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This usually means heart problems, a stroke, cancer or something like that. Supposing you remain perfectly healthy and fit your entire life, and you never get sick or have any health problems, would you just die from nothing at a certain age or would you be able to live indefinitely?--92.251.191.108 (talk) 00:32, 31 March 2010 (UTC)[reply]
No you would not die of "nothing". Imagine an old car wearing out-- eventually something fails that is fatal. "Natural causes" is simply a legal category that means "not trauma, not at the hands of someone else, not drowning or lightning or asphyxia, or poisoning...". In some other contexts, "natural causes" = "we dont know the cause of death but we didnt think anyone killed him and doubt that an autopsy would show anything interesting". alteripse (talk) 00:40, 31 March 2010 (UTC)[reply]
Even if you lived your entire life without a single illness, disease or trauma, there is still evidence to suggest aging still occurs and that aging as an effect on your bodies ability to repair itself. Therefore, even after living a perfectly healthy life, your body will ultimately become unable to repair or replace its damaged cells sufficiently to continue living.
That said, it's definitely not dying from "nothing". You're dying for a reason, the aforementioned one. Regards, --—Cyclonenim | Chat 00:55, 31 March 2010 (UTC)[reply]
The way I see it, the human body is like a pair of pants. When it gets worn-out, it will start to tear - say, for instance, at the knees. And when the pants themselves are strong, you can sew it up. But the more you wear said pants, the harder it will be to patch a hole, because the fabric just isn't strong enough to hold string and will unravel when pulled on. You can put as big a patch on as you want, but it's not going to be holding onto anything if the rest of the pants are the consistency of paper. Now in the world of trousers, you could (in theory) have pants made entirely of patches, but that doesn't really work for human beings... ZigSaw 01:47, 31 March 2010 (UTC)[reply]
Interestingly enough that is one of the major interests in stem cell research. In theory you could patch everything that wears out instead of reproducing but unless you hate sex you will probably not opt for stem cell renewal as reproduction versus reproducing in the old fashioned way. 71.100.3.207 (talk) 02:16, 31 March 2010 (UTC)[reply]
Convieniently, we have an article on aging, and a more biologically oriented article Senescence. These both have a lot of other good articles linked, depending on how interested you are. There are some species that effectively don't age (like Hydras, maybe), but humans and almost everything else cannot live indefinately. See Biological Immortality and Maximum life span. Buddy431 (talk) 04:45, 31 March 2010 (UTC)[reply]
Well, there are some people where it does seem like they "died from nothing". That is, their heart just stops beating. There may be an underlying cause other than old age, but it isn't known. StuRat (talk) 04:48, 31 March 2010 (UTC)[reply]
If their heart stops beating, they didn't die from nothing. I get your point, though. Regards, --—Cyclonenim | Chat 12:37, 31 March 2010 (UTC)[reply]
Rest assured the health and death are completely incompatible features. See senescence. Vranak (talk) 21:11, 31 March 2010 (UTC)[reply]
This is supposed to be a maelstrom. That is to say, a huge destructive whirlpool of chaos and death. I look at this picture and I see calm water. Am I missing something really important, or did someone post a picture of something entirely not a maelstrom? ZigSaw 01:44, 31 March 2010 (UTC)[reply]
"The Moskstraumen (popularly known as the Maelstrom) is a system of tidal eddies and whirlpools, one of the strongest in the world, that forms in a strait..." (The sky view is more illustrative of the point.)71.100.3.207 (talk) 02:19, 31 March 2010 (UTC)[reply]
Maelstroms are not very visible (see this). It is more of a swirling downdraft in the water. --The HighFinSpermWhale 04:15, 31 March 2010 (UTC)[reply]
In addition, our article doesn't say anything about 'chaos' or 'death'. It's better to find out what is actually meant by a maelstrom (or whatever you're thinking of) rather then rely on any pre-existing ideas you may have Nil Einne (talk) 06:22, 31 March 2010 (UTC)[reply]
I have to say — I too was surprised how greatly exaggerated was the power of the great Maelstrom; all I knew of it was from reading Twenty Thousand Leagues Under the Sea, and while I had assumed that the book exaggerated a bit, I didn't know until giving my answer earlier how greatly it exaggerated. Nyttend (talk) 12:14, 31 March 2010 (UTC)[reply]
There is a powerful whirlpool in the Fraser Canyon in BC that is more dramatic, and not something you'd want to get caught in. It's at the Hell's Gate feature that was created when blasting dropped a huge amount of rock into the river, making it much narrower and deeper than it was naturally. I couldn't find a great picture of that whirlpool, but this gives you a good idea of the tumult involved. Vranak (talk) 20:03, 31 March 2010 (UTC)[reply]
I've read that light cones in GR are tilted, however, I thought that light cones were traced out by light emitted from a spacetime point in time. So, shouldn't a gravitational field not only tilt a light cone, but curve it? Thanks:) 66.202.66.78 (talk) 04:34, 31 March 2010 (UTC)[reply]
