he mean by that?) being "changed" (changed how?) would affect only the Pioneer satellites and not, say, all of the planets that are further out from said satellites. Anyway, whatever the merits of this guys theory, it would have to be put into actual astronomical terms before it could be really evaluated. The above description of it is insufficient to make sense of what he's proposing. --Mr.98 (talk) 22:59, 24 February 2011 (UTC)[reply]
Wikipedia has an article on Russell Humphreys, though it doesn't say much about his cosmology. It does say that his cosmology predicts that the Earth has a (proper) age of ~6000 years, which is contradicted by a huge body of evidence, so it's pretty much a non-starter. I might glance through his book if I find it online for free, but in my experience (some of it firsthand!) it's common for people with new theories of everything to imagine that they've explained some unexplained phenomenon on the most amazingly flimsy evidence. I doubt that he has done a single calculation of the magnitude of the supposed "change in the fabric". For what it's worth, the shell theorem applies in general relativity, meaning that the Pioneer anomaly can't be explained by adding or removing distant matter (unless, perhaps, it's done in a non-isotropic way). -- BenRG (talk) 01:53, 25 February 2011 (UTC)[reply]

Natural gas

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How much of it is burned every day? --T H F S W (T · C · E) 23:17, 24 February 2011 (UTC)[reply]

According to BP (boo, hiss), 2940.4 trillion cubic meters was consumed in 2009, which gives an average of just over 8 trillion per day. Clarityfiend (talk) 04:45, 25 February 2011 (UTC)[reply]
Just for clarity, is that at STP or is that in some conditions? With numbers that large, and when dealing with a gas, it makes a big difference as to what conditions it is being counted under? If it is at non-standard conditions, the volume numbers would be less meaningful than, say, mass numbers might be. --Jayron32 05:19, 25 February 2011 (UTC)[reply]
Based on this I don't think that number is accurate. You mean billion (10^9), not trillion (10^12). And Natural gas#Energy content.2C statistics and pricing has info on what the unit means. Ariel. (talk) 05:49, 25 February 2011 (UTC)[reply]
Yes billion is what the source says Nil Einne (talk) 17:33, 25 February 2011 (UTC)[reply]
D'oh. Clarityfiend (talk) 20:52, 25 February 2011 (UTC)[reply]
World production in 2008 was 3.127 trillion cu m[11] so I don't think consumption was nearly 1000 times that. --Colapeninsula (talk) 15:12, 25 February 2011 (UTC)[reply]

infra red radiation

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how does infra red radiation make something warmer — Preceding unsigned comment added by Lufc88 (talkcontribs) 23:30, 24 February 2011 (UTC)[reply]

Most molecules, if not all, absorbs infra red better than wavelengths of EMR. The energy bound within the radiation is converted to kinetic energy of the molecules. Since temperature is a measure of the mean thermal kinetic energy of a system, the temperature is also raised.
Infrared is prefered due to the very nature of molecular bonds. Plasmic Physics (talk) 23:43, 24 February 2011 (UTC)[reply]

Thanks — Preceding unsigned comment added by Lufc88 (talkcontribs) 23:53, 24 February 2011 (UTC)[reply]