Yes, in general the path of a light ray in a gravitational field is curved. When articles talk about (and sometimes illustrate) light cones being tilted, they use "light cone" as a short-hand for the "direction" in spacetime in which a light ray would initially set off if emitted from a given point. You can think of these light cones as first order approximations or linearisations - they tell you about the direction of the tangent to a light ray passing through a given point in spacetime. It is analogous to illustrating a vector field by drawing lots of little arrows representing the direction and magnitude of the field at various points. Gandalf61 (talk) 08:47, 31 March 2010 (UTC)[reply]
Also, it's not really that the light cones tilt. When that seems to happen in illustrations, it just means that the coordinate system chosen happens to be skewed with respect to the real structure of spacetime. It is a true effect of GR that one often cannot choose a coordinate system that aligns with the light cones everythere. But at any particular event you're interested in, you can get the light cone to stand nicely upright, simply by choosing coordinates appropriately. –Henning Makholm (talk) 08:54, 31 March 2010 (UTC)[reply]
Thank you:) I have another, related question: suppose we are in 2+1 dimensions. If this space is flat and we emit out a pulse of light(light going in all directions out from us) it will make a sequence of circles in time. Supposing we are near a massive object, how do those circles deform? To be honest, I don't really know much about GR and am not sure how much sense this question makes. Thank you for any help; if I am going terribly wrong in my thinking, could you point me in the right direction:) Am I making the mistake of thinking in terms of a given set of coords? If so, could you consider how the light propogates in each coordinates, then consider "deformations" in terms of how the light pulse transforms between them? Again, I may be talking nonsense, most of what I work on is far removed from diffgeo and GR 66.202.66.78 (talk) 09:05, 31 March 2010 (UTC)[reply]
My question might be better summed up by asking if each for each null geodesic will be there a light ray in the pulse that will follow it? Essentially, do the circles of light map out these paths? If so, is it only in an open set around the event that the pulse is emitted from? Thanks:) Sorry if I am being confusing/senseless. 66.202.66.78 (talk) 09:37, 31 March 2010 (UTC)[reply]
Not to yammer on; the more I think about it, I'm guessing that the circles would trace out something like the field lines for the electric fields, and that this would end up being the same as the vector field above in terms of information about the gravitational field in the area. I'll stop posting now:) 66.202.66.78 (talk) 09:57, 31 March 2010 (UTC)[reply]
Yes, each null geodetic represents a possible light ray. Assuming, of course, that it does not go through a region of spacetime filled with opaque matter, in which case you need to hypothesize some non-interacting massless particle to travel along the geodesic instead. Or just stipulate that null geodesics are much more interesting than something as annoyingly material as light. :-)
A couple of possible obstacles here. First, an interesting question whose answer I don't know: Does it make sense to do GR in 2+1 spacetime? Certainly one can write down the tensor equations in 2+1 dimensions and look for solutions, but I suspect you will not get an inverse-square law for gravity out of it. The usual simplification here is to say that we're really working in 3+1 dimensions but assume enough symmetry of the situation that we can learn something by studying what happens within a 2+1 (or 1+1) dimensional symmetry surface of the entire spacetime.
Second: Yes, I think you're possibly by assuming that you have a nice enough coordinate system at your disposal. What you do in SR corresponds to intersecting the light front with a succession of spacelike Euclidean hypersurfaces and look at the shape this intersection makes in each "instant" surface. As long as all of your surfaces are parallel, you'll get a nice and meaningful picture that way. However, in GR spacetime does not foliate nicely into a succession of spacelike cuts. Once you choose a coordinate system, you can of course use it to slice up spacetime, but it will be difficult to distinguish whether the answers you get tell you more about your particular choice of coordinates than about anything physically significant. –Henning Makholm (talk) 18:11, 31 March 2010 (UTC)[reply]
You can do GR in 2+1 dimensions, but it's weird. In n spacetime dimensions, the stress-energy tensor has n (n + 1) / 2 components and the gravitational field (Riemann curvature) has n² (n² − 1) / 12 components. The GR field equation equates n (n + 1) / 2 components of the gravitational field to the components of the stress-energy tensor. In 3+1 dimensions, that's 10 of 20, and the other 10 describe the free field (the part that reaches out into the vacuum). In 2+1 dimensions, that's 6 of 6, so there's no free field and no gravitational attraction in the usual sense. There is still a gravitational interaction, but it's purely topological, depending only on how the worldlines are braided together in spacetime. -- BenRG (talk) 01:47, 2 April 2010 (UTC)[reply]
E. vigintioctopunctata or H. vigintioctopunctata?[edit]
Hi all!
This appears to be all a-fuddled. For example, the CSIRO refs here and here agree that the critter has twenty-eight spots, but disagree as to what is the Nomen protectum.
So, the question: is Epilachna vigintioctopunctata the junior synonym of Henosepilachna vigintioctopunctata, or is Henosepilachna vigintioctopunctata the junior synonym of Epilachna vigintioctopunctata?
--Shirt58 (talk) 12:07, 31 March 2010 (UTC)[reply]
The current bit at In The News says that the collisions between two 3.5 TeV proton beams in the Large Hadron Collider are a world's record for the highest energy man-made particle collisions. Seeing that Electronvolt says one 100W lightbulb takes 8000 TeV to work for one second, why is it amazing that these beams have such a comparatively small amount of energy? Is it simply that protons are very small and thus don't need much energy to go extremely fast, or am I missing something else? Nyttend (talk) 12:49, 31 March 2010 (UTC)[reply]
It's a matter of scale (and comparing a composite total to an individual value). A lightbulb may use 8000 tera-eV, but that translates to around 8 trillion electrons each at 110 volts. Each individual electron, then, is at about 110 eV (assuming US mains voltage) The LHC is colliding protons that are individually at several TeV. — Lomn 13:08, 31 March 2010 (UTC)[reply]
Aha; I'd missed that. I thought that it meant that the entire beam had an energy of 3.5 TeV. Nyttend (talk) 13:32, 31 March 2010 (UTC)[reply]
Could I use Galileo and GPS at the same time? Could I use Galileo with normal (Same than I use now) GPS-locator? And, I know Galileo be finished (Maybe) at 2014. Aku506 (talk) 12:55, 31 March 2010 (UTC)[reply]
Galileo and GPS use different frequencies. Asking if you can use both at the same time is the same as asking if you can use two cell phones at the same time. There is no problem. As for using Galileo with a GPS unit, the GPS unit must have Galileo compatibility - which some do, but it is clearly noted on the unit. -- kainaw™ 13:11, 31 March 2010 (UTC)[reply]
Yes, it would be possible to combine Galileo and GPS, just as any two navigational aids can be combined to improve their total performance. However, Galileo isn't the same protocol as GPS. While I'm sure plenty of dual-system receivers will be manufactured, an arbitrary present-day GPS receiver won't use Galileo, just as it won't use digital TV, AM radio, or WiFi. Note that while Galileo claims to have better resolution than GPS, it also has (just as GPS does) a rough-accuracy basic version and a higher-accuracy military (and for Galileo, paid commercial) version. It's not clear that the "better accuracy" claim in our article is comparing real-world basic user performance. We also have an article on GNSS Augmentation (with GNSS being Global Navigation Satellite System, the general term for this sort of thing). — Lomn 13:21, 31 March 2010 (UTC)[reply]
someone sent this info in an email...would this be correct?
Ed says that when food poisoning is reported, the first thing the officials look for is when the 'victim' last ate ONIONS and where those onions came from (in the potato salad?). Ed says it's not the mayonnaise (as long as it's not homemade Mayo) that spoils in the outdoors. It's probably the onions, and if not the onions, it's the POTATOES.
Extended content
I have used an onion which has been left in the fridge, and sometimes I don't use a whole one at one time, so save the other half for later.
Now with this info, I have changed my mind....will buy smaller onions in the future.
I had the wonderful privilege of touring Mullins Food Products, Makers of mayonnaise. Mullins is huge, and is owned by 11 brothers and sisters in the Mullins family. My friend, Jeanne, is the CEO.
Questions about food poisoning came up, and I wanted to share what I learned from a chemist.
The guy who gave us our tour is named Ed. He's one of the brothers Ed is a chemistry expert and is involved in developing most of the sauce formula. He's even developed sauce formula for McDonald's.
Keep in mind that Ed is a food chemistry whiz. During the tour, someone asked if we really needed to worry about mayonnaise. People are always worried that mayonnaise will spoil. Ed's answer will surprise you. Ed said that all commercially- made Mayo is completely safe.
"It doesn't even have to be refrigerated. No harm in refrigerating it, but it's not really necessary." He explained that the pH in mayonnaise is set at a point that bacteria could not survive in that environment. He then talked about the quaint essential picnic, with the bowl of potato salad sitting on the table and how everyone blames the mayonnaise when someone gets sick.
Ed says that when food poisoning is reported, the first thing the officials look for is when the 'victim' last ate ONIONS and where those onions came from (in the potato salad?). Ed says it's not the mayonnaise (as long as it's not homemade Mayo) that spoils in the outdoors. It's probably the onions, and if not the onions, it's the POTATOES.
He explained, onions are a huge magnet for bacteria, especially uncooked onions. You should never plan to keep a portion of a sliced onion.. He says it's not even safe if you put it in a zip-lock bag and put it in your refrigerator.
It's already contaminated enough just by being cut open and out for a bit, that it can be a danger to you (and doubly watch out for those onions you put in your hotdogs at the baseball park!)
Ed says if you take the leftover onion and cook it like crazy you'll probably be okay, but if you slice that left-over onion and put on your sandwich, you're asking for trouble. Both the onions and the moist potato in a potato salad, will attract and grow bacteria faster than any commercial mayonnaise will even begin to break down.
So, how's that for news? Take it for what you will. I (the author) am going to be very careful about my onions from now on. For some reason, I see a lot of credibility coming from a chemist and a company that produces millions of pounds of mayonnaise every year.'
Also, dogs should never eat onions. Their stomachs cannot metabolize onions .Please remember it is dangerous to cut onions and try to use it to cook the next day, it becomes highly poisonous for even a single night and creates Toxic bacteria which may cause Adverse Stomach infections because of excess Bile secretions and even Food poisoning.
I've summarized the relevant question while collapsing the long email, as I found the distinction between you and the email you're quoting confusing. Anyway, as with most circulating emails, Snopes has an entry. Some mayo inhibits bacterial growth, and certainly many places around the world get by on far less refrigeration than the US. Onions, though, aren't attested as some sort of food poisoning magnet. — Lomn 15:12, 31 March 2010 (UTC)[reply]
I was also going to point to Snopes, but I will add that it is true that onions can be toxic to dogs (per my wife, a veterinarian, as well as this source). -- Coneslayer (talk) 15:18, 31 March 2010 (UTC)[reply]
Dandelion bulbs look a lot like onions and are poisonous. A few people a year get poisoned by making that mistake. Onions themselves aren't known for food poisoning to the best of my knowledge. --Tango (talk) 15:45, 31 March 2010 (UTC)[reply]
In particular, the Dandelion article says "Both species are edible in their entirety". --Sean 19:14, 31 March 2010 (UTC)[reply]
I got the first two letters right! Yes, I meant daffodil. I had an image of a daffodil in my mind, I just attached the wrong word to it. Thanks! --Tango (talk) 21:22, 31 March 2010 (UTC)[reply]
I don't think the flesh of onions contracts bacteria any quicker than a hard-boiled egg would. And commercial mayonnaise has preservatives in it. --Cheminterest (talk) 21:02, 31 March 2010 (UTC)[reply]
I often make spaghetti for myself for lunch, and I'll toss little bits of onions into the sauce; I take a few weeks per onion, and I don't do anything more than warm the sauce, and I've never had trouble. Of course, anecdotes don't prove it, but they prove that you won't always be poisoned. Nyttend (talk) 02:11, 1 April 2010 (UTC)[reply]
Which would melt faster, a 4' deep swimming pool that is filled with a solid block of ice, or an 8' deep swimming pool (same area as pool A) which has a 4' thick solid block of ice floating on 4' of near freezing water if they are in equal conditions for sunlight temperature etc...? I am trying to decide if a pond that freezes in winter will thaw faster if it freezes solid then a deeper pond that forms a thick layer of ice on top, all else being equal. Googlemeister (talk) 19:27, 31 March 2010 (UTC)[reply]
This is not a definitive answer, but I believe the answer is "it depends". If the ground underneath the ice/water is warmer or colder than freezing, you'd get a different result. If the water is insulating the ice from the warmer ground (or warming ground, if the heat is being conveyed from the surface via conduction), then presumably the added bulk of cold water that must be warmed would delay melting. For a pond, the water underneath is presumably flowing to a greater or lesser extent, which would probably aid the melting process by eroding the ice from underneath. —ShadowRanger(talk|stalk) 19:43, 31 March 2010 (UTC)[reply]
The one with water probably would. the water can help absorb some of the heat, even without freezing. --Cheminterest (talk) 21:08, 31 March 2010 (UTC)[reply]
In the 8' pool, is the ice covering the entire surface area of the water? Crevasses may also form in the ice as it becomes thinner, and the absorption of heat by the water would accelerate as more of the unfrozen water's surface area is exposed. See also Mpemba effect. ~AH1(TCU) 23:56, 4 April 2010 (UTC)[reply]
If you were to pour enough antimatter into a black hole, could it annihilate the singularity? ScienceApe (talk) 19:33, 31 March 2010 (UTC)[reply]
Yes. But given that (to my knowledge) production of anti-matter creates a matching amount of matter (since matter can be neither created nor destroyed in the grand scheme of things), you'd be producing a black hole's worth of matter in order to annihilate the original black hole. Unless you spread it out really thinly, you'd have another black hole in short order. —ShadowRanger(talk|stalk) 19:38, 31 March 2010 (UTC)[reply]
Antimatter doesn't have "antimass", and so there's no expectation I can see that you produce hydrogen for a given quantity of antihydrogen. For the specific question, though, the answer appears to be "no". A matter-antimatter reaction converts the rest mass into kinetic energy. However, this is where mass–energy equivalence kicks in. Per the nature of black holes, the resultant energy can't escape. As such, the effective mass can't change. See, from m-e equivalence, "if a body gives off the energy L in the form of radiation, its mass diminishes by L/c2". Note that this is consistent with the nature of an event horizon -- once the antimatter passes within the black hole's event horizon, what happens to it is unknowable and irrelevant. — Lomn 19:55, 31 March 2010 (UTC)[reply]
Hmm... How does that work with the description of Hawking Radiation then? My understanding is that the universe regularly creates and annihilates matter and anti-matter "virtual particles". When this happens at the edge of a black hole, the anti-particle gets sucked in to the black hole, while the paired particle escapes as Hawking Radiation. To satisfy conservation of energy/mass, the anti-particle annihilates it's equivalent particle in the black hole, thereby reducing the black hole's mass by the same amount as that of the particle that escaped. I'm not a physicist, so perhaps the distinction between virtual antimatter and real antimatter is messing me up, but wouldn't supplying an enormous amount of antimatter produce a similar reduction in mass? —ShadowRanger(talk|stalk) 21:12, 31 March 2010 (UTC)[reply]
I don't really understand it myself, but virtual particles behave differently to real ones. I think that is why your argument doesn't hold. I'm not sure what the correct argument is, though. --Tango (talk) 21:26, 31 March 2010 (UTC)[reply]
Re: Hawking radiation: the virtual particle thing is that energy is transformed into particles, which typically annihilate back into energy. For HR, one of the two particles (and it doesn't matter whether it's the electron or the positron) escapes, reducing the mass of the black hole by whatever the correct conversion amount is. What happens to the other particle (again, positron or electron) doesn't matter, as it remains hidden behind the event horizon. The reduction in mass results from the loss of a particle but isn't contingent on whether that particle is matter or antimatter. — Lomn 21:34, 31 March 2010 (UTC)[reply]
By the way, according to Antimatter#Fuel, "Known methods of producing antimatter from energy also produce an equal amount of normal matter, so the theoretical limit is that half of the input energy is converted to antimatter." So at least according to the current understanding, creating enough antimatter to match the mass of a black hole would intrinsically produce enough matter to make a black hole as well. And use a god-awful amount of energy to boot. —ShadowRanger(talk|stalk) 23:44, 31 March 2010 (UTC)[reply]
Antimatter has positive mass, so the black hole would get even heavier (and bigger) instead of lighter (and smaller). Dauto (talk) 20:42, 31 March 2010 (UTC)[reply]
See my note above. If the black hole is made of matter, I would think the anti-matter/matter annihilation would reduce the mass. —ShadowRanger(talk|stalk) 21:12, 31 March 2010 (UTC)[reply]
No, because energy has mass - see mass-energy equivalence. The no hair theorem says that a black hole made from matter is indistinguishable from one made from the same mass of antimatter with the same charge and angular momentum - or, indeed, from a black hole made from 50% matter and 50% antimatter. Gandalf61 (talk) 22:44, 31 March 2010 (UTC)[reply]
I'm more than willing to admit to a weakness in my physics on this matter. Although I'm frankly perplexed by the concept of a 50/50 matter/anti-matter black hole. Presumably compressed that densely, a matter/anti-matter reaction would be inevitable? It would produce energy, which presumably could not escape (assuming the mix wasn't actually 50/50 and what remained could maintain the black hole's integrity), but I'm frankly perplexed as to what would happen with that energy. I realize matter and anti-matter both have mass, but I thought the annihilation reaction between them eliminated the mass entirely, in favor of energy (though what form that energy takes is beyond me). —ShadowRanger(talk|stalk) 22:54, 31 March 2010 (UTC)[reply]
Reading further, the energy apparently takes the form of either high energy photons (which have no rest mass) or a new particle-antiparticle pair. So I suppose a matter/antimatter composite black hole would be possible if all particle-antiparticle reactions produced new pairs of particle-antiparticle. It seems like the collision of a matter black hole and an antimatter black hole (of sufficiently similar mass), or an exactly 50/50 black hole might have a different result though. If the annihilation is sufficiently energetic, and the vast majority of the colliding holes is transformed into gamma rays, might it be possible that the instantaneous reduction in mass would allow for the whole thing to self-destruct, leading to the release of some matter and/or antimatter propelled by the energy of the gamma rays? This doesn't work if the gamma rays are considered to have mass for the purposes of holding the black hole together, but I'm stuck in a conundrum here. The gamma rays only have mass when moving, so if they are moving they might produce the mass necessary to keep the black hole intact. But if the black hole remains intact, the gamma rays can't escape, i.e. they cannot move, which means they have no mass and the black hole wouldn't have the necessary mass to stay together. Gah! Driving myself insane. I'm either one step away from figuring out the cause of the Big Bang, or (more likely) I really suck at advanced theoretical physics. —ShadowRanger(talk|stalk) 23:13, 31 March 2010 (UTC)[reply]
Once you reach the singularity, the distinction between matter and antimatter is simply lost, according to current understanding. A black hole has no hair. --Trovatore (talk) 23:20, 31 March 2010 (UTC)[reply]
Yeah, that seems to be what it is saying. I'm a little confused as to why that would be, but then, I'm not Hawking, so I'll shut up now. :-) —ShadowRanger(talk|stalk) 23:48, 31 March 2010 (UTC)[reply]
The products of a matter / antimatter reaction have exactly the same effective mass as the reactants. That's true even if you products are gamma rays. Gamma rays have no rest mass but their relativistic mass matches exactly what you'd predict from E = mc2, and that's what counts for determining gravity (more or less). Mass is energy, energy is mass. The annihilation of matter and anti-matter has no effect on the size of the black hole. The advantage of gamma rays is that they are easily converted to other forms of energy, whereas particles are usually hard to convert to useful energy. So in principle, matter / anti-matter reaction can produce lots of useful energy, but we don't know any easy sources of anti-matter. Dragons flight (talk) 00:02, 1 April 2010 (UTC)[reply]
Like most people here (evidently), I find the Hawking radiation thing perplexing. If you smack a kilogram of hydrogen into a kilo of antihydrogen, you certainly get 2kg times the speed of light squared joules of energy. If that happens inside the event horizon of a black hole, the energy can't escape so as far as an outside observer is concerned - it's overall mass/energy (and therefore it's gravity and general blackholeyness) doesn't change. The explanation of how Hawkins radiation works isn't so simple. Since the virtual particle and virtual anti-particle both have positive mass, it takes energy to create them - and energy is produced when they are destroyed. The "popular science" explanation is that this energy is "temporarily borrowed" from the universe and has to be "repaid" shortly afterwards. That is (to say the least) an unsatisfying explanation - but it's the only one I've heard that I can understand! Anyway - following that rather tricky explanation to it's conclusion - when one of the virtual particles vanishes into the black hole, never to return and the other one doesn't - the annihilation of the original pair of virtual particles doesn't happen. So the energy that was 'borrowed' from "the universe" gets "paid back" inside the black hole...which loses energy as a result. This is a really unsatisfying answer...I wish we had a better one that was reasonably comprehensible. However, what we CAN say for sure (and understand easily) is that it's not that the antiparticle has negative mass or turns into negative energy somehow...that's definitely not what happens because we can easily imagine that the 'normal' matter particle of the pair falls into the hole about 50% of the time and if antiparticles shrank the black hole due to "negative mass", it would all be cancelled out by the normal particles falling in the other half of the time. So the average black holes wouldn't evaporate. Hence, we know the explanation is more tricky than that...even if those here on this reference desk can't explain why that is. SteveBaker (talk) 00:04, 1 April 2010 (UTC)[reply]
I don't purport to understand Hawking radiation either, but: Instead of the virtual-twins-separated-at-birth story popularized by Hawking himself, we could consider a story where a particle inside the black hole tunnels across the horizon. The event horizon is (by one definition) an "outward-moving light front that doesn't get anywhere", so all one needs to escape is to overtake it -- and moving faster than light is one of the things a tunnelling/virtual particle can do, at the cost of having imaginary momentum and a chance of pulling it off that decreases exponentially. This is not extraordinarily fanciful; in quantum electrodynamics you routinely have to consider contributions from worlds where a particle moves from event A to event B faster than c.
It feels plausible to me that tunnelling could actually be mathematically equivalent to the standard story. It would also "explain", heuristically, why a smaller black hole emits more Hawking radiation: As the distance to tunnel through decreases, the probability of success increases exponentially. –Henning Makholm (talk) 01:58, 1 April 2010 (UTC)[reply]
Let me try to clear up a couple of common misconceptions that may be hampering people's understanding.
First it is important to understand the relationship between energy of a particle , its momentum and its rest mass ---- That relationship is given by the equation . If the particle is at rest than that equation reduces to the notorious . Note that the last equation is not telling us that a particle of mass can be converted to energy . It is simply telling us that a particle of mass has energy . The energy is already there but not in a form that can be detected by more conventional means. When a particle meets its antiparticle and they annihilate each other, photons get released. The energy of the photons will be the same energy that was already there in the form of mass-energy. If that happens inside a black hole, the energy of the photons will have as much of a gravitational effect as the particles that were there before them and the total mass of the black hole won't change. Simply, mass creates gravity because it contains energy and all forms of energy do create gravity.
Second it is important to understand what makes virtual particles different from real particles and how does that difference allow for Hawking radiation to leave the black hole ---- Virtual particles do not have to obey the relationship . That means that, unlike real particles, they are allowed for a brief moment to exist and have any energy, unrestricted by the need to satisfy a relationship with a rest mass and a momentum. They are even allowed to have negative energy. Oddly, the technical speak for that is to say that real particles must be on their mass shell while virtual particles may be off shell. Because of that, pairs of particles and antiparticles pop out of the vacuum all the time, one with a positive energy and the other one with a negative energy so that energy conservation is preserved ---- energy conservation is always preserved. There is no need to borrow any energy. The borrowing of energy is one of the worst analogies I've ever seen. It produces much more confusion than any benefit that it might have. These virtual particles live only long enough to ask themselves one single question : "Am I on my mass shell?" For at least one of them the answer is always "no" because it has negative energy and therefore it must vanish back into the oblivion whence it came from dragging with it the other particle. But near a black hole things are different. If the particle with negative energy falls in the black hole than it doesn't need to answer that question anymore because it's already gone to a different kind of oblivion. It has been absorbed by the black hole and since it had negative energy, the black hole becomes a little lighter. The other particle, the one with positive energy, might also fall in the black hole, in which case nothing was accomplished. But if it happens to be on shell and if it happens to have enough energy to climb out of the hole than it may leave the black hole as Hawking radiation. Dauto (talk) 06:03, 1 April 2010 (UTC)[reply]
I would like to add that individual virtual particles are - by definition - impossible to observe. If you ever observe a particle directly, then by definition it must be real. The notion of virtual particles as individual entities is really part of the interpretation of quantum mechanics, rather than part of physics itself. They emerge as intermediate states in Feynman diagrams. Feynman diagrams describe a process of doing mathematical calculations such that one can predict the probability of reaching a specific quantum mechanical end state given a specific initial state. Those probabilities are real and observable. The calculations however require one to do integrations across intermediate states that are assumed to contain unobservable (i.e. virtual) particles which can have arbitrary energy not conforming to the particles' normal momentum-energy relationship. In fact, to predict any physically observable quantity, one doesn't just consider a single possible virtual particle, but one must integrate over all conceivable particle energies and momenta (including negative energy states). Whether one considers a virtual particle to be a real entity or merely some form of mathematical convenience for accounting purposes is a really a philosophical question rather than a physics question. Many physicists do view them as real entities, but personally, I tend to see their strange properties as too bizarre. Hence I prefer to think of them as a mathematical contrivance. In the case of Hawking radiation, the mathematics of virtual particles falling into the black hole can be shown to be the same as the mathematics of real particles quantum tunneling out of the black hole. So again, whether you choose to believe in virtual particles as real and individualized things boils down to a question of philosophical interpretation, rather than one that can be settled by any direct measurement. Dragons flight (talk) 08:03, 1 April 2010 (UTC)[reply]
I find it interesting that you dislike virtual particles because you "tend to see their strange properties as too bizarre" and than you suggest as an alternative 'real' particles tunelling out of the black whole. I guess particles moving faster than the speed of light (or backwards in time if you will) are not too bizarre. What I tell my students about quantum weirdness is "Embrace the weirdness instead on trying to 'understand' it of explain it away as a mathematical artifact. The same thing applies here. Virtual particles have real observable effects and should be considered as real as anything else. There is a long history in physics of second guesses about how real our theoretical constructs really are. Electromagnetic fields, electromagnetic potentials Wave function, quarks, quasi-particles, etc, all were at some point considred by some as simple mathematical artifacts. Ultimately I think this is a pointless philosophical discussion. I say: "If it works, than it's real" Dauto (talk) 17:46, 1 April 2010 (UTC)[reply]
Yep, I do think superluminal quantum tunneling seems more natural / intuitive than talking about one half of a virtual particle pair falling into a black hole and contributing negative mass. Is that so crazy? ;-) Newtonian gravity and the Bohr atom are also useful theoretical constructs, but if you interpret them literally you end up with incorrect models of reality. Eventually, science led us to better theories and new perceptions. Personally, I suspect that the conception of virtual particles as real individualized things will also end up getting displaced by a better understanding eventually. You're right that the math works of course, and that's good enough for now, but a theory that requires the existence of ephemeral, untouchable particles with strange variations in energy and momentum is enough for me to believe that something is missing in our understanding. Given that interpreting the mathematics is more philosophy than science, I choose to adhere to a point of view that feels more natural until there is evidence to the contrary. Dragons flight (talk) 23:36, 1 April 2010 (UTC)[reply]
I think they are both bizarre but I also think that doesn't matter. Being bizarre is not a reason to drop a theory (or an interpretation of a theory). What matters is whether it is consistent and how well it works to explain the experiments, observations and, lets not forget, the gedanken experiments. Dauto (talk) 02:09, 2 April 2010 (UTC)[reply]
The thing that exists is the math describing how these processes work. Virtual particles and quantum tunneling rates are both approximations that hold in many circumstances, but neither is "really" going on. As far as we know, what is happening is that a (wave)function over a quantum mechanical Hilbert space is changing over time in respect to certain rules. Virtual particles are just a way of putting some of these rules into easy boxes for simplified human calculation and tunneling is more the wavefunction slowly leaking out of a black holes than individual particles escaping one at a time. 99.238.19.115 (talk) 03:52, 2 April 2010 (UTC)[reply]
Matter has mass and volume. A black hole's gravitational singularity has zero volume, so it might not be considered matter, as both matter and antimatter consists of particles of finite volume. So, it could be said that once an object gets past the black hole's singularity, what happens to it afterwards doesn't really "matter". :-) ~AH1(TCU) 23:51, 4 April 2010 (UTC)[reply]
Does anyone know what kind of animal this is nds-nl:Ofbeelding:Dier.JPG? Looks like some kind of badger to me... this picture is taken in South Africa (if it helps) Sεrvιεи | T@lk page 20:56, 31 March 2010 (UTC)[reply]
It looks like one. The name doesn't give any clues (looked it up in the translation). Dier just means animal. --Cheminterest (talk) 21:06, 31 March 2010 (UTC)[reply]
If it is a Large Spotted Genet, which it certainly looks like it is, then perhaps this picture should replace the one currently in the article (or at least be added to the article). -Pete5x5 (talk) 00:19, 1 April 2010 (UTC)[reply]
The hurricane article you linked says "...which allows the release of the heat of condensation that powers a tropical cyclone." -- Finlay McWalter • Talk 22:25, 31 March 2010 (UTC)[reply]
Ultimately, the source of all energy for geophysical processes is solar radiation, produced in the sun by solar fusion. Hurricanes can be thought of as the result of a massive heat-transfer from the warm tropicalocean waters into the atmosphere. Nimur (talk) 00:34, 1 April 2010 (UTC)[reply]
The source is the difference in temperature between the water and the (high altitude) air. Warmer water makes for a more powerful hurricane. A hurricane is basically a heat engine. Water can store a lot more heat than land, that's the main reason hurricanes die when they go over land. Ariel. (talk) 01:01, 1 April 2010 (UTC)[reply]
